Optimal. Leaf size=322 \[ \frac {e^2 x^3 \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} F_1\left (\frac {3}{2};-p,3;\frac {5}{2};-\frac {b x^2}{a},\frac {e^2 x^2}{d^2}\right )}{d^5}+\frac {x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} F_1\left (\frac {1}{2};-p,3;\frac {3}{2};-\frac {b x^2}{a},\frac {e^2 x^2}{d^2}\right )}{d^3}-\frac {3 b^2 d^2 e \left (a+b x^2\right )^{p+1} \, _2F_1\left (3,p+1;p+2;\frac {e^2 \left (b x^2+a\right )}{b d^2+a e^2}\right )}{2 (p+1) \left (a e^2+b d^2\right )^3}+\frac {b e \left (a+b x^2\right )^{p+1} \left (2 a e^2+b d^2 (p+1)\right ) \, _2F_1\left (2,p+1;p+2;\frac {e^2 \left (b x^2+a\right )}{b d^2+a e^2}\right )}{4 (p+1) \left (a e^2+b d^2\right )^3}-\frac {d^2 e \left (a+b x^2\right )^{p+1}}{4 \left (d^2-e^2 x^2\right )^2 \left (a e^2+b d^2\right )} \]
[Out]
________________________________________________________________________________________
Rubi [A] time = 0.33, antiderivative size = 322, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 9, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.529, Rules used = {757, 430, 429, 444, 68, 511, 510, 446, 78} \[ \frac {x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} F_1\left (\frac {1}{2};-p,3;\frac {3}{2};-\frac {b x^2}{a},\frac {e^2 x^2}{d^2}\right )}{d^3}+\frac {e^2 x^3 \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} F_1\left (\frac {3}{2};-p,3;\frac {5}{2};-\frac {b x^2}{a},\frac {e^2 x^2}{d^2}\right )}{d^5}-\frac {3 b^2 d^2 e \left (a+b x^2\right )^{p+1} \, _2F_1\left (3,p+1;p+2;\frac {e^2 \left (b x^2+a\right )}{b d^2+a e^2}\right )}{2 (p+1) \left (a e^2+b d^2\right )^3}+\frac {b e \left (a+b x^2\right )^{p+1} \left (2 a e^2+b d^2 (p+1)\right ) \, _2F_1\left (2,p+1;p+2;\frac {e^2 \left (b x^2+a\right )}{b d^2+a e^2}\right )}{4 (p+1) \left (a e^2+b d^2\right )^3}-\frac {d^2 e \left (a+b x^2\right )^{p+1}}{4 \left (d^2-e^2 x^2\right )^2 \left (a e^2+b d^2\right )} \]
Antiderivative was successfully verified.
[In]
[Out]
Rule 68
Rule 78
Rule 429
Rule 430
Rule 444
Rule 446
Rule 510
Rule 511
Rule 757
Rubi steps
\begin {align*} \int \frac {\left (a+b x^2\right )^p}{(d+e x)^3} \, dx &=\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^3 \int \frac {\left (a+b x^2\right )^p}{\left (d^2-e^2 x^2\right )^3} \, dx-\left (3 d^2 e\right ) \int \frac {x \left (a+b x^2\right )^p}{\left (d^2-e^2 x^2\right )^3} \, dx+\left (3 d e^2\right ) \int \frac {x^2 \left (a+b x^2\right )^p}{\left (d^2-e^2 x^2\right )^3} \, dx+e^3 \int \frac {x^3 \left (a+b x^2\right )^p}{\left (-d^2+e^2 x^2\right )^3} \, dx\\ &=-\left (\frac {1}{2} \left (3 d^2 e\right ) \operatorname {Subst}\left (\int \frac {(a+b x)^p}{\left (d^2-e^2 x\right )^3} \, dx,x,x^2\right )\right )+\frac {1}{2} e^3 \operatorname {Subst}\left (\int \frac {x (a+b x)^p}{\left (-d^2+e^2 x\right )^3} \, dx,x,x^2\right )+\left (d^3 \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+\left (3 d e^2 \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\\ &=-\frac {d^2 e \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 {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^3}+\frac {e^2 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^5}-\frac {3 b^2 d^2 e \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 \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 \left (b d^2+a e^2\right )}\\ &=-\frac {d^2 e \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 {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^3}+\frac {e^2 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^5}+\frac {b e \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 \left (b d^2+a e^2\right )^3 (1+p)}-\frac {3 b^2 d^2 e \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 [A] time = 0.11, size = 142, normalized size = 0.44 \[ \frac {\left (a+b x^2\right )^p \left (\frac {e \left (x-\sqrt {-\frac {a}{b}}\right )}{d+e x}\right )^{-p} \left (\frac {e \left (\sqrt {-\frac {a}{b}}+x\right )}{d+e x}\right )^{-p} F_1\left (2-2 p;-p,-p;3-2 p;\frac {d-\sqrt {-\frac {a}{b}} e}{d+e x},\frac {d+\sqrt {-\frac {a}{b}} e}{d+e x}\right )}{2 e (p-1) (d+e x)^2} \]
Warning: Unable to verify antiderivative.
[In]
[Out]
________________________________________________________________________________________
fricas [F] time = 0.82, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b x^{2} + a\right )}^{p}}{e^{3} x^{3} + 3 \, d e^{2} x^{2} + 3 \, d^{2} e x + d^{3}}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x^{2} + a\right )}^{p}}{{\left (e x + d\right )}^{3}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maple [F] time = 0.06, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \,x^{2}+a \right )^{p}}{\left (e x +d \right )^{3}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x^{2} + a\right )}^{p}}{{\left (e x + d\right )}^{3}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (b\,x^2+a\right )}^p}{{\left (d+e\,x\right )}^3} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
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.
[In]
[Out]
________________________________________________________________________________________