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

Optimal. Leaf size=233 \[ \frac{a}{e^2 \left (\frac{e}{x}+f\right )}-\frac{a f}{2 e^2 \left (\frac{e}{x}+f\right )^2}-\frac{b d^2 f \sin \left (c-\frac{d f}{e}\right ) \text{CosIntegral}\left (d \left (\frac{f}{e}+\frac{1}{x}\right )\right )}{2 e^4}-\frac{b d \cos \left (c-\frac{d f}{e}\right ) \text{CosIntegral}\left (d \left (\frac{f}{e}+\frac{1}{x}\right )\right )}{e^3}-\frac{b d^2 f \cos \left (c-\frac{d f}{e}\right ) \text{Si}\left (d \left (\frac{f}{e}+\frac{1}{x}\right )\right )}{2 e^4}+\frac{b d \sin \left (c-\frac{d f}{e}\right ) \text{Si}\left (d \left (\frac{f}{e}+\frac{1}{x}\right )\right )}{e^3}+\frac{b \sin \left (c+\frac{d}{x}\right )}{e^2 \left (\frac{e}{x}+f\right )}-\frac{b f \sin \left (c+\frac{d}{x}\right )}{2 e^2 \left (\frac{e}{x}+f\right )^2}-\frac{b d f \cos \left (c+\frac{d}{x}\right )}{2 e^3 \left (\frac{e}{x}+f\right )} \]

[Out]

-(a*f)/(2*e^2*(f + e/x)^2) + a/(e^2*(f + e/x)) - (b*d*f*Cos[c + d/x])/(2*e^3*(f + e/x)) - (b*d*Cos[c - (d*f)/e
]*CosIntegral[d*(f/e + x^(-1))])/e^3 - (b*d^2*f*CosIntegral[d*(f/e + x^(-1))]*Sin[c - (d*f)/e])/(2*e^4) - (b*f
*Sin[c + d/x])/(2*e^2*(f + e/x)^2) + (b*Sin[c + d/x])/(e^2*(f + e/x)) - (b*d^2*f*Cos[c - (d*f)/e]*SinIntegral[
d*(f/e + x^(-1))])/(2*e^4) + (b*d*Sin[c - (d*f)/e]*SinIntegral[d*(f/e + x^(-1))])/e^3

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Rubi [A]  time = 0.487439, antiderivative size = 233, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 6, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {3431, 3317, 3297, 3303, 3299, 3302} \[ \frac{a}{e^2 \left (\frac{e}{x}+f\right )}-\frac{a f}{2 e^2 \left (\frac{e}{x}+f\right )^2}-\frac{b d^2 f \sin \left (c-\frac{d f}{e}\right ) \text{CosIntegral}\left (d \left (\frac{f}{e}+\frac{1}{x}\right )\right )}{2 e^4}-\frac{b d \cos \left (c-\frac{d f}{e}\right ) \text{CosIntegral}\left (d \left (\frac{f}{e}+\frac{1}{x}\right )\right )}{e^3}-\frac{b d^2 f \cos \left (c-\frac{d f}{e}\right ) \text{Si}\left (d \left (\frac{f}{e}+\frac{1}{x}\right )\right )}{2 e^4}+\frac{b d \sin \left (c-\frac{d f}{e}\right ) \text{Si}\left (d \left (\frac{f}{e}+\frac{1}{x}\right )\right )}{e^3}+\frac{b \sin \left (c+\frac{d}{x}\right )}{e^2 \left (\frac{e}{x}+f\right )}-\frac{b f \sin \left (c+\frac{d}{x}\right )}{2 e^2 \left (\frac{e}{x}+f\right )^2}-\frac{b d f \cos \left (c+\frac{d}{x}\right )}{2 e^3 \left (\frac{e}{x}+f\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*f)/(2*e^2*(f + e/x)^2) + a/(e^2*(f + e/x)) - (b*d*f*Cos[c + d/x])/(2*e^3*(f + e/x)) - (b*d*Cos[c - (d*f)/e
]*CosIntegral[d*(f/e + x^(-1))])/e^3 - (b*d^2*f*CosIntegral[d*(f/e + x^(-1))]*Sin[c - (d*f)/e])/(2*e^4) - (b*f
*Sin[c + d/x])/(2*e^2*(f + e/x)^2) + (b*Sin[c + d/x])/(e^2*(f + e/x)) - (b*d^2*f*Cos[c - (d*f)/e]*SinIntegral[
d*(f/e + x^(-1))])/(2*e^4) + (b*d*Sin[c - (d*f)/e]*SinIntegral[d*(f/e + x^(-1))])/e^3

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]

Rubi steps

\begin{align*} \int \frac{a+b \sin \left (c+\frac{d}{x}\right )}{(e+f x)^3} \, dx &=-\operatorname{Subst}\left (\int \left (-\frac{f (a+b \sin (c+d x))}{e (f+e x)^3}+\frac{a+b \sin (c+d x)}{e (f+e x)^2}\right ) \, dx,x,\frac{1}{x}\right )\\ &=-\frac{\operatorname{Subst}\left (\int \frac{a+b \sin (c+d x)}{(f+e x)^2} \, dx,x,\frac{1}{x}\right )}{e}+\frac{f \operatorname{Subst}\left (\int \frac{a+b \sin (c+d x)}{(f+e x)^3} \, dx,x,\frac{1}{x}\right )}{e}\\ &=-\frac{\operatorname{Subst}\left (\int \left (\frac{a}{(f+e x)^2}+\frac{b \sin (c+d x)}{(f+e x)^2}\right ) \, dx,x,\frac{1}{x}\right )}{e}+\frac{f \operatorname{Subst}\left (\int \left (\frac{a}{(f+e x)^3}+\frac{b \sin (c+d x)}{(f+e x)^3}\right ) \, dx,x,\frac{1}{x}\right )}{e}\\ &=-\frac{a f}{2 e^2 \left (f+\frac{e}{x}\right )^2}+\frac{a}{e^2 \left (f+\frac{e}{x}\right )}-\frac{b \operatorname{Subst}\left (\int \frac{\sin (c+d x)}{(f+e x)^2} \, dx,x,\frac{1}{x}\right )}{e}+\frac{(b f) \operatorname{Subst}\left (\int \frac{\sin (c+d x)}{(f+e x)^3} \, dx,x,\frac{1}{x}\right )}{e}\\ &=-\frac{a f}{2 e^2 \left (f+\frac{e}{x}\right )^2}+\frac{a}{e^2 \left (f+\frac{e}{x}\right )}-\frac{b f \sin \left (c+\frac{d}{x}\right )}{2 e^2 \left (f+\frac{e}{x}\right )^2}+\frac{b \sin \left (c+\frac{d}{x}\right )}{e^2 \left (f+\frac{e}{x}\right )}-\frac{(b d) \operatorname{Subst}\left (\int \frac{\cos (c+d x)}{f+e x} \, dx,x,\frac{1}{x}\right )}{e^2}+\frac{(b d f) \operatorname{Subst}\left (\int \frac{\cos (c+d x)}{(f+e x)^2} \, dx,x,\frac{1}{x}\right )}{2 e^2}\\ &=-\frac{a f}{2 e^2 \left (f+\frac{e}{x}\right )^2}+\frac{a}{e^2 \left (f+\frac{e}{x}\right )}-\frac{b d f \cos \left (c+\frac{d}{x}\right )}{2 e^3 \left (f+\frac{e}{x}\right )}-\frac{b f \sin \left (c+\frac{d}{x}\right )}{2 e^2 \left (f+\frac{e}{x}\right )^2}+\frac{b \sin \left (c+\frac{d}{x}\right )}{e^2 \left (f+\frac{e}{x}\right )}-\frac{\left (b d^2 f\right ) \operatorname{Subst}\left (\int \frac{\sin (c+d x)}{f+e x} \, dx,x,\frac{1}{x}\right )}{2 e^3}-\frac{\left (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^2}+\frac{\left (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^2}\\ &=-\frac{a f}{2 e^2 \left (f+\frac{e}{x}\right )^2}+\frac{a}{e^2 \left (f+\frac{e}{x}\right )}-\frac{b d f \cos \left (c+\frac{d}{x}\right )}{2 e^3 \left (f+\frac{e}{x}\right )}-\frac{b d \cos \left (c-\frac{d f}{e}\right ) \text{Ci}\left (\frac{d \left (f+\frac{e}{x}\right )}{e}\right )}{e^3}-\frac{b f \sin \left (c+\frac{d}{x}\right )}{2 e^2 \left (f+\frac{e}{x}\right )^2}+\frac{b \sin \left (c+\frac{d}{x}\right )}{e^2 \left (f+\frac{e}{x}\right )}+\frac{b d \sin \left (c-\frac{d f}{e}\right ) \text{Si}\left (\frac{d \left (f+\frac{e}{x}\right )}{e}\right )}{e^3}-\frac{\left (b d^2 f \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 )}{2 e^3}-\frac{\left (b d^2 f \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 )}{2 e^3}\\ &=-\frac{a f}{2 e^2 \left (f+\frac{e}{x}\right )^2}+\frac{a}{e^2 \left (f+\frac{e}{x}\right )}-\frac{b d f \cos \left (c+\frac{d}{x}\right )}{2 e^3 \left (f+\frac{e}{x}\right )}-\frac{b d \cos \left (c-\frac{d f}{e}\right ) \text{Ci}\left (\frac{d \left (f+\frac{e}{x}\right )}{e}\right )}{e^3}-\frac{b d^2 f \text{Ci}\left (\frac{d \left (f+\frac{e}{x}\right )}{e}\right ) \sin \left (c-\frac{d f}{e}\right )}{2 e^4}-\frac{b f \sin \left (c+\frac{d}{x}\right )}{2 e^2 \left (f+\frac{e}{x}\right )^2}+\frac{b \sin \left (c+\frac{d}{x}\right )}{e^2 \left (f+\frac{e}{x}\right )}-\frac{b d^2 f \cos \left (c-\frac{d f}{e}\right ) \text{Si}\left (\frac{d \left (f+\frac{e}{x}\right )}{e}\right )}{2 e^4}+\frac{b d \sin \left (c-\frac{d f}{e}\right ) \text{Si}\left (\frac{d \left (f+\frac{e}{x}\right )}{e}\right )}{e^3}\\ \end{align*}

Mathematica [A]  time = 1.8573, size = 151, normalized size = 0.65 \[ -\frac{\frac{e \left (a e^3+b d f^2 x (e+f x) \cos \left (c+\frac{d}{x}\right )-b e f x (2 e+f x) \sin \left (c+\frac{d}{x}\right )\right )}{f (e+f x)^2}+b d \text{CosIntegral}\left (d \left (\frac{f}{e}+\frac{1}{x}\right )\right ) \left (d f \sin \left (c-\frac{d f}{e}\right )+2 e \cos \left (c-\frac{d f}{e}\right )\right )+b d \text{Si}\left (d \left (\frac{f}{e}+\frac{1}{x}\right )\right ) \left (d f \cos \left (c-\frac{d f}{e}\right )-2 e \sin \left (c-\frac{d f}{e}\right )\right )}{2 e^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*d*CosIntegral[d*(f/e + x^(-1))]*(2*e*Cos[c - (d*f)/e] + d*f*Sin[c - (d*f)/e]) + (e*(a*e^3 + b*d*f^2*x*(e +
f*x)*Cos[c + d/x] - b*e*f*x*(2*e + f*x)*Sin[c + d/x]))/(f*(e + f*x)^2) + b*d*(d*f*Cos[c - (d*f)/e] - 2*e*Sin[
c - (d*f)/e])*SinIntegral[d*(f/e + x^(-1))])/(2*e^4)

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Maple [B]  time = 0.02, size = 527, normalized size = 2.3 \begin{align*} -d \left ( -{\frac{a}{{e}^{2}} \left ( e \left ( c+{\frac{d}{x}} \right ) -ce+df \right ) ^{-1}}-{\frac{a \left ( ce-df \right ) }{2\,{e}^{2}} \left ( e \left ( c+{\frac{d}{x}} \right ) -ce+df \right ) ^{-2}}+{\frac{ \left ( ce-df \right ) b}{e} \left ( -{\frac{1}{2\,e}\sin \left ( c+{\frac{d}{x}} \right ) \left ( e \left ( c+{\frac{d}{x}} \right ) -ce+df \right ) ^{-2}}+{\frac{1}{2\,e} \left ( -{\frac{1}{e}\cos \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 ) \cos \left ({\frac{-ce+df}{e}} \right ) }-{\frac{1}{e}{\it Ci} \left ({\frac{d}{x}}+c+{\frac{-ce+df}{e}} \right ) \sin \left ({\frac{-ce+df}{e}} \right ) } \right ) } \right ) } \right ) }+{\frac{b}{e} \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{ca}{2\,e} \left ( e \left ( c+{\frac{d}{x}} \right ) -ce+df \right ) ^{-2}}-cb \left ( -{\frac{1}{2\,e}\sin \left ( c+{\frac{d}{x}} \right ) \left ( e \left ( c+{\frac{d}{x}} \right ) -ce+df \right ) ^{-2}}+{\frac{1}{2\,e} \left ( -{\frac{1}{e}\cos \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 ) \cos \left ({\frac{-ce+df}{e}} \right ) }-{\frac{1}{e}{\it Ci} \left ({\frac{d}{x}}+c+{\frac{-ce+df}{e}} \right ) \sin \left ({\frac{-ce+df}{e}} \right ) } \right ) } \right ) } \right ) \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} b{\left (\int \frac{\sin \left (\frac{c x + d}{x}\right )}{2 \,{\left (f^{3} x^{3} + 3 \, e f^{2} x^{2} + 3 \, e^{2} f x + e^{3}\right )}}\,{d x} + \int \frac{\sin \left (\frac{c x + d}{x}\right )}{2 \,{\left ({\left (f^{3} x^{3} + 3 \, e f^{2} x^{2} + 3 \, e^{2} f x + e^{3}\right )} \cos \left (\frac{c x + d}{x}\right )^{2} +{\left (f^{3} x^{3} + 3 \, e f^{2} x^{2} + 3 \, e^{2} f x + e^{3}\right )} \sin \left (\frac{c x + d}{x}\right )^{2}\right )}}\,{d x}\right )} - \frac{a}{2 \,{\left (f^{3} x^{2} + 2 \, e f^{2} x + e^{2} f\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(c+d/x))/(f*x+e)^3,x, algorithm="maxima")

[Out]

b*(integrate(1/2*sin((c*x + d)/x)/(f^3*x^3 + 3*e*f^2*x^2 + 3*e^2*f*x + e^3), x) + integrate(1/2*sin((c*x + d)/
x)/((f^3*x^3 + 3*e*f^2*x^2 + 3*e^2*f*x + e^3)*cos((c*x + d)/x)^2 + (f^3*x^3 + 3*e*f^2*x^2 + 3*e^2*f*x + e^3)*s
in((c*x + d)/x)^2), x)) - 1/2*a/(f^3*x^2 + 2*e*f^2*x + e^2*f)

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Fricas [A]  time = 1.59395, size = 959, normalized size = 4.12 \begin{align*} -\frac{2 \, a e^{4} + 2 \,{\left ({\left (b d e f^{3} x^{2} + 2 \, b d e^{2} f^{2} x + b d e^{3} f\right )} \operatorname{Ci}\left (\frac{d f x + d e}{e x}\right ) +{\left (b d e f^{3} x^{2} + 2 \, b d e^{2} f^{2} x + b d e^{3} f\right )} \operatorname{Ci}\left (-\frac{d f x + d e}{e x}\right ) +{\left (b d^{2} f^{4} x^{2} + 2 \, b d^{2} e f^{3} x + b d^{2} e^{2} f^{2}\right )} \operatorname{Si}\left (\frac{d f x + d e}{e x}\right )\right )} \cos \left (-\frac{c e - d f}{e}\right ) + 2 \,{\left (b d e f^{3} x^{2} + b d e^{2} f^{2} x\right )} \cos \left (\frac{c x + d}{x}\right ) -{\left ({\left (b d^{2} f^{4} x^{2} + 2 \, b d^{2} e f^{3} x + b d^{2} e^{2} f^{2}\right )} \operatorname{Ci}\left (\frac{d f x + d e}{e x}\right ) +{\left (b d^{2} f^{4} x^{2} + 2 \, b d^{2} e f^{3} x + b d^{2} e^{2} f^{2}\right )} \operatorname{Ci}\left (-\frac{d f x + d e}{e x}\right ) - 4 \,{\left (b d e f^{3} x^{2} + 2 \, b d e^{2} f^{2} x + b d e^{3} f\right )} \operatorname{Si}\left (\frac{d f x + d e}{e x}\right )\right )} \sin \left (-\frac{c e - d f}{e}\right ) - 2 \,{\left (b e^{2} f^{2} x^{2} + 2 \, b e^{3} f x\right )} \sin \left (\frac{c x + d}{x}\right )}{4 \,{\left (e^{4} f^{3} x^{2} + 2 \, e^{5} f^{2} x + e^{6} f\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(c+d/x))/(f*x+e)^3,x, algorithm="fricas")

[Out]

-1/4*(2*a*e^4 + 2*((b*d*e*f^3*x^2 + 2*b*d*e^2*f^2*x + b*d*e^3*f)*cos_integral((d*f*x + d*e)/(e*x)) + (b*d*e*f^
3*x^2 + 2*b*d*e^2*f^2*x + b*d*e^3*f)*cos_integral(-(d*f*x + d*e)/(e*x)) + (b*d^2*f^4*x^2 + 2*b*d^2*e*f^3*x + b
*d^2*e^2*f^2)*sin_integral((d*f*x + d*e)/(e*x)))*cos(-(c*e - d*f)/e) + 2*(b*d*e*f^3*x^2 + b*d*e^2*f^2*x)*cos((
c*x + d)/x) - ((b*d^2*f^4*x^2 + 2*b*d^2*e*f^3*x + b*d^2*e^2*f^2)*cos_integral((d*f*x + d*e)/(e*x)) + (b*d^2*f^
4*x^2 + 2*b*d^2*e*f^3*x + b*d^2*e^2*f^2)*cos_integral(-(d*f*x + d*e)/(e*x)) - 4*(b*d*e*f^3*x^2 + 2*b*d*e^2*f^2
*x + b*d*e^3*f)*sin_integral((d*f*x + d*e)/(e*x)))*sin(-(c*e - d*f)/e) - 2*(b*e^2*f^2*x^2 + 2*b*e^3*f*x)*sin((
c*x + d)/x))/(e^4*f^3*x^2 + 2*e^5*f^2*x + e^6*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))/(f*x+e)**3,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{b \sin \left (c + \frac{d}{x}\right ) + a}{{\left (f x + e\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*sin(c + d/x) + a)/(f*x + e)^3, x)