Optimal. Leaf size=389 \[ -\frac {4 a \left (a^2 (-C)+4 A b^2+5 b^2 C\right ) \sin (c+d x)}{15 b d \left (a^2-b^2\right )^2 (a+b \cos (c+d x))^{3/2}}-\frac {2 \left (a^2 C+A b^2\right ) \sin (c+d x)}{5 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{5/2}}-\frac {4 a \left (4 A b^2-C \left (a^2-5 b^2\right )\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{15 b^2 d \left (a^2-b^2\right )^2 \sqrt {a+b \cos (c+d x)}}+\frac {2 \left (2 a^4 C-a^2 b^2 (23 A+19 C)-3 b^4 (3 A+5 C)\right ) \sin (c+d x)}{15 b d \left (a^2-b^2\right )^3 \sqrt {a+b \cos (c+d x)}}-\frac {2 \left (2 a^4 C-a^2 b^2 (23 A+19 C)-3 b^4 (3 A+5 C)\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{15 b^2 d \left (a^2-b^2\right )^3 \sqrt {\frac {a+b \cos (c+d x)}{a+b}}} \]
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Rubi [A] time = 0.61, antiderivative size = 389, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {3022, 2754, 2752, 2663, 2661, 2655, 2653} \[ \frac {2 \left (-a^2 b^2 (23 A+19 C)+2 a^4 C-3 b^4 (3 A+5 C)\right ) \sin (c+d x)}{15 b d \left (a^2-b^2\right )^3 \sqrt {a+b \cos (c+d x)}}-\frac {4 a \left (a^2 (-C)+4 A b^2+5 b^2 C\right ) \sin (c+d x)}{15 b d \left (a^2-b^2\right )^2 (a+b \cos (c+d x))^{3/2}}-\frac {2 \left (a^2 C+A b^2\right ) \sin (c+d x)}{5 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{5/2}}-\frac {4 a \left (4 A b^2-C \left (a^2-5 b^2\right )\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{15 b^2 d \left (a^2-b^2\right )^2 \sqrt {a+b \cos (c+d x)}}-\frac {2 \left (-a^2 b^2 (23 A+19 C)+2 a^4 C-3 b^4 (3 A+5 C)\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{15 b^2 d \left (a^2-b^2\right )^3 \sqrt {\frac {a+b \cos (c+d x)}{a+b}}} \]
Antiderivative was successfully verified.
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Rule 2653
Rule 2655
Rule 2661
Rule 2663
Rule 2752
Rule 2754
Rule 3022
Rubi steps
\begin {align*} \int \frac {A+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^{7/2}} \, dx &=-\frac {2 \left (A b^2+a^2 C\right ) \sin (c+d x)}{5 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{5/2}}-\frac {2 \int \frac {-\frac {5}{2} a b (A+C)+\frac {1}{2} \left (3 A b^2-2 a^2 C+5 b^2 C\right ) \cos (c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx}{5 b \left (a^2-b^2\right )}\\ &=-\frac {2 \left (A b^2+a^2 C\right ) \sin (c+d x)}{5 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{5/2}}-\frac {4 a \left (4 A b^2-a^2 C+5 b^2 C\right ) \sin (c+d x)}{15 b \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))^{3/2}}+\frac {4 \int \frac {\frac {3}{4} b \left (a^2 (5 A+3 C)+b^2 (3 A+5 C)\right )-\frac {1}{2} a \left (4 A b^2-\left (a^2-5 b^2\right ) C\right ) \cos (c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx}{15 b \left (a^2-b^2\right )^2}\\ &=-\frac {2 \left (A b^2+a^2 C\right ) \sin (c+d x)}{5 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{5/2}}-\frac {4 a \left (4 A b^2-a^2 C+5 b^2 C\right ) \sin (c+d x)}{15 b \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))^{3/2}}+\frac {2 \left (2 a^4 C-3 b^4 (3 A+5 C)-a^2 b^2 (23 A+19 C)\right ) \sin (c+d x)}{15 b \left (a^2-b^2\right )^3 d \sqrt {a+b \cos (c+d x)}}-\frac {8 \int \frac {-\frac {1}{8} a b \left (a^2 (15 A+7 C)+b^2 (17 A+25 C)\right )+\frac {1}{8} \left (2 a^4 C-3 b^4 (3 A+5 C)-a^2 b^2 (23 A+19 C)\right ) \cos (c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx}{15 b \left (a^2-b^2\right )^3}\\ &=-\frac {2 \left (A b^2+a^2 C\right ) \sin (c+d x)}{5 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{5/2}}-\frac {4 a \left (4 A b^2-a^2 C+5 b^2 C\right ) \sin (c+d x)}{15 b \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))^{3/2}}+\frac {2 \left (2 a^4 C-3 b^4 (3 A+5 C)-a^2 b^2 (23 A+19 C)\right ) \sin (c+d x)}{15 b \left (a^2-b^2\right )^3 d \sqrt {a+b \cos (c+d x)}}-\frac {\left (2 a \left (4 A b^2-a^2 C+5 b^2 C\right )\right ) \int \frac {1}{\sqrt {a+b \cos (c+d x)}} \, dx}{15 b^2 \left (a^2-b^2\right )^2}-\frac {\left (2 a^4 C-3 b^4 (3 A+5 C)-a^2 b^2 (23 A+19 C)\right ) \int \sqrt {a+b \cos (c+d x)} \, dx}{15 b^2 \left (a^2-b^2\right )^3}\\ &=-\frac {2 \left (A b^2+a^2 C\right ) \sin (c+d x)}{5 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{5/2}}-\frac {4 a \left (4 A b^2-a^2 C+5 b^2 C\right ) \sin (c+d x)}{15 b \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))^{3/2}}+\frac {2 \left (2 a^4 C-3 b^4 (3 A+5 C)-a^2 b^2 (23 A+19 C)\right ) \sin (c+d x)}{15 b \left (a^2-b^2\right )^3 d \sqrt {a+b \cos (c+d x)}}-\frac {\left (\left (2 a^4 C-3 b^4 (3 A+5 C)-a^2 b^2 (23 A+19 C)\right ) \sqrt {a+b \cos (c+d x)}\right ) \int \sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}} \, dx}{15 b^2 \left (a^2-b^2\right )^3 \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {\left (2 a \left (4 A b^2-a^2 C+5 b^2 C\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}}\right ) \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}} \, dx}{15 b^2 \left (a^2-b^2\right )^2 \sqrt {a+b \cos (c+d x)}}\\ &=-\frac {2 \left (2 a^4 C-3 b^4 (3 A+5 C)-a^2 b^2 (23 A+19 C)\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{15 b^2 \left (a^2-b^2\right )^3 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {4 a \left (4 A b^2-\left (a^2-5 b^2\right ) C\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{15 b^2 \left (a^2-b^2\right )^2 d \sqrt {a+b \cos (c+d x)}}-\frac {2 \left (A b^2+a^2 C\right ) \sin (c+d x)}{5 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{5/2}}-\frac {4 a \left (4 A b^2-a^2 C+5 b^2 C\right ) \sin (c+d x)}{15 b \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))^{3/2}}+\frac {2 \left (2 a^4 C-3 b^4 (3 A+5 C)-a^2 b^2 (23 A+19 C)\right ) \sin (c+d x)}{15 b \left (a^2-b^2\right )^3 d \sqrt {a+b \cos (c+d x)}}\\ \end {align*}
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Mathematica [A] time = 2.76, size = 314, normalized size = 0.81 \[ \frac {2 \left (\frac {\left (\frac {a+b \cos (c+d x)}{a+b}\right )^{5/2} \left (2 a (a-b) \left (C \left (a^2-5 b^2\right )-4 A b^2\right ) F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )+\left (-2 a^4 C+a^2 b^2 (23 A+19 C)+3 b^4 (3 A+5 C)\right ) E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )\right )}{(a-b)^3}+\frac {b \sin (c+d x) \left (-2 a^6 C+68 a^4 A b^2+48 a^4 b^2 C+13 a^2 A b^4+35 a^2 b^4 C-4 a b \left (3 a^4 C-a^2 b^2 (27 A+25 C)-5 b^4 (A+2 C)\right ) \cos (c+d x)+\left (-2 a^4 b^2 C+a^2 b^4 (23 A+19 C)+3 b^6 (3 A+5 C)\right ) \cos (2 (c+d x))+15 A b^6+15 b^6 C\right )}{2 \left (b^2-a^2\right )^3}\right )}{15 b^2 d (a+b \cos (c+d x))^{5/2}} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.95, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} \sqrt {b \cos \left (d x + c\right ) + a}}{b^{4} \cos \left (d x + c\right )^{4} + 4 \, a b^{3} \cos \left (d x + c\right )^{3} + 6 \, a^{2} b^{2} \cos \left (d x + c\right )^{2} + 4 \, a^{3} b \cos \left (d x + c\right ) + a^{4}}, x\right ) \]
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 \[ \int \frac {C \cos \left (d x + c\right )^{2} + A}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {7}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 13.15, size = 1305, normalized size = 3.35 \[ \text {result too large to display} \]
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 \[ \int \frac {C \cos \left (d x + c\right )^{2} + A}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {7}{2}}}\,{d x} \]
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 \[ \int \frac {C\,{\cos \left (c+d\,x\right )}^2+A}{{\left (a+b\,\cos \left (c+d\,x\right )\right )}^{7/2}} \,d x \]
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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