Optimal. Leaf size=137 \[ \frac{2 b^4 \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{5/2}}-\frac{\sec ^3(c+d x) (b-a \sin (c+d x))}{3 d \left (a^2-b^2\right )}+\frac{\sec (c+d x) \left (a \left (2 a^2-5 b^2\right ) \sin (c+d x)+3 b^3\right )}{3 d \left (a^2-b^2\right )^2} \]
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Rubi [A] time = 0.250985, antiderivative size = 137, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {2696, 2866, 12, 2660, 618, 204} \[ \frac{2 b^4 \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{5/2}}-\frac{\sec ^3(c+d x) (b-a \sin (c+d x))}{3 d \left (a^2-b^2\right )}+\frac{\sec (c+d x) \left (a \left (2 a^2-5 b^2\right ) \sin (c+d x)+3 b^3\right )}{3 d \left (a^2-b^2\right )^2} \]
Antiderivative was successfully verified.
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Rule 2696
Rule 2866
Rule 12
Rule 2660
Rule 618
Rule 204
Rubi steps
\begin{align*} \int \frac{\sec ^4(c+d x)}{a+b \sin (c+d x)} \, dx &=-\frac{\sec ^3(c+d x) (b-a \sin (c+d x))}{3 \left (a^2-b^2\right ) d}-\frac{\int \frac{\sec ^2(c+d x) \left (-2 a^2+3 b^2-2 a b \sin (c+d x)\right )}{a+b \sin (c+d x)} \, dx}{3 \left (a^2-b^2\right )}\\ &=-\frac{\sec ^3(c+d x) (b-a \sin (c+d x))}{3 \left (a^2-b^2\right ) d}+\frac{\sec (c+d x) \left (3 b^3+a \left (2 a^2-5 b^2\right ) \sin (c+d x)\right )}{3 \left (a^2-b^2\right )^2 d}+\frac{\int \frac{3 b^4}{a+b \sin (c+d x)} \, dx}{3 \left (a^2-b^2\right )^2}\\ &=-\frac{\sec ^3(c+d x) (b-a \sin (c+d x))}{3 \left (a^2-b^2\right ) d}+\frac{\sec (c+d x) \left (3 b^3+a \left (2 a^2-5 b^2\right ) \sin (c+d x)\right )}{3 \left (a^2-b^2\right )^2 d}+\frac{b^4 \int \frac{1}{a+b \sin (c+d x)} \, dx}{\left (a^2-b^2\right )^2}\\ &=-\frac{\sec ^3(c+d x) (b-a \sin (c+d x))}{3 \left (a^2-b^2\right ) d}+\frac{\sec (c+d x) \left (3 b^3+a \left (2 a^2-5 b^2\right ) \sin (c+d x)\right )}{3 \left (a^2-b^2\right )^2 d}+\frac{\left (2 b^4\right ) \operatorname{Subst}\left (\int \frac{1}{a+2 b x+a x^2} \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{\left (a^2-b^2\right )^2 d}\\ &=-\frac{\sec ^3(c+d x) (b-a \sin (c+d x))}{3 \left (a^2-b^2\right ) d}+\frac{\sec (c+d x) \left (3 b^3+a \left (2 a^2-5 b^2\right ) \sin (c+d x)\right )}{3 \left (a^2-b^2\right )^2 d}-\frac{\left (4 b^4\right ) \operatorname{Subst}\left (\int \frac{1}{-4 \left (a^2-b^2\right )-x^2} \, dx,x,2 b+2 a \tan \left (\frac{1}{2} (c+d x)\right )\right )}{\left (a^2-b^2\right )^2 d}\\ &=\frac{2 b^4 \tan ^{-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 )^{5/2} d}-\frac{\sec ^3(c+d x) (b-a \sin (c+d x))}{3 \left (a^2-b^2\right ) d}+\frac{\sec (c+d x) \left (3 b^3+a \left (2 a^2-5 b^2\right ) \sin (c+d x)\right )}{3 \left (a^2-b^2\right )^2 d}\\ \end{align*}
Mathematica [A] time = 1.26235, size = 202, normalized size = 1.47 \[ \frac{\frac{24 b^4 \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}+\frac{\sec ^3(c+d x) \left (\frac{3}{2} b \left (a^2-7 b^2\right ) \cos (c+d x)+\frac{1}{2} a^2 b \cos (3 (c+d x))-4 a^2 b+6 a^3 \sin (c+d x)+2 a^3 \sin (3 (c+d x))-9 a b^2 \sin (c+d x)-5 a b^2 \sin (3 (c+d x))+6 b^3 \cos (2 (c+d x))-\frac{7}{2} b^3 \cos (3 (c+d x))+10 b^3\right )}{(a-b)^2 (a+b)^2}}{12 d} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.059, size = 270, normalized size = 2. \begin{align*} -{\frac{2}{3\,d \left ( 2\,a+2\,b \right ) } \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-3}}-{\frac{1}{d \left ( 2\,a+2\,b \right ) } \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-2}}-{\frac{a}{d \left ( a+b \right ) ^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}-{\frac{3\,b}{2\,d \left ( a+b \right ) ^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}-{\frac{2}{3\,d \left ( 2\,a-2\,b \right ) } \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-3}}+{\frac{1}{d \left ( 2\,a-2\,b \right ) } \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-2}}-{\frac{a}{d \left ( a-b \right ) ^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}+{\frac{3\,b}{2\,d \left ( a-b \right ) ^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}+2\,{\frac{{b}^{4}}{d \left ( a-b \right ) ^{2} \left ( a+b \right ) ^{2}\sqrt{{a}^{2}-{b}^{2}}}\arctan \left ( 1/2\,{\frac{2\,a\tan \left ( 1/2\,dx+c/2 \right ) +2\,b}{\sqrt{{a}^{2}-{b}^{2}}}} \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 [A] time = 2.80367, size = 1027, normalized size = 7.5 \begin{align*} \left [-\frac{3 \, \sqrt{-a^{2} + b^{2}} b^{4} \cos \left (d x + c\right )^{3} \log \left (\frac{{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2} + 2 \,{\left (a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + b \cos \left (d x + c\right )\right )} \sqrt{-a^{2} + b^{2}}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}\right ) + 2 \, a^{4} b - 4 \, a^{2} b^{3} + 2 \, b^{5} - 6 \,{\left (a^{2} b^{3} - b^{5}\right )} \cos \left (d x + c\right )^{2} - 2 \,{\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4} +{\left (2 \, a^{5} - 7 \, a^{3} b^{2} + 5 \, a b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{6 \,{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} d \cos \left (d x + c\right )^{3}}, -\frac{3 \, \sqrt{a^{2} - b^{2}} b^{4} \arctan \left (-\frac{a \sin \left (d x + c\right ) + b}{\sqrt{a^{2} - b^{2}} \cos \left (d x + c\right )}\right ) \cos \left (d x + c\right )^{3} + a^{4} b - 2 \, a^{2} b^{3} + b^{5} - 3 \,{\left (a^{2} b^{3} - b^{5}\right )} \cos \left (d x + c\right )^{2} -{\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4} +{\left (2 \, a^{5} - 7 \, a^{3} b^{2} + 5 \, a b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{3 \,{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} d \cos \left (d x + c\right )^{3}}\right ] \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 ^{4}{\left (c + d x \right )}}{a + b \sin{\left (c + d x \right )}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [B] time = 1.13423, size = 369, normalized size = 2.69 \begin{align*} \frac{2 \,{\left (\frac{3 \,{\left (\pi \left \lfloor \frac{d x + c}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (a\right ) + \arctan \left (\frac{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + b}{\sqrt{a^{2} - b^{2}}}\right )\right )} b^{4}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt{a^{2} - b^{2}}} - \frac{3 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 6 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 3 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 6 \, b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 2 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 8 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 6 \, b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 3 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 6 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - a^{2} b + 4 \, b^{3}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )}{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}^{3}}\right )}}{3 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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