Optimal. Leaf size=273 \[ \frac {2 a \sin (c+d x) \left (3 a^2 (3 A+5 C)+35 a b B+24 A b^2\right )}{15 d \sqrt {\cos (c+d x)}}+\frac {2 F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (a^3 B+3 a^2 b (A+3 C)+9 a b^2 B+b^3 (3 A+C)\right )}{3 d}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (a^3 (3 A+5 C)+15 a^2 b B+15 a b^2 (A-C)-5 b^3 B\right )}{5 d}-\frac {2 b^2 \sin (c+d x) \sqrt {\cos (c+d x)} (5 a B+9 A b-5 b C)}{15 d}+\frac {2 (5 a B+6 A b) \sin (c+d x) (a+b \cos (c+d x))^2}{15 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^3}{5 d \cos ^{\frac {5}{2}}(c+d x)} \]
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Rubi [A] time = 0.82, antiderivative size = 273, 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, 3023, 2748, 2641, 2639} \[ \frac {2 F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (3 a^2 b (A+3 C)+a^3 B+9 a b^2 B+b^3 (3 A+C)\right )}{3 d}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (a^3 (3 A+5 C)+15 a^2 b B+15 a b^2 (A-C)-5 b^3 B\right )}{5 d}+\frac {2 a \sin (c+d x) \left (3 a^2 (3 A+5 C)+35 a b B+24 A b^2\right )}{15 d \sqrt {\cos (c+d x)}}-\frac {2 b^2 \sin (c+d x) \sqrt {\cos (c+d x)} (5 a B+9 A b-5 b C)}{15 d}+\frac {2 (5 a B+6 A b) \sin (c+d x) (a+b \cos (c+d x))^2}{15 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^3}{5 d \cos ^{\frac {5}{2}}(c+d x)} \]
Antiderivative was successfully verified.
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Rule 2639
Rule 2641
Rule 2748
Rule 3023
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 {7}{2}}(c+d x)} \, dx &=\frac {2 A (a+b \cos (c+d x))^3 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2}{5} \int \frac {(a+b \cos (c+d x))^2 \left (\frac {1}{2} (6 A b+5 a B)+\frac {1}{2} (3 a A+5 b B+5 a C) \cos (c+d x)-\frac {1}{2} b (3 A-5 C) \cos ^2(c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx\\ &=\frac {2 (6 A b+5 a B) (a+b \cos (c+d x))^2 \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 A (a+b \cos (c+d x))^3 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {4}{15} \int \frac {(a+b \cos (c+d x)) \left (\frac {1}{4} \left (24 A b^2+35 a b B+3 a^2 (3 A+5 C)\right )+\frac {1}{4} \left (5 a^2 B+15 b^2 B+6 a b (A+5 C)\right ) \cos (c+d x)-\frac {3}{4} b (9 A b+5 a B-5 b C) \cos ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx\\ &=\frac {2 a \left (24 A b^2+35 a b B+3 a^2 (3 A+5 C)\right ) \sin (c+d x)}{15 d \sqrt {\cos (c+d x)}}+\frac {2 (6 A b+5 a B) (a+b \cos (c+d x))^2 \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 A (a+b \cos (c+d x))^3 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}-\frac {8}{15} \int \frac {\frac {1}{8} \left (-24 A b^3-5 a^3 B-50 a b^2 B-15 a^2 b (A+3 C)\right )+\frac {3}{8} \left (15 a^2 b B-5 b^3 B+15 a b^2 (A-C)+a^3 (3 A+5 C)\right ) \cos (c+d x)+\frac {3}{8} b^2 (9 A b+5 a B-5 b C) \cos ^2(c+d x)}{\sqrt {\cos (c+d x)}} \, dx\\ &=\frac {2 a \left (24 A b^2+35 a b B+3 a^2 (3 A+5 C)\right ) \sin (c+d x)}{15 d \sqrt {\cos (c+d x)}}-\frac {2 b^2 (9 A b+5 a B-5 b C) \sqrt {\cos (c+d x)} \sin (c+d x)}{15 d}+\frac {2 (6 A b+5 a B) (a+b \cos (c+d x))^2 \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 A (a+b \cos (c+d x))^3 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}-\frac {16}{45} \int \frac {-\frac {15}{16} \left (a^3 B+9 a b^2 B+b^3 (3 A+C)+3 a^2 b (A+3 C)\right )+\frac {9}{16} \left (15 a^2 b B-5 b^3 B+15 a b^2 (A-C)+a^3 (3 A+5 C)\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)}} \, dx\\ &=\frac {2 a \left (24 A b^2+35 a b B+3 a^2 (3 A+5 C)\right ) \sin (c+d x)}{15 d \sqrt {\cos (c+d x)}}-\frac {2 b^2 (9 A b+5 a B-5 b C) \sqrt {\cos (c+d x)} \sin (c+d x)}{15 d}+\frac {2 (6 A b+5 a B) (a+b \cos (c+d x))^2 \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 A (a+b \cos (c+d x))^3 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}-\frac {1}{3} \left (-a^3 B-9 a b^2 B-b^3 (3 A+C)-3 a^2 b (A+3 C)\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx-\frac {1}{5} \left (15 a^2 b B-5 b^3 B+15 a b^2 (A-C)+a^3 (3 A+5 C)\right ) \int \sqrt {\cos (c+d x)} \, dx\\ &=-\frac {2 \left (15 a^2 b B-5 b^3 B+15 a b^2 (A-C)+a^3 (3 A+5 C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 \left (a^3 B+9 a b^2 B+b^3 (3 A+C)+3 a^2 b (A+3 C)\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d}+\frac {2 a \left (24 A b^2+35 a b B+3 a^2 (3 A+5 C)\right ) \sin (c+d x)}{15 d \sqrt {\cos (c+d x)}}-\frac {2 b^2 (9 A b+5 a B-5 b C) \sqrt {\cos (c+d x)} \sin (c+d x)}{15 d}+\frac {2 (6 A b+5 a B) (a+b \cos (c+d x))^2 \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 A (a+b \cos (c+d x))^3 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\\ \end {align*}
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Mathematica [A] time = 2.23, size = 248, normalized size = 0.91 \[ \frac {9 a^3 A \sin (2 (c+d x))+6 a^3 A \tan (c+d x)+10 a^3 B \sin (c+d x)+15 a^3 C \sin (2 (c+d x))+30 a^2 A b \sin (c+d x)+45 a^2 b B \sin (2 (c+d x))+10 \cos ^{\frac {3}{2}}(c+d x) F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (a^3 B+3 a^2 b (A+3 C)+9 a b^2 B+b^3 (3 A+C)\right )-6 \cos ^{\frac {3}{2}}(c+d x) E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (a^3 (3 A+5 C)+15 a^2 b B+15 a b^2 (A-C)-5 b^3 B\right )+45 a A b^2 \sin (2 (c+d x))+10 b^3 C \sin (c+d x) \cos ^2(c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.65, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {C b^{3} \cos \left (d x + c\right )^{5} + {\left (3 \, C a b^{2} + B b^{3}\right )} \cos \left (d x + c\right )^{4} + A a^{3} + {\left (3 \, C a^{2} b + 3 \, B a b^{2} + A b^{3}\right )} \cos \left (d x + c\right )^{3} + {\left (C a^{3} + 3 \, B a^{2} b + 3 \, A a b^{2}\right )} \cos \left (d x + c\right )^{2} + {\left (B a^{3} + 3 \, A a^{2} b\right )} \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{\frac {7}{2}}}, x\right ) \]
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 \[ \int \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{3}}{\cos \left (d x + c\right )^{\frac {7}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 9.67, size = 1419, normalized size = 5.20 \[ \text {result too large to display} \]
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 \[ \int \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{3}}{\cos \left (d x + c\right )^{\frac {7}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 5.38, size = 414, normalized size = 1.52 \[ \frac {2\,\left (B\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )\,b^3+3\,B\,a\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )\,b^2\right )}{d}+\frac {C\,b^3\,\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^3\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {6\,C\,a\,b^2\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {6\,C\,a^2\,b\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,A\,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\,B\,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 {2\,C\,a^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 )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {6\,A\,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\,A\,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\,B\,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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