Optimal. Leaf size=413 \[ -\frac {b^3 \cos \left (a-\frac {b c}{d}\right ) \text {Ci}\left (\frac {b c}{d}+b x\right )}{48 d^4}-\frac {9 b^3 \cos \left (3 a-\frac {3 b c}{d}\right ) \text {Ci}\left (\frac {3 b c}{d}+3 b x\right )}{32 d^4}+\frac {125 b^3 \cos \left (5 a-\frac {5 b c}{d}\right ) \text {Ci}\left (\frac {5 b c}{d}+5 b x\right )}{96 d^4}+\frac {b^3 \sin \left (a-\frac {b c}{d}\right ) \text {Si}\left (\frac {b c}{d}+b x\right )}{48 d^4}+\frac {9 b^3 \sin \left (3 a-\frac {3 b c}{d}\right ) \text {Si}\left (\frac {3 b c}{d}+3 b x\right )}{32 d^4}-\frac {125 b^3 \sin \left (5 a-\frac {5 b c}{d}\right ) \text {Si}\left (\frac {5 b c}{d}+5 b x\right )}{96 d^4}+\frac {b^2 \sin (a+b x)}{48 d^3 (c+d x)}+\frac {3 b^2 \sin (3 a+3 b x)}{32 d^3 (c+d x)}-\frac {25 b^2 \sin (5 a+5 b x)}{96 d^3 (c+d x)}-\frac {b \cos (a+b x)}{48 d^2 (c+d x)^2}-\frac {b \cos (3 a+3 b x)}{32 d^2 (c+d x)^2}+\frac {5 b \cos (5 a+5 b x)}{96 d^2 (c+d x)^2}-\frac {\sin (a+b x)}{24 d (c+d x)^3}-\frac {\sin (3 a+3 b x)}{48 d (c+d x)^3}+\frac {\sin (5 a+5 b x)}{48 d (c+d x)^3} \]
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Rubi [A] time = 0.59, antiderivative size = 413, normalized size of antiderivative = 1.00, number of steps used = 20, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {4406, 3297, 3303, 3299, 3302} \[ -\frac {b^3 \cos \left (a-\frac {b c}{d}\right ) \text {CosIntegral}\left (\frac {b c}{d}+b x\right )}{48 d^4}-\frac {9 b^3 \cos \left (3 a-\frac {3 b c}{d}\right ) \text {CosIntegral}\left (\frac {3 b c}{d}+3 b x\right )}{32 d^4}+\frac {125 b^3 \cos \left (5 a-\frac {5 b c}{d}\right ) \text {CosIntegral}\left (\frac {5 b c}{d}+5 b x\right )}{96 d^4}+\frac {b^3 \sin \left (a-\frac {b c}{d}\right ) \text {Si}\left (\frac {b c}{d}+b x\right )}{48 d^4}+\frac {9 b^3 \sin \left (3 a-\frac {3 b c}{d}\right ) \text {Si}\left (\frac {3 b c}{d}+3 b x\right )}{32 d^4}-\frac {125 b^3 \sin \left (5 a-\frac {5 b c}{d}\right ) \text {Si}\left (\frac {5 b c}{d}+5 b x\right )}{96 d^4}+\frac {b^2 \sin (a+b x)}{48 d^3 (c+d x)}+\frac {3 b^2 \sin (3 a+3 b x)}{32 d^3 (c+d x)}-\frac {25 b^2 \sin (5 a+5 b x)}{96 d^3 (c+d x)}-\frac {b \cos (a+b x)}{48 d^2 (c+d x)^2}-\frac {b \cos (3 a+3 b x)}{32 d^2 (c+d x)^2}+\frac {5 b \cos (5 a+5 b x)}{96 d^2 (c+d x)^2}-\frac {\sin (a+b x)}{24 d (c+d x)^3}-\frac {\sin (3 a+3 b x)}{48 d (c+d x)^3}+\frac {\sin (5 a+5 b x)}{48 d (c+d x)^3} \]
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 ^2(a+b x) \sin ^3(a+b x)}{(c+d x)^4} \, dx &=\int \left (\frac {\sin (a+b x)}{8 (c+d x)^4}+\frac {\sin (3 a+3 b x)}{16 (c+d x)^4}-\frac {\sin (5 a+5 b x)}{16 (c+d x)^4}\right ) \, dx\\ &=\frac {1}{16} \int \frac {\sin (3 a+3 b x)}{(c+d x)^4} \, dx-\frac {1}{16} \int \frac {\sin (5 a+5 b x)}{(c+d x)^4} \, dx+\frac {1}{8} \int \frac {\sin (a+b x)}{(c+d x)^4} \, dx\\ &=-\frac {\sin (a+b x)}{24 d (c+d x)^3}-\frac {\sin (3 a+3 b x)}{48 d (c+d x)^3}+\frac {\sin (5 a+5 b x)}{48 d (c+d x)^3}+\frac {b \int \frac {\cos (a+b x)}{(c+d x)^3} \, dx}{24 d}+\frac {b \int \frac {\cos (3 a+3 b x)}{(c+d x)^3} \, dx}{16 d}-\frac {(5 b) \int \frac {\cos (5 a+5 b x)}{(c+d x)^3} \, dx}{48 d}\\ &=-\frac {b \cos (a+b x)}{48 d^2 (c+d x)^2}-\frac {b \cos (3 a+3 b x)}{32 d^2 (c+d x)^2}+\frac {5 b \cos (5 a+5 b x)}{96 d^2 (c+d x)^2}-\frac {\sin (a+b x)}{24 d (c+d x)^3}-\frac {\sin (3 a+3 b x)}{48 d (c+d x)^3}+\frac {\sin (5 a+5 b x)}{48 d (c+d x)^3}-\frac {b^2 \int \frac {\sin (a+b x)}{(c+d x)^2} \, dx}{48 d^2}-\frac {\left (3 b^2\right ) \int \frac {\sin (3 a+3 b x)}{(c+d x)^2} \, dx}{32 d^2}+\frac {\left (25 b^2\right ) \int \frac {\sin (5 a+5 b x)}{(c+d x)^2} \, dx}{96 d^2}\\ &=-\frac {b \cos (a+b x)}{48 d^2 (c+d x)^2}-\frac {b \cos (3 a+3 b x)}{32 d^2 (c+d x)^2}+\frac {5 b \cos (5 a+5 b x)}{96 d^2 (c+d x)^2}-\frac {\sin (a+b x)}{24 d (c+d x)^3}+\frac {b^2 \sin (a+b x)}{48 d^3 (c+d x)}-\frac {\sin (3 a+3 b x)}{48 d (c+d x)^3}+\frac {3 b^2 \sin (3 a+3 b x)}{32 d^3 (c+d x)}+\frac {\sin (5 a+5 b x)}{48 d (c+d x)^3}-\frac {25 b^2 \sin (5 a+5 b x)}{96 d^3 (c+d x)}-\frac {b^3 \int \frac {\cos (a+b x)}{c+d x} \, dx}{48 d^3}-\frac {\left (9 b^3\right ) \int \frac {\cos (3 a+3 b x)}{c+d x} \, dx}{32 d^3}+\frac {\left (125 b^3\right ) \int \frac {\cos (5 a+5 b x)}{c+d x} \, dx}{96 d^3}\\ &=-\frac {b \cos (a+b x)}{48 d^2 (c+d x)^2}-\frac {b \cos (3 a+3 b x)}{32 d^2 (c+d x)^2}+\frac {5 b \cos (5 a+5 b x)}{96 d^2 (c+d x)^2}-\frac {\sin (a+b x)}{24 d (c+d x)^3}+\frac {b^2 \sin (a+b x)}{48 d^3 (c+d x)}-\frac {\sin (3 a+3 b x)}{48 d (c+d x)^3}+\frac {3 b^2 \sin (3 a+3 b x)}{32 d^3 (c+d x)}+\frac {\sin (5 a+5 b x)}{48 d (c+d x)^3}-\frac {25 b^2 \sin (5 a+5 b x)}{96 d^3 (c+d x)}+\frac {\left (125 b^3 \cos \left (5 a-\frac {5 b c}{d}\right )\right ) \int \frac {\cos \left (\frac {5 b c}{d}+5 b x\right )}{c+d x} \, dx}{96 d^3}-\frac {\left (9 b^3 \cos \left (3 a-\frac {3 b c}{d}\right )\right ) \int \frac {\cos \left (\frac {3 b c}{d}+3 b x\right )}{c+d x} \, dx}{32 d^3}-\frac {\left (b^3 \cos \left (a-\frac {b c}{d}\right )\right ) \int \frac {\cos \left (\frac {b c}{d}+b x\right )}{c+d x} \, dx}{48 d^3}-\frac {\left (125 b^3 \sin \left (5 a-\frac {5 b c}{d}\right )\right ) \int \frac {\sin \left (\frac {5 b c}{d}+5 b x\right )}{c+d x} \, dx}{96 d^3}+\frac {\left (9 b^3 \sin \left (3 a-\frac {3 b c}{d}\right )\right ) \int \frac {\sin \left (\frac {3 b c}{d}+3 b x\right )}{c+d x} \, dx}{32 d^3}+\frac {\left (b^3 \sin \left (a-\frac {b c}{d}\right )\right ) \int \frac {\sin \left (\frac {b c}{d}+b x\right )}{c+d x} \, dx}{48 d^3}\\ &=-\frac {b \cos (a+b x)}{48 d^2 (c+d x)^2}-\frac {b \cos (3 a+3 b x)}{32 d^2 (c+d x)^2}+\frac {5 b \cos (5 a+5 b x)}{96 d^2 (c+d x)^2}-\frac {b^3 \cos \left (a-\frac {b c}{d}\right ) \text {Ci}\left (\frac {b c}{d}+b x\right )}{48 d^4}-\frac {9 b^3 \cos \left (3 a-\frac {3 b c}{d}\right ) \text {Ci}\left (\frac {3 b c}{d}+3 b x\right )}{32 d^4}+\frac {125 b^3 \cos \left (5 a-\frac {5 b c}{d}\right ) \text {Ci}\left (\frac {5 b c}{d}+5 b x\right )}{96 d^4}-\frac {\sin (a+b x)}{24 d (c+d x)^3}+\frac {b^2 \sin (a+b x)}{48 d^3 (c+d x)}-\frac {\sin (3 a+3 b x)}{48 d (c+d x)^3}+\frac {3 b^2 \sin (3 a+3 b x)}{32 d^3 (c+d x)}+\frac {\sin (5 a+5 b x)}{48 d (c+d x)^3}-\frac {25 b^2 \sin (5 a+5 b x)}{96 d^3 (c+d x)}+\frac {b^3 \sin \left (a-\frac {b c}{d}\right ) \text {Si}\left (\frac {b c}{d}+b x\right )}{48 d^4}+\frac {9 b^3 \sin \left (3 a-\frac {3 b c}{d}\right ) \text {Si}\left (\frac {3 b c}{d}+3 b x\right )}{32 d^4}-\frac {125 b^3 \sin \left (5 a-\frac {5 b c}{d}\right ) \text {Si}\left (\frac {5 b c}{d}+5 b x\right )}{96 d^4}\\ \end {align*}
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Mathematica [A] time = 2.90, size = 457, normalized size = 1.11 \[ \frac {-27 b^3 (c+d x)^3 \left (\cos \left (3 a-\frac {3 b c}{d}\right ) \text {Ci}\left (\frac {3 b (c+d x)}{d}\right )-\sin \left (3 a-\frac {3 b c}{d}\right ) \text {Si}\left (\frac {3 b (c+d x)}{d}\right )\right )+125 b^3 (c+d x)^3 \left (\cos \left (5 a-\frac {5 b c}{d}\right ) \text {Ci}\left (\frac {5 b (c+d x)}{d}\right )-\sin \left (5 a-\frac {5 b c}{d}\right ) \text {Si}\left (\frac {5 b (c+d x)}{d}\right )\right )-d \cos (3 b x) \left (3 b d \cos (3 a) (c+d x)-\sin (3 a) \left (9 b^2 (c+d x)^2-2 d^2\right )\right )+d \cos (5 b x) \left (5 b d \cos (5 a) (c+d x)-\sin (5 a) \left (25 b^2 (c+d x)^2-2 d^2\right )\right )+d \sin (3 b x) \left (\cos (3 a) \left (9 b^2 (c+d x)^2-2 d^2\right )+3 b d \sin (3 a) (c+d x)\right )-d \sin (5 b x) \left (\cos (5 a) \left (25 b^2 (c+d x)^2-2 d^2\right )+5 b d \sin (5 a) (c+d x)\right )-2 \left (b^3 (c+d x)^3 \left (\cos \left (a-\frac {b c}{d}\right ) \text {Ci}\left (b \left (\frac {c}{d}+x\right )\right )-\sin \left (a-\frac {b c}{d}\right ) \text {Si}\left (b \left (\frac {c}{d}+x\right )\right )\right )+d \cos (b x) \left (b d \cos (a) (c+d x)-\sin (a) \left (b^2 (c+d x)^2-2 d^2\right )\right )-d \sin (b x) \left (\cos (a) \left (b^2 (c+d x)^2-2 d^2\right )+b d \sin (a) (c+d x)\right )\right )}{96 d^4 (c+d x)^3} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.62, size = 824, normalized size = 2.00 \[ \frac {160 \, {\left (b d^{3} x + b c d^{2}\right )} \cos \left (b x + a\right )^{5} - 224 \, {\left (b d^{3} x + b c d^{2}\right )} \cos \left (b x + a\right )^{3} - 250 \, {\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 {5 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {Si}\left (\frac {5 \, {\left (b d x + b c\right )}}{d}\right ) + 54 \, {\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 {3 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {Si}\left (\frac {3 \, {\left (b d x + b c\right )}}{d}\right ) + 4 \, {\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 {b c - a d}{d}\right ) \operatorname {Si}\left (\frac {b d x + b c}{d}\right ) + 64 \, {\left (b d^{3} x + b c d^{2}\right )} \cos \left (b x + a\right ) - 2 \, {\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 {b d x + b c}{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 {b d x + b c}{d}\right )\right )} \cos \left (-\frac {b c - a d}{d}\right ) - 27 \, {\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 {3 \, {\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 {3 \, {\left (b d x + b c\right )}}{d}\right )\right )} \cos \left (-\frac {3 \, {\left (b c - a d\right )}}{d}\right ) + 125 \, {\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 {5 \, {\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 {5 \, {\left (b d x + b c\right )}}{d}\right )\right )} \cos \left (-\frac {5 \, {\left (b c - a d\right )}}{d}\right ) - 32 \, {\left (2 \, b^{2} d^{3} x^{2} + 4 \, b^{2} c d^{2} x + 2 \, b^{2} c^{2} d + {\left (25 \, b^{2} d^{3} x^{2} + 50 \, b^{2} c d^{2} x + 25 \, b^{2} c^{2} d - 2 \, d^{3}\right )} \cos \left (b x + a\right )^{4} - {\left (21 \, b^{2} d^{3} x^{2} + 42 \, b^{2} c d^{2} x + 21 \, b^{2} c^{2} d - 2 \, d^{3}\right )} \cos \left (b x + a\right )^{2}\right )} \sin \left (b x + a\right )}{192 \, {\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 = 0.03, size = 580, normalized size = 1.40 \[ \frac {-\frac {b^{4} \left (-\frac {5 \sin \left (5 b x +5 a \right )}{3 \left (\left (b x +a \right ) d -d a +c b \right )^{3} d}+\frac {-\frac {25 \cos \left (5 b x +5 a \right )}{6 \left (\left (b x +a \right ) d -d a +c b \right )^{2} d}-\frac {25 \left (-\frac {5 \sin \left (5 b x +5 a \right )}{\left (\left (b x +a \right ) d -d a +c b \right ) d}+\frac {\frac {25 \Si \left (5 b x +5 a +\frac {-5 d a +5 c b}{d}\right ) \sin \left (\frac {-5 d a +5 c b}{d}\right )}{d}+\frac {25 \Ci \left (5 b x +5 a +\frac {-5 d a +5 c b}{d}\right ) \cos \left (\frac {-5 d a +5 c b}{d}\right )}{d}}{d}\right )}{6 d}}{d}\right )}{80}+\frac {b^{4} \left (-\frac {\sin \left (b x +a \right )}{3 \left (\left (b x +a \right ) d -d a +c b \right )^{3} d}+\frac {-\frac {\cos \left (b x +a \right )}{2 \left (\left (b x +a \right ) d -d a +c b \right )^{2} d}-\frac {-\frac {\sin \left (b x +a \right )}{\left (\left (b x +a \right ) d -d a +c b \right ) d}+\frac {\frac {\Si \left (b x +a +\frac {-d a +c b}{d}\right ) \sin \left (\frac {-d a +c b}{d}\right )}{d}+\frac {\Ci \left (b x +a +\frac {-d a +c b}{d}\right ) \cos \left (\frac {-d a +c b}{d}\right )}{d}}{d}}{2 d}}{3 d}\right )}{8}+\frac {b^{4} \left (-\frac {\sin \left (3 b x +3 a \right )}{\left (\left (b x +a \right ) d -d a +c b \right )^{3} d}+\frac {-\frac {3 \cos \left (3 b x +3 a \right )}{2 \left (\left (b x +a \right ) d -d a +c b \right )^{2} d}-\frac {3 \left (-\frac {3 \sin \left (3 b x +3 a \right )}{\left (\left (b x +a \right ) d -d a +c b \right ) d}+\frac {\frac {9 \Si \left (3 b x +3 a +\frac {-3 d a +3 c b}{d}\right ) \sin \left (\frac {-3 d a +3 c b}{d}\right )}{d}+\frac {9 \Ci \left (3 b x +3 a +\frac {-3 d a +3 c b}{d}\right ) \cos \left (\frac {-3 d a +3 c b}{d}\right )}{d}}{d}\right )}{2 d}}{d}\right )}{48}}{b} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [C] time = 1.10, size = 523, normalized size = 1.27 \[ \frac {b^{4} {\left (-2 i \, E_{4}\left (\frac {i \, b c + i \, {\left (b x + a\right )} d - i \, a d}{d}\right ) + 2 i \, E_{4}\left (-\frac {i \, b c + i \, {\left (b x + a\right )} d - i \, a d}{d}\right )\right )} \cos \left (-\frac {b c - a d}{d}\right ) + b^{4} {\left (-i \, E_{4}\left (\frac {3 i \, b c + 3 i \, {\left (b x + a\right )} d - 3 i \, a d}{d}\right ) + i \, E_{4}\left (-\frac {3 i \, b c + 3 i \, {\left (b x + a\right )} d - 3 i \, a d}{d}\right )\right )} \cos \left (-\frac {3 \, {\left (b c - a d\right )}}{d}\right ) + b^{4} {\left (i \, E_{4}\left (\frac {5 i \, b c + 5 i \, {\left (b x + a\right )} d - 5 i \, a d}{d}\right ) - i \, E_{4}\left (-\frac {5 i \, b c + 5 i \, {\left (b x + a\right )} d - 5 i \, a d}{d}\right )\right )} \cos \left (-\frac {5 \, {\left (b c - a d\right )}}{d}\right ) - 2 \, b^{4} {\left (E_{4}\left (\frac {i \, b c + i \, {\left (b x + a\right )} d - i \, a d}{d}\right ) + E_{4}\left (-\frac {i \, b c + i \, {\left (b x + a\right )} d - i \, a d}{d}\right )\right )} \sin \left (-\frac {b c - a d}{d}\right ) - b^{4} {\left (E_{4}\left (\frac {3 i \, b c + 3 i \, {\left (b x + a\right )} d - 3 i \, a d}{d}\right ) + E_{4}\left (-\frac {3 i \, b c + 3 i \, {\left (b x + a\right )} d - 3 i \, a d}{d}\right )\right )} \sin \left (-\frac {3 \, {\left (b c - a d\right )}}{d}\right ) + b^{4} {\left (E_{4}\left (\frac {5 i \, b c + 5 i \, {\left (b x + a\right )} d - 5 i \, a d}{d}\right ) + E_{4}\left (-\frac {5 i \, b c + 5 i \, {\left (b x + a\right )} d - 5 i \, a d}{d}\right )\right )} \sin \left (-\frac {5 \, {\left (b c - a d\right )}}{d}\right )}{32 \, {\left (b^{3} c^{3} d - 3 \, a b^{2} c^{2} d^{2} + 3 \, a^{2} b c d^{3} + {\left (b x + a\right )}^{3} d^{4} - a^{3} d^{4} + 3 \, {\left (b c d^{3} - a d^{4}\right )} {\left (b x + a\right )}^{2} + 3 \, {\left (b^{2} c^{2} d^{2} - 2 \, a b c d^{3} + a^{2} d^{4}\right )} {\left (b x + a\right )}\right )} b} \]
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 \[ \int \frac {{\cos \left (a+b\,x\right )}^2\,{\sin \left (a+b\,x\right )}^3}{{\left (c+d\,x\right )}^4} \,d x \]
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 \[ \int \frac {\sin ^{3}{\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{\left (c + d x\right )^{4}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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