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

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

[Out]

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

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Rubi [A]  time = 0.51, antiderivative size = 225, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 8, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, 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 31

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

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 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 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 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])

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*}

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Mathematica [A]  time = 1.02, size = 353, normalized size = 1.57 \[ -\frac {a^2 \left (4 c^2 f^2 \sin \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (\frac {2 f (c+d x)}{d}\right )+4 c^2 f^2 \cos \left (e-\frac {c f}{d}\right ) \text {Si}\left (f \left (\frac {c}{d}+x\right )\right )+4 f^2 (c+d x)^2 \text {Ci}\left (f \left (\frac {c}{d}+x\right )\right ) \sin \left (e-\frac {c f}{d}\right )-4 f^2 (c+d x)^2 \text {Ci}\left (\frac {2 f (c+d x)}{d}\right ) \cos \left (2 e-\frac {2 c f}{d}\right )+4 d^2 f^2 x^2 \sin \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (\frac {2 f (c+d x)}{d}\right )+4 d^2 f^2 x^2 \cos \left (e-\frac {c f}{d}\right ) \text {Si}\left (f \left (\frac {c}{d}+x\right )\right )+8 c d f^2 x \sin \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (\frac {2 f (c+d x)}{d}\right )+8 c d f^2 x \cos \left (e-\frac {c f}{d}\right ) \text {Si}\left (f \left (\frac {c}{d}+x\right )\right )+2 c d f \sin (2 (e+f x))+4 c d f \cos (e+f x)+4 d^2 \sin (e+f x)+2 d^2 f x \sin (2 (e+f x))+4 d^2 f x \cos (e+f x)-d^2 \cos (2 (e+f x))+3 d^2\right )}{4 d^3 (c+d x)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*(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*Co
s[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]*Si
nIntegral[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
]*SinIntegral[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]))/(d^3*(c + d*x)^2)

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fricas [B]  time = 0.86, size = 475, normalized size = 2.11 \[ \frac {a^{2} d^{2} \cos \left (f x + e\right )^{2} - 2 \, a^{2} d^{2} + 2 \, {\left (a^{2} d^{2} f^{2} x^{2} + 2 \, a^{2} c d f^{2} x + a^{2} c^{2} f^{2}\right )} \sin \left (-\frac {2 \, {\left (d e - c f\right )}}{d}\right ) \operatorname {Si}\left (\frac {2 \, {\left (d f x + c f\right )}}{d}\right ) - 2 \, {\left (a^{2} d^{2} f^{2} x^{2} + 2 \, a^{2} c d f^{2} x + a^{2} c^{2} f^{2}\right )} \cos \left (-\frac {d e - c f}{d}\right ) \operatorname {Si}\left (\frac {d f x + c f}{d}\right ) - 2 \, {\left (a^{2} d^{2} f x + a^{2} c d f\right )} \cos \left (f x + e\right ) + {\left ({\left (a^{2} d^{2} f^{2} x^{2} + 2 \, a^{2} c d f^{2} x + a^{2} c^{2} f^{2}\right )} \operatorname {Ci}\left (\frac {2 \, {\left (d f x + c f\right )}}{d}\right ) + {\left (a^{2} d^{2} f^{2} x^{2} + 2 \, a^{2} c d f^{2} x + a^{2} c^{2} f^{2}\right )} \operatorname {Ci}\left (-\frac {2 \, {\left (d f x + c f\right )}}{d}\right )\right )} \cos \left (-\frac {2 \, {\left (d e - c f\right )}}{d}\right ) - 2 \, {\left (a^{2} d^{2} + {\left (a^{2} d^{2} f x + a^{2} c d f\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right ) + {\left ({\left (a^{2} d^{2} f^{2} x^{2} + 2 \, a^{2} c d f^{2} x + a^{2} c^{2} f^{2}\right )} \operatorname {Ci}\left (\frac {d f x + c f}{d}\right ) + {\left (a^{2} d^{2} f^{2} x^{2} + 2 \, a^{2} c d f^{2} x + a^{2} c^{2} f^{2}\right )} \operatorname {Ci}\left (-\frac {d f x + c f}{d}\right )\right )} \sin \left (-\frac {d e - c f}{d}\right )}{2 \, {\left (d^{5} x^{2} + 2 \, c d^{4} x + c^{2} d^{3}\right )}} \]

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

[Out]

Timed out

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maple [A]  time = 0.05, size = 347, normalized size = 1.54 \[ \frac {-\frac {3 a^{2} f^{3}}{4 \left (\left (f x +e \right ) d +c f -d e \right )^{2} d}-\frac {a^{2} f^{3} \left (-\frac {\cos \left (2 f x +2 e \right )}{\left (\left (f x +e \right ) d +c f -d e \right )^{2} d}-\frac {-\frac {2 \sin \left (2 f x +2 e \right )}{\left (\left (f x +e \right ) d +c f -d e \right ) d}+\frac {\frac {4 \Si \left (2 f x +2 e +\frac {2 c f -2 d e}{d}\right ) \sin \left (\frac {2 c f -2 d e}{d}\right )}{d}+\frac {4 \Ci \left (2 f x +2 e +\frac {2 c f -2 d e}{d}\right ) \cos \left (\frac {2 c f -2 d e}{d}\right )}{d}}{d}}{d}\right )}{4}+2 a^{2} f^{3} \left (-\frac {\sin \left (f x +e \right )}{2 \left (\left (f x +e \right ) d +c f -d e \right )^{2} d}+\frac {-\frac {\cos \left (f x +e \right )}{\left (\left (f x +e \right ) d +c f -d e \right ) d}-\frac {\frac {\Si \left (f x +e +\frac {c f -d e}{d}\right ) \cos \left (\frac {c f -d e}{d}\right )}{d}-\frac {\Ci \left (f x +e +\frac {c f -d e}{d}\right ) \sin \left (\frac {c f -d e}{d}\right )}{d}}{d}}{2 d}\right )}{f} \]

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 = 1.09, size = 475, normalized size = 2.11 \[ -\frac {\frac {32 \, a^{2} f^{3}}{{\left (f x + e\right )}^{2} d^{3} + d^{3} e^{2} - 2 \, c d^{2} e f + c^{2} d f^{2} - 2 \, {\left (d^{3} e - c d^{2} f\right )} {\left (f x + e\right )}} - \frac {64 \, {\left (f^{3} {\left (-i \, E_{3}\left (\frac {i \, {\left (f x + e\right )} d - i \, d e + i \, c f}{d}\right ) + i \, E_{3}\left (-\frac {i \, {\left (f x + e\right )} d - i \, d e + i \, c f}{d}\right )\right )} \cos \left (-\frac {d e - c f}{d}\right ) + f^{3} {\left (E_{3}\left (\frac {i \, {\left (f x + e\right )} d - i \, d e + i \, c f}{d}\right ) + E_{3}\left (-\frac {i \, {\left (f x + e\right )} d - i \, d e + i \, c f}{d}\right )\right )} \sin \left (-\frac {d e - c f}{d}\right )\right )} a^{2}}{{\left (f x + e\right )}^{2} d^{3} + d^{3} e^{2} - 2 \, c d^{2} e f + c^{2} d f^{2} - 2 \, {\left (d^{3} e - c d^{2} f\right )} {\left (f x + e\right )}} - \frac {{\left (16 \, f^{3} {\left (E_{3}\left (\frac {2 i \, {\left (f x + e\right )} d - 2 i \, d e + 2 i \, c f}{d}\right ) + E_{3}\left (-\frac {2 i \, {\left (f x + e\right )} d - 2 i \, d e + 2 i \, c f}{d}\right )\right )} \cos \left (-\frac {2 \, {\left (d e - c f\right )}}{d}\right ) + f^{3} {\left (16 i \, E_{3}\left (\frac {2 i \, {\left (f x + e\right )} d - 2 i \, d e + 2 i \, c f}{d}\right ) - 16 i \, E_{3}\left (-\frac {2 i \, {\left (f x + e\right )} d - 2 i \, d e + 2 i \, c f}{d}\right )\right )} \sin \left (-\frac {2 \, {\left (d e - c f\right )}}{d}\right ) - 16 \, f^{3}\right )} a^{2}}{{\left (f x + e\right )}^{2} d^{3} + d^{3} e^{2} - 2 \, c d^{2} e f + c^{2} d f^{2} - 2 \, {\left (d^{3} e - c d^{2} f\right )} {\left (f x + e\right )}}}{64 \, f} \]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((a + a*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 \[ a^{2} \left (\int \frac {2 \sin {\left (e + f x \right )}}{c^{3} + 3 c^{2} d x + 3 c d^{2} x^{2} + d^{3} x^{3}}\, dx + \int \frac {\sin ^{2}{\left (e + f x \right )}}{c^{3} + 3 c^{2} d x + 3 c d^{2} x^{2} + d^{3} x^{3}}\, dx + \int \frac {1}{c^{3} + 3 c^{2} d x + 3 c d^{2} x^{2} + d^{3} x^{3}}\, dx\right ) \]

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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