Optimal. Leaf size=359 \[ \frac{a^4 b^3 x \left (a^2 \sec ^2(d+e x)+2 a b \sec (d+e x)+b^2\right )^{3/2}}{\left (a^2 \sec (d+e x)+a b\right )^3}+\frac{a^5 \left (3 a^2+5 b^2\right ) \tan (d+e x) \sec (d+e x) \left (a^2 \sec ^2(d+e x)+2 a b \sec (d+e x)+b^2\right )^{3/2}}{6 e \left (a^2 \sec (d+e x)+a b\right )^3}+\frac{a^4 b \left (11 a^2+8 b^2\right ) \tan (d+e x) \left (a^2 \sec ^2(d+e x)+2 a b \sec (d+e x)+b^2\right )^{3/2}}{3 e \left (a^2 \sec (d+e x)+a b\right )^3}+\frac{b \tan (d+e x) \left (a^2 b+a^3 \sec (d+e x)\right )^2 \left (a^2 \sec ^2(d+e x)+2 a b \sec (d+e x)+b^2\right )^{3/2}}{3 e \left (a^2 \sec (d+e x)+a b\right )^3}+\frac{\left (9 a^2 b^2+a^4+2 b^4\right ) \tanh ^{-1}(\sin (d+e x)) \left (a^2 \sec ^2(d+e x)+2 a b \sec (d+e x)+b^2\right )^{3/2}}{2 e (a \sec (d+e x)+b)^3} \]
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Rubi [A] time = 0.287014, antiderivative size = 359, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.146, Rules used = {4174, 3918, 4048, 3770, 3767, 8} \[ \frac{a^4 b^3 x \left (a^2 \sec ^2(d+e x)+2 a b \sec (d+e x)+b^2\right )^{3/2}}{\left (a^2 \sec (d+e x)+a b\right )^3}+\frac{a^5 \left (3 a^2+5 b^2\right ) \tan (d+e x) \sec (d+e x) \left (a^2 \sec ^2(d+e x)+2 a b \sec (d+e x)+b^2\right )^{3/2}}{6 e \left (a^2 \sec (d+e x)+a b\right )^3}+\frac{a^4 b \left (11 a^2+8 b^2\right ) \tan (d+e x) \left (a^2 \sec ^2(d+e x)+2 a b \sec (d+e x)+b^2\right )^{3/2}}{3 e \left (a^2 \sec (d+e x)+a b\right )^3}+\frac{b \tan (d+e x) \left (a^2 b+a^3 \sec (d+e x)\right )^2 \left (a^2 \sec ^2(d+e x)+2 a b \sec (d+e x)+b^2\right )^{3/2}}{3 e \left (a^2 \sec (d+e x)+a b\right )^3}+\frac{\left (9 a^2 b^2+a^4+2 b^4\right ) \tanh ^{-1}(\sin (d+e x)) \left (a^2 \sec ^2(d+e x)+2 a b \sec (d+e x)+b^2\right )^{3/2}}{2 e (a \sec (d+e x)+b)^3} \]
Antiderivative was successfully verified.
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Rule 4174
Rule 3918
Rule 4048
Rule 3770
Rule 3767
Rule 8
Rubi steps
\begin{align*} \int (a+b \sec (d+e x)) \left (b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x)\right )^{3/2} \, dx &=\frac{\left (b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x)\right )^{3/2} \int \left (2 a b+2 a^2 \sec (d+e x)\right )^3 (a+b \sec (d+e x)) \, dx}{\left (2 a b+2 a^2 \sec (d+e x)\right )^3}\\ &=\frac{b \left (a^2 b+a^3 \sec (d+e x)\right )^2 \left (b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x)\right )^{3/2} \tan (d+e x)}{3 e \left (a b+a^2 \sec (d+e x)\right )^3}+\frac{\left (b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x)\right )^{3/2} \int \left (2 a b+2 a^2 \sec (d+e x)\right ) \left (12 a^3 b^2+4 a^2 b \left (8 a^2+3 b^2\right ) \sec (d+e x)+4 a^3 \left (3 a^2+5 b^2\right ) \sec ^2(d+e x)\right ) \, dx}{3 \left (2 a b+2 a^2 \sec (d+e x)\right )^3}\\ &=\frac{a^5 \left (3 a^2+5 b^2\right ) \sec (d+e x) \left (b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x)\right )^{3/2} \tan (d+e x)}{6 e \left (a b+a^2 \sec (d+e x)\right )^3}+\frac{b \left (a^2 b+a^3 \sec (d+e x)\right )^2 \left (b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x)\right )^{3/2} \tan (d+e x)}{3 e \left (a b+a^2 \sec (d+e x)\right )^3}+\frac{\left (b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x)\right )^{3/2} \int \left (48 a^4 b^3+24 a^3 \left (a^4+9 a^2 b^2+2 b^4\right ) \sec (d+e x)+16 a^4 b \left (11 a^2+8 b^2\right ) \sec ^2(d+e x)\right ) \, dx}{6 \left (2 a b+2 a^2 \sec (d+e x)\right )^3}\\ &=\frac{a^4 b^3 x \left (b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x)\right )^{3/2}}{\left (a b+a^2 \sec (d+e x)\right )^3}+\frac{a^5 \left (3 a^2+5 b^2\right ) \sec (d+e x) \left (b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x)\right )^{3/2} \tan (d+e x)}{6 e \left (a b+a^2 \sec (d+e x)\right )^3}+\frac{b \left (a^2 b+a^3 \sec (d+e x)\right )^2 \left (b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x)\right )^{3/2} \tan (d+e x)}{3 e \left (a b+a^2 \sec (d+e x)\right )^3}+\frac{\left (8 a^4 b \left (11 a^2+8 b^2\right ) \left (b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x)\right )^{3/2}\right ) \int \sec ^2(d+e x) \, dx}{3 \left (2 a b+2 a^2 \sec (d+e x)\right )^3}+\frac{\left (4 a^3 \left (a^4+9 a^2 b^2+2 b^4\right ) \left (b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x)\right )^{3/2}\right ) \int \sec (d+e x) \, dx}{\left (2 a b+2 a^2 \sec (d+e x)\right )^3}\\ &=\frac{\left (a^4+9 a^2 b^2+2 b^4\right ) \tanh ^{-1}(\sin (d+e x)) \left (b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x)\right )^{3/2}}{2 e (b+a \sec (d+e x))^3}+\frac{a^4 b^3 x \left (b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x)\right )^{3/2}}{\left (a b+a^2 \sec (d+e x)\right )^3}+\frac{a^5 \left (3 a^2+5 b^2\right ) \sec (d+e x) \left (b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x)\right )^{3/2} \tan (d+e x)}{6 e \left (a b+a^2 \sec (d+e x)\right )^3}+\frac{b \left (a^2 b+a^3 \sec (d+e x)\right )^2 \left (b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x)\right )^{3/2} \tan (d+e x)}{3 e \left (a b+a^2 \sec (d+e x)\right )^3}-\frac{\left (8 a^4 b \left (11 a^2+8 b^2\right ) \left (b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x)\right )^{3/2}\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (d+e x))}{3 e \left (2 a b+2 a^2 \sec (d+e x)\right )^3}\\ &=\frac{\left (a^4+9 a^2 b^2+2 b^4\right ) \tanh ^{-1}(\sin (d+e x)) \left (b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x)\right )^{3/2}}{2 e (b+a \sec (d+e x))^3}+\frac{a^4 b^3 x \left (b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x)\right )^{3/2}}{\left (a b+a^2 \sec (d+e x)\right )^3}+\frac{a^4 b \left (11 a^2+8 b^2\right ) \left (b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x)\right )^{3/2} \tan (d+e x)}{3 e \left (a b+a^2 \sec (d+e x)\right )^3}+\frac{a^5 \left (3 a^2+5 b^2\right ) \sec (d+e x) \left (b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x)\right )^{3/2} \tan (d+e x)}{6 e \left (a b+a^2 \sec (d+e x)\right )^3}+\frac{b \left (a^2 b+a^3 \sec (d+e x)\right )^2 \left (b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x)\right )^{3/2} \tan (d+e x)}{3 e \left (a b+a^2 \sec (d+e x)\right )^3}\\ \end{align*}
Mathematica [A] time = 0.837774, size = 128, normalized size = 0.36 \[ \frac{\cos (d+e x) \sqrt{(a \sec (d+e x)+b)^2} \left (3 \left (9 a^2 b^2+a^4+2 b^4\right ) \tanh ^{-1}(\sin (d+e x))+3 a \tan (d+e x) \left (a \left (a^2+3 b^2\right ) \sec (d+e x)+8 a^2 b+6 b^3\right )+2 a^3 b \tan ^3(d+e x)+6 a b^3 e x\right )}{6 e (a+b \cos (d+e x))} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.264, size = 387, normalized size = 1.1 \begin{align*}{\frac{1}{6\,e \left ( b\cos \left ( ex+d \right ) +a \right ) ^{3}} \left ( 3\,\ln \left ({\frac{\sin \left ( ex+d \right ) +1-\cos \left ( ex+d \right ) }{\sin \left ( ex+d \right ) }} \right ) \left ( \cos \left ( ex+d \right ) \right ) ^{3}{a}^{4}+27\,\ln \left ({\frac{\sin \left ( ex+d \right ) +1-\cos \left ( ex+d \right ) }{\sin \left ( ex+d \right ) }} \right ) \left ( \cos \left ( ex+d \right ) \right ) ^{3}{a}^{2}{b}^{2}+6\,\ln \left ({\frac{\sin \left ( ex+d \right ) +1-\cos \left ( ex+d \right ) }{\sin \left ( ex+d \right ) }} \right ) \left ( \cos \left ( ex+d \right ) \right ) ^{3}{b}^{4}-3\,\ln \left ( -{\frac{\cos \left ( ex+d \right ) -1+\sin \left ( ex+d \right ) }{\sin \left ( ex+d \right ) }} \right ) \left ( \cos \left ( ex+d \right ) \right ) ^{3}{a}^{4}-27\,\ln \left ( -{\frac{\cos \left ( ex+d \right ) -1+\sin \left ( ex+d \right ) }{\sin \left ( ex+d \right ) }} \right ) \left ( \cos \left ( ex+d \right ) \right ) ^{3}{a}^{2}{b}^{2}-6\,\ln \left ( -{\frac{\cos \left ( ex+d \right ) -1+\sin \left ( ex+d \right ) }{\sin \left ( ex+d \right ) }} \right ) \left ( \cos \left ( ex+d \right ) \right ) ^{3}{b}^{4}+6\, \left ( \cos \left ( ex+d \right ) \right ) ^{3} \left ( ex+d \right ) a{b}^{3}+22\,\sin \left ( ex+d \right ) \left ( \cos \left ( ex+d \right ) \right ) ^{2}{a}^{3}b+18\,\sin \left ( ex+d \right ) \left ( \cos \left ( ex+d \right ) \right ) ^{2}a{b}^{3}+3\,\sin \left ( ex+d \right ) \cos \left ( ex+d \right ){a}^{4}+9\,\sin \left ( ex+d \right ) \cos \left ( ex+d \right ){a}^{2}{b}^{2}+2\,{a}^{3}b\sin \left ( ex+d \right ) \right ) \left ({\frac{ \left ( b\cos \left ( ex+d \right ) +a \right ) ^{2}}{ \left ( \cos \left ( ex+d \right ) \right ) ^{2}}} \right ) ^{{\frac{3}{2}}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.66739, size = 594, normalized size = 1.65 \begin{align*} \frac{3 \,{\left (4 \, b^{3} \arctan \left (\frac{\sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1}\right ) +{\left (a^{3} + 6 \, a b^{2}\right )} \log \left (\frac{\sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1} + 1\right ) -{\left (a^{3} + 6 \, a b^{2}\right )} \log \left (\frac{\sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1} - 1\right ) - \frac{2 \,{\left (\frac{{\left (a^{3} + 6 \, a^{2} b\right )} \sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1} + \frac{{\left (a^{3} - 6 \, a^{2} b\right )} \sin \left (e x + d\right )^{3}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{3}}\right )}}{\frac{2 \, \sin \left (e x + d\right )^{2}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{2}} - \frac{\sin \left (e x + d\right )^{4}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{4}} - 1}\right )} a +{\left (3 \,{\left (3 \, a^{2} b + 2 \, b^{3}\right )} \log \left (\frac{\sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1} + 1\right ) - 3 \,{\left (3 \, a^{2} b + 2 \, b^{3}\right )} \log \left (\frac{\sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1} - 1\right ) - \frac{2 \,{\left (\frac{3 \,{\left (2 \, a^{3} + 3 \, a^{2} b + 6 \, a b^{2}\right )} \sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1} - \frac{4 \,{\left (a^{3} + 9 \, a b^{2}\right )} \sin \left (e x + d\right )^{3}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{3}} + \frac{3 \,{\left (2 \, a^{3} - 3 \, a^{2} b + 6 \, a b^{2}\right )} \sin \left (e x + d\right )^{5}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{5}}\right )}}{\frac{3 \, \sin \left (e x + d\right )^{2}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{2}} - \frac{3 \, \sin \left (e x + d\right )^{4}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{4}} + \frac{\sin \left (e x + d\right )^{6}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{6}} - 1}\right )} b}{6 \, e} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 2.20469, size = 394, normalized size = 1.1 \begin{align*} \frac{12 \, a b^{3} e x \cos \left (e x + d\right )^{3} + 3 \,{\left (a^{4} + 9 \, a^{2} b^{2} + 2 \, b^{4}\right )} \cos \left (e x + d\right )^{3} \log \left (\sin \left (e x + d\right ) + 1\right ) - 3 \,{\left (a^{4} + 9 \, a^{2} b^{2} + 2 \, b^{4}\right )} \cos \left (e x + d\right )^{3} \log \left (-\sin \left (e x + d\right ) + 1\right ) + 2 \,{\left (2 \, a^{3} b + 2 \,{\left (11 \, a^{3} b + 9 \, a b^{3}\right )} \cos \left (e x + d\right )^{2} + 3 \,{\left (a^{4} + 3 \, a^{2} b^{2}\right )} \cos \left (e x + d\right )\right )} \sin \left (e x + d\right )}{12 \, e \cos \left (e x + d\right )^{3}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a + b \sec{\left (d + e x \right )}\right ) \left (\left (a \sec{\left (d + e x \right )} + b\right )^{2}\right )^{\frac{3}{2}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.66667, size = 880, normalized size = 2.45 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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