Optimal. Leaf size=483 \[ \frac {7 e^{9/2} \left (5 a^2-2 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {e \cos (c+d x)}}{\sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{8 b^{9/2} d \sqrt [4]{b^2-a^2}}-\frac {7 e^{9/2} \left (5 a^2-2 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {e \cos (c+d x)}}{\sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{8 b^{9/2} d \sqrt [4]{b^2-a^2}}+\frac {7 a e^5 \left (5 a^2-2 b^2\right ) \sqrt {\cos (c+d x)} \Pi \left (\frac {2 b}{b-\sqrt {b^2-a^2}};\left .\frac {1}{2} (c+d x)\right |2\right )}{8 b^5 d \left (b-\sqrt {b^2-a^2}\right ) \sqrt {e \cos (c+d x)}}+\frac {7 a e^5 \left (5 a^2-2 b^2\right ) \sqrt {\cos (c+d x)} \Pi \left (\frac {2 b}{b+\sqrt {b^2-a^2}};\left .\frac {1}{2} (c+d x)\right |2\right )}{8 b^5 d \left (\sqrt {b^2-a^2}+b\right ) \sqrt {e \cos (c+d x)}}-\frac {35 a e^4 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \cos (c+d x)}}{4 b^4 d \sqrt {\cos (c+d x)}}-\frac {7 e^3 (e \cos (c+d x))^{3/2} (5 a+2 b \sin (c+d x))}{12 b^3 d (a+b \sin (c+d x))}-\frac {e (e \cos (c+d x))^{7/2}}{2 b d (a+b \sin (c+d x))^2} \]
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Rubi [A] time = 1.08, antiderivative size = 483, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 12, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.480, Rules used = {2693, 2863, 2867, 2640, 2639, 2701, 2807, 2805, 329, 298, 205, 208} \[ \frac {7 e^{9/2} \left (5 a^2-2 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {e \cos (c+d x)}}{\sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{8 b^{9/2} d \sqrt [4]{b^2-a^2}}-\frac {7 e^{9/2} \left (5 a^2-2 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {e \cos (c+d x)}}{\sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{8 b^{9/2} d \sqrt [4]{b^2-a^2}}+\frac {7 a e^5 \left (5 a^2-2 b^2\right ) \sqrt {\cos (c+d x)} \Pi \left (\frac {2 b}{b-\sqrt {b^2-a^2}};\left .\frac {1}{2} (c+d x)\right |2\right )}{8 b^5 d \left (b-\sqrt {b^2-a^2}\right ) \sqrt {e \cos (c+d x)}}+\frac {7 a e^5 \left (5 a^2-2 b^2\right ) \sqrt {\cos (c+d x)} \Pi \left (\frac {2 b}{b+\sqrt {b^2-a^2}};\left .\frac {1}{2} (c+d x)\right |2\right )}{8 b^5 d \left (\sqrt {b^2-a^2}+b\right ) \sqrt {e \cos (c+d x)}}-\frac {7 e^3 (e \cos (c+d x))^{3/2} (5 a+2 b \sin (c+d x))}{12 b^3 d (a+b \sin (c+d x))}-\frac {35 a e^4 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \cos (c+d x)}}{4 b^4 d \sqrt {\cos (c+d x)}}-\frac {e (e \cos (c+d x))^{7/2}}{2 b d (a+b \sin (c+d x))^2} \]
Antiderivative was successfully verified.
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Rule 205
Rule 208
Rule 298
Rule 329
Rule 2639
Rule 2640
Rule 2693
Rule 2701
Rule 2805
Rule 2807
Rule 2863
Rule 2867
Rubi steps
\begin {align*} \int \frac {(e \cos (c+d x))^{9/2}}{(a+b \sin (c+d x))^3} \, dx &=-\frac {e (e \cos (c+d x))^{7/2}}{2 b d (a+b \sin (c+d x))^2}-\frac {\left (7 e^2\right ) \int \frac {(e \cos (c+d x))^{5/2} \sin (c+d x)}{(a+b \sin (c+d x))^2} \, dx}{4 b}\\ &=-\frac {e (e \cos (c+d x))^{7/2}}{2 b d (a+b \sin (c+d x))^2}-\frac {7 e^3 (e \cos (c+d x))^{3/2} (5 a+2 b \sin (c+d x))}{12 b^3 d (a+b \sin (c+d x))}+\frac {\left (7 e^4\right ) \int \frac {\sqrt {e \cos (c+d x)} \left (-b-\frac {5}{2} a \sin (c+d x)\right )}{a+b \sin (c+d x)} \, dx}{4 b^3}\\ &=-\frac {e (e \cos (c+d x))^{7/2}}{2 b d (a+b \sin (c+d x))^2}-\frac {7 e^3 (e \cos (c+d x))^{3/2} (5 a+2 b \sin (c+d x))}{12 b^3 d (a+b \sin (c+d x))}-\frac {\left (35 a e^4\right ) \int \sqrt {e \cos (c+d x)} \, dx}{8 b^4}+\frac {\left (7 \left (5 a^2-2 b^2\right ) e^4\right ) \int \frac {\sqrt {e \cos (c+d x)}}{a+b \sin (c+d x)} \, dx}{8 b^4}\\ &=-\frac {e (e \cos (c+d x))^{7/2}}{2 b d (a+b \sin (c+d x))^2}-\frac {7 e^3 (e \cos (c+d x))^{3/2} (5 a+2 b \sin (c+d x))}{12 b^3 d (a+b \sin (c+d x))}-\frac {\left (7 a \left (5 a^2-2 b^2\right ) e^5\right ) \int \frac {1}{\sqrt {e \cos (c+d x)} \left (\sqrt {-a^2+b^2}-b \cos (c+d x)\right )} \, dx}{16 b^5}+\frac {\left (7 a \left (5 a^2-2 b^2\right ) e^5\right ) \int \frac {1}{\sqrt {e \cos (c+d x)} \left (\sqrt {-a^2+b^2}+b \cos (c+d x)\right )} \, dx}{16 b^5}+\frac {\left (7 \left (5 a^2-2 b^2\right ) e^5\right ) \operatorname {Subst}\left (\int \frac {\sqrt {x}}{\left (a^2-b^2\right ) e^2+b^2 x^2} \, dx,x,e \cos (c+d x)\right )}{8 b^3 d}-\frac {\left (35 a e^4 \sqrt {e \cos (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx}{8 b^4 \sqrt {\cos (c+d x)}}\\ &=-\frac {35 a e^4 \sqrt {e \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 b^4 d \sqrt {\cos (c+d x)}}-\frac {e (e \cos (c+d x))^{7/2}}{2 b d (a+b \sin (c+d x))^2}-\frac {7 e^3 (e \cos (c+d x))^{3/2} (5 a+2 b \sin (c+d x))}{12 b^3 d (a+b \sin (c+d x))}+\frac {\left (7 \left (5 a^2-2 b^2\right ) e^5\right ) \operatorname {Subst}\left (\int \frac {x^2}{\left (a^2-b^2\right ) e^2+b^2 x^4} \, dx,x,\sqrt {e \cos (c+d x)}\right )}{4 b^3 d}-\frac {\left (7 a \left (5 a^2-2 b^2\right ) e^5 \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}{16 b^5 \sqrt {e \cos (c+d x)}}+\frac {\left (7 a \left (5 a^2-2 b^2\right ) e^5 \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}{16 b^5 \sqrt {e \cos (c+d x)}}\\ &=-\frac {35 a e^4 \sqrt {e \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 b^4 d \sqrt {\cos (c+d x)}}+\frac {7 a \left (5 a^2-2 b^2\right ) e^5 \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 )}{8 b^5 \left (b-\sqrt {-a^2+b^2}\right ) d \sqrt {e \cos (c+d x)}}+\frac {7 a \left (5 a^2-2 b^2\right ) e^5 \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 )}{8 b^5 \left (b+\sqrt {-a^2+b^2}\right ) d \sqrt {e \cos (c+d x)}}-\frac {e (e \cos (c+d x))^{7/2}}{2 b d (a+b \sin (c+d x))^2}-\frac {7 e^3 (e \cos (c+d x))^{3/2} (5 a+2 b \sin (c+d x))}{12 b^3 d (a+b \sin (c+d x))}-\frac {\left (7 \left (5 a^2-2 b^2\right ) e^5\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 )}{8 b^4 d}+\frac {\left (7 \left (5 a^2-2 b^2\right ) e^5\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 )}{8 b^4 d}\\ &=\frac {7 \left (5 a^2-2 b^2\right ) e^{9/2} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {e \cos (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{8 b^{9/2} \sqrt [4]{-a^2+b^2} d}-\frac {7 \left (5 a^2-2 b^2\right ) e^{9/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {e \cos (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{8 b^{9/2} \sqrt [4]{-a^2+b^2} d}-\frac {35 a e^4 \sqrt {e \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 b^4 d \sqrt {\cos (c+d x)}}+\frac {7 a \left (5 a^2-2 b^2\right ) e^5 \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 )}{8 b^5 \left (b-\sqrt {-a^2+b^2}\right ) d \sqrt {e \cos (c+d x)}}+\frac {7 a \left (5 a^2-2 b^2\right ) e^5 \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 )}{8 b^5 \left (b+\sqrt {-a^2+b^2}\right ) d \sqrt {e \cos (c+d x)}}-\frac {e (e \cos (c+d x))^{7/2}}{2 b d (a+b \sin (c+d x))^2}-\frac {7 e^3 (e \cos (c+d x))^{3/2} (5 a+2 b \sin (c+d x))}{12 b^3 d (a+b \sin (c+d x))}\\ \end {align*}
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Mathematica [C] time = 25.99, size = 777, normalized size = 1.61 \[ \frac {(e \cos (c+d x))^{9/2} \left (\frac {28 \sin (c+d x) \left (a+b \sqrt {\sin ^2(c+d x)}\right ) \left (\frac {a \cos ^{\frac {3}{2}}(c+d x) F_1\left (\frac {3}{4};\frac {1}{2},1;\frac {7}{4};\cos ^2(c+d x),\frac {b^2 \cos ^2(c+d x)}{b^2-a^2}\right )}{3 \left (a^2-b^2\right )}+\frac {\left (\frac {1}{8}+\frac {i}{8}\right ) \left (-\log \left (-(1+i) \sqrt {b} \sqrt [4]{b^2-a^2} \sqrt {\cos (c+d x)}+\sqrt {b^2-a^2}+i b \cos (c+d x)\right )+\log \left ((1+i) \sqrt {b} \sqrt [4]{b^2-a^2} \sqrt {\cos (c+d x)}+\sqrt {b^2-a^2}+i b \cos (c+d x)\right )+2 \tan ^{-1}\left (1-\frac {(1+i) \sqrt {b} \sqrt {\cos (c+d x)}}{\sqrt [4]{b^2-a^2}}\right )-2 \tan ^{-1}\left (1+\frac {(1+i) \sqrt {b} \sqrt {\cos (c+d x)}}{\sqrt [4]{b^2-a^2}}\right )\right )}{\sqrt {b} \sqrt [4]{b^2-a^2}}\right )}{b^2 \sqrt {\sin ^2(c+d x)} (a+b \sin (c+d x))}+\frac {35 a \left (a+b \sqrt {\sin ^2(c+d x)}\right ) \left (8 b^{5/2} \cos ^{\frac {3}{2}}(c+d x) F_1\left (\frac {3}{4};-\frac {1}{2},1;\frac {7}{4};\cos ^2(c+d x),\frac {b^2 \cos ^2(c+d x)}{b^2-a^2}\right )+3 \sqrt {2} a \left (a^2-b^2\right )^{3/4} \left (-\log \left (-\sqrt {2} \sqrt {b} \sqrt [4]{a^2-b^2} \sqrt {\cos (c+d x)}+\sqrt {a^2-b^2}+b \cos (c+d x)\right )+\log \left (\sqrt {2} \sqrt {b} \sqrt [4]{a^2-b^2} \sqrt {\cos (c+d x)}+\sqrt {a^2-b^2}+b \cos (c+d x)\right )+2 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {b} \sqrt {\cos (c+d x)}}{\sqrt [4]{a^2-b^2}}\right )-2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {b} \sqrt {\cos (c+d x)}}{\sqrt [4]{a^2-b^2}}+1\right )\right )\right )}{12 b^{9/2} \left (b^2-a^2\right ) (a+b \sin (c+d x))}+\frac {4 \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x)}{b^3 (a+b \sin (c+d x))^2}-\frac {22 a \cos ^{\frac {3}{2}}(c+d x)}{b^3 (a+b \sin (c+d x))}-\frac {16 \cos ^{\frac {3}{2}}(c+d x)}{3 b^3}\right )}{8 d \cos ^{\frac {9}{2}}(c+d x)} \]
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 = 29.46, size = 85489, normalized size = 177.00 \[ \text {output too large to display} \]
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 \[ \int \frac {\left (e \cos \left (d x + c\right )\right )^{\frac {9}{2}}}{{\left (b \sin \left (d x + c\right ) + a\right )}^{3}}\,{d x} \]
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 \[ \int \frac {{\left (e\,\cos \left (c+d\,x\right )\right )}^{9/2}}{{\left (a+b\,\sin \left (c+d\,x\right )\right )}^3} \,d x \]
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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