Optimal. Leaf size=184 \[ \frac{a \left (50 a^2 b^2+4 a^4+19 b^4\right ) \tan (d+e x)}{6 e}+\frac{b \left (56 a^2 b^2+19 a^4+8 b^4\right ) \tanh ^{-1}(\sin (d+e x))}{8 e}+\frac{a^2 b \left (41 a^2+26 b^2\right ) \tan (d+e x) \sec (d+e x)}{24 e}+\frac{\left (4 a^2+7 b^2\right ) \tan (d+e x) \left (a^2 \sec (d+e x)+a b\right )^2}{12 a e}+\frac{b \tan (d+e x) \left (a^2 \sec (d+e x)+a b\right )^3}{4 a^2 e}+a b^4 x \]
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Rubi [A] time = 0.429861, antiderivative size = 184, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 39, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.18, Rules used = {4172, 3918, 4056, 4048, 3770, 3767, 8} \[ \frac{a \left (50 a^2 b^2+4 a^4+19 b^4\right ) \tan (d+e x)}{6 e}+\frac{b \left (56 a^2 b^2+19 a^4+8 b^4\right ) \tanh ^{-1}(\sin (d+e x))}{8 e}+\frac{a^2 b \left (41 a^2+26 b^2\right ) \tan (d+e x) \sec (d+e x)}{24 e}+\frac{\left (4 a^2+7 b^2\right ) \tan (d+e x) \left (a^2 \sec (d+e x)+a b\right )^2}{12 a e}+\frac{b \tan (d+e x) \left (a^2 \sec (d+e x)+a b\right )^3}{4 a^2 e}+a b^4 x \]
Antiderivative was successfully verified.
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Rule 4172
Rule 3918
Rule 4056
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 )^2 \, dx &=\frac{\int \left (2 a b+2 a^2 \sec (d+e x)\right )^4 (a+b \sec (d+e x)) \, dx}{16 a^4}\\ &=\frac{b \left (a b+a^2 \sec (d+e x)\right )^3 \tan (d+e x)}{4 a^2 e}+\frac{\int \left (2 a b+2 a^2 \sec (d+e x)\right )^2 \left (16 a^3 b^2+4 a^2 b \left (11 a^2+4 b^2\right ) \sec (d+e x)+4 a^3 \left (4 a^2+7 b^2\right ) \sec ^2(d+e x)\right ) \, dx}{64 a^4}\\ &=\frac{\left (4 a^2+7 b^2\right ) \left (a b+a^2 \sec (d+e x)\right )^2 \tan (d+e x)}{12 a e}+\frac{b \left (a b+a^2 \sec (d+e x)\right )^3 \tan (d+e x)}{4 a^2 e}+\frac{\int \left (2 a b+2 a^2 \sec (d+e x)\right ) \left (96 a^4 b^3+8 a^3 \left (8 a^4+59 a^2 b^2+12 b^4\right ) \sec (d+e x)+8 a^4 b \left (41 a^2+26 b^2\right ) \sec ^2(d+e x)\right ) \, dx}{192 a^4}\\ &=\frac{a^2 b \left (41 a^2+26 b^2\right ) \sec (d+e x) \tan (d+e x)}{24 e}+\frac{\left (4 a^2+7 b^2\right ) \left (a b+a^2 \sec (d+e x)\right )^2 \tan (d+e x)}{12 a e}+\frac{b \left (a b+a^2 \sec (d+e x)\right )^3 \tan (d+e x)}{4 a^2 e}+\frac{\int \left (384 a^5 b^4+48 a^4 b \left (19 a^4+56 a^2 b^2+8 b^4\right ) \sec (d+e x)+64 a^5 \left (4 a^4+50 a^2 b^2+19 b^4\right ) \sec ^2(d+e x)\right ) \, dx}{384 a^4}\\ &=a b^4 x+\frac{a^2 b \left (41 a^2+26 b^2\right ) \sec (d+e x) \tan (d+e x)}{24 e}+\frac{\left (4 a^2+7 b^2\right ) \left (a b+a^2 \sec (d+e x)\right )^2 \tan (d+e x)}{12 a e}+\frac{b \left (a b+a^2 \sec (d+e x)\right )^3 \tan (d+e x)}{4 a^2 e}+\frac{1}{8} \left (b \left (19 a^4+56 a^2 b^2+8 b^4\right )\right ) \int \sec (d+e x) \, dx+\frac{1}{6} \left (a \left (4 a^4+50 a^2 b^2+19 b^4\right )\right ) \int \sec ^2(d+e x) \, dx\\ &=a b^4 x+\frac{b \left (19 a^4+56 a^2 b^2+8 b^4\right ) \tanh ^{-1}(\sin (d+e x))}{8 e}+\frac{a^2 b \left (41 a^2+26 b^2\right ) \sec (d+e x) \tan (d+e x)}{24 e}+\frac{\left (4 a^2+7 b^2\right ) \left (a b+a^2 \sec (d+e x)\right )^2 \tan (d+e x)}{12 a e}+\frac{b \left (a b+a^2 \sec (d+e x)\right )^3 \tan (d+e x)}{4 a^2 e}-\frac{\left (a \left (4 a^4+50 a^2 b^2+19 b^4\right )\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (d+e x))}{6 e}\\ &=a b^4 x+\frac{b \left (19 a^4+56 a^2 b^2+8 b^4\right ) \tanh ^{-1}(\sin (d+e x))}{8 e}+\frac{a \left (4 a^4+50 a^2 b^2+19 b^4\right ) \tan (d+e x)}{6 e}+\frac{a^2 b \left (41 a^2+26 b^2\right ) \sec (d+e x) \tan (d+e x)}{24 e}+\frac{\left (4 a^2+7 b^2\right ) \left (a b+a^2 \sec (d+e x)\right )^2 \tan (d+e x)}{12 a e}+\frac{b \left (a b+a^2 \sec (d+e x)\right )^3 \tan (d+e x)}{4 a^2 e}\\ \end{align*}
Mathematica [A] time = 0.80197, size = 130, normalized size = 0.71 \[ \frac{8 a^3 \left (a^2+4 b^2\right ) \tan ^3(d+e x)+3 b \left (56 a^2 b^2+19 a^4+8 b^4\right ) \tanh ^{-1}(\sin (d+e x))+3 a \tan (d+e x) \left (a b \left (19 a^2+24 b^2\right ) \sec (d+e x)+8 \left (10 a^2 b^2+a^4+4 b^4\right )+2 a^3 b \sec ^3(d+e x)\right )+24 a b^4 e x}{24 e} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.073, size = 246, normalized size = 1.3 \begin{align*} a{b}^{4}x+{\frac{a{b}^{4}d}{e}}+7\,{\frac{{a}^{2}{b}^{3}\ln \left ( \sec \left ( ex+d \right ) +\tan \left ( ex+d \right ) \right ) }{e}}+{\frac{26\,{a}^{3}{b}^{2}\tan \left ( ex+d \right ) }{3\,e}}+{\frac{19\,{a}^{4}b\sec \left ( ex+d \right ) \tan \left ( ex+d \right ) }{8\,e}}+{\frac{19\,{a}^{4}b\ln \left ( \sec \left ( ex+d \right ) +\tan \left ( ex+d \right ) \right ) }{8\,e}}+{\frac{2\,{a}^{5}\tan \left ( ex+d \right ) }{3\,e}}+{\frac{{a}^{5}\tan \left ( ex+d \right ) \left ( \sec \left ( ex+d \right ) \right ) ^{2}}{3\,e}}+{\frac{{b}^{5}\ln \left ( \sec \left ( ex+d \right ) +\tan \left ( ex+d \right ) \right ) }{e}}+4\,{\frac{a{b}^{4}\tan \left ( ex+d \right ) }{e}}+3\,{\frac{{a}^{2}{b}^{3}\sec \left ( ex+d \right ) \tan \left ( ex+d \right ) }{e}}+{\frac{4\,{a}^{3}{b}^{2}\tan \left ( ex+d \right ) \left ( \sec \left ( ex+d \right ) \right ) ^{2}}{3\,e}}+{\frac{{a}^{4}b\tan \left ( ex+d \right ) \left ( \sec \left ( ex+d \right ) \right ) ^{3}}{4\,e}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.03341, size = 404, normalized size = 2.2 \begin{align*} \frac{16 \,{\left (\tan \left (e x + d\right )^{3} + 3 \, \tan \left (e x + d\right )\right )} a^{5} + 64 \,{\left (\tan \left (e x + d\right )^{3} + 3 \, \tan \left (e x + d\right )\right )} a^{3} b^{2} + 48 \,{\left (e x + d\right )} a b^{4} - 3 \, a^{4} b{\left (\frac{2 \,{\left (3 \, \sin \left (e x + d\right )^{3} - 5 \, \sin \left (e x + d\right )\right )}}{\sin \left (e x + d\right )^{4} - 2 \, \sin \left (e x + d\right )^{2} + 1} - 3 \, \log \left (\sin \left (e x + d\right ) + 1\right ) + 3 \, \log \left (\sin \left (e x + d\right ) - 1\right )\right )} - 48 \, a^{4} b{\left (\frac{2 \, \sin \left (e x + d\right )}{\sin \left (e x + d\right )^{2} - 1} - \log \left (\sin \left (e x + d\right ) + 1\right ) + \log \left (\sin \left (e x + d\right ) - 1\right )\right )} - 72 \, a^{2} b^{3}{\left (\frac{2 \, \sin \left (e x + d\right )}{\sin \left (e x + d\right )^{2} - 1} - \log \left (\sin \left (e x + d\right ) + 1\right ) + \log \left (\sin \left (e x + d\right ) - 1\right )\right )} + 192 \, a^{2} b^{3} \log \left (\sec \left (e x + d\right ) + \tan \left (e x + d\right )\right ) + 48 \, b^{5} \log \left (\sec \left (e x + d\right ) + \tan \left (e x + d\right )\right ) + 288 \, a^{3} b^{2} \tan \left (e x + d\right ) + 192 \, a b^{4} \tan \left (e x + d\right )}{48 \, e} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.97682, size = 481, normalized size = 2.61 \begin{align*} \frac{48 \, a b^{4} e x \cos \left (e x + d\right )^{4} + 3 \,{\left (19 \, a^{4} b + 56 \, a^{2} b^{3} + 8 \, b^{5}\right )} \cos \left (e x + d\right )^{4} \log \left (\sin \left (e x + d\right ) + 1\right ) - 3 \,{\left (19 \, a^{4} b + 56 \, a^{2} b^{3} + 8 \, b^{5}\right )} \cos \left (e x + d\right )^{4} \log \left (-\sin \left (e x + d\right ) + 1\right ) + 2 \,{\left (6 \, a^{4} b + 16 \,{\left (a^{5} + 13 \, a^{3} b^{2} + 6 \, a b^{4}\right )} \cos \left (e x + d\right )^{3} + 3 \,{\left (19 \, a^{4} b + 24 \, a^{2} b^{3}\right )} \cos \left (e x + d\right )^{2} + 8 \,{\left (a^{5} + 4 \, a^{3} b^{2}\right )} \cos \left (e x + d\right )\right )} \sin \left (e x + d\right )}{48 \, e \cos \left (e x + d\right )^{4}} \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 (a \sec{\left (d + e x \right )} + b\right )^{4}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [B] time = 1.28002, size = 635, normalized size = 3.45 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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