Optimal. Leaf size=328 \[ -\frac {\left (8 a^4-29 a^2 b^2+15 b^4\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a^3 \left (a^2-b^2\right )^2 d}+\frac {b \left (11 a^2-5 b^2\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a^2 \left (a^2-b^2\right )^2 d}-\frac {b \left (35 a^4-38 a^2 b^2+15 b^4\right ) \Pi \left (\frac {2 b}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a^3 (a-b)^2 (a+b)^3 d}+\frac {\left (8 a^4-29 a^2 b^2+15 b^4\right ) \sin (c+d x)}{4 a^3 \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)}}+\frac {b^2 \sin (c+d x)}{2 a \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^2}+\frac {b^2 \left (11 a^2-5 b^2\right ) \sin (c+d x)}{4 a^2 \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \]
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Rubi [A]
time = 0.70, antiderivative size = 328, normalized size of antiderivative = 1.00, number of steps
used = 8, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {2881, 3134,
3138, 2719, 3081, 2720, 2884} \begin {gather*} \frac {b \left (11 a^2-5 b^2\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a^2 d \left (a^2-b^2\right )^2}+\frac {b^2 \left (11 a^2-5 b^2\right ) \sin (c+d x)}{4 a^2 d \left (a^2-b^2\right )^2 \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}+\frac {b^2 \sin (c+d x)}{2 a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^2}-\frac {\left (8 a^4-29 a^2 b^2+15 b^4\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a^3 d \left (a^2-b^2\right )^2}-\frac {b \left (35 a^4-38 a^2 b^2+15 b^4\right ) \Pi \left (\frac {2 b}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a^3 d (a-b)^2 (a+b)^3}+\frac {\left (8 a^4-29 a^2 b^2+15 b^4\right ) \sin (c+d x)}{4 a^3 d \left (a^2-b^2\right )^2 \sqrt {\cos (c+d x)}} \end {gather*}
Antiderivative was successfully verified.
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Rule 2719
Rule 2720
Rule 2881
Rule 2884
Rule 3081
Rule 3134
Rule 3138
Rubi steps
\begin {align*} \int \frac {1}{\cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^3} \, dx &=\frac {b^2 \sin (c+d x)}{2 a \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^2}+\frac {\int \frac {\frac {1}{2} \left (4 a^2-5 b^2\right )-2 a b \cos (c+d x)+\frac {3}{2} b^2 \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^2} \, dx}{2 a \left (a^2-b^2\right )}\\ &=\frac {b^2 \sin (c+d x)}{2 a \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^2}+\frac {b^2 \left (11 a^2-5 b^2\right ) \sin (c+d x)}{4 a^2 \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}+\frac {\int \frac {\frac {1}{4} \left (8 a^4-29 a^2 b^2+15 b^4\right )-a b \left (4 a^2-b^2\right ) \cos (c+d x)+\frac {1}{4} b^2 \left (11 a^2-5 b^2\right ) \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))} \, dx}{2 a^2 \left (a^2-b^2\right )^2}\\ &=\frac {\left (8 a^4-29 a^2 b^2+15 b^4\right ) \sin (c+d x)}{4 a^3 \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)}}+\frac {b^2 \sin (c+d x)}{2 a \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^2}+\frac {b^2 \left (11 a^2-5 b^2\right ) \sin (c+d x)}{4 a^2 \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}+\frac {\int \frac {-\frac {3}{8} b \left (8 a^4-11 a^2 b^2+5 b^4\right )-\frac {1}{2} a \left (2 a^4-10 a^2 b^2+5 b^4\right ) \cos (c+d x)-\frac {1}{8} b \left (8 a^4-29 a^2 b^2+15 b^4\right ) \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{a^3 \left (a^2-b^2\right )^2}\\ &=\frac {\left (8 a^4-29 a^2 b^2+15 b^4\right ) \sin (c+d x)}{4 a^3 \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)}}+\frac {b^2 \sin (c+d x)}{2 a \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^2}+\frac {b^2 \left (11 a^2-5 b^2\right ) \sin (c+d x)}{4 a^2 \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}-\frac {\int \frac {\frac {3}{8} b^2 \left (8 a^4-11 a^2 b^2+5 b^4\right )-\frac {1}{8} a b^3 \left (11 a^2-5 b^2\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{a^3 b \left (a^2-b^2\right )^2}-\frac {\left (8 a^4-29 a^2 b^2+15 b^4\right ) \int \sqrt {\cos (c+d x)} \, dx}{8 a^3 \left (a^2-b^2\right )^2}\\ &=-\frac {\left (8 a^4-29 a^2 b^2+15 b^4\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a^3 \left (a^2-b^2\right )^2 d}+\frac {\left (8 a^4-29 a^2 b^2+15 b^4\right ) \sin (c+d x)}{4 a^3 \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)}}+\frac {b^2 \sin (c+d x)}{2 a \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^2}+\frac {b^2 \left (11 a^2-5 b^2\right ) \sin (c+d x)}{4 a^2 \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}+\frac {\left (b \left (11 a^2-5 b^2\right )\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{8 a^2 \left (a^2-b^2\right )^2}-\frac {\left (b \left (35 a^4-38 a^2 b^2+15 b^4\right )\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{8 a^3 \left (a^2-b^2\right )^2}\\ &=-\frac {\left (8 a^4-29 a^2 b^2+15 b^4\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a^3 \left (a^2-b^2\right )^2 d}+\frac {b \left (11 a^2-5 b^2\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a^2 \left (a^2-b^2\right )^2 d}-\frac {b \left (35 a^4-38 a^2 b^2+15 b^4\right ) \Pi \left (\frac {2 b}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a^3 (a-b)^2 (a+b)^3 d}+\frac {\left (8 a^4-29 a^2 b^2+15 b^4\right ) \sin (c+d x)}{4 a^3 \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)}}+\frac {b^2 \sin (c+d x)}{2 a \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^2}+\frac {b^2 \left (11 a^2-5 b^2\right ) \sin (c+d x)}{4 a^2 \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}\\ \end {align*}
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Mathematica [A]
time = 13.60, size = 334, normalized size = 1.02 \begin {gather*} \frac {-\frac {\frac {2 \left (56 a^4 b-95 a^2 b^3+45 b^5\right ) \Pi \left (\frac {2 b}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a+b}+\frac {8 a \left (2 a^4-10 a^2 b^2+5 b^4\right ) \left (2 F\left (\left .\frac {1}{2} (c+d x)\right |2\right )-\frac {2 a \Pi \left (\frac {2 b}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a+b}\right )}{b}+\frac {2 \left (8 a^4-29 a^2 b^2+15 b^4\right ) \left (-2 a b E\left (\left .\text {ArcSin}\left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+2 a (a+b) F\left (\left .\text {ArcSin}\left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+\left (-2 a^2+b^2\right ) \Pi \left (-\frac {b}{a};\left .\text {ArcSin}\left (\sqrt {\cos (c+d x)}\right )\right |-1\right )\right ) \sin (c+d x)}{a b \sqrt {\sin ^2(c+d x)}}}{(a-b)^2 (a+b)^2}+4 \sqrt {\cos (c+d x)} \left (\frac {b^3 \left (-15 a^3+9 a b^2+\left (-13 a^2 b+7 b^3\right ) \cos (c+d x)\right ) \sin (c+d x)}{\left (a^2-b^2\right )^2 (a+b \cos (c+d x))^2}+8 \tan (c+d x)\right )}{16 a^3 d} \end {gather*}
Antiderivative was successfully verified.
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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1964\) vs.
\(2(388)=776\).
time = 0.86, size = 1965, normalized size = 5.99
method | result | size |
default | \(\text {Expression too large to display}\) | \(1965\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}
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 \begin {gather*} \text {could not integrate} \end {gather*}
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 \begin {gather*} \int \frac {1}{{\cos \left (c+d\,x\right )}^{3/2}\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^3} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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