Optimal. Leaf size=154 \[ \frac {2 \left (a^2 (A+3 C)+b^2 (3 A+C)\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d}-\frac {4 a b (A-C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{3 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {8 a A b \sin (c+d x)}{3 d \sqrt {\cos (c+d x)}}-\frac {2 b^2 (A-C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d} \]
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Rubi [A] time = 0.40, antiderivative size = 154, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {3048, 3031, 3023, 2748, 2641, 2639} \[ \frac {2 \left (a^2 (A+3 C)+b^2 (3 A+C)\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d}-\frac {4 a b (A-C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{3 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {8 a A b \sin (c+d x)}{3 d \sqrt {\cos (c+d x)}}-\frac {2 b^2 (A-C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d} \]
Antiderivative was successfully verified.
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Rule 2639
Rule 2641
Rule 2748
Rule 3023
Rule 3031
Rule 3048
Rubi steps
\begin {align*} \int \frac {(a+b \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx &=\frac {2 A (a+b \cos (c+d x))^2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2}{3} \int \frac {(a+b \cos (c+d x)) \left (2 A b+\frac {1}{2} a (A+3 C) \cos (c+d x)-\frac {3}{2} b (A-C) \cos ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx\\ &=\frac {8 a A b \sin (c+d x)}{3 d \sqrt {\cos (c+d x)}}+\frac {2 A (a+b \cos (c+d x))^2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}-\frac {4}{3} \int \frac {\frac {1}{4} \left (-4 A b^2-a^2 (A+3 C)\right )+\frac {3}{2} a b (A-C) \cos (c+d x)+\frac {3}{4} b^2 (A-C) \cos ^2(c+d x)}{\sqrt {\cos (c+d x)}} \, dx\\ &=\frac {8 a A b \sin (c+d x)}{3 d \sqrt {\cos (c+d x)}}-\frac {2 b^2 (A-C) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 d}+\frac {2 A (a+b \cos (c+d x))^2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}-\frac {8}{9} \int \frac {-\frac {3}{8} \left (b^2 (3 A+C)+a^2 (A+3 C)\right )+\frac {9}{4} a b (A-C) \cos (c+d x)}{\sqrt {\cos (c+d x)}} \, dx\\ &=\frac {8 a A b \sin (c+d x)}{3 d \sqrt {\cos (c+d x)}}-\frac {2 b^2 (A-C) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 d}+\frac {2 A (a+b \cos (c+d x))^2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}-(2 a b (A-C)) \int \sqrt {\cos (c+d x)} \, dx-\frac {1}{3} \left (-b^2 (3 A+C)-a^2 (A+3 C)\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx\\ &=-\frac {4 a b (A-C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {2 \left (b^2 (3 A+C)+a^2 (A+3 C)\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d}+\frac {8 a A b \sin (c+d x)}{3 d \sqrt {\cos (c+d x)}}-\frac {2 b^2 (A-C) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 d}+\frac {2 A (a+b \cos (c+d x))^2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\\ \end {align*}
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Mathematica [A] time = 1.46, size = 108, normalized size = 0.70 \[ \frac {2 \left (a^2 (A+3 C)+b^2 (3 A+C)\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )+\frac {2 a^2 A \tan (c+d x)+12 a A b \sin (c+d x)+b^2 C \sin (2 (c+d x))}{\sqrt {\cos (c+d x)}}+12 a b (C-A) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.56, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {C b^{2} \cos \left (d x + c\right )^{4} + 2 \, C a b \cos \left (d x + c\right )^{3} + 2 \, A a b \cos \left (d x + c\right ) + A a^{2} + {\left (C a^{2} + A b^{2}\right )} \cos \left (d x + c\right )^{2}}{\cos \left (d x + c\right )^{\frac {5}{2}}}, x\right ) \]
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 \[ \int \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{2}}{\cos \left (d x + c\right )^{\frac {5}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 2.33, size = 871, normalized size = 5.66 \[ \text {result too large to display} \]
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 \[ \int \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{2}}{\cos \left (d x + c\right )^{\frac {5}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 3.19, size = 185, normalized size = 1.20 \[ \frac {C\,b^2\,\left (\frac {2\,\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )}{3}+\frac {2\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{3}\right )}{d}+\frac {2\,A\,b^2\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,C\,a^2\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {4\,C\,a\,b\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,A\,a^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{3\,d\,{\cos \left (c+d\,x\right )}^{3/2}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {4\,A\,a\,b\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \]
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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