Optimal. Leaf size=117 \[ \frac{2 (d+e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}+\frac{15 d+16 e x}{15 d^5 \sqrt{d^2-e^2 x^2}}+\frac{5 d+8 e x}{15 d^3 \left (d^2-e^2 x^2\right )^{3/2}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right )}{d^5} \]
[Out]
________________________________________________________________________________________
Rubi [A] time = 0.158799, antiderivative size = 117, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {1805, 823, 12, 266, 63, 208} \[ \frac{2 (d+e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}+\frac{15 d+16 e x}{15 d^5 \sqrt{d^2-e^2 x^2}}+\frac{5 d+8 e x}{15 d^3 \left (d^2-e^2 x^2\right )^{3/2}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right )}{d^5} \]
Antiderivative was successfully verified.
[In]
[Out]
Rule 1805
Rule 823
Rule 12
Rule 266
Rule 63
Rule 208
Rubi steps
\begin{align*} \int \frac{(d+e x)^2}{x \left (d^2-e^2 x^2\right )^{7/2}} \, dx &=\frac{2 (d+e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}-\frac{\int \frac{-5 d^2-8 d e x}{x \left (d^2-e^2 x^2\right )^{5/2}} \, dx}{5 d^2}\\ &=\frac{2 (d+e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}+\frac{5 d+8 e x}{15 d^3 \left (d^2-e^2 x^2\right )^{3/2}}-\frac{\int \frac{-15 d^4 e^2-16 d^3 e^3 x}{x \left (d^2-e^2 x^2\right )^{3/2}} \, dx}{15 d^6 e^2}\\ &=\frac{2 (d+e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}+\frac{5 d+8 e x}{15 d^3 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{15 d+16 e x}{15 d^5 \sqrt{d^2-e^2 x^2}}-\frac{\int -\frac{15 d^6 e^4}{x \sqrt{d^2-e^2 x^2}} \, dx}{15 d^{10} e^4}\\ &=\frac{2 (d+e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}+\frac{5 d+8 e x}{15 d^3 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{15 d+16 e x}{15 d^5 \sqrt{d^2-e^2 x^2}}+\frac{\int \frac{1}{x \sqrt{d^2-e^2 x^2}} \, dx}{d^4}\\ &=\frac{2 (d+e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}+\frac{5 d+8 e x}{15 d^3 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{15 d+16 e x}{15 d^5 \sqrt{d^2-e^2 x^2}}+\frac{\operatorname{Subst}\left (\int \frac{1}{x \sqrt{d^2-e^2 x}} \, dx,x,x^2\right )}{2 d^4}\\ &=\frac{2 (d+e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}+\frac{5 d+8 e x}{15 d^3 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{15 d+16 e x}{15 d^5 \sqrt{d^2-e^2 x^2}}-\frac{\operatorname{Subst}\left (\int \frac{1}{\frac{d^2}{e^2}-\frac{x^2}{e^2}} \, dx,x,\sqrt{d^2-e^2 x^2}\right )}{d^4 e^2}\\ &=\frac{2 (d+e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}+\frac{5 d+8 e x}{15 d^3 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{15 d+16 e x}{15 d^5 \sqrt{d^2-e^2 x^2}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right )}{d^5}\\ \end{align*}
Mathematica [C] time = 0.0470325, size = 81, normalized size = 0.69 \[ \frac{3 d^5 \, _2F_1\left (-\frac{5}{2},1;-\frac{3}{2};1-\frac{e^2 x^2}{d^2}\right )-40 d^2 e^3 x^3+30 d^4 e x+3 d^5+16 e^5 x^5}{15 d^5 \left (d^2-e^2 x^2\right )^{5/2}} \]
Antiderivative was successfully verified.
[In]
[Out]
________________________________________________________________________________________
Maple [A] time = 0.058, size = 160, normalized size = 1.4 \begin{align*}{\frac{2}{5} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{-{\frac{5}{2}}}}+{\frac{2\,ex}{5\,d} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{-{\frac{5}{2}}}}+{\frac{8\,ex}{15\,{d}^{3}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{-{\frac{3}{2}}}}+{\frac{16\,ex}{15\,{d}^{5}}{\frac{1}{\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}}}+{\frac{1}{3\,{d}^{2}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{-{\frac{3}{2}}}}+{\frac{1}{{d}^{4}}{\frac{1}{\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}}}-{\frac{1}{{d}^{4}}\ln \left ({\frac{1}{x} \left ( 2\,{d}^{2}+2\,\sqrt{{d}^{2}}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}} \right ) } \right ){\frac{1}{\sqrt{{d}^{2}}}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Fricas [A] time = 2.02699, size = 350, normalized size = 2.99 \begin{align*} \frac{26 \, e^{4} x^{4} - 52 \, d e^{3} x^{3} + 52 \, d^{3} e x - 26 \, d^{4} + 15 \,{\left (e^{4} x^{4} - 2 \, d e^{3} x^{3} + 2 \, d^{3} e x - d^{4}\right )} \log \left (-\frac{d - \sqrt{-e^{2} x^{2} + d^{2}}}{x}\right ) -{\left (16 \, e^{3} x^{3} - 17 \, d e^{2} x^{2} - 22 \, d^{2} e x + 26 \, d^{3}\right )} \sqrt{-e^{2} x^{2} + d^{2}}}{15 \,{\left (d^{5} e^{4} x^{4} - 2 \, d^{6} e^{3} x^{3} + 2 \, d^{8} e x - d^{9}\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (d + e x\right )^{2}}{x \left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac{7}{2}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Giac [A] time = 1.15945, size = 159, normalized size = 1.36 \begin{align*} -\frac{\sqrt{-x^{2} e^{2} + d^{2}}{\left ({\left ({\left ({\left (x{\left (\frac{16 \, x e^{5}}{d^{5}} + \frac{15 \, e^{4}}{d^{4}}\right )} - \frac{40 \, e^{3}}{d^{3}}\right )} x - \frac{35 \, e^{2}}{d^{2}}\right )} x + \frac{30 \, e}{d}\right )} x + 26\right )}}{15 \,{\left (x^{2} e^{2} - d^{2}\right )}^{3}} - \frac{\log \left (\frac{{\left | -2 \, d e - 2 \, \sqrt{-x^{2} e^{2} + d^{2}} e \right |} e^{\left (-2\right )}}{2 \,{\left | x \right |}}\right )}{d^{5}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]