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

Optimal. Leaf size=195 \[ \frac{a^2}{e \left (\frac{e}{x}+f\right )}-\frac{2 a b d \cos \left (c-\frac{d f}{e}\right ) \text{CosIntegral}\left (d \left (\frac{f}{e}+\frac{1}{x}\right )\right )}{e^2}+\frac{2 a b d \sin \left (c-\frac{d f}{e}\right ) \text{Si}\left (d \left (\frac{f}{e}+\frac{1}{x}\right )\right )}{e^2}+\frac{2 a b \sin \left (c+\frac{d}{x}\right )}{e \left (\frac{e}{x}+f\right )}-\frac{b^2 d \sin \left (2 c-\frac{2 d f}{e}\right ) \text{CosIntegral}\left (2 d \left (\frac{f}{e}+\frac{1}{x}\right )\right )}{e^2}-\frac{b^2 d \cos \left (2 c-\frac{2 d f}{e}\right ) \text{Si}\left (2 d \left (\frac{f}{e}+\frac{1}{x}\right )\right )}{e^2}+\frac{b^2 \sin ^2\left (c+\frac{d}{x}\right )}{e \left (\frac{e}{x}+f\right )} \]

[Out]

a^2/(e*(f + e/x)) - (2*a*b*d*Cos[c - (d*f)/e]*CosIntegral[d*(f/e + x^(-1))])/e^2 - (b^2*d*CosIntegral[2*d*(f/e
+ x^(-1))]*Sin[2*c - (2*d*f)/e])/e^2 + (2*a*b*Sin[c + d/x])/(e*(f + e/x)) + (b^2*Sin[c + d/x]^2)/(e*(f + e/x)
) + (2*a*b*d*Sin[c - (d*f)/e]*SinIntegral[d*(f/e + x^(-1))])/e^2 - (b^2*d*Cos[2*c - (2*d*f)/e]*SinIntegral[2*d
*(f/e + x^(-1))])/e^2

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Rubi [A]  time = 0.391029, antiderivative size = 195, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 8, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.364, Rules used = {3431, 3317, 3297, 3303, 3299, 3302, 3313, 12} \[ \frac{a^2}{e \left (\frac{e}{x}+f\right )}-\frac{2 a b d \cos \left (c-\frac{d f}{e}\right ) \text{CosIntegral}\left (d \left (\frac{f}{e}+\frac{1}{x}\right )\right )}{e^2}+\frac{2 a b d \sin \left (c-\frac{d f}{e}\right ) \text{Si}\left (d \left (\frac{f}{e}+\frac{1}{x}\right )\right )}{e^2}+\frac{2 a b \sin \left (c+\frac{d}{x}\right )}{e \left (\frac{e}{x}+f\right )}-\frac{b^2 d \sin \left (2 c-\frac{2 d f}{e}\right ) \text{CosIntegral}\left (2 d \left (\frac{f}{e}+\frac{1}{x}\right )\right )}{e^2}-\frac{b^2 d \cos \left (2 c-\frac{2 d f}{e}\right ) \text{Si}\left (2 d \left (\frac{f}{e}+\frac{1}{x}\right )\right )}{e^2}+\frac{b^2 \sin ^2\left (c+\frac{d}{x}\right )}{e \left (\frac{e}{x}+f\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

a^2/(e*(f + e/x)) - (2*a*b*d*Cos[c - (d*f)/e]*CosIntegral[d*(f/e + x^(-1))])/e^2 - (b^2*d*CosIntegral[2*d*(f/e
+ x^(-1))]*Sin[2*c - (2*d*f)/e])/e^2 + (2*a*b*Sin[c + d/x])/(e*(f + e/x)) + (b^2*Sin[c + d/x]^2)/(e*(f + e/x)
) + (2*a*b*d*Sin[c - (d*f)/e]*SinIntegral[d*(f/e + x^(-1))])/e^2 - (b^2*d*Cos[2*c - (2*d*f)/e]*SinIntegral[2*d
*(f/e + x^(-1))])/e^2

Rule 3431

Int[((g_.) + (h_.)*(x_))^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*((e_.) + (f_.)*(x_))^(n_)])^(p_.), x_Symbol] :
> Dist[1/(n*f), Subst[Int[ExpandIntegrand[(a + b*Sin[c + d*x])^p, x^(1/n - 1)*(g - (e*h)/f + (h*x^(1/n))/f)^m,
x], x], x, (e + f*x)^n], x] /; FreeQ[{a, b, c, d, e, f, g, h, m}, x] && IGtQ[p, 0] && IntegerQ[1/n]

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{\left (a+b \sin \left (c+\frac{d}{x}\right )\right )^2}{(e+f x)^2} \, dx &=-\operatorname{Subst}\left (\int \frac{(a+b \sin (c+d x))^2}{(f+e x)^2} \, dx,x,\frac{1}{x}\right )\\ &=-\operatorname{Subst}\left (\int \left (\frac{a^2}{(f+e x)^2}+\frac{2 a b \sin (c+d x)}{(f+e x)^2}+\frac{b^2 \sin ^2(c+d x)}{(f+e x)^2}\right ) \, dx,x,\frac{1}{x}\right )\\ &=\frac{a^2}{e \left (f+\frac{e}{x}\right )}-(2 a b) \operatorname{Subst}\left (\int \frac{\sin (c+d x)}{(f+e x)^2} \, dx,x,\frac{1}{x}\right )-b^2 \operatorname{Subst}\left (\int \frac{\sin ^2(c+d x)}{(f+e x)^2} \, dx,x,\frac{1}{x}\right )\\ &=\frac{a^2}{e \left (f+\frac{e}{x}\right )}+\frac{2 a b \sin \left (c+\frac{d}{x}\right )}{e \left (f+\frac{e}{x}\right )}+\frac{b^2 \sin ^2\left (c+\frac{d}{x}\right )}{e \left (f+\frac{e}{x}\right )}-\frac{(2 a b d) \operatorname{Subst}\left (\int \frac{\cos (c+d x)}{f+e x} \, dx,x,\frac{1}{x}\right )}{e}-\frac{\left (2 b^2 d\right ) \operatorname{Subst}\left (\int \frac{\sin (2 c+2 d x)}{2 (f+e x)} \, dx,x,\frac{1}{x}\right )}{e}\\ &=\frac{a^2}{e \left (f+\frac{e}{x}\right )}+\frac{2 a b \sin \left (c+\frac{d}{x}\right )}{e \left (f+\frac{e}{x}\right )}+\frac{b^2 \sin ^2\left (c+\frac{d}{x}\right )}{e \left (f+\frac{e}{x}\right )}-\frac{\left (b^2 d\right ) \operatorname{Subst}\left (\int \frac{\sin (2 c+2 d x)}{f+e x} \, dx,x,\frac{1}{x}\right )}{e}-\frac{\left (2 a b d \cos \left (c-\frac{d f}{e}\right )\right ) \operatorname{Subst}\left (\int \frac{\cos \left (\frac{d f}{e}+d x\right )}{f+e x} \, dx,x,\frac{1}{x}\right )}{e}+\frac{\left (2 a b d \sin \left (c-\frac{d f}{e}\right )\right ) \operatorname{Subst}\left (\int \frac{\sin \left (\frac{d f}{e}+d x\right )}{f+e x} \, dx,x,\frac{1}{x}\right )}{e}\\ &=\frac{a^2}{e \left (f+\frac{e}{x}\right )}-\frac{2 a b d \cos \left (c-\frac{d f}{e}\right ) \text{Ci}\left (\frac{d \left (f+\frac{e}{x}\right )}{e}\right )}{e^2}+\frac{2 a b \sin \left (c+\frac{d}{x}\right )}{e \left (f+\frac{e}{x}\right )}+\frac{b^2 \sin ^2\left (c+\frac{d}{x}\right )}{e \left (f+\frac{e}{x}\right )}+\frac{2 a b d \sin \left (c-\frac{d f}{e}\right ) \text{Si}\left (\frac{d \left (f+\frac{e}{x}\right )}{e}\right )}{e^2}-\frac{\left (b^2 d \cos \left (2 c-\frac{2 d f}{e}\right )\right ) \operatorname{Subst}\left (\int \frac{\sin \left (\frac{2 d f}{e}+2 d x\right )}{f+e x} \, dx,x,\frac{1}{x}\right )}{e}-\frac{\left (b^2 d \sin \left (2 c-\frac{2 d f}{e}\right )\right ) \operatorname{Subst}\left (\int \frac{\cos \left (\frac{2 d f}{e}+2 d x\right )}{f+e x} \, dx,x,\frac{1}{x}\right )}{e}\\ &=\frac{a^2}{e \left (f+\frac{e}{x}\right )}-\frac{2 a b d \cos \left (c-\frac{d f}{e}\right ) \text{Ci}\left (\frac{d \left (f+\frac{e}{x}\right )}{e}\right )}{e^2}-\frac{b^2 d \text{Ci}\left (\frac{2 d \left (f+\frac{e}{x}\right )}{e}\right ) \sin \left (2 c-\frac{2 d f}{e}\right )}{e^2}+\frac{2 a b \sin \left (c+\frac{d}{x}\right )}{e \left (f+\frac{e}{x}\right )}+\frac{b^2 \sin ^2\left (c+\frac{d}{x}\right )}{e \left (f+\frac{e}{x}\right )}+\frac{2 a b d \sin \left (c-\frac{d f}{e}\right ) \text{Si}\left (\frac{d \left (f+\frac{e}{x}\right )}{e}\right )}{e^2}-\frac{b^2 d \cos \left (2 c-\frac{2 d f}{e}\right ) \text{Si}\left (\frac{2 d \left (f+\frac{e}{x}\right )}{e}\right )}{e^2}\\ \end{align*}

Mathematica [A]  time = 1.45129, size = 263, normalized size = 1.35 \[ -\frac{2 a^2 e^2+4 a b d f (e+f x) \cos \left (c-\frac{d f}{e}\right ) \text{CosIntegral}\left (d \left (\frac{f}{e}+\frac{1}{x}\right )\right )-4 a b d f^2 x \sin \left (c-\frac{d f}{e}\right ) \text{Si}\left (d \left (\frac{f}{e}+\frac{1}{x}\right )\right )-4 a b d e f \sin \left (c-\frac{d f}{e}\right ) \text{Si}\left (d \left (\frac{f}{e}+\frac{1}{x}\right )\right )-4 a b e f x \sin \left (c+\frac{d}{x}\right )+2 b^2 d f (e+f x) \sin \left (2 c-\frac{2 d f}{e}\right ) \text{CosIntegral}\left (2 d \left (\frac{f}{e}+\frac{1}{x}\right )\right )+2 b^2 d f^2 x \cos \left (2 c-\frac{2 d f}{e}\right ) \text{Si}\left (2 d \left (\frac{f}{e}+\frac{1}{x}\right )\right )+2 b^2 d e f \cos \left (2 c-\frac{2 d f}{e}\right ) \text{Si}\left (2 d \left (\frac{f}{e}+\frac{1}{x}\right )\right )+b^2 e f x \cos \left (2 \left (c+\frac{d}{x}\right )\right )+b^2 e^2}{2 e^2 f (e+f x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(2*a^2*e^2 + b^2*e^2 + b^2*e*f*x*Cos[2*(c + d/x)] + 4*a*b*d*f*(e + f*x)*Cos[c - (d*f)/e]*CosIntegral[d*(f/e +
x^(-1))] + 2*b^2*d*f*(e + f*x)*CosIntegral[2*d*(f/e + x^(-1))]*Sin[2*c - (2*d*f)/e] - 4*a*b*e*f*x*Sin[c + d/x
] - 4*a*b*d*e*f*Sin[c - (d*f)/e]*SinIntegral[d*(f/e + x^(-1))] - 4*a*b*d*f^2*x*Sin[c - (d*f)/e]*SinIntegral[d*
(f/e + x^(-1))] + 2*b^2*d*e*f*Cos[2*c - (2*d*f)/e]*SinIntegral[2*d*(f/e + x^(-1))] + 2*b^2*d*f^2*x*Cos[2*c - (
2*d*f)/e]*SinIntegral[2*d*(f/e + x^(-1))])/(2*e^2*f*(e + f*x))

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Maple [A]  time = 0.029, size = 308, normalized size = 1.6 \begin{align*} -d \left ( -{\frac{{a}^{2}}{e} \left ( e \left ( c+{\frac{d}{x}} \right ) -ce+df \right ) ^{-1}}+2\,ab \left ( -{\frac{1}{e}\sin \left ( c+{\frac{d}{x}} \right ) \left ( e \left ( c+{\frac{d}{x}} \right ) -ce+df \right ) ^{-1}}+{\frac{1}{e} \left ({\frac{1}{e}{\it Si} \left ({\frac{d}{x}}+c+{\frac{-ce+df}{e}} \right ) \sin \left ({\frac{-ce+df}{e}} \right ) }+{\frac{1}{e}{\it Ci} \left ({\frac{d}{x}}+c+{\frac{-ce+df}{e}} \right ) \cos \left ({\frac{-ce+df}{e}} \right ) } \right ) } \right ) -{\frac{{b}^{2}}{2\,e} \left ( e \left ( c+{\frac{d}{x}} \right ) -ce+df \right ) ^{-1}}-{\frac{{b}^{2}}{4} \left ( -2\,{\frac{1}{e}\cos \left ( 2\,{\frac{d}{x}}+2\,c \right ) \left ( e \left ( c+{\frac{d}{x}} \right ) -ce+df \right ) ^{-1}}-2\,{\frac{1}{e} \left ( 2\,{\frac{1}{e}{\it Si} \left ( 2\,{\frac{d}{x}}+2\,c+2\,{\frac{-ce+df}{e}} \right ) \cos \left ( 2\,{\frac{-ce+df}{e}} \right ) }-2\,{\frac{1}{e}{\it Ci} \left ( 2\,{\frac{d}{x}}+2\,c+2\,{\frac{-ce+df}{e}} \right ) \sin \left ( 2\,{\frac{-ce+df}{e}} \right ) } \right ) } \right ) } \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [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(c+d/x))^2/(f*x+e)^2,x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 1.69288, size = 805, normalized size = 4.13 \begin{align*} -\frac{2 \, b^{2} e f x \cos \left (\frac{c x + d}{x}\right )^{2} - 4 \, a b e f x \sin \left (\frac{c x + d}{x}\right ) - b^{2} e f x +{\left (2 \, a^{2} + b^{2}\right )} e^{2} + 2 \,{\left (b^{2} d f^{2} x + b^{2} d e f\right )} \cos \left (-\frac{2 \,{\left (c e - d f\right )}}{e}\right ) \operatorname{Si}\left (\frac{2 \,{\left (d f x + d e\right )}}{e x}\right ) + 4 \,{\left (a b d f^{2} x + a b d e f\right )} \sin \left (-\frac{c e - d f}{e}\right ) \operatorname{Si}\left (\frac{d f x + d e}{e x}\right ) + 2 \,{\left ({\left (a b d f^{2} x + a b d e f\right )} \operatorname{Ci}\left (\frac{d f x + d e}{e x}\right ) +{\left (a b d f^{2} x + a b d e f\right )} \operatorname{Ci}\left (-\frac{d f x + d e}{e x}\right )\right )} \cos \left (-\frac{c e - d f}{e}\right ) -{\left ({\left (b^{2} d f^{2} x + b^{2} d e f\right )} \operatorname{Ci}\left (\frac{2 \,{\left (d f x + d e\right )}}{e x}\right ) +{\left (b^{2} d f^{2} x + b^{2} d e f\right )} \operatorname{Ci}\left (-\frac{2 \,{\left (d f x + d e\right )}}{e x}\right )\right )} \sin \left (-\frac{2 \,{\left (c e - d f\right )}}{e}\right )}{2 \,{\left (e^{2} f^{2} x + e^{3} f\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [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(c+d/x))**2/(f*x+e)**2,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \sin \left (c + \frac{d}{x}\right ) + a\right )}^{2}}{{\left (f x + e\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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