Optimal. Leaf size=285 \[ -\frac {2 \left (6 a^2 C-5 b^2 (7 A+5 C)\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{105 b d}-\frac {2 \left (a^2-b^2\right ) \left (-6 a^2 C+35 A b^2+25 b^2 C\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{105 b^2 d \sqrt {a+b \cos (c+d x)}}+\frac {4 a \left (-3 a^2 C+70 A b^2+41 b^2 C\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{105 b^2 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {2 C \sin (c+d x) (a+b \cos (c+d x))^{5/2}}{7 b d}-\frac {4 a C \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{35 b d} \]
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Rubi [A] time = 0.47, antiderivative size = 285, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {3024, 2753, 2752, 2663, 2661, 2655, 2653} \[ -\frac {2 \left (6 a^2 C-5 b^2 (7 A+5 C)\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{105 b d}-\frac {2 \left (a^2-b^2\right ) \left (-6 a^2 C+35 A b^2+25 b^2 C\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{105 b^2 d \sqrt {a+b \cos (c+d x)}}+\frac {4 a \left (-3 a^2 C+70 A b^2+41 b^2 C\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{105 b^2 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {2 C \sin (c+d x) (a+b \cos (c+d x))^{5/2}}{7 b d}-\frac {4 a C \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{35 b d} \]
Antiderivative was successfully verified.
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Rule 2653
Rule 2655
Rule 2661
Rule 2663
Rule 2752
Rule 2753
Rule 3024
Rubi steps
\begin {align*} \int (a+b \cos (c+d x))^{3/2} \left (A+C \cos ^2(c+d x)\right ) \, dx &=\frac {2 C (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{7 b d}+\frac {2 \int (a+b \cos (c+d x))^{3/2} \left (\frac {1}{2} b (7 A+5 C)-a C \cos (c+d x)\right ) \, dx}{7 b}\\ &=-\frac {4 a C (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{35 b d}+\frac {2 C (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{7 b d}+\frac {4 \int \sqrt {a+b \cos (c+d x)} \left (\frac {1}{4} a b (35 A+19 C)-\frac {1}{4} \left (6 a^2 C-5 b^2 (7 A+5 C)\right ) \cos (c+d x)\right ) \, dx}{35 b}\\ &=-\frac {2 \left (6 a^2 C-5 b^2 (7 A+5 C)\right ) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{105 b d}-\frac {4 a C (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{35 b d}+\frac {2 C (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{7 b d}+\frac {8 \int \frac {\frac {1}{8} b \left (5 b^2 (7 A+5 C)+3 a^2 (35 A+17 C)\right )+\frac {1}{4} a \left (70 A b^2-3 a^2 C+41 b^2 C\right ) \cos (c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx}{105 b}\\ &=-\frac {2 \left (6 a^2 C-5 b^2 (7 A+5 C)\right ) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{105 b d}-\frac {4 a C (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{35 b d}+\frac {2 C (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{7 b d}-\frac {\left (\left (a^2-b^2\right ) \left (35 A b^2-6 a^2 C+25 b^2 C\right )\right ) \int \frac {1}{\sqrt {a+b \cos (c+d x)}} \, dx}{105 b^2}+\frac {\left (2 a \left (70 A b^2-3 a^2 C+41 b^2 C\right )\right ) \int \sqrt {a+b \cos (c+d x)} \, dx}{105 b^2}\\ &=-\frac {2 \left (6 a^2 C-5 b^2 (7 A+5 C)\right ) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{105 b d}-\frac {4 a C (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{35 b d}+\frac {2 C (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{7 b d}+\frac {\left (2 a \left (70 A b^2-3 a^2 C+41 b^2 C\right ) \sqrt {a+b \cos (c+d x)}\right ) \int \sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}} \, dx}{105 b^2 \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {\left (\left (a^2-b^2\right ) \left (35 A b^2-6 a^2 C+25 b^2 C\right ) \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}{105 b^2 \sqrt {a+b \cos (c+d x)}}\\ &=\frac {4 a \left (70 A b^2-3 a^2 C+41 b^2 C\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{105 b^2 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {2 \left (a^2-b^2\right ) \left (35 A b^2-6 a^2 C+25 b^2 C\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 )}{105 b^2 d \sqrt {a+b \cos (c+d x)}}-\frac {2 \left (6 a^2 C-5 b^2 (7 A+5 C)\right ) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{105 b d}-\frac {4 a C (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{35 b d}+\frac {2 C (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{7 b d}\\ \end {align*}
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Mathematica [A] time = 0.86, size = 224, normalized size = 0.79 \[ \frac {2 b \sin (c+d x) (a+b \cos (c+d x)) \left (6 a^2 C+48 a b C \cos (c+d x)+70 A b^2+15 b^2 C \cos (2 (c+d x))+65 b^2 C\right )+4 \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \left (b^2 \left (3 a^2 (35 A+17 C)+5 b^2 (7 A+5 C)\right ) F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )-2 a \left (3 a^2 C-70 A b^2-41 b^2 C\right ) \left ((a+b) E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )-a F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )\right )\right )}{210 b^2 d \sqrt {a+b \cos (c+d x)}} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.91, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (C b \cos \left (d x + c\right )^{3} + C a \cos \left (d x + c\right )^{2} + A b \cos \left (d x + c\right ) + A a\right )} \sqrt {b \cos \left (d x + c\right ) + a}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 2.55, size = 1131, normalized size = 3.97 \[ \text {result too large to display} \]
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 \[ \int {\left (C \cos \left (d x + c\right )^{2} + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {3}{2}}\,{d x} \]
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 \[ \int \left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^{3/2} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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