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

Optimal. Leaf size=245 \[ -\frac{a^2}{2 d (c+d x)^2}-\frac{a b f^2 \text{CosIntegral}\left (\frac{c f}{d}+f x\right ) \sin \left (e-\frac{c f}{d}\right )}{d^3}-\frac{a b f^2 \cos \left (e-\frac{c f}{d}\right ) \text{Si}\left (x f+\frac{c f}{d}\right )}{d^3}-\frac{a b f \cos (e+f x)}{d^2 (c+d x)}-\frac{a b \sin (e+f x)}{d (c+d x)^2}+\frac{b^2 f^2 \text{CosIntegral}\left (\frac{2 c f}{d}+2 f x\right ) \cos \left (2 e-\frac{2 c f}{d}\right )}{d^3}-\frac{b^2 f^2 \sin \left (2 e-\frac{2 c f}{d}\right ) \text{Si}\left (2 x f+\frac{2 c f}{d}\right )}{d^3}-\frac{b^2 f \sin (e+f x) \cos (e+f x)}{d^2 (c+d x)}-\frac{b^2 \sin ^2(e+f x)}{2 d (c+d x)^2} \]

[Out]

-a^2/(2*d*(c + d*x)^2) - (a*b*f*Cos[e + f*x])/(d^2*(c + d*x)) + (b^2*f^2*Cos[2*e - (2*c*f)/d]*CosIntegral[(2*c
*f)/d + 2*f*x])/d^3 - (a*b*f^2*CosIntegral[(c*f)/d + f*x]*Sin[e - (c*f)/d])/d^3 - (a*b*Sin[e + f*x])/(d*(c + d
*x)^2) - (b^2*f*Cos[e + f*x]*Sin[e + f*x])/(d^2*(c + d*x)) - (b^2*Sin[e + f*x]^2)/(2*d*(c + d*x)^2) - (a*b*f^2
*Cos[e - (c*f)/d]*SinIntegral[(c*f)/d + f*x])/d^3 - (b^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 = 0.424202, antiderivative size = 245, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 8, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {3317, 3297, 3303, 3299, 3302, 3314, 31, 3312} \[ -\frac{a^2}{2 d (c+d x)^2}-\frac{a b f^2 \text{CosIntegral}\left (\frac{c f}{d}+f x\right ) \sin \left (e-\frac{c f}{d}\right )}{d^3}-\frac{a b f^2 \cos \left (e-\frac{c f}{d}\right ) \text{Si}\left (x f+\frac{c f}{d}\right )}{d^3}-\frac{a b f \cos (e+f x)}{d^2 (c+d x)}-\frac{a b \sin (e+f x)}{d (c+d x)^2}+\frac{b^2 f^2 \text{CosIntegral}\left (\frac{2 c f}{d}+2 f x\right ) \cos \left (2 e-\frac{2 c f}{d}\right )}{d^3}-\frac{b^2 f^2 \sin \left (2 e-\frac{2 c f}{d}\right ) \text{Si}\left (2 x f+\frac{2 c f}{d}\right )}{d^3}-\frac{b^2 f \sin (e+f x) \cos (e+f x)}{d^2 (c+d x)}-\frac{b^2 \sin ^2(e+f x)}{2 d (c+d x)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3317

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

Rule 3297

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

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 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 31

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

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+b \sin (e+f x))^2}{(c+d x)^3} \, dx &=\int \left (\frac{a^2}{(c+d x)^3}+\frac{2 a b \sin (e+f x)}{(c+d x)^3}+\frac{b^2 \sin ^2(e+f x)}{(c+d x)^3}\right ) \, dx\\ &=-\frac{a^2}{2 d (c+d x)^2}+(2 a b) \int \frac{\sin (e+f x)}{(c+d x)^3} \, dx+b^2 \int \frac{\sin ^2(e+f x)}{(c+d x)^3} \, dx\\ &=-\frac{a^2}{2 d (c+d x)^2}-\frac{a b \sin (e+f x)}{d (c+d x)^2}-\frac{b^2 f \cos (e+f x) \sin (e+f x)}{d^2 (c+d x)}-\frac{b^2 \sin ^2(e+f x)}{2 d (c+d x)^2}+\frac{(a b f) \int \frac{\cos (e+f x)}{(c+d x)^2} \, dx}{d}+\frac{\left (b^2 f^2\right ) \int \frac{1}{c+d x} \, dx}{d^2}-\frac{\left (2 b^2 f^2\right ) \int \frac{\sin ^2(e+f x)}{c+d x} \, dx}{d^2}\\ &=-\frac{a^2}{2 d (c+d x)^2}-\frac{a b f \cos (e+f x)}{d^2 (c+d x)}+\frac{b^2 f^2 \log (c+d x)}{d^3}-\frac{a b \sin (e+f x)}{d (c+d x)^2}-\frac{b^2 f \cos (e+f x) \sin (e+f x)}{d^2 (c+d x)}-\frac{b^2 \sin ^2(e+f x)}{2 d (c+d x)^2}-\frac{\left (a b f^2\right ) \int \frac{\sin (e+f x)}{c+d x} \, dx}{d^2}-\frac{\left (2 b^2 f^2\right ) \int \left (\frac{1}{2 (c+d x)}-\frac{\cos (2 e+2 f x)}{2 (c+d x)}\right ) \, dx}{d^2}\\ &=-\frac{a^2}{2 d (c+d x)^2}-\frac{a b f \cos (e+f x)}{d^2 (c+d x)}-\frac{a b \sin (e+f x)}{d (c+d x)^2}-\frac{b^2 f \cos (e+f x) \sin (e+f x)}{d^2 (c+d x)}-\frac{b^2 \sin ^2(e+f x)}{2 d (c+d x)^2}+\frac{\left (b^2 f^2\right ) \int \frac{\cos (2 e+2 f x)}{c+d x} \, dx}{d^2}-\frac{\left (a b 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 b 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}{2 d (c+d x)^2}-\frac{a b f \cos (e+f x)}{d^2 (c+d x)}-\frac{a b f^2 \text{Ci}\left (\frac{c f}{d}+f x\right ) \sin \left (e-\frac{c f}{d}\right )}{d^3}-\frac{a b \sin (e+f x)}{d (c+d x)^2}-\frac{b^2 f \cos (e+f x) \sin (e+f x)}{d^2 (c+d x)}-\frac{b^2 \sin ^2(e+f x)}{2 d (c+d x)^2}-\frac{a b f^2 \cos \left (e-\frac{c f}{d}\right ) \text{Si}\left (\frac{c f}{d}+f x\right )}{d^3}+\frac{\left (b^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 (b^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{a^2}{2 d (c+d x)^2}-\frac{a b f \cos (e+f x)}{d^2 (c+d x)}+\frac{b^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 b f^2 \text{Ci}\left (\frac{c f}{d}+f x\right ) \sin \left (e-\frac{c f}{d}\right )}{d^3}-\frac{a b \sin (e+f x)}{d (c+d x)^2}-\frac{b^2 f \cos (e+f x) \sin (e+f x)}{d^2 (c+d x)}-\frac{b^2 \sin ^2(e+f x)}{2 d (c+d x)^2}-\frac{a b f^2 \cos \left (e-\frac{c f}{d}\right ) \text{Si}\left (\frac{c f}{d}+f x\right )}{d^3}-\frac{b^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 = 1.22926, size = 395, normalized size = 1.61 \[ -\frac{2 a^2 d^2+4 a b c^2 f^2 \cos \left (e-\frac{c f}{d}\right ) \text{Si}\left (f \left (\frac{c}{d}+x\right )\right )+4 a b f^2 (c+d x)^2 \text{CosIntegral}\left (f \left (\frac{c}{d}+x\right )\right ) \sin \left (e-\frac{c f}{d}\right )+4 a b d^2 f^2 x^2 \cos \left (e-\frac{c f}{d}\right ) \text{Si}\left (f \left (\frac{c}{d}+x\right )\right )+8 a b c d f^2 x \cos \left (e-\frac{c f}{d}\right ) \text{Si}\left (f \left (\frac{c}{d}+x\right )\right )+4 a b c d f \cos (e+f x)+4 a b d^2 \sin (e+f x)+4 a b d^2 f x \cos (e+f x)+4 b^2 c^2 f^2 \sin \left (2 e-\frac{2 c f}{d}\right ) \text{Si}\left (\frac{2 f (c+d x)}{d}\right )-4 b^2 f^2 (c+d x)^2 \text{CosIntegral}\left (\frac{2 f (c+d x)}{d}\right ) \cos \left (2 e-\frac{2 c f}{d}\right )+4 b^2 d^2 f^2 x^2 \sin \left (2 e-\frac{2 c f}{d}\right ) \text{Si}\left (\frac{2 f (c+d x)}{d}\right )+8 b^2 c d f^2 x \sin \left (2 e-\frac{2 c f}{d}\right ) \text{Si}\left (\frac{2 f (c+d x)}{d}\right )+2 b^2 c d f \sin (2 (e+f x))+2 b^2 d^2 f x \sin (2 (e+f x))-b^2 d^2 \cos (2 (e+f x))+b^2 d^2}{4 d^3 (c+d x)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(2*a^2*d^2 + b^2*d^2 + 4*a*b*c*d*f*Cos[e + f*x] + 4*a*b*d^2*f*x*Cos[e + f*x] - b^2*d^2*Cos[2*(e + f*x)] - 4*b
^2*f^2*(c + d*x)^2*Cos[2*e - (2*c*f)/d]*CosIntegral[(2*f*(c + d*x))/d] + 4*a*b*f^2*(c + d*x)^2*CosIntegral[f*(
c/d + x)]*Sin[e - (c*f)/d] + 4*a*b*d^2*Sin[e + f*x] + 2*b^2*c*d*f*Sin[2*(e + f*x)] + 2*b^2*d^2*f*x*Sin[2*(e +
f*x)] + 4*a*b*c^2*f^2*Cos[e - (c*f)/d]*SinIntegral[f*(c/d + x)] + 8*a*b*c*d*f^2*x*Cos[e - (c*f)/d]*SinIntegral
[f*(c/d + x)] + 4*a*b*d^2*f^2*x^2*Cos[e - (c*f)/d]*SinIntegral[f*(c/d + x)] + 4*b^2*c^2*f^2*Sin[2*e - (2*c*f)/
d]*SinIntegral[(2*f*(c + d*x))/d] + 8*b^2*c*d*f^2*x*Sin[2*e - (2*c*f)/d]*SinIntegral[(2*f*(c + d*x))/d] + 4*b^
2*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 = 374, normalized size = 1.5 \begin{align*}{\frac{1}{f} \left ( -{\frac{{a}^{2}{f}^{3}}{2\, \left ( \left ( fx+e \right ) d+cf-de \right ) ^{2}d}}+2\,{f}^{3}ab \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 ) -{\frac{{f}^{3}{b}^{2}}{4\, \left ( \left ( fx+e \right ) d+cf-de \right ) ^{2}d}}-{\frac{{f}^{3}{b}^{2}}{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 ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [C]  time = 1.87744, size = 640, normalized size = 2.61 \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 b}{{\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 )} b^{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+b*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*b/((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)*b^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 [A]  time = 2.50881, size = 1057, normalized size = 4.31 \begin{align*} \frac{b^{2} d^{2} \cos \left (f x + e\right )^{2} -{\left (a^{2} + b^{2}\right )} d^{2} + 2 \,{\left (b^{2} d^{2} f^{2} x^{2} + 2 \, b^{2} c d f^{2} x + b^{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 b d^{2} f^{2} x^{2} + 2 \, a b c d f^{2} x + a b 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 b d^{2} f x + a b c d f\right )} \cos \left (f x + e\right ) +{\left ({\left (b^{2} d^{2} f^{2} x^{2} + 2 \, b^{2} c d f^{2} x + b^{2} c^{2} f^{2}\right )} \operatorname{Ci}\left (\frac{2 \,{\left (d f x + c f\right )}}{d}\right ) +{\left (b^{2} d^{2} f^{2} x^{2} + 2 \, b^{2} c d f^{2} x + b^{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 b d^{2} +{\left (b^{2} d^{2} f x + b^{2} c d f\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right ) +{\left ({\left (a b d^{2} f^{2} x^{2} + 2 \, a b c d f^{2} x + a b c^{2} f^{2}\right )} \operatorname{Ci}\left (\frac{d f x + c f}{d}\right ) +{\left (a b d^{2} f^{2} x^{2} + 2 \, a b c d f^{2} x + a b 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+b*sin(f*x+e))^2/(d*x+c)^3,x, algorithm="fricas")

[Out]

1/2*(b^2*d^2*cos(f*x + e)^2 - (a^2 + b^2)*d^2 + 2*(b^2*d^2*f^2*x^2 + 2*b^2*c*d*f^2*x + b^2*c^2*f^2)*sin(-2*(d*
e - c*f)/d)*sin_integral(2*(d*f*x + c*f)/d) - 2*(a*b*d^2*f^2*x^2 + 2*a*b*c*d*f^2*x + a*b*c^2*f^2)*cos(-(d*e -
c*f)/d)*sin_integral((d*f*x + c*f)/d) - 2*(a*b*d^2*f*x + a*b*c*d*f)*cos(f*x + e) + ((b^2*d^2*f^2*x^2 + 2*b^2*c
*d*f^2*x + b^2*c^2*f^2)*cos_integral(2*(d*f*x + c*f)/d) + (b^2*d^2*f^2*x^2 + 2*b^2*c*d*f^2*x + b^2*c^2*f^2)*co
s_integral(-2*(d*f*x + c*f)/d))*cos(-2*(d*e - c*f)/d) - 2*(a*b*d^2 + (b^2*d^2*f*x + b^2*c*d*f)*cos(f*x + e))*s
in(f*x + e) + ((a*b*d^2*f^2*x^2 + 2*a*b*c*d*f^2*x + a*b*c^2*f^2)*cos_integral((d*f*x + c*f)/d) + (a*b*d^2*f^2*
x^2 + 2*a*b*c*d*f^2*x + a*b*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*} \int \frac{\left (a + b \sin{\left (e + f x \right )}\right )^{2}}{\left (c + d x\right )^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: NotImplementedError