Optimal. Leaf size=571 \[ -\frac {15 a e^{11/2} \left (7 a^2-6 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {e \cos (c+d x)}}{\sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{16 b^{11/2} d \left (b^2-a^2\right )^{3/4}}-\frac {15 a e^{11/2} \left (7 a^2-6 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {e \cos (c+d x)}}{\sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{16 b^{11/2} d \left (b^2-a^2\right )^{3/4}}-\frac {5 e^6 \left (21 a^2-4 b^2\right ) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{8 b^6 d \sqrt {e \cos (c+d x)}}+\frac {15 a^2 e^6 \left (7 a^2-6 b^2\right ) \sqrt {\cos (c+d x)} \Pi \left (\frac {2 b}{b-\sqrt {b^2-a^2}};\left .\frac {1}{2} (c+d x)\right |2\right )}{16 b^6 d \left (a^2-b \left (b-\sqrt {b^2-a^2}\right )\right ) \sqrt {e \cos (c+d x)}}+\frac {15 a^2 e^6 \left (7 a^2-6 b^2\right ) \sqrt {\cos (c+d x)} \Pi \left (\frac {2 b}{b+\sqrt {b^2-a^2}};\left .\frac {1}{2} (c+d x)\right |2\right )}{16 b^6 d \left (a^2-b \left (\sqrt {b^2-a^2}+b\right )\right ) \sqrt {e \cos (c+d x)}}-\frac {5 e^5 \sqrt {e \cos (c+d x)} \left (21 a^2+14 a b \sin (c+d x)-4 b^2\right )}{8 b^5 d (a+b \sin (c+d x))}-\frac {e^3 (e \cos (c+d x))^{5/2} (7 a+4 b \sin (c+d x))}{4 b^3 d (a+b \sin (c+d x))^2}-\frac {e (e \cos (c+d x))^{9/2}}{3 b d (a+b \sin (c+d x))^3} \]
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Rubi [A] time = 1.38, antiderivative size = 571, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 12, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.480, Rules used = {2693, 2863, 2867, 2642, 2641, 2702, 2807, 2805, 329, 212, 208, 205} \[ -\frac {5 e^5 \sqrt {e \cos (c+d x)} \left (21 a^2+14 a b \sin (c+d x)-4 b^2\right )}{8 b^5 d (a+b \sin (c+d x))}-\frac {15 a e^{11/2} \left (7 a^2-6 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {e \cos (c+d x)}}{\sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{16 b^{11/2} d \left (b^2-a^2\right )^{3/4}}-\frac {15 a e^{11/2} \left (7 a^2-6 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {e \cos (c+d x)}}{\sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{16 b^{11/2} d \left (b^2-a^2\right )^{3/4}}-\frac {5 e^6 \left (21 a^2-4 b^2\right ) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{8 b^6 d \sqrt {e \cos (c+d x)}}+\frac {15 a^2 e^6 \left (7 a^2-6 b^2\right ) \sqrt {\cos (c+d x)} \Pi \left (\frac {2 b}{b-\sqrt {b^2-a^2}};\left .\frac {1}{2} (c+d x)\right |2\right )}{16 b^6 d \left (a^2-b \left (b-\sqrt {b^2-a^2}\right )\right ) \sqrt {e \cos (c+d x)}}+\frac {15 a^2 e^6 \left (7 a^2-6 b^2\right ) \sqrt {\cos (c+d x)} \Pi \left (\frac {2 b}{b+\sqrt {b^2-a^2}};\left .\frac {1}{2} (c+d x)\right |2\right )}{16 b^6 d \left (a^2-b \left (\sqrt {b^2-a^2}+b\right )\right ) \sqrt {e \cos (c+d x)}}-\frac {e^3 (e \cos (c+d x))^{5/2} (7 a+4 b \sin (c+d x))}{4 b^3 d (a+b \sin (c+d x))^2}-\frac {e (e \cos (c+d x))^{9/2}}{3 b d (a+b \sin (c+d x))^3} \]
Antiderivative was successfully verified.
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Rule 205
Rule 208
Rule 212
Rule 329
Rule 2641
Rule 2642
Rule 2693
Rule 2702
Rule 2805
Rule 2807
Rule 2863
Rule 2867
Rubi steps
\begin {align*} \int \frac {(e \cos (c+d x))^{11/2}}{(a+b \sin (c+d x))^4} \, dx &=-\frac {e (e \cos (c+d x))^{9/2}}{3 b d (a+b \sin (c+d x))^3}-\frac {\left (3 e^2\right ) \int \frac {(e \cos (c+d x))^{7/2} \sin (c+d x)}{(a+b \sin (c+d x))^3} \, dx}{2 b}\\ &=-\frac {e (e \cos (c+d x))^{9/2}}{3 b d (a+b \sin (c+d x))^3}-\frac {e^3 (e \cos (c+d x))^{5/2} (7 a+4 b \sin (c+d x))}{4 b^3 d (a+b \sin (c+d x))^2}+\frac {\left (5 e^4\right ) \int \frac {(e \cos (c+d x))^{3/2} \left (-2 b-\frac {7}{2} a \sin (c+d x)\right )}{(a+b \sin (c+d x))^2} \, dx}{4 b^3}\\ &=-\frac {e (e \cos (c+d x))^{9/2}}{3 b d (a+b \sin (c+d x))^3}-\frac {e^3 (e \cos (c+d x))^{5/2} (7 a+4 b \sin (c+d x))}{4 b^3 d (a+b \sin (c+d x))^2}-\frac {5 e^5 \sqrt {e \cos (c+d x)} \left (21 a^2-4 b^2+14 a b \sin (c+d x)\right )}{8 b^5 d (a+b \sin (c+d x))}-\frac {\left (5 e^6\right ) \int \frac {\frac {7 a b}{2}+\frac {1}{4} \left (21 a^2-4 b^2\right ) \sin (c+d x)}{\sqrt {e \cos (c+d x)} (a+b \sin (c+d x))} \, dx}{4 b^5}\\ &=-\frac {e (e \cos (c+d x))^{9/2}}{3 b d (a+b \sin (c+d x))^3}-\frac {e^3 (e \cos (c+d x))^{5/2} (7 a+4 b \sin (c+d x))}{4 b^3 d (a+b \sin (c+d x))^2}-\frac {5 e^5 \sqrt {e \cos (c+d x)} \left (21 a^2-4 b^2+14 a b \sin (c+d x)\right )}{8 b^5 d (a+b \sin (c+d x))}+\frac {\left (15 a \left (7 a^2-6 b^2\right ) e^6\right ) \int \frac {1}{\sqrt {e \cos (c+d x)} (a+b \sin (c+d x))} \, dx}{16 b^6}-\frac {\left (5 \left (21 a^2-4 b^2\right ) e^6\right ) \int \frac {1}{\sqrt {e \cos (c+d x)}} \, dx}{16 b^6}\\ &=-\frac {e (e \cos (c+d x))^{9/2}}{3 b d (a+b \sin (c+d x))^3}-\frac {e^3 (e \cos (c+d x))^{5/2} (7 a+4 b \sin (c+d x))}{4 b^3 d (a+b \sin (c+d x))^2}-\frac {5 e^5 \sqrt {e \cos (c+d x)} \left (21 a^2-4 b^2+14 a b \sin (c+d x)\right )}{8 b^5 d (a+b \sin (c+d x))}-\frac {\left (15 a^2 \left (7 a^2-6 b^2\right ) e^6\right ) \int \frac {1}{\sqrt {e \cos (c+d x)} \left (\sqrt {-a^2+b^2}-b \cos (c+d x)\right )} \, dx}{32 b^6 \sqrt {-a^2+b^2}}-\frac {\left (15 a^2 \left (7 a^2-6 b^2\right ) e^6\right ) \int \frac {1}{\sqrt {e \cos (c+d x)} \left (\sqrt {-a^2+b^2}+b \cos (c+d x)\right )} \, dx}{32 b^6 \sqrt {-a^2+b^2}}+\frac {\left (15 a \left (7 a^2-6 b^2\right ) e^7\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {x} \left (\left (a^2-b^2\right ) e^2+b^2 x^2\right )} \, dx,x,e \cos (c+d x)\right )}{16 b^5 d}-\frac {\left (5 \left (21 a^2-4 b^2\right ) e^6 \sqrt {\cos (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{16 b^6 \sqrt {e \cos (c+d x)}}\\ &=-\frac {5 \left (21 a^2-4 b^2\right ) e^6 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{8 b^6 d \sqrt {e \cos (c+d x)}}-\frac {e (e \cos (c+d x))^{9/2}}{3 b d (a+b \sin (c+d x))^3}-\frac {e^3 (e \cos (c+d x))^{5/2} (7 a+4 b \sin (c+d x))}{4 b^3 d (a+b \sin (c+d x))^2}-\frac {5 e^5 \sqrt {e \cos (c+d x)} \left (21 a^2-4 b^2+14 a b \sin (c+d x)\right )}{8 b^5 d (a+b \sin (c+d x))}+\frac {\left (15 a \left (7 a^2-6 b^2\right ) e^7\right ) \operatorname {Subst}\left (\int \frac {1}{\left (a^2-b^2\right ) e^2+b^2 x^4} \, dx,x,\sqrt {e \cos (c+d x)}\right )}{8 b^5 d}-\frac {\left (15 a^2 \left (7 a^2-6 b^2\right ) e^6 \sqrt {\cos (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)} \left (\sqrt {-a^2+b^2}-b \cos (c+d x)\right )} \, dx}{32 b^6 \sqrt {-a^2+b^2} \sqrt {e \cos (c+d x)}}-\frac {\left (15 a^2 \left (7 a^2-6 b^2\right ) e^6 \sqrt {\cos (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)} \left (\sqrt {-a^2+b^2}+b \cos (c+d x)\right )} \, dx}{32 b^6 \sqrt {-a^2+b^2} \sqrt {e \cos (c+d x)}}\\ &=-\frac {5 \left (21 a^2-4 b^2\right ) e^6 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{8 b^6 d \sqrt {e \cos (c+d x)}}+\frac {15 a^2 \left (7 a^2-6 b^2\right ) e^6 \sqrt {\cos (c+d x)} \Pi \left (\frac {2 b}{b-\sqrt {-a^2+b^2}};\left .\frac {1}{2} (c+d x)\right |2\right )}{16 b^6 \left (a^2-b \left (b-\sqrt {-a^2+b^2}\right )\right ) d \sqrt {e \cos (c+d x)}}-\frac {15 a^2 \left (7 a^2-6 b^2\right ) e^6 \sqrt {\cos (c+d x)} \Pi \left (\frac {2 b}{b+\sqrt {-a^2+b^2}};\left .\frac {1}{2} (c+d x)\right |2\right )}{16 b^6 \sqrt {-a^2+b^2} \left (b+\sqrt {-a^2+b^2}\right ) d \sqrt {e \cos (c+d x)}}-\frac {e (e \cos (c+d x))^{9/2}}{3 b d (a+b \sin (c+d x))^3}-\frac {e^3 (e \cos (c+d x))^{5/2} (7 a+4 b \sin (c+d x))}{4 b^3 d (a+b \sin (c+d x))^2}-\frac {5 e^5 \sqrt {e \cos (c+d x)} \left (21 a^2-4 b^2+14 a b \sin (c+d x)\right )}{8 b^5 d (a+b \sin (c+d x))}-\frac {\left (15 a \left (7 a^2-6 b^2\right ) e^6\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {-a^2+b^2} e-b x^2} \, dx,x,\sqrt {e \cos (c+d x)}\right )}{16 b^5 \sqrt {-a^2+b^2} d}-\frac {\left (15 a \left (7 a^2-6 b^2\right ) e^6\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {-a^2+b^2} e+b x^2} \, dx,x,\sqrt {e \cos (c+d x)}\right )}{16 b^5 \sqrt {-a^2+b^2} d}\\ &=-\frac {15 a \left (7 a^2-6 b^2\right ) e^{11/2} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {e \cos (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{16 b^{11/2} \left (-a^2+b^2\right )^{3/4} d}-\frac {15 a \left (7 a^2-6 b^2\right ) e^{11/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {e \cos (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{16 b^{11/2} \left (-a^2+b^2\right )^{3/4} d}-\frac {5 \left (21 a^2-4 b^2\right ) e^6 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{8 b^6 d \sqrt {e \cos (c+d x)}}+\frac {15 a^2 \left (7 a^2-6 b^2\right ) e^6 \sqrt {\cos (c+d x)} \Pi \left (\frac {2 b}{b-\sqrt {-a^2+b^2}};\left .\frac {1}{2} (c+d x)\right |2\right )}{16 b^6 \left (a^2-b \left (b-\sqrt {-a^2+b^2}\right )\right ) d \sqrt {e \cos (c+d x)}}-\frac {15 a^2 \left (7 a^2-6 b^2\right ) e^6 \sqrt {\cos (c+d x)} \Pi \left (\frac {2 b}{b+\sqrt {-a^2+b^2}};\left .\frac {1}{2} (c+d x)\right |2\right )}{16 b^6 \sqrt {-a^2+b^2} \left (b+\sqrt {-a^2+b^2}\right ) d \sqrt {e \cos (c+d x)}}-\frac {e (e \cos (c+d x))^{9/2}}{3 b d (a+b \sin (c+d x))^3}-\frac {e^3 (e \cos (c+d x))^{5/2} (7 a+4 b \sin (c+d x))}{4 b^3 d (a+b \sin (c+d x))^2}-\frac {5 e^5 \sqrt {e \cos (c+d x)} \left (21 a^2-4 b^2+14 a b \sin (c+d x)\right )}{8 b^5 d (a+b \sin (c+d x))}\\ \end {align*}
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Mathematica [C] time = 26.50, size = 2020, normalized size = 3.54 \[ \text {Result too large to show} \]
Warning: Unable to verify antiderivative.
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fricas [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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giac [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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maple [C] time = 111.92, size = 144252, normalized size = 252.63 \[ \text {output too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [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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mupad [F(-1)] time = 0.00, size = -1, normalized size = -0.00 \[ \text {Hanged} \]
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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