Optimal. Leaf size=265 \[ \frac {a^3 d^2 \sin \left (c-\frac {a d}{b}\right ) \text {Ci}\left (x d+\frac {a d}{b}\right )}{2 b^6}+\frac {a^3 d^2 \cos \left (c-\frac {a d}{b}\right ) \text {Si}\left (x d+\frac {a d}{b}\right )}{2 b^6}+\frac {a^3 d \cos (c+d x)}{2 b^5 (a+b x)}+\frac {a^3 \sin (c+d x)}{2 b^4 (a+b x)^2}+\frac {3 a^2 d \cos \left (c-\frac {a d}{b}\right ) \text {Ci}\left (x d+\frac {a d}{b}\right )}{b^5}-\frac {3 a^2 d \sin \left (c-\frac {a d}{b}\right ) \text {Si}\left (x d+\frac {a d}{b}\right )}{b^5}-\frac {3 a^2 \sin (c+d x)}{b^4 (a+b x)}-\frac {3 a \sin \left (c-\frac {a d}{b}\right ) \text {Ci}\left (x d+\frac {a d}{b}\right )}{b^4}-\frac {3 a \cos \left (c-\frac {a d}{b}\right ) \text {Si}\left (x d+\frac {a d}{b}\right )}{b^4}-\frac {\cos (c+d x)}{b^3 d} \]
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Rubi [A] time = 0.61, antiderivative size = 265, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 6, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {6742, 2638, 3297, 3303, 3299, 3302} \[ \frac {a^3 d^2 \sin \left (c-\frac {a d}{b}\right ) \text {CosIntegral}\left (\frac {a d}{b}+d x\right )}{2 b^6}+\frac {3 a^2 d \cos \left (c-\frac {a d}{b}\right ) \text {CosIntegral}\left (\frac {a d}{b}+d x\right )}{b^5}+\frac {a^3 d^2 \cos \left (c-\frac {a d}{b}\right ) \text {Si}\left (x d+\frac {a d}{b}\right )}{2 b^6}-\frac {3 a^2 d \sin \left (c-\frac {a d}{b}\right ) \text {Si}\left (x d+\frac {a d}{b}\right )}{b^5}+\frac {a^3 \sin (c+d x)}{2 b^4 (a+b x)^2}-\frac {3 a^2 \sin (c+d x)}{b^4 (a+b x)}+\frac {a^3 d \cos (c+d x)}{2 b^5 (a+b x)}-\frac {3 a \sin \left (c-\frac {a d}{b}\right ) \text {CosIntegral}\left (\frac {a d}{b}+d x\right )}{b^4}-\frac {3 a \cos \left (c-\frac {a d}{b}\right ) \text {Si}\left (x d+\frac {a d}{b}\right )}{b^4}-\frac {\cos (c+d x)}{b^3 d} \]
Antiderivative was successfully verified.
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Rule 2638
Rule 3297
Rule 3299
Rule 3302
Rule 3303
Rule 6742
Rubi steps
\begin {align*} \int \frac {x^3 \sin (c+d x)}{(a+b x)^3} \, dx &=\int \left (\frac {\sin (c+d x)}{b^3}-\frac {a^3 \sin (c+d x)}{b^3 (a+b x)^3}+\frac {3 a^2 \sin (c+d x)}{b^3 (a+b x)^2}-\frac {3 a \sin (c+d x)}{b^3 (a+b x)}\right ) \, dx\\ &=\frac {\int \sin (c+d x) \, dx}{b^3}-\frac {(3 a) \int \frac {\sin (c+d x)}{a+b x} \, dx}{b^3}+\frac {\left (3 a^2\right ) \int \frac {\sin (c+d x)}{(a+b x)^2} \, dx}{b^3}-\frac {a^3 \int \frac {\sin (c+d x)}{(a+b x)^3} \, dx}{b^3}\\ &=-\frac {\cos (c+d x)}{b^3 d}+\frac {a^3 \sin (c+d x)}{2 b^4 (a+b x)^2}-\frac {3 a^2 \sin (c+d x)}{b^4 (a+b x)}+\frac {\left (3 a^2 d\right ) \int \frac {\cos (c+d x)}{a+b x} \, dx}{b^4}-\frac {\left (a^3 d\right ) \int \frac {\cos (c+d x)}{(a+b x)^2} \, dx}{2 b^4}-\frac {\left (3 a \cos \left (c-\frac {a d}{b}\right )\right ) \int \frac {\sin \left (\frac {a d}{b}+d x\right )}{a+b x} \, dx}{b^3}-\frac {\left (3 a \sin \left (c-\frac {a d}{b}\right )\right ) \int \frac {\cos \left (\frac {a d}{b}+d x\right )}{a+b x} \, dx}{b^3}\\ &=-\frac {\cos (c+d x)}{b^3 d}+\frac {a^3 d \cos (c+d x)}{2 b^5 (a+b x)}-\frac {3 a \text {Ci}\left (\frac {a d}{b}+d x\right ) \sin \left (c-\frac {a d}{b}\right )}{b^4}+\frac {a^3 \sin (c+d x)}{2 b^4 (a+b x)^2}-\frac {3 a^2 \sin (c+d x)}{b^4 (a+b x)}-\frac {3 a \cos \left (c-\frac {a d}{b}\right ) \text {Si}\left (\frac {a d}{b}+d x\right )}{b^4}+\frac {\left (a^3 d^2\right ) \int \frac {\sin (c+d x)}{a+b x} \, dx}{2 b^5}+\frac {\left (3 a^2 d \cos \left (c-\frac {a d}{b}\right )\right ) \int \frac {\cos \left (\frac {a d}{b}+d x\right )}{a+b x} \, dx}{b^4}-\frac {\left (3 a^2 d \sin \left (c-\frac {a d}{b}\right )\right ) \int \frac {\sin \left (\frac {a d}{b}+d x\right )}{a+b x} \, dx}{b^4}\\ &=-\frac {\cos (c+d x)}{b^3 d}+\frac {a^3 d \cos (c+d x)}{2 b^5 (a+b x)}+\frac {3 a^2 d \cos \left (c-\frac {a d}{b}\right ) \text {Ci}\left (\frac {a d}{b}+d x\right )}{b^5}-\frac {3 a \text {Ci}\left (\frac {a d}{b}+d x\right ) \sin \left (c-\frac {a d}{b}\right )}{b^4}+\frac {a^3 \sin (c+d x)}{2 b^4 (a+b x)^2}-\frac {3 a^2 \sin (c+d x)}{b^4 (a+b x)}-\frac {3 a \cos \left (c-\frac {a d}{b}\right ) \text {Si}\left (\frac {a d}{b}+d x\right )}{b^4}-\frac {3 a^2 d \sin \left (c-\frac {a d}{b}\right ) \text {Si}\left (\frac {a d}{b}+d x\right )}{b^5}+\frac {\left (a^3 d^2 \cos \left (c-\frac {a d}{b}\right )\right ) \int \frac {\sin \left (\frac {a d}{b}+d x\right )}{a+b x} \, dx}{2 b^5}+\frac {\left (a^3 d^2 \sin \left (c-\frac {a d}{b}\right )\right ) \int \frac {\cos \left (\frac {a d}{b}+d x\right )}{a+b x} \, dx}{2 b^5}\\ &=-\frac {\cos (c+d x)}{b^3 d}+\frac {a^3 d \cos (c+d x)}{2 b^5 (a+b x)}+\frac {3 a^2 d \cos \left (c-\frac {a d}{b}\right ) \text {Ci}\left (\frac {a d}{b}+d x\right )}{b^5}-\frac {3 a \text {Ci}\left (\frac {a d}{b}+d x\right ) \sin \left (c-\frac {a d}{b}\right )}{b^4}+\frac {a^3 d^2 \text {Ci}\left (\frac {a d}{b}+d x\right ) \sin \left (c-\frac {a d}{b}\right )}{2 b^6}+\frac {a^3 \sin (c+d x)}{2 b^4 (a+b x)^2}-\frac {3 a^2 \sin (c+d x)}{b^4 (a+b x)}-\frac {3 a \cos \left (c-\frac {a d}{b}\right ) \text {Si}\left (\frac {a d}{b}+d x\right )}{b^4}+\frac {a^3 d^2 \cos \left (c-\frac {a d}{b}\right ) \text {Si}\left (\frac {a d}{b}+d x\right )}{2 b^6}-\frac {3 a^2 d \sin \left (c-\frac {a d}{b}\right ) \text {Si}\left (\frac {a d}{b}+d x\right )}{b^5}\\ \end {align*}
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Mathematica [A] time = 1.12, size = 235, normalized size = 0.89 \[ -\frac {-a d (a+b x)^2 \left (\text {Ci}\left (d \left (\frac {a}{b}+x\right )\right ) \left (\left (a^2 d^2-6 b^2\right ) \sin \left (c-\frac {a d}{b}\right )+6 a b d \cos \left (c-\frac {a d}{b}\right )\right )+\text {Si}\left (d \left (\frac {a}{b}+x\right )\right ) \left (\left (a^2 d^2-6 b^2\right ) \cos \left (c-\frac {a d}{b}\right )-6 a b d \sin \left (c-\frac {a d}{b}\right )\right )\right )+b \cos (d x) \left (a^2 b d \sin (c) (5 a+6 b x)-\cos (c) (a+b x) \left (a^3 d^2-2 a b^2-2 b^3 x\right )\right )+b \sin (d x) \left (\sin (c) (a+b x) \left (a^3 d^2-2 a b^2-2 b^3 x\right )+a^2 b d \cos (c) (5 a+6 b x)\right )}{2 b^6 d (a+b x)^2} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.70, size = 515, normalized size = 1.94 \[ \frac {2 \, {\left (a^{4} b d^{2} - 2 \, b^{5} x^{2} - 2 \, a^{2} b^{3} + {\left (a^{3} b^{2} d^{2} - 4 \, a b^{4}\right )} x\right )} \cos \left (d x + c\right ) + 2 \, {\left (3 \, {\left (a^{2} b^{3} d^{2} x^{2} + 2 \, a^{3} b^{2} d^{2} x + a^{4} b d^{2}\right )} \operatorname {Ci}\left (\frac {b d x + a d}{b}\right ) + 3 \, {\left (a^{2} b^{3} d^{2} x^{2} + 2 \, a^{3} b^{2} d^{2} x + a^{4} b d^{2}\right )} \operatorname {Ci}\left (-\frac {b d x + a d}{b}\right ) + {\left (a^{5} d^{3} - 6 \, a^{3} b^{2} d + {\left (a^{3} b^{2} d^{3} - 6 \, a b^{4} d\right )} x^{2} + 2 \, {\left (a^{4} b d^{3} - 6 \, a^{2} b^{3} d\right )} x\right )} \operatorname {Si}\left (\frac {b d x + a d}{b}\right )\right )} \cos \left (-\frac {b c - a d}{b}\right ) - 2 \, {\left (6 \, a^{2} b^{3} d x + 5 \, a^{3} b^{2} d\right )} \sin \left (d x + c\right ) - {\left ({\left (a^{5} d^{3} - 6 \, a^{3} b^{2} d + {\left (a^{3} b^{2} d^{3} - 6 \, a b^{4} d\right )} x^{2} + 2 \, {\left (a^{4} b d^{3} - 6 \, a^{2} b^{3} d\right )} x\right )} \operatorname {Ci}\left (\frac {b d x + a d}{b}\right ) + {\left (a^{5} d^{3} - 6 \, a^{3} b^{2} d + {\left (a^{3} b^{2} d^{3} - 6 \, a b^{4} d\right )} x^{2} + 2 \, {\left (a^{4} b d^{3} - 6 \, a^{2} b^{3} d\right )} x\right )} \operatorname {Ci}\left (-\frac {b d x + a d}{b}\right ) - 12 \, {\left (a^{2} b^{3} d^{2} x^{2} + 2 \, a^{3} b^{2} d^{2} x + a^{4} b d^{2}\right )} \operatorname {Si}\left (\frac {b d x + a d}{b}\right )\right )} \sin \left (-\frac {b c - a d}{b}\right )}{4 \, {\left (b^{8} d x^{2} + 2 \, a b^{7} d x + a^{2} b^{6} d\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [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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maple [B] time = 0.04, size = 1208, normalized size = 4.56 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [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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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^3\,\sin \left (c+d\,x\right )}{{\left (a+b\,x\right )}^3} \,d x \]
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 \[ \int \frac {x^{3} \sin {\left (c + d x \right )}}{\left (a + b x\right )^{3}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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