3.106 \(\int \frac{(a+a \sin (e+f x))^2}{(c+d x)^3} \, dx\)

Optimal. Leaf size=225 \[ -\frac{a^2 f^2 \text{CosIntegral}\left (\frac{c f}{d}+f x\right ) \sin \left (e-\frac{c f}{d}\right )}{d^3}+\frac{a^2 f^2 \text{CosIntegral}\left (\frac{2 c f}{d}+2 f x\right ) \cos \left (2 e-\frac{2 c f}{d}\right )}{d^3}-\frac{a^2 f^2 \sin \left (2 e-\frac{2 c f}{d}\right ) \text{Si}\left (2 x f+\frac{2 c f}{d}\right )}{d^3}-\frac{a^2 f^2 \cos \left (e-\frac{c f}{d}\right ) \text{Si}\left (x f+\frac{c f}{d}\right )}{d^3}-\frac{4 a^2 f \sin ^3\left (\frac{e}{2}+\frac{f x}{2}+\frac{\pi }{4}\right ) \cos \left (\frac{e}{2}+\frac{f x}{2}+\frac{\pi }{4}\right )}{d^2 (c+d x)}-\frac{2 a^2 \sin ^4\left (\frac{e}{2}+\frac{f x}{2}+\frac{\pi }{4}\right )}{d (c+d x)^2} \]

[Out]

(a^2*f^2*Cos[2*e - (2*c*f)/d]*CosIntegral[(2*c*f)/d + 2*f*x])/d^3 - (a^2*f^2*CosIntegral[(c*f)/d + f*x]*Sin[e
- (c*f)/d])/d^3 - (4*a^2*f*Cos[e/2 + Pi/4 + (f*x)/2]*Sin[e/2 + Pi/4 + (f*x)/2]^3)/(d^2*(c + d*x)) - (2*a^2*Sin
[e/2 + Pi/4 + (f*x)/2]^4)/(d*(c + d*x)^2) - (a^2*f^2*Cos[e - (c*f)/d]*SinIntegral[(c*f)/d + f*x])/d^3 - (a^2*f
^2*Sin[2*e - (2*c*f)/d]*SinIntegral[(2*c*f)/d + 2*f*x])/d^3

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Rubi [A]  time = 0.506181, antiderivative size = 225, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 8, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {3318, 3314, 3309, 31, 3303, 3299, 3302, 3312} \[ -\frac{a^2 f^2 \text{CosIntegral}\left (\frac{c f}{d}+f x\right ) \sin \left (e-\frac{c f}{d}\right )}{d^3}+\frac{a^2 f^2 \text{CosIntegral}\left (\frac{2 c f}{d}+2 f x\right ) \cos \left (2 e-\frac{2 c f}{d}\right )}{d^3}-\frac{a^2 f^2 \sin \left (2 e-\frac{2 c f}{d}\right ) \text{Si}\left (2 x f+\frac{2 c f}{d}\right )}{d^3}-\frac{a^2 f^2 \cos \left (e-\frac{c f}{d}\right ) \text{Si}\left (x f+\frac{c f}{d}\right )}{d^3}-\frac{4 a^2 f \sin ^3\left (\frac{e}{2}+\frac{f x}{2}+\frac{\pi }{4}\right ) \cos \left (\frac{e}{2}+\frac{f x}{2}+\frac{\pi }{4}\right )}{d^2 (c+d x)}-\frac{2 a^2 \sin ^4\left (\frac{e}{2}+\frac{f x}{2}+\frac{\pi }{4}\right )}{d (c+d x)^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[e + f*x])^2/(c + d*x)^3,x]

[Out]

(a^2*f^2*Cos[2*e - (2*c*f)/d]*CosIntegral[(2*c*f)/d + 2*f*x])/d^3 - (a^2*f^2*CosIntegral[(c*f)/d + f*x]*Sin[e
- (c*f)/d])/d^3 - (4*a^2*f*Cos[e/2 + Pi/4 + (f*x)/2]*Sin[e/2 + Pi/4 + (f*x)/2]^3)/(d^2*(c + d*x)) - (2*a^2*Sin
[e/2 + Pi/4 + (f*x)/2]^4)/(d*(c + d*x)^2) - (a^2*f^2*Cos[e - (c*f)/d]*SinIntegral[(c*f)/d + f*x])/d^3 - (a^2*f
^2*Sin[2*e - (2*c*f)/d]*SinIntegral[(2*c*f)/d + 2*f*x])/d^3

Rule 3318

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

Rule 3314

Int[((c_.) + (d_.)*(x_))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[((c + d*x)^(m + 1)*(b*Si
n[e + f*x])^n)/(d*(m + 1)), x] + (Dist[(b^2*f^2*n*(n - 1))/(d^2*(m + 1)*(m + 2)), Int[(c + d*x)^(m + 2)*(b*Sin
[e + f*x])^(n - 2), x], x] - Dist[(f^2*n^2)/(d^2*(m + 1)*(m + 2)), Int[(c + d*x)^(m + 2)*(b*Sin[e + f*x])^n, x
], x] - Simp[(b*f*n*(c + d*x)^(m + 2)*Cos[e + f*x]*(b*Sin[e + f*x])^(n - 1))/(d^2*(m + 1)*(m + 2)), x]) /; Fre
eQ[{b, c, d, e, f}, x] && GtQ[n, 1] && LtQ[m, -2]

Rule 3309

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 3303

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Dist[Cos[(d*e - c*f)/d], Int[Sin[(c*f)/d + f*x]
/(c + d*x), x], x] + Dist[Sin[(d*e - c*f)/d], Int[Cos[(c*f)/d + f*x]/(c + d*x), x], x] /; FreeQ[{c, d, e, f},
x] && NeQ[d*e - c*f, 0]

Rule 3299

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[SinIntegral[e + f*x]/d, x] /; FreeQ[{c, d,
e, f}, x] && EqQ[d*e - c*f, 0]

Rule 3302

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CosIntegral[e - Pi/2 + f*x]/d, x] /; FreeQ
[{c, d, e, f}, x] && EqQ[d*(e - Pi/2) - c*f, 0]

Rule 3312

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin
[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 1])
)

Rubi steps

\begin{align*} \int \frac{(a+a \sin (e+f x))^2}{(c+d x)^3} \, dx &=\left (4 a^2\right ) \int \frac{\sin ^4\left (\frac{1}{2} \left (e+\frac{\pi }{2}\right )+\frac{f x}{2}\right )}{(c+d x)^3} \, dx\\ &=-\frac{4 a^2 f \cos \left (\frac{e}{2}+\frac{\pi }{4}+\frac{f x}{2}\right ) \sin ^3\left (\frac{e}{2}+\frac{\pi }{4}+\frac{f x}{2}\right )}{d^2 (c+d x)}-\frac{2 a^2 \sin ^4\left (\frac{e}{2}+\frac{\pi }{4}+\frac{f x}{2}\right )}{d (c+d x)^2}+\frac{\left (6 a^2 f^2\right ) \int \frac{\sin ^2\left (\frac{e}{2}+\frac{\pi }{4}+\frac{f x}{2}\right )}{c+d x} \, dx}{d^2}-\frac{\left (8 a^2 f^2\right ) \int \frac{\sin ^4\left (\frac{e}{2}+\frac{\pi }{4}+\frac{f x}{2}\right )}{c+d x} \, dx}{d^2}\\ &=-\frac{4 a^2 f \cos \left (\frac{e}{2}+\frac{\pi }{4}+\frac{f x}{2}\right ) \sin ^3\left (\frac{e}{2}+\frac{\pi }{4}+\frac{f x}{2}\right )}{d^2 (c+d x)}-\frac{2 a^2 \sin ^4\left (\frac{e}{2}+\frac{\pi }{4}+\frac{f x}{2}\right )}{d (c+d x)^2}+\frac{\left (3 a^2 f^2\right ) \int \frac{1}{c+d x} \, dx}{d^2}-\frac{\left (3 a^2 f^2\right ) \int \frac{\cos \left (2 \left (\frac{e}{2}+\frac{\pi }{4}\right )+f x\right )}{c+d x} \, dx}{d^2}-\frac{\left (8 a^2 f^2\right ) \int \left (\frac{3}{8 (c+d x)}-\frac{\cos (2 e+2 f x)}{8 (c+d x)}+\frac{\sin (e+f x)}{2 (c+d x)}\right ) \, dx}{d^2}\\ &=-\frac{4 a^2 f \cos \left (\frac{e}{2}+\frac{\pi }{4}+\frac{f x}{2}\right ) \sin ^3\left (\frac{e}{2}+\frac{\pi }{4}+\frac{f x}{2}\right )}{d^2 (c+d x)}-\frac{2 a^2 \sin ^4\left (\frac{e}{2}+\frac{\pi }{4}+\frac{f x}{2}\right )}{d (c+d x)^2}+\frac{\left (a^2 f^2\right ) \int \frac{\cos (2 e+2 f x)}{c+d x} \, dx}{d^2}-\frac{\left (4 a^2 f^2\right ) \int \frac{\sin (e+f x)}{c+d x} \, dx}{d^2}+\frac{\left (3 a^2 f^2 \cos \left (e-\frac{c f}{d}\right )\right ) \int \frac{\sin \left (\frac{c f}{d}+f x\right )}{c+d x} \, dx}{d^2}+\frac{\left (3 a^2 f^2 \sin \left (e-\frac{c f}{d}\right )\right ) \int \frac{\cos \left (\frac{c f}{d}+f x\right )}{c+d x} \, dx}{d^2}\\ &=\frac{3 a^2 f^2 \text{Ci}\left (\frac{c f}{d}+f x\right ) \sin \left (e-\frac{c f}{d}\right )}{d^3}-\frac{4 a^2 f \cos \left (\frac{e}{2}+\frac{\pi }{4}+\frac{f x}{2}\right ) \sin ^3\left (\frac{e}{2}+\frac{\pi }{4}+\frac{f x}{2}\right )}{d^2 (c+d x)}-\frac{2 a^2 \sin ^4\left (\frac{e}{2}+\frac{\pi }{4}+\frac{f x}{2}\right )}{d (c+d x)^2}+\frac{3 a^2 f^2 \cos \left (e-\frac{c f}{d}\right ) \text{Si}\left (\frac{c f}{d}+f x\right )}{d^3}+\frac{\left (a^2 f^2 \cos \left (2 e-\frac{2 c f}{d}\right )\right ) \int \frac{\cos \left (\frac{2 c f}{d}+2 f x\right )}{c+d x} \, dx}{d^2}-\frac{\left (4 a^2 f^2 \cos \left (e-\frac{c f}{d}\right )\right ) \int \frac{\sin \left (\frac{c f}{d}+f x\right )}{c+d x} \, dx}{d^2}-\frac{\left (a^2 f^2 \sin \left (2 e-\frac{2 c f}{d}\right )\right ) \int \frac{\sin \left (\frac{2 c f}{d}+2 f x\right )}{c+d x} \, dx}{d^2}-\frac{\left (4 a^2 f^2 \sin \left (e-\frac{c f}{d}\right )\right ) \int \frac{\cos \left (\frac{c f}{d}+f x\right )}{c+d x} \, dx}{d^2}\\ &=\frac{a^2 f^2 \cos \left (2 e-\frac{2 c f}{d}\right ) \text{Ci}\left (\frac{2 c f}{d}+2 f x\right )}{d^3}-\frac{a^2 f^2 \text{Ci}\left (\frac{c f}{d}+f x\right ) \sin \left (e-\frac{c f}{d}\right )}{d^3}-\frac{4 a^2 f \cos \left (\frac{e}{2}+\frac{\pi }{4}+\frac{f x}{2}\right ) \sin ^3\left (\frac{e}{2}+\frac{\pi }{4}+\frac{f x}{2}\right )}{d^2 (c+d x)}-\frac{2 a^2 \sin ^4\left (\frac{e}{2}+\frac{\pi }{4}+\frac{f x}{2}\right )}{d (c+d x)^2}-\frac{a^2 f^2 \cos \left (e-\frac{c f}{d}\right ) \text{Si}\left (\frac{c f}{d}+f x\right )}{d^3}-\frac{a^2 f^2 \sin \left (2 e-\frac{2 c f}{d}\right ) \text{Si}\left (\frac{2 c f}{d}+2 f x\right )}{d^3}\\ \end{align*}

Mathematica [A]  time = 0.900699, size = 353, normalized size = 1.57 \[ -\frac{a^2 \left (4 c^2 f^2 \sin \left (2 e-\frac{2 c f}{d}\right ) \text{Si}\left (\frac{2 f (c+d x)}{d}\right )+4 c^2 f^2 \cos \left (e-\frac{c f}{d}\right ) \text{Si}\left (f \left (\frac{c}{d}+x\right )\right )+4 f^2 (c+d x)^2 \text{CosIntegral}\left (f \left (\frac{c}{d}+x\right )\right ) \sin \left (e-\frac{c f}{d}\right )-4 f^2 (c+d x)^2 \text{CosIntegral}\left (\frac{2 f (c+d x)}{d}\right ) \cos \left (2 e-\frac{2 c f}{d}\right )+4 d^2 f^2 x^2 \sin \left (2 e-\frac{2 c f}{d}\right ) \text{Si}\left (\frac{2 f (c+d x)}{d}\right )+4 d^2 f^2 x^2 \cos \left (e-\frac{c f}{d}\right ) \text{Si}\left (f \left (\frac{c}{d}+x\right )\right )+8 c d f^2 x \sin \left (2 e-\frac{2 c f}{d}\right ) \text{Si}\left (\frac{2 f (c+d x)}{d}\right )+8 c d f^2 x \cos \left (e-\frac{c f}{d}\right ) \text{Si}\left (f \left (\frac{c}{d}+x\right )\right )+2 c d f \sin (2 (e+f x))+4 c d f \cos (e+f x)+4 d^2 \sin (e+f x)+2 d^2 f x \sin (2 (e+f x))+4 d^2 f x \cos (e+f x)-d^2 \cos (2 (e+f x))+3 d^2\right )}{4 d^3 (c+d x)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sin[e + f*x])^2/(c + d*x)^3,x]

[Out]

-(a^2*(3*d^2 + 4*c*d*f*Cos[e + f*x] + 4*d^2*f*x*Cos[e + f*x] - d^2*Cos[2*(e + f*x)] - 4*f^2*(c + d*x)^2*Cos[2*
e - (2*c*f)/d]*CosIntegral[(2*f*(c + d*x))/d] + 4*f^2*(c + d*x)^2*CosIntegral[f*(c/d + x)]*Sin[e - (c*f)/d] +
4*d^2*Sin[e + f*x] + 2*c*d*f*Sin[2*(e + f*x)] + 2*d^2*f*x*Sin[2*(e + f*x)] + 4*c^2*f^2*Cos[e - (c*f)/d]*SinInt
egral[f*(c/d + x)] + 8*c*d*f^2*x*Cos[e - (c*f)/d]*SinIntegral[f*(c/d + x)] + 4*d^2*f^2*x^2*Cos[e - (c*f)/d]*Si
nIntegral[f*(c/d + x)] + 4*c^2*f^2*Sin[2*e - (2*c*f)/d]*SinIntegral[(2*f*(c + d*x))/d] + 8*c*d*f^2*x*Sin[2*e -
(2*c*f)/d]*SinIntegral[(2*f*(c + d*x))/d] + 4*d^2*f^2*x^2*Sin[2*e - (2*c*f)/d]*SinIntegral[(2*f*(c + d*x))/d]
))/(4*d^3*(c + d*x)^2)

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Maple [A]  time = 0.023, size = 347, normalized size = 1.5 \begin{align*}{\frac{1}{f} \left ( -{\frac{3\,{a}^{2}{f}^{3}}{4\, \left ( \left ( fx+e \right ) d+cf-de \right ) ^{2}d}}-{\frac{{a}^{2}{f}^{3}}{4} \left ( -{\frac{\cos \left ( 2\,fx+2\,e \right ) }{ \left ( \left ( fx+e \right ) d+cf-de \right ) ^{2}d}}-{\frac{1}{d} \left ( -2\,{\frac{\sin \left ( 2\,fx+2\,e \right ) }{ \left ( \left ( fx+e \right ) d+cf-de \right ) d}}+2\,{\frac{1}{d} \left ( 2\,{\frac{1}{d}{\it Si} \left ( 2\,fx+2\,e+2\,{\frac{cf-de}{d}} \right ) \sin \left ( 2\,{\frac{cf-de}{d}} \right ) }+2\,{\frac{1}{d}{\it Ci} \left ( 2\,fx+2\,e+2\,{\frac{cf-de}{d}} \right ) \cos \left ( 2\,{\frac{cf-de}{d}} \right ) } \right ) } \right ) } \right ) }+2\,{a}^{2}{f}^{3} \left ( -1/2\,{\frac{\sin \left ( fx+e \right ) }{ \left ( \left ( fx+e \right ) d+cf-de \right ) ^{2}d}}+1/2\,{\frac{1}{d} \left ( -{\frac{\cos \left ( fx+e \right ) }{ \left ( \left ( fx+e \right ) d+cf-de \right ) d}}-{\frac{1}{d} \left ({\frac{1}{d}{\it Si} \left ( fx+e+{\frac{cf-de}{d}} \right ) \cos \left ({\frac{cf-de}{d}} \right ) }-{\frac{1}{d}{\it Ci} \left ( fx+e+{\frac{cf-de}{d}} \right ) \sin \left ({\frac{cf-de}{d}} \right ) } \right ) } \right ) } \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(f*x+e))^2/(d*x+c)^3,x)

[Out]

1/f*(-3/4*a^2*f^3/((f*x+e)*d+c*f-d*e)^2/d-1/4*a^2*f^3*(-cos(2*f*x+2*e)/((f*x+e)*d+c*f-d*e)^2/d-(-2*sin(2*f*x+2
*e)/((f*x+e)*d+c*f-d*e)/d+2*(2*Si(2*f*x+2*e+2*(c*f-d*e)/d)*sin(2*(c*f-d*e)/d)/d+2*Ci(2*f*x+2*e+2*(c*f-d*e)/d)*
cos(2*(c*f-d*e)/d)/d)/d)/d)+2*a^2*f^3*(-1/2*sin(f*x+e)/((f*x+e)*d+c*f-d*e)^2/d+1/2*(-cos(f*x+e)/((f*x+e)*d+c*f
-d*e)/d-(Si(f*x+e+(c*f-d*e)/d)*cos((c*f-d*e)/d)/d-Ci(f*x+e+(c*f-d*e)/d)*sin((c*f-d*e)/d)/d)/d)/d))

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Maxima [C]  time = 1.95089, size = 641, normalized size = 2.85 \begin{align*} -\frac{\frac{32 \, a^{2} f^{3}}{{\left (f x + e\right )}^{2} d^{3} + d^{3} e^{2} - 2 \, c d^{2} e f + c^{2} d f^{2} - 2 \,{\left (d^{3} e - c d^{2} f\right )}{\left (f x + e\right )}} - \frac{64 \,{\left (f^{3}{\left (-i \, E_{3}\left (\frac{i \,{\left (f x + e\right )} d - i \, d e + i \, c f}{d}\right ) + i \, E_{3}\left (-\frac{i \,{\left (f x + e\right )} d - i \, d e + i \, c f}{d}\right )\right )} \cos \left (-\frac{d e - c f}{d}\right ) + f^{3}{\left (E_{3}\left (\frac{i \,{\left (f x + e\right )} d - i \, d e + i \, c f}{d}\right ) + E_{3}\left (-\frac{i \,{\left (f x + e\right )} d - i \, d e + i \, c f}{d}\right )\right )} \sin \left (-\frac{d e - c f}{d}\right )\right )} a^{2}}{{\left (f x + e\right )}^{2} d^{3} + d^{3} e^{2} - 2 \, c d^{2} e f + c^{2} d f^{2} - 2 \,{\left (d^{3} e - c d^{2} f\right )}{\left (f x + e\right )}} - \frac{{\left (16 \, f^{3}{\left (E_{3}\left (\frac{2 i \,{\left (f x + e\right )} d - 2 i \, d e + 2 i \, c f}{d}\right ) + E_{3}\left (-\frac{2 i \,{\left (f x + e\right )} d - 2 i \, d e + 2 i \, c f}{d}\right )\right )} \cos \left (-\frac{2 \,{\left (d e - c f\right )}}{d}\right ) + f^{3}{\left (16 i \, E_{3}\left (\frac{2 i \,{\left (f x + e\right )} d - 2 i \, d e + 2 i \, c f}{d}\right ) - 16 i \, E_{3}\left (-\frac{2 i \,{\left (f x + e\right )} d - 2 i \, d e + 2 i \, c f}{d}\right )\right )} \sin \left (-\frac{2 \,{\left (d e - c f\right )}}{d}\right ) - 16 \, f^{3}\right )} a^{2}}{{\left (f x + e\right )}^{2} d^{3} + d^{3} e^{2} - 2 \, c d^{2} e f + c^{2} d f^{2} - 2 \,{\left (d^{3} e - c d^{2} f\right )}{\left (f x + e\right )}}}{64 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^2/(d*x+c)^3,x, algorithm="maxima")

[Out]

-1/64*(32*a^2*f^3/((f*x + e)^2*d^3 + d^3*e^2 - 2*c*d^2*e*f + c^2*d*f^2 - 2*(d^3*e - c*d^2*f)*(f*x + e)) - 64*(
f^3*(-I*exp_integral_e(3, (I*(f*x + e)*d - I*d*e + I*c*f)/d) + I*exp_integral_e(3, -(I*(f*x + e)*d - I*d*e + I
*c*f)/d))*cos(-(d*e - c*f)/d) + f^3*(exp_integral_e(3, (I*(f*x + e)*d - I*d*e + I*c*f)/d) + exp_integral_e(3,
-(I*(f*x + e)*d - I*d*e + I*c*f)/d))*sin(-(d*e - c*f)/d))*a^2/((f*x + e)^2*d^3 + d^3*e^2 - 2*c*d^2*e*f + c^2*d
*f^2 - 2*(d^3*e - c*d^2*f)*(f*x + e)) - (16*f^3*(exp_integral_e(3, (2*I*(f*x + e)*d - 2*I*d*e + 2*I*c*f)/d) +
exp_integral_e(3, -(2*I*(f*x + e)*d - 2*I*d*e + 2*I*c*f)/d))*cos(-2*(d*e - c*f)/d) + f^3*(16*I*exp_integral_e(
3, (2*I*(f*x + e)*d - 2*I*d*e + 2*I*c*f)/d) - 16*I*exp_integral_e(3, -(2*I*(f*x + e)*d - 2*I*d*e + 2*I*c*f)/d)
)*sin(-2*(d*e - c*f)/d) - 16*f^3)*a^2/((f*x + e)^2*d^3 + d^3*e^2 - 2*c*d^2*e*f + c^2*d*f^2 - 2*(d^3*e - c*d^2*
f)*(f*x + e)))/f

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Fricas [B]  time = 2.07093, size = 1049, normalized size = 4.66 \begin{align*} \frac{a^{2} d^{2} \cos \left (f x + e\right )^{2} - 2 \, a^{2} d^{2} + 2 \,{\left (a^{2} d^{2} f^{2} x^{2} + 2 \, a^{2} c d f^{2} x + a^{2} c^{2} f^{2}\right )} \sin \left (-\frac{2 \,{\left (d e - c f\right )}}{d}\right ) \operatorname{Si}\left (\frac{2 \,{\left (d f x + c f\right )}}{d}\right ) - 2 \,{\left (a^{2} d^{2} f^{2} x^{2} + 2 \, a^{2} c d f^{2} x + a^{2} c^{2} f^{2}\right )} \cos \left (-\frac{d e - c f}{d}\right ) \operatorname{Si}\left (\frac{d f x + c f}{d}\right ) - 2 \,{\left (a^{2} d^{2} f x + a^{2} c d f\right )} \cos \left (f x + e\right ) +{\left ({\left (a^{2} d^{2} f^{2} x^{2} + 2 \, a^{2} c d f^{2} x + a^{2} c^{2} f^{2}\right )} \operatorname{Ci}\left (\frac{2 \,{\left (d f x + c f\right )}}{d}\right ) +{\left (a^{2} d^{2} f^{2} x^{2} + 2 \, a^{2} c d f^{2} x + a^{2} c^{2} f^{2}\right )} \operatorname{Ci}\left (-\frac{2 \,{\left (d f x + c f\right )}}{d}\right )\right )} \cos \left (-\frac{2 \,{\left (d e - c f\right )}}{d}\right ) - 2 \,{\left (a^{2} d^{2} +{\left (a^{2} d^{2} f x + a^{2} c d f\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right ) +{\left ({\left (a^{2} d^{2} f^{2} x^{2} + 2 \, a^{2} c d f^{2} x + a^{2} c^{2} f^{2}\right )} \operatorname{Ci}\left (\frac{d f x + c f}{d}\right ) +{\left (a^{2} d^{2} f^{2} x^{2} + 2 \, a^{2} c d f^{2} x + a^{2} c^{2} f^{2}\right )} \operatorname{Ci}\left (-\frac{d f x + c f}{d}\right )\right )} \sin \left (-\frac{d e - c f}{d}\right )}{2 \,{\left (d^{5} x^{2} + 2 \, c d^{4} x + c^{2} d^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^2/(d*x+c)^3,x, algorithm="fricas")

[Out]

1/2*(a^2*d^2*cos(f*x + e)^2 - 2*a^2*d^2 + 2*(a^2*d^2*f^2*x^2 + 2*a^2*c*d*f^2*x + a^2*c^2*f^2)*sin(-2*(d*e - c*
f)/d)*sin_integral(2*(d*f*x + c*f)/d) - 2*(a^2*d^2*f^2*x^2 + 2*a^2*c*d*f^2*x + a^2*c^2*f^2)*cos(-(d*e - c*f)/d
)*sin_integral((d*f*x + c*f)/d) - 2*(a^2*d^2*f*x + a^2*c*d*f)*cos(f*x + e) + ((a^2*d^2*f^2*x^2 + 2*a^2*c*d*f^2
*x + a^2*c^2*f^2)*cos_integral(2*(d*f*x + c*f)/d) + (a^2*d^2*f^2*x^2 + 2*a^2*c*d*f^2*x + a^2*c^2*f^2)*cos_inte
gral(-2*(d*f*x + c*f)/d))*cos(-2*(d*e - c*f)/d) - 2*(a^2*d^2 + (a^2*d^2*f*x + a^2*c*d*f)*cos(f*x + e))*sin(f*x
+ e) + ((a^2*d^2*f^2*x^2 + 2*a^2*c*d*f^2*x + a^2*c^2*f^2)*cos_integral((d*f*x + c*f)/d) + (a^2*d^2*f^2*x^2 +
2*a^2*c*d*f^2*x + a^2*c^2*f^2)*cos_integral(-(d*f*x + c*f)/d))*sin(-(d*e - c*f)/d))/(d^5*x^2 + 2*c*d^4*x + c^2
*d^3)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} a^{2} \left (\int \frac{2 \sin{\left (e + f x \right )}}{c^{3} + 3 c^{2} d x + 3 c d^{2} x^{2} + d^{3} x^{3}}\, dx + \int \frac{\sin ^{2}{\left (e + f x \right )}}{c^{3} + 3 c^{2} d x + 3 c d^{2} x^{2} + d^{3} x^{3}}\, dx + \int \frac{1}{c^{3} + 3 c^{2} d x + 3 c d^{2} x^{2} + d^{3} x^{3}}\, dx\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))**2/(d*x+c)**3,x)

[Out]

a**2*(Integral(2*sin(e + f*x)/(c**3 + 3*c**2*d*x + 3*c*d**2*x**2 + d**3*x**3), x) + Integral(sin(e + f*x)**2/(
c**3 + 3*c**2*d*x + 3*c*d**2*x**2 + d**3*x**3), x) + Integral(1/(c**3 + 3*c**2*d*x + 3*c*d**2*x**2 + d**3*x**3
), x))

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: NotImplementedError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^2/(d*x+c)^3,x, algorithm="giac")

[Out]

Exception raised: NotImplementedError