Optimal. Leaf size=287 \[ -\frac {b^3 \cos \left (2 a-\frac {2 b c}{d}\right ) \text {Ci}\left (\frac {2 b c}{d}+2 b x\right )}{3 d^4}-\frac {4 b^3 \cos \left (4 a-\frac {4 b c}{d}\right ) \text {Ci}\left (\frac {4 b c}{d}+4 b x\right )}{3 d^4}+\frac {b^3 \sin \left (2 a-\frac {2 b c}{d}\right ) \text {Si}\left (\frac {2 b c}{d}+2 b x\right )}{3 d^4}+\frac {4 b^3 \sin \left (4 a-\frac {4 b c}{d}\right ) \text {Si}\left (\frac {4 b c}{d}+4 b x\right )}{3 d^4}+\frac {b^2 \sin (2 a+2 b x)}{6 d^3 (c+d x)}+\frac {b^2 \sin (4 a+4 b x)}{3 d^3 (c+d x)}-\frac {b \cos (2 a+2 b x)}{12 d^2 (c+d x)^2}-\frac {b \cos (4 a+4 b x)}{12 d^2 (c+d x)^2}-\frac {\sin (2 a+2 b x)}{12 d (c+d x)^3}-\frac {\sin (4 a+4 b x)}{24 d (c+d x)^3} \]
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Rubi [A] time = 0.45, antiderivative size = 287, normalized size of antiderivative = 1.00, number of steps used = 14, 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} \[ -\frac {b^3 \cos \left (2 a-\frac {2 b c}{d}\right ) \text {CosIntegral}\left (\frac {2 b c}{d}+2 b x\right )}{3 d^4}-\frac {4 b^3 \cos \left (4 a-\frac {4 b c}{d}\right ) \text {CosIntegral}\left (\frac {4 b c}{d}+4 b x\right )}{3 d^4}+\frac {b^3 \sin \left (2 a-\frac {2 b c}{d}\right ) \text {Si}\left (\frac {2 b c}{d}+2 b x\right )}{3 d^4}+\frac {4 b^3 \sin \left (4 a-\frac {4 b c}{d}\right ) \text {Si}\left (\frac {4 b c}{d}+4 b x\right )}{3 d^4}+\frac {b^2 \sin (2 a+2 b x)}{6 d^3 (c+d x)}+\frac {b^2 \sin (4 a+4 b x)}{3 d^3 (c+d x)}-\frac {b \cos (2 a+2 b x)}{12 d^2 (c+d x)^2}-\frac {b \cos (4 a+4 b x)}{12 d^2 (c+d x)^2}-\frac {\sin (2 a+2 b x)}{12 d (c+d x)^3}-\frac {\sin (4 a+4 b x)}{24 d (c+d x)^3} \]
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 ^3(a+b x) \sin (a+b x)}{(c+d x)^4} \, dx &=\int \left (\frac {\sin (2 a+2 b x)}{4 (c+d x)^4}+\frac {\sin (4 a+4 b x)}{8 (c+d x)^4}\right ) \, dx\\ &=\frac {1}{8} \int \frac {\sin (4 a+4 b x)}{(c+d x)^4} \, dx+\frac {1}{4} \int \frac {\sin (2 a+2 b x)}{(c+d x)^4} \, dx\\ &=-\frac {\sin (2 a+2 b x)}{12 d (c+d x)^3}-\frac {\sin (4 a+4 b x)}{24 d (c+d x)^3}+\frac {b \int \frac {\cos (2 a+2 b x)}{(c+d x)^3} \, dx}{6 d}+\frac {b \int \frac {\cos (4 a+4 b x)}{(c+d x)^3} \, dx}{6 d}\\ &=-\frac {b \cos (2 a+2 b x)}{12 d^2 (c+d x)^2}-\frac {b \cos (4 a+4 b x)}{12 d^2 (c+d x)^2}-\frac {\sin (2 a+2 b x)}{12 d (c+d x)^3}-\frac {\sin (4 a+4 b x)}{24 d (c+d x)^3}-\frac {b^2 \int \frac {\sin (2 a+2 b x)}{(c+d x)^2} \, dx}{6 d^2}-\frac {b^2 \int \frac {\sin (4 a+4 b x)}{(c+d x)^2} \, dx}{3 d^2}\\ &=-\frac {b \cos (2 a+2 b x)}{12 d^2 (c+d x)^2}-\frac {b \cos (4 a+4 b x)}{12 d^2 (c+d x)^2}-\frac {\sin (2 a+2 b x)}{12 d (c+d x)^3}+\frac {b^2 \sin (2 a+2 b x)}{6 d^3 (c+d x)}-\frac {\sin (4 a+4 b x)}{24 d (c+d x)^3}+\frac {b^2 \sin (4 a+4 b x)}{3 d^3 (c+d x)}-\frac {b^3 \int \frac {\cos (2 a+2 b x)}{c+d x} \, dx}{3 d^3}-\frac {\left (4 b^3\right ) \int \frac {\cos (4 a+4 b x)}{c+d x} \, dx}{3 d^3}\\ &=-\frac {b \cos (2 a+2 b x)}{12 d^2 (c+d x)^2}-\frac {b \cos (4 a+4 b x)}{12 d^2 (c+d x)^2}-\frac {\sin (2 a+2 b x)}{12 d (c+d x)^3}+\frac {b^2 \sin (2 a+2 b x)}{6 d^3 (c+d x)}-\frac {\sin (4 a+4 b x)}{24 d (c+d x)^3}+\frac {b^2 \sin (4 a+4 b x)}{3 d^3 (c+d x)}-\frac {\left (4 b^3 \cos \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}{3 d^3}-\frac {\left (b^3 \cos \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}{3 d^3}+\frac {\left (4 b^3 \sin \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}{3 d^3}+\frac {\left (b^3 \sin \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}{3 d^3}\\ &=-\frac {b \cos (2 a+2 b x)}{12 d^2 (c+d x)^2}-\frac {b \cos (4 a+4 b x)}{12 d^2 (c+d x)^2}-\frac {b^3 \cos \left (2 a-\frac {2 b c}{d}\right ) \text {Ci}\left (\frac {2 b c}{d}+2 b x\right )}{3 d^4}-\frac {4 b^3 \cos \left (4 a-\frac {4 b c}{d}\right ) \text {Ci}\left (\frac {4 b c}{d}+4 b x\right )}{3 d^4}-\frac {\sin (2 a+2 b x)}{12 d (c+d x)^3}+\frac {b^2 \sin (2 a+2 b x)}{6 d^3 (c+d x)}-\frac {\sin (4 a+4 b x)}{24 d (c+d x)^3}+\frac {b^2 \sin (4 a+4 b x)}{3 d^3 (c+d x)}+\frac {b^3 \sin \left (2 a-\frac {2 b c}{d}\right ) \text {Si}\left (\frac {2 b c}{d}+2 b x\right )}{3 d^4}+\frac {4 b^3 \sin \left (4 a-\frac {4 b c}{d}\right ) \text {Si}\left (\frac {4 b c}{d}+4 b x\right )}{3 d^4}\\ \end {align*}
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Mathematica [A] time = 2.49, size = 316, normalized size = 1.10 \[ -\frac {8 b^3 (c+d x)^3 \left (\cos \left (2 a-\frac {2 b c}{d}\right ) \text {Ci}\left (\frac {2 b (c+d x)}{d}\right )-\sin \left (2 a-\frac {2 b c}{d}\right ) \text {Si}\left (\frac {2 b (c+d x)}{d}\right )\right )+32 b^3 (c+d x)^3 \left (\cos \left (4 a-\frac {4 b c}{d}\right ) \text {Ci}\left (\frac {4 b (c+d x)}{d}\right )-\sin \left (4 a-\frac {4 b c}{d}\right ) \text {Si}\left (\frac {4 b (c+d x)}{d}\right )\right )+2 d \cos (2 b x) \left (\sin (2 a) \left (d^2-2 b^2 (c+d x)^2\right )+b d \cos (2 a) (c+d x)\right )+d \cos (4 b x) \left (\sin (4 a) \left (d^2-8 b^2 (c+d x)^2\right )+2 b d \cos (4 a) (c+d x)\right )-2 d \sin (2 b x) \left (\cos (2 a) \left (2 b^2 (c+d x)^2-d^2\right )+b d \sin (2 a) (c+d x)\right )-d \sin (4 b x) \left (\cos (4 a) \left (8 b^2 (c+d x)^2-d^2\right )+2 b d \sin (4 a) (c+d x)\right )}{24 d^4 (c+d x)^3} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.70, size = 568, normalized size = 1.98 \[ -\frac {4 \, {\left (b d^{3} x + b c d^{2}\right )} \cos \left (b x + a\right )^{4} - 3 \, {\left (b d^{3} x + b c d^{2}\right )} \cos \left (b x + a\right )^{2} - 8 \, {\left (b^{3} d^{3} x^{3} + 3 \, b^{3} c d^{2} x^{2} + 3 \, b^{3} c^{2} d x + b^{3} c^{3}\right )} \sin \left (-\frac {4 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {Si}\left (\frac {4 \, {\left (b d x + b c\right )}}{d}\right ) - 2 \, {\left (b^{3} d^{3} x^{3} + 3 \, b^{3} c d^{2} x^{2} + 3 \, b^{3} c^{2} d x + b^{3} c^{3}\right )} \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {Si}\left (\frac {2 \, {\left (b d x + b c\right )}}{d}\right ) + {\left ({\left (b^{3} d^{3} x^{3} + 3 \, b^{3} c d^{2} x^{2} + 3 \, b^{3} c^{2} d x + b^{3} c^{3}\right )} \operatorname {Ci}\left (\frac {2 \, {\left (b d x + b c\right )}}{d}\right ) + {\left (b^{3} d^{3} x^{3} + 3 \, b^{3} c d^{2} x^{2} + 3 \, b^{3} c^{2} d x + b^{3} c^{3}\right )} \operatorname {Ci}\left (-\frac {2 \, {\left (b d x + b c\right )}}{d}\right )\right )} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) + 4 \, {\left ({\left (b^{3} d^{3} x^{3} + 3 \, b^{3} c d^{2} x^{2} + 3 \, b^{3} c^{2} d x + b^{3} c^{3}\right )} \operatorname {Ci}\left (\frac {4 \, {\left (b d x + b c\right )}}{d}\right ) + {\left (b^{3} d^{3} x^{3} + 3 \, b^{3} c d^{2} x^{2} + 3 \, b^{3} c^{2} d x + b^{3} c^{3}\right )} \operatorname {Ci}\left (-\frac {4 \, {\left (b d x + b c\right )}}{d}\right )\right )} \cos \left (-\frac {4 \, {\left (b c - a d\right )}}{d}\right ) - 2 \, {\left ({\left (8 \, b^{2} d^{3} x^{2} + 16 \, b^{2} c d^{2} x + 8 \, b^{2} c^{2} d - d^{3}\right )} \cos \left (b x + a\right )^{3} - 3 \, {\left (b^{2} d^{3} x^{2} + 2 \, b^{2} c d^{2} x + b^{2} c^{2} d\right )} \cos \left (b x + a\right )\right )} \sin \left (b x + a\right )}{6 \, {\left (d^{7} x^{3} + 3 \, c d^{6} x^{2} + 3 \, c^{2} d^{5} x + c^{3} d^{4}\right )}} \]
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 = 404, normalized size = 1.41 \[ \frac {\frac {b^{4} \left (-\frac {4 \sin \left (4 b x +4 a \right )}{3 \left (\left (b x +a \right ) d -d a +c b \right )^{3} d}+\frac {-\frac {8 \cos \left (4 b x +4 a \right )}{3 \left (\left (b x +a \right ) d -d a +c b \right )^{2} d}-\frac {8 \left (-\frac {4 \sin \left (4 b x +4 a \right )}{\left (\left (b x +a \right ) d -d a +c b \right ) d}+\frac {\frac {16 \Si \left (4 b x +4 a +\frac {-4 d a +4 c b}{d}\right ) \sin \left (\frac {-4 d a +4 c b}{d}\right )}{d}+\frac {16 \Ci \left (4 b x +4 a +\frac {-4 d a +4 c b}{d}\right ) \cos \left (\frac {-4 d a +4 c b}{d}\right )}{d}}{d}\right )}{3 d}}{d}\right )}{32}+\frac {b^{4} \left (-\frac {2 \sin \left (2 b x +2 a \right )}{3 \left (\left (b x +a \right ) d -d a +c b \right )^{3} d}+\frac {-\frac {2 \cos \left (2 b x +2 a \right )}{3 \left (\left (b x +a \right ) d -d a +c b \right )^{2} d}-\frac {2 \left (-\frac {2 \sin \left (2 b x +2 a \right )}{\left (\left (b x +a \right ) d -d a +c b \right ) d}+\frac {\frac {4 \Si \left (2 b x +2 a +\frac {-2 d a +2 c b}{d}\right ) \sin \left (\frac {-2 d a +2 c b}{d}\right )}{d}+\frac {4 \Ci \left (2 b x +2 a +\frac {-2 d a +2 c b}{d}\right ) \cos \left (\frac {-2 d a +2 c b}{d}\right )}{d}}{d}\right )}{3 d}}{d}\right )}{8}}{b} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [C] time = 1.90, size = 386, normalized size = 1.34 \[ -\frac {b^{4} {\left (2 i \, E_{4}\left (\frac {2 i \, b c + 2 i \, {\left (b x + a\right )} d - 2 i \, a d}{d}\right ) - 2 i \, E_{4}\left (-\frac {2 i \, b c + 2 i \, {\left (b x + a\right )} d - 2 i \, a d}{d}\right )\right )} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) + b^{4} {\left (i \, E_{4}\left (\frac {4 i \, b c + 4 i \, {\left (b x + a\right )} d - 4 i \, a d}{d}\right ) - i \, E_{4}\left (-\frac {4 i \, b c + 4 i \, {\left (b x + a\right )} d - 4 i \, a d}{d}\right )\right )} \cos \left (-\frac {4 \, {\left (b c - a d\right )}}{d}\right ) + 2 \, b^{4} {\left (E_{4}\left (\frac {2 i \, b c + 2 i \, {\left (b x + a\right )} d - 2 i \, a d}{d}\right ) + E_{4}\left (-\frac {2 i \, b c + 2 i \, {\left (b x + a\right )} d - 2 i \, a d}{d}\right )\right )} \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) + b^{4} {\left (E_{4}\left (\frac {4 i \, b c + 4 i \, {\left (b x + a\right )} d - 4 i \, a d}{d}\right ) + E_{4}\left (-\frac {4 i \, b c + 4 i \, {\left (b x + a\right )} d - 4 i \, a d}{d}\right )\right )} \sin \left (-\frac {4 \, {\left (b c - a d\right )}}{d}\right )}{16 \, {\left (b^{3} c^{3} d - 3 \, a b^{2} c^{2} d^{2} + 3 \, a^{2} b c d^{3} + {\left (b x + a\right )}^{3} d^{4} - a^{3} d^{4} + 3 \, {\left (b c d^{3} - a d^{4}\right )} {\left (b x + a\right )}^{2} + 3 \, {\left (b^{2} c^{2} d^{2} - 2 \, a b c d^{3} + a^{2} d^{4}\right )} {\left (b x + a\right )}\right )} b} \]
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 \[ \int \frac {{\cos \left (a+b\,x\right )}^3\,\sin \left (a+b\,x\right )}{{\left (c+d\,x\right )}^4} \,d x \]
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 \[ \int \frac {\sin {\left (a + b x \right )} \cos ^{3}{\left (a + b x \right )}}{\left (c + d x\right )^{4}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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