Optimal. Leaf size=172 \[ \frac{2 \left (a^2 B+2 a A b+3 b^2 B\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}-\frac{2 \left (3 a^2 A+10 a b B+5 A b^2\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 \left (3 a^2 A+10 a b B+5 A b^2\right ) \sin (c+d x)}{5 d \sqrt{\cos (c+d x)}}+\frac{2 a^2 A \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2 a (a B+2 A b) \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x)} \]
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Rubi [A] time = 0.350534, antiderivative size = 172, 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 = {2988, 3021, 2748, 2636, 2639, 2641} \[ \frac{2 \left (a^2 B+2 a A b+3 b^2 B\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}-\frac{2 \left (3 a^2 A+10 a b B+5 A b^2\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 \left (3 a^2 A+10 a b B+5 A b^2\right ) \sin (c+d x)}{5 d \sqrt{\cos (c+d x)}}+\frac{2 a^2 A \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2 a (a B+2 A b) \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x)} \]
Antiderivative was successfully verified.
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Rule 2988
Rule 3021
Rule 2748
Rule 2636
Rule 2639
Rule 2641
Rubi steps
\begin{align*} \int \frac{(a+b \cos (c+d x))^2 (A+B \cos (c+d x))}{\cos ^{\frac{7}{2}}(c+d x)} \, dx &=\frac{2 a^2 A \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)}-\frac{2}{5} \int \frac{-\frac{5}{2} a (2 A b+a B)-\frac{1}{2} \left (3 a^2 A+5 A b^2+10 a b B\right ) \cos (c+d x)-\frac{5}{2} b^2 B \cos ^2(c+d x)}{\cos ^{\frac{5}{2}}(c+d x)} \, dx\\ &=\frac{2 a^2 A \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2 a (2 A b+a B) \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x)}-\frac{4}{15} \int \frac{-\frac{3}{4} \left (3 a^2 A+5 A b^2+10 a b B\right )-\frac{5}{4} \left (2 a A b+a^2 B+3 b^2 B\right ) \cos (c+d x)}{\cos ^{\frac{3}{2}}(c+d x)} \, dx\\ &=\frac{2 a^2 A \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2 a (2 A b+a B) \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x)}-\frac{1}{5} \left (-3 a^2 A-5 A b^2-10 a b B\right ) \int \frac{1}{\cos ^{\frac{3}{2}}(c+d x)} \, dx-\frac{1}{3} \left (-2 a A b-a^2 B-3 b^2 B\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx\\ &=\frac{2 \left (2 a A b+a^2 B+3 b^2 B\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}+\frac{2 a^2 A \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2 a (2 A b+a B) \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 \left (3 a^2 A+5 A b^2+10 a b B\right ) \sin (c+d x)}{5 d \sqrt{\cos (c+d x)}}-\frac{1}{5} \left (3 a^2 A+5 A b^2+10 a b B\right ) \int \sqrt{\cos (c+d x)} \, dx\\ &=-\frac{2 \left (3 a^2 A+5 A b^2+10 a b B\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 \left (2 a A b+a^2 B+3 b^2 B\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}+\frac{2 a^2 A \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2 a (2 A b+a B) \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 \left (3 a^2 A+5 A b^2+10 a b B\right ) \sin (c+d x)}{5 d \sqrt{\cos (c+d x)}}\\ \end{align*}
Mathematica [A] time = 1.05638, size = 175, normalized size = 1.02 \[ \frac{10 \left (a^2 B+2 a A b+3 b^2 B\right ) \cos ^{\frac{3}{2}}(c+d x) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )-6 \left (3 a^2 A+10 a b B+5 A b^2\right ) \cos ^{\frac{3}{2}}(c+d x) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )+9 a^2 A \sin (2 (c+d x))+6 a^2 A \tan (c+d x)+10 a^2 B \sin (c+d x)+20 a A b \sin (c+d x)+30 a b B \sin (2 (c+d x))+15 A b^2 \sin (2 (c+d x))}{15 d \cos ^{\frac{3}{2}}(c+d x)} \]
Antiderivative was successfully verified.
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Maple [B] time = 10.05, size = 750, normalized size = 4.4 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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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 )}^{2}}{\cos \left (d x + c\right )^{\frac{7}{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^{2} \cos \left (d x + c\right )^{3} + A a^{2} +{\left (2 \, B a b + A b^{2}\right )} \cos \left (d x + c\right )^{2} +{\left (B a^{2} + 2 \, A a b\right )} \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{\frac{7}{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 )}^{2}}{\cos \left (d x + c\right )^{\frac{7}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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