3.271 \(\int \frac{(e+f x) \sec (c+d x)}{a+a \sin (c+d x)} \, dx\)

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

[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)

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Rubi [A]  time = 0.138839, antiderivative size = 172, normalized size of antiderivative = 1., 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 = {4531, 4185, 4181, 2279, 2391, 4409, 3767, 8} \[ \frac{i f \text{PolyLog}\left (2,-i e^{i (c+d x)}\right )}{2 a d^2}-\frac{i f \text{PolyLog}\left (2,i e^{i (c+d x)}\right )}{2 a d^2}+\frac{f \tan (c+d x)}{2 a d^2}-\frac{f \sec (c+d x)}{2 a d^2}-\frac{i (e+f x) \tan ^{-1}\left (e^{i (c+d x)}\right )}{a d}-\frac{(e+f x) \sec ^2(c+d x)}{2 a d}+\frac{(e+f x) \tan (c+d x) \sec (c+d x)}{2 a d} \]

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 4531

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]

Rule 4185

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] && G
tQ[n, 1] && NeQ[n, 2]

Rule 4181

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 2279

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 2391

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 4409

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 3767

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 8

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

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 \operatorname{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) \operatorname{Subst}\left (\int \frac{\log (1-i x)}{x} \, dx,x,e^{i (c+d x)}\right )}{2 a d^2}-\frac{(i f) \operatorname{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*}

Mathematica [B]  time = 2.93253, size = 655, normalized size = 3.81 \[ -\frac{-\frac{f \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2 \left ((-1)^{3/4} (c+d x)^2+\frac{4 i \text{PolyLog}\left (2,-i e^{i (c+d x)}\right )-3 i \pi (c+d x)-4 \pi \log \left (1+e^{-i (c+d x)}\right )+2 (-2 c-2 d x+\pi ) \log \left (1+i e^{i (c+d x)}\right )-2 \pi \log \left (\sin \left (\frac{1}{4} (2 c+2 d x-\pi )\right )\right )+4 \pi \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )}{\sqrt{2}}\right )}{\sqrt{2}}+\frac{f \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2 \left (\sqrt [4]{-1} (c+d x)^2+\frac{4 i \text{PolyLog}\left (2,i e^{i (c+d x)}\right )-i \pi (c+d x)-4 \pi \log \left (1+e^{-i (c+d x)}\right )-2 (2 c+2 d x+\pi ) \log \left (1-i e^{i (c+d x)}\right )+2 \pi \log \left (\sin \left (\frac{1}{4} (2 c+2 d x+\pi )\right )\right )+4 \pi \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )}{\sqrt{2}}\right )}{\sqrt{2}}+(c+d x) (c f-d (2 e+f x)) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2+d e \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2 \left (2 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+c+d x\right )+d e \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2 \left (-2 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+c+d x\right )-4 f \sin \left (\frac{1}{2} (c+d x)\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )-c f \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2 \left (2 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+c+d x\right )-c f \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2 \left (-2 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+c+d x\right )+2 d (e+f x)}{4 a d^2 (\sin (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(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] + 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 - (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)*Po
lyLog[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])/(4*a*d^2*(1 + Sin[c + d*x]))

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Maple [B]  time = 0.214, size = 303, normalized size = 1.8 \begin{align*}{\frac{-i \left ( dfx{{\rm e}^{i \left ( dx+c \right ) }}+de{{\rm e}^{i \left ( dx+c \right ) }}+f-if{{\rm e}^{i \left ( dx+c \right ) }} \right ) }{{d}^{2} \left ({{\rm e}^{i \left ( dx+c \right ) }}+i \right ) ^{2}a}}-{\frac{e\ln \left ({{\rm e}^{i \left ( dx+c \right ) }}-i \right ) }{2\,da}}+{\frac{e\ln \left ({{\rm e}^{i \left ( dx+c \right ) }}+i \right ) }{2\,da}}-{\frac{f\ln \left ( 1+i{{\rm e}^{i \left ( dx+c \right ) }} \right ) x}{2\,da}}-{\frac{f\ln \left ( 1+i{{\rm e}^{i \left ( dx+c \right ) }} \right ) c}{2\,a{d}^{2}}}+{\frac{{\frac{i}{2}}f{\it polylog} \left ( 2,-i{{\rm e}^{i \left ( dx+c \right ) }} \right ) }{a{d}^{2}}}+{\frac{f\ln \left ( 1-i{{\rm e}^{i \left ( dx+c \right ) }} \right ) x}{2\,da}}+{\frac{f\ln \left ( 1-i{{\rm e}^{i \left ( dx+c \right ) }} \right ) c}{2\,a{d}^{2}}}-{\frac{{\frac{i}{2}}f{\it polylog} \left ( 2,i{{\rm e}^{i \left ( dx+c \right ) }} \right ) }{a{d}^{2}}}+{\frac{cf\ln \left ({{\rm e}^{i \left ( dx+c \right ) }}-i \right ) }{2\,a{d}^{2}}}-{\frac{cf\ln \left ({{\rm e}^{i \left ( dx+c \right ) }}+i \right ) }{2\,a{d}^{2}}} \end{align*}

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)

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

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Maxima [B]  time = 1.62525, size = 986, normalized size = 5.73 \begin{align*} \text{result too large to display} \end{align*}

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) + 4*I*d*e*cos(d*x + c) + 2*I*d*e*sin(2*d*x + 2*c) - 4*d*e*sin(d*x + c) - 2*d*e)*arcta
n2(sin(d*x + c) + 1, cos(d*x + c)) - (2*d*e*cos(2*d*x + 2*c) + 4*I*d*e*cos(d*x + c) + 2*I*d*e*sin(2*d*x + 2*c)
- 4*d*e*sin(d*x + c) - 2*d*e)*arctan2(sin(d*x + c) - 1, cos(d*x + c)) - (2*d*f*x*cos(2*d*x + 2*c) + 4*I*d*f*x
*cos(d*x + c) + 2*I*d*f*x*sin(2*d*x + 2*c) - 4*d*f*x*sin(d*x + c) - 2*d*f*x)*arctan2(cos(d*x + c), sin(d*x + c
) + 1) - (2*d*f*x*cos(2*d*x + 2*c) + 4*I*d*f*x*cos(d*x + c) + 2*I*d*f*x*sin(2*d*x + 2*c) - 4*d*f*x*sin(d*x + c
) - 2*d*f*x)*arctan2(cos(d*x + c), -sin(d*x + c) + 1) - (4*d*f*x + 4*d*e - 4*I*f)*cos(d*x + c) - (2*f*cos(2*d*
x + 2*c) + 4*I*f*cos(d*x + c) + 2*I*f*sin(2*d*x + 2*c) - 4*f*sin(d*x + c) - 2*f)*dilog(I*e^(I*d*x + I*c)) + (2
*f*cos(2*d*x + 2*c) + 4*I*f*cos(d*x + c) + 2*I*f*sin(2*d*x + 2*c) - 4*f*sin(d*x + c) - 2*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 + 2*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)*s
in(2*d*x + 2*c) + (-2*I*d*f*x - 2*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 - 4*I*d*e - 4*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*sin(d*x + c) + 4*I*a*d^2)

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Fricas [B]  time = 2.11148, size = 1339, normalized size = 7.78 \begin{align*} -\frac{2 \, d f x + 2 \, d e + 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 ) -{\left (d e - c f +{\left (d e - c f\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 e - c f +{\left (d e - c f\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 (d e - c f +{\left (d e - c f\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 e - c f +{\left (d e - c f\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{align*}

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*d*e + 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)) - (d*e - c*f + (d*e - c*f)*si
n(d*x + c))*log(cos(d*x + c) + I*sin(d*x + c) + I) + (d*e - c*f + (d*e - c*f)*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) - (d*e - c*f + (d*e - c*f)*sin(d*x + c))*log(-cos(d*x + c) + I*sin(d*x + c) + I)
+ (d*e - c*f + (d*e - c*f)*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., size = 0, normalized size = 0. \begin{align*} \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{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (f x + e\right )} \sec \left (d x + c\right )}{a \sin \left (d x + c\right ) + a}\,{d x} \end{align*}

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)