Optimal. Leaf size=202 \[ \frac{2 \left (9 a^2 A b+3 a^3 B+3 a b^2 B+A b^3\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}-\frac{2 \left (5 a^3 A-15 a^2 b B-15 a A b^2-3 b^3 B\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}-\frac{2 b \left (6 a^2 A-3 a b B-A b^2\right ) \sin (c+d x) \sqrt{\cos (c+d x)}}{3 d}-\frac{2 b^2 (5 a A-b B) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{5 d}+\frac{2 a A \sin (c+d x) (a+b \cos (c+d x))^2}{d \sqrt{\cos (c+d x)}} \]
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Rubi [A] time = 0.463768, antiderivative size = 202, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {2989, 3033, 3023, 2748, 2641, 2639} \[ \frac{2 \left (9 a^2 A b+3 a^3 B+3 a b^2 B+A b^3\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}-\frac{2 \left (5 a^3 A-15 a^2 b B-15 a A b^2-3 b^3 B\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}-\frac{2 b \left (6 a^2 A-3 a b B-A b^2\right ) \sin (c+d x) \sqrt{\cos (c+d x)}}{3 d}-\frac{2 b^2 (5 a A-b B) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{5 d}+\frac{2 a A \sin (c+d x) (a+b \cos (c+d x))^2}{d \sqrt{\cos (c+d x)}} \]
Antiderivative was successfully verified.
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Rule 2989
Rule 3033
Rule 3023
Rule 2748
Rule 2641
Rule 2639
Rubi steps
\begin{align*} \int \frac{(a+b \cos (c+d x))^3 (A+B \cos (c+d x))}{\cos ^{\frac{3}{2}}(c+d x)} \, dx &=\frac{2 a A (a+b \cos (c+d x))^2 \sin (c+d x)}{d \sqrt{\cos (c+d x)}}+2 \int \frac{(a+b \cos (c+d x)) \left (\frac{1}{2} a (5 A b+a B)-\frac{1}{2} \left (a^2 A-A b^2-2 a b B\right ) \cos (c+d x)-\frac{1}{2} b (5 a A-b B) \cos ^2(c+d x)\right )}{\sqrt{\cos (c+d x)}} \, dx\\ &=-\frac{2 b^2 (5 a A-b B) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac{2 a A (a+b \cos (c+d x))^2 \sin (c+d x)}{d \sqrt{\cos (c+d x)}}+\frac{4}{5} \int \frac{\frac{5}{4} a^2 (5 A b+a B)-\frac{1}{4} \left (5 a^3 A-15 a A b^2-15 a^2 b B-3 b^3 B\right ) \cos (c+d x)-\frac{5}{4} b \left (6 a^2 A-A b^2-3 a b B\right ) \cos ^2(c+d x)}{\sqrt{\cos (c+d x)}} \, dx\\ &=-\frac{2 b \left (6 a^2 A-A b^2-3 a b B\right ) \sqrt{\cos (c+d x)} \sin (c+d x)}{3 d}-\frac{2 b^2 (5 a A-b B) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac{2 a A (a+b \cos (c+d x))^2 \sin (c+d x)}{d \sqrt{\cos (c+d x)}}+\frac{8}{15} \int \frac{\frac{5}{8} \left (9 a^2 A b+A b^3+3 a^3 B+3 a b^2 B\right )-\frac{3}{8} \left (5 a^3 A-15 a A b^2-15 a^2 b B-3 b^3 B\right ) \cos (c+d x)}{\sqrt{\cos (c+d x)}} \, dx\\ &=-\frac{2 b \left (6 a^2 A-A b^2-3 a b B\right ) \sqrt{\cos (c+d x)} \sin (c+d x)}{3 d}-\frac{2 b^2 (5 a A-b B) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac{2 a A (a+b \cos (c+d x))^2 \sin (c+d x)}{d \sqrt{\cos (c+d x)}}+\frac{1}{3} \left (9 a^2 A b+A b^3+3 a^3 B+3 a b^2 B\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx+\frac{1}{5} \left (-5 a^3 A+15 a A b^2+15 a^2 b B+3 b^3 B\right ) \int \sqrt{\cos (c+d x)} \, dx\\ &=-\frac{2 \left (5 a^3 A-15 a A b^2-15 a^2 b B-3 b^3 B\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 \left (9 a^2 A b+A b^3+3 a^3 B+3 a b^2 B\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}-\frac{2 b \left (6 a^2 A-A b^2-3 a b B\right ) \sqrt{\cos (c+d x)} \sin (c+d x)}{3 d}-\frac{2 b^2 (5 a A-b B) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac{2 a A (a+b \cos (c+d x))^2 \sin (c+d x)}{d \sqrt{\cos (c+d x)}}\\ \end{align*}
Mathematica [A] time = 1.08531, size = 150, normalized size = 0.74 \[ \frac{10 \left (9 a^2 A b+3 a^3 B+3 a b^2 B+A b^3\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )+\left (-30 a^3 A+90 a^2 b B+90 a A b^2+18 b^3 B\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )+\frac{\sin (c+d x) \left (3 \left (10 a^3 A+b^3 B \cos (2 (c+d x))+b^3 B\right )+10 b^2 (3 a B+A b) \cos (c+d x)\right )}{\sqrt{\cos (c+d x)}}}{15 d} \]
Antiderivative was successfully verified.
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Maple [B] time = 3.497, size = 867, normalized size = 4.3 \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] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B \cos \left (d x + c\right ) + A\right )}{\left (b \cos \left (d x + c\right ) + a\right )}^{3}}{\cos \left (d x + c\right )^{\frac{3}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{B b^{3} \cos \left (d x + c\right )^{4} + A a^{3} +{\left (3 \, B a b^{2} + A b^{3}\right )} \cos \left (d x + c\right )^{3} + 3 \,{\left (B a^{2} b + A a b^{2}\right )} \cos \left (d x + c\right )^{2} +{\left (B a^{3} + 3 \, A a^{2} b\right )} \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{\frac{3}{2}}}, x\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 [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B \cos \left (d x + c\right ) + A\right )}{\left (b \cos \left (d x + c\right ) + a\right )}^{3}}{\cos \left (d x + c\right )^{\frac{3}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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