Optimal. Leaf size=193 \[ \frac{10 a b^4 \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{7/2}}+\frac{b \sec ^3(c+d x)}{d \left (a^2-b^2\right ) (a+b \sin (c+d x))}-\frac{\sec ^3(c+d x) \left (5 a b-\left (a^2+4 b^2\right ) \sin (c+d x)\right )}{3 d \left (a^2-b^2\right )^2}+\frac{\sec (c+d x) \left (\left (-9 a^2 b^2+2 a^4-8 b^4\right ) \sin (c+d x)+15 a b^3\right )}{3 d \left (a^2-b^2\right )^3} \]
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Rubi [A] time = 0.36678, antiderivative size = 193, 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 = {2694, 2866, 12, 2660, 618, 204} \[ \frac{10 a b^4 \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{7/2}}+\frac{b \sec ^3(c+d x)}{d \left (a^2-b^2\right ) (a+b \sin (c+d x))}-\frac{\sec ^3(c+d x) \left (5 a b-\left (a^2+4 b^2\right ) \sin (c+d x)\right )}{3 d \left (a^2-b^2\right )^2}+\frac{\sec (c+d x) \left (\left (-9 a^2 b^2+2 a^4-8 b^4\right ) \sin (c+d x)+15 a b^3\right )}{3 d \left (a^2-b^2\right )^3} \]
Antiderivative was successfully verified.
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Rule 2694
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))^2} \, dx &=\frac{b \sec ^3(c+d x)}{\left (a^2-b^2\right ) d (a+b \sin (c+d x))}+\frac{\int \frac{\sec ^4(c+d x) (-a+4 b \sin (c+d x))}{a+b \sin (c+d x)} \, dx}{-a^2+b^2}\\ &=\frac{b \sec ^3(c+d x)}{\left (a^2-b^2\right ) d (a+b \sin (c+d x))}-\frac{\sec ^3(c+d x) \left (5 a b-\left (a^2+4 b^2\right ) \sin (c+d x)\right )}{3 \left (a^2-b^2\right )^2 d}+\frac{\int \frac{\sec ^2(c+d x) \left (a \left (2 a^2-7 b^2\right )+2 b \left (a^2+4 b^2\right ) \sin (c+d x)\right )}{a+b \sin (c+d x)} \, dx}{3 \left (a^2-b^2\right )^2}\\ &=\frac{b \sec ^3(c+d x)}{\left (a^2-b^2\right ) d (a+b \sin (c+d x))}-\frac{\sec ^3(c+d x) \left (5 a b-\left (a^2+4 b^2\right ) \sin (c+d x)\right )}{3 \left (a^2-b^2\right )^2 d}+\frac{\sec (c+d x) \left (15 a b^3+\left (2 a^4-9 a^2 b^2-8 b^4\right ) \sin (c+d x)\right )}{3 \left (a^2-b^2\right )^3 d}-\frac{\int -\frac{15 a b^4}{a+b \sin (c+d x)} \, dx}{3 \left (a^2-b^2\right )^3}\\ &=\frac{b \sec ^3(c+d x)}{\left (a^2-b^2\right ) d (a+b \sin (c+d x))}-\frac{\sec ^3(c+d x) \left (5 a b-\left (a^2+4 b^2\right ) \sin (c+d x)\right )}{3 \left (a^2-b^2\right )^2 d}+\frac{\sec (c+d x) \left (15 a b^3+\left (2 a^4-9 a^2 b^2-8 b^4\right ) \sin (c+d x)\right )}{3 \left (a^2-b^2\right )^3 d}+\frac{\left (5 a b^4\right ) \int \frac{1}{a+b \sin (c+d x)} \, dx}{\left (a^2-b^2\right )^3}\\ &=\frac{b \sec ^3(c+d x)}{\left (a^2-b^2\right ) d (a+b \sin (c+d x))}-\frac{\sec ^3(c+d x) \left (5 a b-\left (a^2+4 b^2\right ) \sin (c+d x)\right )}{3 \left (a^2-b^2\right )^2 d}+\frac{\sec (c+d x) \left (15 a b^3+\left (2 a^4-9 a^2 b^2-8 b^4\right ) \sin (c+d x)\right )}{3 \left (a^2-b^2\right )^3 d}+\frac{\left (10 a 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 )^3 d}\\ &=\frac{b \sec ^3(c+d x)}{\left (a^2-b^2\right ) d (a+b \sin (c+d x))}-\frac{\sec ^3(c+d x) \left (5 a b-\left (a^2+4 b^2\right ) \sin (c+d x)\right )}{3 \left (a^2-b^2\right )^2 d}+\frac{\sec (c+d x) \left (15 a b^3+\left (2 a^4-9 a^2 b^2-8 b^4\right ) \sin (c+d x)\right )}{3 \left (a^2-b^2\right )^3 d}-\frac{\left (20 a 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 )^3 d}\\ &=\frac{10 a 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 )^{7/2} d}+\frac{b \sec ^3(c+d x)}{\left (a^2-b^2\right ) d (a+b \sin (c+d x))}-\frac{\sec ^3(c+d x) \left (5 a b-\left (a^2+4 b^2\right ) \sin (c+d x)\right )}{3 \left (a^2-b^2\right )^2 d}+\frac{\sec (c+d x) \left (15 a b^3+\left (2 a^4-9 a^2 b^2-8 b^4\right ) \sin (c+d x)\right )}{3 \left (a^2-b^2\right )^3 d}\\ \end{align*}
Mathematica [A] time = 1.9094, size = 336, normalized size = 1.74 \[ \frac{\frac{120 a b^4 \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{7/2}}+\frac{12 b^5 \cos (c+d x)}{(a-b)^3 (a+b)^3 (a+b \sin (c+d x))}+\frac{4 (2 a+5 b) \sin \left (\frac{1}{2} (c+d x)\right )}{(a+b)^3 \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}+\frac{2 \sin \left (\frac{1}{2} (c+d x)\right )}{(a+b)^2 \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^3}+\frac{4 (2 a-5 b) \sin \left (\frac{1}{2} (c+d x)\right )}{(a-b)^3 \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}+\frac{1}{(a+b)^2 \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}-\frac{1}{(a-b)^2 \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}+\frac{2 \sin \left (\frac{1}{2} (c+d x)\right )}{(a-b)^2 \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^3}}{12 d} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.118, size = 370, normalized size = 1.9 \begin{align*} -{\frac{1}{3\,d \left ( a+b \right ) ^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-3}}-{\frac{1}{2\,d \left ( a+b \right ) ^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-2}}-{\frac{a}{d \left ( a+b \right ) ^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}-2\,{\frac{b}{d \left ( a+b \right ) ^{3} \left ( \tan \left ( 1/2\,dx+c/2 \right ) -1 \right ) }}-{\frac{1}{3\,d \left ( a-b \right ) ^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-3}}+{\frac{1}{2\,d \left ( a-b \right ) ^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-2}}-{\frac{a}{d \left ( a-b \right ) ^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}+2\,{\frac{b}{d \left ( a-b \right ) ^{3} \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) }}+2\,{\frac{{b}^{6}\tan \left ( 1/2\,dx+c/2 \right ) }{d \left ( a-b \right ) ^{3} \left ( a+b \right ) ^{3} \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a+2\,\tan \left ( 1/2\,dx+c/2 \right ) b+a \right ) a}}+2\,{\frac{{b}^{5}}{d \left ( a-b \right ) ^{3} \left ( a+b \right ) ^{3} \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a+2\,\tan \left ( 1/2\,dx+c/2 \right ) b+a \right ) }}+10\,{\frac{a{b}^{4}}{d \left ( a-b \right ) ^{3} \left ( a+b \right ) ^{3}\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 = 3.38134, size = 1728, normalized size = 8.95 \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 ^{4}{\left (c + d x \right )}}{\left (a + b \sin{\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 [B] time = 1.14284, size = 576, normalized size = 2.98 \begin{align*} \frac{2 \,{\left (\frac{15 \,{\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 )} a b^{4}}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} \sqrt{a^{2} - b^{2}}} + \frac{3 \,{\left (b^{6} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + a b^{5}\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} + 2 \, b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + a\right )}} - \frac{3 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 9 \, a^{2} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 6 \, b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 6 \, a^{3} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 18 \, a b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 2 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 18 \, a^{2} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 8 \, b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 24 \, a b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 3 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 9 \, a^{2} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 6 \, b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 2 \, a^{3} b + 14 \, a b^{3}}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\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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