3.4.12 \(\int \frac {(e+f x) \sec ^2(c+d x)}{a+b \sin (c+d x)} \, dx\) [312]

Optimal. Leaf size=349 \[ \frac {b f \tanh ^{-1}(\sin (c+d x))}{\left (a^2-b^2\right ) d^2}+\frac {i b^2 (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d}-\frac {i b^2 (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d}+\frac {a f \log (\cos (c+d x))}{\left (a^2-b^2\right ) d^2}+\frac {b^2 f \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d^2}-\frac {b^2 f \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d^2}-\frac {b (e+f x) \sec (c+d x)}{\left (a^2-b^2\right ) d}+\frac {a (e+f x) \tan (c+d x)}{\left (a^2-b^2\right ) d} \]

[Out]

b*f*arctanh(sin(d*x+c))/(a^2-b^2)/d^2+a*f*ln(cos(d*x+c))/(a^2-b^2)/d^2+I*b^2*(f*x+e)*ln(1-I*b*exp(I*(d*x+c))/(
a-(a^2-b^2)^(1/2)))/(a^2-b^2)^(3/2)/d-I*b^2*(f*x+e)*ln(1-I*b*exp(I*(d*x+c))/(a+(a^2-b^2)^(1/2)))/(a^2-b^2)^(3/
2)/d+b^2*f*polylog(2,I*b*exp(I*(d*x+c))/(a-(a^2-b^2)^(1/2)))/(a^2-b^2)^(3/2)/d^2-b^2*f*polylog(2,I*b*exp(I*(d*
x+c))/(a+(a^2-b^2)^(1/2)))/(a^2-b^2)^(3/2)/d^2-b*(f*x+e)*sec(d*x+c)/(a^2-b^2)/d+a*(f*x+e)*tan(d*x+c)/(a^2-b^2)
/d

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Rubi [A]
time = 0.56, antiderivative size = 349, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 11, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.423, Rules used = {4629, 3404, 2296, 2221, 2317, 2438, 6874, 4269, 3556, 4494, 3855} \begin {gather*} \frac {b^2 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 )^{3/2}}-\frac {b^2 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 )^{3/2}}+\frac {b f \tanh ^{-1}(\sin (c+d x))}{d^2 \left (a^2-b^2\right )}+\frac {a f \log (\cos (c+d x))}{d^2 \left (a^2-b^2\right )}+\frac {i b^2 (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 )^{3/2}}-\frac {i b^2 (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 )^{3/2}}+\frac {a (e+f x) \tan (c+d x)}{d \left (a^2-b^2\right )}-\frac {b (e+f x) \sec (c+d x)}{d \left (a^2-b^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((e + f*x)*Sec[c + d*x]^2)/(a + b*Sin[c + d*x]),x]

[Out]

(b*f*ArcTanh[Sin[c + d*x]])/((a^2 - b^2)*d^2) + (I*b^2*(e + f*x)*Log[1 - (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 -
b^2])])/((a^2 - b^2)^(3/2)*d) - (I*b^2*(e + f*x)*Log[1 - (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/((a^2
- b^2)^(3/2)*d) + (a*f*Log[Cos[c + d*x]])/((a^2 - b^2)*d^2) + (b^2*f*PolyLog[2, (I*b*E^(I*(c + d*x)))/(a - Sqr
t[a^2 - b^2])])/((a^2 - b^2)^(3/2)*d^2) - (b^2*f*PolyLog[2, (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/((a^
2 - b^2)^(3/2)*d^2) - (b*(e + f*x)*Sec[c + d*x])/((a^2 - b^2)*d) + (a*(e + f*x)*Tan[c + d*x])/((a^2 - b^2)*d)

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 2296

Int[((F_)^(u_)*((f_.) + (g_.)*(x_))^(m_.))/((a_.) + (b_.)*(F_)^(u_) + (c_.)*(F_)^(v_)), x_Symbol] :> With[{q =
Rt[b^2 - 4*a*c, 2]}, Dist[2*(c/q), Int[(f + g*x)^m*(F^u/(b - q + 2*c*F^u)), x], x] - Dist[2*(c/q), Int[(f + g
*x)^m*(F^u/(b + q + 2*c*F^u)), x], x]] /; FreeQ[{F, a, b, c, f, g}, x] && EqQ[v, 2*u] && LinearQ[u, x] && NeQ[
b^2 - 4*a*c, 0] && 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 3404

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

Rule 3556

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3855

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 4269

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

Rule 4494

Int[((c_.) + (d_.)*(x_))^(m_.)*Sec[(a_.) + (b_.)*(x_)]^(n_.)*Tan[(a_.) + (b_.)*(x_)]^(p_.), x_Symbol] :> Simp[
(c + d*x)^m*(Sec[a + b*x]^n/(b*n)), x] - Dist[d*(m/(b*n)), Int[(c + d*x)^(m - 1)*Sec[a + b*x]^n, x], x] /; Fre
eQ[{a, b, c, d, n}, x] && EqQ[p, 1] && GtQ[m, 0]

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 ^2(c+d x)}{a+b \sin (c+d x)} \, dx &=\frac {\int (e+f x) \sec ^2(c+d x) (a-b \sin (c+d x)) \, dx}{a^2-b^2}-\frac {b^2 \int \frac {e+f x}{a+b \sin (c+d x)} \, dx}{a^2-b^2}\\ &=\frac {\int \left (a (e+f x) \sec ^2(c+d x)-b (e+f x) \sec (c+d x) \tan (c+d x)\right ) \, dx}{a^2-b^2}-\frac {\left (2 b^2\right ) \int \frac {e^{i (c+d x)} (e+f x)}{i b+2 a e^{i (c+d x)}-i b e^{2 i (c+d x)}} \, dx}{a^2-b^2}\\ &=\frac {\left (2 i b^3\right ) \int \frac {e^{i (c+d x)} (e+f x)}{2 a-2 \sqrt {a^2-b^2}-2 i b e^{i (c+d x)}} \, dx}{\left (a^2-b^2\right )^{3/2}}-\frac {\left (2 i b^3\right ) \int \frac {e^{i (c+d x)} (e+f x)}{2 a+2 \sqrt {a^2-b^2}-2 i b e^{i (c+d x)}} \, dx}{\left (a^2-b^2\right )^{3/2}}+\frac {a \int (e+f x) \sec ^2(c+d x) \, dx}{a^2-b^2}-\frac {b \int (e+f x) \sec (c+d x) \tan (c+d x) \, dx}{a^2-b^2}\\ &=\frac {i b^2 (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 )^{3/2} d}-\frac {i b^2 (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 )^{3/2} d}-\frac {b (e+f x) \sec (c+d x)}{\left (a^2-b^2\right ) d}+\frac {a (e+f x) \tan (c+d x)}{\left (a^2-b^2\right ) d}-\frac {\left (i b^2 f\right ) \int \log \left (1-\frac {2 i b e^{i (c+d x)}}{2 a-2 \sqrt {a^2-b^2}}\right ) \, dx}{\left (a^2-b^2\right )^{3/2} d}+\frac {\left (i b^2 f\right ) \int \log \left (1-\frac {2 i b e^{i (c+d x)}}{2 a+2 \sqrt {a^2-b^2}}\right ) \, dx}{\left (a^2-b^2\right )^{3/2} d}-\frac {(a f) \int \tan (c+d x) \, dx}{\left (a^2-b^2\right ) d}+\frac {(b f) \int \sec (c+d x) \, dx}{\left (a^2-b^2\right ) d}\\ &=\frac {b f \tanh ^{-1}(\sin (c+d x))}{\left (a^2-b^2\right ) d^2}+\frac {i b^2 (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 )^{3/2} d}-\frac {i b^2 (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 )^{3/2} d}+\frac {a f \log (\cos (c+d x))}{\left (a^2-b^2\right ) d^2}-\frac {b (e+f x) \sec (c+d x)}{\left (a^2-b^2\right ) d}+\frac {a (e+f x) \tan (c+d x)}{\left (a^2-b^2\right ) d}-\frac {\left (b^2 f\right ) \text {Subst}\left (\int \frac {\log \left (1-\frac {2 i b x}{2 a-2 \sqrt {a^2-b^2}}\right )}{x} \, dx,x,e^{i (c+d x)}\right )}{\left (a^2-b^2\right )^{3/2} d^2}+\frac {\left (b^2 f\right ) \text {Subst}\left (\int \frac {\log \left (1-\frac {2 i b x}{2 a+2 \sqrt {a^2-b^2}}\right )}{x} \, dx,x,e^{i (c+d x)}\right )}{\left (a^2-b^2\right )^{3/2} d^2}\\ &=\frac {b f \tanh ^{-1}(\sin (c+d x))}{\left (a^2-b^2\right ) d^2}+\frac {i b^2 (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 )^{3/2} d}-\frac {i b^2 (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 )^{3/2} d}+\frac {a f \log (\cos (c+d x))}{\left (a^2-b^2\right ) d^2}+\frac {b^2 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 )^{3/2} d^2}-\frac {b^2 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 )^{3/2} d^2}-\frac {b (e+f x) \sec (c+d x)}{\left (a^2-b^2\right ) d}+\frac {a (e+f x) \tan (c+d x)}{\left (a^2-b^2\right ) d}\\ \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. \(842\) vs. \(2(349)=698\).
time = 5.57, size = 842, normalized size = 2.41 \begin {gather*} \frac {\frac {b d (e+f x)}{-a^2+b^2}+\frac {f \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}{a+b}+\frac {f \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}{a-b}+\frac {b^2 d (e+f x) \left (\frac {2 (d e-c f) \tan ^{-1}\left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}-\frac {i f \left (\log \left (1-i \tan \left (\frac {1}{2} (c+d x)\right )\right ) \log \left (\frac {b+\sqrt {-a^2+b^2}+a \tan \left (\frac {1}{2} (c+d x)\right )}{-i a+b+\sqrt {-a^2+b^2}}\right )+\text {Li}_2\left (\frac {a \left (1-i \tan \left (\frac {1}{2} (c+d x)\right )\right )}{a+i \left (b+\sqrt {-a^2+b^2}\right )}\right )\right )}{\sqrt {-a^2+b^2}}+\frac {i f \left (\log \left (1+i \tan \left (\frac {1}{2} (c+d x)\right )\right ) \log \left (\frac {b+\sqrt {-a^2+b^2}+a \tan \left (\frac {1}{2} (c+d x)\right )}{i a+b+\sqrt {-a^2+b^2}}\right )+\text {Li}_2\left (\frac {a \left (1+i \tan \left (\frac {1}{2} (c+d x)\right )\right )}{a-i \left (b+\sqrt {-a^2+b^2}\right )}\right )\right )}{\sqrt {-a^2+b^2}}+\frac {i f \left (\log \left (1-i \tan \left (\frac {1}{2} (c+d x)\right )\right ) \log \left (\frac {-b+\sqrt {-a^2+b^2}-a \tan \left (\frac {1}{2} (c+d x)\right )}{i a-b+\sqrt {-a^2+b^2}}\right )+\text {Li}_2\left (\frac {a \left (i+\tan \left (\frac {1}{2} (c+d x)\right )\right )}{i a-b+\sqrt {-a^2+b^2}}\right )\right )}{\sqrt {-a^2+b^2}}-\frac {i f \left (\log \left (1+i \tan \left (\frac {1}{2} (c+d x)\right )\right ) \log \left (\frac {b-\sqrt {-a^2+b^2}+a \tan \left (\frac {1}{2} (c+d x)\right )}{i a+b-\sqrt {-a^2+b^2}}\right )+\text {Li}_2\left (\frac {a+i a \tan \left (\frac {1}{2} (c+d x)\right )}{a+i \left (-b+\sqrt {-a^2+b^2}\right )}\right )\right )}{\sqrt {-a^2+b^2}}\right )}{\left (-a^2+b^2\right ) \left (d e-c f+i f \log \left (1-i \tan \left (\frac {1}{2} (c+d x)\right )\right )-i f \log \left (1+i \tan \left (\frac {1}{2} (c+d x)\right )\right )\right )}+\frac {d (e+f x) \sin \left (\frac {1}{2} (c+d x)\right )}{(a+b) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {d (e+f x) \sin \left (\frac {1}{2} (c+d x)\right )}{(a-b) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}}{d^2} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[((e + f*x)*Sec[c + d*x]^2)/(a + b*Sin[c + d*x]),x]

[Out]

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

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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. 1541 vs. \(2 (319 ) = 638\).
time = 0.33, size = 1542, normalized size = 4.42

method	result	size
risch	\(\text {Expression too large to display}\)	\(1542\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f*x+e)*sec(d*x+c)^2/(a+b*sin(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

2*(f*x+e)*(-I*a+b*exp(I*(d*x+c)))/d/(-a^2+b^2)/(1+exp(2*I*(d*x+c)))-4/(a^2-b^2)/d^2*b^2*f/(4*a+4*b)*ln(exp(I*(
d*x+c))-I)-4/(a^2-b^2)/d^2*b^2*f/(4*a-4*b)*ln(exp(I*(d*x+c))+I)-I/(a^2-b^2)^(3/2)/d^2*b^2*f/(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)))*a^2*c-I/(a^2-b^2)^(3/2)/d^2*b^4*f/(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-2/(a^2-b^2)/d^2*a*f*ln(exp(I*(d*x+c)))+1/(a^2-b^2)
^(3/2)/d^2*b^4*f/(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/(a^2-b^2)^(3
/2)/d^2*b^4*f/(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/(a^2-b^2)/d^2*a
^2*f/(4*a+4*b)*ln(exp(I*(d*x+c))-I)+4/(a^2-b^2)/d^2*a^2*f/(4*a-4*b)*ln(exp(I*(d*x+c))+I)+2*I/(a^2-b^2)/d*b^4*e
/(a-b)/(a+b)/(-a^2+b^2)^(1/2)*arctan(1/2*(2*I*b*exp(I*(d*x+c))-2*a)/(-a^2+b^2)^(1/2))-I/(a^2-b^2)^(3/2)/d*b^2*
f/(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)))*a^2*x-2*I/(a^2-b^2)/d*b^2*e/(a-b
)/(a+b)/(-a^2+b^2)^(1/2)*arctan(1/2*(2*I*b*exp(I*(d*x+c))-2*a)/(-a^2+b^2)^(1/2))*a^2-I/(a^2-b^2)^(3/2)/d*b^4*f
/(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+I/(a^2-b^2)^(3/2)/d*b^2*f/(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)))*a^2*x+2*I/(a^2-b^2)/d^2*b^2*c*f/(a-b)/(
a+b)/(-a^2+b^2)^(1/2)*arctan(1/2*(2*I*b*exp(I*(d*x+c))-2*a)/(-a^2+b^2)^(1/2))*a^2+1/(a^2-b^2)^(3/2)/d^2*b^2*f/
(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)))*a^2-2*I/(a^2-b^2)/d^2*b^4*c*f/(
a-b)/(a+b)/(-a^2+b^2)^(1/2)*arctan(1/2*(2*I*b*exp(I*(d*x+c))-2*a)/(-a^2+b^2)^(1/2))+I/(a^2-b^2)^(3/2)/d^2*b^4*
f/(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/(a^2-b^2)^(3/2)/d^2*b^2*f/(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)))*a^2*c+I/(a^2-b^2)^(3/2)/d*b^4*f/(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/(a^2-b^2)^(3/2)/d^2*b^2*f/(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)))*a^2

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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)^2/(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. 1273 vs. \(2 (315) = 630\).
time = 0.58, size = 1273, normalized size = 3.65 \begin {gather*} -\frac {i \, b^{3} f \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} \cos \left (d x + c\right ) {\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^{3} f \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} \cos \left (d x + c\right ) {\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^{3} f \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} \cos \left (d x + c\right ) {\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^{3} f \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} \cos \left (d x + c\right ) {\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 ) + 2 \, {\left (a^{2} b - b^{3}\right )} d f x - {\left (a^{3} + a^{2} b - a b^{2} - b^{3}\right )} f \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (a^{3} - a^{2} b - a b^{2} + b^{3}\right )} f \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) - {\left (b^{3} c f - b^{3} d e\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} \cos \left (d x + c\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^{3} c f - b^{3} d e\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} \cos \left (d x + c\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^{3} c f - b^{3} d e\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} \cos \left (d x + c\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^{3} c f - b^{3} d e\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} \cos \left (d x + c\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^{3} d f x + b^{3} c f\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} \cos \left (d x + c\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^{3} d f x + b^{3} c f\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} \cos \left (d x + c\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^{3} d f x + b^{3} c f\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} \cos \left (d x + c\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^{3} d f x + b^{3} c f\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} \cos \left (d x + c\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 ) + 2 \, {\left (a^{2} b - b^{3}\right )} d e - 2 \, {\left ({\left (a^{3} - a b^{2}\right )} d f x + {\left (a^{3} - a b^{2}\right )} d e\right )} \sin \left (d x + c\right )}{2 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} d^{2} \cos \left (d x + c\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)*sec(d*x+c)^2/(a+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/2*(I*b^3*f*sqrt(-(a^2 - b^2)/b^2)*cos(d*x + c)*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^3*f*sqrt(-(a^2 - b^2)/b^2)*cos(d*x + c)*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^
3*f*sqrt(-(a^2 - b^2)/b^2)*cos(d*x + c)*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^3*f*sqrt(-(a^2 - b^2)/b^2)*cos(d*x + c)*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) + 2*(a^2*b - b
^3)*d*f*x - (a^3 + a^2*b - a*b^2 - b^3)*f*cos(d*x + c)*log(sin(d*x + c) + 1) - (a^3 - a^2*b - a*b^2 + b^3)*f*c
os(d*x + c)*log(-sin(d*x + c) + 1) - (b^3*c*f - b^3*d*e)*sqrt(-(a^2 - b^2)/b^2)*cos(d*x + c)*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^3*c*f - b^3*d*e)*sqrt(-(a^2 - b^2)/b^2)*co
s(d*x + c)*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^3*c*f - b^3*d*
e)*sqrt(-(a^2 - b^2)/b^2)*cos(d*x + c)*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^3*c*f - b^3*d*e)*sqrt(-(a^2 - b^2)/b^2)*cos(d*x + c)*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^3*d*f*x + b^3*c*f)*sqrt(-(a^2 - b^2)/b^2)*cos(d*x + c)*log(-(I*a*c
os(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^3*d*f*x
+ b^3*c*f)*sqrt(-(a^2 - b^2)/b^2)*cos(d*x + c)*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^3*d*f*x + b^3*c*f)*sqrt(-(a^2 - b^2)/b^2)*cos(d*x + c)*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^3*d*f*x + b^3*c*f)*sqrt(-(a^2 - b^2)/b^2)*cos(d*x + c)*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) + 2*(a^2*b - b^3)*d*e - 2*((a^3 - a*b^2)*d*f*x + (a^3
- a*b^2)*d*e)*sin(d*x + c))/((a^4 - 2*a^2*b^2 + b^4)*d^2*cos(d*x + c))

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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 ^{2}{\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)**2/(a+b*sin(d*x+c)),x)

[Out]

Integral((e + f*x)*sec(c + d*x)**2/(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)^2/(a+b*sin(d*x+c)),x, algorithm="giac")

[Out]

integrate((f*x + e)*sec(d*x + c)^2/(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)^2*(a + b*sin(c + d*x))),x)

[Out]

\text{Hanged}

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