Optimal. Leaf size=219 \[ -\frac{8 b \left (5 a^2+20 a b+16 b^2\right ) \sec (e+f x)}{15 a^5 f \sqrt{a+b \sec ^2(e+f x)}}-\frac{4 b \left (5 a^2+20 a b+16 b^2\right ) \sec (e+f x)}{15 a^4 f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac{\left (5 a^2+20 a b+16 b^2\right ) \cos (e+f x)}{5 a^3 f \left (a+b \sec ^2(e+f x)\right )^{3/2}}+\frac{2 (5 a+4 b) \cos ^3(e+f x)}{15 a^2 f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac{\cos ^5(e+f x)}{5 a f \left (a+b \sec ^2(e+f x)\right )^{3/2}} \]
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Rubi [A] time = 0.215292, antiderivative size = 219, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {4134, 462, 453, 271, 192, 191} \[ -\frac{8 b \left (5 a^2+20 a b+16 b^2\right ) \sec (e+f x)}{15 a^5 f \sqrt{a+b \sec ^2(e+f x)}}-\frac{4 b \left (5 a^2+20 a b+16 b^2\right ) \sec (e+f x)}{15 a^4 f \left (a+b \sec ^2(e+f x)\right )^{3/2}}+\frac{2 (5 a+4 b) \cos ^3(e+f x)}{15 a^2 f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac{\left (\frac{4 b (5 a+4 b)}{a^2}+5\right ) \cos (e+f x)}{5 a f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac{\cos ^5(e+f x)}{5 a f \left (a+b \sec ^2(e+f x)\right )^{3/2}} \]
Antiderivative was successfully verified.
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Rule 4134
Rule 462
Rule 453
Rule 271
Rule 192
Rule 191
Rubi steps
\begin{align*} \int \frac{\sin ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (-1+x^2\right )^2}{x^6 \left (a+b x^2\right )^{5/2}} \, dx,x,\sec (e+f x)\right )}{f}\\ &=-\frac{\cos ^5(e+f x)}{5 a f \left (a+b \sec ^2(e+f x)\right )^{3/2}}+\frac{\operatorname{Subst}\left (\int \frac{-2 (5 a+4 b)+5 a x^2}{x^4 \left (a+b x^2\right )^{5/2}} \, dx,x,\sec (e+f x)\right )}{5 a f}\\ &=\frac{2 (5 a+4 b) \cos ^3(e+f x)}{15 a^2 f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac{\cos ^5(e+f x)}{5 a f \left (a+b \sec ^2(e+f x)\right )^{3/2}}+\frac{\left (5 a^2+20 a b+16 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{x^2 \left (a+b x^2\right )^{5/2}} \, dx,x,\sec (e+f x)\right )}{5 a^2 f}\\ &=-\frac{\left (5 a^2+20 a b+16 b^2\right ) \cos (e+f x)}{5 a^3 f \left (a+b \sec ^2(e+f x)\right )^{3/2}}+\frac{2 (5 a+4 b) \cos ^3(e+f x)}{15 a^2 f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac{\cos ^5(e+f x)}{5 a f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac{\left (4 b \left (5 a^2+20 a b+16 b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\left (a+b x^2\right )^{5/2}} \, dx,x,\sec (e+f x)\right )}{5 a^3 f}\\ &=-\frac{\left (5 a^2+20 a b+16 b^2\right ) \cos (e+f x)}{5 a^3 f \left (a+b \sec ^2(e+f x)\right )^{3/2}}+\frac{2 (5 a+4 b) \cos ^3(e+f x)}{15 a^2 f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac{\cos ^5(e+f x)}{5 a f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac{4 b \left (5 a^2+20 a b+16 b^2\right ) \sec (e+f x)}{15 a^4 f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac{\left (8 b \left (5 a^2+20 a b+16 b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\left (a+b x^2\right )^{3/2}} \, dx,x,\sec (e+f x)\right )}{15 a^4 f}\\ &=-\frac{\left (5 a^2+20 a b+16 b^2\right ) \cos (e+f x)}{5 a^3 f \left (a+b \sec ^2(e+f x)\right )^{3/2}}+\frac{2 (5 a+4 b) \cos ^3(e+f x)}{15 a^2 f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac{\cos ^5(e+f x)}{5 a f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac{4 b \left (5 a^2+20 a b+16 b^2\right ) \sec (e+f x)}{15 a^4 f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac{8 b \left (5 a^2+20 a b+16 b^2\right ) \sec (e+f x)}{15 a^5 f \sqrt{a+b \sec ^2(e+f x)}}\\ \end{align*}
Mathematica [A] time = 3.41962, size = 182, normalized size = 0.83 \[ -\frac{\sec ^5(e+f x) (a \cos (2 (e+f x))+a+2 b) \left (12 a^2 \left (7 a^2+64 a b+64 b^2\right ) \cos (4 (e+f x))+48 a \left (150 a^2 b+11 a^3+384 a b^2+256 b^3\right ) \cos (2 (e+f x))+22784 a^2 b^2-32 a^3 b \cos (6 (e+f x))+6400 a^3 b-16 a^4 \cos (6 (e+f x))+3 a^4 \cos (8 (e+f x))+425 a^4+32768 a b^3+16384 b^4\right )}{3840 a^5 f \left (a+b \sec ^2(e+f x)\right )^{5/2}} \]
Antiderivative was successfully verified.
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Maple [A] time = 2.01, size = 229, normalized size = 1.1 \begin{align*}{\frac{{a}^{2}\sqrt{4} \left ( a+b \right ) ^{7} \left ( b+a \left ( \cos \left ( fx+e \right ) \right ) ^{2} \right ) \left ( 3\, \left ( \cos \left ( fx+e \right ) \right ) ^{8}{a}^{4}-10\, \left ( \cos \left ( fx+e \right ) \right ) ^{6}{a}^{4}-8\, \left ( \cos \left ( fx+e \right ) \right ) ^{6}{a}^{3}b+15\, \left ( \cos \left ( fx+e \right ) \right ) ^{4}{a}^{4}+60\, \left ( \cos \left ( fx+e \right ) \right ) ^{4}{a}^{3}b+48\, \left ( \cos \left ( fx+e \right ) \right ) ^{4}{a}^{2}{b}^{2}+60\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}{a}^{3}b+240\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}{a}^{2}{b}^{2}+192\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}a{b}^{3}+40\,{a}^{2}{b}^{2}+160\,a{b}^{3}+128\,{b}^{4} \right ) }{30\,f \left ( \cos \left ( fx+e \right ) \right ) ^{5}} \left ( \sqrt{-ab}+a \right ) ^{-7} \left ( \sqrt{-ab}-a \right ) ^{-7} \left ({\frac{b+a \left ( \cos \left ( fx+e \right ) \right ) ^{2}}{ \left ( \cos \left ( fx+e \right ) \right ) ^{2}}} \right ) ^{-{\frac{5}{2}}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.01714, size = 451, normalized size = 2.06 \begin{align*} -\frac{\frac{15 \, \sqrt{a + \frac{b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{a^{3}} - \frac{10 \,{\left ({\left (a + \frac{b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac{3}{2}} \cos \left (f x + e\right )^{3} - 9 \, \sqrt{a + \frac{b}{\cos \left (f x + e\right )^{2}}} b \cos \left (f x + e\right )\right )}}{a^{4}} + \frac{3 \,{\left (a + \frac{b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac{5}{2}} \cos \left (f x + e\right )^{5} - 20 \,{\left (a + \frac{b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac{3}{2}} b \cos \left (f x + e\right )^{3} + 90 \, \sqrt{a + \frac{b}{\cos \left (f x + e\right )^{2}}} b^{2} \cos \left (f x + e\right )}{a^{5}} + \frac{5 \,{\left (6 \,{\left (a + \frac{b}{\cos \left (f x + e\right )^{2}}\right )} b \cos \left (f x + e\right )^{2} - b^{2}\right )}}{{\left (a + \frac{b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac{3}{2}} a^{3} \cos \left (f x + e\right )^{3}} + \frac{10 \,{\left (9 \,{\left (a + \frac{b}{\cos \left (f x + e\right )^{2}}\right )} b^{2} \cos \left (f x + e\right )^{2} - b^{3}\right )}}{{\left (a + \frac{b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac{3}{2}} a^{4} \cos \left (f x + e\right )^{3}} + \frac{5 \,{\left (12 \,{\left (a + \frac{b}{\cos \left (f x + e\right )^{2}}\right )} b^{3} \cos \left (f x + e\right )^{2} - b^{4}\right )}}{{\left (a + \frac{b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac{3}{2}} a^{5} \cos \left (f x + e\right )^{3}}}{15 \, f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.43799, size = 441, normalized size = 2.01 \begin{align*} -\frac{{\left (3 \, a^{4} \cos \left (f x + e\right )^{9} - 2 \,{\left (5 \, a^{4} + 4 \, a^{3} b\right )} \cos \left (f x + e\right )^{7} + 3 \,{\left (5 \, a^{4} + 20 \, a^{3} b + 16 \, a^{2} b^{2}\right )} \cos \left (f x + e\right )^{5} + 12 \,{\left (5 \, a^{3} b + 20 \, a^{2} b^{2} + 16 \, a b^{3}\right )} \cos \left (f x + e\right )^{3} + 8 \,{\left (5 \, a^{2} b^{2} + 20 \, a b^{3} + 16 \, b^{4}\right )} \cos \left (f x + e\right )\right )} \sqrt{\frac{a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{15 \,{\left (a^{7} f \cos \left (f x + e\right )^{4} + 2 \, a^{6} b f \cos \left (f x + e\right )^{2} + a^{5} b^{2} f\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sin \left (f x + e\right )^{5}}{{\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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