Optimal. Leaf size=754 \[ \frac{3 e^2 x \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} F_1\left (\frac{1}{2};-p,1;\frac{3}{2};-\frac{b x^2}{a},\frac{e^2 x^2}{d^2}\right )}{d^5}+\frac{2 e^2 x \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} F_1\left (\frac{1}{2};-p,2;\frac{3}{2};-\frac{b x^2}{a},\frac{e^2 x^2}{d^2}\right )}{d^5}+\frac{e^2 x \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} F_1\left (\frac{1}{2};-p,3;\frac{3}{2};-\frac{b x^2}{a},\frac{e^2 x^2}{d^2}\right )}{d^5}+\frac{2 e^4 x^3 \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} F_1\left (\frac{3}{2};-p,2;\frac{5}{2};-\frac{b x^2}{a},\frac{e^2 x^2}{d^2}\right )}{3 d^7}+\frac{e^4 x^3 \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} F_1\left (\frac{3}{2};-p,3;\frac{5}{2};-\frac{b x^2}{a},\frac{e^2 x^2}{d^2}\right )}{d^7}-\frac{3 b^2 e^3 \left (a+b x^2\right )^{p+1} \, _2F_1\left (3,p+1;p+2;\frac{e^2 \left (b x^2+a\right )}{b d^2+a e^2}\right )}{2 (p+1) \left (a e^2+b d^2\right )^3}-\frac{3 e^3 \left (a+b x^2\right )^{p+1} \, _2F_1\left (1,p+1;p+2;\frac{e^2 \left (b x^2+a\right )}{b d^2+a e^2}\right )}{2 d^4 (p+1) \left (a e^2+b d^2\right )}+\frac{b e^3 \left (a+b x^2\right )^{p+1} \left (2 a e^2+b d^2 (p+1)\right ) \, _2F_1\left (2,p+1;p+2;\frac{e^2 \left (b x^2+a\right )}{b d^2+a e^2}\right )}{4 d^2 (p+1) \left (a e^2+b d^2\right )^3}-\frac{2 b e^3 \left (a+b x^2\right )^{p+1} \, _2F_1\left (2,p+1;p+2;\frac{e^2 \left (b x^2+a\right )}{b d^2+a e^2}\right )}{d^2 (p+1) \left (a e^2+b d^2\right )^2}-\frac{e^3 \left (a+b x^2\right )^{p+1}}{4 \left (d^2-e^2 x^2\right )^2 \left (a e^2+b d^2\right )}+\frac{3 e \left (a+b x^2\right )^{p+1} \, _2F_1\left (1,p+1;p+2;\frac{b x^2}{a}+1\right )}{2 a d^4 (p+1)}-\frac{\left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} \, _2F_1\left (-\frac{1}{2},-p;\frac{1}{2};-\frac{b x^2}{a}\right )}{d^3 x} \]
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Rubi [A] time = 0.859768, antiderivative size = 754, normalized size of antiderivative = 1., number of steps used = 31, number of rules used = 14, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.7, Rules used = {961, 365, 364, 266, 65, 757, 430, 429, 444, 68, 511, 510, 446, 78} \[ \frac{3 e^2 x \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} F_1\left (\frac{1}{2};-p,1;\frac{3}{2};-\frac{b x^2}{a},\frac{e^2 x^2}{d^2}\right )}{d^5}+\frac{2 e^2 x \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} F_1\left (\frac{1}{2};-p,2;\frac{3}{2};-\frac{b x^2}{a},\frac{e^2 x^2}{d^2}\right )}{d^5}+\frac{e^2 x \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} F_1\left (\frac{1}{2};-p,3;\frac{3}{2};-\frac{b x^2}{a},\frac{e^2 x^2}{d^2}\right )}{d^5}+\frac{2 e^4 x^3 \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} F_1\left (\frac{3}{2};-p,2;\frac{5}{2};-\frac{b x^2}{a},\frac{e^2 x^2}{d^2}\right )}{3 d^7}+\frac{e^4 x^3 \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} F_1\left (\frac{3}{2};-p,3;\frac{5}{2};-\frac{b x^2}{a},\frac{e^2 x^2}{d^2}\right )}{d^7}-\frac{3 b^2 e^3 \left (a+b x^2\right )^{p+1} \, _2F_1\left (3,p+1;p+2;\frac{e^2 \left (b x^2+a\right )}{b d^2+a e^2}\right )}{2 (p+1) \left (a e^2+b d^2\right )^3}-\frac{3 e^3 \left (a+b x^2\right )^{p+1} \, _2F_1\left (1,p+1;p+2;\frac{e^2 \left (b x^2+a\right )}{b d^2+a e^2}\right )}{2 d^4 (p+1) \left (a e^2+b d^2\right )}+\frac{b e^3 \left (a+b x^2\right )^{p+1} \left (2 a e^2+b d^2 (p+1)\right ) \, _2F_1\left (2,p+1;p+2;\frac{e^2 \left (b x^2+a\right )}{b d^2+a e^2}\right )}{4 d^2 (p+1) \left (a e^2+b d^2\right )^3}-\frac{2 b e^3 \left (a+b x^2\right )^{p+1} \, _2F_1\left (2,p+1;p+2;\frac{e^2 \left (b x^2+a\right )}{b d^2+a e^2}\right )}{d^2 (p+1) \left (a e^2+b d^2\right )^2}-\frac{e^3 \left (a+b x^2\right )^{p+1}}{4 \left (d^2-e^2 x^2\right )^2 \left (a e^2+b d^2\right )}+\frac{3 e \left (a+b x^2\right )^{p+1} \, _2F_1\left (1,p+1;p+2;\frac{b x^2}{a}+1\right )}{2 a d^4 (p+1)}-\frac{\left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} \, _2F_1\left (-\frac{1}{2},-p;\frac{1}{2};-\frac{b x^2}{a}\right )}{d^3 x} \]
Antiderivative was successfully verified.
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Rule 961
Rule 365
Rule 364
Rule 266
Rule 65
Rule 757
Rule 430
Rule 429
Rule 444
Rule 68
Rule 511
Rule 510
Rule 446
Rule 78
Rubi steps
\begin{align*} \int \frac{\left (a+b x^2\right )^p}{x^2 (d+e x)^3} \, dx &=\int \left (\frac{\left (a+b x^2\right )^p}{d^3 x^2}-\frac{3 e \left (a+b x^2\right )^p}{d^4 x}+\frac{e^2 \left (a+b x^2\right )^p}{d^2 (d+e x)^3}+\frac{2 e^2 \left (a+b x^2\right )^p}{d^3 (d+e x)^2}+\frac{3 e^2 \left (a+b x^2\right )^p}{d^4 (d+e x)}\right ) \, dx\\ &=\frac{\int \frac{\left (a+b x^2\right )^p}{x^2} \, dx}{d^3}-\frac{(3 e) \int \frac{\left (a+b x^2\right )^p}{x} \, dx}{d^4}+\frac{\left (3 e^2\right ) \int \frac{\left (a+b x^2\right )^p}{d+e x} \, dx}{d^4}+\frac{\left (2 e^2\right ) \int \frac{\left (a+b x^2\right )^p}{(d+e x)^2} \, dx}{d^3}+\frac{e^2 \int \frac{\left (a+b x^2\right )^p}{(d+e x)^3} \, dx}{d^2}\\ &=-\frac{(3 e) \operatorname{Subst}\left (\int \frac{(a+b x)^p}{x} \, dx,x,x^2\right )}{2 d^4}+\frac{\left (3 e^2\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}{d^4}+\frac{\left (2 e^2\right ) \int \left (\frac{d^2 \left (a+b x^2\right )^p}{\left (d^2-e^2 x^2\right )^2}-\frac{2 d e x \left (a+b x^2\right )^p}{\left (d^2-e^2 x^2\right )^2}+\frac{e^2 x^2 \left (a+b x^2\right )^p}{\left (-d^2+e^2 x^2\right )^2}\right ) \, dx}{d^3}+\frac{e^2 \int \left (\frac{d^3 \left (a+b x^2\right )^p}{\left (d^2-e^2 x^2\right )^3}-\frac{3 d^2 e x \left (a+b x^2\right )^p}{\left (d^2-e^2 x^2\right )^3}+\frac{3 d e^2 x^2 \left (a+b x^2\right )^p}{\left (d^2-e^2 x^2\right )^3}+\frac{e^3 x^3 \left (a+b x^2\right )^p}{\left (-d^2+e^2 x^2\right )^3}\right ) \, dx}{d^2}+\frac{\left (\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}{x^2} \, dx}{d^3}\\ &=-\frac{\left (a+b x^2\right )^p \left (1+\frac{b x^2}{a}\right )^{-p} \, _2F_1\left (-\frac{1}{2},-p;\frac{1}{2};-\frac{b x^2}{a}\right )}{d^3 x}+\frac{3 e \left (a+b x^2\right )^{1+p} \, _2F_1\left (1,1+p;2+p;1+\frac{b x^2}{a}\right )}{2 a d^4 (1+p)}+\frac{\left (3 e^2\right ) \int \frac{\left (a+b x^2\right )^p}{d^2-e^2 x^2} \, dx}{d^3}+\frac{\left (2 e^2\right ) \int \frac{\left (a+b x^2\right )^p}{\left (d^2-e^2 x^2\right )^2} \, dx}{d}+\left (d e^2\right ) \int \frac{\left (a+b x^2\right )^p}{\left (d^2-e^2 x^2\right )^3} \, dx-\left (3 e^3\right ) \int \frac{x \left (a+b x^2\right )^p}{\left (d^2-e^2 x^2\right )^3} \, dx+\frac{\left (3 e^3\right ) \int \frac{x \left (a+b x^2\right )^p}{-d^2+e^2 x^2} \, dx}{d^4}-\frac{\left (4 e^3\right ) \int \frac{x \left (a+b x^2\right )^p}{\left (d^2-e^2 x^2\right )^2} \, dx}{d^2}+\frac{\left (2 e^4\right ) \int \frac{x^2 \left (a+b x^2\right )^p}{\left (-d^2+e^2 x^2\right )^2} \, dx}{d^3}+\frac{\left (3 e^4\right ) \int \frac{x^2 \left (a+b x^2\right )^p}{\left (d^2-e^2 x^2\right )^3} \, dx}{d}+\frac{e^5 \int \frac{x^3 \left (a+b x^2\right )^p}{\left (-d^2+e^2 x^2\right )^3} \, dx}{d^2}\\ &=-\frac{\left (a+b x^2\right )^p \left (1+\frac{b x^2}{a}\right )^{-p} \, _2F_1\left (-\frac{1}{2},-p;\frac{1}{2};-\frac{b x^2}{a}\right )}{d^3 x}+\frac{3 e \left (a+b x^2\right )^{1+p} \, _2F_1\left (1,1+p;2+p;1+\frac{b x^2}{a}\right )}{2 a d^4 (1+p)}-\frac{1}{2} \left (3 e^3\right ) \operatorname{Subst}\left (\int \frac{(a+b x)^p}{\left (d^2-e^2 x\right )^3} \, dx,x,x^2\right )+\frac{\left (3 e^3\right ) \operatorname{Subst}\left (\int \frac{(a+b x)^p}{-d^2+e^2 x} \, dx,x,x^2\right )}{2 d^4}-\frac{\left (2 e^3\right ) \operatorname{Subst}\left (\int \frac{(a+b x)^p}{\left (d^2-e^2 x\right )^2} \, dx,x,x^2\right )}{d^2}+\frac{e^5 \operatorname{Subst}\left (\int \frac{x (a+b x)^p}{\left (-d^2+e^2 x\right )^3} \, dx,x,x^2\right )}{2 d^2}+\frac{\left (3 e^2 \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}{d^3}+\frac{\left (2 e^2 \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}{\left (d^2-e^2 x^2\right )^2} \, dx}{d}+\left (d e^2 \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}{\left (d^2-e^2 x^2\right )^3} \, dx+\frac{\left (2 e^4 \left (a+b x^2\right )^p \left (1+\frac{b x^2}{a}\right )^{-p}\right ) \int \frac{x^2 \left (1+\frac{b x^2}{a}\right )^p}{\left (-d^2+e^2 x^2\right )^2} \, dx}{d^3}+\frac{\left (3 e^4 \left (a+b x^2\right )^p \left (1+\frac{b x^2}{a}\right )^{-p}\right ) \int \frac{x^2 \left (1+\frac{b x^2}{a}\right )^p}{\left (d^2-e^2 x^2\right )^3} \, dx}{d}\\ &=-\frac{e^3 \left (a+b x^2\right )^{1+p}}{4 \left (b d^2+a e^2\right ) \left (d^2-e^2 x^2\right )^2}+\frac{3 e^2 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 )}{d^5}+\frac{2 e^2 x \left (a+b x^2\right )^p \left (1+\frac{b x^2}{a}\right )^{-p} F_1\left (\frac{1}{2};-p,2;\frac{3}{2};-\frac{b x^2}{a},\frac{e^2 x^2}{d^2}\right )}{d^5}+\frac{e^2 x \left (a+b x^2\right )^p \left (1+\frac{b x^2}{a}\right )^{-p} F_1\left (\frac{1}{2};-p,3;\frac{3}{2};-\frac{b x^2}{a},\frac{e^2 x^2}{d^2}\right )}{d^5}+\frac{2 e^4 x^3 \left (a+b x^2\right )^p \left (1+\frac{b x^2}{a}\right )^{-p} F_1\left (\frac{3}{2};-p,2;\frac{5}{2};-\frac{b x^2}{a},\frac{e^2 x^2}{d^2}\right )}{3 d^7}+\frac{e^4 x^3 \left (a+b x^2\right )^p \left (1+\frac{b x^2}{a}\right )^{-p} F_1\left (\frac{3}{2};-p,3;\frac{5}{2};-\frac{b x^2}{a},\frac{e^2 x^2}{d^2}\right )}{d^7}-\frac{\left (a+b x^2\right )^p \left (1+\frac{b x^2}{a}\right )^{-p} \, _2F_1\left (-\frac{1}{2},-p;\frac{1}{2};-\frac{b x^2}{a}\right )}{d^3 x}-\frac{3 e^3 \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 )}{2 d^4 \left (b d^2+a e^2\right ) (1+p)}+\frac{3 e \left (a+b x^2\right )^{1+p} \, _2F_1\left (1,1+p;2+p;1+\frac{b x^2}{a}\right )}{2 a d^4 (1+p)}-\frac{2 b e^3 \left (a+b x^2\right )^{1+p} \, _2F_1\left (2,1+p;2+p;\frac{e^2 \left (a+b x^2\right )}{b d^2+a e^2}\right )}{d^2 \left (b d^2+a e^2\right )^2 (1+p)}-\frac{3 b^2 e^3 \left (a+b x^2\right )^{1+p} \, _2F_1\left (3,1+p;2+p;\frac{e^2 \left (a+b x^2\right )}{b d^2+a e^2}\right )}{2 \left (b d^2+a e^2\right )^3 (1+p)}+\frac{\left (e^3 \left (2 a e^2+b d^2 (1+p)\right )\right ) \operatorname{Subst}\left (\int \frac{(a+b x)^p}{\left (-d^2+e^2 x\right )^2} \, dx,x,x^2\right )}{4 d^2 \left (b d^2+a e^2\right )}\\ &=-\frac{e^3 \left (a+b x^2\right )^{1+p}}{4 \left (b d^2+a e^2\right ) \left (d^2-e^2 x^2\right )^2}+\frac{3 e^2 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 )}{d^5}+\frac{2 e^2 x \left (a+b x^2\right )^p \left (1+\frac{b x^2}{a}\right )^{-p} F_1\left (\frac{1}{2};-p,2;\frac{3}{2};-\frac{b x^2}{a},\frac{e^2 x^2}{d^2}\right )}{d^5}+\frac{e^2 x \left (a+b x^2\right )^p \left (1+\frac{b x^2}{a}\right )^{-p} F_1\left (\frac{1}{2};-p,3;\frac{3}{2};-\frac{b x^2}{a},\frac{e^2 x^2}{d^2}\right )}{d^5}+\frac{2 e^4 x^3 \left (a+b x^2\right )^p \left (1+\frac{b x^2}{a}\right )^{-p} F_1\left (\frac{3}{2};-p,2;\frac{5}{2};-\frac{b x^2}{a},\frac{e^2 x^2}{d^2}\right )}{3 d^7}+\frac{e^4 x^3 \left (a+b x^2\right )^p \left (1+\frac{b x^2}{a}\right )^{-p} F_1\left (\frac{3}{2};-p,3;\frac{5}{2};-\frac{b x^2}{a},\frac{e^2 x^2}{d^2}\right )}{d^7}-\frac{\left (a+b x^2\right )^p \left (1+\frac{b x^2}{a}\right )^{-p} \, _2F_1\left (-\frac{1}{2},-p;\frac{1}{2};-\frac{b x^2}{a}\right )}{d^3 x}-\frac{3 e^3 \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 )}{2 d^4 \left (b d^2+a e^2\right ) (1+p)}+\frac{3 e \left (a+b x^2\right )^{1+p} \, _2F_1\left (1,1+p;2+p;1+\frac{b x^2}{a}\right )}{2 a d^4 (1+p)}-\frac{2 b e^3 \left (a+b x^2\right )^{1+p} \, _2F_1\left (2,1+p;2+p;\frac{e^2 \left (a+b x^2\right )}{b d^2+a e^2}\right )}{d^2 \left (b d^2+a e^2\right )^2 (1+p)}+\frac{b e^3 \left (2 a e^2+b d^2 (1+p)\right ) \left (a+b x^2\right )^{1+p} \, _2F_1\left (2,1+p;2+p;\frac{e^2 \left (a+b x^2\right )}{b d^2+a e^2}\right )}{4 d^2 \left (b d^2+a e^2\right )^3 (1+p)}-\frac{3 b^2 e^3 \left (a+b x^2\right )^{1+p} \, _2F_1\left (3,1+p;2+p;\frac{e^2 \left (a+b x^2\right )}{b d^2+a e^2}\right )}{2 \left (b d^2+a e^2\right )^3 (1+p)}\\ \end{align*}
Mathematica [F] time = 0.17533, size = 0, normalized size = 0. \[ \int \frac{\left (a+b x^2\right )^p}{x^2 (d+e x)^3} \, dx \]
Verification is Not applicable to the result.
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Maple [F] time = 0.793, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( b{x}^{2}+a \right ) ^{p}}{{x}^{2} \left ( ex+d \right ) ^{3}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x^{2} + a\right )}^{p}}{{\left (e x + d\right )}^{3} x^{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (b x^{2} + a\right )}^{p}}{e^{3} x^{5} + 3 \, d e^{2} x^{4} + 3 \, d^{2} e x^{3} + d^{3} x^{2}}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x^{2} + a\right )}^{p}}{{\left (e x + d\right )}^{3} x^{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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