Optimal. Leaf size=213 \[ -\frac {\left (a+b x^2\right )^{p+1} \left (a e^2+b d^2 p\right ) \, _2F_1\left (1,p+1;p+2;\frac {b x^2}{a}+1\right )}{2 a^2 d^3 (p+1)}+\frac {e \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} F_1\left (-\frac {1}{2};-p,1;\frac {1}{2};-\frac {b x^2}{a},\frac {e^2 x^2}{d^2}\right )}{d^2 x}+\frac {e^4 \left (a+b x^2\right )^{p+1} \, _2F_1\left (1,p+1;p+2;\frac {e^2 \left (b x^2+a\right )}{b d^2+a e^2}\right )}{2 d^3 (p+1) \left (a e^2+b d^2\right )}-\frac {\left (a+b x^2\right )^{p+1}}{2 a d x^2} \]
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Rubi [A] time = 0.24, antiderivative size = 213, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {959, 446, 103, 156, 65, 68, 511, 510} \[ -\frac {\left (a+b x^2\right )^{p+1} \left (a e^2+b d^2 p\right ) \, _2F_1\left (1,p+1;p+2;\frac {b x^2}{a}+1\right )}{2 a^2 d^3 (p+1)}+\frac {e \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} F_1\left (-\frac {1}{2};-p,1;\frac {1}{2};-\frac {b x^2}{a},\frac {e^2 x^2}{d^2}\right )}{d^2 x}+\frac {e^4 \left (a+b x^2\right )^{p+1} \, _2F_1\left (1,p+1;p+2;\frac {e^2 \left (b x^2+a\right )}{b d^2+a e^2}\right )}{2 d^3 (p+1) \left (a e^2+b d^2\right )}-\frac {\left (a+b x^2\right )^{p+1}}{2 a d x^2} \]
Antiderivative was successfully verified.
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Rule 65
Rule 68
Rule 103
Rule 156
Rule 446
Rule 510
Rule 511
Rule 959
Rubi steps
\begin {align*} \int \frac {\left (a+b x^2\right )^p}{x^3 (d+e x)} \, dx &=d \int \frac {\left (a+b x^2\right )^p}{x^3 \left (d^2-e^2 x^2\right )} \, dx-e \int \frac {\left (a+b x^2\right )^p}{x^2 \left (d^2-e^2 x^2\right )} \, dx\\ &=\frac {1}{2} d \operatorname {Subst}\left (\int \frac {(a+b x)^p}{x^2 \left (d^2-e^2 x\right )} \, dx,x,x^2\right )-\left (e \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 \left (d^2-e^2 x^2\right )} \, dx\\ &=-\frac {\left (a+b x^2\right )^{1+p}}{2 a d x^2}+\frac {e \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 {1}{2};-\frac {b x^2}{a},\frac {e^2 x^2}{d^2}\right )}{d^2 x}-\frac {\operatorname {Subst}\left (\int \frac {(a+b x)^p \left (-a e^2-b d^2 p+b e^2 p x\right )}{x \left (d^2-e^2 x\right )} \, dx,x,x^2\right )}{2 a d}\\ &=-\frac {\left (a+b x^2\right )^{1+p}}{2 a d x^2}+\frac {e \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 {1}{2};-\frac {b x^2}{a},\frac {e^2 x^2}{d^2}\right )}{d^2 x}+\frac {e^4 \operatorname {Subst}\left (\int \frac {(a+b x)^p}{d^2-e^2 x} \, dx,x,x^2\right )}{2 d^3}+\frac {\left (a e^2+b d^2 p\right ) \operatorname {Subst}\left (\int \frac {(a+b x)^p}{x} \, dx,x,x^2\right )}{2 a d^3}\\ &=-\frac {\left (a+b x^2\right )^{1+p}}{2 a d x^2}+\frac {e \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 {1}{2};-\frac {b x^2}{a},\frac {e^2 x^2}{d^2}\right )}{d^2 x}+\frac {e^4 \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^3 \left (b d^2+a e^2\right ) (1+p)}-\frac {\left (a e^2+b d^2 p\right ) \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^2 d^3 (1+p)}\\ \end {align*}
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Mathematica [A] time = 0.28, size = 256, normalized size = 1.20 \[ \frac {\left (a+b x^2\right )^p \left (-\frac {e^2 \left (\frac {e \left (x-\sqrt {-\frac {a}{b}}\right )}{d+e x}\right )^{-p} \left (\frac {e \left (\sqrt {-\frac {a}{b}}+x\right )}{d+e x}\right )^{-p} F_1\left (-2 p;-p,-p;1-2 p;\frac {d-\sqrt {-\frac {a}{b}} e}{d+e x},\frac {d+\sqrt {-\frac {a}{b}} e}{d+e x}\right )}{p}+\left (\frac {a}{b x^2}+1\right )^{-p} \left (\frac {d^2 \, _2F_1\left (1-p,-p;2-p;-\frac {a}{b x^2}\right )}{(p-1) x^2}+\frac {e^2 \, _2F_1\left (-p,-p;1-p;-\frac {a}{b x^2}\right )}{p}\right )+\frac {2 d e \left (\frac {b x^2}{a}+1\right )^{-p} \, _2F_1\left (-\frac {1}{2},-p;\frac {1}{2};-\frac {b x^2}{a}\right )}{x}\right )}{2 d^3} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.85, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b x^{2} + a\right )}^{p}}{e x^{4} + d x^{3}}, x\right ) \]
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 \[ \int \frac {{\left (b x^{2} + a\right )}^{p}}{{\left (e x + d\right )} x^{3}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.07, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \,x^{2}+a \right )^{p}}{\left (e x +d \right ) x^{3}}\, dx \]
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 \[ \int \frac {{\left (b x^{2} + a\right )}^{p}}{{\left (e x + d\right )} x^{3}}\,{d x} \]
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 \[ \int \frac {{\left (b\,x^2+a\right )}^p}{x^3\,\left (d+e\,x\right )} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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