Optimal. Leaf size=192 \[ -\frac{b \left (4 a^2+b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{d (a-b)^{7/2} (a+b)^{7/2}}+\frac{a \left (2 a^2+13 b^2\right ) \tan (c+d x)}{6 d \left (a^2-b^2\right )^3 (a+b \sec (c+d x))}+\frac{\left (2 a^2+3 b^2\right ) \tan (c+d x)}{6 d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))^2}+\frac{a \tan (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^3} \]
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Rubi [A] time = 0.307318, antiderivative size = 192, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {3836, 4003, 12, 3831, 2659, 208} \[ -\frac{b \left (4 a^2+b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{d (a-b)^{7/2} (a+b)^{7/2}}+\frac{a \left (2 a^2+13 b^2\right ) \tan (c+d x)}{6 d \left (a^2-b^2\right )^3 (a+b \sec (c+d x))}+\frac{\left (2 a^2+3 b^2\right ) \tan (c+d x)}{6 d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))^2}+\frac{a \tan (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^3} \]
Antiderivative was successfully verified.
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Rule 3836
Rule 4003
Rule 12
Rule 3831
Rule 2659
Rule 208
Rubi steps
\begin{align*} \int \frac{\sec ^2(c+d x)}{(a+b \sec (c+d x))^4} \, dx &=\frac{a \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 b+2 a \sec (c+d x))}{(a+b \sec (c+d x))^3} \, dx}{3 \left (a^2-b^2\right )}\\ &=\frac{a \tan (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}+\frac{\left (2 a^2+3 b^2\right ) \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 (10 a b-\left (2 a^2+3 b^2\right ) \sec (c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx}{6 \left (a^2-b^2\right )^2}\\ &=\frac{a \tan (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}+\frac{\left (2 a^2+3 b^2\right ) \tan (c+d x)}{6 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}+\frac{a \left (2 a^2+13 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 b \left (4 a^2+b^2\right ) \sec (c+d x)}{a+b \sec (c+d x)} \, dx}{6 \left (a^2-b^2\right )^3}\\ &=\frac{a \tan (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}+\frac{\left (2 a^2+3 b^2\right ) \tan (c+d x)}{6 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}+\frac{a \left (2 a^2+13 b^2\right ) \tan (c+d x)}{6 \left (a^2-b^2\right )^3 d (a+b \sec (c+d x))}-\frac{\left (b \left (4 a^2+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{a \tan (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}+\frac{\left (2 a^2+3 b^2\right ) \tan (c+d x)}{6 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}+\frac{a \left (2 a^2+13 b^2\right ) \tan (c+d x)}{6 \left (a^2-b^2\right )^3 d (a+b \sec (c+d x))}-\frac{\left (4 a^2+b^2\right ) \int \frac{1}{1+\frac{a \cos (c+d x)}{b}} \, dx}{2 \left (a^2-b^2\right )^3}\\ &=\frac{a \tan (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}+\frac{\left (2 a^2+3 b^2\right ) \tan (c+d x)}{6 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}+\frac{a \left (2 a^2+13 b^2\right ) \tan (c+d x)}{6 \left (a^2-b^2\right )^3 d (a+b \sec (c+d x))}-\frac{\left (4 a^2+b^2\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 )}{\left (a^2-b^2\right )^3 d}\\ &=-\frac{b \left (4 a^2+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{a \tan (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}+\frac{\left (2 a^2+3 b^2\right ) \tan (c+d x)}{6 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}+\frac{a \left (2 a^2+13 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 = 1.20827, size = 164, normalized size = 0.85 \[ \frac{\frac{\sin (c+d x) \left (a \left (10 a^2 b^2+6 a^4-b^4\right ) \cos ^2(c+d x)-3 b \left (-9 a^2 b^2-2 a^4+b^4\right ) \cos (c+d x)+2 a^3 b^2+13 a b^4\right )}{(a-b)^3 (a+b)^3 (a \cos (c+d x)+b)^3}+\frac{6 b \left (4 a^2+b^2\right ) \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{7/2}}}{6 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.06, size = 297, normalized size = 1.6 \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 ( 2\,{a}^{3}+2\,{a}^{2}b+6\,a{b}^{2}+{b}^{3} \right ) \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 ( 3\,{a}^{2}+7\,{b}^{2} \right ) a \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 ( 2\,{a}^{3}-2\,{a}^{2}b+6\,a{b}^{2}-{b}^{3} \right ) \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{b \left ( 4\,{a}^{2}+{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 = 2.19169, size = 1978, normalized size = 10.3 \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 ^{2}{\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 = 1.35889, size = 582, normalized size = 3.03 \begin{align*} -\frac{\frac{3 \,{\left (4 \, a^{2} b + b^{3}\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{6 \, a^{5} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 6 \, a^{4} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 12 \, a^{3} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 27 \, a^{2} b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 12 \, a b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 3 \, b^{5} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 12 \, a^{5} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 16 \, a^{3} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 28 \, a b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 6 \, a^{5} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 6 \, a^{4} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 12 \, a^{3} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 27 \, a^{2} b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 12 \, a b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 3 \, 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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