3.4.8 \(\int \frac {(e+f x) \sec (c+d x)}{a+b \sin (c+d x)} \, dx\) [308]

Optimal. Leaf size=413 \[ -\frac {2 i a (e+f x) \tan ^{-1}\left (e^{i (c+d x)}\right )}{\left (a^2-b^2\right ) d}-\frac {b (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right ) d}-\frac {b (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right ) d}+\frac {b (e+f x) \log \left (1+e^{2 i (c+d x)}\right )}{\left (a^2-b^2\right ) d}+\frac {i a f \text {Li}_2\left (-i e^{i (c+d x)}\right )}{\left (a^2-b^2\right ) d^2}-\frac {i a f \text {Li}_2\left (i e^{i (c+d x)}\right )}{\left (a^2-b^2\right ) d^2}+\frac {i b f \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right ) d^2}+\frac {i b f \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right ) d^2}-\frac {i b f \text {Li}_2\left (-e^{2 i (c+d x)}\right )}{2 \left (a^2-b^2\right ) d^2} \]

[Out]

-2*I*a*(f*x+e)*arctan(exp(I*(d*x+c)))/(a^2-b^2)/d+b*(f*x+e)*ln(1+exp(2*I*(d*x+c)))/(a^2-b^2)/d-b*(f*x+e)*ln(1-
I*b*exp(I*(d*x+c))/(a-(a^2-b^2)^(1/2)))/(a^2-b^2)/d-b*(f*x+e)*ln(1-I*b*exp(I*(d*x+c))/(a+(a^2-b^2)^(1/2)))/(a^
2-b^2)/d+I*a*f*polylog(2,-I*exp(I*(d*x+c)))/(a^2-b^2)/d^2-I*a*f*polylog(2,I*exp(I*(d*x+c)))/(a^2-b^2)/d^2-1/2*
I*b*f*polylog(2,-exp(2*I*(d*x+c)))/(a^2-b^2)/d^2+I*b*f*polylog(2,I*b*exp(I*(d*x+c))/(a-(a^2-b^2)^(1/2)))/(a^2-
b^2)/d^2+I*b*f*polylog(2,I*b*exp(I*(d*x+c))/(a+(a^2-b^2)^(1/2)))/(a^2-b^2)/d^2

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Rubi [A]
time = 0.43, antiderivative size = 413, normalized size of antiderivative = 1.00, number of steps used = 19, number of rules used = 8, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {4629, 4615, 2221, 2317, 2438, 6874, 4266, 3800} \begin {gather*} \frac {i a f \text {PolyLog}\left (2,-i e^{i (c+d x)}\right )}{d^2 \left (a^2-b^2\right )}-\frac {i a f \text {PolyLog}\left (2,i e^{i (c+d x)}\right )}{d^2 \left (a^2-b^2\right )}+\frac {i b f \text {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{d^2 \left (a^2-b^2\right )}+\frac {i b f \text {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{d^2 \left (a^2-b^2\right )}-\frac {i b f \text {PolyLog}\left (2,-e^{2 i (c+d x)}\right )}{2 d^2 \left (a^2-b^2\right )}-\frac {2 i a (e+f x) \text {ArcTan}\left (e^{i (c+d x)}\right )}{d \left (a^2-b^2\right )}-\frac {b (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{d \left (a^2-b^2\right )}-\frac {b (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{d \left (a^2-b^2\right )}+\frac {b (e+f x) \log \left (1+e^{2 i (c+d x)}\right )}{d \left (a^2-b^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((e + f*x)*Sec[c + d*x])/(a + b*Sin[c + d*x]),x]

[Out]

((-2*I)*a*(e + f*x)*ArcTan[E^(I*(c + d*x))])/((a^2 - b^2)*d) - (b*(e + f*x)*Log[1 - (I*b*E^(I*(c + d*x)))/(a -
Sqrt[a^2 - b^2])])/((a^2 - b^2)*d) - (b*(e + f*x)*Log[1 - (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/((a^2
- b^2)*d) + (b*(e + f*x)*Log[1 + E^((2*I)*(c + d*x))])/((a^2 - b^2)*d) + (I*a*f*PolyLog[2, (-I)*E^(I*(c + d*x
))])/((a^2 - b^2)*d^2) - (I*a*f*PolyLog[2, I*E^(I*(c + d*x))])/((a^2 - b^2)*d^2) + (I*b*f*PolyLog[2, (I*b*E^(I
*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/((a^2 - b^2)*d^2) + (I*b*f*PolyLog[2, (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2
- b^2])])/((a^2 - b^2)*d^2) - ((I/2)*b*f*PolyLog[2, -E^((2*I)*(c + d*x))])/((a^2 - b^2)*d^2)

Rule 2221

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x]
- Dist[d*(m/(b*f*g*n*Log[F])), Int[(c + d*x)^(m - 1)*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2317

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2438

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2, (-c)*e*x^n]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 3800

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[I*((c + d*x)^(m + 1)/(d*(m + 1))), x
] - Dist[2*I, Int[(c + d*x)^m*(E^(2*I*(e + f*x))/(1 + E^(2*I*(e + f*x)))), x], x] /; FreeQ[{c, d, e, f}, x] &&
IGtQ[m, 0]

Rule 4266

Int[csc[(e_.) + Pi*(k_.) + (f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[-2*(c + d*x)^m*(ArcTanh[E
^(I*k*Pi)*E^(I*(e + f*x))]/f), x] + (-Dist[d*(m/f), Int[(c + d*x)^(m - 1)*Log[1 - E^(I*k*Pi)*E^(I*(e + f*x))],
x], x] + Dist[d*(m/f), Int[(c + d*x)^(m - 1)*Log[1 + E^(I*k*Pi)*E^(I*(e + f*x))], x], x]) /; FreeQ[{c, d, e,
f}, x] && IntegerQ[2*k] && IGtQ[m, 0]

Rule 4615

Int[(Cos[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_))^(m_.))/((a_) + (b_.)*Sin[(c_.) + (d_.)*(x_)]), x_Symbol] :>
Simp[(-I)*((e + f*x)^(m + 1)/(b*f*(m + 1))), x] + (Int[(e + f*x)^m*(E^(I*(c + d*x))/(a - Rt[a^2 - b^2, 2] - I*
b*E^(I*(c + d*x)))), x] + Int[(e + f*x)^m*(E^(I*(c + d*x))/(a + Rt[a^2 - b^2, 2] - I*b*E^(I*(c + d*x)))), x])
/; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[m, 0] && PosQ[a^2 - b^2]

Rule 4629

Int[(((e_.) + (f_.)*(x_))^(m_.)*Sec[(c_.) + (d_.)*(x_)]^(n_.))/((a_) + (b_.)*Sin[(c_.) + (d_.)*(x_)]), x_Symbo
l] :> Dist[-b^2/(a^2 - b^2), Int[(e + f*x)^m*(Sec[c + d*x]^(n - 2)/(a + b*Sin[c + d*x])), x], x] + Dist[1/(a^2
- b^2), Int[(e + f*x)^m*Sec[c + d*x]^n*(a - b*Sin[c + d*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[m
, 0] && NeQ[a^2 - b^2, 0] && IGtQ[n, 0]

Rule 6874

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {align*} \int \frac {(e+f x) \sec (c+d x)}{a+b \sin (c+d x)} \, dx &=\frac {\int (e+f x) \sec (c+d x) (a-b \sin (c+d x)) \, dx}{a^2-b^2}-\frac {b^2 \int \frac {(e+f x) \cos (c+d x)}{a+b \sin (c+d x)} \, dx}{a^2-b^2}\\ &=\frac {i b (e+f x)^2}{2 \left (a^2-b^2\right ) f}+\frac {\int (a (e+f x) \sec (c+d x)-b (e+f x) \tan (c+d x)) \, dx}{a^2-b^2}-\frac {b^2 \int \frac {e^{i (c+d x)} (e+f x)}{a-\sqrt {a^2-b^2}-i b e^{i (c+d x)}} \, dx}{a^2-b^2}-\frac {b^2 \int \frac {e^{i (c+d x)} (e+f x)}{a+\sqrt {a^2-b^2}-i b e^{i (c+d x)}} \, dx}{a^2-b^2}\\ &=\frac {i b (e+f x)^2}{2 \left (a^2-b^2\right ) f}-\frac {b (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right ) d}-\frac {b (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right ) d}+\frac {a \int (e+f x) \sec (c+d x) \, dx}{a^2-b^2}-\frac {b \int (e+f x) \tan (c+d x) \, dx}{a^2-b^2}+\frac {(b f) \int \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right ) \, dx}{\left (a^2-b^2\right ) d}+\frac {(b f) \int \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right ) \, dx}{\left (a^2-b^2\right ) d}\\ &=-\frac {2 i a (e+f x) \tan ^{-1}\left (e^{i (c+d x)}\right )}{\left (a^2-b^2\right ) d}-\frac {b (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right ) d}-\frac {b (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right ) d}+\frac {(2 i b) \int \frac {e^{2 i (c+d x)} (e+f x)}{1+e^{2 i (c+d x)}} \, dx}{a^2-b^2}-\frac {(i b f) \text {Subst}\left (\int \frac {\log \left (1-\frac {i b x}{a-\sqrt {a^2-b^2}}\right )}{x} \, dx,x,e^{i (c+d x)}\right )}{\left (a^2-b^2\right ) d^2}-\frac {(i b f) \text {Subst}\left (\int \frac {\log \left (1-\frac {i b x}{a+\sqrt {a^2-b^2}}\right )}{x} \, dx,x,e^{i (c+d x)}\right )}{\left (a^2-b^2\right ) d^2}-\frac {(a f) \int \log \left (1-i e^{i (c+d x)}\right ) \, dx}{\left (a^2-b^2\right ) d}+\frac {(a f) \int \log \left (1+i e^{i (c+d x)}\right ) \, dx}{\left (a^2-b^2\right ) d}\\ &=-\frac {2 i a (e+f x) \tan ^{-1}\left (e^{i (c+d x)}\right )}{\left (a^2-b^2\right ) d}-\frac {b (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right ) d}-\frac {b (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right ) d}+\frac {b (e+f x) \log \left (1+e^{2 i (c+d x)}\right )}{\left (a^2-b^2\right ) d}+\frac {i b f \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right ) d^2}+\frac {i b f \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right ) d^2}+\frac {(i a f) \text {Subst}\left (\int \frac {\log (1-i x)}{x} \, dx,x,e^{i (c+d x)}\right )}{\left (a^2-b^2\right ) d^2}-\frac {(i a f) \text {Subst}\left (\int \frac {\log (1+i x)}{x} \, dx,x,e^{i (c+d x)}\right )}{\left (a^2-b^2\right ) d^2}-\frac {(b f) \int \log \left (1+e^{2 i (c+d x)}\right ) \, dx}{\left (a^2-b^2\right ) d}\\ &=-\frac {2 i a (e+f x) \tan ^{-1}\left (e^{i (c+d x)}\right )}{\left (a^2-b^2\right ) d}-\frac {b (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right ) d}-\frac {b (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right ) d}+\frac {b (e+f x) \log \left (1+e^{2 i (c+d x)}\right )}{\left (a^2-b^2\right ) d}+\frac {i a f \text {Li}_2\left (-i e^{i (c+d x)}\right )}{\left (a^2-b^2\right ) d^2}-\frac {i a f \text {Li}_2\left (i e^{i (c+d x)}\right )}{\left (a^2-b^2\right ) d^2}+\frac {i b f \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right ) d^2}+\frac {i b f \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right ) d^2}+\frac {(i b f) \text {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{2 i (c+d x)}\right )}{2 \left (a^2-b^2\right ) d^2}\\ &=-\frac {2 i a (e+f x) \tan ^{-1}\left (e^{i (c+d x)}\right )}{\left (a^2-b^2\right ) d}-\frac {b (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right ) d}-\frac {b (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right ) d}+\frac {b (e+f x) \log \left (1+e^{2 i (c+d x)}\right )}{\left (a^2-b^2\right ) d}+\frac {i a f \text {Li}_2\left (-i e^{i (c+d x)}\right )}{\left (a^2-b^2\right ) d^2}-\frac {i a f \text {Li}_2\left (i e^{i (c+d x)}\right )}{\left (a^2-b^2\right ) d^2}+\frac {i b f \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right ) d^2}+\frac {i b f \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right ) d^2}-\frac {i b f \text {Li}_2\left (-e^{2 i (c+d x)}\right )}{2 \left (a^2-b^2\right ) d^2}\\ \end {align*}

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Mathematica [B] Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(2743\) vs. \(2(413)=826\).
time = 14.86, size = 2743, normalized size = 6.64 \begin {gather*} \text {Result too large to show} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[((e + f*x)*Sec[c + d*x])/(a + b*Sin[c + d*x]),x]

[Out]

((d*e + d*f*x)*(((-I)*b*(d*e + d*f*x)^2)/f + 2*(a - b)*(d*e - c*f)*Log[1 - Tan[(c + d*x)/2]] - 4*b*(d*e + d*f*
x)*Log[(-2*I)/(-I + Tan[(c + d*x)/2])] - 2*(a + b)*(d*e - c*f)*Log[1 + Tan[(c + d*x)/2]] - (4*I)*b*f*PolyLog[2
, -Cos[c + d*x] + I*Sin[c + d*x]] + (2*I)*(a + b)*f*(Log[1 + I*Tan[(c + d*x)/2]]*Log[(1/2 - I/2)*(1 + Tan[(c +
d*x)/2])] + PolyLog[2, ((1 + I) - (1 - I)*Tan[(c + d*x)/2])/2]) - (2*I)*(a + b)*f*(Log[1 - I*Tan[(c + d*x)/2]
]*Log[(1/2 + I/2)*(1 + Tan[(c + d*x)/2])] + PolyLog[2, (-1/2 - I/2)*(I + Tan[(c + d*x)/2])]) + (2*I)*(a - b)*f
*(Log[1 - I*Tan[(c + d*x)/2]]*Log[(-1/2 + I/2)*(-1 + Tan[(c + d*x)/2])] + PolyLog[2, ((1 + I) + (1 - I)*Tan[(c
+ d*x)/2])/2]) - (2*I)*(a - b)*f*(Log[1 + I*Tan[(c + d*x)/2]]*Log[(-1/2 - I/2)*(-1 + Tan[(c + d*x)/2])] + Pol
yLog[2, ((1 - I) + (1 + I)*Tan[(c + d*x)/2])/2]))*(a - b*Sin[c + d*x]))/((a^2 - b^2)*d^2*(-2*a*d*e + 2*a*c*f -
(2*I)*a*f*Log[1 - I*Tan[(c + d*x)/2]] + (2*I)*a*f*Log[1 + I*Tan[(c + d*x)/2]] + 4*b*f*Cos[c + d*x]*(Log[1 + C
os[c + d*x] - I*Sin[c + d*x]] - Log[(-2*I)/(-I + Tan[(c + d*x)/2])]) + b*(d*e - c*f + f*(c + d*x))*Sec[(c + d*
x)/2]*Sin[(3*(c + d*x))/2] + b*d*e*Tan[(c + d*x)/2] - b*c*f*Tan[(c + d*x)/2] - b*f*(c + d*x)*Tan[(c + d*x)/2]
+ (2*I)*b*f*Log[1 - I*Tan[(c + d*x)/2]]*Tan[(c + d*x)/2] - (2*I)*b*f*Log[1 + I*Tan[(c + d*x)/2]]*Tan[(c + d*x)
/2])) + ((f*(c + d*x)^2 + (2*I)*d*e*Log[Sec[(c + d*x)/2]^2] - (2*I)*c*f*Log[Sec[(c + d*x)/2]^2] - (2*I)*d*e*Lo
g[Sec[(c + d*x)/2]^2*(a + b*Sin[c + d*x])] + (2*I)*c*f*Log[Sec[(c + d*x)/2]^2*(a + b*Sin[c + d*x])] - (4*I)*f*
(c + d*x)*Log[(-2*I)/(-I + Tan[(c + d*x)/2])] - 2*f*Log[1 + I*Tan[(c + d*x)/2]]*Log[(b - Sqrt[-a^2 + b^2] + a*
Tan[(c + d*x)/2])/(I*a + b - Sqrt[-a^2 + b^2])] + 2*f*Log[1 - I*Tan[(c + d*x)/2]]*Log[-((b - Sqrt[-a^2 + b^2]
+ a*Tan[(c + d*x)/2])/(I*a - b + Sqrt[-a^2 + b^2]))] + 2*f*Log[1 - I*Tan[(c + d*x)/2]]*Log[(b + Sqrt[-a^2 + b^
2] + a*Tan[(c + d*x)/2])/((-I)*a + b + Sqrt[-a^2 + b^2])] - 2*f*Log[1 + I*Tan[(c + d*x)/2]]*Log[(b + Sqrt[-a^2
+ b^2] + a*Tan[(c + d*x)/2])/(I*a + b + Sqrt[-a^2 + b^2])] + 4*f*PolyLog[2, -Cos[c + d*x] + I*Sin[c + d*x]] +
2*f*PolyLog[2, (a*(1 - I*Tan[(c + d*x)/2]))/(a + I*(b + Sqrt[-a^2 + b^2]))] - 2*f*PolyLog[2, (a*(1 + I*Tan[(c
+ d*x)/2]))/(a - I*(b + Sqrt[-a^2 + b^2]))] + 2*f*PolyLog[2, (a*(I + Tan[(c + d*x)/2]))/(I*a - b + Sqrt[-a^2
+ b^2])] - 2*f*PolyLog[2, (a + I*a*Tan[(c + d*x)/2])/(a + I*(-b + Sqrt[-a^2 + b^2]))])*(-((b^2*e*Cos[c + d*x])
/((a^2 - b^2)*(a + b*Sin[c + d*x]))) + (b^2*c*f*Cos[c + d*x])/((a^2 - b^2)*d*(a + b*Sin[c + d*x])) - (b^2*f*(c
+ d*x)*Cos[c + d*x])/((a^2 - b^2)*d*(a + b*Sin[c + d*x]))))/(d*(2*f*(c + d*x) - (4*I)*f*Log[(-2*I)/(-I + Tan[
(c + d*x)/2])] - (4*f*Log[1 + Cos[c + d*x] - I*Sin[c + d*x]]*(I*Cos[c + d*x] + Sin[c + d*x]))/(-Cos[c + d*x] +
I*Sin[c + d*x]) + (I*f*Log[1 - (a*(1 - I*Tan[(c + d*x)/2]))/(a + I*(b + Sqrt[-a^2 + b^2]))]*Sec[(c + d*x)/2]^
2)/(1 - I*Tan[(c + d*x)/2]) - (I*f*Log[-((b - Sqrt[-a^2 + b^2] + a*Tan[(c + d*x)/2])/(I*a - b + Sqrt[-a^2 + b^
2]))]*Sec[(c + d*x)/2]^2)/(1 - I*Tan[(c + d*x)/2]) - (I*f*Log[(b + Sqrt[-a^2 + b^2] + a*Tan[(c + d*x)/2])/((-I
)*a + b + Sqrt[-a^2 + b^2])]*Sec[(c + d*x)/2]^2)/(1 - I*Tan[(c + d*x)/2]) + (I*f*Log[1 - (a*(1 + I*Tan[(c + d*
x)/2]))/(a - I*(b + Sqrt[-a^2 + b^2]))]*Sec[(c + d*x)/2]^2)/(1 + I*Tan[(c + d*x)/2]) - (I*f*Log[(b - Sqrt[-a^2
+ b^2] + a*Tan[(c + d*x)/2])/(I*a + b - Sqrt[-a^2 + b^2])]*Sec[(c + d*x)/2]^2)/(1 + I*Tan[(c + d*x)/2]) - (I*
f*Log[(b + Sqrt[-a^2 + b^2] + a*Tan[(c + d*x)/2])/(I*a + b + Sqrt[-a^2 + b^2])]*Sec[(c + d*x)/2]^2)/(1 + I*Tan
[(c + d*x)/2]) + (2*I)*d*e*Tan[(c + d*x)/2] - (2*I)*c*f*Tan[(c + d*x)/2] + ((2*I)*f*(c + d*x)*Sec[(c + d*x)/2]
^2)/(-I + Tan[(c + d*x)/2]) - (f*Log[1 - (a*(I + Tan[(c + d*x)/2]))/(I*a - b + Sqrt[-a^2 + b^2])]*Sec[(c + d*x
)/2]^2)/(I + Tan[(c + d*x)/2]) + (I*a*f*Log[1 - (a + I*a*Tan[(c + d*x)/2])/(a + I*(-b + Sqrt[-a^2 + b^2]))]*Se
c[(c + d*x)/2]^2)/(a + I*a*Tan[(c + d*x)/2]) + (a*f*Log[1 - I*Tan[(c + d*x)/2]]*Sec[(c + d*x)/2]^2)/(b - Sqrt[
-a^2 + b^2] + a*Tan[(c + d*x)/2]) - (a*f*Log[1 + I*Tan[(c + d*x)/2]]*Sec[(c + d*x)/2]^2)/(b - Sqrt[-a^2 + b^2]
+ a*Tan[(c + d*x)/2]) + (a*f*Log[1 - I*Tan[(c + d*x)/2]]*Sec[(c + d*x)/2]^2)/(b + Sqrt[-a^2 + b^2] + a*Tan[(c
+ d*x)/2]) - (a*f*Log[1 + I*Tan[(c + d*x)/2]]*Sec[(c + d*x)/2]^2)/(b + Sqrt[-a^2 + b^2] + a*Tan[(c + d*x)/2])
- ((2*I)*d*e*Cos[(c + d*x)/2]^2*(b*Cos[c + d*x]*Sec[(c + d*x)/2]^2 + Sec[(c + d*x)/2]^2*(a + b*Sin[c + d*x])*
Tan[(c + d*x)/2]))/(a + b*Sin[c + d*x]) + ((2*I)*c*f*Cos[(c + d*x)/2]^2*(b*Cos[c + d*x]*Sec[(c + d*x)/2]^2 + S
ec[(c + d*x)/2]^2*(a + b*Sin[c + d*x])*Tan[(c + d*x)/2]))/(a + b*Sin[c + d*x])))

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Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 860 vs. \(2 (373 ) = 746\).
time = 0.21, size = 861, normalized size = 2.08

method	result	size
risch	\(-\frac {e b \ln \left (i b \,{\mathrm e}^{2 i \left (d x +c \right )}-2 a \,{\mathrm e}^{i \left (d x +c \right )}-i b \right )}{d \left (a -b \right ) \left (a +b \right )}-\frac {4 e \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d \left (4 a +4 b \right )}+\frac {4 e \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d \left (4 a -4 b \right )}-\frac {4 i f \dilog \left (-i {\mathrm e}^{i \left (d x +c \right )}\right )}{d^{2} \left (4 a +4 b \right )}-\frac {4 i f \dilog \left (-i \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )\right )}{d^{2} \left (4 a -4 b \right )}-\frac {4 i f \ln \left (-i \left (i-{\mathrm e}^{i \left (d x +c \right )}\right )\right ) \ln \left (-i {\mathrm e}^{i \left (d x +c \right )}\right )}{d^{2} \left (4 a +4 b \right )}+\frac {i f b \dilog \left (\frac {i b \,{\mathrm e}^{i \left (d x +c \right )}+\sqrt {a^{2}-b^{2}}-a}{-a +\sqrt {a^{2}-b^{2}}}\right )}{d^{2} \left (a -b \right ) \left (a +b \right )}-\frac {4 f \ln \left (-i \left (i-{\mathrm e}^{i \left (d x +c \right )}\right )\right ) x}{d \left (4 a +4 b \right )}-\frac {4 f \ln \left (-i \left (i-{\mathrm e}^{i \left (d x +c \right )}\right )\right ) c}{d^{2} \left (4 a +4 b \right )}-\frac {f b \ln \left (-\frac {i b \,{\mathrm e}^{i \left (d x +c \right )}-\sqrt {a^{2}-b^{2}}-a}{a +\sqrt {a^{2}-b^{2}}}\right ) x}{d \left (a -b \right ) \left (a +b \right )}-\frac {f b \ln \left (-\frac {i b \,{\mathrm e}^{i \left (d x +c \right )}-\sqrt {a^{2}-b^{2}}-a}{a +\sqrt {a^{2}-b^{2}}}\right ) c}{d^{2} \left (a -b \right ) \left (a +b \right )}+\frac {4 f \ln \left (-i \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )\right ) x}{d \left (4 a -4 b \right )}+\frac {4 f \ln \left (-i \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )\right ) c}{d^{2} \left (4 a -4 b \right )}-\frac {f b \ln \left (\frac {i b \,{\mathrm e}^{i \left (d x +c \right )}+\sqrt {a^{2}-b^{2}}-a}{-a +\sqrt {a^{2}-b^{2}}}\right ) x}{d \left (a -b \right ) \left (a +b \right )}-\frac {f b \ln \left (\frac {i b \,{\mathrm e}^{i \left (d x +c \right )}+\sqrt {a^{2}-b^{2}}-a}{-a +\sqrt {a^{2}-b^{2}}}\right ) c}{d^{2} \left (a -b \right ) \left (a +b \right )}+\frac {i f b \dilog \left (-\frac {i b \,{\mathrm e}^{i \left (d x +c \right )}-\sqrt {a^{2}-b^{2}}-a}{a +\sqrt {a^{2}-b^{2}}}\right )}{d^{2} \left (a -b \right ) \left (a +b \right )}+\frac {f c b \ln \left (i b \,{\mathrm e}^{2 i \left (d x +c \right )}-2 a \,{\mathrm e}^{i \left (d x +c \right )}-i b \right )}{d^{2} \left (a -b \right ) \left (a +b \right )}+\frac {4 f c \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d^{2} \left (4 a +4 b \right )}-\frac {4 f c \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d^{2} \left (4 a -4 b \right )}\)	\(861\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f*x+e)*sec(d*x+c)/(a+b*sin(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

-1/d*e*b/(a-b)/(a+b)*ln(I*b*exp(2*I*(d*x+c))-2*a*exp(I*(d*x+c))-I*b)-4/d*e/(4*a+4*b)*ln(exp(I*(d*x+c))-I)+4/d*
e/(4*a-4*b)*ln(exp(I*(d*x+c))+I)-4*I/d^2*f/(4*a+4*b)*dilog(-I*exp(I*(d*x+c)))+I/d^2*f*b/(a-b)/(a+b)*dilog((I*b
*exp(I*(d*x+c))+(a^2-b^2)^(1/2)-a)/(-a+(a^2-b^2)^(1/2)))-4*I/d^2*f/(4*a+4*b)*ln(-I*(I-exp(I*(d*x+c))))*ln(-I*e
xp(I*(d*x+c)))-4*I/d^2*f/(4*a-4*b)*dilog(-I*(exp(I*(d*x+c))+I))-4/d*f/(4*a+4*b)*ln(-I*(I-exp(I*(d*x+c))))*x-4/
d^2*f/(4*a+4*b)*ln(-I*(I-exp(I*(d*x+c))))*c-1/d*f*b/(a-b)/(a+b)*ln(-(I*b*exp(I*(d*x+c))-(a^2-b^2)^(1/2)-a)/(a+
(a^2-b^2)^(1/2)))*x-1/d^2*f*b/(a-b)/(a+b)*ln(-(I*b*exp(I*(d*x+c))-(a^2-b^2)^(1/2)-a)/(a+(a^2-b^2)^(1/2)))*c+4/
d*f/(4*a-4*b)*ln(-I*(exp(I*(d*x+c))+I))*x+4/d^2*f/(4*a-4*b)*ln(-I*(exp(I*(d*x+c))+I))*c-1/d*f*b/(a-b)/(a+b)*ln
((I*b*exp(I*(d*x+c))+(a^2-b^2)^(1/2)-a)/(-a+(a^2-b^2)^(1/2)))*x-1/d^2*f*b/(a-b)/(a+b)*ln((I*b*exp(I*(d*x+c))+(
a^2-b^2)^(1/2)-a)/(-a+(a^2-b^2)^(1/2)))*c+I/d^2*f*b/(a-b)/(a+b)*dilog(-(I*b*exp(I*(d*x+c))-(a^2-b^2)^(1/2)-a)/
(a+(a^2-b^2)^(1/2)))+1/d^2*f*c*b/(a-b)/(a+b)*ln(I*b*exp(2*I*(d*x+c))-2*a*exp(I*(d*x+c))-I*b)+4/d^2*f*c/(4*a+4*
b)*ln(exp(I*(d*x+c))-I)-4/d^2*f*c/(4*a-4*b)*ln(exp(I*(d*x+c))+I)

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)*sec(d*x+c)/(a+b*sin(d*x+c)),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?`
for more de

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Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1193 vs. \(2 (360) = 720\).
time = 0.63, size = 1193, normalized size = 2.89 \begin {gather*} -\frac {-i \, b f {\rm Li}_2\left (\frac {i \, a \cos \left (d x + c\right ) - a \sin \left (d x + c\right ) + {\left (b \cos \left (d x + c\right ) + i \, b \sin \left (d x + c\right )\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} - b}{b} + 1\right ) - i \, b f {\rm Li}_2\left (\frac {i \, a \cos \left (d x + c\right ) - a \sin \left (d x + c\right ) - {\left (b \cos \left (d x + c\right ) + i \, b \sin \left (d x + c\right )\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} - b}{b} + 1\right ) + i \, b f {\rm Li}_2\left (\frac {-i \, a \cos \left (d x + c\right ) - a \sin \left (d x + c\right ) + {\left (b \cos \left (d x + c\right ) - i \, b \sin \left (d x + c\right )\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} - b}{b} + 1\right ) + i \, b f {\rm Li}_2\left (\frac {-i \, a \cos \left (d x + c\right ) - a \sin \left (d x + c\right ) - {\left (b \cos \left (d x + c\right ) - i \, b \sin \left (d x + c\right )\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} - b}{b} + 1\right ) + i \, {\left (a - b\right )} f {\rm Li}_2\left (i \, \cos \left (d x + c\right ) + \sin \left (d x + c\right )\right ) + i \, {\left (a + b\right )} f {\rm Li}_2\left (i \, \cos \left (d x + c\right ) - \sin \left (d x + c\right )\right ) - i \, {\left (a - b\right )} f {\rm Li}_2\left (-i \, \cos \left (d x + c\right ) + \sin \left (d x + c\right )\right ) - i \, {\left (a + b\right )} f {\rm Li}_2\left (-i \, \cos \left (d x + c\right ) - \sin \left (d x + c\right )\right ) - {\left (b c f - b d e\right )} \log \left (2 \, b \cos \left (d x + c\right ) + 2 i \, b \sin \left (d x + c\right ) + 2 \, b \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} + 2 i \, a\right ) - {\left (b c f - b d e\right )} \log \left (2 \, b \cos \left (d x + c\right ) - 2 i \, b \sin \left (d x + c\right ) + 2 \, b \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} - 2 i \, a\right ) - {\left (b c f - b d e\right )} \log \left (-2 \, b \cos \left (d x + c\right ) + 2 i \, b \sin \left (d x + c\right ) + 2 \, b \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} + 2 i \, a\right ) - {\left (b c f - b d e\right )} \log \left (-2 \, b \cos \left (d x + c\right ) - 2 i \, b \sin \left (d x + c\right ) + 2 \, b \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} - 2 i \, a\right ) + {\left (b d f x + b c f\right )} \log \left (-\frac {i \, a \cos \left (d x + c\right ) - a \sin \left (d x + c\right ) + {\left (b \cos \left (d x + c\right ) + i \, b \sin \left (d x + c\right )\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} - b}{b}\right ) + {\left (b d f x + b c f\right )} \log \left (-\frac {i \, a \cos \left (d x + c\right ) - a \sin \left (d x + c\right ) - {\left (b \cos \left (d x + c\right ) + i \, b \sin \left (d x + c\right )\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} - b}{b}\right ) + {\left (b d f x + b c f\right )} \log \left (-\frac {-i \, a \cos \left (d x + c\right ) - a \sin \left (d x + c\right ) + {\left (b \cos \left (d x + c\right ) - i \, b \sin \left (d x + c\right )\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} - b}{b}\right ) + {\left (b d f x + b c f\right )} \log \left (-\frac {-i \, a \cos \left (d x + c\right ) - a \sin \left (d x + c\right ) - {\left (b \cos \left (d x + c\right ) - i \, b \sin \left (d x + c\right )\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} - b}{b}\right ) + {\left ({\left (a + b\right )} c f - {\left (a + b\right )} d e\right )} \log \left (\cos \left (d x + c\right ) + i \, \sin \left (d x + c\right ) + i\right ) - {\left ({\left (a - b\right )} c f - {\left (a - b\right )} d e\right )} \log \left (\cos \left (d x + c\right ) - i \, \sin \left (d x + c\right ) + i\right ) - {\left ({\left (a + b\right )} d f x + {\left (a + b\right )} c f\right )} \log \left (i \, \cos \left (d x + c\right ) + \sin \left (d x + c\right ) + 1\right ) + {\left ({\left (a - b\right )} d f x + {\left (a - b\right )} c f\right )} \log \left (i \, \cos \left (d x + c\right ) - \sin \left (d x + c\right ) + 1\right ) - {\left ({\left (a + b\right )} d f x + {\left (a + b\right )} c f\right )} \log \left (-i \, \cos \left (d x + c\right ) + \sin \left (d x + c\right ) + 1\right ) + {\left ({\left (a - b\right )} d f x + {\left (a - b\right )} c f\right )} \log \left (-i \, \cos \left (d x + c\right ) - \sin \left (d x + c\right ) + 1\right ) + {\left ({\left (a + b\right )} c f - {\left (a + b\right )} d e\right )} \log \left (-\cos \left (d x + c\right ) + i \, \sin \left (d x + c\right ) + i\right ) - {\left ({\left (a - b\right )} c f - {\left (a - b\right )} d e\right )} \log \left (-\cos \left (d x + c\right ) - i \, \sin \left (d x + c\right ) + i\right )}{2 \, {\left (a^{2} - b^{2}\right )} d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)*sec(d*x+c)/(a+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/2*(-I*b*f*dilog((I*a*cos(d*x + c) - a*sin(d*x + c) + (b*cos(d*x + c) + I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/
b^2) - b)/b + 1) - I*b*f*dilog((I*a*cos(d*x + c) - a*sin(d*x + c) - (b*cos(d*x + c) + I*b*sin(d*x + c))*sqrt(-
(a^2 - b^2)/b^2) - b)/b + 1) + I*b*f*dilog((-I*a*cos(d*x + c) - a*sin(d*x + c) + (b*cos(d*x + c) - I*b*sin(d*x
+ c))*sqrt(-(a^2 - b^2)/b^2) - b)/b + 1) + I*b*f*dilog((-I*a*cos(d*x + c) - a*sin(d*x + c) - (b*cos(d*x + c)
- I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) - b)/b + 1) + I*(a - b)*f*dilog(I*cos(d*x + c) + sin(d*x + c)) + I*
(a + b)*f*dilog(I*cos(d*x + c) - sin(d*x + c)) - I*(a - b)*f*dilog(-I*cos(d*x + c) + sin(d*x + c)) - I*(a + b)
*f*dilog(-I*cos(d*x + c) - sin(d*x + c)) - (b*c*f - b*d*e)*log(2*b*cos(d*x + c) + 2*I*b*sin(d*x + c) + 2*b*sqr
t(-(a^2 - b^2)/b^2) + 2*I*a) - (b*c*f - b*d*e)*log(2*b*cos(d*x + c) - 2*I*b*sin(d*x + c) + 2*b*sqrt(-(a^2 - b^
2)/b^2) - 2*I*a) - (b*c*f - b*d*e)*log(-2*b*cos(d*x + c) + 2*I*b*sin(d*x + c) + 2*b*sqrt(-(a^2 - b^2)/b^2) + 2
*I*a) - (b*c*f - b*d*e)*log(-2*b*cos(d*x + c) - 2*I*b*sin(d*x + c) + 2*b*sqrt(-(a^2 - b^2)/b^2) - 2*I*a) + (b*
d*f*x + b*c*f)*log(-(I*a*cos(d*x + c) - a*sin(d*x + c) + (b*cos(d*x + c) + I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)
/b^2) - b)/b) + (b*d*f*x + b*c*f)*log(-(I*a*cos(d*x + c) - a*sin(d*x + c) - (b*cos(d*x + c) + I*b*sin(d*x + c)
)*sqrt(-(a^2 - b^2)/b^2) - b)/b) + (b*d*f*x + b*c*f)*log(-(-I*a*cos(d*x + c) - a*sin(d*x + c) + (b*cos(d*x + c
) - I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) - b)/b) + (b*d*f*x + b*c*f)*log(-(-I*a*cos(d*x + c) - a*sin(d*x +
c) - (b*cos(d*x + c) - I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) - b)/b) + ((a + b)*c*f - (a + b)*d*e)*log(cos
(d*x + c) + I*sin(d*x + c) + I) - ((a - b)*c*f - (a - b)*d*e)*log(cos(d*x + c) - I*sin(d*x + c) + I) - ((a + b
)*d*f*x + (a + b)*c*f)*log(I*cos(d*x + c) + sin(d*x + c) + 1) + ((a - b)*d*f*x + (a - b)*c*f)*log(I*cos(d*x +
c) - sin(d*x + c) + 1) - ((a + b)*d*f*x + (a + b)*c*f)*log(-I*cos(d*x + c) + sin(d*x + c) + 1) + ((a - b)*d*f*
x + (a - b)*c*f)*log(-I*cos(d*x + c) - sin(d*x + c) + 1) + ((a + b)*c*f - (a + b)*d*e)*log(-cos(d*x + c) + I*s
in(d*x + c) + I) - ((a - b)*c*f - (a - b)*d*e)*log(-cos(d*x + c) - I*sin(d*x + c) + I))/((a^2 - b^2)*d^2)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (e + f x\right ) \sec {\left (c + d x \right )}}{a + b \sin {\left (c + d x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)*sec(d*x+c)/(a+b*sin(d*x+c)),x)

[Out]

Integral((e + f*x)*sec(c + d*x)/(a + b*sin(c + d*x)), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)*sec(d*x+c)/(a+b*sin(d*x+c)),x, algorithm="giac")

[Out]

integrate((f*x + e)*sec(d*x + c)/(b*sin(d*x + c) + a), x)

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Mupad [F(-1)]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \text {Hanged} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e + f*x)/(cos(c + d*x)*(a + b*sin(c + d*x))),x)

[Out]

\text{Hanged}

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