Optimal. Leaf size=229 \[ \frac {b^2 \sin \left (4 a-\frac {4 b c}{d}\right ) \text {Ci}\left (\frac {4 b c}{d}+4 b x\right )}{d^3}-\frac {b^2 \sin \left (2 a-\frac {2 b c}{d}\right ) \text {Ci}\left (\frac {2 b c}{d}+2 b x\right )}{2 d^3}-\frac {b^2 \cos \left (2 a-\frac {2 b c}{d}\right ) \text {Si}\left (\frac {2 b c}{d}+2 b x\right )}{2 d^3}+\frac {b^2 \cos \left (4 a-\frac {4 b c}{d}\right ) \text {Si}\left (\frac {4 b c}{d}+4 b x\right )}{d^3}-\frac {b \cos (2 a+2 b x)}{4 d^2 (c+d x)}+\frac {b \cos (4 a+4 b x)}{4 d^2 (c+d x)}-\frac {\sin (2 a+2 b x)}{8 d (c+d x)^2}+\frac {\sin (4 a+4 b x)}{16 d (c+d x)^2} \]
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Rubi [A] time = 0.34, antiderivative size = 229, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {4406, 3297, 3303, 3299, 3302} \[ \frac {b^2 \sin \left (4 a-\frac {4 b c}{d}\right ) \text {CosIntegral}\left (\frac {4 b c}{d}+4 b x\right )}{d^3}-\frac {b^2 \sin \left (2 a-\frac {2 b c}{d}\right ) \text {CosIntegral}\left (\frac {2 b c}{d}+2 b x\right )}{2 d^3}-\frac {b^2 \cos \left (2 a-\frac {2 b c}{d}\right ) \text {Si}\left (\frac {2 b c}{d}+2 b x\right )}{2 d^3}+\frac {b^2 \cos \left (4 a-\frac {4 b c}{d}\right ) \text {Si}\left (\frac {4 b c}{d}+4 b x\right )}{d^3}-\frac {b \cos (2 a+2 b x)}{4 d^2 (c+d x)}+\frac {b \cos (4 a+4 b x)}{4 d^2 (c+d x)}-\frac {\sin (2 a+2 b x)}{8 d (c+d x)^2}+\frac {\sin (4 a+4 b x)}{16 d (c+d x)^2} \]
Antiderivative was successfully verified.
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Rule 3297
Rule 3299
Rule 3302
Rule 3303
Rule 4406
Rubi steps
\begin {align*} \int \frac {\cos (a+b x) \sin ^3(a+b x)}{(c+d x)^3} \, dx &=\int \left (\frac {\sin (2 a+2 b x)}{4 (c+d x)^3}-\frac {\sin (4 a+4 b x)}{8 (c+d x)^3}\right ) \, dx\\ &=-\left (\frac {1}{8} \int \frac {\sin (4 a+4 b x)}{(c+d x)^3} \, dx\right )+\frac {1}{4} \int \frac {\sin (2 a+2 b x)}{(c+d x)^3} \, dx\\ &=-\frac {\sin (2 a+2 b x)}{8 d (c+d x)^2}+\frac {\sin (4 a+4 b x)}{16 d (c+d x)^2}+\frac {b \int \frac {\cos (2 a+2 b x)}{(c+d x)^2} \, dx}{4 d}-\frac {b \int \frac {\cos (4 a+4 b x)}{(c+d x)^2} \, dx}{4 d}\\ &=-\frac {b \cos (2 a+2 b x)}{4 d^2 (c+d x)}+\frac {b \cos (4 a+4 b x)}{4 d^2 (c+d x)}-\frac {\sin (2 a+2 b x)}{8 d (c+d x)^2}+\frac {\sin (4 a+4 b x)}{16 d (c+d x)^2}-\frac {b^2 \int \frac {\sin (2 a+2 b x)}{c+d x} \, dx}{2 d^2}+\frac {b^2 \int \frac {\sin (4 a+4 b x)}{c+d x} \, dx}{d^2}\\ &=-\frac {b \cos (2 a+2 b x)}{4 d^2 (c+d x)}+\frac {b \cos (4 a+4 b x)}{4 d^2 (c+d x)}-\frac {\sin (2 a+2 b x)}{8 d (c+d x)^2}+\frac {\sin (4 a+4 b x)}{16 d (c+d x)^2}+\frac {\left (b^2 \cos \left (4 a-\frac {4 b c}{d}\right )\right ) \int \frac {\sin \left (\frac {4 b c}{d}+4 b x\right )}{c+d x} \, dx}{d^2}-\frac {\left (b^2 \cos \left (2 a-\frac {2 b c}{d}\right )\right ) \int \frac {\sin \left (\frac {2 b c}{d}+2 b x\right )}{c+d x} \, dx}{2 d^2}+\frac {\left (b^2 \sin \left (4 a-\frac {4 b c}{d}\right )\right ) \int \frac {\cos \left (\frac {4 b c}{d}+4 b x\right )}{c+d x} \, dx}{d^2}-\frac {\left (b^2 \sin \left (2 a-\frac {2 b c}{d}\right )\right ) \int \frac {\cos \left (\frac {2 b c}{d}+2 b x\right )}{c+d x} \, dx}{2 d^2}\\ &=-\frac {b \cos (2 a+2 b x)}{4 d^2 (c+d x)}+\frac {b \cos (4 a+4 b x)}{4 d^2 (c+d x)}+\frac {b^2 \text {Ci}\left (\frac {4 b c}{d}+4 b x\right ) \sin \left (4 a-\frac {4 b c}{d}\right )}{d^3}-\frac {b^2 \text {Ci}\left (\frac {2 b c}{d}+2 b x\right ) \sin \left (2 a-\frac {2 b c}{d}\right )}{2 d^3}-\frac {\sin (2 a+2 b x)}{8 d (c+d x)^2}+\frac {\sin (4 a+4 b x)}{16 d (c+d x)^2}-\frac {b^2 \cos \left (2 a-\frac {2 b c}{d}\right ) \text {Si}\left (\frac {2 b c}{d}+2 b x\right )}{2 d^3}+\frac {b^2 \cos \left (4 a-\frac {4 b c}{d}\right ) \text {Si}\left (\frac {4 b c}{d}+4 b x\right )}{d^3}\\ \end {align*}
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Mathematica [A] time = 2.76, size = 199, normalized size = 0.87 \[ \frac {-2 \left (4 b^2 \sin \left (2 a-\frac {2 b c}{d}\right ) \text {Ci}\left (\frac {2 b (c+d x)}{d}\right )+4 b^2 \cos \left (2 a-\frac {2 b c}{d}\right ) \text {Si}\left (\frac {2 b (c+d x)}{d}\right )+\frac {d (2 b (c+d x) \cos (2 (a+b x))+d \sin (2 (a+b x)))}{(c+d x)^2}\right )+16 b^2 \sin \left (4 a-\frac {4 b c}{d}\right ) \text {Ci}\left (\frac {4 b (c+d x)}{d}\right )+16 b^2 \cos \left (4 a-\frac {4 b c}{d}\right ) \text {Si}\left (\frac {4 b (c+d x)}{d}\right )+\frac {d (4 b (c+d x) \cos (4 (a+b x))+d \sin (4 (a+b x)))}{(c+d x)^2}}{16 d^3} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.65, size = 423, normalized size = 1.85 \[ \frac {8 \, {\left (b d^{2} x + b c d\right )} \cos \left (b x + a\right )^{4} + 2 \, b d^{2} x + 2 \, b c d - 10 \, {\left (b d^{2} x + b c d\right )} \cos \left (b x + a\right )^{2} + 4 \, {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \cos \left (-\frac {4 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {Si}\left (\frac {4 \, {\left (b d x + b c\right )}}{d}\right ) - 2 \, {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {Si}\left (\frac {2 \, {\left (b d x + b c\right )}}{d}\right ) + 2 \, {\left (d^{2} \cos \left (b x + a\right )^{3} - d^{2} \cos \left (b x + a\right )\right )} \sin \left (b x + a\right ) - {\left ({\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \operatorname {Ci}\left (\frac {2 \, {\left (b d x + b c\right )}}{d}\right ) + {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \operatorname {Ci}\left (-\frac {2 \, {\left (b d x + b c\right )}}{d}\right )\right )} \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) + 2 \, {\left ({\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \operatorname {Ci}\left (\frac {4 \, {\left (b d x + b c\right )}}{d}\right ) + {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \operatorname {Ci}\left (-\frac {4 \, {\left (b d x + b c\right )}}{d}\right )\right )} \sin \left (-\frac {4 \, {\left (b c - a d\right )}}{d}\right )}{4 \, {\left (d^{5} x^{2} + 2 \, c d^{4} x + c^{2} d^{3}\right )}} \]
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 [A] time = 0.02, size = 329, normalized size = 1.44 \[ \frac {-\frac {b^{3} \left (-\frac {2 \sin \left (4 b x +4 a \right )}{\left (\left (b x +a \right ) d -d a +c b \right )^{2} d}+\frac {-\frac {8 \cos \left (4 b x +4 a \right )}{\left (\left (b x +a \right ) d -d a +c b \right ) d}-\frac {8 \left (\frac {4 \Si \left (4 b x +4 a +\frac {-4 d a +4 c b}{d}\right ) \cos \left (\frac {-4 d a +4 c b}{d}\right )}{d}-\frac {4 \Ci \left (4 b x +4 a +\frac {-4 d a +4 c b}{d}\right ) \sin \left (\frac {-4 d a +4 c b}{d}\right )}{d}\right )}{d}}{d}\right )}{32}+\frac {b^{3} \left (-\frac {\sin \left (2 b x +2 a \right )}{\left (\left (b x +a \right ) d -d a +c b \right )^{2} d}+\frac {-\frac {2 \cos \left (2 b x +2 a \right )}{\left (\left (b x +a \right ) d -d a +c b \right ) d}-\frac {2 \left (\frac {2 \Si \left (2 b x +2 a +\frac {-2 d a +2 c b}{d}\right ) \cos \left (\frac {-2 d a +2 c b}{d}\right )}{d}-\frac {2 \Ci \left (2 b x +2 a +\frac {-2 d a +2 c b}{d}\right ) \sin \left (\frac {-2 d a +2 c b}{d}\right )}{d}\right )}{d}}{d}\right )}{8}}{b} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [C] time = 0.68, size = 336, normalized size = 1.47 \[ \frac {b^{3} {\left (-2 i \, E_{3}\left (\frac {2 i \, b c + 2 i \, {\left (b x + a\right )} d - 2 i \, a d}{d}\right ) + 2 i \, E_{3}\left (-\frac {2 i \, b c + 2 i \, {\left (b x + a\right )} d - 2 i \, a d}{d}\right )\right )} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) + b^{3} {\left (i \, E_{3}\left (\frac {4 i \, b c + 4 i \, {\left (b x + a\right )} d - 4 i \, a d}{d}\right ) - i \, E_{3}\left (-\frac {4 i \, b c + 4 i \, {\left (b x + a\right )} d - 4 i \, a d}{d}\right )\right )} \cos \left (-\frac {4 \, {\left (b c - a d\right )}}{d}\right ) - 2 \, b^{3} {\left (E_{3}\left (\frac {2 i \, b c + 2 i \, {\left (b x + a\right )} d - 2 i \, a d}{d}\right ) + E_{3}\left (-\frac {2 i \, b c + 2 i \, {\left (b x + a\right )} d - 2 i \, a d}{d}\right )\right )} \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) + b^{3} {\left (E_{3}\left (\frac {4 i \, b c + 4 i \, {\left (b x + a\right )} d - 4 i \, a d}{d}\right ) + E_{3}\left (-\frac {4 i \, b c + 4 i \, {\left (b x + a\right )} d - 4 i \, a d}{d}\right )\right )} \sin \left (-\frac {4 \, {\left (b c - a d\right )}}{d}\right )}{16 \, {\left (b^{2} c^{2} d - 2 \, a b c d^{2} + {\left (b x + a\right )}^{2} d^{3} + a^{2} d^{3} + 2 \, {\left (b c d^{2} - a d^{3}\right )} {\left (b x + a\right )}\right )} b} \]
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 \[ \int \frac {\cos \left (a+b\,x\right )\,{\sin \left (a+b\,x\right )}^3}{{\left (c+d\,x\right )}^3} \,d x \]
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 \[ \int \frac {\sin ^{3}{\left (a + b x \right )} \cos {\left (a + b x \right )}}{\left (c + d x\right )^{3}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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