Optimal. Leaf size=197 \[ -\frac{2 b^6 \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{7/2}}-\frac{\sec ^5(c+d x) (b-a \sin (c+d x))}{5 d \left (a^2-b^2\right )}+\frac{\sec ^3(c+d x) \left (a \left (4 a^2-9 b^2\right ) \sin (c+d x)+5 b^3\right )}{15 d \left (a^2-b^2\right )^2}-\frac{\sec (c+d x) \left (15 b^5-a \left (-26 a^2 b^2+8 a^4+33 b^4\right ) \sin (c+d x)\right )}{15 d \left (a^2-b^2\right )^3} \]
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Rubi [A] time = 0.495377, antiderivative size = 197, 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 = {2696, 2866, 12, 2660, 618, 204} \[ -\frac{2 b^6 \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{7/2}}-\frac{\sec ^5(c+d x) (b-a \sin (c+d x))}{5 d \left (a^2-b^2\right )}+\frac{\sec ^3(c+d x) \left (a \left (4 a^2-9 b^2\right ) \sin (c+d x)+5 b^3\right )}{15 d \left (a^2-b^2\right )^2}-\frac{\sec (c+d x) \left (15 b^5-a \left (-26 a^2 b^2+8 a^4+33 b^4\right ) \sin (c+d x)\right )}{15 d \left (a^2-b^2\right )^3} \]
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 ^6(c+d x)}{a+b \sin (c+d x)} \, dx &=-\frac{\sec ^5(c+d x) (b-a \sin (c+d x))}{5 \left (a^2-b^2\right ) d}-\frac{\int \frac{\sec ^4(c+d x) \left (-4 a^2+5 b^2-4 a b \sin (c+d x)\right )}{a+b \sin (c+d x)} \, dx}{5 \left (a^2-b^2\right )}\\ &=-\frac{\sec ^5(c+d x) (b-a \sin (c+d x))}{5 \left (a^2-b^2\right ) d}+\frac{\sec ^3(c+d x) \left (5 b^3+a \left (4 a^2-9 b^2\right ) \sin (c+d x)\right )}{15 \left (a^2-b^2\right )^2 d}+\frac{\int \frac{\sec ^2(c+d x) \left (8 a^4-18 a^2 b^2+15 b^4+2 a b \left (4 a^2-9 b^2\right ) \sin (c+d x)\right )}{a+b \sin (c+d x)} \, dx}{15 \left (a^2-b^2\right )^2}\\ &=-\frac{\sec ^5(c+d x) (b-a \sin (c+d x))}{5 \left (a^2-b^2\right ) d}+\frac{\sec ^3(c+d x) \left (5 b^3+a \left (4 a^2-9 b^2\right ) \sin (c+d x)\right )}{15 \left (a^2-b^2\right )^2 d}-\frac{\sec (c+d x) \left (15 b^5-a \left (8 a^4-26 a^2 b^2+33 b^4\right ) \sin (c+d x)\right )}{15 \left (a^2-b^2\right )^3 d}-\frac{\int \frac{15 b^6}{a+b \sin (c+d x)} \, dx}{15 \left (a^2-b^2\right )^3}\\ &=-\frac{\sec ^5(c+d x) (b-a \sin (c+d x))}{5 \left (a^2-b^2\right ) d}+\frac{\sec ^3(c+d x) \left (5 b^3+a \left (4 a^2-9 b^2\right ) \sin (c+d x)\right )}{15 \left (a^2-b^2\right )^2 d}-\frac{\sec (c+d x) \left (15 b^5-a \left (8 a^4-26 a^2 b^2+33 b^4\right ) \sin (c+d x)\right )}{15 \left (a^2-b^2\right )^3 d}-\frac{b^6 \int \frac{1}{a+b \sin (c+d x)} \, dx}{\left (a^2-b^2\right )^3}\\ &=-\frac{\sec ^5(c+d x) (b-a \sin (c+d x))}{5 \left (a^2-b^2\right ) d}+\frac{\sec ^3(c+d x) \left (5 b^3+a \left (4 a^2-9 b^2\right ) \sin (c+d x)\right )}{15 \left (a^2-b^2\right )^2 d}-\frac{\sec (c+d x) \left (15 b^5-a \left (8 a^4-26 a^2 b^2+33 b^4\right ) \sin (c+d x)\right )}{15 \left (a^2-b^2\right )^3 d}-\frac{\left (2 b^6\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{\sec ^5(c+d x) (b-a \sin (c+d x))}{5 \left (a^2-b^2\right ) d}+\frac{\sec ^3(c+d x) \left (5 b^3+a \left (4 a^2-9 b^2\right ) \sin (c+d x)\right )}{15 \left (a^2-b^2\right )^2 d}-\frac{\sec (c+d x) \left (15 b^5-a \left (8 a^4-26 a^2 b^2+33 b^4\right ) \sin (c+d x)\right )}{15 \left (a^2-b^2\right )^3 d}+\frac{\left (4 b^6\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{2 b^6 \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{\sec ^5(c+d x) (b-a \sin (c+d x))}{5 \left (a^2-b^2\right ) d}+\frac{\sec ^3(c+d x) \left (5 b^3+a \left (4 a^2-9 b^2\right ) \sin (c+d x)\right )}{15 \left (a^2-b^2\right )^2 d}-\frac{\sec (c+d x) \left (15 b^5-a \left (8 a^4-26 a^2 b^2+33 b^4\right ) \sin (c+d x)\right )}{15 \left (a^2-b^2\right )^3 d}\\ \end{align*}
Mathematica [A] time = 2.54924, size = 370, normalized size = 1.88 \[ \frac{\sec ^5(c+d x) \left (-1600 a^3 b^2 \sin (c+d x)-1040 a^3 b^2 \sin (3 (c+d x))-208 a^3 b^2 \sin (5 (c+d x))-190 a^2 b^3 \cos (3 (c+d x))-38 a^2 b^3 \cos (5 (c+d x))+10 b \left (-38 a^2 b^2+9 a^4+149 b^4\right ) \cos (c+d x)+320 b^3 \left (a^2-4 b^2\right ) \cos (2 (c+d x))+1088 a^2 b^3+45 a^4 b \cos (3 (c+d x))+9 a^4 b \cos (5 (c+d x))-384 a^4 b+640 a^5 \sin (c+d x)+320 a^5 \sin (3 (c+d x))+64 a^5 \sin (5 (c+d x))+1200 a b^4 \sin (c+d x)+1080 a b^4 \sin (3 (c+d x))+264 a b^4 \sin (5 (c+d x))+745 b^5 \cos (3 (c+d x))-240 b^5 \cos (4 (c+d x))+149 b^5 \cos (5 (c+d x))-1424 b^5\right )}{1920 d (a-b)^3 (a+b)^3}-\frac{2 b^6 \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{7/2}} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.063, size = 525, normalized size = 2.7 \begin{align*} \text{result too large to display} \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 = 3.28354, size = 1482, normalized size = 7.52 \begin{align*} \left [\frac{15 \, \sqrt{-a^{2} + b^{2}} b^{6} \cos \left (d x + c\right )^{5} \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 ) - 6 \, a^{6} b + 18 \, a^{4} b^{3} - 18 \, a^{2} b^{5} + 6 \, b^{7} - 30 \,{\left (a^{2} b^{5} - b^{7}\right )} \cos \left (d x + c\right )^{4} + 10 \,{\left (a^{4} b^{3} - 2 \, a^{2} b^{5} + b^{7}\right )} \cos \left (d x + c\right )^{2} + 2 \,{\left (3 \, a^{7} - 9 \, a^{5} b^{2} + 9 \, a^{3} b^{4} - 3 \, a b^{6} +{\left (8 \, a^{7} - 34 \, a^{5} b^{2} + 59 \, a^{3} b^{4} - 33 \, a b^{6}\right )} \cos \left (d x + c\right )^{4} +{\left (4 \, a^{7} - 17 \, a^{5} b^{2} + 22 \, a^{3} b^{4} - 9 \, a b^{6}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{30 \,{\left (a^{8} - 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} - 4 \, a^{2} b^{6} + b^{8}\right )} d \cos \left (d x + c\right )^{5}}, \frac{15 \, \sqrt{a^{2} - b^{2}} b^{6} \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 )^{5} - 3 \, a^{6} b + 9 \, a^{4} b^{3} - 9 \, a^{2} b^{5} + 3 \, b^{7} - 15 \,{\left (a^{2} b^{5} - b^{7}\right )} \cos \left (d x + c\right )^{4} + 5 \,{\left (a^{4} b^{3} - 2 \, a^{2} b^{5} + b^{7}\right )} \cos \left (d x + c\right )^{2} +{\left (3 \, a^{7} - 9 \, a^{5} b^{2} + 9 \, a^{3} b^{4} - 3 \, a b^{6} +{\left (8 \, a^{7} - 34 \, a^{5} b^{2} + 59 \, a^{3} b^{4} - 33 \, a b^{6}\right )} \cos \left (d x + c\right )^{4} +{\left (4 \, a^{7} - 17 \, a^{5} b^{2} + 22 \, a^{3} b^{4} - 9 \, a b^{6}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{15 \,{\left (a^{8} - 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} - 4 \, a^{2} b^{6} + b^{8}\right )} d \cos \left (d x + c\right )^{5}}\right ] \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [B] time = 1.16254, size = 788, normalized size = 4. \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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