Optimal. Leaf size=184 \[ \frac{a \left (2 a^2+3 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{d (a-b)^{7/2} (a+b)^{7/2}}-\frac{b \left (11 a^2+4 b^2\right ) \tan (c+d x)}{6 d \left (a^2-b^2\right )^3 (a+b \sec (c+d x))}-\frac{5 a b \tan (c+d x)}{6 d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))^2}-\frac{b \tan (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^3} \]
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Rubi [A] time = 0.307289, antiderivative size = 184, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.316, Rules used = {3833, 4003, 12, 3831, 2659, 208} \[ \frac{a \left (2 a^2+3 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{d (a-b)^{7/2} (a+b)^{7/2}}-\frac{b \left (11 a^2+4 b^2\right ) \tan (c+d x)}{6 d \left (a^2-b^2\right )^3 (a+b \sec (c+d x))}-\frac{5 a b \tan (c+d x)}{6 d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))^2}-\frac{b \tan (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^3} \]
Antiderivative was successfully verified.
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Rule 3833
Rule 4003
Rule 12
Rule 3831
Rule 2659
Rule 208
Rubi steps
\begin{align*} \int \frac{\sec (c+d x)}{(a+b \sec (c+d x))^4} \, dx &=-\frac{b \tan (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}-\frac{\int \frac{\sec (c+d x) (-3 a+2 b \sec (c+d x))}{(a+b \sec (c+d x))^3} \, dx}{3 \left (a^2-b^2\right )}\\ &=-\frac{b \tan (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}-\frac{5 a b \tan (c+d x)}{6 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}+\frac{\int \frac{\sec (c+d x) \left (2 \left (3 a^2+2 b^2\right )-5 a b \sec (c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx}{6 \left (a^2-b^2\right )^2}\\ &=-\frac{b \tan (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}-\frac{5 a b \tan (c+d x)}{6 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}-\frac{b \left (11 a^2+4 b^2\right ) \tan (c+d x)}{6 \left (a^2-b^2\right )^3 d (a+b \sec (c+d x))}-\frac{\int -\frac{3 a \left (2 a^2+3 b^2\right ) \sec (c+d x)}{a+b \sec (c+d x)} \, dx}{6 \left (a^2-b^2\right )^3}\\ &=-\frac{b \tan (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}-\frac{5 a b \tan (c+d x)}{6 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}-\frac{b \left (11 a^2+4 b^2\right ) \tan (c+d x)}{6 \left (a^2-b^2\right )^3 d (a+b \sec (c+d x))}+\frac{\left (a \left (2 a^2+3 b^2\right )\right ) \int \frac{\sec (c+d x)}{a+b \sec (c+d x)} \, dx}{2 \left (a^2-b^2\right )^3}\\ &=-\frac{b \tan (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}-\frac{5 a b \tan (c+d x)}{6 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}-\frac{b \left (11 a^2+4 b^2\right ) \tan (c+d x)}{6 \left (a^2-b^2\right )^3 d (a+b \sec (c+d x))}+\frac{\left (a \left (2 a^2+3 b^2\right )\right ) \int \frac{1}{1+\frac{a \cos (c+d x)}{b}} \, dx}{2 b \left (a^2-b^2\right )^3}\\ &=-\frac{b \tan (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}-\frac{5 a b \tan (c+d x)}{6 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}-\frac{b \left (11 a^2+4 b^2\right ) \tan (c+d x)}{6 \left (a^2-b^2\right )^3 d (a+b \sec (c+d x))}+\frac{\left (a \left (2 a^2+3 b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{1+\frac{a}{b}+\left (1-\frac{a}{b}\right ) x^2} \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{b \left (a^2-b^2\right )^3 d}\\ &=\frac{a \left (2 a^2+3 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{(a-b)^{7/2} (a+b)^{7/2} d}-\frac{b \tan (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}-\frac{5 a b \tan (c+d x)}{6 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}-\frac{b \left (11 a^2+4 b^2\right ) \tan (c+d x)}{6 \left (a^2-b^2\right )^3 d (a+b \sec (c+d x))}\\ \end{align*}
Mathematica [A] time = 1.10894, size = 163, normalized size = 0.89 \[ -\frac{\frac{b \sin (c+d x) \left (6 a b \left (9 a^2+b^2\right ) \cos (c+d x)+\left (-5 a^2 b^2+18 a^4+2 b^4\right ) \cos (2 (c+d x))+17 a^2 b^2+18 a^4+10 b^4\right )}{(a \cos (c+d x)+b)^3}+\frac{12 a \left (2 a^2+3 b^2\right ) \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}}{12 d (a-b)^3 (a+b)^3} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.061, size = 284, normalized size = 1.5 \begin{align*}{\frac{1}{d} \left ( -2\,{\frac{1}{ \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a- \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}b-a-b \right ) ^{3}} \left ( -1/2\,{\frac{ \left ( 6\,{a}^{2}+3\,ab+2\,{b}^{2} \right ) b \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{5}}{ \left ( a-b \right ) \left ({a}^{3}+3\,{a}^{2}b+3\,a{b}^{2}+{b}^{3} \right ) }}+2/3\,{\frac{ \left ( 9\,{a}^{2}+{b}^{2} \right ) b \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}}{ \left ({a}^{2}+2\,ab+{b}^{2} \right ) \left ({a}^{2}-2\,ab+{b}^{2} \right ) }}-1/2\,{\frac{ \left ( 6\,{a}^{2}-3\,ab+2\,{b}^{2} \right ) b\tan \left ( 1/2\,dx+c/2 \right ) }{ \left ( a+b \right ) \left ({a}^{3}-3\,{a}^{2}b+3\,a{b}^{2}-{b}^{3} \right ) }} \right ) }+{\frac{a \left ( 2\,{a}^{2}+3\,{b}^{2} \right ) }{{a}^{6}-3\,{a}^{4}{b}^{2}+3\,{a}^{2}{b}^{4}-{b}^{6}}{\it Artanh} \left ({(a-b)\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ){\frac{1}{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}}} \right ){\frac{1}{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}}} \right ) } \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 = 2.25008, size = 1974, normalized size = 10.73 \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] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec{\left (c + d x \right )}}{\left (a + b \sec{\left (c + d x \right )}\right )^{4}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [B] time = 1.34987, size = 544, normalized size = 2.96 \begin{align*} -\frac{\frac{3 \,{\left (2 \, a^{3} + 3 \, a b^{2}\right )}{\left (\pi \left \lfloor \frac{d x + c}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (2 \, a - 2 \, b\right ) + \arctan \left (\frac{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{\sqrt{-a^{2} + b^{2}}}\right )\right )}}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} \sqrt{-a^{2} + b^{2}}} - \frac{18 \, a^{4} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 27 \, a^{3} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 6 \, a^{2} b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 3 \, a b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 6 \, b^{5} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 36 \, a^{4} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 32 \, a^{2} b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 4 \, b^{5} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 18 \, a^{4} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 27 \, a^{3} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 6 \, a^{2} b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 3 \, a b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 6 \, b^{5} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )}{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a - b\right )}^{3}}}{3 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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