Optimal. Leaf size=111 \[ \frac {3 x}{4}+\frac {3 \cos (a+b x) \sin (a+b x)}{4 b}+\frac {\cos ^3(a+b x) \sin (a+b x)}{2 b}+\frac {2 \cos ^5(a+b x) \sin (a+b x)}{5 b}-\frac {12 \cos ^7(a+b x) \sin (a+b x)}{5 b}-\frac {32 \cos ^7(a+b x) \sin ^3(a+b x)}{5 b} \]
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Rubi [A]
time = 0.10, antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps
used = 7, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4373, 2648,
2715, 8} \begin {gather*} -\frac {32 \sin ^3(a+b x) \cos ^7(a+b x)}{5 b}-\frac {12 \sin (a+b x) \cos ^7(a+b x)}{5 b}+\frac {2 \sin (a+b x) \cos ^5(a+b x)}{5 b}+\frac {\sin (a+b x) \cos ^3(a+b x)}{2 b}+\frac {3 \sin (a+b x) \cos (a+b x)}{4 b}+\frac {3 x}{4} \end {gather*}
Antiderivative was successfully verified.
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Rule 8
Rule 2648
Rule 2715
Rule 4373
Rubi steps
\begin {align*} \int \csc ^2(a+b x) \sin ^6(2 a+2 b x) \, dx &=64 \int \cos ^6(a+b x) \sin ^4(a+b x) \, dx\\ &=-\frac {32 \cos ^7(a+b x) \sin ^3(a+b x)}{5 b}+\frac {96}{5} \int \cos ^6(a+b x) \sin ^2(a+b x) \, dx\\ &=-\frac {12 \cos ^7(a+b x) \sin (a+b x)}{5 b}-\frac {32 \cos ^7(a+b x) \sin ^3(a+b x)}{5 b}+\frac {12}{5} \int \cos ^6(a+b x) \, dx\\ &=\frac {2 \cos ^5(a+b x) \sin (a+b x)}{5 b}-\frac {12 \cos ^7(a+b x) \sin (a+b x)}{5 b}-\frac {32 \cos ^7(a+b x) \sin ^3(a+b x)}{5 b}+2 \int \cos ^4(a+b x) \, dx\\ &=\frac {\cos ^3(a+b x) \sin (a+b x)}{2 b}+\frac {2 \cos ^5(a+b x) \sin (a+b x)}{5 b}-\frac {12 \cos ^7(a+b x) \sin (a+b x)}{5 b}-\frac {32 \cos ^7(a+b x) \sin ^3(a+b x)}{5 b}+\frac {3}{2} \int \cos ^2(a+b x) \, dx\\ &=\frac {3 \cos (a+b x) \sin (a+b x)}{4 b}+\frac {\cos ^3(a+b x) \sin (a+b x)}{2 b}+\frac {2 \cos ^5(a+b x) \sin (a+b x)}{5 b}-\frac {12 \cos ^7(a+b x) \sin (a+b x)}{5 b}-\frac {32 \cos ^7(a+b x) \sin ^3(a+b x)}{5 b}+\frac {3 \int 1 \, dx}{4}\\ &=\frac {3 x}{4}+\frac {3 \cos (a+b x) \sin (a+b x)}{4 b}+\frac {\cos ^3(a+b x) \sin (a+b x)}{2 b}+\frac {2 \cos ^5(a+b x) \sin (a+b x)}{5 b}-\frac {12 \cos ^7(a+b x) \sin (a+b x)}{5 b}-\frac {32 \cos ^7(a+b x) \sin ^3(a+b x)}{5 b}\\ \end {align*}
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Mathematica [A]
time = 0.22, size = 62, normalized size = 0.56 \begin {gather*} \frac {120 b x+20 \sin (2 (a+b x))-40 \sin (4 (a+b x))-10 \sin (6 (a+b x))+5 \sin (8 (a+b x))+2 \sin (10 (a+b x))}{160 b} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.10, size = 83, normalized size = 0.75
method | result | size |
risch | \(\frac {3 x}{4}+\frac {\sin \left (10 x b +10 a \right )}{80 b}+\frac {\sin \left (8 x b +8 a \right )}{32 b}-\frac {\sin \left (6 x b +6 a \right )}{16 b}-\frac {\sin \left (4 x b +4 a \right )}{4 b}+\frac {\sin \left (2 x b +2 a \right )}{8 b}\) | \(75\) |
default | \(\frac {-\frac {32 \left (\sin ^{3}\left (x b +a \right )\right ) \left (\cos ^{7}\left (x b +a \right )\right )}{5}-\frac {12 \sin \left (x b +a \right ) \left (\cos ^{7}\left (x b +a \right )\right )}{5}+\frac {2 \left (\cos ^{5}\left (x b +a \right )+\frac {5 \left (\cos ^{3}\left (x b +a \right )\right )}{4}+\frac {15 \cos \left (x b +a \right )}{8}\right ) \sin \left (x b +a \right )}{5}+\frac {3 x b}{4}+\frac {3 a}{4}}{b}\) | \(83\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.28, size = 65, normalized size = 0.59 \begin {gather*} \frac {120 \, b x + 2 \, \sin \left (10 \, b x + 10 \, a\right ) + 5 \, \sin \left (8 \, b x + 8 \, a\right ) - 10 \, \sin \left (6 \, b x + 6 \, a\right ) - 40 \, \sin \left (4 \, b x + 4 \, a\right ) + 20 \, \sin \left (2 \, b x + 2 \, a\right )}{160 \, b} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 2.89, size = 66, normalized size = 0.59 \begin {gather*} \frac {15 \, b x + {\left (128 \, \cos \left (b x + a\right )^{9} - 176 \, \cos \left (b x + a\right )^{7} + 8 \, \cos \left (b x + a\right )^{5} + 10 \, \cos \left (b x + a\right )^{3} + 15 \, \cos \left (b x + a\right )\right )} \sin \left (b x + a\right )}{20 \, b} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.42, size = 75, normalized size = 0.68 \begin {gather*} \frac {15 \, b x + 15 \, a + \frac {15 \, \tan \left (b x + a\right )^{9} + 70 \, \tan \left (b x + a\right )^{7} + 128 \, \tan \left (b x + a\right )^{5} - 70 \, \tan \left (b x + a\right )^{3} - 15 \, \tan \left (b x + a\right )}{{\left (\tan \left (b x + a\right )^{2} + 1\right )}^{5}}}{20 \, b} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 1.69, size = 109, normalized size = 0.98 \begin {gather*} \frac {3\,x}{4}+\frac {\frac {3\,{\mathrm {tan}\left (a+b\,x\right )}^9}{4}+\frac {7\,{\mathrm {tan}\left (a+b\,x\right )}^7}{2}+\frac {32\,{\mathrm {tan}\left (a+b\,x\right )}^5}{5}-\frac {7\,{\mathrm {tan}\left (a+b\,x\right )}^3}{2}-\frac {3\,\mathrm {tan}\left (a+b\,x\right )}{4}}{b\,\left ({\mathrm {tan}\left (a+b\,x\right )}^{10}+5\,{\mathrm {tan}\left (a+b\,x\right )}^8+10\,{\mathrm {tan}\left (a+b\,x\right )}^6+10\,{\mathrm {tan}\left (a+b\,x\right )}^4+5\,{\mathrm {tan}\left (a+b\,x\right )}^2+1\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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