Optimal. Leaf size=254 \[ a^2 e x+\frac {1}{2} a^2 f x^2+a b d^2 f \sin (c) \text {Ci}\left (\frac {d}{x}\right )-2 a b d e \cos (c) \text {Ci}\left (\frac {d}{x}\right )+a b d^2 f \cos (c) \text {Si}\left (\frac {d}{x}\right )+2 a b d e \sin (c) \text {Si}\left (\frac {d}{x}\right )+2 a b e x \sin \left (c+\frac {d}{x}\right )+a b f x^2 \sin \left (c+\frac {d}{x}\right )+a b d f x \cos \left (c+\frac {d}{x}\right )-b^2 d^2 f \cos (2 c) \text {Ci}\left (\frac {2 d}{x}\right )-b^2 d e \sin (2 c) \text {Ci}\left (\frac {2 d}{x}\right )+b^2 d^2 f \sin (2 c) \text {Si}\left (\frac {2 d}{x}\right )-b^2 d e \cos (2 c) \text {Si}\left (\frac {2 d}{x}\right )+b^2 e x \sin ^2\left (c+\frac {d}{x}\right )+\frac {1}{2} b^2 f x^2 \sin ^2\left (c+\frac {d}{x}\right )+b^2 d f x \sin \left (c+\frac {d}{x}\right ) \cos \left (c+\frac {d}{x}\right ) \]
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Rubi [A] time = 0.62, antiderivative size = 254, normalized size of antiderivative = 1.00, number of steps used = 27, number of rules used = 11, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.550, Rules used = {3431, 3317, 3297, 3303, 3299, 3302, 3314, 29, 3312, 3313, 12} \[ a^2 e x+\frac {1}{2} a^2 f x^2+a b d^2 f \sin (c) \text {CosIntegral}\left (\frac {d}{x}\right )-2 a b d e \cos (c) \text {CosIntegral}\left (\frac {d}{x}\right )+a b d^2 f \cos (c) \text {Si}\left (\frac {d}{x}\right )+2 a b d e \sin (c) \text {Si}\left (\frac {d}{x}\right )+2 a b e x \sin \left (c+\frac {d}{x}\right )+a b f x^2 \sin \left (c+\frac {d}{x}\right )+a b d f x \cos \left (c+\frac {d}{x}\right )-b^2 d^2 f \cos (2 c) \text {CosIntegral}\left (\frac {2 d}{x}\right )-b^2 d e \sin (2 c) \text {CosIntegral}\left (\frac {2 d}{x}\right )+b^2 d^2 f \sin (2 c) \text {Si}\left (\frac {2 d}{x}\right )-b^2 d e \cos (2 c) \text {Si}\left (\frac {2 d}{x}\right )+b^2 e x \sin ^2\left (c+\frac {d}{x}\right )+\frac {1}{2} b^2 f x^2 \sin ^2\left (c+\frac {d}{x}\right )+b^2 d f x \sin \left (c+\frac {d}{x}\right ) \cos \left (c+\frac {d}{x}\right ) \]
Antiderivative was successfully verified.
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Rule 12
Rule 29
Rule 3297
Rule 3299
Rule 3302
Rule 3303
Rule 3312
Rule 3313
Rule 3314
Rule 3317
Rule 3431
Rubi steps
\begin {align*} \int (e+f x) \left (a+b \sin \left (c+\frac {d}{x}\right )\right )^2 \, dx &=-\operatorname {Subst}\left (\int \left (\frac {f (a+b \sin (c+d x))^2}{x^3}+\frac {e (a+b \sin (c+d x))^2}{x^2}\right ) \, dx,x,\frac {1}{x}\right )\\ &=-\left (e \operatorname {Subst}\left (\int \frac {(a+b \sin (c+d x))^2}{x^2} \, dx,x,\frac {1}{x}\right )\right )-f \operatorname {Subst}\left (\int \frac {(a+b \sin (c+d x))^2}{x^3} \, dx,x,\frac {1}{x}\right )\\ &=-\left (e \operatorname {Subst}\left (\int \left (\frac {a^2}{x^2}+\frac {2 a b \sin (c+d x)}{x^2}+\frac {b^2 \sin ^2(c+d x)}{x^2}\right ) \, dx,x,\frac {1}{x}\right )\right )-f \operatorname {Subst}\left (\int \left (\frac {a^2}{x^3}+\frac {2 a b \sin (c+d x)}{x^3}+\frac {b^2 \sin ^2(c+d x)}{x^3}\right ) \, dx,x,\frac {1}{x}\right )\\ &=a^2 e x+\frac {1}{2} a^2 f x^2-(2 a b e) \operatorname {Subst}\left (\int \frac {\sin (c+d x)}{x^2} \, dx,x,\frac {1}{x}\right )-\left (b^2 e\right ) \operatorname {Subst}\left (\int \frac {\sin ^2(c+d x)}{x^2} \, dx,x,\frac {1}{x}\right )-(2 a b f) \operatorname {Subst}\left (\int \frac {\sin (c+d x)}{x^3} \, dx,x,\frac {1}{x}\right )-\left (b^2 f\right ) \operatorname {Subst}\left (\int \frac {\sin ^2(c+d x)}{x^3} \, dx,x,\frac {1}{x}\right )\\ &=a^2 e x+\frac {1}{2} a^2 f x^2+2 a b e x \sin \left (c+\frac {d}{x}\right )+a b f x^2 \sin \left (c+\frac {d}{x}\right )+b^2 d f x \cos \left (c+\frac {d}{x}\right ) \sin \left (c+\frac {d}{x}\right )+b^2 e x \sin ^2\left (c+\frac {d}{x}\right )+\frac {1}{2} b^2 f x^2 \sin ^2\left (c+\frac {d}{x}\right )-(2 a b d e) \operatorname {Subst}\left (\int \frac {\cos (c+d x)}{x} \, dx,x,\frac {1}{x}\right )-\left (2 b^2 d e\right ) \operatorname {Subst}\left (\int \frac {\sin (2 c+2 d x)}{2 x} \, dx,x,\frac {1}{x}\right )-(a b d f) \operatorname {Subst}\left (\int \frac {\cos (c+d x)}{x^2} \, dx,x,\frac {1}{x}\right )-\left (b^2 d^2 f\right ) \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,\frac {1}{x}\right )+\left (2 b^2 d^2 f\right ) \operatorname {Subst}\left (\int \frac {\sin ^2(c+d x)}{x} \, dx,x,\frac {1}{x}\right )\\ &=a^2 e x+\frac {1}{2} a^2 f x^2+a b d f x \cos \left (c+\frac {d}{x}\right )+b^2 d^2 f \log (x)+2 a b e x \sin \left (c+\frac {d}{x}\right )+a b f x^2 \sin \left (c+\frac {d}{x}\right )+b^2 d f x \cos \left (c+\frac {d}{x}\right ) \sin \left (c+\frac {d}{x}\right )+b^2 e x \sin ^2\left (c+\frac {d}{x}\right )+\frac {1}{2} b^2 f x^2 \sin ^2\left (c+\frac {d}{x}\right )-\left (b^2 d e\right ) \operatorname {Subst}\left (\int \frac {\sin (2 c+2 d x)}{x} \, dx,x,\frac {1}{x}\right )+\left (a b d^2 f\right ) \operatorname {Subst}\left (\int \frac {\sin (c+d x)}{x} \, dx,x,\frac {1}{x}\right )+\left (2 b^2 d^2 f\right ) \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 )-(2 a b d e \cos (c)) \operatorname {Subst}\left (\int \frac {\cos (d x)}{x} \, dx,x,\frac {1}{x}\right )+(2 a b d e \sin (c)) \operatorname {Subst}\left (\int \frac {\sin (d x)}{x} \, dx,x,\frac {1}{x}\right )\\ &=a^2 e x+\frac {1}{2} a^2 f x^2+a b d f x \cos \left (c+\frac {d}{x}\right )-2 a b d e \cos (c) \text {Ci}\left (\frac {d}{x}\right )+2 a b e x \sin \left (c+\frac {d}{x}\right )+a b f x^2 \sin \left (c+\frac {d}{x}\right )+b^2 d f x \cos \left (c+\frac {d}{x}\right ) \sin \left (c+\frac {d}{x}\right )+b^2 e x \sin ^2\left (c+\frac {d}{x}\right )+\frac {1}{2} b^2 f x^2 \sin ^2\left (c+\frac {d}{x}\right )+2 a b d e \sin (c) \text {Si}\left (\frac {d}{x}\right )-\left (b^2 d^2 f\right ) \operatorname {Subst}\left (\int \frac {\cos (2 c+2 d x)}{x} \, dx,x,\frac {1}{x}\right )+\left (a b d^2 f \cos (c)\right ) \operatorname {Subst}\left (\int \frac {\sin (d x)}{x} \, dx,x,\frac {1}{x}\right )-\left (b^2 d e \cos (2 c)\right ) \operatorname {Subst}\left (\int \frac {\sin (2 d x)}{x} \, dx,x,\frac {1}{x}\right )+\left (a b d^2 f \sin (c)\right ) \operatorname {Subst}\left (\int \frac {\cos (d x)}{x} \, dx,x,\frac {1}{x}\right )-\left (b^2 d e \sin (2 c)\right ) \operatorname {Subst}\left (\int \frac {\cos (2 d x)}{x} \, dx,x,\frac {1}{x}\right )\\ &=a^2 e x+\frac {1}{2} a^2 f x^2+a b d f x \cos \left (c+\frac {d}{x}\right )-2 a b d e \cos (c) \text {Ci}\left (\frac {d}{x}\right )+a b d^2 f \text {Ci}\left (\frac {d}{x}\right ) \sin (c)-b^2 d e \text {Ci}\left (\frac {2 d}{x}\right ) \sin (2 c)+2 a b e x \sin \left (c+\frac {d}{x}\right )+a b f x^2 \sin \left (c+\frac {d}{x}\right )+b^2 d f x \cos \left (c+\frac {d}{x}\right ) \sin \left (c+\frac {d}{x}\right )+b^2 e x \sin ^2\left (c+\frac {d}{x}\right )+\frac {1}{2} b^2 f x^2 \sin ^2\left (c+\frac {d}{x}\right )+a b d^2 f \cos (c) \text {Si}\left (\frac {d}{x}\right )+2 a b d e \sin (c) \text {Si}\left (\frac {d}{x}\right )-b^2 d e \cos (2 c) \text {Si}\left (\frac {2 d}{x}\right )-\left (b^2 d^2 f \cos (2 c)\right ) \operatorname {Subst}\left (\int \frac {\cos (2 d x)}{x} \, dx,x,\frac {1}{x}\right )+\left (b^2 d^2 f \sin (2 c)\right ) \operatorname {Subst}\left (\int \frac {\sin (2 d x)}{x} \, dx,x,\frac {1}{x}\right )\\ &=a^2 e x+\frac {1}{2} a^2 f x^2+a b d f x \cos \left (c+\frac {d}{x}\right )-2 a b d e \cos (c) \text {Ci}\left (\frac {d}{x}\right )-b^2 d^2 f \cos (2 c) \text {Ci}\left (\frac {2 d}{x}\right )+a b d^2 f \text {Ci}\left (\frac {d}{x}\right ) \sin (c)-b^2 d e \text {Ci}\left (\frac {2 d}{x}\right ) \sin (2 c)+2 a b e x \sin \left (c+\frac {d}{x}\right )+a b f x^2 \sin \left (c+\frac {d}{x}\right )+b^2 d f x \cos \left (c+\frac {d}{x}\right ) \sin \left (c+\frac {d}{x}\right )+b^2 e x \sin ^2\left (c+\frac {d}{x}\right )+\frac {1}{2} b^2 f x^2 \sin ^2\left (c+\frac {d}{x}\right )+a b d^2 f \cos (c) \text {Si}\left (\frac {d}{x}\right )+2 a b d e \sin (c) \text {Si}\left (\frac {d}{x}\right )-b^2 d e \cos (2 c) \text {Si}\left (\frac {2 d}{x}\right )+b^2 d^2 f \sin (2 c) \text {Si}\left (\frac {2 d}{x}\right )\\ \end {align*}
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Mathematica [A] time = 0.56, size = 252, normalized size = 0.99 \[ \frac {1}{4} \left (4 a^2 e x+2 a^2 f x^2+4 a b d \text {Ci}\left (\frac {d}{x}\right ) (d f \sin (c)-2 e \cos (c))+4 a b d^2 f \cos (c) \text {Si}\left (\frac {d}{x}\right )+8 a b d e \sin (c) \text {Si}\left (\frac {d}{x}\right )+8 a b e x \sin \left (c+\frac {d}{x}\right )+4 a b f x^2 \sin \left (c+\frac {d}{x}\right )+4 a b d f x \cos \left (c+\frac {d}{x}\right )-4 b^2 d \text {Ci}\left (\frac {2 d}{x}\right ) (d f \cos (2 c)+e \sin (2 c))+4 b^2 d^2 f \sin (2 c) \text {Si}\left (\frac {2 d}{x}\right )-4 b^2 d e \cos (2 c) \text {Si}\left (\frac {2 d}{x}\right )-2 b^2 e x \cos \left (2 \left (c+\frac {d}{x}\right )\right )-b^2 f x^2 \cos \left (2 \left (c+\frac {d}{x}\right )\right )+2 b^2 d f x \sin \left (2 \left (c+\frac {d}{x}\right )\right )+2 b^2 e x+b^2 f x^2\right ) \]
Antiderivative was successfully verified.
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fricas [A] time = 0.62, size = 300, normalized size = 1.18 \[ a b d f x \cos \left (\frac {c x + d}{x}\right ) + \frac {1}{2} \, {\left (a^{2} + b^{2}\right )} f x^{2} + {\left (a^{2} + b^{2}\right )} e x - \frac {1}{2} \, {\left (b^{2} f x^{2} + 2 \, b^{2} e x\right )} \cos \left (\frac {c x + d}{x}\right )^{2} - \frac {1}{2} \, {\left (b^{2} d^{2} f \operatorname {Ci}\left (\frac {2 \, d}{x}\right ) + b^{2} d^{2} f \operatorname {Ci}\left (-\frac {2 \, d}{x}\right ) + 2 \, b^{2} d e \operatorname {Si}\left (\frac {2 \, d}{x}\right )\right )} \cos \left (2 \, c\right ) + {\left (a b d^{2} f \operatorname {Si}\left (\frac {d}{x}\right ) - a b d e \operatorname {Ci}\left (\frac {d}{x}\right ) - a b d e \operatorname {Ci}\left (-\frac {d}{x}\right )\right )} \cos \relax (c) + \frac {1}{2} \, {\left (2 \, b^{2} d^{2} f \operatorname {Si}\left (\frac {2 \, d}{x}\right ) - b^{2} d e \operatorname {Ci}\left (\frac {2 \, d}{x}\right ) - b^{2} d e \operatorname {Ci}\left (-\frac {2 \, d}{x}\right )\right )} \sin \left (2 \, c\right ) + \frac {1}{2} \, {\left (a b d^{2} f \operatorname {Ci}\left (\frac {d}{x}\right ) + a b d^{2} f \operatorname {Ci}\left (-\frac {d}{x}\right ) + 4 \, a b d e \operatorname {Si}\left (\frac {d}{x}\right )\right )} \sin \relax (c) + {\left (b^{2} d f x \cos \left (\frac {c x + d}{x}\right ) + a b f x^{2} + 2 \, a b e x\right )} \sin \left (\frac {c x + d}{x}\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 1.07, size = 1145, normalized size = 4.51 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.10, size = 265, normalized size = 1.04 \[ -d \left (-\frac {a^{2} e x}{d}-\frac {a^{2} f \,x^{2}}{2 d}+2 a b e \left (-\frac {\sin \left (c +\frac {d}{x}\right ) x}{d}-\Si \left (\frac {d}{x}\right ) \sin \relax (c )+\Ci \left (\frac {d}{x}\right ) \cos \relax (c )\right )+2 a b d f \left (-\frac {\sin \left (c +\frac {d}{x}\right ) x^{2}}{2 d^{2}}-\frac {\cos \left (c +\frac {d}{x}\right ) x}{2 d}-\frac {\Si \left (\frac {d}{x}\right ) \cos \relax (c )}{2}-\frac {\Ci \left (\frac {d}{x}\right ) \sin \relax (c )}{2}\right )-\frac {b^{2} e x}{2 d}-\frac {b^{2} e \left (-\frac {2 \cos \left (\frac {2 d}{x}+2 c \right ) x}{d}-4 \Si \left (\frac {2 d}{x}\right ) \cos \left (2 c \right )-4 \Ci \left (\frac {2 d}{x}\right ) \sin \left (2 c \right )\right )}{4}-\frac {b^{2} f \,x^{2}}{4 d}-\frac {b^{2} d f \left (-\frac {\cos \left (\frac {2 d}{x}+2 c \right ) x^{2}}{d^{2}}+\frac {2 \sin \left (\frac {2 d}{x}+2 c \right ) x}{d}+4 \Si \left (\frac {2 d}{x}\right ) \sin \left (2 c \right )-4 \Ci \left (\frac {2 d}{x}\right ) \cos \left (2 c \right )\right )}{4}\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [C] time = 0.56, size = 322, normalized size = 1.27 \[ \frac {1}{2} \, a^{2} f x^{2} - {\left ({\left ({\left ({\rm Ei}\left (\frac {i \, d}{x}\right ) + {\rm Ei}\left (-\frac {i \, d}{x}\right )\right )} \cos \relax (c) - {\left (-i \, {\rm Ei}\left (\frac {i \, d}{x}\right ) + i \, {\rm Ei}\left (-\frac {i \, d}{x}\right )\right )} \sin \relax (c)\right )} d - 2 \, x \sin \left (\frac {c x + d}{x}\right )\right )} a b e - \frac {1}{2} \, {\left ({\left ({\left (-i \, {\rm Ei}\left (\frac {2 i \, d}{x}\right ) + i \, {\rm Ei}\left (-\frac {2 i \, d}{x}\right )\right )} \cos \left (2 \, c\right ) + {\left ({\rm Ei}\left (\frac {2 i \, d}{x}\right ) + {\rm Ei}\left (-\frac {2 i \, d}{x}\right )\right )} \sin \left (2 \, c\right )\right )} d + x \cos \left (\frac {2 \, {\left (c x + d\right )}}{x}\right ) - x\right )} b^{2} e + \frac {1}{2} \, {\left ({\left ({\left (-i \, {\rm Ei}\left (\frac {i \, d}{x}\right ) + i \, {\rm Ei}\left (-\frac {i \, d}{x}\right )\right )} \cos \relax (c) + {\left ({\rm Ei}\left (\frac {i \, d}{x}\right ) + {\rm Ei}\left (-\frac {i \, d}{x}\right )\right )} \sin \relax (c)\right )} d^{2} + 2 \, d x \cos \left (\frac {c x + d}{x}\right ) + 2 \, x^{2} \sin \left (\frac {c x + d}{x}\right )\right )} a b f - \frac {1}{4} \, {\left ({\left (2 \, {\left ({\rm Ei}\left (\frac {2 i \, d}{x}\right ) + {\rm Ei}\left (-\frac {2 i \, d}{x}\right )\right )} \cos \left (2 \, c\right ) - {\left (-2 i \, {\rm Ei}\left (\frac {2 i \, d}{x}\right ) + 2 i \, {\rm Ei}\left (-\frac {2 i \, d}{x}\right )\right )} \sin \left (2 \, c\right )\right )} d^{2} + x^{2} \cos \left (\frac {2 \, {\left (c x + d\right )}}{x}\right ) - 2 \, d x \sin \left (\frac {2 \, {\left (c x + d\right )}}{x}\right ) - x^{2}\right )} b^{2} f + a^{2} e x \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \left (e+f\,x\right )\,{\left (a+b\,\sin \left (c+\frac {d}{x}\right )\right )}^2 \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \sin {\left (c + \frac {d}{x} \right )}\right )^{2} \left (e + f x\right )\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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