Optimal. Leaf size=312 \[ -\frac {e^{9/2} \text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a^2 d}+\frac {e^{9/2} \text {ArcTan}\left (1+\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a^2 d}+\frac {e^{9/2} \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} a^2 d}-\frac {e^{9/2} \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} a^2 d}+\frac {4 e^4 \cos (c+d x) E\left (\left .c-\frac {\pi }{4}+d x\right |2\right ) \sqrt {e \tan (c+d x)}}{a^2 d \sqrt {\sin (2 c+2 d x)}}+\frac {2 e^3 (e \tan (c+d x))^{3/2}}{3 a^2 d}-\frac {4 e^3 \cos (c+d x) (e \tan (c+d x))^{3/2}}{a^2 d} \]
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Rubi [A]
time = 0.30, antiderivative size = 312, normalized size of antiderivative = 1.00, number of steps
used = 20, number of rules used = 16, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.640, Rules used = {3973,
3971, 3557, 335, 303, 1176, 631, 210, 1179, 642, 2693, 2695, 2652, 2719, 2687, 32}
\begin {gather*} -\frac {e^{9/2} \text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a^2 d}+\frac {e^{9/2} \text {ArcTan}\left (\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}+1\right )}{\sqrt {2} a^2 d}+\frac {e^{9/2} \log \left (\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} a^2 d}-\frac {e^{9/2} \log \left (\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} a^2 d}+\frac {4 e^4 \cos (c+d x) E\left (\left .c+d x-\frac {\pi }{4}\right |2\right ) \sqrt {e \tan (c+d x)}}{a^2 d \sqrt {\sin (2 c+2 d x)}}+\frac {2 e^3 (e \tan (c+d x))^{3/2}}{3 a^2 d}-\frac {4 e^3 \cos (c+d x) (e \tan (c+d x))^{3/2}}{a^2 d} \end {gather*}
Antiderivative was successfully verified.
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Rule 32
Rule 210
Rule 303
Rule 335
Rule 631
Rule 642
Rule 1176
Rule 1179
Rule 2652
Rule 2687
Rule 2693
Rule 2695
Rule 2719
Rule 3557
Rule 3971
Rule 3973
Rubi steps
\begin {align*} \int \frac {(e \tan (c+d x))^{9/2}}{(a+a \sec (c+d x))^2} \, dx &=\frac {e^4 \int (-a+a \sec (c+d x))^2 \sqrt {e \tan (c+d x)} \, dx}{a^4}\\ &=\frac {e^4 \int \left (a^2 \sqrt {e \tan (c+d x)}-2 a^2 \sec (c+d x) \sqrt {e \tan (c+d x)}+a^2 \sec ^2(c+d x) \sqrt {e \tan (c+d x)}\right ) \, dx}{a^4}\\ &=\frac {e^4 \int \sqrt {e \tan (c+d x)} \, dx}{a^2}+\frac {e^4 \int \sec ^2(c+d x) \sqrt {e \tan (c+d x)} \, dx}{a^2}-\frac {\left (2 e^4\right ) \int \sec (c+d x) \sqrt {e \tan (c+d x)} \, dx}{a^2}\\ &=-\frac {4 e^3 \cos (c+d x) (e \tan (c+d x))^{3/2}}{a^2 d}+\frac {\left (4 e^4\right ) \int \cos (c+d x) \sqrt {e \tan (c+d x)} \, dx}{a^2}+\frac {e^4 \text {Subst}\left (\int \sqrt {e x} \, dx,x,\tan (c+d x)\right )}{a^2 d}+\frac {e^5 \text {Subst}\left (\int \frac {\sqrt {x}}{e^2+x^2} \, dx,x,e \tan (c+d x)\right )}{a^2 d}\\ &=\frac {2 e^3 (e \tan (c+d x))^{3/2}}{3 a^2 d}-\frac {4 e^3 \cos (c+d x) (e \tan (c+d x))^{3/2}}{a^2 d}+\frac {\left (2 e^5\right ) \text {Subst}\left (\int \frac {x^2}{e^2+x^4} \, dx,x,\sqrt {e \tan (c+d x)}\right )}{a^2 d}+\frac {\left (4 e^4 \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^2 \sqrt {\sin (c+d x)}}\\ &=\frac {2 e^3 (e \tan (c+d x))^{3/2}}{3 a^2 d}-\frac {4 e^3 \cos (c+d x) (e \tan (c+d x))^{3/2}}{a^2 d}-\frac {e^5 \text {Subst}\left (\int \frac {e-x^2}{e^2+x^4} \, dx,x,\sqrt {e \tan (c+d x)}\right )}{a^2 d}+\frac {e^5 \text {Subst}\left (\int \frac {e+x^2}{e^2+x^4} \, dx,x,\sqrt {e \tan (c+d x)}\right )}{a^2 d}+\frac {\left (4 e^4 \cos (c+d x) \sqrt {e \tan (c+d x)}\right ) \int \sqrt {\sin (2 c+2 d x)} \, dx}{a^2 \sqrt {\sin (2 c+2 d x)}}\\ &=\frac {4 e^4 \cos (c+d x) E\left (\left .c-\frac {\pi }{4}+d x\right |2\right ) \sqrt {e \tan (c+d x)}}{a^2 d \sqrt {\sin (2 c+2 d x)}}+\frac {2 e^3 (e \tan (c+d x))^{3/2}}{3 a^2 d}-\frac {4 e^3 \cos (c+d x) (e \tan (c+d x))^{3/2}}{a^2 d}+\frac {e^{9/2} \text {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^2 d}+\frac {e^{9/2} \text {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^2 d}+\frac {e^5 \text {Subst}\left (\int \frac {1}{e-\sqrt {2} \sqrt {e} x+x^2} \, dx,x,\sqrt {e \tan (c+d x)}\right )}{2 a^2 d}+\frac {e^5 \text {Subst}\left (\int \frac {1}{e+\sqrt {2} \sqrt {e} x+x^2} \, dx,x,\sqrt {e \tan (c+d x)}\right )}{2 a^2 d}\\ &=\frac {e^{9/2} \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} a^2 d}-\frac {e^{9/2} \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} a^2 d}+\frac {4 e^4 \cos (c+d x) E\left (\left .c-\frac {\pi }{4}+d x\right |2\right ) \sqrt {e \tan (c+d x)}}{a^2 d \sqrt {\sin (2 c+2 d x)}}+\frac {2 e^3 (e \tan (c+d x))^{3/2}}{3 a^2 d}-\frac {4 e^3 \cos (c+d x) (e \tan (c+d x))^{3/2}}{a^2 d}+\frac {e^{9/2} \text {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^2 d}-\frac {e^{9/2} \text {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^2 d}\\ &=-\frac {e^{9/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a^2 d}+\frac {e^{9/2} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a^2 d}+\frac {e^{9/2} \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} a^2 d}-\frac {e^{9/2} \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} a^2 d}+\frac {4 e^4 \cos (c+d x) E\left (\left .c-\frac {\pi }{4}+d x\right |2\right ) \sqrt {e \tan (c+d x)}}{a^2 d \sqrt {\sin (2 c+2 d x)}}+\frac {2 e^3 (e \tan (c+d x))^{3/2}}{3 a^2 d}-\frac {4 e^3 \cos (c+d x) (e \tan (c+d x))^{3/2}}{a^2 d}\\ \end {align*}
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Mathematica [F]
time = 4.37, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {(e \tan (c+d x))^{9/2}}{(a+a \sec (c+d x))^2} \, dx \end {gather*}
Verification is not applicable to the result.
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Maple [C] Result contains complex when optimal does not.
time = 0.29, size = 1504, normalized size = 4.82
method | result | size |
default | \(\text {Expression too large to display}\) | \(1504\) |
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 \begin {gather*} \text {Failed to integrate} \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(-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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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 {{\cos \left (c+d\,x\right )}^2\,{\left (e\,\mathrm {tan}\left (c+d\,x\right )\right )}^{9/2}}{a^2\,{\left (\cos \left (c+d\,x\right )+1\right )}^2} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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