Optimal. Leaf size=235 \[ \frac{\left (-5 a^2 b^2+15 a^4-2 b^4\right ) \cos (c+d x)}{15 b^5 d}-\frac{2 a^4 \sqrt{a^2-b^2} \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{b^6 d}+\frac{\left (5 a^2-b^2\right ) \sin ^2(c+d x) \cos (c+d x)}{15 b^3 d}-\frac{a \left (4 a^2-b^2\right ) \sin (c+d x) \cos (c+d x)}{8 b^4 d}+\frac{a x \left (-4 a^2 b^2+8 a^4-b^4\right )}{8 b^6}-\frac{a \sin ^3(c+d x) \cos (c+d x)}{4 b^2 d}+\frac{\sin ^4(c+d x) \cos (c+d x)}{5 b d} \]
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Rubi [A] time = 0.91264, antiderivative size = 235, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 8, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.276, Rules used = {2889, 3050, 3049, 3023, 2735, 2660, 618, 204} \[ \frac{\left (-5 a^2 b^2+15 a^4-2 b^4\right ) \cos (c+d x)}{15 b^5 d}-\frac{2 a^4 \sqrt{a^2-b^2} \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{b^6 d}+\frac{\left (5 a^2-b^2\right ) \sin ^2(c+d x) \cos (c+d x)}{15 b^3 d}-\frac{a \left (4 a^2-b^2\right ) \sin (c+d x) \cos (c+d x)}{8 b^4 d}+\frac{a x \left (-4 a^2 b^2+8 a^4-b^4\right )}{8 b^6}-\frac{a \sin ^3(c+d x) \cos (c+d x)}{4 b^2 d}+\frac{\sin ^4(c+d x) \cos (c+d x)}{5 b d} \]
Antiderivative was successfully verified.
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Rule 2889
Rule 3050
Rule 3049
Rule 3023
Rule 2735
Rule 2660
Rule 618
Rule 204
Rubi steps
\begin{align*} \int \frac{\cos ^2(c+d x) \sin ^4(c+d x)}{a+b \sin (c+d x)} \, dx &=\int \frac{\sin ^4(c+d x) \left (1-\sin ^2(c+d x)\right )}{a+b \sin (c+d x)} \, dx\\ &=\frac{\cos (c+d x) \sin ^4(c+d x)}{5 b d}+\frac{\int \frac{\sin ^3(c+d x) \left (-4 a+b \sin (c+d x)+5 a \sin ^2(c+d x)\right )}{a+b \sin (c+d x)} \, dx}{5 b}\\ &=-\frac{a \cos (c+d x) \sin ^3(c+d x)}{4 b^2 d}+\frac{\cos (c+d x) \sin ^4(c+d x)}{5 b d}+\frac{\int \frac{\sin ^2(c+d x) \left (15 a^2-a b \sin (c+d x)-4 \left (5 a^2-b^2\right ) \sin ^2(c+d x)\right )}{a+b \sin (c+d x)} \, dx}{20 b^2}\\ &=\frac{\left (5 a^2-b^2\right ) \cos (c+d x) \sin ^2(c+d x)}{15 b^3 d}-\frac{a \cos (c+d x) \sin ^3(c+d x)}{4 b^2 d}+\frac{\cos (c+d x) \sin ^4(c+d x)}{5 b d}+\frac{\int \frac{\sin (c+d x) \left (-8 a \left (5 a^2-b^2\right )+b \left (5 a^2+8 b^2\right ) \sin (c+d x)+15 a \left (4 a^2-b^2\right ) \sin ^2(c+d x)\right )}{a+b \sin (c+d x)} \, dx}{60 b^3}\\ &=-\frac{a \left (4 a^2-b^2\right ) \cos (c+d x) \sin (c+d x)}{8 b^4 d}+\frac{\left (5 a^2-b^2\right ) \cos (c+d x) \sin ^2(c+d x)}{15 b^3 d}-\frac{a \cos (c+d x) \sin ^3(c+d x)}{4 b^2 d}+\frac{\cos (c+d x) \sin ^4(c+d x)}{5 b d}+\frac{\int \frac{15 a^2 \left (4 a^2-b^2\right )-a b \left (20 a^2-b^2\right ) \sin (c+d x)-8 \left (15 a^4-5 a^2 b^2-2 b^4\right ) \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx}{120 b^4}\\ &=\frac{\left (15 a^4-5 a^2 b^2-2 b^4\right ) \cos (c+d x)}{15 b^5 d}-\frac{a \left (4 a^2-b^2\right ) \cos (c+d x) \sin (c+d x)}{8 b^4 d}+\frac{\left (5 a^2-b^2\right ) \cos (c+d x) \sin ^2(c+d x)}{15 b^3 d}-\frac{a \cos (c+d x) \sin ^3(c+d x)}{4 b^2 d}+\frac{\cos (c+d x) \sin ^4(c+d x)}{5 b d}+\frac{\int \frac{15 a^2 b \left (4 a^2-b^2\right )+15 a \left (8 a^4-4 a^2 b^2-b^4\right ) \sin (c+d x)}{a+b \sin (c+d x)} \, dx}{120 b^5}\\ &=\frac{a \left (8 a^4-4 a^2 b^2-b^4\right ) x}{8 b^6}+\frac{\left (15 a^4-5 a^2 b^2-2 b^4\right ) \cos (c+d x)}{15 b^5 d}-\frac{a \left (4 a^2-b^2\right ) \cos (c+d x) \sin (c+d x)}{8 b^4 d}+\frac{\left (5 a^2-b^2\right ) \cos (c+d x) \sin ^2(c+d x)}{15 b^3 d}-\frac{a \cos (c+d x) \sin ^3(c+d x)}{4 b^2 d}+\frac{\cos (c+d x) \sin ^4(c+d x)}{5 b d}-\frac{\left (a^4 \left (a^2-b^2\right )\right ) \int \frac{1}{a+b \sin (c+d x)} \, dx}{b^6}\\ &=\frac{a \left (8 a^4-4 a^2 b^2-b^4\right ) x}{8 b^6}+\frac{\left (15 a^4-5 a^2 b^2-2 b^4\right ) \cos (c+d x)}{15 b^5 d}-\frac{a \left (4 a^2-b^2\right ) \cos (c+d x) \sin (c+d x)}{8 b^4 d}+\frac{\left (5 a^2-b^2\right ) \cos (c+d x) \sin ^2(c+d x)}{15 b^3 d}-\frac{a \cos (c+d x) \sin ^3(c+d x)}{4 b^2 d}+\frac{\cos (c+d x) \sin ^4(c+d x)}{5 b d}-\frac{\left (2 a^4 \left (a^2-b^2\right )\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 )}{b^6 d}\\ &=\frac{a \left (8 a^4-4 a^2 b^2-b^4\right ) x}{8 b^6}+\frac{\left (15 a^4-5 a^2 b^2-2 b^4\right ) \cos (c+d x)}{15 b^5 d}-\frac{a \left (4 a^2-b^2\right ) \cos (c+d x) \sin (c+d x)}{8 b^4 d}+\frac{\left (5 a^2-b^2\right ) \cos (c+d x) \sin ^2(c+d x)}{15 b^3 d}-\frac{a \cos (c+d x) \sin ^3(c+d x)}{4 b^2 d}+\frac{\cos (c+d x) \sin ^4(c+d x)}{5 b d}+\frac{\left (4 a^4 \left (a^2-b^2\right )\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 )}{b^6 d}\\ &=\frac{a \left (8 a^4-4 a^2 b^2-b^4\right ) x}{8 b^6}-\frac{2 a^4 \sqrt{a^2-b^2} \tan ^{-1}\left (\frac{b+a \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{b^6 d}+\frac{\left (15 a^4-5 a^2 b^2-2 b^4\right ) \cos (c+d x)}{15 b^5 d}-\frac{a \left (4 a^2-b^2\right ) \cos (c+d x) \sin (c+d x)}{8 b^4 d}+\frac{\left (5 a^2-b^2\right ) \cos (c+d x) \sin ^2(c+d x)}{15 b^3 d}-\frac{a \cos (c+d x) \sin ^3(c+d x)}{4 b^2 d}+\frac{\cos (c+d x) \sin ^4(c+d x)}{5 b d}\\ \end{align*}
Mathematica [A] time = 1.74478, size = 177, normalized size = 0.75 \[ \frac{15 a \left (4 \left (-4 a^2 b^2+8 a^4-b^4\right ) (c+d x)-8 a^2 b^2 \sin (2 (c+d x))+b^4 \sin (4 (c+d x))\right )-60 b \left (2 a^2 b^2-8 a^4+b^4\right ) \cos (c+d x)-10 \left (4 a^2 b^3+b^5\right ) \cos (3 (c+d x))-960 a^4 \sqrt{a^2-b^2} \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )+6 b^5 \cos (5 (c+d x))}{480 b^6 d} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.092, size = 871, normalized size = 3.7 \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 = 1.79199, size = 996, normalized size = 4.24 \begin{align*} \left [\frac{24 \, b^{5} \cos \left (d x + c\right )^{5} + 120 \, a^{4} b \cos \left (d x + c\right ) + 60 \, \sqrt{-a^{2} + b^{2}} a^{4} \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 ) - 40 \,{\left (a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{3} + 15 \,{\left (8 \, a^{5} - 4 \, a^{3} b^{2} - a b^{4}\right )} d x + 15 \,{\left (2 \, a b^{4} \cos \left (d x + c\right )^{3} -{\left (4 \, a^{3} b^{2} + a b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{120 \, b^{6} d}, \frac{24 \, b^{5} \cos \left (d x + c\right )^{5} + 120 \, a^{4} b \cos \left (d x + c\right ) + 120 \, \sqrt{a^{2} - b^{2}} a^{4} \arctan \left (-\frac{a \sin \left (d x + c\right ) + b}{\sqrt{a^{2} - b^{2}} \cos \left (d x + c\right )}\right ) - 40 \,{\left (a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{3} + 15 \,{\left (8 \, a^{5} - 4 \, a^{3} b^{2} - a b^{4}\right )} d x + 15 \,{\left (2 \, a b^{4} \cos \left (d x + c\right )^{3} -{\left (4 \, a^{3} b^{2} + a b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{120 \, b^{6} d}\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 = 1.24936, size = 630, normalized size = 2.68 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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