Optimal. Leaf size=360 \[ \frac{b^6 \sin (c+d x)}{a d \left (a^2-b^2\right )^3 (a \cos (c+d x)+b)}-\frac{2 b^7 \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^2 d (a-b)^{7/2} (a+b)^{7/2}}-\frac{4 b^5 \left (3 a^2-b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^2 d (a-b)^{7/2} (a+b)^{7/2}}+\frac{x}{a^2}+\frac{(3 a+5 b) \sin (c+d x)}{4 d (a+b)^3 (1-\cos (c+d x))}-\frac{\sin (c+d x)}{12 d (a+b)^2 (1-\cos (c+d x))}+\frac{\sin (c+d x)}{12 d (a-b)^2 (\cos (c+d x)+1)}-\frac{(3 a-5 b) \sin (c+d x)}{4 d (a-b)^3 (\cos (c+d x)+1)}-\frac{\sin (c+d x)}{12 d (a+b)^2 (1-\cos (c+d x))^2}+\frac{\sin (c+d x)}{12 d (a-b)^2 (\cos (c+d x)+1)^2} \]
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Rubi [A] time = 0.567013, antiderivative size = 360, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 8, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.381, Rules used = {3898, 2897, 2650, 2648, 2664, 12, 2659, 208} \[ \frac{b^6 \sin (c+d x)}{a d \left (a^2-b^2\right )^3 (a \cos (c+d x)+b)}-\frac{2 b^7 \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^2 d (a-b)^{7/2} (a+b)^{7/2}}-\frac{4 b^5 \left (3 a^2-b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^2 d (a-b)^{7/2} (a+b)^{7/2}}+\frac{x}{a^2}+\frac{(3 a+5 b) \sin (c+d x)}{4 d (a+b)^3 (1-\cos (c+d x))}-\frac{\sin (c+d x)}{12 d (a+b)^2 (1-\cos (c+d x))}+\frac{\sin (c+d x)}{12 d (a-b)^2 (\cos (c+d x)+1)}-\frac{(3 a-5 b) \sin (c+d x)}{4 d (a-b)^3 (\cos (c+d x)+1)}-\frac{\sin (c+d x)}{12 d (a+b)^2 (1-\cos (c+d x))^2}+\frac{\sin (c+d x)}{12 d (a-b)^2 (\cos (c+d x)+1)^2} \]
Antiderivative was successfully verified.
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Rule 3898
Rule 2897
Rule 2650
Rule 2648
Rule 2664
Rule 12
Rule 2659
Rule 208
Rubi steps
\begin{align*} \int \frac{\cot ^4(c+d x)}{(a+b \sec (c+d x))^2} \, dx &=\int \frac{\cos ^2(c+d x) \cot ^4(c+d x)}{(b+a \cos (c+d x))^2} \, dx\\ &=\int \left (\frac{1}{a^2}+\frac{1}{4 (a-b)^2 (-1-\cos (c+d x))^2}+\frac{3 a-5 b}{4 (a-b)^3 (-1-\cos (c+d x))}+\frac{1}{4 (a+b)^2 (1-\cos (c+d x))^2}+\frac{-3 a-5 b}{4 (a+b)^3 (1-\cos (c+d x))}+\frac{b^6}{a^2 \left (a^2-b^2\right )^2 (-b-a \cos (c+d x))^2}+\frac{2 b^5 \left (3 a^2-b^2\right )}{a^2 \left (a^2-b^2\right )^3 (-b-a \cos (c+d x))}\right ) \, dx\\ &=\frac{x}{a^2}+\frac{(3 a-5 b) \int \frac{1}{-1-\cos (c+d x)} \, dx}{4 (a-b)^3}+\frac{\int \frac{1}{(-1-\cos (c+d x))^2} \, dx}{4 (a-b)^2}+\frac{\int \frac{1}{(1-\cos (c+d x))^2} \, dx}{4 (a+b)^2}-\frac{(3 a+5 b) \int \frac{1}{1-\cos (c+d x)} \, dx}{4 (a+b)^3}+\frac{b^6 \int \frac{1}{(-b-a \cos (c+d x))^2} \, dx}{a^2 \left (a^2-b^2\right )^2}+\frac{\left (2 b^5 \left (3 a^2-b^2\right )\right ) \int \frac{1}{-b-a \cos (c+d x)} \, dx}{a^2 \left (a^2-b^2\right )^3}\\ &=\frac{x}{a^2}-\frac{\sin (c+d x)}{12 (a+b)^2 d (1-\cos (c+d x))^2}+\frac{(3 a+5 b) \sin (c+d x)}{4 (a+b)^3 d (1-\cos (c+d x))}+\frac{\sin (c+d x)}{12 (a-b)^2 d (1+\cos (c+d x))^2}-\frac{(3 a-5 b) \sin (c+d x)}{4 (a-b)^3 d (1+\cos (c+d x))}+\frac{b^6 \sin (c+d x)}{a \left (a^2-b^2\right )^3 d (b+a \cos (c+d x))}-\frac{\int \frac{1}{-1-\cos (c+d x)} \, dx}{12 (a-b)^2}+\frac{\int \frac{1}{1-\cos (c+d x)} \, dx}{12 (a+b)^2}+\frac{b^6 \int \frac{b}{-b-a \cos (c+d x)} \, dx}{a^2 \left (a^2-b^2\right )^3}+\frac{\left (4 b^5 \left (3 a^2-b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{-a-b+(a-b) x^2} \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{a^2 \left (a^2-b^2\right )^3 d}\\ &=\frac{x}{a^2}-\frac{4 b^5 \left (3 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^2 (a-b)^{7/2} (a+b)^{7/2} d}-\frac{\sin (c+d x)}{12 (a+b)^2 d (1-\cos (c+d x))^2}-\frac{\sin (c+d x)}{12 (a+b)^2 d (1-\cos (c+d x))}+\frac{(3 a+5 b) \sin (c+d x)}{4 (a+b)^3 d (1-\cos (c+d x))}+\frac{\sin (c+d x)}{12 (a-b)^2 d (1+\cos (c+d x))^2}-\frac{(3 a-5 b) \sin (c+d x)}{4 (a-b)^3 d (1+\cos (c+d x))}+\frac{\sin (c+d x)}{12 (a-b)^2 d (1+\cos (c+d x))}+\frac{b^6 \sin (c+d x)}{a \left (a^2-b^2\right )^3 d (b+a \cos (c+d x))}+\frac{b^7 \int \frac{1}{-b-a \cos (c+d x)} \, dx}{a^2 \left (a^2-b^2\right )^3}\\ &=\frac{x}{a^2}-\frac{4 b^5 \left (3 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^2 (a-b)^{7/2} (a+b)^{7/2} d}-\frac{\sin (c+d x)}{12 (a+b)^2 d (1-\cos (c+d x))^2}-\frac{\sin (c+d x)}{12 (a+b)^2 d (1-\cos (c+d x))}+\frac{(3 a+5 b) \sin (c+d x)}{4 (a+b)^3 d (1-\cos (c+d x))}+\frac{\sin (c+d x)}{12 (a-b)^2 d (1+\cos (c+d x))^2}-\frac{(3 a-5 b) \sin (c+d x)}{4 (a-b)^3 d (1+\cos (c+d x))}+\frac{\sin (c+d x)}{12 (a-b)^2 d (1+\cos (c+d x))}+\frac{b^6 \sin (c+d x)}{a \left (a^2-b^2\right )^3 d (b+a \cos (c+d x))}+\frac{\left (2 b^7\right ) \operatorname{Subst}\left (\int \frac{1}{-a-b+(a-b) x^2} \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{a^2 \left (a^2-b^2\right )^3 d}\\ &=\frac{x}{a^2}-\frac{2 b^7 \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^2 (a-b)^{7/2} (a+b)^{7/2} d}-\frac{4 b^5 \left (3 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^2 (a-b)^{7/2} (a+b)^{7/2} d}-\frac{\sin (c+d x)}{12 (a+b)^2 d (1-\cos (c+d x))^2}-\frac{\sin (c+d x)}{12 (a+b)^2 d (1-\cos (c+d x))}+\frac{(3 a+5 b) \sin (c+d x)}{4 (a+b)^3 d (1-\cos (c+d x))}+\frac{\sin (c+d x)}{12 (a-b)^2 d (1+\cos (c+d x))^2}-\frac{(3 a-5 b) \sin (c+d x)}{4 (a-b)^3 d (1+\cos (c+d x))}+\frac{\sin (c+d x)}{12 (a-b)^2 d (1+\cos (c+d x))}+\frac{b^6 \sin (c+d x)}{a \left (a^2-b^2\right )^3 d (b+a \cos (c+d x))}\\ \end{align*}
Mathematica [A] time = 2.4544, size = 303, normalized size = 0.84 \[ \frac{\sec ^2(c+d x) (a \cos (c+d x)+b) \left (-\frac{48 b^5 \left (b^2-6 a^2\right ) (a \cos (c+d x)+b) \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{a^2 \left (a^2-b^2\right )^{7/2}}+\frac{24 (c+d x) (a \cos (c+d x)+b)}{a^2}+\frac{24 b^6 \sin (c+d x)}{a (a-b)^3 (a+b)^3}+\frac{4 (7 b-4 a) \tan \left (\frac{1}{2} (c+d x)\right ) (a \cos (c+d x)+b)}{(a-b)^3}+\frac{4 (4 a+7 b) \cot \left (\frac{1}{2} (c+d x)\right ) (a \cos (c+d x)+b)}{(a+b)^3}-\frac{\cot \left (\frac{1}{2} (c+d x)\right ) \csc ^2\left (\frac{1}{2} (c+d x)\right ) (a \cos (c+d x)+b)}{(a+b)^2}+\frac{\tan \left (\frac{1}{2} (c+d x)\right ) \sec ^2\left (\frac{1}{2} (c+d x)\right ) (a \cos (c+d x)+b)}{(a-b)^2}\right )}{24 d (a+b \sec (c+d x))^2} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.099, size = 416, normalized size = 1.2 \begin{align*}{\frac{a}{24\,d \left ({a}^{2}-2\,ab+{b}^{2} \right ) \left ( a-b \right ) } \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}-{\frac{b}{24\,d \left ({a}^{2}-2\,ab+{b}^{2} \right ) \left ( a-b \right ) } \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}-{\frac{5\,a}{8\,d \left ({a}^{2}-2\,ab+{b}^{2} \right ) \left ( a-b \right ) }\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+{\frac{9\,b}{8\,d \left ({a}^{2}-2\,ab+{b}^{2} \right ) \left ( a-b \right ) }\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+2\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{d{a}^{2}}}-2\,{\frac{{b}^{6}\tan \left ( 1/2\,dx+c/2 \right ) }{d \left ( a+b \right ) ^{3} \left ( a-b \right ) ^{3}a \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 ) }}-12\,{\frac{{b}^{5}}{d \left ( a+b \right ) ^{3} \left ( a-b \right ) ^{3}\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}{\it Artanh} \left ({\frac{ \left ( a-b \right ) \tan \left ( 1/2\,dx+c/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) }+2\,{\frac{{b}^{7}}{d \left ( a+b \right ) ^{3} \left ( a-b \right ) ^{3}{a}^{2}\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}{\it Artanh} \left ({\frac{ \left ( a-b \right ) \tan \left ( 1/2\,dx+c/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) }-{\frac{1}{24\,d \left ( a+b \right ) ^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-3}}+{\frac{5\,a}{8\,d \left ( a+b \right ) ^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}}+{\frac{9\,b}{8\,d \left ( a+b \right ) ^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}} \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 = 1.37913, size = 3191, normalized size = 8.86 \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{\cot ^{4}{\left (c + d x \right )}}{\left (a + b \sec{\left (c + d x \right )}\right )^{2}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.38352, size = 657, normalized size = 1.82 \begin{align*} -\frac{\frac{48 \, b^{6} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{{\left (a^{7} - 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} - a 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 )}} - \frac{48 \,{\left (6 \, a^{2} b^{5} - b^{7}\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^{8} - 3 \, a^{6} b^{2} + 3 \, a^{4} b^{4} - a^{2} b^{6}\right )} \sqrt{-a^{2} + b^{2}}} - \frac{a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 4 \, a^{3} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 6 \, a^{2} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 4 \, a b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 15 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 72 \, a^{3} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 126 \, a^{2} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 96 \, a b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 27 \, b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{6} - 6 \, a^{5} b + 15 \, a^{4} b^{2} - 20 \, a^{3} b^{3} + 15 \, a^{2} b^{4} - 6 \, a b^{5} + b^{6}} - \frac{24 \,{\left (d x + c\right )}}{a^{2}} - \frac{15 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 27 \, b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a - b}{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3}}}{24 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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