Optimal. Leaf size=254 \[ \frac {(3 A-43 B+115 C) \tan ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{16 \sqrt {2} a^{5/2} d}+\frac {(2 B-5 C) \sin ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{a^{5/2} d}+\frac {(3 A-11 B+35 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{16 a^2 d \sqrt {a \cos (c+d x)+a}}-\frac {(A-B+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}+\frac {(A+7 B-15 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{16 a d (a \cos (c+d x)+a)^{3/2}} \]
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Rubi [A] time = 0.87, antiderivative size = 254, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.178, Rules used = {3041, 2977, 2983, 2982, 2782, 205, 2774, 216} \[ \frac {(3 A-11 B+35 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{16 a^2 d \sqrt {a \cos (c+d x)+a}}+\frac {(3 A-43 B+115 C) \tan ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{16 \sqrt {2} a^{5/2} d}+\frac {(2 B-5 C) \sin ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{a^{5/2} d}-\frac {(A-B+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}+\frac {(A+7 B-15 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{16 a d (a \cos (c+d x)+a)^{3/2}} \]
Antiderivative was successfully verified.
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Rule 205
Rule 216
Rule 2774
Rule 2782
Rule 2977
Rule 2982
Rule 2983
Rule 3041
Rubi steps
\begin {align*} \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^{5/2}} \, dx &=-\frac {(A-B+C) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}+\frac {\int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (\frac {1}{2} a (3 A+5 B-5 C)+a (A-B+5 C) \cos (c+d x)\right )}{(a+a \cos (c+d x))^{3/2}} \, dx}{4 a^2}\\ &=-\frac {(A-B+C) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}+\frac {(A+7 B-15 C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac {\int \frac {\sqrt {\cos (c+d x)} \left (\frac {3}{4} a^2 (A+7 B-15 C)+\frac {1}{2} a^2 (3 A-11 B+35 C) \cos (c+d x)\right )}{\sqrt {a+a \cos (c+d x)}} \, dx}{8 a^4}\\ &=-\frac {(A-B+C) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}+\frac {(A+7 B-15 C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac {(3 A-11 B+35 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{16 a^2 d \sqrt {a+a \cos (c+d x)}}+\frac {\int \frac {\frac {1}{4} a^3 (3 A-11 B+35 C)+4 a^3 (2 B-5 C) \cos (c+d x)}{\sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}} \, dx}{8 a^5}\\ &=-\frac {(A-B+C) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}+\frac {(A+7 B-15 C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac {(3 A-11 B+35 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{16 a^2 d \sqrt {a+a \cos (c+d x)}}+\frac {(2 B-5 C) \int \frac {\sqrt {a+a \cos (c+d x)}}{\sqrt {\cos (c+d x)}} \, dx}{2 a^3}+\frac {(3 A-43 B+115 C) \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}} \, dx}{32 a^2}\\ &=-\frac {(A-B+C) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}+\frac {(A+7 B-15 C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac {(3 A-11 B+35 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{16 a^2 d \sqrt {a+a \cos (c+d x)}}-\frac {(2 B-5 C) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-\frac {x^2}{a}}} \, dx,x,-\frac {a \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{a^3 d}-\frac {(3 A-43 B+115 C) \operatorname {Subst}\left (\int \frac {1}{2 a^2+a x^2} \, dx,x,-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\right )}{16 a d}\\ &=\frac {(2 B-5 C) \sin ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{a^{5/2} d}+\frac {(3 A-43 B+115 C) \tan ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {(A-B+C) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}+\frac {(A+7 B-15 C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac {(3 A-11 B+35 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{16 a^2 d \sqrt {a+a \cos (c+d x)}}\\ \end {align*}
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Mathematica [C] time = 4.10, size = 434, normalized size = 1.71 \[ \frac {\cos ^5\left (\frac {1}{2} (c+d x)\right ) \left (\sqrt {\cos (c+d x)} \tan \left (\frac {1}{2} (c+d x)\right ) \sec ^3\left (\frac {1}{2} (c+d x)\right ) ((7 A-15 B+55 C) \cos (c+d x)+3 A-11 B+8 C \cos (2 (c+d x))+43 C)+\frac {\sqrt {2} e^{\frac {1}{2} i (c+d x)} \sqrt {e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right )} \left (-i \sqrt {2} (3 A-43 B+115 C) \log \left (1+e^{i (c+d x)}\right )+3 i \sqrt {2} A \log \left (\sqrt {2} \sqrt {1+e^{2 i (c+d x)}}-e^{i (c+d x)}+1\right )-16 i (2 B-5 C) \sinh ^{-1}\left (e^{i (c+d x)}\right )+32 i B \log \left (1+\sqrt {1+e^{2 i (c+d x)}}\right )-43 i \sqrt {2} B \log \left (\sqrt {2} \sqrt {1+e^{2 i (c+d x)}}-e^{i (c+d x)}+1\right )+32 B d x-80 i C \log \left (1+\sqrt {1+e^{2 i (c+d x)}}\right )+115 i \sqrt {2} C \log \left (\sqrt {2} \sqrt {1+e^{2 i (c+d x)}}-e^{i (c+d x)}+1\right )-80 C d x\right )}{\sqrt {1+e^{2 i (c+d x)}}}\right )}{8 d (a (\cos (c+d x)+1))^{5/2}} \]
Antiderivative was successfully verified.
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fricas [A] time = 122.48, size = 319, normalized size = 1.26 \[ -\frac {\sqrt {2} {\left ({\left (3 \, A - 43 \, B + 115 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (3 \, A - 43 \, B + 115 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (3 \, A - 43 \, B + 115 \, C\right )} \cos \left (d x + c\right ) + 3 \, A - 43 \, B + 115 \, C\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right ) - 2 \, {\left (16 \, C \cos \left (d x + c\right )^{2} + {\left (7 \, A - 15 \, B + 55 \, C\right )} \cos \left (d x + c\right ) + 3 \, A - 11 \, B + 35 \, C\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + 32 \, {\left ({\left (2 \, B - 5 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (2 \, B - 5 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (2 \, B - 5 \, C\right )} \cos \left (d x + c\right ) + 2 \, B - 5 \, C\right )} \sqrt {a} \arctan \left (\frac {\sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right )}{32 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \]
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 \[ \int \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{\frac {3}{2}}}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.37, size = 881, normalized size = 3.47 \[ \text {result too large to display} \]
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 \[ \int \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{\frac {3}{2}}}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\cos \left (c+d\,x\right )}^{3/2}\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right )}{{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{5/2}} \,d x \]
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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