Optimal. Leaf size=294 \[ \frac {2 a \sin (c+d x) \left (5 a^2 (5 A+7 C)+63 a b B+24 A b^2\right )}{105 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (a^3 (5 A+7 C)+21 a^2 b B+21 a b^2 (A+3 C)+21 b^3 B\right )}{21 d}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (3 a^3 B+3 a^2 b (3 A+5 C)+15 a b^2 B+5 b^3 (A-C)\right )}{5 d}+\frac {2 \sin (c+d x) \left (21 a^3 B+21 a^2 b (3 A+5 C)+98 a b^2 B+24 A b^3\right )}{35 d \sqrt {\cos (c+d x)}}+\frac {2 (7 a B+6 A b) \sin (c+d x) (a+b \cos (c+d x))^2}{35 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^3}{7 d \cos ^{\frac {7}{2}}(c+d x)} \]
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Rubi [A] time = 0.85, antiderivative size = 294, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.140, Rules used = {3047, 3031, 3021, 2748, 2641, 2639} \[ \frac {2 F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (a^3 (5 A+7 C)+21 a^2 b B+21 a b^2 (A+3 C)+21 b^3 B\right )}{21 d}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (3 a^2 b (3 A+5 C)+3 a^3 B+15 a b^2 B+5 b^3 (A-C)\right )}{5 d}+\frac {2 a \sin (c+d x) \left (5 a^2 (5 A+7 C)+63 a b B+24 A b^2\right )}{105 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 \sin (c+d x) \left (21 a^2 b (3 A+5 C)+21 a^3 B+98 a b^2 B+24 A b^3\right )}{35 d \sqrt {\cos (c+d x)}}+\frac {2 (7 a B+6 A b) \sin (c+d x) (a+b \cos (c+d x))^2}{35 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^3}{7 d \cos ^{\frac {7}{2}}(c+d x)} \]
Antiderivative was successfully verified.
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Rule 2639
Rule 2641
Rule 2748
Rule 3021
Rule 3031
Rule 3047
Rubi steps
\begin {align*} \int \frac {(a+b \cos (c+d x))^3 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {9}{2}}(c+d x)} \, dx &=\frac {2 A (a+b \cos (c+d x))^3 \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {2}{7} \int \frac {(a+b \cos (c+d x))^2 \left (\frac {1}{2} (6 A b+7 a B)+\frac {1}{2} (5 a A+7 b B+7 a C) \cos (c+d x)-\frac {1}{2} b (A-7 C) \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx\\ &=\frac {2 (6 A b+7 a B) (a+b \cos (c+d x))^2 \sin (c+d x)}{35 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 A (a+b \cos (c+d x))^3 \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {4}{35} \int \frac {(a+b \cos (c+d x)) \left (\frac {1}{4} \left (24 A b^2+63 a b B+5 a^2 (5 A+7 C)\right )+\frac {1}{4} \left (38 a A b+21 a^2 B+35 b^2 B+70 a b C\right ) \cos (c+d x)-\frac {1}{4} b (11 A b+7 a B-35 b C) \cos ^2(c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx\\ &=\frac {2 a \left (24 A b^2+63 a b B+5 a^2 (5 A+7 C)\right ) \sin (c+d x)}{105 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 (6 A b+7 a B) (a+b \cos (c+d x))^2 \sin (c+d x)}{35 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 A (a+b \cos (c+d x))^3 \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}-\frac {8}{105} \int \frac {-\frac {3}{8} \left (24 A b^3+21 a^3 B+98 a b^2 B+21 a^2 b (3 A+5 C)\right )-\frac {5}{8} \left (21 a^2 b B+21 b^3 B+21 a b^2 (A+3 C)+a^3 (5 A+7 C)\right ) \cos (c+d x)+\frac {3}{8} b^2 (11 A b+7 a B-35 b C) \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x)} \, dx\\ &=\frac {2 a \left (24 A b^2+63 a b B+5 a^2 (5 A+7 C)\right ) \sin (c+d x)}{105 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (24 A b^3+21 a^3 B+98 a b^2 B+21 a^2 b (3 A+5 C)\right ) \sin (c+d x)}{35 d \sqrt {\cos (c+d x)}}+\frac {2 (6 A b+7 a B) (a+b \cos (c+d x))^2 \sin (c+d x)}{35 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 A (a+b \cos (c+d x))^3 \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}-\frac {16}{105} \int \frac {-\frac {5}{16} \left (21 a^2 b B+21 b^3 B+21 a b^2 (A+3 C)+a^3 (5 A+7 C)\right )+\frac {21}{16} \left (3 a^3 B+15 a b^2 B+5 b^3 (A-C)+3 a^2 b (3 A+5 C)\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)}} \, dx\\ &=\frac {2 a \left (24 A b^2+63 a b B+5 a^2 (5 A+7 C)\right ) \sin (c+d x)}{105 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (24 A b^3+21 a^3 B+98 a b^2 B+21 a^2 b (3 A+5 C)\right ) \sin (c+d x)}{35 d \sqrt {\cos (c+d x)}}+\frac {2 (6 A b+7 a B) (a+b \cos (c+d x))^2 \sin (c+d x)}{35 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 A (a+b \cos (c+d x))^3 \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}-\frac {1}{5} \left (3 a^3 B+15 a b^2 B+5 b^3 (A-C)+3 a^2 b (3 A+5 C)\right ) \int \sqrt {\cos (c+d x)} \, dx-\frac {1}{21} \left (-21 a^2 b B-21 b^3 B-21 a b^2 (A+3 C)-a^3 (5 A+7 C)\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx\\ &=-\frac {2 \left (3 a^3 B+15 a b^2 B+5 b^3 (A-C)+3 a^2 b (3 A+5 C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 \left (21 a^2 b B+21 b^3 B+21 a b^2 (A+3 C)+a^3 (5 A+7 C)\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}+\frac {2 a \left (24 A b^2+63 a b B+5 a^2 (5 A+7 C)\right ) \sin (c+d x)}{105 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (24 A b^3+21 a^3 B+98 a b^2 B+21 a^2 b (3 A+5 C)\right ) \sin (c+d x)}{35 d \sqrt {\cos (c+d x)}}+\frac {2 (6 A b+7 a B) (a+b \cos (c+d x))^2 \sin (c+d x)}{35 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 A (a+b \cos (c+d x))^3 \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}\\ \end {align*}
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Mathematica [A] time = 4.96, size = 251, normalized size = 0.85 \[ \frac {2 \left (\frac {15 a^3 A \sin (c+d x)}{\cos ^{\frac {7}{2}}(c+d x)}+\frac {5 a \sin (c+d x) \left (a^2 (5 A+7 C)+21 a b B+21 A b^2\right )}{\cos ^{\frac {3}{2}}(c+d x)}+\frac {21 a^2 (a B+3 A b) \sin (c+d x)}{\cos ^{\frac {5}{2}}(c+d x)}+5 F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (a^3 (5 A+7 C)+21 a^2 b B+21 a b^2 (A+3 C)+21 b^3 B\right )-21 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (3 a^3 B+3 a^2 b (3 A+5 C)+15 a b^2 B+5 b^3 (A-C)\right )+\frac {21 \sin (c+d x) \left (3 a^3 B+3 a^2 b (3 A+5 C)+15 a b^2 B+5 A b^3\right )}{\sqrt {\cos (c+d x)}}\right )}{105 d} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.76, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {C b^{3} \cos \left (d x + c\right )^{5} + {\left (3 \, C a b^{2} + B b^{3}\right )} \cos \left (d x + c\right )^{4} + A a^{3} + {\left (3 \, C a^{2} b + 3 \, B a b^{2} + A b^{3}\right )} \cos \left (d x + c\right )^{3} + {\left (C a^{3} + 3 \, B a^{2} b + 3 \, A a b^{2}\right )} \cos \left (d x + c\right )^{2} + {\left (B a^{3} + 3 \, A a^{2} b\right )} \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{\frac {9}{2}}}, x\right ) \]
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 \[ \int \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{3}}{\cos \left (d x + c\right )^{\frac {9}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 10.06, size = 1205, normalized size = 4.10 \[ \text {result too large to display} \]
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 \[ \int \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{3}}{\cos \left (d x + c\right )^{\frac {9}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 7.09, size = 442, normalized size = 1.50 \[ \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 {\frac {2\,A\,a^3\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {7}{4},\frac {1}{2};\ -\frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{7}+2\,A\,b^3\,{\cos \left (c+d\,x\right )}^3\,\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 )+2\,A\,a\,b^2\,{\cos \left (c+d\,x\right )}^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 )+\frac {6\,A\,a^2\,b\,\cos \left (c+d\,x\right )\,\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 )}^{7/2}\,\sqrt {1-{\cos \left (c+d\,x\right )}^2}}+\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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