Optimal. Leaf size=70 \[ \frac {15 \sec (a+b x)}{8 b}-\frac {15 \tanh ^{-1}(\cos (a+b x))}{8 b}-\frac {\csc ^4(a+b x) \sec (a+b x)}{4 b}-\frac {5 \csc ^2(a+b x) \sec (a+b x)}{8 b} \]
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Rubi [A] time = 0.04, antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {2622, 288, 321, 207} \[ \frac {15 \sec (a+b x)}{8 b}-\frac {15 \tanh ^{-1}(\cos (a+b x))}{8 b}-\frac {\csc ^4(a+b x) \sec (a+b x)}{4 b}-\frac {5 \csc ^2(a+b x) \sec (a+b x)}{8 b} \]
Antiderivative was successfully verified.
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Rule 207
Rule 288
Rule 321
Rule 2622
Rubi steps
\begin {align*} \int \csc ^5(a+b x) \sec ^2(a+b x) \, dx &=\frac {\operatorname {Subst}\left (\int \frac {x^6}{\left (-1+x^2\right )^3} \, dx,x,\sec (a+b x)\right )}{b}\\ &=-\frac {\csc ^4(a+b x) \sec (a+b x)}{4 b}+\frac {5 \operatorname {Subst}\left (\int \frac {x^4}{\left (-1+x^2\right )^2} \, dx,x,\sec (a+b x)\right )}{4 b}\\ &=-\frac {5 \csc ^2(a+b x) \sec (a+b x)}{8 b}-\frac {\csc ^4(a+b x) \sec (a+b x)}{4 b}+\frac {15 \operatorname {Subst}\left (\int \frac {x^2}{-1+x^2} \, dx,x,\sec (a+b x)\right )}{8 b}\\ &=\frac {15 \sec (a+b x)}{8 b}-\frac {5 \csc ^2(a+b x) \sec (a+b x)}{8 b}-\frac {\csc ^4(a+b x) \sec (a+b x)}{4 b}+\frac {15 \operatorname {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\sec (a+b x)\right )}{8 b}\\ &=-\frac {15 \tanh ^{-1}(\cos (a+b x))}{8 b}+\frac {15 \sec (a+b x)}{8 b}-\frac {5 \csc ^2(a+b x) \sec (a+b x)}{8 b}-\frac {\csc ^4(a+b x) \sec (a+b x)}{4 b}\\ \end {align*}
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Mathematica [A] time = 3.99, size = 129, normalized size = 1.84 \[ -\frac {\csc ^4\left (\frac {1}{2} (a+b x)\right )+14 \csc ^2\left (\frac {1}{2} (a+b x)\right )+\frac {\sec ^2\left (\frac {1}{2} (a+b x)\right ) \left (-14 \tan ^2\left (\frac {1}{2} (a+b x)\right )+\cos (a+b x) \left (\sec ^4\left (\frac {1}{2} (a+b x)\right )-8 \left (-15 \log \left (\sin \left (\frac {1}{2} (a+b x)\right )\right )+15 \log \left (\cos \left (\frac {1}{2} (a+b x)\right )\right )+8\right )\right )+78\right )}{\tan ^2\left (\frac {1}{2} (a+b x)\right )-1}}{64 b} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.46, size = 132, normalized size = 1.89 \[ \frac {30 \, \cos \left (b x + a\right )^{4} - 50 \, \cos \left (b x + a\right )^{2} - 15 \, {\left (\cos \left (b x + a\right )^{5} - 2 \, \cos \left (b x + a\right )^{3} + \cos \left (b x + a\right )\right )} \log \left (\frac {1}{2} \, \cos \left (b x + a\right ) + \frac {1}{2}\right ) + 15 \, {\left (\cos \left (b x + a\right )^{5} - 2 \, \cos \left (b x + a\right )^{3} + \cos \left (b x + a\right )\right )} \log \left (-\frac {1}{2} \, \cos \left (b x + a\right ) + \frac {1}{2}\right ) + 16}{16 \, {\left (b \cos \left (b x + a\right )^{5} - 2 \, b \cos \left (b x + a\right )^{3} + b \cos \left (b x + a\right )\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.23, size = 163, normalized size = 2.33 \[ \frac {\frac {{\left (\frac {16 \, {\left (\cos \left (b x + a\right ) - 1\right )}}{\cos \left (b x + a\right ) + 1} - \frac {90 \, {\left (\cos \left (b x + a\right ) - 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}} - 1\right )} {\left (\cos \left (b x + a\right ) + 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) - 1\right )}^{2}} - \frac {16 \, {\left (\cos \left (b x + a\right ) - 1\right )}}{\cos \left (b x + a\right ) + 1} + \frac {{\left (\cos \left (b x + a\right ) - 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}} + \frac {128}{\frac {\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} + 1} + 60 \, \log \left (\frac {{\left | -\cos \left (b x + a\right ) + 1 \right |}}{{\left | \cos \left (b x + a\right ) + 1 \right |}}\right )}{64 \, b} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.03, size = 78, normalized size = 1.11 \[ -\frac {1}{4 b \sin \left (b x +a \right )^{4} \cos \left (b x +a \right )}-\frac {5}{8 b \sin \left (b x +a \right )^{2} \cos \left (b x +a \right )}+\frac {15}{8 b \cos \left (b x +a \right )}+\frac {15 \ln \left (\csc \left (b x +a \right )-\cot \left (b x +a \right )\right )}{8 b} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.34, size = 79, normalized size = 1.13 \[ \frac {\frac {2 \, {\left (15 \, \cos \left (b x + a\right )^{4} - 25 \, \cos \left (b x + a\right )^{2} + 8\right )}}{\cos \left (b x + a\right )^{5} - 2 \, \cos \left (b x + a\right )^{3} + \cos \left (b x + a\right )} - 15 \, \log \left (\cos \left (b x + a\right ) + 1\right ) + 15 \, \log \left (\cos \left (b x + a\right ) - 1\right )}{16 \, b} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.39, size = 66, normalized size = 0.94 \[ \frac {\frac {15\,{\cos \left (a+b\,x\right )}^4}{8}-\frac {25\,{\cos \left (a+b\,x\right )}^2}{8}+1}{b\,\left ({\cos \left (a+b\,x\right )}^5-2\,{\cos \left (a+b\,x\right )}^3+\cos \left (a+b\,x\right )\right )}-\frac {15\,\mathrm {atanh}\left (\cos \left (a+b\,x\right )\right )}{8\,b} \]
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 \[ \int \frac {\sec ^{2}{\left (a + b x \right )}}{\sin ^{5}{\left (a + b x \right )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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