3.3.96 \(\int \frac {(a+b \sin (c+\frac {d}{x}))^2}{e+f x} \, dx\) [296]

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

[Out]

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

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Rubi [A]
time = 0.45, antiderivative size = 255, normalized size of antiderivative = 1.00, 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 = {3512, 3398, 3384, 3380, 3383, 3393} \begin {gather*} \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} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(b^2*Cos[2*c - (2*d*f)/e]*CosIntegral[2*d*(f/e + x^(-1))])/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 3380

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 3383

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 3384

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 3393

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 3398

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 3512

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]

Rubi steps

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

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Mathematica [A]
time = 0.24, size = 195, normalized size = 0.76 \begin {gather*} \frac {-b^2 \cos \left (2 c-\frac {2 d f}{e}\right ) \text {Ci}\left (2 d \left (\frac {f}{e}+\frac {1}{x}\right )\right )+b^2 \cos (2 c) \text {Ci}\left (\frac {2 d}{x}\right )+2 a^2 \log (e+f x)+b^2 \log (e+f x)-4 a b \text {Ci}\left (\frac {d}{x}\right ) \sin (c)+4 a b \text {Ci}\left (d \left (\frac {f}{e}+\frac {1}{x}\right )\right ) \sin \left (c-\frac {d f}{e}\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 )+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 )-4 a b \cos (c) \text {Si}\left (\frac {d}{x}\right )-b^2 \sin (2 c) \text {Si}\left (\frac {2 d}{x}\right )}{2 f} \end {gather*}

Antiderivative was successfully verified.

[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 = 0.13, size = 354, normalized size = 1.39

method	result	size
risch	\(-\frac {i a b \,{\mathrm e}^{-\frac {i \left (c e -d f \right )}{e}} \expIntegral \left (1, \frac {i d}{x}+i c -\frac {i \left (c e -d f \right )}{e}\right )}{f}+\frac {i a b \expIntegral \left (1, \frac {i d}{x}\right ) {\mathrm e}^{-i c}}{f}+\frac {\ln \left (f x +e \right ) a^{2}}{f}+\frac {\ln \left (f x +e \right ) b^{2}}{2 f}+\frac {b^{2} {\mathrm e}^{-\frac {2 i \left (c e -d f \right )}{e}} \expIntegral \left (1, \frac {2 i d}{x}+2 i c -\frac {2 i \left (c e -d f \right )}{e}\right )}{4 f}-\frac {b^{2} \expIntegral \left (1, \frac {2 i d}{x}\right ) {\mathrm e}^{-2 i c}}{4 f}+\frac {b^{2} {\mathrm e}^{\frac {2 i \left (c e -d f \right )}{e}} \expIntegral \left (1, -\frac {2 i d}{x}-2 i c -\frac {2 \left (-i c e +i d f \right )}{e}\right )}{4 f}-\frac {b^{2} \expIntegral \left (1, -\frac {2 i d}{x}\right ) {\mathrm e}^{2 i c}}{4 f}+\frac {i a b \,{\mathrm e}^{\frac {i \left (c e -d f \right )}{e}} \expIntegral \left (1, -\frac {i d}{x}-i c -\frac {-i c e +i d f}{e}\right )}{f}-\frac {i a b \expIntegral \left (1, -\frac {i d}{x}\right ) {\mathrm e}^{i c}}{f}\)	\(325\)
derivativedivides	\(-d \left (\frac {a^{2} \ln \left (\frac {d}{x}\right )}{f d}-\frac {a^{2} \ln \left (-c e +d f +e \left (c +\frac {d}{x}\right )\right )}{f d}+\frac {2 a b \left (\sinIntegral \left (\frac {d}{x}\right ) \cos \left (c \right )+\cosineIntegral \left (\frac {d}{x}\right ) \sin \left (c \right )\right )}{f d}-\frac {2 a b e \left (-\frac {\sinIntegral \left (-\frac {d}{x}-c -\frac {-c e +d f}{e}\right ) \cos \left (\frac {-c e +d f}{e}\right )}{e}-\frac {\cosineIntegral \left (\frac {d}{x}+c +\frac {-c e +d f}{e}\right ) \sin \left (\frac {-c e +d f}{e}\right )}{e}\right )}{f d}+\frac {b^{2} \ln \left (\frac {d}{x}\right )}{2 f d}-\frac {b^{2} \ln \left (-c e +d f +e \left (c +\frac {d}{x}\right )\right )}{2 f d}-\frac {b^{2} \left (-2 \sinIntegral \left (\frac {2 d}{x}\right ) \sin \left (2 c \right )+2 \cosineIntegral \left (\frac {2 d}{x}\right ) \cos \left (2 c \right )\right )}{4 f d}+\frac {b^{2} e \left (-\frac {2 \sinIntegral \left (-\frac {2 d}{x}-2 c -\frac {2 \left (-c e +d f \right )}{e}\right ) \sin \left (\frac {-2 c e +2 d f}{e}\right )}{e}+\frac {2 \cosineIntegral \left (\frac {2 d}{x}+2 c +\frac {-2 c e +2 d f}{e}\right ) \cos \left (\frac {-2 c e +2 d f}{e}\right )}{e}\right )}{4 f d}\right )\)	\(354\)
default	\(-d \left (\frac {a^{2} \ln \left (\frac {d}{x}\right )}{f d}-\frac {a^{2} \ln \left (-c e +d f +e \left (c +\frac {d}{x}\right )\right )}{f d}+\frac {2 a b \left (\sinIntegral \left (\frac {d}{x}\right ) \cos \left (c \right )+\cosineIntegral \left (\frac {d}{x}\right ) \sin \left (c \right )\right )}{f d}-\frac {2 a b e \left (-\frac {\sinIntegral \left (-\frac {d}{x}-c -\frac {-c e +d f}{e}\right ) \cos \left (\frac {-c e +d f}{e}\right )}{e}-\frac {\cosineIntegral \left (\frac {d}{x}+c +\frac {-c e +d f}{e}\right ) \sin \left (\frac {-c e +d f}{e}\right )}{e}\right )}{f d}+\frac {b^{2} \ln \left (\frac {d}{x}\right )}{2 f d}-\frac {b^{2} \ln \left (-c e +d f +e \left (c +\frac {d}{x}\right )\right )}{2 f d}-\frac {b^{2} \left (-2 \sinIntegral \left (\frac {2 d}{x}\right ) \sin \left (2 c \right )+2 \cosineIntegral \left (\frac {2 d}{x}\right ) \cos \left (2 c \right )\right )}{4 f d}+\frac {b^{2} e \left (-\frac {2 \sinIntegral \left (-\frac {2 d}{x}-2 c -\frac {2 \left (-c e +d f \right )}{e}\right ) \sin \left (\frac {-2 c e +2 d f}{e}\right )}{e}+\frac {2 \cosineIntegral \left (\frac {2 d}{x}+2 c +\frac {-2 c e +2 d f}{e}\right ) \cos \left (\frac {-2 c e +2 d f}{e}\right )}{e}\right )}{4 f d}\right )\)	\(354\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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]

a^2*log(f*x + e)/f - 1/2*(2*b^2*f*integrate(1/4*cos(2*(c*x + d)/x)/((f*x + e)*cos(2*(c*x + d)/x)^2 + (f*x + e)
*sin(2*(c*x + d)/x)^2), x) + 2*b^2*f*integrate(1/4*cos(2*(c*x + d)/x)/(f*x + e), x) - 2*a*b*f*integrate(sin((c
*x + d)/x)/((f*x + e)*cos((c*x + d)/x)^2 + (f*x + e)*sin((c*x + d)/x)^2), x) - 2*a*b*f*integrate(sin((c*x + d)
/x)/(f*x + e), x) - b^2*log(f*x + e))/f

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Fricas [A]
time = 0.39, size = 289, normalized size = 1.13 \begin {gather*} \frac {2 \, b^{2} \sin \left (-2 \, {\left (d f - c e\right )} e^{\left (-1\right )}\right ) \operatorname {Si}\left (\frac {2 \, {\left (d f x + d e\right )} e^{\left (-1\right )}}{x}\right ) + 8 \, a b \cos \left (-{\left (d f - c e\right )} e^{\left (-1\right )}\right ) \operatorname {Si}\left (\frac {{\left (d f x + d e\right )} e^{\left (-1\right )}}{x}\right ) - 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 ) - {\left (b^{2} \operatorname {Ci}\left (\frac {2 \, {\left (d f x + d e\right )} e^{\left (-1\right )}}{x}\right ) + b^{2} \operatorname {Ci}\left (-\frac {2 \, {\left (d f x + d e\right )} e^{\left (-1\right )}}{x}\right )\right )} \cos \left (-2 \, {\left (d f - c e\right )} e^{\left (-1\right )}\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 ) + 2 \, {\left (2 \, a^{2} + b^{2}\right )} \log \left (f x + e\right ) + 4 \, {\left (a b \operatorname {Ci}\left (\frac {{\left (d f x + d e\right )} e^{\left (-1\right )}}{x}\right ) + a b \operatorname {Ci}\left (-\frac {{\left (d f x + d e\right )} e^{\left (-1\right )}}{x}\right )\right )} \sin \left (-{\left (d f - c e\right )} e^{\left (-1\right )}\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 \, f} \end {gather*}

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a + b \sin {\left (c + \frac {d}{x} \right )}\right )^{2}}{e + f x}\, dx \end {gather*}

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 [A]
time = 3.88, size = 368, normalized size = 1.44 \begin {gather*} -\frac {b^{2} d \cos \left (-2 \, {\left (d f - c e\right )} e^{\left (-1\right )}\right ) \operatorname {Ci}\left (2 \, {\left (d f - c e + \frac {{\left (c x + d\right )} e}{x}\right )} e^{\left (-1\right )}\right ) - b^{2} d \cos \left (2 \, c\right ) \operatorname {Ci}\left (-2 \, c + \frac {2 \, {\left (c x + d\right )}}{x}\right ) - 4 \, a b d \operatorname {Ci}\left ({\left (d f - c e + \frac {{\left (c x + d\right )} e}{x}\right )} e^{\left (-1\right )}\right ) \sin \left (-{\left (d f - c e\right )} e^{\left (-1\right )}\right ) + 4 \, a b d \operatorname {Ci}\left (-c + \frac {c x + d}{x}\right ) \sin \left (c\right ) + 4 \, a b d \cos \left (-{\left (d f - c e\right )} e^{\left (-1\right )}\right ) \operatorname {Si}\left (-{\left (d f - c e + \frac {{\left (c x + d\right )} e}{x}\right )} e^{\left (-1\right )}\right ) + b^{2} d \sin \left (-2 \, {\left (d f - c e\right )} e^{\left (-1\right )}\right ) \operatorname {Si}\left (-2 \, {\left (d f - c e + \frac {{\left (c x + d\right )} e}{x}\right )} e^{\left (-1\right )}\right ) - b^{2} d \sin \left (2 \, c\right ) \operatorname {Si}\left (2 \, c - \frac {2 \, {\left (c x + d\right )}}{x}\right ) - 4 \, a b d \cos \left (c\right ) \operatorname {Si}\left (c - \frac {c x + d}{x}\right ) - 2 \, a^{2} d \log \left (-d f + c e - \frac {{\left (c x + d\right )} e}{x}\right ) - b^{2} d \log \left (-d f + c e - \frac {{\left (c x + d\right )} e}{x}\right ) + 2 \, a^{2} d \log \left (c - \frac {c x + d}{x}\right ) + b^{2} d \log \left (c - \frac {c x + d}{x}\right )}{2 \, d f} \end {gather*}

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]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (a+b\,\sin \left (c+\frac {d}{x}\right )\right )}^2}{e+f\,x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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