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

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

[Out]

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

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Rubi [A]  time = 0.42, antiderivative size = 245, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 8, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, 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 31

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

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 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 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 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])
)

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 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])

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*}

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Mathematica [A]  time = 1.30, size = 395, normalized size = 1.61 \[ -\frac {2 a^2 d^2+4 a b c^2 f^2 \cos \left (e-\frac {c f}{d}\right ) \text {Si}\left (f \left (\frac {c}{d}+x\right )\right )+4 a b f^2 (c+d x)^2 \text {Ci}\left (f \left (\frac {c}{d}+x\right )\right ) \sin \left (e-\frac {c f}{d}\right )+4 a b d^2 f^2 x^2 \cos \left (e-\frac {c f}{d}\right ) \text {Si}\left (f \left (\frac {c}{d}+x\right )\right )+8 a b c d f^2 x \cos \left (e-\frac {c f}{d}\right ) \text {Si}\left (f \left (\frac {c}{d}+x\right )\right )+4 a b c d f \cos (e+f x)+4 a b d^2 \sin (e+f x)+4 a b d^2 f x \cos (e+f x)+4 b^2 c^2 f^2 \sin \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (\frac {2 f (c+d x)}{d}\right )-4 b^2 f^2 (c+d x)^2 \text {Ci}\left (\frac {2 f (c+d x)}{d}\right ) \cos \left (2 e-\frac {2 c f}{d}\right )+4 b^2 d^2 f^2 x^2 \sin \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (\frac {2 f (c+d x)}{d}\right )+8 b^2 c d f^2 x \sin \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (\frac {2 f (c+d x)}{d}\right )+2 b^2 c d f \sin (2 (e+f x))+2 b^2 d^2 f x \sin (2 (e+f x))-b^2 d^2 \cos (2 (e+f x))+b^2 d^2}{4 d^3 (c+d x)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*(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]*SinInte
gral[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])/(d^3*(c + d*x)^2)

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fricas [A]  time = 0.84, size = 467, normalized size = 1.91 \[ \frac {b^{2} d^{2} \cos \left (f x + e\right )^{2} - {\left (a^{2} + b^{2}\right )} d^{2} + 2 \, {\left (b^{2} d^{2} f^{2} x^{2} + 2 \, b^{2} c d f^{2} x + b^{2} c^{2} f^{2}\right )} \sin \left (-\frac {2 \, {\left (d e - c f\right )}}{d}\right ) \operatorname {Si}\left (\frac {2 \, {\left (d f x + c f\right )}}{d}\right ) - 2 \, {\left (a b d^{2} f^{2} x^{2} + 2 \, a b c d f^{2} x + a b c^{2} f^{2}\right )} \cos \left (-\frac {d e - c f}{d}\right ) \operatorname {Si}\left (\frac {d f x + c f}{d}\right ) - 2 \, {\left (a b d^{2} f x + a b c d f\right )} \cos \left (f x + e\right ) + {\left ({\left (b^{2} d^{2} f^{2} x^{2} + 2 \, b^{2} c d f^{2} x + b^{2} c^{2} f^{2}\right )} \operatorname {Ci}\left (\frac {2 \, {\left (d f x + c f\right )}}{d}\right ) + {\left (b^{2} d^{2} f^{2} x^{2} + 2 \, b^{2} c d f^{2} x + b^{2} c^{2} f^{2}\right )} \operatorname {Ci}\left (-\frac {2 \, {\left (d f x + c f\right )}}{d}\right )\right )} \cos \left (-\frac {2 \, {\left (d e - c f\right )}}{d}\right ) - 2 \, {\left (a b d^{2} + {\left (b^{2} d^{2} f x + b^{2} c d f\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right ) + {\left ({\left (a b d^{2} f^{2} x^{2} + 2 \, a b c d f^{2} x + a b c^{2} f^{2}\right )} \operatorname {Ci}\left (\frac {d f x + c f}{d}\right ) + {\left (a b d^{2} f^{2} x^{2} + 2 \, a b c d f^{2} x + a b c^{2} f^{2}\right )} \operatorname {Ci}\left (-\frac {d f x + c f}{d}\right )\right )} \sin \left (-\frac {d e - c f}{d}\right )}{2 \, {\left (d^{5} x^{2} + 2 \, c d^{4} x + c^{2} d^{3}\right )}} \]

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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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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]

Timed out

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maple [A]  time = 0.04, size = 374, normalized size = 1.53 \[ \frac {-\frac {a^{2} f^{3}}{2 \left (\left (f x +e \right ) d +c f -d e \right )^{2} d}+2 f^{3} a b \left (-\frac {\sin \left (f x +e \right )}{2 \left (\left (f x +e \right ) d +c f -d e \right )^{2} d}+\frac {-\frac {\cos \left (f x +e \right )}{\left (\left (f x +e \right ) d +c f -d e \right ) d}-\frac {\frac {\Si \left (f x +e +\frac {c f -d e}{d}\right ) \cos \left (\frac {c f -d e}{d}\right )}{d}-\frac {\Ci \left (f x +e +\frac {c f -d e}{d}\right ) \sin \left (\frac {c f -d e}{d}\right )}{d}}{d}}{2 d}\right )-\frac {f^{3} b^{2}}{4 \left (\left (f x +e \right ) d +c f -d e \right )^{2} d}-\frac {f^{3} b^{2} \left (-\frac {\cos \left (2 f x +2 e \right )}{\left (\left (f x +e \right ) d +c f -d e \right )^{2} d}-\frac {-\frac {2 \sin \left (2 f x +2 e \right )}{\left (\left (f x +e \right ) d +c f -d e \right ) d}+\frac {\frac {4 \Si \left (2 f x +2 e +\frac {2 c f -2 d e}{d}\right ) \sin \left (\frac {2 c f -2 d e}{d}\right )}{d}+\frac {4 \Ci \left (2 f x +2 e +\frac {2 c f -2 d e}{d}\right ) \cos \left (\frac {2 c f -2 d e}{d}\right )}{d}}{d}}{d}\right )}{4}}{f} \]

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 = 0.72, size = 474, normalized size = 1.93 \[ -\frac {\frac {32 \, a^{2} f^{3}}{{\left (f x + e\right )}^{2} d^{3} + d^{3} e^{2} - 2 \, c d^{2} e f + c^{2} d f^{2} - 2 \, {\left (d^{3} e - c d^{2} f\right )} {\left (f x + e\right )}} - \frac {64 \, {\left (f^{3} {\left (-i \, E_{3}\left (\frac {i \, {\left (f x + e\right )} d - i \, d e + i \, c f}{d}\right ) + i \, E_{3}\left (-\frac {i \, {\left (f x + e\right )} d - i \, d e + i \, c f}{d}\right )\right )} \cos \left (-\frac {d e - c f}{d}\right ) + f^{3} {\left (E_{3}\left (\frac {i \, {\left (f x + e\right )} d - i \, d e + i \, c f}{d}\right ) + E_{3}\left (-\frac {i \, {\left (f x + e\right )} d - i \, d e + i \, c f}{d}\right )\right )} \sin \left (-\frac {d e - c f}{d}\right )\right )} a b}{{\left (f x + e\right )}^{2} d^{3} + d^{3} e^{2} - 2 \, c d^{2} e f + c^{2} d f^{2} - 2 \, {\left (d^{3} e - c d^{2} f\right )} {\left (f x + e\right )}} - \frac {{\left (16 \, f^{3} {\left (E_{3}\left (\frac {2 i \, {\left (f x + e\right )} d - 2 i \, d e + 2 i \, c f}{d}\right ) + E_{3}\left (-\frac {2 i \, {\left (f x + e\right )} d - 2 i \, d e + 2 i \, c f}{d}\right )\right )} \cos \left (-\frac {2 \, {\left (d e - c f\right )}}{d}\right ) + f^{3} {\left (16 i \, E_{3}\left (\frac {2 i \, {\left (f x + e\right )} d - 2 i \, d e + 2 i \, c f}{d}\right ) - 16 i \, E_{3}\left (-\frac {2 i \, {\left (f x + e\right )} d - 2 i \, d e + 2 i \, c f}{d}\right )\right )} \sin \left (-\frac {2 \, {\left (d e - c f\right )}}{d}\right ) - 16 \, f^{3}\right )} b^{2}}{{\left (f x + e\right )}^{2} d^{3} + d^{3} e^{2} - 2 \, c d^{2} e f + c^{2} d f^{2} - 2 \, {\left (d^{3} e - c d^{2} f\right )} {\left (f x + e\right )}}}{64 \, f} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a+b\,\sin \left (e+f\,x\right )\right )}^2}{{\left (c+d\,x\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b \sin {\left (e + f x \right )}\right )^{2}}{\left (c + d x\right )^{3}}\, dx \]

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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