Optimal. Leaf size=345 \[ -\frac {2 a^2 \sin (c+d x) \cos ^2(c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}+\frac {2 \left (2 a^2-b^2\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 b^3 d \left (a^2-b^2\right )}+\frac {2 \left (16 a^4-16 a^2 b^2-b^4\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{3 b^4 d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}-\frac {8 a \left (4 a^4-7 a^2 b^2+2 b^4\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{3 b^4 d \left (a^2-b^2\right )^2 \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {4 a^3 \left (3 a^2-5 b^2\right ) \sin (c+d x)}{3 b^3 d \left (a^2-b^2\right )^2 \sqrt {a+b \cos (c+d x)}} \]
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Rubi [A] time = 0.57, antiderivative size = 345, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {2792, 3031, 3023, 2752, 2663, 2661, 2655, 2653} \[ -\frac {2 a^2 \sin (c+d x) \cos ^2(c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}+\frac {4 a^3 \left (3 a^2-5 b^2\right ) \sin (c+d x)}{3 b^3 d \left (a^2-b^2\right )^2 \sqrt {a+b \cos (c+d x)}}+\frac {2 \left (2 a^2-b^2\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 b^3 d \left (a^2-b^2\right )}+\frac {2 \left (-16 a^2 b^2+16 a^4-b^4\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{3 b^4 d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}-\frac {8 a \left (-7 a^2 b^2+4 a^4+2 b^4\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{3 b^4 d \left (a^2-b^2\right )^2 \sqrt {\frac {a+b \cos (c+d x)}{a+b}}} \]
Antiderivative was successfully verified.
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Rule 2653
Rule 2655
Rule 2661
Rule 2663
Rule 2752
Rule 2792
Rule 3023
Rule 3031
Rubi steps
\begin {align*} \int \frac {\cos ^4(c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx &=-\frac {2 a^2 \cos ^2(c+d x) \sin (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}-\frac {2 \int \frac {\cos (c+d x) \left (2 a^2-\frac {3}{2} a b \cos (c+d x)-\frac {3}{2} \left (2 a^2-b^2\right ) \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^{3/2}} \, dx}{3 b \left (a^2-b^2\right )}\\ &=-\frac {2 a^2 \cos ^2(c+d x) \sin (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}+\frac {4 a^3 \left (3 a^2-5 b^2\right ) \sin (c+d x)}{3 b^3 \left (a^2-b^2\right )^2 d \sqrt {a+b \cos (c+d x)}}-\frac {4 \int \frac {\frac {1}{2} a^2 b \left (3 a^2-5 b^2\right )+\frac {1}{2} a \left (6 a^4-11 a^2 b^2+3 b^4\right ) \cos (c+d x)-\frac {3}{4} b \left (a^2-b^2\right ) \left (2 a^2-b^2\right ) \cos ^2(c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx}{3 b^3 \left (a^2-b^2\right )^2}\\ &=-\frac {2 a^2 \cos ^2(c+d x) \sin (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}+\frac {4 a^3 \left (3 a^2-5 b^2\right ) \sin (c+d x)}{3 b^3 \left (a^2-b^2\right )^2 d \sqrt {a+b \cos (c+d x)}}+\frac {2 \left (2 a^2-b^2\right ) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{3 b^3 \left (a^2-b^2\right ) d}-\frac {8 \int \frac {\frac {3}{8} b^2 \left (4 a^4-7 a^2 b^2-b^4\right )+\frac {3}{2} a b \left (4 a^4-7 a^2 b^2+2 b^4\right ) \cos (c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx}{9 b^4 \left (a^2-b^2\right )^2}\\ &=-\frac {2 a^2 \cos ^2(c+d x) \sin (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}+\frac {4 a^3 \left (3 a^2-5 b^2\right ) \sin (c+d x)}{3 b^3 \left (a^2-b^2\right )^2 d \sqrt {a+b \cos (c+d x)}}+\frac {2 \left (2 a^2-b^2\right ) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{3 b^3 \left (a^2-b^2\right ) d}+\frac {\left (16 a^4-16 a^2 b^2-b^4\right ) \int \frac {1}{\sqrt {a+b \cos (c+d x)}} \, dx}{3 b^4 \left (a^2-b^2\right )}-\frac {\left (4 a \left (4 a^4-7 a^2 b^2+2 b^4\right )\right ) \int \sqrt {a+b \cos (c+d x)} \, dx}{3 b^4 \left (a^2-b^2\right )^2}\\ &=-\frac {2 a^2 \cos ^2(c+d x) \sin (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}+\frac {4 a^3 \left (3 a^2-5 b^2\right ) \sin (c+d x)}{3 b^3 \left (a^2-b^2\right )^2 d \sqrt {a+b \cos (c+d x)}}+\frac {2 \left (2 a^2-b^2\right ) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{3 b^3 \left (a^2-b^2\right ) d}-\frac {\left (4 a \left (4 a^4-7 a^2 b^2+2 b^4\right ) \sqrt {a+b \cos (c+d x)}\right ) \int \sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}} \, dx}{3 b^4 \left (a^2-b^2\right )^2 \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {\left (\left (16 a^4-16 a^2 b^2-b^4\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}{3 b^4 \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\\ &=-\frac {8 a \left (4 a^4-7 a^2 b^2+2 b^4\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{3 b^4 \left (a^2-b^2\right )^2 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {2 \left (16 a^4-16 a^2 b^2-b^4\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 )}{3 b^4 \left (a^2-b^2\right ) d \sqrt {a+b \cos (c+d x)}}-\frac {2 a^2 \cos ^2(c+d x) \sin (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}+\frac {4 a^3 \left (3 a^2-5 b^2\right ) \sin (c+d x)}{3 b^3 \left (a^2-b^2\right )^2 d \sqrt {a+b \cos (c+d x)}}+\frac {2 \left (2 a^2-b^2\right ) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{3 b^3 \left (a^2-b^2\right ) d}\\ \end {align*}
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Mathematica [A] time = 1.64, size = 237, normalized size = 0.69 \[ \frac {2 \left (\frac {b \sin (c+d x) \left (16 a^6-25 a^4 b^2+\left (b^3-a^2 b\right )^2 \cos (2 (c+d x))+4 a b \left (5 a^4-8 a^2 b^2+b^4\right ) \cos (c+d x)+b^6\right )}{2 \left (a^2-b^2\right )^2}+\frac {\left (\frac {a+b \cos (c+d x)}{a+b}\right )^{3/2} \left (\left (16 a^5-16 a^4 b-16 a^3 b^2+16 a^2 b^3-a b^4+b^5\right ) F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )-4 \left (4 a^5-7 a^3 b^2+2 a b^4\right ) E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )\right )}{(a-b)^2}\right )}{3 b^4 d (a+b \cos (c+d x))^{3/2}} \]
Antiderivative was successfully verified.
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fricas [F] time = 1.64, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {b \cos \left (d x + c\right ) + a} \cos \left (d x + c\right )^{4}}{b^{3} \cos \left (d x + c\right )^{3} + 3 \, a b^{2} \cos \left (d x + c\right )^{2} + 3 \, a^{2} b \cos \left (d x + c\right ) + a^{3}}, x\right ) \]
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 \[ \int \frac {\cos \left (d x + c\right )^{4}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 3.38, size = 1291, normalized size = 3.74 \[ \text {result too large to display} \]
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 \[ \int \frac {\cos \left (d x + c\right )^{4}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}\,{d x} \]
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 \[ \int \frac {{\cos \left (c+d\,x\right )}^4}{{\left (a+b\,\cos \left (c+d\,x\right )\right )}^{5/2}} \,d x \]
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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