Optimal. Leaf size=149 \[ \frac{\sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)+b}}\right )}{f}-\frac{\cot ^5(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}{5 f (a+b)}-\frac{2 (5 a+4 b) \cot ^3(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}{15 f (a+b)^2}-\frac{\cot (e+f x) \sqrt{a+b \tan ^2(e+f x)+b}}{f} \]
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Rubi [A] time = 0.14012, antiderivative size = 149, 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 = {4132, 462, 451, 277, 217, 206} \[ \frac{\sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)+b}}\right )}{f}-\frac{\cot ^5(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}{5 f (a+b)}-\frac{2 (5 a+4 b) \cot ^3(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}{15 f (a+b)^2}-\frac{\cot (e+f x) \sqrt{a+b \tan ^2(e+f x)+b}}{f} \]
Antiderivative was successfully verified.
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Rule 4132
Rule 462
Rule 451
Rule 277
Rule 217
Rule 206
Rubi steps
\begin{align*} \int \csc ^6(e+f x) \sqrt{a+b \sec ^2(e+f x)} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (1+x^2\right )^2 \sqrt{a+b+b x^2}}{x^6} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{\cot ^5(e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}{5 (a+b) f}+\frac{\operatorname{Subst}\left (\int \frac{\sqrt{a+b+b x^2} \left (2 (5 a+4 b)+5 (a+b) x^2\right )}{x^4} \, dx,x,\tan (e+f x)\right )}{5 (a+b) f}\\ &=-\frac{2 (5 a+4 b) \cot ^3(e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}{15 (a+b)^2 f}-\frac{\cot ^5(e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}{5 (a+b) f}+\frac{\operatorname{Subst}\left (\int \frac{\sqrt{a+b+b x^2}}{x^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{\cot (e+f x) \sqrt{a+b+b \tan ^2(e+f x)}}{f}-\frac{2 (5 a+4 b) \cot ^3(e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}{15 (a+b)^2 f}-\frac{\cot ^5(e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}{5 (a+b) f}+\frac{b \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{\cot (e+f x) \sqrt{a+b+b \tan ^2(e+f x)}}{f}-\frac{2 (5 a+4 b) \cot ^3(e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}{15 (a+b)^2 f}-\frac{\cot ^5(e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}{5 (a+b) f}+\frac{b \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{\tan (e+f x)}{\sqrt{a+b+b \tan ^2(e+f x)}}\right )}{f}\\ &=\frac{\sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b+b \tan ^2(e+f x)}}\right )}{f}-\frac{\cot (e+f x) \sqrt{a+b+b \tan ^2(e+f x)}}{f}-\frac{2 (5 a+4 b) \cot ^3(e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}{15 (a+b)^2 f}-\frac{\cot ^5(e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}{5 (a+b) f}\\ \end{align*}
Mathematica [C] time = 9.55979, size = 941, normalized size = 6.32 \[ \frac{\sqrt{2} \csc ^5(e+f x) \sec (e+f x) \sqrt{b \sec ^2(e+f x)+a} \left (1-\frac{a \sin ^2(e+f x)}{a+b}\right ) \left (\frac{24 a^2 b^2 \text{Hypergeometric2F1}\left (2,2,\frac{3}{2},-\frac{b \tan ^2(e+f x)}{a+b}\right ) \tan ^2(e+f x) \sin ^6(e+f x)}{(a+b)^3}+\frac{8 a^2 b^2 \text{HypergeometricPFQ}\left (\{2,2,2\},\left \{1,\frac{3}{2}\right \},-\frac{b \tan ^2(e+f x)}{a+b}\right ) \tan ^2(e+f x) \sin ^6(e+f x)}{(a+b)^3}+\frac{24 a^2 b \text{Hypergeometric2F1}\left (2,2,\frac{3}{2},-\frac{b \tan ^2(e+f x)}{a+b}\right ) \sin ^6(e+f x)}{(a+b)^2}+\frac{8 a^2 b \text{HypergeometricPFQ}\left (\{2,2,2\},\left \{1,\frac{3}{2}\right \},-\frac{b \tan ^2(e+f x)}{a+b}\right ) \sin ^6(e+f x)}{(a+b)^2}+\frac{8 a^2 b \sin ^{-1}\left (\sqrt{-\frac{b \tan ^2(e+f x)}{a+b}}\right ) \sin ^6(e+f x)}{(a+b)^2 \sqrt{-\frac{b \sec ^2(e+f x) \left (-a \sin ^2(e+f x)+a+b\right ) \tan ^2(e+f x)}{(a+b)^2}}}-\frac{8 a^2 \cos ^2(e+f x) \sin ^4(e+f x)}{a+b}-\frac{8 a b^2 \text{Hypergeometric2F1}\left (2,2,\frac{3}{2},-\frac{b \tan ^2(e+f x)}{a+b}\right ) \tan ^2(e+f x) \sin ^4(e+f x)}{(a+b)^2}-\frac{16 a b^2 \text{HypergeometricPFQ}\left (\{2,2,2\},\left \{1,\frac{3}{2}\right \},-\frac{b \tan ^2(e+f x)}{a+b}\right ) \tan ^2(e+f x) \sin ^4(e+f x)}{(a+b)^2}-\frac{8 a b \text{Hypergeometric2F1}\left (2,2,\frac{3}{2},-\frac{b \tan ^2(e+f x)}{a+b}\right ) \sin ^4(e+f x)}{a+b}-\frac{16 a b \text{HypergeometricPFQ}\left (\{2,2,2\},\left \{1,\frac{3}{2}\right \},-\frac{b \tan ^2(e+f x)}{a+b}\right ) \sin ^4(e+f x)}{a+b}-4 a \cos ^2(e+f x) \sin ^2(e+f x)-\frac{16 b^2 \text{Hypergeometric2F1}\left (2,2,\frac{3}{2},-\frac{b \tan ^2(e+f x)}{a+b}\right ) \tan ^2(e+f x) \sin ^2(e+f x)}{a+b}+\frac{8 b^2 \text{HypergeometricPFQ}\left (\{2,2,2\},\left \{1,\frac{3}{2}\right \},-\frac{b \tan ^2(e+f x)}{a+b}\right ) \tan ^2(e+f x) \sin ^2(e+f x)}{a+b}-16 b \text{Hypergeometric2F1}\left (2,2,\frac{3}{2},-\frac{b \tan ^2(e+f x)}{a+b}\right ) \sin ^2(e+f x)+8 b \text{HypergeometricPFQ}\left (\{2,2,2\},\left \{1,\frac{3}{2}\right \},-\frac{b \tan ^2(e+f x)}{a+b}\right ) \sin ^2(e+f x)-\frac{4 a \sin ^{-1}\left (\sqrt{-\frac{b \tan ^2(e+f x)}{a+b}}\right ) \cos ^2(e+f x) \sqrt{-\frac{b \tan ^2(e+f x)}{a+b}} \sin ^2(e+f x)}{\sqrt{\frac{b \sec ^2(e+f x)+a}{a+b}}}+\frac{3 b \sin ^{-1}\left (\sqrt{-\frac{b \tan ^2(e+f x)}{a+b}}\right ) \sin ^2(e+f x)}{\sqrt{-\frac{b \sec ^2(e+f x) \left (-a \sin ^2(e+f x)+a+b\right ) \tan ^2(e+f x)}{(a+b)^2}}}-3 (a+b) \cos ^2(e+f x)\right )}{15 f \sqrt{\cos (2 e+2 f x) a+a+2 b} \sqrt{-a \sin ^2(e+f x)+a+b}} \]
Warning: Unable to verify antiderivative.
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Maple [C] time = 0.58, size = 8587, normalized size = 57.6 \begin{align*} \text{output too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 4.45319, size = 1639, normalized size = 11. \begin{align*} \text{result too large to display} \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 \sqrt{b \sec \left (f x + e\right )^{2} + a} \csc \left (f x + e\right )^{6}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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