3.3.71 \(\int \frac {(e+f x) \sec (c+d x)}{a+a \sin (c+d x)} \, dx\) [271]

Optimal. Leaf size=172 \[ -\frac {i (e+f x) \tan ^{-1}\left (e^{i (c+d x)}\right )}{a d}+\frac {i f \text {Li}_2\left (-i e^{i (c+d x)}\right )}{2 a d^2}-\frac {i f \text {Li}_2\left (i e^{i (c+d x)}\right )}{2 a d^2}-\frac {f \sec (c+d x)}{2 a d^2}-\frac {(e+f x) \sec ^2(c+d x)}{2 a d}+\frac {f \tan (c+d x)}{2 a d^2}+\frac {(e+f x) \sec (c+d x) \tan (c+d x)}{2 a d} \]

[Out]

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

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Rubi [A]
time = 0.10, antiderivative size = 172, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 8, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {4627, 4270, 4266, 2317, 2438, 4494, 3852, 8} \begin {gather*} \frac {i f \text {PolyLog}\left (2,-i e^{i (c+d x)}\right )}{2 a d^2}-\frac {i f \text {PolyLog}\left (2,i e^{i (c+d x)}\right )}{2 a d^2}-\frac {i (e+f x) \text {ArcTan}\left (e^{i (c+d x)}\right )}{a d}+\frac {f \tan (c+d x)}{2 a d^2}-\frac {f \sec (c+d x)}{2 a d^2}-\frac {(e+f x) \sec ^2(c+d x)}{2 a d}+\frac {(e+f x) \tan (c+d x) \sec (c+d x)}{2 a d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-I)*(e + f*x)*ArcTan[E^(I*(c + d*x))])/(a*d) + ((I/2)*f*PolyLog[2, (-I)*E^(I*(c + d*x))])/(a*d^2) - ((I/2)*f
*PolyLog[2, I*E^(I*(c + d*x))])/(a*d^2) - (f*Sec[c + d*x])/(2*a*d^2) - ((e + f*x)*Sec[c + d*x]^2)/(2*a*d) + (f
*Tan[c + d*x])/(2*a*d^2) + ((e + f*x)*Sec[c + d*x]*Tan[c + d*x])/(2*a*d)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

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 3852

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 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 4270

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

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 4627

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

Rubi steps

\begin {align*} \int \frac {(e+f x) \sec (c+d x)}{a+a \sin (c+d x)} \, dx &=\frac {\int (e+f x) \sec ^3(c+d x) \, dx}{a}-\frac {\int (e+f x) \sec ^2(c+d x) \tan (c+d x) \, dx}{a}\\ &=-\frac {f \sec (c+d x)}{2 a d^2}-\frac {(e+f x) \sec ^2(c+d x)}{2 a d}+\frac {(e+f x) \sec (c+d x) \tan (c+d x)}{2 a d}+\frac {\int (e+f x) \sec (c+d x) \, dx}{2 a}+\frac {f \int \sec ^2(c+d x) \, dx}{2 a d}\\ &=-\frac {i (e+f x) \tan ^{-1}\left (e^{i (c+d x)}\right )}{a d}-\frac {f \sec (c+d x)}{2 a d^2}-\frac {(e+f x) \sec ^2(c+d x)}{2 a d}+\frac {(e+f x) \sec (c+d x) \tan (c+d x)}{2 a d}-\frac {f \text {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{2 a d^2}-\frac {f \int \log \left (1-i e^{i (c+d x)}\right ) \, dx}{2 a d}+\frac {f \int \log \left (1+i e^{i (c+d x)}\right ) \, dx}{2 a d}\\ &=-\frac {i (e+f x) \tan ^{-1}\left (e^{i (c+d x)}\right )}{a d}-\frac {f \sec (c+d x)}{2 a d^2}-\frac {(e+f x) \sec ^2(c+d x)}{2 a d}+\frac {f \tan (c+d x)}{2 a d^2}+\frac {(e+f x) \sec (c+d x) \tan (c+d x)}{2 a d}+\frac {(i f) \text {Subst}\left (\int \frac {\log (1-i x)}{x} \, dx,x,e^{i (c+d x)}\right )}{2 a d^2}-\frac {(i f) \text {Subst}\left (\int \frac {\log (1+i x)}{x} \, dx,x,e^{i (c+d x)}\right )}{2 a d^2}\\ &=-\frac {i (e+f x) \tan ^{-1}\left (e^{i (c+d x)}\right )}{a d}+\frac {i f \text {Li}_2\left (-i e^{i (c+d x)}\right )}{2 a d^2}-\frac {i f \text {Li}_2\left (i e^{i (c+d x)}\right )}{2 a d^2}-\frac {f \sec (c+d x)}{2 a d^2}-\frac {(e+f x) \sec ^2(c+d x)}{2 a d}+\frac {f \tan (c+d x)}{2 a d^2}+\frac {(e+f x) \sec (c+d x) \tan (c+d x)}{2 a 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. \(655\) vs. \(2(172)=344\).
time = 2.96, size = 655, normalized size = 3.81 \begin {gather*} -\frac {2 d (e+f x)-4 f \sin \left (\frac {1}{2} (c+d x)\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+(c+d x) (c f-d (2 e+f x)) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2+d e \left (c+d x+2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2-c f \left (c+d x+2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2+d e \left (c+d x-2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2-c f \left (c+d x-2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2-\frac {f \left ((-1)^{3/4} (c+d x)^2+\frac {-3 i \pi (c+d x)-4 \pi \log \left (1+e^{-i (c+d x)}\right )+2 (-2 c+\pi -2 d x) \log \left (1+i e^{i (c+d x)}\right )+4 \pi \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-2 \pi \log \left (\sin \left (\frac {1}{4} (2 c-\pi +2 d x)\right )\right )+4 i \text {Li}_2\left (-i e^{i (c+d x)}\right )}{\sqrt {2}}\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}{\sqrt {2}}+\frac {f \left (\sqrt [4]{-1} (c+d x)^2+\frac {-i \pi (c+d x)-4 \pi \log \left (1+e^{-i (c+d x)}\right )-2 (2 c+\pi +2 d x) \log \left (1-i e^{i (c+d x)}\right )+4 \pi \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+2 \pi \log \left (\sin \left (\frac {1}{4} (2 c+\pi +2 d x)\right )\right )+4 i \text {Li}_2\left (i e^{i (c+d x)}\right )}{\sqrt {2}}\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}{\sqrt {2}}}{4 a d^2 (1+\sin (c+d x))} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*(2*d*(e + f*x) - 4*f*Sin[(c + d*x)/2]*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]) + (c + d*x)*(c*f - d*(2*e + f
*x))*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2 + d*e*(c + d*x + 2*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]])*(Cos
[(c + d*x)/2] + Sin[(c + d*x)/2])^2 - c*f*(c + d*x + 2*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]])*(Cos[(c + d*x
)/2] + Sin[(c + d*x)/2])^2 + d*e*(c + d*x - 2*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]])*(Cos[(c + d*x)/2] + Si
n[(c + d*x)/2])^2 - c*f*(c + d*x - 2*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]])*(Cos[(c + d*x)/2] + Sin[(c + d*
x)/2])^2 - (f*((-1)^(3/4)*(c + d*x)^2 + ((-3*I)*Pi*(c + d*x) - 4*Pi*Log[1 + E^((-I)*(c + d*x))] + 2*(-2*c + Pi
- 2*d*x)*Log[1 + I*E^(I*(c + d*x))] + 4*Pi*Log[Cos[(c + d*x)/2]] - 2*Pi*Log[Sin[(2*c - Pi + 2*d*x)/4]] + (4*I
)*PolyLog[2, (-I)*E^(I*(c + d*x))])/Sqrt[2])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2)/Sqrt[2] + (f*((-1)^(1/4)
*(c + d*x)^2 + ((-I)*Pi*(c + d*x) - 4*Pi*Log[1 + E^((-I)*(c + d*x))] - 2*(2*c + Pi + 2*d*x)*Log[1 - I*E^(I*(c
+ d*x))] + 4*Pi*Log[Cos[(c + d*x)/2]] + 2*Pi*Log[Sin[(2*c + Pi + 2*d*x)/4]] + (4*I)*PolyLog[2, I*E^(I*(c + d*x
))])/Sqrt[2])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2)/Sqrt[2])/(a*d^2*(1 + 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. 302 vs. \(2 (149 ) = 298\).
time = 0.20, size = 303, normalized size = 1.76

method	result	size
risch	\(-\frac {i \left (d f x \,{\mathrm e}^{i \left (d x +c \right )}+d e \,{\mathrm e}^{i \left (d x +c \right )}+f -i f \,{\mathrm e}^{i \left (d x +c \right )}\right )}{d^{2} \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{2} a}-\frac {e \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 a d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) e}{2 d a}-\frac {f \ln \left (1+i {\mathrm e}^{i \left (d x +c \right )}\right ) x}{2 a d}-\frac {f \ln \left (1+i {\mathrm e}^{i \left (d x +c \right )}\right ) c}{2 a \,d^{2}}+\frac {i f \polylog \left (2, -i {\mathrm e}^{i \left (d x +c \right )}\right )}{2 a \,d^{2}}+\frac {f \ln \left (1-i {\mathrm e}^{i \left (d x +c \right )}\right ) x}{2 d a}+\frac {f \ln \left (1-i {\mathrm e}^{i \left (d x +c \right )}\right ) c}{2 d^{2} a}-\frac {i f \polylog \left (2, i {\mathrm e}^{i \left (d x +c \right )}\right )}{2 a \,d^{2}}+\frac {f c \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 a \,d^{2}}-\frac {f c \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 d^{2} a}\)	\(303\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 725 vs. \(2 (144) = 288\).
time = 0.41, size = 725, normalized size = 4.22 \begin {gather*} \frac {2 \, {\left (d e \cos \left (2 \, d x + 2 \, c\right ) + 2 i \, d e \cos \left (d x + c\right ) + i \, d e \sin \left (2 \, d x + 2 \, c\right ) - 2 \, d e \sin \left (d x + c\right ) - d e\right )} \arctan \left (\sin \left (d x + c\right ) + 1, \cos \left (d x + c\right )\right ) - 2 \, {\left (d e \cos \left (2 \, d x + 2 \, c\right ) + 2 i \, d e \cos \left (d x + c\right ) + i \, d e \sin \left (2 \, d x + 2 \, c\right ) - 2 \, d e \sin \left (d x + c\right ) - d e\right )} \arctan \left (\sin \left (d x + c\right ) - 1, \cos \left (d x + c\right )\right ) - 2 \, {\left (d f x \cos \left (2 \, d x + 2 \, c\right ) + 2 i \, d f x \cos \left (d x + c\right ) + i \, d f x \sin \left (2 \, d x + 2 \, c\right ) - 2 \, d f x \sin \left (d x + c\right ) - d f x\right )} \arctan \left (\cos \left (d x + c\right ), \sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (d f x \cos \left (2 \, d x + 2 \, c\right ) + 2 i \, d f x \cos \left (d x + c\right ) + i \, d f x \sin \left (2 \, d x + 2 \, c\right ) - 2 \, d f x \sin \left (d x + c\right ) - d f x\right )} \arctan \left (\cos \left (d x + c\right ), -\sin \left (d x + c\right ) + 1\right ) - 4 \, {\left (d f x + d e - i \, f\right )} \cos \left (d x + c\right ) - 2 \, {\left (f \cos \left (2 \, d x + 2 \, c\right ) + 2 i \, f \cos \left (d x + c\right ) + i \, f \sin \left (2 \, d x + 2 \, c\right ) - 2 \, f \sin \left (d x + c\right ) - f\right )} {\rm Li}_2\left (i \, e^{\left (i \, d x + i \, c\right )}\right ) + 2 \, {\left (f \cos \left (2 \, d x + 2 \, c\right ) + 2 i \, f \cos \left (d x + c\right ) + i \, f \sin \left (2 \, d x + 2 \, c\right ) - 2 \, f \sin \left (d x + c\right ) - f\right )} {\rm Li}_2\left (-i \, e^{\left (i \, d x + i \, c\right )}\right ) + {\left (i \, d f x + i \, d e + {\left (-i \, d f x - i \, d e\right )} \cos \left (2 \, d x + 2 \, c\right ) + 2 \, {\left (d f x + d e\right )} \cos \left (d x + c\right ) + {\left (d f x + d e\right )} \sin \left (2 \, d x + 2 \, c\right ) - 2 \, {\left (-i \, d f x - i \, d e\right )} \sin \left (d x + c\right )\right )} \log \left (\cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} + 2 \, \sin \left (d x + c\right ) + 1\right ) + {\left (-i \, d f x - i \, d e + {\left (i \, d f x + i \, d e\right )} \cos \left (2 \, d x + 2 \, c\right ) - 2 \, {\left (d f x + d e\right )} \cos \left (d x + c\right ) - {\left (d f x + d e\right )} \sin \left (2 \, d x + 2 \, c\right ) - 2 \, {\left (i \, d f x + i \, d e\right )} \sin \left (d x + c\right )\right )} \log \left (\cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right ) + 1\right ) - 4 \, {\left (i \, d f x + i \, d e + f\right )} \sin \left (d x + c\right ) - 4 \, f}{-4 i \, a d^{2} \cos \left (2 \, d x + 2 \, c\right ) + 8 \, a d^{2} \cos \left (d x + c\right ) + 4 \, a d^{2} \sin \left (2 \, d x + 2 \, c\right ) + 8 i \, a d^{2} \sin \left (d x + c\right ) + 4 i \, a d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*(d*e*cos(2*d*x + 2*c) + 2*I*d*e*cos(d*x + c) + I*d*e*sin(2*d*x + 2*c) - 2*d*e*sin(d*x + c) - d*e)*arctan2(s
in(d*x + c) + 1, cos(d*x + c)) - 2*(d*e*cos(2*d*x + 2*c) + 2*I*d*e*cos(d*x + c) + I*d*e*sin(2*d*x + 2*c) - 2*d
*e*sin(d*x + c) - d*e)*arctan2(sin(d*x + c) - 1, cos(d*x + c)) - 2*(d*f*x*cos(2*d*x + 2*c) + 2*I*d*f*x*cos(d*x
+ c) + I*d*f*x*sin(2*d*x + 2*c) - 2*d*f*x*sin(d*x + c) - d*f*x)*arctan2(cos(d*x + c), sin(d*x + c) + 1) - 2*(
d*f*x*cos(2*d*x + 2*c) + 2*I*d*f*x*cos(d*x + c) + I*d*f*x*sin(2*d*x + 2*c) - 2*d*f*x*sin(d*x + c) - d*f*x)*arc
tan2(cos(d*x + c), -sin(d*x + c) + 1) - 4*(d*f*x + d*e - I*f)*cos(d*x + c) - 2*(f*cos(2*d*x + 2*c) + 2*I*f*cos
(d*x + c) + I*f*sin(2*d*x + 2*c) - 2*f*sin(d*x + c) - f)*dilog(I*e^(I*d*x + I*c)) + 2*(f*cos(2*d*x + 2*c) + 2*
I*f*cos(d*x + c) + I*f*sin(2*d*x + 2*c) - 2*f*sin(d*x + c) - f)*dilog(-I*e^(I*d*x + I*c)) + (I*d*f*x + I*d*e +
(-I*d*f*x - I*d*e)*cos(2*d*x + 2*c) + 2*(d*f*x + d*e)*cos(d*x + c) + (d*f*x + d*e)*sin(2*d*x + 2*c) - 2*(-I*d
*f*x - I*d*e)*sin(d*x + c))*log(cos(d*x + c)^2 + sin(d*x + c)^2 + 2*sin(d*x + c) + 1) + (-I*d*f*x - I*d*e + (I
*d*f*x + I*d*e)*cos(2*d*x + 2*c) - 2*(d*f*x + d*e)*cos(d*x + c) - (d*f*x + d*e)*sin(2*d*x + 2*c) - 2*(I*d*f*x
+ I*d*e)*sin(d*x + c))*log(cos(d*x + c)^2 + sin(d*x + c)^2 - 2*sin(d*x + c) + 1) - 4*(I*d*f*x + I*d*e + f)*sin
(d*x + c) - 4*f)/(-4*I*a*d^2*cos(2*d*x + 2*c) + 8*a*d^2*cos(d*x + c) + 4*a*d^2*sin(2*d*x + 2*c) + 8*I*a*d^2*si
n(d*x + c) + 4*I*a*d^2)

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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. 517 vs. \(2 (147) = 294\).
time = 0.41, size = 517, normalized size = 3.01 \begin {gather*} -\frac {2 \, d f x + 2 \, f \cos \left (d x + c\right ) - {\left (-i \, f \sin \left (d x + c\right ) - i \, f\right )} {\rm Li}_2\left (i \, \cos \left (d x + c\right ) + \sin \left (d x + c\right )\right ) - {\left (-i \, f \sin \left (d x + c\right ) - i \, f\right )} {\rm Li}_2\left (i \, \cos \left (d x + c\right ) - \sin \left (d x + c\right )\right ) - {\left (i \, f \sin \left (d x + c\right ) + i \, f\right )} {\rm Li}_2\left (-i \, \cos \left (d x + c\right ) + \sin \left (d x + c\right )\right ) - {\left (i \, f \sin \left (d x + c\right ) + i \, f\right )} {\rm Li}_2\left (-i \, \cos \left (d x + c\right ) - \sin \left (d x + c\right )\right ) + 2 \, d e + {\left (c f - d e + {\left (c f - d e\right )} \sin \left (d x + c\right )\right )} \log \left (\cos \left (d x + c\right ) + i \, \sin \left (d x + c\right ) + i\right ) - {\left (c f - d e + {\left (c f - d e\right )} \sin \left (d x + c\right )\right )} \log \left (\cos \left (d x + c\right ) - i \, \sin \left (d x + c\right ) + i\right ) - {\left (d f x + c f + {\left (d f x + c f\right )} \sin \left (d x + c\right )\right )} \log \left (i \, \cos \left (d x + c\right ) + \sin \left (d x + c\right ) + 1\right ) + {\left (d f x + c f + {\left (d f x + c f\right )} \sin \left (d x + c\right )\right )} \log \left (i \, \cos \left (d x + c\right ) - \sin \left (d x + c\right ) + 1\right ) - {\left (d f x + c f + {\left (d f x + c f\right )} \sin \left (d x + c\right )\right )} \log \left (-i \, \cos \left (d x + c\right ) + \sin \left (d x + c\right ) + 1\right ) + {\left (d f x + c f + {\left (d f x + c f\right )} \sin \left (d x + c\right )\right )} \log \left (-i \, \cos \left (d x + c\right ) - \sin \left (d x + c\right ) + 1\right ) + {\left (c f - d e + {\left (c f - d e\right )} \sin \left (d x + c\right )\right )} \log \left (-\cos \left (d x + c\right ) + i \, \sin \left (d x + c\right ) + i\right ) - {\left (c f - d e + {\left (c f - d e\right )} \sin \left (d x + c\right )\right )} \log \left (-\cos \left (d x + c\right ) - i \, \sin \left (d x + c\right ) + i\right )}{4 \, {\left (a d^{2} \sin \left (d x + c\right ) + a d^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(2*d*f*x + 2*f*cos(d*x + c) - (-I*f*sin(d*x + c) - I*f)*dilog(I*cos(d*x + c) + sin(d*x + c)) - (-I*f*sin(
d*x + c) - I*f)*dilog(I*cos(d*x + c) - sin(d*x + c)) - (I*f*sin(d*x + c) + I*f)*dilog(-I*cos(d*x + c) + sin(d*
x + c)) - (I*f*sin(d*x + c) + I*f)*dilog(-I*cos(d*x + c) - sin(d*x + c)) + 2*d*e + (c*f - d*e + (c*f - d*e)*si
n(d*x + c))*log(cos(d*x + c) + I*sin(d*x + c) + I) - (c*f - d*e + (c*f - d*e)*sin(d*x + c))*log(cos(d*x + c) -
I*sin(d*x + c) + I) - (d*f*x + c*f + (d*f*x + c*f)*sin(d*x + c))*log(I*cos(d*x + c) + sin(d*x + c) + 1) + (d*
f*x + c*f + (d*f*x + c*f)*sin(d*x + c))*log(I*cos(d*x + c) - sin(d*x + c) + 1) - (d*f*x + c*f + (d*f*x + c*f)*
sin(d*x + c))*log(-I*cos(d*x + c) + sin(d*x + c) + 1) + (d*f*x + c*f + (d*f*x + c*f)*sin(d*x + c))*log(-I*cos(
d*x + c) - sin(d*x + c) + 1) + (c*f - d*e + (c*f - d*e)*sin(d*x + c))*log(-cos(d*x + c) + I*sin(d*x + c) + I)
- (c*f - d*e + (c*f - d*e)*sin(d*x + c))*log(-cos(d*x + c) - I*sin(d*x + c) + I))/(a*d^2*sin(d*x + c) + a*d^2)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {e \sec {\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx + \int \frac {f x \sec {\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx}{a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(e*sec(c + d*x)/(sin(c + d*x) + 1), x) + Integral(f*x*sec(c + d*x)/(sin(c + d*x) + 1), x))/a

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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+a*sin(d*x+c)),x, algorithm="giac")

[Out]

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

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Mupad [F(-1)]
time = 0.00, size = -1, normalized size = -0.01 \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 + a*sin(c + d*x))),x)

[Out]

\text{Hanged}

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