Optimal. Leaf size=281 \[ -\frac {d^2 \left (a+b x^2\right )^{1+p}}{e \left (b d^2+a e^2\right ) (d+e x)}-\frac {2 \left (a e^2+b d^2 (1+p)\right ) x \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} F_1\left (\frac {1}{2};-p,1;\frac {3}{2};-\frac {b x^2}{a},\frac {e^2 x^2}{d^2}\right )}{e^2 \left (b d^2+a e^2\right )}+\frac {\left (a e^2+2 b d^2 (1+p)\right ) x \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \, _2F_1\left (\frac {1}{2},-p;\frac {3}{2};-\frac {b x^2}{a}\right )}{e^2 \left (b d^2+a e^2\right )}+\frac {d \left (a e^2+b d^2 (1+p)\right ) \left (a+b x^2\right )^{1+p} \, _2F_1\left (1,1+p;2+p;\frac {e^2 \left (a+b x^2\right )}{b d^2+a e^2}\right )}{e \left (b d^2+a e^2\right )^2 (1+p)} \]
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Rubi [A]
time = 0.21, antiderivative size = 277, normalized size of antiderivative = 0.99, number of steps
used = 10, number of rules used = 9, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.450, Rules used = {1665, 858,
252, 251, 771, 441, 440, 455, 70} \begin {gather*} -\frac {2 x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (a e^2+b d^2 (p+1)\right ) F_1\left (\frac {1}{2};-p,1;\frac {3}{2};-\frac {b x^2}{a},\frac {e^2 x^2}{d^2}\right )}{e^2 \left (a e^2+b d^2\right )}+\frac {x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (a+\frac {2 b d^2 (p+1)}{e^2}\right ) \, _2F_1\left (\frac {1}{2},-p;\frac {3}{2};-\frac {b x^2}{a}\right )}{a e^2+b d^2}+\frac {d \left (a+b x^2\right )^{p+1} \left (a e^2+b d^2 (p+1)\right ) \, _2F_1\left (1,p+1;p+2;\frac {e^2 \left (b x^2+a\right )}{b d^2+a e^2}\right )}{e (p+1) \left (a e^2+b d^2\right )^2}-\frac {d^2 \left (a+b x^2\right )^{p+1}}{e (d+e x) \left (a e^2+b d^2\right )} \end {gather*}
Antiderivative was successfully verified.
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Rule 70
Rule 251
Rule 252
Rule 440
Rule 441
Rule 455
Rule 771
Rule 858
Rule 1665
Rubi steps
\begin {align*} \int \frac {x^2 \left (a+b x^2\right )^p}{(d+e x)^2} \, dx &=-\frac {d^2 \left (a+b x^2\right )^{1+p}}{e \left (b d^2+a e^2\right ) (d+e x)}-\frac {\int \frac {\left (a d-\frac {\left (a e^2+2 b d^2 (1+p)\right ) x}{e}\right ) \left (a+b x^2\right )^p}{d+e x} \, dx}{b d^2+a e^2}\\ &=-\frac {d^2 \left (a+b x^2\right )^{1+p}}{e \left (b d^2+a e^2\right ) (d+e x)}-\frac {\left (2 d \left (a e^2+b d^2 (1+p)\right )\right ) \int \frac {\left (a+b x^2\right )^p}{d+e x} \, dx}{e^2 \left (b d^2+a e^2\right )}+\frac {\left (a+\frac {2 b d^2 (1+p)}{e^2}\right ) \int \left (a+b x^2\right )^p \, dx}{b d^2+a e^2}\\ &=-\frac {d^2 \left (a+b x^2\right )^{1+p}}{e \left (b d^2+a e^2\right ) (d+e x)}-\frac {\left (2 d \left (a e^2+b d^2 (1+p)\right )\right ) \int \left (\frac {d \left (a+b x^2\right )^p}{d^2-e^2 x^2}+\frac {e x \left (a+b x^2\right )^p}{-d^2+e^2 x^2}\right ) \, dx}{e^2 \left (b d^2+a e^2\right )}+\frac {\left (\left (a+\frac {2 b d^2 (1+p)}{e^2}\right ) \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p}\right ) \int \left (1+\frac {b x^2}{a}\right )^p \, dx}{b d^2+a e^2}\\ &=-\frac {d^2 \left (a+b x^2\right )^{1+p}}{e \left (b d^2+a e^2\right ) (d+e x)}+\frac {\left (a+\frac {2 b d^2 (1+p)}{e^2}\right ) x \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \, _2F_1\left (\frac {1}{2},-p;\frac {3}{2};-\frac {b x^2}{a}\right )}{b d^2+a e^2}-\frac {\left (2 d^2 \left (a e^2+b d^2 (1+p)\right )\right ) \int \frac {\left (a+b x^2\right )^p}{d^2-e^2 x^2} \, dx}{e^2 \left (b d^2+a e^2\right )}-\frac {\left (2 d \left (a e^2+b d^2 (1+p)\right )\right ) \int \frac {x \left (a+b x^2\right )^p}{-d^2+e^2 x^2} \, dx}{e \left (b d^2+a e^2\right )}\\ &=-\frac {d^2 \left (a+b x^2\right )^{1+p}}{e \left (b d^2+a e^2\right ) (d+e x)}+\frac {\left (a+\frac {2 b d^2 (1+p)}{e^2}\right ) x \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \, _2F_1\left (\frac {1}{2},-p;\frac {3}{2};-\frac {b x^2}{a}\right )}{b d^2+a e^2}-\frac {\left (d \left (a e^2+b d^2 (1+p)\right )\right ) \text {Subst}\left (\int \frac {(a+b x)^p}{-d^2+e^2 x} \, dx,x,x^2\right )}{e \left (b d^2+a e^2\right )}-\frac {\left (2 d^2 \left (a e^2+b d^2 (1+p)\right ) \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p}\right ) \int \frac {\left (1+\frac {b x^2}{a}\right )^p}{d^2-e^2 x^2} \, dx}{e^2 \left (b d^2+a e^2\right )}\\ &=-\frac {d^2 \left (a+b x^2\right )^{1+p}}{e \left (b d^2+a e^2\right ) (d+e x)}-\frac {2 \left (a e^2+b d^2 (1+p)\right ) x \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} F_1\left (\frac {1}{2};-p,1;\frac {3}{2};-\frac {b x^2}{a},\frac {e^2 x^2}{d^2}\right )}{e^2 \left (b d^2+a e^2\right )}+\frac {\left (a+\frac {2 b d^2 (1+p)}{e^2}\right ) x \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \, _2F_1\left (\frac {1}{2},-p;\frac {3}{2};-\frac {b x^2}{a}\right )}{b d^2+a e^2}+\frac {d \left (a e^2+b d^2 (1+p)\right ) \left (a+b x^2\right )^{1+p} \, _2F_1\left (1,1+p;2+p;\frac {e^2 \left (a+b x^2\right )}{b d^2+a e^2}\right )}{e \left (b d^2+a e^2\right )^2 (1+p)}\\ \end {align*}
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Mathematica [A]
time = 0.42, size = 300, normalized size = 1.07 \begin {gather*} \frac {\left (a+b x^2\right )^p \left (\frac {d^2 \left (\frac {e \left (-\sqrt {-\frac {a}{b}}+x\right )}{d+e x}\right )^{-p} \left (\frac {e \left (\sqrt {-\frac {a}{b}}+x\right )}{d+e x}\right )^{-p} F_1\left (1-2 p;-p,-p;2-2 p;\frac {d-\sqrt {-\frac {a}{b}} e}{d+e x},\frac {d+\sqrt {-\frac {a}{b}} e}{d+e x}\right )}{(-1+2 p) (d+e x)}-\frac {d \left (\frac {e \left (-\sqrt {-\frac {a}{b}}+x\right )}{d+e x}\right )^{-p} \left (\frac {e \left (\sqrt {-\frac {a}{b}}+x\right )}{d+e x}\right )^{-p} F_1\left (-2 p;-p,-p;1-2 p;\frac {d-\sqrt {-\frac {a}{b}} e}{d+e x},\frac {d+\sqrt {-\frac {a}{b}} e}{d+e x}\right )}{p}+e x \left (1+\frac {b x^2}{a}\right )^{-p} \, _2F_1\left (\frac {1}{2},-p;\frac {3}{2};-\frac {b x^2}{a}\right )\right )}{e^3} \end {gather*}
Antiderivative was successfully verified.
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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {x^{2} \left (b \,x^{2}+a \right )^{p}}{\left (e x +d \right )^{2}}\, dx\]
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 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [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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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 {x^2\,{\left (b\,x^2+a\right )}^p}{{\left (d+e\,x\right )}^2} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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