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

Optimal. Leaf size=183 \[ -\frac{a^2}{d (c+d x)}+\frac{2 a b f \text{CosIntegral}\left (\frac{c f}{d}+f x\right ) \cos \left (e-\frac{c f}{d}\right )}{d^2}-\frac{2 a b f \sin \left (e-\frac{c f}{d}\right ) \text{Si}\left (x f+\frac{c f}{d}\right )}{d^2}-\frac{2 a b \sin (e+f x)}{d (c+d x)}+\frac{b^2 f \text{CosIntegral}\left (\frac{2 c f}{d}+2 f x\right ) \sin \left (2 e-\frac{2 c f}{d}\right )}{d^2}+\frac{b^2 f \cos \left (2 e-\frac{2 c f}{d}\right ) \text{Si}\left (2 x f+\frac{2 c f}{d}\right )}{d^2}-\frac{b^2 \sin ^2(e+f x)}{d (c+d x)} \]

[Out]

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

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Rubi [A]  time = 0.334011, antiderivative size = 183, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 7, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.35, Rules used = {3317, 3297, 3303, 3299, 3302, 3313, 12} \[ -\frac{a^2}{d (c+d x)}+\frac{2 a b f \text{CosIntegral}\left (\frac{c f}{d}+f x\right ) \cos \left (e-\frac{c f}{d}\right )}{d^2}-\frac{2 a b f \sin \left (e-\frac{c f}{d}\right ) \text{Si}\left (x f+\frac{c f}{d}\right )}{d^2}-\frac{2 a b \sin (e+f x)}{d (c+d x)}+\frac{b^2 f \text{CosIntegral}\left (\frac{2 c f}{d}+2 f x\right ) \sin \left (2 e-\frac{2 c f}{d}\right )}{d^2}+\frac{b^2 f \cos \left (2 e-\frac{2 c f}{d}\right ) \text{Si}\left (2 x f+\frac{2 c f}{d}\right )}{d^2}-\frac{b^2 \sin ^2(e+f x)}{d (c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rubi steps

\begin{align*} \int \frac{(a+b \sin (e+f x))^2}{(c+d x)^2} \, dx &=\int \left (\frac{a^2}{(c+d x)^2}+\frac{2 a b \sin (e+f x)}{(c+d x)^2}+\frac{b^2 \sin ^2(e+f x)}{(c+d x)^2}\right ) \, dx\\ &=-\frac{a^2}{d (c+d x)}+(2 a b) \int \frac{\sin (e+f x)}{(c+d x)^2} \, dx+b^2 \int \frac{\sin ^2(e+f x)}{(c+d x)^2} \, dx\\ &=-\frac{a^2}{d (c+d x)}-\frac{2 a b \sin (e+f x)}{d (c+d x)}-\frac{b^2 \sin ^2(e+f x)}{d (c+d x)}+\frac{(2 a b f) \int \frac{\cos (e+f x)}{c+d x} \, dx}{d}+\frac{\left (2 b^2 f\right ) \int \frac{\sin (2 e+2 f x)}{2 (c+d x)} \, dx}{d}\\ &=-\frac{a^2}{d (c+d x)}-\frac{2 a b \sin (e+f x)}{d (c+d x)}-\frac{b^2 \sin ^2(e+f x)}{d (c+d x)}+\frac{\left (b^2 f\right ) \int \frac{\sin (2 e+2 f x)}{c+d x} \, dx}{d}+\frac{\left (2 a b 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 (2 a b 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{a^2}{d (c+d x)}+\frac{2 a b f \cos \left (e-\frac{c f}{d}\right ) \text{Ci}\left (\frac{c f}{d}+f x\right )}{d^2}-\frac{2 a b \sin (e+f x)}{d (c+d x)}-\frac{b^2 \sin ^2(e+f x)}{d (c+d x)}-\frac{2 a b f \sin \left (e-\frac{c f}{d}\right ) \text{Si}\left (\frac{c f}{d}+f x\right )}{d^2}+\frac{\left (b^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 (b^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{a^2}{d (c+d x)}+\frac{2 a b f \cos \left (e-\frac{c f}{d}\right ) \text{Ci}\left (\frac{c f}{d}+f x\right )}{d^2}+\frac{b^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{2 a b \sin (e+f x)}{d (c+d x)}-\frac{b^2 \sin ^2(e+f x)}{d (c+d x)}-\frac{2 a b f \sin \left (e-\frac{c f}{d}\right ) \text{Si}\left (\frac{c f}{d}+f x\right )}{d^2}+\frac{b^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 = 0.596356, size = 232, normalized size = 1.27 \[ \frac{-2 a^2 d+4 a b f (c+d x) \text{CosIntegral}\left (f \left (\frac{c}{d}+x\right )\right ) \cos \left (e-\frac{c f}{d}\right )-4 a b c f \sin \left (e-\frac{c f}{d}\right ) \text{Si}\left (f \left (\frac{c}{d}+x\right )\right )-4 a b d f x \sin \left (e-\frac{c f}{d}\right ) \text{Si}\left (f \left (\frac{c}{d}+x\right )\right )-4 a b d \sin (e+f x)+2 b^2 f (c+d x) \text{CosIntegral}\left (\frac{2 f (c+d x)}{d}\right ) \sin \left (2 e-\frac{2 c f}{d}\right )+2 b^2 c f \cos \left (2 e-\frac{2 c f}{d}\right ) \text{Si}\left (\frac{2 f (c+d x)}{d}\right )+2 b^2 d f x \cos \left (2 e-\frac{2 c f}{d}\right ) \text{Si}\left (\frac{2 f (c+d x)}{d}\right )+b^2 d \cos (2 (e+f x))-b^2 d}{2 d^2 (c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.018, size = 301, normalized size = 1.6 \begin{align*}{\frac{1}{f} \left ( -{\frac{{a}^{2}{f}^{2}}{ \left ( \left ( fx+e \right ) d+cf-de \right ) d}}+2\,{f}^{2}ab \left ( -{\frac{\sin \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 ) \sin \left ({\frac{cf-de}{d}} \right ) }+{\frac{1}{d}{\it Ci} \left ( fx+e+{\frac{cf-de}{d}} \right ) \cos \left ({\frac{cf-de}{d}} \right ) } \right ) } \right ) -{\frac{{f}^{2}{b}^{2}}{ \left ( 2\, \left ( fx+e \right ) d+2\,cf-2\,de \right ) d}}-{\frac{{f}^{2}{b}^{2}}{4} \left ( -2\,{\frac{\cos \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 ) \cos \left ( 2\,{\frac{cf-de}{d}} \right ) }-2\,{\frac{1}{d}{\it Ci} \left ( 2\,fx+2\,e+2\,{\frac{cf-de}{d}} \right ) \sin \left ( 2\,{\frac{cf-de}{d}} \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)^2,x)

[Out]

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

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Maxima [C]  time = 1.50062, size = 498, normalized size = 2.72 \begin{align*} -\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 b}{{\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 )} b^{2}}{{\left (f x + e\right )} d^{2} - d^{2} e + c d f}}{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)^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*b/((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)*b^2/((f*x + e)*d^2 - d^2*e + c*d*f))/f

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Fricas [A]  time = 2.26381, size = 693, normalized size = 3.79 \begin{align*} \frac{2 \, b^{2} d \cos \left (f x + e\right )^{2} - 4 \, a b d \sin \left (f x + e\right ) + 2 \,{\left (b^{2} d f x + b^{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 b d f x + a b 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 (a^{2} + b^{2}\right )} d + 2 \,{\left ({\left (a b d f x + a b c f\right )} \operatorname{Ci}\left (\frac{d f x + c f}{d}\right ) +{\left (a b d f x + a b 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 (b^{2} d f x + b^{2} c f\right )} \operatorname{Ci}\left (\frac{2 \,{\left (d f x + c f\right )}}{d}\right ) +{\left (b^{2} d f x + b^{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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*b^2*d*cos(f*x + e)^2 - 4*a*b*d*sin(f*x + e) + 2*(b^2*d*f*x + b^2*c*f)*cos(-2*(d*e - c*f)/d)*sin_integra
l(2*(d*f*x + c*f)/d) + 4*(a*b*d*f*x + a*b*c*f)*sin(-(d*e - c*f)/d)*sin_integral((d*f*x + c*f)/d) - 2*(a^2 + b^
2)*d + 2*((a*b*d*f*x + a*b*c*f)*cos_integral((d*f*x + c*f)/d) + (a*b*d*f*x + a*b*c*f)*cos_integral(-(d*f*x + c
*f)/d))*cos(-(d*e - c*f)/d) - ((b^2*d*f*x + b^2*c*f)*cos_integral(2*(d*f*x + c*f)/d) + (b^2*d*f*x + b^2*c*f)*c
os_integral(-2*(d*f*x + c*f)/d))*sin(-2*(d*e - c*f)/d))/(d^3*x + c*d^2)

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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 )^{2}}\, 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)**2,x)

[Out]

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

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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out