Integrand size = 25, antiderivative size = 152 \[ \int \frac {(a+a \sin (c+d x))^4}{(e \cos (c+d x))^{5/2}} \, dx=-\frac {10 a^4 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d e^2 \sqrt {e \cos (c+d x)}}-\frac {10 a^4 \sqrt {e \cos (c+d x)} \sin (c+d x)}{d e^3}+\frac {4 a^7 (e \cos (c+d x))^{9/2}}{3 d e^7 (a-a \sin (c+d x))^3}+\frac {12 a^8 (e \cos (c+d x))^{5/2}}{d e^5 \left (a^4-a^4 \sin (c+d x)\right )} \]
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Time = 0.16 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2749, 2759, 2715, 2721, 2720} \[ \int \frac {(a+a \sin (c+d x))^4}{(e \cos (c+d x))^{5/2}} \, dx=\frac {4 a^7 (e \cos (c+d x))^{9/2}}{3 d e^7 (a-a \sin (c+d x))^3}-\frac {10 a^4 \sin (c+d x) \sqrt {e \cos (c+d x)}}{d e^3}-\frac {10 a^4 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d e^2 \sqrt {e \cos (c+d x)}}+\frac {12 a^8 (e \cos (c+d x))^{5/2}}{d e^5 \left (a^4-a^4 \sin (c+d x)\right )} \]
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Rule 2715
Rule 2720
Rule 2721
Rule 2749
Rule 2759
Rubi steps \begin{align*} \text {integral}& = \frac {a^8 \int \frac {(e \cos (c+d x))^{11/2}}{(a-a \sin (c+d x))^4} \, dx}{e^8} \\ & = \frac {4 a^7 (e \cos (c+d x))^{9/2}}{3 d e^7 (a-a \sin (c+d x))^3}-\frac {\left (3 a^6\right ) \int \frac {(e \cos (c+d x))^{7/2}}{(a-a \sin (c+d x))^2} \, dx}{e^6} \\ & = \frac {4 a^7 (e \cos (c+d x))^{9/2}}{3 d e^7 (a-a \sin (c+d x))^3}+\frac {12 a^6 (e \cos (c+d x))^{5/2}}{d e^5 \left (a^2-a^2 \sin (c+d x)\right )}-\frac {\left (15 a^4\right ) \int (e \cos (c+d x))^{3/2} \, dx}{e^4} \\ & = -\frac {10 a^4 \sqrt {e \cos (c+d x)} \sin (c+d x)}{d e^3}+\frac {4 a^7 (e \cos (c+d x))^{9/2}}{3 d e^7 (a-a \sin (c+d x))^3}+\frac {12 a^6 (e \cos (c+d x))^{5/2}}{d e^5 \left (a^2-a^2 \sin (c+d x)\right )}-\frac {\left (5 a^4\right ) \int \frac {1}{\sqrt {e \cos (c+d x)}} \, dx}{e^2} \\ & = -\frac {10 a^4 \sqrt {e \cos (c+d x)} \sin (c+d x)}{d e^3}+\frac {4 a^7 (e \cos (c+d x))^{9/2}}{3 d e^7 (a-a \sin (c+d x))^3}+\frac {12 a^6 (e \cos (c+d x))^{5/2}}{d e^5 \left (a^2-a^2 \sin (c+d x)\right )}-\frac {\left (5 a^4 \sqrt {\cos (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{e^2 \sqrt {e \cos (c+d x)}} \\ & = -\frac {10 a^4 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d e^2 \sqrt {e \cos (c+d x)}}-\frac {10 a^4 \sqrt {e \cos (c+d x)} \sin (c+d x)}{d e^3}+\frac {4 a^7 (e \cos (c+d x))^{9/2}}{3 d e^7 (a-a \sin (c+d x))^3}+\frac {12 a^6 (e \cos (c+d x))^{5/2}}{d e^5 \left (a^2-a^2 \sin (c+d x)\right )} \\ \end{align*}
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.04 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.43 \[ \int \frac {(a+a \sin (c+d x))^4}{(e \cos (c+d x))^{5/2}} \, dx=\frac {16 \sqrt [4]{2} a^4 \operatorname {Hypergeometric2F1}\left (-\frac {9}{4},-\frac {3}{4},\frac {1}{4},\frac {1}{2} (1-\sin (c+d x))\right ) (1+\sin (c+d x))^{3/4}}{3 d e (e \cos (c+d x))^{3/2}} \]
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Time = 6.01 (sec) , antiderivative size = 263, normalized size of antiderivative = 1.73
method | result | size |
default | \(-\frac {2 \left (8 \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-8 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-30 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+48 \left (\sin ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+18 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+15 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-48 \left (\sin ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+20 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{4}}{3 \left (2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e +e}\, e^{2} d}\) | \(263\) |
parts | \(\text {Expression too large to display}\) | \(835\) |
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Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.11 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.03 \[ \int \frac {(a+a \sin (c+d x))^4}{(e \cos (c+d x))^{5/2}} \, dx=-\frac {15 \, {\left (-i \, \sqrt {2} a^{4} \sin \left (d x + c\right ) + i \, \sqrt {2} a^{4}\right )} \sqrt {e} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 15 \, {\left (i \, \sqrt {2} a^{4} \sin \left (d x + c\right ) - i \, \sqrt {2} a^{4}\right )} \sqrt {e} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 2 \, {\left (a^{4} \cos \left (d x + c\right )^{2} - 11 \, a^{4} \sin \left (d x + c\right ) + 19 \, a^{4}\right )} \sqrt {e \cos \left (d x + c\right )}}{3 \, {\left (d e^{3} \sin \left (d x + c\right ) - d e^{3}\right )}} \]
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Timed out. \[ \int \frac {(a+a \sin (c+d x))^4}{(e \cos (c+d x))^{5/2}} \, dx=\text {Timed out} \]
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\[ \int \frac {(a+a \sin (c+d x))^4}{(e \cos (c+d x))^{5/2}} \, dx=\int { \frac {{\left (a \sin \left (d x + c\right ) + a\right )}^{4}}{\left (e \cos \left (d x + c\right )\right )^{\frac {5}{2}}} \,d x } \]
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\[ \int \frac {(a+a \sin (c+d x))^4}{(e \cos (c+d x))^{5/2}} \, dx=\int { \frac {{\left (a \sin \left (d x + c\right ) + a\right )}^{4}}{\left (e \cos \left (d x + c\right )\right )^{\frac {5}{2}}} \,d x } \]
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Timed out. \[ \int \frac {(a+a \sin (c+d x))^4}{(e \cos (c+d x))^{5/2}} \, dx=\int \frac {{\left (a+a\,\sin \left (c+d\,x\right )\right )}^4}{{\left (e\,\cos \left (c+d\,x\right )\right )}^{5/2}} \,d x \]
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