Optimal. Leaf size=204 \[ \frac {2 a \left (3 a^2 B+15 a b C+14 b^2 B\right ) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}+\frac {2 a^2 (5 a C+9 b B) \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (a^3 C+3 a^2 b B+9 a b^2 C+3 b^3 B\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d}-\frac {2 \left (3 a^3 B+15 a^2 b C+15 a b^2 B-5 b^3 C\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 a B \sin (c+d x) (a+b \cos (c+d x))^2}{5 d \cos ^{\frac {5}{2}}(c+d x)} \]
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Rubi [A] time = 0.55, antiderivative size = 204, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 42, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3029, 2989, 3031, 3021, 2748, 2641, 2639} \[ \frac {2 \left (3 a^2 b B+a^3 C+9 a b^2 C+3 b^3 B\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d}-\frac {2 \left (15 a^2 b C+3 a^3 B+15 a b^2 B-5 b^3 C\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 a \left (3 a^2 B+15 a b C+14 b^2 B\right ) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}+\frac {2 a^2 (5 a C+9 b B) \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 a B \sin (c+d x) (a+b \cos (c+d x))^2}{5 d \cos ^{\frac {5}{2}}(c+d x)} \]
Antiderivative was successfully verified.
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Rule 2639
Rule 2641
Rule 2748
Rule 2989
Rule 3021
Rule 3029
Rule 3031
Rubi steps
\begin {align*} \int \frac {(a+b \cos (c+d x))^3 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {9}{2}}(c+d x)} \, dx &=\int \frac {(a+b \cos (c+d x))^3 (B+C \cos (c+d x))}{\cos ^{\frac {7}{2}}(c+d x)} \, dx\\ &=\frac {2 a B (a+b \cos (c+d x))^2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2}{5} \int \frac {(a+b \cos (c+d x)) \left (\frac {1}{2} a (9 b B+5 a C)+\frac {1}{2} \left (3 a^2 B+5 b^2 B+10 a b C\right ) \cos (c+d x)-\frac {1}{2} b (a B-5 b C) \cos ^2(c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx\\ &=\frac {2 a^2 (9 b B+5 a C) \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 a B (a+b \cos (c+d x))^2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}-\frac {4}{15} \int \frac {-\frac {3}{4} a \left (3 a^2 B+14 b^2 B+15 a b C\right )-\frac {5}{4} \left (3 a^2 b B+3 b^3 B+a^3 C+9 a b^2 C\right ) \cos (c+d x)+\frac {3}{4} b^2 (a B-5 b C) \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x)} \, dx\\ &=\frac {2 a^2 (9 b B+5 a C) \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 a \left (3 a^2 B+14 b^2 B+15 a b C\right ) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}+\frac {2 a B (a+b \cos (c+d x))^2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}-\frac {8}{15} \int \frac {-\frac {5}{8} \left (3 a^2 b B+3 b^3 B+a^3 C+9 a b^2 C\right )+\frac {3}{8} \left (3 a^3 B+15 a b^2 B+15 a^2 b C-5 b^3 C\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)}} \, dx\\ &=\frac {2 a^2 (9 b B+5 a C) \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 a \left (3 a^2 B+14 b^2 B+15 a b C\right ) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}+\frac {2 a B (a+b \cos (c+d x))^2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}-\frac {1}{3} \left (-3 a^2 b B-3 b^3 B-a^3 C-9 a b^2 C\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx-\frac {1}{5} \left (3 a^3 B+15 a b^2 B+15 a^2 b C-5 b^3 C\right ) \int \sqrt {\cos (c+d x)} \, dx\\ &=-\frac {2 \left (3 a^3 B+15 a b^2 B+15 a^2 b C-5 b^3 C\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 \left (3 a^2 b B+3 b^3 B+a^3 C+9 a b^2 C\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d}+\frac {2 a^2 (9 b B+5 a C) \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 a \left (3 a^2 B+14 b^2 B+15 a b C\right ) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}+\frac {2 a B (a+b \cos (c+d x))^2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\\ \end {align*}
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Mathematica [A] time = 2.18, size = 176, normalized size = 0.86 \[ \frac {6 a^3 B \tan (c+d x)+9 a \left (a^2 B+5 a b C+5 b^2 B\right ) \sin (2 (c+d x))+10 a^2 (a C+3 b B) \sin (c+d x)+10 \left (a^3 C+3 a^2 b B+9 a b^2 C+3 b^3 B\right ) \cos ^{\frac {3}{2}}(c+d x) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )-6 \left (3 a^3 B+15 a^2 b C+15 a b^2 B-5 b^3 C\right ) \cos ^{\frac {3}{2}}(c+d x) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d \cos ^{\frac {3}{2}}(c+d x)} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.48, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {C b^{3} \cos \left (d x + c\right )^{4} + B a^{3} + {\left (3 \, C a b^{2} + B b^{3}\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (C a^{2} b + B a b^{2}\right )} \cos \left (d x + c\right )^{2} + {\left (C a^{3} + 3 \, B a^{2} b\right )} \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{\frac {7}{2}}}, x\right ) \]
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 \[ \int \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{3}}{\cos \left (d x + c\right )^{\frac {9}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 7.48, size = 997, normalized size = 4.89 \[ \text {result too large to display} \]
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 \[ \int \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{3}}{\cos \left (d x + c\right )^{\frac {9}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 4.96, size = 291, normalized size = 1.43 \[ \frac {2\,\left (C\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )\,b^3+3\,C\,a\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )\,b^2\right )}{d}+\frac {2\,B\,b^3\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,B\,a^3\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {5}{4},\frac {1}{2};\ -\frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{5\,d\,{\cos \left (c+d\,x\right )}^{5/2}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {2\,C\,a^3\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{3\,d\,{\cos \left (c+d\,x\right )}^{3/2}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {6\,B\,a\,b^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {2\,B\,a^2\,b\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,{\cos \left (c+d\,x\right )}^{3/2}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {6\,C\,a^2\,b\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \]
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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