Optimal. Leaf size=201 \[ -\frac {30 e^{13/4} \left (\sqrt {d}+\sqrt {e} x\right ) \sqrt {\frac {d+e x^2}{\left (\sqrt {d}+\sqrt {e} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{e} \sqrt {x}}{\sqrt [4]{d}}\right )|\frac {1}{2}\right )}{1001 d^{13/4} \sqrt {d+e x^2}}-\frac {60 e^{5/2} \sqrt {d+e x^2}}{1001 d^3 x^{3/2}}+\frac {36 e^{3/2} \sqrt {d+e x^2}}{1001 d^2 x^{7/2}}-\frac {4 \sqrt {e} \sqrt {d+e x^2}}{143 d x^{11/2}}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{13 x^{13/2}} \]
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Rubi [A] time = 0.10, antiderivative size = 201, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {6221, 325, 329, 220} \[ -\frac {60 e^{5/2} \sqrt {d+e x^2}}{1001 d^3 x^{3/2}}+\frac {36 e^{3/2} \sqrt {d+e x^2}}{1001 d^2 x^{7/2}}-\frac {30 e^{13/4} \left (\sqrt {d}+\sqrt {e} x\right ) \sqrt {\frac {d+e x^2}{\left (\sqrt {d}+\sqrt {e} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{e} \sqrt {x}}{\sqrt [4]{d}}\right )|\frac {1}{2}\right )}{1001 d^{13/4} \sqrt {d+e x^2}}-\frac {4 \sqrt {e} \sqrt {d+e x^2}}{143 d x^{11/2}}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{13 x^{13/2}} \]
Antiderivative was successfully verified.
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Rule 220
Rule 325
Rule 329
Rule 6221
Rubi steps
\begin {align*} \int \frac {\tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{x^{15/2}} \, dx &=-\frac {2 \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{13 x^{13/2}}+\frac {1}{13} \left (2 \sqrt {e}\right ) \int \frac {1}{x^{13/2} \sqrt {d+e x^2}} \, dx\\ &=-\frac {4 \sqrt {e} \sqrt {d+e x^2}}{143 d x^{11/2}}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{13 x^{13/2}}-\frac {\left (18 e^{3/2}\right ) \int \frac {1}{x^{9/2} \sqrt {d+e x^2}} \, dx}{143 d}\\ &=-\frac {4 \sqrt {e} \sqrt {d+e x^2}}{143 d x^{11/2}}+\frac {36 e^{3/2} \sqrt {d+e x^2}}{1001 d^2 x^{7/2}}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{13 x^{13/2}}+\frac {\left (90 e^{5/2}\right ) \int \frac {1}{x^{5/2} \sqrt {d+e x^2}} \, dx}{1001 d^2}\\ &=-\frac {4 \sqrt {e} \sqrt {d+e x^2}}{143 d x^{11/2}}+\frac {36 e^{3/2} \sqrt {d+e x^2}}{1001 d^2 x^{7/2}}-\frac {60 e^{5/2} \sqrt {d+e x^2}}{1001 d^3 x^{3/2}}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{13 x^{13/2}}-\frac {\left (30 e^{7/2}\right ) \int \frac {1}{\sqrt {x} \sqrt {d+e x^2}} \, dx}{1001 d^3}\\ &=-\frac {4 \sqrt {e} \sqrt {d+e x^2}}{143 d x^{11/2}}+\frac {36 e^{3/2} \sqrt {d+e x^2}}{1001 d^2 x^{7/2}}-\frac {60 e^{5/2} \sqrt {d+e x^2}}{1001 d^3 x^{3/2}}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{13 x^{13/2}}-\frac {\left (60 e^{7/2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {d+e x^4}} \, dx,x,\sqrt {x}\right )}{1001 d^3}\\ &=-\frac {4 \sqrt {e} \sqrt {d+e x^2}}{143 d x^{11/2}}+\frac {36 e^{3/2} \sqrt {d+e x^2}}{1001 d^2 x^{7/2}}-\frac {60 e^{5/2} \sqrt {d+e x^2}}{1001 d^3 x^{3/2}}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{13 x^{13/2}}-\frac {30 e^{13/4} \left (\sqrt {d}+\sqrt {e} x\right ) \sqrt {\frac {d+e x^2}{\left (\sqrt {d}+\sqrt {e} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{e} \sqrt {x}}{\sqrt [4]{d}}\right )|\frac {1}{2}\right )}{1001 d^{13/4} \sqrt {d+e x^2}}\\ \end {align*}
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Mathematica [C] time = 0.47, size = 163, normalized size = 0.81 \[ \frac {2 \left (-\frac {30 e^4 x^{15/2} \sqrt {\frac {i \sqrt {d}}{\sqrt {e}}} \sqrt {\frac {d}{e x^2}+1} F\left (\left .i \sinh ^{-1}\left (\frac {\sqrt {\frac {i \sqrt {d}}{\sqrt {e}}}}{\sqrt {x}}\right )\right |-1\right )}{d^{7/2} \sqrt {d+e x^2}}-\frac {2 \sqrt {e} x \sqrt {d+e x^2} \left (7 d^2-9 d e x^2+15 e^2 x^4\right )}{d^3}-77 \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )\right )}{1001 x^{13/2}} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.52, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\operatorname {artanh}\left (\frac {\sqrt {e} x}{\sqrt {e x^{2} + d}}\right )}{x^{\frac {15}{2}}}, x\right ) \]
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 \[ \int \frac {\operatorname {artanh}\left (\frac {\sqrt {e} x}{\sqrt {e x^{2} + d}}\right )}{x^{\frac {15}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F(-2)] time = 180.00, size = 0, normalized size = 0.00 \[ \int \frac {\arctanh \left (\frac {x \sqrt {e}}{\sqrt {e \,x^{2}+d}}\right )}{x^{\frac {15}{2}}}\, dx \]
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 \[ 2 \, d \sqrt {e} \int -\frac {\sqrt {e x^{2} + d} x}{13 \, {\left ({\left (e^{2} x^{4} + d e x^{2}\right )} x^{\frac {15}{2}} - {\left (e x^{2} + d\right )} e^{\left (\log \left (e x^{2} + d\right ) + \frac {15}{2} \, \log \relax (x)\right )}\right )}}\,{d x} - \frac {\log \left (\sqrt {e} x + \sqrt {e x^{2} + d}\right )}{13 \, x^{\frac {13}{2}}} + \frac {\log \left (-\sqrt {e} x + \sqrt {e x^{2} + d}\right )}{13 \, x^{\frac {13}{2}}} \]
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 \[ \int \frac {\mathrm {atanh}\left (\frac {\sqrt {e}\,x}{\sqrt {e\,x^2+d}}\right )}{x^{15/2}} \,d x \]
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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