Optimal. Leaf size=214 \[ -\frac{2 e (g x)^{m+2} \left (d^2-e^2 x^2\right )^p \left (1-\frac{e^2 x^2}{d^2}\right )^{-p} \, _2F_1\left (\frac{m+2}{2},2-p;\frac{m+4}{2};\frac{e^2 x^2}{d^2}\right )}{d^3 g^2 (m+2)}-\frac{2 (m+p) (g x)^{m+1} \left (d^2-e^2 x^2\right )^p \left (1-\frac{e^2 x^2}{d^2}\right )^{-p} \, _2F_1\left (\frac{m+1}{2},2-p;\frac{m+3}{2};\frac{e^2 x^2}{d^2}\right )}{d^2 g (m+1) (-m-2 p+1)}+\frac{(g x)^{m+1} \left (d^2-e^2 x^2\right )^{p-1}}{g (-m-2 p+1)} \]
[Out]
________________________________________________________________________________________
Rubi [A] time = 0.223438, antiderivative size = 214, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.185, Rules used = {852, 1809, 808, 365, 364} \[ -\frac{2 e (g x)^{m+2} \left (d^2-e^2 x^2\right )^p \left (1-\frac{e^2 x^2}{d^2}\right )^{-p} \, _2F_1\left (\frac{m+2}{2},2-p;\frac{m+4}{2};\frac{e^2 x^2}{d^2}\right )}{d^3 g^2 (m+2)}-\frac{2 (m+p) (g x)^{m+1} \left (d^2-e^2 x^2\right )^p \left (1-\frac{e^2 x^2}{d^2}\right )^{-p} \, _2F_1\left (\frac{m+1}{2},2-p;\frac{m+3}{2};\frac{e^2 x^2}{d^2}\right )}{d^2 g (m+1) (-m-2 p+1)}+\frac{(g x)^{m+1} \left (d^2-e^2 x^2\right )^{p-1}}{g (-m-2 p+1)} \]
Antiderivative was successfully verified.
[In]
[Out]
Rule 852
Rule 1809
Rule 808
Rule 365
Rule 364
Rubi steps
\begin{align*} \int \frac{(g x)^m \left (d^2-e^2 x^2\right )^p}{(d+e x)^2} \, dx &=\int (g x)^m (d-e x)^2 \left (d^2-e^2 x^2\right )^{-2+p} \, dx\\ &=\frac{(g x)^{1+m} \left (d^2-e^2 x^2\right )^{-1+p}}{g (1-m-2 p)}+\frac{\int (g x)^m \left (-2 d^2 e^2 (m+p)-2 d e^3 (1-m-2 p) x\right ) \left (d^2-e^2 x^2\right )^{-2+p} \, dx}{e^2 (1-m-2 p)}\\ &=\frac{(g x)^{1+m} \left (d^2-e^2 x^2\right )^{-1+p}}{g (1-m-2 p)}-\frac{(2 d e) \int (g x)^{1+m} \left (d^2-e^2 x^2\right )^{-2+p} \, dx}{g}-\frac{\left (2 d^2 (m+p)\right ) \int (g x)^m \left (d^2-e^2 x^2\right )^{-2+p} \, dx}{1-m-2 p}\\ &=\frac{(g x)^{1+m} \left (d^2-e^2 x^2\right )^{-1+p}}{g (1-m-2 p)}-\frac{\left (2 e \left (d^2-e^2 x^2\right )^p \left (1-\frac{e^2 x^2}{d^2}\right )^{-p}\right ) \int (g x)^{1+m} \left (1-\frac{e^2 x^2}{d^2}\right )^{-2+p} \, dx}{d^3 g}-\frac{\left (2 (m+p) \left (d^2-e^2 x^2\right )^p \left (1-\frac{e^2 x^2}{d^2}\right )^{-p}\right ) \int (g x)^m \left (1-\frac{e^2 x^2}{d^2}\right )^{-2+p} \, dx}{d^2 (1-m-2 p)}\\ &=\frac{(g x)^{1+m} \left (d^2-e^2 x^2\right )^{-1+p}}{g (1-m-2 p)}-\frac{2 (m+p) (g x)^{1+m} \left (d^2-e^2 x^2\right )^p \left (1-\frac{e^2 x^2}{d^2}\right )^{-p} \, _2F_1\left (\frac{1+m}{2},2-p;\frac{3+m}{2};\frac{e^2 x^2}{d^2}\right )}{d^2 g (1+m) (1-m-2 p)}-\frac{2 e (g x)^{2+m} \left (d^2-e^2 x^2\right )^p \left (1-\frac{e^2 x^2}{d^2}\right )^{-p} \, _2F_1\left (\frac{2+m}{2},2-p;\frac{4+m}{2};\frac{e^2 x^2}{d^2}\right )}{d^3 g^2 (2+m)}\\ \end{align*}
Mathematica [A] time = 0.110775, size = 180, normalized size = 0.84 \[ \frac{x (g x)^m \left (d^2-e^2 x^2\right )^p \left (1-\frac{e^2 x^2}{d^2}\right )^{-p} \left (d^2 \left (m^2+5 m+6\right ) \, _2F_1\left (\frac{m+1}{2},2-p;\frac{m+3}{2};\frac{e^2 x^2}{d^2}\right )-e (m+1) x \left (2 d (m+3) \, _2F_1\left (\frac{m+2}{2},2-p;\frac{m+4}{2};\frac{e^2 x^2}{d^2}\right )-e (m+2) x \, _2F_1\left (\frac{m+3}{2},2-p;\frac{m+5}{2};\frac{e^2 x^2}{d^2}\right )\right )\right )}{d^4 (m+1) (m+2) (m+3)} \]
Antiderivative was successfully verified.
[In]
[Out]
________________________________________________________________________________________
Maple [F] time = 0.694, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( gx \right ) ^{m} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{p}}{ \left ( ex+d \right ) ^{2}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (-e^{2} x^{2} + d^{2}\right )}^{p} \left (g x\right )^{m}}{{\left (e x + d\right )}^{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (-e^{2} x^{2} + d^{2}\right )}^{p} \left (g x\right )^{m}}{e^{2} x^{2} + 2 \, d e x + d^{2}}, x\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 (g x\right )^{m} \left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{p}}{\left (d + e x\right )^{2}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (-e^{2} x^{2} + d^{2}\right )}^{p} \left (g x\right )^{m}}{{\left (e x + d\right )}^{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]