3.298 \(\int \frac{(a+b \sin (c+\frac{d}{x}))^2}{(e+f x)^3} \, dx\)

Optimal. Leaf size=470 \[ \frac{a^2}{e^2 \left (\frac{e}{x}+f\right )}-\frac{a^2 f}{2 e^2 \left (\frac{e}{x}+f\right )^2}-\frac{a b d^2 f \sin \left (c-\frac{d f}{e}\right ) \text{CosIntegral}\left (d \left (\frac{f}{e}+\frac{1}{x}\right )\right )}{e^4}-\frac{2 a b d \cos \left (c-\frac{d f}{e}\right ) \text{CosIntegral}\left (d \left (\frac{f}{e}+\frac{1}{x}\right )\right )}{e^3}-\frac{a b d^2 f \cos \left (c-\frac{d f}{e}\right ) \text{Si}\left (d \left (\frac{f}{e}+\frac{1}{x}\right )\right )}{e^4}+\frac{2 a b d \sin \left (c-\frac{d f}{e}\right ) \text{Si}\left (d \left (\frac{f}{e}+\frac{1}{x}\right )\right )}{e^3}+\frac{2 a b \sin \left (c+\frac{d}{x}\right )}{e^2 \left (\frac{e}{x}+f\right )}-\frac{a b f \sin \left (c+\frac{d}{x}\right )}{e^2 \left (\frac{e}{x}+f\right )^2}-\frac{a b d f \cos \left (c+\frac{d}{x}\right )}{e^3 \left (\frac{e}{x}+f\right )}+\frac{b^2 d^2 f \cos \left (2 c-\frac{2 d f}{e}\right ) \text{CosIntegral}\left (2 d \left (\frac{f}{e}+\frac{1}{x}\right )\right )}{e^4}-\frac{b^2 d \sin \left (2 c-\frac{2 d f}{e}\right ) \text{CosIntegral}\left (2 d \left (\frac{f}{e}+\frac{1}{x}\right )\right )}{e^3}-\frac{b^2 d^2 f \sin \left (2 c-\frac{2 d f}{e}\right ) \text{Si}\left (2 d \left (\frac{f}{e}+\frac{1}{x}\right )\right )}{e^4}-\frac{b^2 d \cos \left (2 c-\frac{2 d f}{e}\right ) \text{Si}\left (2 d \left (\frac{f}{e}+\frac{1}{x}\right )\right )}{e^3}+\frac{b^2 \sin ^2\left (c+\frac{d}{x}\right )}{e^2 \left (\frac{e}{x}+f\right )}-\frac{b^2 f \sin ^2\left (c+\frac{d}{x}\right )}{2 e^2 \left (\frac{e}{x}+f\right )^2}-\frac{b^2 d f \sin \left (c+\frac{d}{x}\right ) \cos \left (c+\frac{d}{x}\right )}{e^3 \left (\frac{e}{x}+f\right )} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.957261, antiderivative size = 470, normalized size of antiderivative = 1., number of steps used = 27, number of rules used = 11, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {3431, 3317, 3297, 3303, 3299, 3302, 3314, 31, 3312, 3313, 12} \[ \frac{a^2}{e^2 \left (\frac{e}{x}+f\right )}-\frac{a^2 f}{2 e^2 \left (\frac{e}{x}+f\right )^2}-\frac{a b d^2 f \sin \left (c-\frac{d f}{e}\right ) \text{CosIntegral}\left (d \left (\frac{f}{e}+\frac{1}{x}\right )\right )}{e^4}-\frac{2 a b d \cos \left (c-\frac{d f}{e}\right ) \text{CosIntegral}\left (d \left (\frac{f}{e}+\frac{1}{x}\right )\right )}{e^3}-\frac{a b d^2 f \cos \left (c-\frac{d f}{e}\right ) \text{Si}\left (d \left (\frac{f}{e}+\frac{1}{x}\right )\right )}{e^4}+\frac{2 a b d \sin \left (c-\frac{d f}{e}\right ) \text{Si}\left (d \left (\frac{f}{e}+\frac{1}{x}\right )\right )}{e^3}+\frac{2 a b \sin \left (c+\frac{d}{x}\right )}{e^2 \left (\frac{e}{x}+f\right )}-\frac{a b f \sin \left (c+\frac{d}{x}\right )}{e^2 \left (\frac{e}{x}+f\right )^2}-\frac{a b d f \cos \left (c+\frac{d}{x}\right )}{e^3 \left (\frac{e}{x}+f\right )}+\frac{b^2 d^2 f \cos \left (2 c-\frac{2 d f}{e}\right ) \text{CosIntegral}\left (2 d \left (\frac{f}{e}+\frac{1}{x}\right )\right )}{e^4}-\frac{b^2 d \sin \left (2 c-\frac{2 d f}{e}\right ) \text{CosIntegral}\left (2 d \left (\frac{f}{e}+\frac{1}{x}\right )\right )}{e^3}-\frac{b^2 d^2 f \sin \left (2 c-\frac{2 d f}{e}\right ) \text{Si}\left (2 d \left (\frac{f}{e}+\frac{1}{x}\right )\right )}{e^4}-\frac{b^2 d \cos \left (2 c-\frac{2 d f}{e}\right ) \text{Si}\left (2 d \left (\frac{f}{e}+\frac{1}{x}\right )\right )}{e^3}+\frac{b^2 \sin ^2\left (c+\frac{d}{x}\right )}{e^2 \left (\frac{e}{x}+f\right )}-\frac{b^2 f \sin ^2\left (c+\frac{d}{x}\right )}{2 e^2 \left (\frac{e}{x}+f\right )^2}-\frac{b^2 d f \sin \left (c+\frac{d}{x}\right ) \cos \left (c+\frac{d}{x}\right )}{e^3 \left (\frac{e}{x}+f\right )} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sin[c + d/x])^2/(e + f*x)^3,x]

[Out]

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

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 3314

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

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])
)

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

Mathematica [A]  time = 3.46648, size = 740, normalized size = 1.57 \[ -\frac{2 a^2 e^4+4 a b d f (e+f x)^2 \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 )+4 a b d^2 e^2 f^2 \cos \left (c-\frac{d f}{e}\right ) \text{Si}\left (d \left (\frac{f}{e}+\frac{1}{x}\right )\right )+4 a b d^2 f^4 x^2 \cos \left (c-\frac{d f}{e}\right ) \text{Si}\left (d \left (\frac{f}{e}+\frac{1}{x}\right )\right )+8 a b d^2 e f^3 x \cos \left (c-\frac{d f}{e}\right ) \text{Si}\left (d \left (\frac{f}{e}+\frac{1}{x}\right )\right )-16 a b d e^2 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 e^2 f^2 x^2 \sin \left (c+\frac{d}{x}\right )+4 a b d e^2 f^2 x \cos \left (c+\frac{d}{x}\right )-8 a b d e^3 f \sin \left (c-\frac{d f}{e}\right ) \text{Si}\left (d \left (\frac{f}{e}+\frac{1}{x}\right )\right )-8 a b e^3 f x \sin \left (c+\frac{d}{x}\right )-8 a b d e f^3 x^2 \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^3 x^2 \cos \left (c+\frac{d}{x}\right )-4 b^2 d f (e+f x)^2 \text{CosIntegral}\left (2 d \left (\frac{f}{e}+\frac{1}{x}\right )\right ) \left (d f \cos \left (2 c-\frac{2 d f}{e}\right )-e \sin \left (2 c-\frac{2 d f}{e}\right )\right )+4 b^2 d^2 e^2 f^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 )+4 b^2 d^2 f^4 x^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 )+8 b^2 d^2 e f^3 x \sin \left (2 c-\frac{2 d f}{e}\right ) \text{Si}\left (2 d \left (\frac{f}{e}+\frac{1}{x}\right )\right )+8 b^2 d e^2 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 )+b^2 e^2 f^2 x^2 \cos \left (2 \left (c+\frac{d}{x}\right )\right )+2 b^2 d e^2 f^2 x \sin \left (2 \left (c+\frac{d}{x}\right )\right )+4 b^2 d e^3 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 )+2 b^2 e^3 f x \cos \left (2 \left (c+\frac{d}{x}\right )\right )+4 b^2 d e f^3 x^2 \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^3 x^2 \sin \left (2 \left (c+\frac{d}{x}\right )\right )+b^2 e^4}{4 e^4 f (e+f x)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sin[c + d/x])^2/(e + f*x)^3,x]

[Out]

-(2*a^2*e^4 + b^2*e^4 + 4*a*b*d*e^2*f^2*x*Cos[c + d/x] + 4*a*b*d*e*f^3*x^2*Cos[c + d/x] + 2*b^2*e^3*f*x*Cos[2*
(c + d/x)] + b^2*e^2*f^2*x^2*Cos[2*(c + d/x)] - 4*b^2*d*f*(e + f*x)^2*CosIntegral[2*d*(f/e + x^(-1))]*(d*f*Cos
[2*c - (2*d*f)/e] - e*Sin[2*c - (2*d*f)/e]) + 4*a*b*d*f*(e + f*x)^2*CosIntegral[d*(f/e + x^(-1))]*(2*e*Cos[c -
(d*f)/e] + d*f*Sin[c - (d*f)/e]) - 8*a*b*e^3*f*x*Sin[c + d/x] - 4*a*b*e^2*f^2*x^2*Sin[c + d/x] + 2*b^2*d*e^2*
f^2*x*Sin[2*(c + d/x)] + 2*b^2*d*e*f^3*x^2*Sin[2*(c + d/x)] + 4*a*b*d^2*e^2*f^2*Cos[c - (d*f)/e]*SinIntegral[d
*(f/e + x^(-1))] + 8*a*b*d^2*e*f^3*x*Cos[c - (d*f)/e]*SinIntegral[d*(f/e + x^(-1))] + 4*a*b*d^2*f^4*x^2*Cos[c
- (d*f)/e]*SinIntegral[d*(f/e + x^(-1))] - 8*a*b*d*e^3*f*Sin[c - (d*f)/e]*SinIntegral[d*(f/e + x^(-1))] - 16*a
*b*d*e^2*f^2*x*Sin[c - (d*f)/e]*SinIntegral[d*(f/e + x^(-1))] - 8*a*b*d*e*f^3*x^2*Sin[c - (d*f)/e]*SinIntegral
[d*(f/e + x^(-1))] + 4*b^2*d*e^3*f*Cos[2*c - (2*d*f)/e]*SinIntegral[2*d*(f/e + x^(-1))] + 8*b^2*d*e^2*f^2*x*Co
s[2*c - (2*d*f)/e]*SinIntegral[2*d*(f/e + x^(-1))] + 4*b^2*d*e*f^3*x^2*Cos[2*c - (2*d*f)/e]*SinIntegral[2*d*(f
/e + x^(-1))] + 4*b^2*d^2*e^2*f^2*Sin[2*c - (2*d*f)/e]*SinIntegral[2*d*(f/e + x^(-1))] + 8*b^2*d^2*e*f^3*x*Sin
[2*c - (2*d*f)/e]*SinIntegral[2*d*(f/e + x^(-1))] + 4*b^2*d^2*f^4*x^2*Sin[2*c - (2*d*f)/e]*SinIntegral[2*d*(f/
e + x^(-1))])/(4*e^4*f*(e + f*x)^2)

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Maple [B]  time = 0.038, size = 1124, normalized size = 2.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sin(c+d/x))^2/(f*x+e)^3,x)

[Out]

-d*(-a^2/e^2/(e*(c+d/x)-c*e+d*f)-1/2*a^2*(c*e-d*f)/e^2/(e*(c+d/x)-c*e+d*f)^2+2*(c*e-d*f)/e*a*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-C
i(d/x+c+(-c*e+d*f)/e)*sin((-c*e+d*f)/e)/e)/e)/e)+2*a*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*b^2/e^2/(e*(c+d/x)-c*e+d*f)-1/4*b
^2*(c*e-d*f)/e^2/(e*(c+d/x)-c*e+d*f)^2-1/4*b^2*(c*e-d*f)/e*(-cos(2*d/x+2*c)/(e*(c+d/x)-c*e+d*f)^2/e-(-2*sin(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)*sin(2*(-c*e+d*f)/e)/e+2*Ci(2*d/x+2*c+2*(-c*e+
d*f)/e)*cos(2*(-c*e+d*f)/e)/e)/e)/e)-1/4*b^2/e*(-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)*sin(2*(-c*e+d*f)/e)/e)/e)+1/2*c*a^2/(e*(c+d/x
)-c*e+d*f)^2/e-2*c*b*a*(-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)+1/4*b^2*c/(e*(c+d/x)-c*
e+d*f)^2/e+1/4*b^2*c*(-cos(2*d/x+2*c)/(e*(c+d/x)-c*e+d*f)^2/e-(-2*sin(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)*sin(2*(-c*e+d*f)/e)/e+2*Ci(2*d/x+2*c+2*(-c*e+d*f)/e)*cos(2*(-c*e+d*f)/e)/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)^3,x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 2.06944, size = 2053, normalized size = 4.37 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(c+d/x))^2/(f*x+e)^3,x, algorithm="fricas")

[Out]

1/4*(b^2*e^2*f^2*x^2 + 2*b^2*e^3*f*x - (2*a^2 + b^2)*e^4 - 2*(b^2*e^2*f^2*x^2 + 2*b^2*e^3*f*x)*cos((c*x + d)/x
)^2 - 4*((a*b*d*e*f^3*x^2 + 2*a*b*d*e^2*f^2*x + a*b*d*e^3*f)*cos_integral((d*f*x + d*e)/(e*x)) + (a*b*d*e*f^3*
x^2 + 2*a*b*d*e^2*f^2*x + a*b*d*e^3*f)*cos_integral(-(d*f*x + d*e)/(e*x)) + (a*b*d^2*f^4*x^2 + 2*a*b*d^2*e*f^3
*x + a*b*d^2*e^2*f^2)*sin_integral((d*f*x + d*e)/(e*x)))*cos(-(c*e - d*f)/e) + 2*((b^2*d^2*f^4*x^2 + 2*b^2*d^2
*e*f^3*x + b^2*d^2*e^2*f^2)*cos_integral(2*(d*f*x + d*e)/(e*x)) + (b^2*d^2*f^4*x^2 + 2*b^2*d^2*e*f^3*x + b^2*d
^2*e^2*f^2)*cos_integral(-2*(d*f*x + d*e)/(e*x)) - 2*(b^2*d*e*f^3*x^2 + 2*b^2*d*e^2*f^2*x + b^2*d*e^3*f)*sin_i
ntegral(2*(d*f*x + d*e)/(e*x)))*cos(-2*(c*e - d*f)/e) - 4*(a*b*d*e*f^3*x^2 + a*b*d*e^2*f^2*x)*cos((c*x + d)/x)
+ 2*((a*b*d^2*f^4*x^2 + 2*a*b*d^2*e*f^3*x + a*b*d^2*e^2*f^2)*cos_integral((d*f*x + d*e)/(e*x)) + (a*b*d^2*f^4
*x^2 + 2*a*b*d^2*e*f^3*x + a*b*d^2*e^2*f^2)*cos_integral(-(d*f*x + d*e)/(e*x)) - 4*(a*b*d*e*f^3*x^2 + 2*a*b*d*
e^2*f^2*x + a*b*d*e^3*f)*sin_integral((d*f*x + d*e)/(e*x)))*sin(-(c*e - d*f)/e) + 2*((b^2*d*e*f^3*x^2 + 2*b^2*
d*e^2*f^2*x + b^2*d*e^3*f)*cos_integral(2*(d*f*x + d*e)/(e*x)) + (b^2*d*e*f^3*x^2 + 2*b^2*d*e^2*f^2*x + b^2*d*
e^3*f)*cos_integral(-2*(d*f*x + d*e)/(e*x)) + 2*(b^2*d^2*f^4*x^2 + 2*b^2*d^2*e*f^3*x + b^2*d^2*e^2*f^2)*sin_in
tegral(2*(d*f*x + d*e)/(e*x)))*sin(-2*(c*e - d*f)/e) + 4*(a*b*e^2*f^2*x^2 + 2*a*b*e^3*f*x - (b^2*d*e*f^3*x^2 +
b^2*d*e^2*f^2*x)*cos((c*x + d)/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))**2/(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{{\left (b \sin \left (c + \frac{d}{x}\right ) + a\right )}^{2}}{{\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))^2/(f*x+e)^3,x, algorithm="giac")

[Out]

integrate((b*sin(c + d/x) + a)^2/(f*x + e)^3, x)