Optimal. Leaf size=321 \[ \frac {8 a b \left (a^2 (A+3 C)+b^2 (3 A+C)\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d}-\frac {2 b^2 \left (3 a^2 (3 A+5 C)+b^2 (59 A-3 C)\right ) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{15 d}+\frac {2 \left (a^2 (3 A+5 C)+16 A b^2\right ) \sin (c+d x) (a+b \cos (c+d x))^2}{5 d \sqrt {\cos (c+d x)}}-\frac {4 a b \left (3 a^2 (3 A+5 C)+2 b^2 (33 A-5 C)\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{15 d}-\frac {2 \left (a^4 (3 A+5 C)+30 a^2 b^2 (A-C)-b^4 (5 A+3 C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^4}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {16 A b \sin (c+d x) (a+b \cos (c+d x))^3}{15 d \cos ^{\frac {3}{2}}(c+d x)} \]
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Rubi [A] time = 1.22, antiderivative size = 321, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3048, 3047, 3033, 3023, 2748, 2641, 2639} \[ \frac {8 a b \left (a^2 (A+3 C)+b^2 (3 A+C)\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d}-\frac {2 \left (30 a^2 b^2 (A-C)+a^4 (3 A+5 C)-b^4 (5 A+3 C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}-\frac {2 b^2 \left (3 a^2 (3 A+5 C)+b^2 (59 A-3 C)\right ) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{15 d}+\frac {2 \left (a^2 (3 A+5 C)+16 A b^2\right ) \sin (c+d x) (a+b \cos (c+d x))^2}{5 d \sqrt {\cos (c+d x)}}-\frac {4 a b \left (3 a^2 (3 A+5 C)+2 b^2 (33 A-5 C)\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{15 d}+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^4}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {16 A b \sin (c+d x) (a+b \cos (c+d x))^3}{15 d \cos ^{\frac {3}{2}}(c+d x)} \]
Antiderivative was successfully verified.
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Rule 2639
Rule 2641
Rule 2748
Rule 3023
Rule 3033
Rule 3047
Rule 3048
Rubi steps
\begin {align*} \int \frac {(a+b \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx &=\frac {2 A (a+b \cos (c+d x))^4 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2}{5} \int \frac {(a+b \cos (c+d x))^3 \left (4 A b+\frac {1}{2} a (3 A+5 C) \cos (c+d x)-\frac {5}{2} b (A-C) \cos ^2(c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx\\ &=\frac {16 A b (a+b \cos (c+d x))^3 \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 A (a+b \cos (c+d x))^4 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {4}{15} \int \frac {(a+b \cos (c+d x))^2 \left (\frac {3}{4} \left (16 A b^2+a^2 (3 A+5 C)\right )+\frac {1}{2} a b (A+15 C) \cos (c+d x)-\frac {5}{4} b^2 (11 A-3 C) \cos ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx\\ &=\frac {2 \left (16 A b^2+a^2 (3 A+5 C)\right ) (a+b \cos (c+d x))^2 \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}+\frac {16 A b (a+b \cos (c+d x))^3 \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 A (a+b \cos (c+d x))^4 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {8}{15} \int \frac {(a+b \cos (c+d x)) \left (\frac {1}{4} b \left (96 A b^2+a^2 (19 A+45 C)\right )-\frac {1}{8} a \left (b^2 (101 A-45 C)+3 a^2 (3 A+5 C)\right ) \cos (c+d x)-\frac {5}{8} b \left (b^2 (59 A-3 C)+3 a^2 (3 A+5 C)\right ) \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx\\ &=-\frac {2 b^2 \left (b^2 (59 A-3 C)+3 a^2 (3 A+5 C)\right ) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac {2 \left (16 A b^2+a^2 (3 A+5 C)\right ) (a+b \cos (c+d x))^2 \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}+\frac {16 A b (a+b \cos (c+d x))^3 \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 A (a+b \cos (c+d x))^4 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {16}{75} \int \frac {\frac {5}{8} a b \left (96 A b^2+a^2 (19 A+45 C)\right )-\frac {15}{16} \left (30 a^2 b^2 (A-C)-b^4 (5 A+3 C)+a^4 (3 A+5 C)\right ) \cos (c+d x)-\frac {15}{8} a b \left (2 b^2 (33 A-5 C)+3 a^2 (3 A+5 C)\right ) \cos ^2(c+d x)}{\sqrt {\cos (c+d x)}} \, dx\\ &=-\frac {4 a b \left (2 b^2 (33 A-5 C)+3 a^2 (3 A+5 C)\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{15 d}-\frac {2 b^2 \left (b^2 (59 A-3 C)+3 a^2 (3 A+5 C)\right ) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac {2 \left (16 A b^2+a^2 (3 A+5 C)\right ) (a+b \cos (c+d x))^2 \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}+\frac {16 A b (a+b \cos (c+d x))^3 \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 A (a+b \cos (c+d x))^4 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {32}{225} \int \frac {\frac {75}{8} a b \left (b^2 (3 A+C)+a^2 (A+3 C)\right )-\frac {45}{32} \left (30 a^2 b^2 (A-C)-b^4 (5 A+3 C)+a^4 (3 A+5 C)\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)}} \, dx\\ &=-\frac {4 a b \left (2 b^2 (33 A-5 C)+3 a^2 (3 A+5 C)\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{15 d}-\frac {2 b^2 \left (b^2 (59 A-3 C)+3 a^2 (3 A+5 C)\right ) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac {2 \left (16 A b^2+a^2 (3 A+5 C)\right ) (a+b \cos (c+d x))^2 \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}+\frac {16 A b (a+b \cos (c+d x))^3 \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 A (a+b \cos (c+d x))^4 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {1}{3} \left (4 a b \left (b^2 (3 A+C)+a^2 (A+3 C)\right )\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx+\frac {1}{5} \left (-30 a^2 b^2 (A-C)+b^4 (5 A+3 C)-a^4 (3 A+5 C)\right ) \int \sqrt {\cos (c+d x)} \, dx\\ &=-\frac {2 \left (30 a^2 b^2 (A-C)-b^4 (5 A+3 C)+a^4 (3 A+5 C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {8 a b \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 {4 a b \left (2 b^2 (33 A-5 C)+3 a^2 (3 A+5 C)\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{15 d}-\frac {2 b^2 \left (b^2 (59 A-3 C)+3 a^2 (3 A+5 C)\right ) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac {2 \left (16 A b^2+a^2 (3 A+5 C)\right ) (a+b \cos (c+d x))^2 \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}+\frac {16 A b (a+b \cos (c+d x))^3 \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 A (a+b \cos (c+d x))^4 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\\ \end {align*}
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Mathematica [A] time = 1.65, size = 233, normalized size = 0.73 \[ \frac {9 a^4 A \sin (2 (c+d x))+6 a^4 A \tan (c+d x)+15 a^4 C \sin (2 (c+d x))+40 a^3 A b \sin (c+d x)+40 a b \left (a^2 (A+3 C)+b^2 (3 A+C)\right ) \cos ^{\frac {3}{2}}(c+d x) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )+90 a^2 A b^2 \sin (2 (c+d x))-6 \left (a^4 (3 A+5 C)+30 a^2 b^2 (A-C)-b^4 (5 A+3 C)\right ) \cos ^{\frac {3}{2}}(c+d x) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+40 a b^3 C \sin (c+d x) \cos ^2(c+d x)+6 b^4 C \sin (c+d x) \cos ^3(c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.86, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {C b^{4} \cos \left (d x + c\right )^{6} + 4 \, C a b^{3} \cos \left (d x + c\right )^{5} + 4 \, A a^{3} b \cos \left (d x + c\right ) + A a^{4} + {\left (6 \, C a^{2} b^{2} + A b^{4}\right )} \cos \left (d x + c\right )^{4} + 4 \, {\left (C a^{3} b + A a b^{3}\right )} \cos \left (d x + c\right )^{3} + {\left (C a^{4} + 6 \, A a^{2} b^{2}\right )} \cos \left (d x + c\right )^{2}}{\cos \left (d x + c\right )^{\frac {7}{2}}}, x\right ) \]
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 \[ \int \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{4}}{\cos \left (d x + c\right )^{\frac {7}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 8.85, size = 1622, normalized size = 5.05 \[ \text {result too large to display} \]
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 \[ \int \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{4}}{\cos \left (d x + c\right )^{\frac {7}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 4.99, size = 355, normalized size = 1.11 \[ \frac {2\,A\,b^4\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {8\,A\,a\,b^3\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {8\,C\,a^3\,b\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {4\,C\,a\,b^3\,\left (\frac {2\,\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )}{3}+\frac {2\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{3}\right )}{d}+\frac {12\,C\,a^2\,b^2\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,A\,a^4\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {5}{4},\frac {1}{2};\ -\frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{5\,d\,{\cos \left (c+d\,x\right )}^{5/2}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {2\,C\,a^4\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {2\,C\,b^4\,{\cos \left (c+d\,x\right )}^{7/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {7}{4};\ \frac {11}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{7\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {8\,A\,a^3\,b\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{3\,d\,{\cos \left (c+d\,x\right )}^{3/2}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {12\,A\,a^2\,b^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \]
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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