Optimal. Leaf size=345 \[ -\frac{2 a^2 \sin (c+d x) \cos ^2(c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}+\frac{4 a^3 \left (3 a^2-5 b^2\right ) \sin (c+d x)}{3 b^3 d \left (a^2-b^2\right )^2 \sqrt{a+b \cos (c+d x)}}+\frac{2 \left (2 a^2-b^2\right ) \sin (c+d x) \sqrt{a+b \cos (c+d x)}}{3 b^3 d \left (a^2-b^2\right )}+\frac{2 \left (-16 a^2 b^2+16 a^4-b^4\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{3 b^4 d \left (a^2-b^2\right ) \sqrt{a+b \cos (c+d x)}}-\frac{8 a \left (-7 a^2 b^2+4 a^4+2 b^4\right ) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{3 b^4 d \left (a^2-b^2\right )^2 \sqrt{\frac{a+b \cos (c+d x)}{a+b}}} \]
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Rubi [A] time = 0.570046, antiderivative size = 345, normalized size of antiderivative = 1., 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 2792
Rule 3031
Rule 3023
Rule 2752
Rule 2663
Rule 2661
Rule 2655
Rule 2653
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*}
Mathematica [A] time = 1.56711, size = 237, normalized size = 0.69 \[ \frac{2 \left (\frac{b \sin (c+d x) \left (4 a b \left (-8 a^2 b^2+5 a^4+b^4\right ) \cos (c+d x)+\left (b^3-a^2 b\right )^2 \cos (2 (c+d x))-25 a^4 b^2+16 a^6+b^6\right )}{2 \left (a^2-b^2\right )^2}+\frac{\left (\frac{a+b \cos (c+d x)}{a+b}\right )^{3/2} \left (\left (-16 a^3 b^2+16 a^2 b^3-16 a^4 b+16 a^5-a b^4+b^5\right ) F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )-4 \left (-7 a^3 b^2+4 a^5+2 a b^4\right ) E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )\right )}{(a-b)^2}\right )}{3 b^4 d (a+b \cos (c+d x))^{3/2}} \]
Antiderivative was successfully verified.
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Maple [B] time = 13.386, size = 1291, normalized size = 3.7 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos \left (d x + c\right )^{4}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\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 ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos \left (d x + c\right )^{4}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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