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

Optimal. Leaf size=255 \[ \frac{a^2 \log \left (\frac{e}{x}+f\right )}{f}+\frac{a^2 \log (x)}{f}+\frac{2 a b \sin \left (c-\frac{d f}{e}\right ) \text{CosIntegral}\left (d \left (\frac{f}{e}+\frac{1}{x}\right )\right )}{f}-\frac{2 a b \sin (c) \text{CosIntegral}\left (\frac{d}{x}\right )}{f}+\frac{2 a b \cos \left (c-\frac{d f}{e}\right ) \text{Si}\left (d \left (\frac{f}{e}+\frac{1}{x}\right )\right )}{f}-\frac{2 a b \cos (c) \text{Si}\left (\frac{d}{x}\right )}{f}-\frac{b^2 \cos \left (2 c-\frac{2 d f}{e}\right ) \text{CosIntegral}\left (2 d \left (\frac{f}{e}+\frac{1}{x}\right )\right )}{2 f}+\frac{b^2 \cos (2 c) \text{CosIntegral}\left (\frac{2 d}{x}\right )}{2 f}+\frac{b^2 \sin \left (2 c-\frac{2 d f}{e}\right ) \text{Si}\left (2 d \left (\frac{f}{e}+\frac{1}{x}\right )\right )}{2 f}-\frac{b^2 \sin (2 c) \text{Si}\left (\frac{2 d}{x}\right )}{2 f}+\frac{b^2 \log \left (\frac{e}{x}+f\right )}{2 f}+\frac{b^2 \log (x)}{2 f} \]

[Out]

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

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Rubi [A]  time = 0.66276, antiderivative size = 255, normalized size of antiderivative = 1., number of steps used = 22, number of rules used = 6, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {3431, 3317, 3303, 3299, 3302, 3312} \[ \frac{a^2 \log \left (\frac{e}{x}+f\right )}{f}+\frac{a^2 \log (x)}{f}+\frac{2 a b \sin \left (c-\frac{d f}{e}\right ) \text{CosIntegral}\left (d \left (\frac{f}{e}+\frac{1}{x}\right )\right )}{f}-\frac{2 a b \sin (c) \text{CosIntegral}\left (\frac{d}{x}\right )}{f}+\frac{2 a b \cos \left (c-\frac{d f}{e}\right ) \text{Si}\left (d \left (\frac{f}{e}+\frac{1}{x}\right )\right )}{f}-\frac{2 a b \cos (c) \text{Si}\left (\frac{d}{x}\right )}{f}-\frac{b^2 \cos \left (2 c-\frac{2 d f}{e}\right ) \text{CosIntegral}\left (2 d \left (\frac{f}{e}+\frac{1}{x}\right )\right )}{2 f}+\frac{b^2 \cos (2 c) \text{CosIntegral}\left (\frac{2 d}{x}\right )}{2 f}+\frac{b^2 \sin \left (2 c-\frac{2 d f}{e}\right ) \text{Si}\left (2 d \left (\frac{f}{e}+\frac{1}{x}\right )\right )}{2 f}-\frac{b^2 \sin (2 c) \text{Si}\left (\frac{2 d}{x}\right )}{2 f}+\frac{b^2 \log \left (\frac{e}{x}+f\right )}{2 f}+\frac{b^2 \log (x)}{2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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])
)

Rubi steps

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

Mathematica [A]  time = 0.400668, size = 195, normalized size = 0.76 \[ \frac{2 a^2 \log (e+f x)+4 a b \sin \left (c-\frac{d f}{e}\right ) \text{CosIntegral}\left (d \left (\frac{f}{e}+\frac{1}{x}\right )\right )-4 a b \sin (c) \text{CosIntegral}\left (\frac{d}{x}\right )+4 a b \cos \left (c-\frac{d f}{e}\right ) \text{Si}\left (d \left (\frac{f}{e}+\frac{1}{x}\right )\right )-4 a b \cos (c) \text{Si}\left (\frac{d}{x}\right )-b^2 \cos \left (2 c-\frac{2 d f}{e}\right ) \text{CosIntegral}\left (2 d \left (\frac{f}{e}+\frac{1}{x}\right )\right )+b^2 \cos (2 c) \text{CosIntegral}\left (\frac{2 d}{x}\right )+b^2 \sin \left (2 c-\frac{2 d f}{e}\right ) \text{Si}\left (2 d \left (\frac{f}{e}+\frac{1}{x}\right )\right )-b^2 \sin (2 c) \text{Si}\left (\frac{2 d}{x}\right )+b^2 \log (e+f x)}{2 f} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [A]  time = 0.034, size = 321, normalized size = 1.3 \begin{align*} -{\frac{{a}^{2}}{f}\ln \left ({\frac{d}{x}} \right ) }+{\frac{{a}^{2}}{f}\ln \left ( e \left ( c+{\frac{d}{x}} \right ) -ce+df \right ) }+2\,{\frac{ab}{f}{\it Si} \left ({\frac{d}{x}}+c+{\frac{-ce+df}{e}} \right ) \cos \left ({\frac{-ce+df}{e}} \right ) }-2\,{\frac{ab}{f}{\it Ci} \left ({\frac{d}{x}}+c+{\frac{-ce+df}{e}} \right ) \sin \left ({\frac{-ce+df}{e}} \right ) }-2\,{\frac{ab\sin \left ( c \right ) }{f}{\it Ci} \left ({\frac{d}{x}} \right ) }-2\,{\frac{ab\cos \left ( c \right ) }{f}{\it Si} \left ({\frac{d}{x}} \right ) }-{\frac{{b}^{2}}{2\,f}\ln \left ({\frac{d}{x}} \right ) }+{\frac{{b}^{2}}{2\,f}\ln \left ( e \left ( c+{\frac{d}{x}} \right ) -ce+df \right ) }-{\frac{{b}^{2}}{2\,f}{\it Si} \left ( 2\,{\frac{d}{x}}+2\,c+2\,{\frac{-ce+df}{e}} \right ) \sin \left ( 2\,{\frac{-ce+df}{e}} \right ) }-{\frac{{b}^{2}}{2\,f}{\it Ci} \left ( 2\,{\frac{d}{x}}+2\,c+2\,{\frac{-ce+df}{e}} \right ) \cos \left ( 2\,{\frac{-ce+df}{e}} \right ) }+{\frac{{b}^{2}\cos \left ( 2\,c \right ) }{2\,f}{\it Ci} \left ( 2\,{\frac{d}{x}} \right ) }-{\frac{{b}^{2}\sin \left ( 2\,c \right ) }{2\,f}{\it Si} \left ( 2\,{\frac{d}{x}} \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),x)

[Out]

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

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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),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 1.47742, size = 776, normalized size = 3.04 \begin{align*} -\frac{2 \, b^{2} \sin \left (2 \, c\right ) \operatorname{Si}\left (\frac{2 \, d}{x}\right ) + 8 \, a b \cos \left (c\right ) \operatorname{Si}\left (\frac{d}{x}\right ) + 2 \, b^{2} \sin \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 ) - 8 \, a b \cos \left (-\frac{c e - d f}{e}\right ) \operatorname{Si}\left (\frac{d f x + d e}{e x}\right ) -{\left (b^{2} \operatorname{Ci}\left (\frac{2 \, d}{x}\right ) + b^{2} \operatorname{Ci}\left (-\frac{2 \, d}{x}\right )\right )} \cos \left (2 \, c\right ) +{\left (b^{2} \operatorname{Ci}\left (\frac{2 \,{\left (d f x + d e\right )}}{e x}\right ) + b^{2} \operatorname{Ci}\left (-\frac{2 \,{\left (d f x + d e\right )}}{e x}\right )\right )} \cos \left (-\frac{2 \,{\left (c e - d f\right )}}{e}\right ) - 2 \,{\left (2 \, a^{2} + b^{2}\right )} \log \left (f x + e\right ) + 4 \,{\left (a b \operatorname{Ci}\left (\frac{d}{x}\right ) + a b \operatorname{Ci}\left (-\frac{d}{x}\right )\right )} \sin \left (c\right ) + 4 \,{\left (a b \operatorname{Ci}\left (\frac{d f x + d e}{e x}\right ) + a b \operatorname{Ci}\left (-\frac{d f x + d e}{e x}\right )\right )} \sin \left (-\frac{c e - d f}{e}\right )}{4 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(2*b^2*sin(2*c)*sin_integral(2*d/x) + 8*a*b*cos(c)*sin_integral(d/x) + 2*b^2*sin(-2*(c*e - d*f)/e)*sin_in
tegral(2*(d*f*x + d*e)/(e*x)) - 8*a*b*cos(-(c*e - d*f)/e)*sin_integral((d*f*x + d*e)/(e*x)) - (b^2*cos_integra
l(2*d/x) + b^2*cos_integral(-2*d/x))*cos(2*c) + (b^2*cos_integral(2*(d*f*x + d*e)/(e*x)) + b^2*cos_integral(-2
*(d*f*x + d*e)/(e*x)))*cos(-2*(c*e - d*f)/e) - 2*(2*a^2 + b^2)*log(f*x + e) + 4*(a*b*cos_integral(d/x) + a*b*c
os_integral(-d/x))*sin(c) + 4*(a*b*cos_integral((d*f*x + d*e)/(e*x)) + a*b*cos_integral(-(d*f*x + d*e)/(e*x)))
*sin(-(c*e - d*f)/e))/f

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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}}{f x + e}\,{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),x, algorithm="giac")

[Out]

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