Optimal. Leaf size=194 \[ \frac {3 \cos \left (\frac {b}{d}\right ) (b c-a d) \text {Ci}\left (\frac {b c-a d}{d (c+d x)}\right )}{4 d^2}-\frac {3 \cos \left (\frac {3 b}{d}\right ) (b c-a d) \text {Ci}\left (\frac {3 (b c-a d)}{d (c+d x)}\right )}{4 d^2}+\frac {3 \sin \left (\frac {b}{d}\right ) (b c-a d) \text {Si}\left (\frac {b c-a d}{d (c+d x)}\right )}{4 d^2}-\frac {3 \sin \left (\frac {3 b}{d}\right ) (b c-a d) \text {Si}\left (\frac {3 (b c-a d)}{d (c+d x)}\right )}{4 d^2}+\frac {(c+d x) \sin ^3\left (\frac {a+b x}{c+d x}\right )}{d} \]
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Rubi [A] time = 0.32, antiderivative size = 194, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 5, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {4563, 3313, 3303, 3299, 3302} \[ \frac {3 \cos \left (\frac {b}{d}\right ) (b c-a d) \text {CosIntegral}\left (\frac {b c-a d}{d (c+d x)}\right )}{4 d^2}-\frac {3 \cos \left (\frac {3 b}{d}\right ) (b c-a d) \text {CosIntegral}\left (\frac {3 (b c-a d)}{d (c+d x)}\right )}{4 d^2}+\frac {3 \sin \left (\frac {b}{d}\right ) (b c-a d) \text {Si}\left (\frac {b c-a d}{d (c+d x)}\right )}{4 d^2}-\frac {3 \sin \left (\frac {3 b}{d}\right ) (b c-a d) \text {Si}\left (\frac {3 (b c-a d)}{d (c+d x)}\right )}{4 d^2}+\frac {(c+d x) \sin ^3\left (\frac {a+b x}{c+d x}\right )}{d} \]
Antiderivative was successfully verified.
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Rule 3299
Rule 3302
Rule 3303
Rule 3313
Rule 4563
Rubi steps
\begin {align*} \int \sin ^3\left (\frac {a+b x}{c+d x}\right ) \, dx &=-\frac {\operatorname {Subst}\left (\int \frac {\sin ^3\left (\frac {b}{d}-\frac {(b c-a d) x}{d}\right )}{x^2} \, dx,x,\frac {1}{c+d x}\right )}{d}\\ &=\frac {(c+d x) \sin ^3\left (\frac {a+b x}{c+d x}\right )}{d}+\frac {(3 (b c-a d)) \operatorname {Subst}\left (\int \left (-\frac {\cos \left (\frac {3 b}{d}-\frac {3 (b c-a d) x}{d}\right )}{4 x}+\frac {\cos \left (\frac {b}{d}-\frac {(b c-a d) x}{d}\right )}{4 x}\right ) \, dx,x,\frac {1}{c+d x}\right )}{d^2}\\ &=\frac {(c+d x) \sin ^3\left (\frac {a+b x}{c+d x}\right )}{d}-\frac {(3 (b c-a d)) \operatorname {Subst}\left (\int \frac {\cos \left (\frac {3 b}{d}-\frac {3 (b c-a d) x}{d}\right )}{x} \, dx,x,\frac {1}{c+d x}\right )}{4 d^2}+\frac {(3 (b c-a d)) \operatorname {Subst}\left (\int \frac {\cos \left (\frac {b}{d}-\frac {(b c-a d) x}{d}\right )}{x} \, dx,x,\frac {1}{c+d x}\right )}{4 d^2}\\ &=\frac {(c+d x) \sin ^3\left (\frac {a+b x}{c+d x}\right )}{d}+\frac {\left (3 (b c-a d) \cos \left (\frac {b}{d}\right )\right ) \operatorname {Subst}\left (\int \frac {\cos \left (\frac {(b c-a d) x}{d}\right )}{x} \, dx,x,\frac {1}{c+d x}\right )}{4 d^2}-\frac {\left (3 (b c-a d) \cos \left (\frac {3 b}{d}\right )\right ) \operatorname {Subst}\left (\int \frac {\cos \left (\frac {3 (b c-a d) x}{d}\right )}{x} \, dx,x,\frac {1}{c+d x}\right )}{4 d^2}+\frac {\left (3 (b c-a d) \sin \left (\frac {b}{d}\right )\right ) \operatorname {Subst}\left (\int \frac {\sin \left (\frac {(b c-a d) x}{d}\right )}{x} \, dx,x,\frac {1}{c+d x}\right )}{4 d^2}-\frac {\left (3 (b c-a d) \sin \left (\frac {3 b}{d}\right )\right ) \operatorname {Subst}\left (\int \frac {\sin \left (\frac {3 (b c-a d) x}{d}\right )}{x} \, dx,x,\frac {1}{c+d x}\right )}{4 d^2}\\ &=\frac {3 (b c-a d) \cos \left (\frac {b}{d}\right ) \text {Ci}\left (\frac {b c-a d}{d (c+d x)}\right )}{4 d^2}-\frac {3 (b c-a d) \cos \left (\frac {3 b}{d}\right ) \text {Ci}\left (\frac {3 (b c-a d)}{d (c+d x)}\right )}{4 d^2}+\frac {(c+d x) \sin ^3\left (\frac {a+b x}{c+d x}\right )}{d}+\frac {3 (b c-a d) \sin \left (\frac {b}{d}\right ) \text {Si}\left (\frac {b c-a d}{d (c+d x)}\right )}{4 d^2}-\frac {3 (b c-a d) \sin \left (\frac {3 b}{d}\right ) \text {Si}\left (\frac {3 (b c-a d)}{d (c+d x)}\right )}{4 d^2}\\ \end {align*}
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Mathematica [C] time = 7.71, size = 657, normalized size = 3.39 \[ -\frac {3 \left (a c d-b c^2\right ) \left (\frac {i \left (1+e^{\frac {2 i b}{d}}\right ) \left (e^{\frac {2 i b c}{d (c+d x)}}-e^{\frac {2 i a}{c+d x}}\right ) \exp \left (-\frac {i (a d+2 b c+b d x)}{d (c+d x)}\right )}{4 (b c-a d)}-\frac {i \left (-1+e^{\frac {2 i b}{d}}\right ) \left (e^{\frac {2 i a}{c+d x}}+e^{\frac {2 i b c}{d (c+d x)}}\right ) \exp \left (-\frac {i (a d+2 b c+b d x)}{d (c+d x)}\right )}{4 (b c-a d)}\right )}{4 d}+\frac {3 \left (a c d-b c^2\right ) \left (\frac {i \left (1+e^{\frac {6 i b}{d}}\right ) \left (e^{\frac {6 i b c}{d (c+d x)}}-e^{\frac {6 i a}{c+d x}}\right ) \exp \left (-\frac {3 i (a d+2 b c+b d x)}{d (c+d x)}\right )}{12 (b c-a d)}-\frac {i \left (-1+e^{\frac {6 i b}{d}}\right ) \left (e^{\frac {6 i a}{c+d x}}+e^{\frac {6 i b c}{d (c+d x)}}\right ) \exp \left (-\frac {3 i (a d+2 b c+b d x)}{d (c+d x)}\right )}{12 (b c-a d)}\right )}{4 d}+\frac {3 (a d-b c) \left (-\cos \left (\frac {b}{d}\right ) \text {Ci}\left (\frac {a d-b c}{d (c+d x)}\right )+\cos \left (\frac {3 b}{d}\right ) \text {Ci}\left (\frac {3 (a d-b c)}{d (c+d x)}\right )+\sin \left (\frac {b}{d}\right ) \text {Si}\left (\frac {a d-b c}{d (c+d x)}\right )-\sin \left (\frac {3 b}{d}\right ) \text {Si}\left (\frac {3 (a d-b c)}{d (c+d x)}\right )\right )}{4 d^2}+\frac {3}{4} x \sin \left (\frac {b}{d}\right ) \cos \left (\frac {a d-b c}{d (c+d x)}\right )-\frac {1}{4} x \sin \left (\frac {3 b}{d}\right ) \cos \left (\frac {3 (a d-b c)}{d (c+d x)}\right )+\frac {3}{4} x \cos \left (\frac {b}{d}\right ) \sin \left (\frac {a d-b c}{d (c+d x)}\right )-\frac {1}{4} x \cos \left (\frac {3 b}{d}\right ) \sin \left (\frac {3 (a d-b c)}{d (c+d x)}\right ) \]
Warning: Unable to verify antiderivative.
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fricas [A] time = 1.14, size = 277, normalized size = 1.43 \[ -\frac {6 \, {\left (b c - a d\right )} \sin \left (\frac {b}{d}\right ) \operatorname {Si}\left (-\frac {b c - a d}{d^{2} x + c d}\right ) - 6 \, {\left (b c - a d\right )} \sin \left (\frac {3 \, b}{d}\right ) \operatorname {Si}\left (-\frac {3 \, {\left (b c - a d\right )}}{d^{2} x + c d}\right ) + 3 \, {\left ({\left (b c - a d\right )} \operatorname {Ci}\left (\frac {3 \, {\left (b c - a d\right )}}{d^{2} x + c d}\right ) + {\left (b c - a d\right )} \operatorname {Ci}\left (-\frac {3 \, {\left (b c - a d\right )}}{d^{2} x + c d}\right )\right )} \cos \left (\frac {3 \, b}{d}\right ) - 3 \, {\left ({\left (b c - a d\right )} \operatorname {Ci}\left (\frac {b c - a d}{d^{2} x + c d}\right ) + {\left (b c - a d\right )} \operatorname {Ci}\left (-\frac {b c - a d}{d^{2} x + c d}\right )\right )} \cos \left (\frac {b}{d}\right ) - 8 \, {\left (d^{2} x - {\left (d^{2} x + c d\right )} \cos \left (\frac {b x + a}{d x + c}\right )^{2} + c d\right )} \sin \left (\frac {b x + a}{d x + c}\right )}{8 \, d^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 177.53, size = 1239, normalized size = 6.39 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.07, size = 295, normalized size = 1.52 \[ -\frac {\left (d a -c b \right ) \left (-\frac {d^{2} \left (-\frac {3 \sin \left (\frac {3 d a -3 c b}{d \left (d x +c \right )}+\frac {3 b}{d}\right )}{\left (\left (\frac {b}{d}+\frac {d a -c b}{d \left (d x +c \right )}\right ) d -b \right ) d}+\frac {-\frac {9 \Si \left (\frac {3 d a -3 c b}{d \left (d x +c \right )}\right ) \sin \left (\frac {3 b}{d}\right )}{d}+\frac {9 \Ci \left (\frac {3 d a -3 c b}{d \left (d x +c \right )}\right ) \cos \left (\frac {3 b}{d}\right )}{d}}{d}\right )}{12}+\frac {3 d^{2} \left (-\frac {\sin \left (\frac {b}{d}+\frac {d a -c b}{d \left (d x +c \right )}\right )}{\left (\left (\frac {b}{d}+\frac {d a -c b}{d \left (d x +c \right )}\right ) d -b \right ) d}+\frac {-\frac {\Si \left (\frac {d a -c b}{d \left (d x +c \right )}\right ) \sin \left (\frac {b}{d}\right )}{d}+\frac {\Ci \left (\frac {d a -c b}{d \left (d x +c \right )}\right ) \cos \left (\frac {b}{d}\right )}{d}}{d}\right )}{4}\right )}{d^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \sin \left (\frac {b x + a}{d x + c}\right )^{3}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int {\sin \left (\frac {a+b\,x}{c+d\,x}\right )}^3 \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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