Optimal. Leaf size=77 \[ -\frac {3 E\left (\left .a+b x-\frac {\pi }{4}\right |2\right )}{10 b}-\frac {\cos ^2(a+b x)}{5 b \sin ^{\frac {5}{2}}(2 a+2 b x)}-\frac {3 \cos (2 a+2 b x)}{10 b \sqrt {\sin (2 a+2 b x)}} \]
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Rubi [A] time = 0.05, antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {4295, 2636, 2639} \[ -\frac {3 E\left (\left .a+b x-\frac {\pi }{4}\right |2\right )}{10 b}-\frac {\cos ^2(a+b x)}{5 b \sin ^{\frac {5}{2}}(2 a+2 b x)}-\frac {3 \cos (2 a+2 b x)}{10 b \sqrt {\sin (2 a+2 b x)}} \]
Antiderivative was successfully verified.
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Rule 2636
Rule 2639
Rule 4295
Rubi steps
\begin {align*} \int \frac {\cos ^2(a+b x)}{\sin ^{\frac {7}{2}}(2 a+2 b x)} \, dx &=-\frac {\cos ^2(a+b x)}{5 b \sin ^{\frac {5}{2}}(2 a+2 b x)}+\frac {3}{10} \int \frac {1}{\sin ^{\frac {3}{2}}(2 a+2 b x)} \, dx\\ &=-\frac {\cos ^2(a+b x)}{5 b \sin ^{\frac {5}{2}}(2 a+2 b x)}-\frac {3 \cos (2 a+2 b x)}{10 b \sqrt {\sin (2 a+2 b x)}}-\frac {3}{10} \int \sqrt {\sin (2 a+2 b x)} \, dx\\ &=-\frac {3 E\left (\left .a-\frac {\pi }{4}+b x\right |2\right )}{10 b}-\frac {\cos ^2(a+b x)}{5 b \sin ^{\frac {5}{2}}(2 a+2 b x)}-\frac {3 \cos (2 a+2 b x)}{10 b \sqrt {\sin (2 a+2 b x)}}\\ \end {align*}
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Mathematica [A] time = 0.56, size = 64, normalized size = 0.83 \[ \frac {\frac {2 (-6 \cos (2 (a+b x))+3 \cos (4 (a+b x))+1) \cot (a+b x)}{\sin ^{\frac {3}{2}}(2 (a+b x))}-12 E\left (\left .a+b x-\frac {\pi }{4}\right |2\right )}{40 b} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.55, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\cos \left (b x + a\right )^{2} \sqrt {\sin \left (2 \, b x + 2 \, a\right )}}{\cos \left (2 \, b x + 2 \, a\right )^{4} - 2 \, \cos \left (2 \, b x + 2 \, a\right )^{2} + 1}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos \left (b x + a\right )^{2}}{\sin \left (2 \, b x + 2 \, a\right )^{\frac {7}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 212.85, size = 227, normalized size = 2.95 \[ \frac {\sqrt {2}\, \left (-\frac {8 \sqrt {2}}{5 \sin \left (2 b x +2 a \right )^{\frac {5}{2}}}+\frac {4 \sqrt {2}\, \left (6 \sqrt {1+\sin \left (2 b x +2 a \right )}\, \sqrt {-2 \sin \left (2 b x +2 a \right )+2}\, \sqrt {-\sin \left (2 b x +2 a \right )}\, \left (\sin ^{2}\left (2 b x +2 a \right )\right ) \EllipticE \left (\sqrt {1+\sin \left (2 b x +2 a \right )}, \frac {\sqrt {2}}{2}\right )-3 \sqrt {1+\sin \left (2 b x +2 a \right )}\, \sqrt {-2 \sin \left (2 b x +2 a \right )+2}\, \sqrt {-\sin \left (2 b x +2 a \right )}\, \left (\sin ^{2}\left (2 b x +2 a \right )\right ) \EllipticF \left (\sqrt {1+\sin \left (2 b x +2 a \right )}, \frac {\sqrt {2}}{2}\right )+6 \left (\sin ^{4}\left (2 b x +2 a \right )\right )-4 \left (\sin ^{2}\left (2 b x +2 a \right )\right )-2\right )}{5 \sin \left (2 b x +2 a \right )^{\frac {5}{2}} \cos \left (2 b x +2 a \right )}\right )}{32 b} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos \left (b x + a\right )^{2}}{\sin \left (2 \, b x + 2 \, a\right )^{\frac {7}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\cos \left (a+b\,x\right )}^2}{{\sin \left (2\,a+2\,b\,x\right )}^{7/2}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [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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