Optimal. Leaf size=285 \[ \frac {e^{5/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a d}-\frac {e^{5/2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}+1\right )}{\sqrt {2} a d}-\frac {e^{5/2} \log \left (\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} a d}+\frac {e^{5/2} \log \left (\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} a d}-\frac {2 e^2 \cos (c+d x) E\left (\left .c+d x-\frac {\pi }{4}\right |2\right ) \sqrt {e \tan (c+d x)}}{a d \sqrt {\sin (2 c+2 d x)}}+\frac {2 e \cos (c+d x) (e \tan (c+d x))^{3/2}}{a d} \]
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Rubi [A] time = 0.33, antiderivative size = 285, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 14, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.560, Rules used = {3888, 3884, 3476, 329, 297, 1162, 617, 204, 1165, 628, 2613, 2615, 2572, 2639} \[ \frac {e^{5/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a d}-\frac {e^{5/2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}+1\right )}{\sqrt {2} a d}-\frac {e^{5/2} \log \left (\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} a d}+\frac {e^{5/2} \log \left (\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} a d}-\frac {2 e^2 \cos (c+d x) E\left (\left .c+d x-\frac {\pi }{4}\right |2\right ) \sqrt {e \tan (c+d x)}}{a d \sqrt {\sin (2 c+2 d x)}}+\frac {2 e \cos (c+d x) (e \tan (c+d x))^{3/2}}{a d} \]
Antiderivative was successfully verified.
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Rule 204
Rule 297
Rule 329
Rule 617
Rule 628
Rule 1162
Rule 1165
Rule 2572
Rule 2613
Rule 2615
Rule 2639
Rule 3476
Rule 3884
Rule 3888
Rubi steps
\begin {align*} \int \frac {(e \tan (c+d x))^{5/2}}{a+a \sec (c+d x)} \, dx &=\frac {e^2 \int (-a+a \sec (c+d x)) \sqrt {e \tan (c+d x)} \, dx}{a^2}\\ &=-\frac {e^2 \int \sqrt {e \tan (c+d x)} \, dx}{a}+\frac {e^2 \int \sec (c+d x) \sqrt {e \tan (c+d x)} \, dx}{a}\\ &=\frac {2 e \cos (c+d x) (e \tan (c+d x))^{3/2}}{a d}-\frac {\left (2 e^2\right ) \int \cos (c+d x) \sqrt {e \tan (c+d x)} \, dx}{a}-\frac {e^3 \operatorname {Subst}\left (\int \frac {\sqrt {x}}{e^2+x^2} \, dx,x,e \tan (c+d x)\right )}{a d}\\ &=\frac {2 e \cos (c+d x) (e \tan (c+d x))^{3/2}}{a d}-\frac {\left (2 e^3\right ) \operatorname {Subst}\left (\int \frac {x^2}{e^2+x^4} \, dx,x,\sqrt {e \tan (c+d x)}\right )}{a d}-\frac {\left (2 e^2 \sqrt {\cos (c+d x)} \sqrt {e \tan (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \sqrt {\sin (c+d x)} \, dx}{a \sqrt {\sin (c+d x)}}\\ &=\frac {2 e \cos (c+d x) (e \tan (c+d x))^{3/2}}{a d}+\frac {e^3 \operatorname {Subst}\left (\int \frac {e-x^2}{e^2+x^4} \, dx,x,\sqrt {e \tan (c+d x)}\right )}{a d}-\frac {e^3 \operatorname {Subst}\left (\int \frac {e+x^2}{e^2+x^4} \, dx,x,\sqrt {e \tan (c+d x)}\right )}{a d}-\frac {\left (2 e^2 \cos (c+d x) \sqrt {e \tan (c+d x)}\right ) \int \sqrt {\sin (2 c+2 d x)} \, dx}{a \sqrt {\sin (2 c+2 d x)}}\\ &=-\frac {2 e^2 \cos (c+d x) E\left (\left .c-\frac {\pi }{4}+d x\right |2\right ) \sqrt {e \tan (c+d x)}}{a d \sqrt {\sin (2 c+2 d x)}}+\frac {2 e \cos (c+d x) (e \tan (c+d x))^{3/2}}{a d}-\frac {e^{5/2} \operatorname {Subst}\left (\int \frac {\sqrt {2} \sqrt {e}+2 x}{-e-\sqrt {2} \sqrt {e} x-x^2} \, dx,x,\sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} a d}-\frac {e^{5/2} \operatorname {Subst}\left (\int \frac {\sqrt {2} \sqrt {e}-2 x}{-e+\sqrt {2} \sqrt {e} x-x^2} \, dx,x,\sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} a d}-\frac {e^3 \operatorname {Subst}\left (\int \frac {1}{e-\sqrt {2} \sqrt {e} x+x^2} \, dx,x,\sqrt {e \tan (c+d x)}\right )}{2 a d}-\frac {e^3 \operatorname {Subst}\left (\int \frac {1}{e+\sqrt {2} \sqrt {e} x+x^2} \, dx,x,\sqrt {e \tan (c+d x)}\right )}{2 a d}\\ &=-\frac {e^{5/2} \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} a d}+\frac {e^{5/2} \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} a d}-\frac {2 e^2 \cos (c+d x) E\left (\left .c-\frac {\pi }{4}+d x\right |2\right ) \sqrt {e \tan (c+d x)}}{a d \sqrt {\sin (2 c+2 d x)}}+\frac {2 e \cos (c+d x) (e \tan (c+d x))^{3/2}}{a d}-\frac {e^{5/2} \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a d}+\frac {e^{5/2} \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a d}\\ &=\frac {e^{5/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a d}-\frac {e^{5/2} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a d}-\frac {e^{5/2} \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} a d}+\frac {e^{5/2} \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} a d}-\frac {2 e^2 \cos (c+d x) E\left (\left .c-\frac {\pi }{4}+d x\right |2\right ) \sqrt {e \tan (c+d x)}}{a d \sqrt {\sin (2 c+2 d x)}}+\frac {2 e \cos (c+d x) (e \tan (c+d x))^{3/2}}{a d}\\ \end {align*}
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Mathematica [C] time = 5.51, size = 105, normalized size = 0.37 \[ \frac {4 \cos ^2\left (\frac {1}{2} (c+d x)\right ) \csc (c+d x) \left (\sqrt {\sec ^2(c+d x)}+1\right ) (e \tan (c+d x))^{5/2} \left (\, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-\tan ^2(c+d x)\right )-\, _2F_1\left (\frac {3}{4},1;\frac {7}{4};-\tan ^2(c+d x)\right )\right )}{3 a d (\sec (c+d x)+1)^2} \]
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] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (e \tan \left (d x + c\right )\right )^{\frac {5}{2}}}{a \sec \left (d x + c\right ) + a}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 2.09, size = 1419, normalized size = 4.98 \[ \text {result too large to display} \]
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 \[ \int \frac {\left (e \tan \left (d x + c\right )\right )^{\frac {5}{2}}}{a \sec \left (d x + c\right ) + a}\,{d x} \]
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 \[ \int \frac {\cos \left (c+d\,x\right )\,{\left (e\,\mathrm {tan}\left (c+d\,x\right )\right )}^{5/2}}{a\,\left (\cos \left (c+d\,x\right )+1\right )} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\left (e \tan {\left (c + d x \right )}\right )^{\frac {5}{2}}}{\sec {\left (c + d x \right )} + 1}\, dx}{a} \]
Verification of antiderivative is not currently implemented for this CAS.
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