Optimal. Leaf size=225 \[ -\frac {2 \left (a^2-b^2\right ) (3 a B+5 A b) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{15 b d \sqrt {a+b \cos (c+d x)}}+\frac {2 \left (3 a^2 B+20 a A b+9 b^2 B\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{15 b d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {2 (3 a B+5 A b) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{15 d}+\frac {2 B \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{5 d} \]
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Rubi [A] time = 0.35, antiderivative size = 225, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {2753, 2752, 2663, 2661, 2655, 2653} \[ -\frac {2 \left (a^2-b^2\right ) (3 a B+5 A b) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{15 b d \sqrt {a+b \cos (c+d x)}}+\frac {2 \left (3 a^2 B+20 a A b+9 b^2 B\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{15 b d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {2 (3 a B+5 A b) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{15 d}+\frac {2 B \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{5 d} \]
Antiderivative was successfully verified.
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Rule 2653
Rule 2655
Rule 2661
Rule 2663
Rule 2752
Rule 2753
Rubi steps
\begin {align*} \int (a+b \cos (c+d x))^{3/2} (A+B \cos (c+d x)) \, dx &=\frac {2 B (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{5 d}+\frac {2}{5} \int \sqrt {a+b \cos (c+d x)} \left (\frac {1}{2} (5 a A+3 b B)+\frac {1}{2} (5 A b+3 a B) \cos (c+d x)\right ) \, dx\\ &=\frac {2 (5 A b+3 a B) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{15 d}+\frac {2 B (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{5 d}+\frac {4}{15} \int \frac {\frac {1}{4} \left (15 a^2 A+5 A b^2+12 a b B\right )+\frac {1}{4} \left (20 a A b+3 a^2 B+9 b^2 B\right ) \cos (c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx\\ &=\frac {2 (5 A b+3 a B) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{15 d}+\frac {2 B (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{5 d}-\frac {\left (\left (a^2-b^2\right ) (5 A b+3 a B)\right ) \int \frac {1}{\sqrt {a+b \cos (c+d x)}} \, dx}{15 b}+\frac {\left (20 a A b+3 a^2 B+9 b^2 B\right ) \int \sqrt {a+b \cos (c+d x)} \, dx}{15 b}\\ &=\frac {2 (5 A b+3 a B) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{15 d}+\frac {2 B (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{5 d}+\frac {\left (\left (20 a A b+3 a^2 B+9 b^2 B\right ) \sqrt {a+b \cos (c+d x)}\right ) \int \sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}} \, dx}{15 b \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {\left (\left (a^2-b^2\right ) (5 A b+3 a B) \sqrt {\frac {a+b \cos (c+d x)}{a+b}}\right ) \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}} \, dx}{15 b \sqrt {a+b \cos (c+d x)}}\\ &=\frac {2 \left (20 a A b+3 a^2 B+9 b^2 B\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{15 b d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {2 \left (a^2-b^2\right ) (5 A b+3 a B) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{15 b d \sqrt {a+b \cos (c+d x)}}+\frac {2 (5 A b+3 a B) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{15 d}+\frac {2 B (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{5 d}\\ \end {align*}
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Mathematica [A] time = 0.78, size = 203, normalized size = 0.90 \[ \frac {2 \left (b \left (15 a^2 A+12 a b B+5 A b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )+\left (3 a^2 B+20 a A b+9 b^2 B\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \left ((a+b) E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )-a F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )\right )+b \sin (c+d x) (a+b \cos (c+d x)) (6 a B+5 A b+3 b B \cos (c+d x))\right )}{15 b d \sqrt {a+b \cos (c+d x)}} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.74, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (B b \cos \left (d x + c\right )^{2} + A a + {\left (B a + A b\right )} \cos \left (d x + c\right )\right )} \sqrt {b \cos \left (d x + c\right ) + a}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (B \cos \left (d x + c\right ) + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {3}{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 1.65, size = 993, normalized size = 4.41 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (B \cos \left (d x + c\right ) + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {3}{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \left (A+B\,\cos \left (c+d\,x\right )\right )\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^{3/2} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (A + B \cos {\left (c + d x \right )}\right ) \left (a + b \cos {\left (c + d x \right )}\right )^{\frac {3}{2}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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