Optimal. Leaf size=166 \[ -\frac {3 e \left (d^2-e^2 x^2\right )^{p-1} \, _2F_1\left (1,p-1;p;1-\frac {e^2 x^2}{d^2}\right )}{2 d^2 (1-p)}-\frac {2 e \left (d^2-e^2 x^2\right )^{p-2}}{2-p}-\frac {d \left (d^2-e^2 x^2\right )^{p-2}}{x}+\frac {2 e^2 (4-p) x \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \left (d^2-e^2 x^2\right )^p \, _2F_1\left (\frac {1}{2},3-p;\frac {3}{2};\frac {e^2 x^2}{d^2}\right )}{d^5} \]
[Out]
________________________________________________________________________________________
Rubi [A] time = 0.23, antiderivative size = 166, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {852, 1807, 1652, 446, 79, 65, 12, 246, 245} \[ -\frac {3 e \left (d^2-e^2 x^2\right )^{p-1} \, _2F_1\left (1,p-1;p;1-\frac {e^2 x^2}{d^2}\right )}{2 d^2 (1-p)}+\frac {2 e^2 (4-p) x \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \left (d^2-e^2 x^2\right )^p \, _2F_1\left (\frac {1}{2},3-p;\frac {3}{2};\frac {e^2 x^2}{d^2}\right )}{d^5}-\frac {2 e \left (d^2-e^2 x^2\right )^{p-2}}{2-p}-\frac {d \left (d^2-e^2 x^2\right )^{p-2}}{x} \]
Antiderivative was successfully verified.
[In]
[Out]
Rule 12
Rule 65
Rule 79
Rule 245
Rule 246
Rule 446
Rule 852
Rule 1652
Rule 1807
Rubi steps
\begin {align*} \int \frac {\left (d^2-e^2 x^2\right )^p}{x^2 (d+e x)^3} \, dx &=\int \frac {(d-e x)^3 \left (d^2-e^2 x^2\right )^{-3+p}}{x^2} \, dx\\ &=-\frac {d \left (d^2-e^2 x^2\right )^{-2+p}}{x}-\frac {\int \frac {\left (d^2-e^2 x^2\right )^{-3+p} \left (3 d^4 e-2 d^3 e^2 (4-p) x+d^2 e^3 x^2\right )}{x} \, dx}{d^2}\\ &=-\frac {d \left (d^2-e^2 x^2\right )^{-2+p}}{x}-\frac {\int -2 d^3 e^2 (4-p) \left (d^2-e^2 x^2\right )^{-3+p} \, dx}{d^2}-\frac {\int \frac {\left (d^2-e^2 x^2\right )^{-3+p} \left (3 d^4 e+d^2 e^3 x^2\right )}{x} \, dx}{d^2}\\ &=-\frac {d \left (d^2-e^2 x^2\right )^{-2+p}}{x}-\frac {\operatorname {Subst}\left (\int \frac {\left (d^2-e^2 x\right )^{-3+p} \left (3 d^4 e+d^2 e^3 x\right )}{x} \, dx,x,x^2\right )}{2 d^2}+\left (2 d e^2 (4-p)\right ) \int \left (d^2-e^2 x^2\right )^{-3+p} \, dx\\ &=-\frac {2 e \left (d^2-e^2 x^2\right )^{-2+p}}{2-p}-\frac {d \left (d^2-e^2 x^2\right )^{-2+p}}{x}-\frac {1}{2} (3 e) \operatorname {Subst}\left (\int \frac {\left (d^2-e^2 x\right )^{-2+p}}{x} \, dx,x,x^2\right )+\frac {\left (2 e^2 (4-p) \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p}\right ) \int \left (1-\frac {e^2 x^2}{d^2}\right )^{-3+p} \, dx}{d^5}\\ &=-\frac {2 e \left (d^2-e^2 x^2\right )^{-2+p}}{2-p}-\frac {d \left (d^2-e^2 x^2\right )^{-2+p}}{x}+\frac {2 e^2 (4-p) x \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}{2},3-p;\frac {3}{2};\frac {e^2 x^2}{d^2}\right )}{d^5}-\frac {3 e \left (d^2-e^2 x^2\right )^{-1+p} \, _2F_1\left (1,-1+p;p;1-\frac {e^2 x^2}{d^2}\right )}{2 d^2 (1-p)}\\ \end {align*}
________________________________________________________________________________________
Mathematica [A] time = 0.43, size = 280, normalized size = 1.69 \[ \frac {\left (d^2-e^2 x^2\right )^p \left (-\frac {12 d e \left (1-\frac {d^2}{e^2 x^2}\right )^{-p} \, _2F_1\left (-p,-p;1-p;\frac {d^2}{e^2 x^2}\right )}{p}-\frac {8 d^2 \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \, _2F_1\left (-\frac {1}{2},-p;\frac {1}{2};\frac {e^2 x^2}{d^2}\right )}{x}+\frac {3 e 2^{p+2} (e x-d) \left (\frac {e x}{d}+1\right )^{-p} \, _2F_1\left (1-p,p+1;p+2;\frac {d-e x}{2 d}\right )}{p+1}+\frac {e 2^{p+2} (e x-d) \left (\frac {e x}{d}+1\right )^{-p} \, _2F_1\left (2-p,p+1;p+2;\frac {d-e x}{2 d}\right )}{p+1}+\frac {e 2^p (e x-d) \left (\frac {e x}{d}+1\right )^{-p} \, _2F_1\left (3-p,p+1;p+2;\frac {d-e x}{2 d}\right )}{p+1}\right )}{8 d^5} \]
Antiderivative was successfully verified.
[In]
[Out]
________________________________________________________________________________________
fricas [F] time = 0.84, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{p}}{e^{3} x^{5} + 3 \, d e^{2} x^{4} + 3 \, d^{2} e x^{3} + d^{3} x^{2}}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{p}}{{\left (e x + d\right )}^{3} x^{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maple [F] time = 0.07, size = 0, normalized size = 0.00 \[ \int \frac {\left (-e^{2} x^{2}+d^{2}\right )^{p}}{\left (e x +d \right )^{3} x^{2}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{p}}{{\left (e x + d\right )}^{3} x^{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (d^2-e^2\,x^2\right )}^p}{x^2\,{\left (d+e\,x\right )}^3} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{p}}{x^{2} \left (d + e x\right )^{3}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________