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

Optimal. Leaf size=162 \[ \frac {a^2 f \text {Ci}\left (2 x f+\frac {2 c f}{d}\right ) \sin \left (2 e-\frac {2 c f}{d}\right )}{d^2}+\frac {2 a^2 f \text {Ci}\left (x f+\frac {c f}{d}\right ) \cos \left (e-\frac {c f}{d}\right )}{d^2}-\frac {2 a^2 f \sin \left (e-\frac {c f}{d}\right ) \text {Si}\left (x f+\frac {c f}{d}\right )}{d^2}+\frac {a^2 f \cos \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (2 x f+\frac {2 c f}{d}\right )}{d^2}-\frac {4 a^2 \sin ^4\left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right )}{d (c+d x)} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.33, antiderivative size = 162, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3318, 3313, 3303, 3299, 3302} \[ \frac {a^2 f \text {CosIntegral}\left (\frac {2 c f}{d}+2 f x\right ) \sin \left (2 e-\frac {2 c f}{d}\right )}{d^2}+\frac {2 a^2 f \text {CosIntegral}\left (\frac {c f}{d}+f x\right ) \cos \left (e-\frac {c f}{d}\right )}{d^2}-\frac {2 a^2 f \sin \left (e-\frac {c f}{d}\right ) \text {Si}\left (x f+\frac {c f}{d}\right )}{d^2}+\frac {a^2 f \cos \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (2 x f+\frac {2 c f}{d}\right )}{d^2}-\frac {4 a^2 \sin ^4\left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right )}{d (c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 3313

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

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

________________________________________________________________________________________

Mathematica [A]  time = 0.63, size = 206, normalized size = 1.27 \[ \frac {a^2 \left (2 f (c+d x) \text {Ci}\left (\frac {2 f (c+d x)}{d}\right ) \sin \left (2 e-\frac {2 c f}{d}\right )+4 f (c+d x) \text {Ci}\left (f \left (\frac {c}{d}+x\right )\right ) \cos \left (e-\frac {c f}{d}\right )-4 d f x \sin \left (e-\frac {c f}{d}\right ) \text {Si}\left (f \left (\frac {c}{d}+x\right )\right )-4 c f \sin \left (e-\frac {c f}{d}\right ) \text {Si}\left (f \left (\frac {c}{d}+x\right )\right )+2 d f x \cos \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (\frac {2 f (c+d x)}{d}\right )+2 c f \cos \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (\frac {2 f (c+d x)}{d}\right )-4 d \sin (e+f x)+d \cos (2 (e+f x))-3 d\right )}{2 d^2 (c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.70, size = 284, normalized size = 1.75 \[ \frac {2 \, a^{2} d \cos \left (f x + e\right )^{2} - 4 \, a^{2} d \sin \left (f x + e\right ) - 4 \, a^{2} d + 2 \, {\left (a^{2} d f x + a^{2} c f\right )} \cos \left (-\frac {2 \, {\left (d e - c f\right )}}{d}\right ) \operatorname {Si}\left (\frac {2 \, {\left (d f x + c f\right )}}{d}\right ) + 4 \, {\left (a^{2} d f x + a^{2} c f\right )} \sin \left (-\frac {d e - c f}{d}\right ) \operatorname {Si}\left (\frac {d f x + c f}{d}\right ) + 2 \, {\left ({\left (a^{2} d f x + a^{2} c f\right )} \operatorname {Ci}\left (\frac {d f x + c f}{d}\right ) + {\left (a^{2} d f x + a^{2} c f\right )} \operatorname {Ci}\left (-\frac {d f x + c f}{d}\right )\right )} \cos \left (-\frac {d e - c f}{d}\right ) - {\left ({\left (a^{2} d f x + a^{2} c f\right )} \operatorname {Ci}\left (\frac {2 \, {\left (d f x + c f\right )}}{d}\right ) + {\left (a^{2} d f x + a^{2} c f\right )} \operatorname {Ci}\left (-\frac {2 \, {\left (d f x + c f\right )}}{d}\right )\right )} \sin \left (-\frac {2 \, {\left (d e - c f\right )}}{d}\right )}{2 \, {\left (d^{3} x + c d^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*a^2*d*cos(f*x + e)^2 - 4*a^2*d*sin(f*x + e) - 4*a^2*d + 2*(a^2*d*f*x + a^2*c*f)*cos(-2*(d*e - c*f)/d)*s
in_integral(2*(d*f*x + c*f)/d) + 4*(a^2*d*f*x + a^2*c*f)*sin(-(d*e - c*f)/d)*sin_integral((d*f*x + c*f)/d) + 2
*((a^2*d*f*x + a^2*c*f)*cos_integral((d*f*x + c*f)/d) + (a^2*d*f*x + a^2*c*f)*cos_integral(-(d*f*x + c*f)/d))*
cos(-(d*e - c*f)/d) - ((a^2*d*f*x + a^2*c*f)*cos_integral(2*(d*f*x + c*f)/d) + (a^2*d*f*x + a^2*c*f)*cos_integ
ral(-2*(d*f*x + c*f)/d))*sin(-2*(d*e - c*f)/d))/(d^3*x + c*d^2)

________________________________________________________________________________________

giac [B]  time = 0.78, size = 1134, normalized size = 7.00 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.05, size = 274, normalized size = 1.69 \[ \frac {-\frac {3 a^{2} f^{2}}{2 \left (\left (f x +e \right ) d +c f -d e \right ) d}-\frac {a^{2} f^{2} \left (-\frac {2 \cos \left (2 f x +2 e \right )}{\left (\left (f x +e \right ) d +c f -d e \right ) d}-\frac {2 \left (\frac {2 \Si \left (2 f x +2 e +\frac {2 c f -2 d e}{d}\right ) \cos \left (\frac {2 c f -2 d e}{d}\right )}{d}-\frac {2 \Ci \left (2 f x +2 e +\frac {2 c f -2 d e}{d}\right ) \sin \left (\frac {2 c f -2 d e}{d}\right )}{d}\right )}{d}\right )}{4}+2 a^{2} f^{2} \left (-\frac {\sin \left (f x +e \right )}{\left (\left (f x +e \right ) d +c f -d e \right ) d}+\frac {\frac {\Si \left (f x +e +\frac {c f -d e}{d}\right ) \sin \left (\frac {c f -d e}{d}\right )}{d}+\frac {\Ci \left (f x +e +\frac {c f -d e}{d}\right ) \cos \left (\frac {c f -d e}{d}\right )}{d}}{d}\right )}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [C]  time = 0.61, size = 370, normalized size = 2.28 \[ -\frac {\frac {64 \, a^{2} f^{2}}{{\left (f x + e\right )} d^{2} - d^{2} e + c d f} - \frac {64 \, {\left (f^{2} {\left (-i \, E_{2}\left (\frac {i \, {\left (f x + e\right )} d - i \, d e + i \, c f}{d}\right ) + i \, E_{2}\left (-\frac {i \, {\left (f x + e\right )} d - i \, d e + i \, c f}{d}\right )\right )} \cos \left (-\frac {d e - c f}{d}\right ) + f^{2} {\left (E_{2}\left (\frac {i \, {\left (f x + e\right )} d - i \, d e + i \, c f}{d}\right ) + E_{2}\left (-\frac {i \, {\left (f x + e\right )} d - i \, d e + i \, c f}{d}\right )\right )} \sin \left (-\frac {d e - c f}{d}\right )\right )} a^{2}}{{\left (f x + e\right )} d^{2} - d^{2} e + c d f} - \frac {{\left (16 \, f^{2} {\left (E_{2}\left (\frac {2 i \, {\left (f x + e\right )} d - 2 i \, d e + 2 i \, c f}{d}\right ) + E_{2}\left (-\frac {2 i \, {\left (f x + e\right )} d - 2 i \, d e + 2 i \, c f}{d}\right )\right )} \cos \left (-\frac {2 \, {\left (d e - c f\right )}}{d}\right ) + f^{2} {\left (16 i \, E_{2}\left (\frac {2 i \, {\left (f x + e\right )} d - 2 i \, d e + 2 i \, c f}{d}\right ) - 16 i \, E_{2}\left (-\frac {2 i \, {\left (f x + e\right )} d - 2 i \, d e + 2 i \, c f}{d}\right )\right )} \sin \left (-\frac {2 \, {\left (d e - c f\right )}}{d}\right ) - 32 \, f^{2}\right )} a^{2}}{{\left (f x + e\right )} d^{2} - d^{2} e + c d f}}{64 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+a\,\sin \left (e+f\,x\right )\right )}^2}{{\left (c+d\,x\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a^{2} \left (\int \frac {2 \sin {\left (e + f x \right )}}{c^{2} + 2 c d x + d^{2} x^{2}}\, dx + \int \frac {\sin ^{2}{\left (e + f x \right )}}{c^{2} + 2 c d x + d^{2} x^{2}}\, dx + \int \frac {1}{c^{2} + 2 c d x + d^{2} x^{2}}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________