Optimal. Leaf size=159 \[ -\frac {d^3 \left (a+b x^2\right )^{p+1}}{a x}+\frac {d x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (3 a e^2+b d^2 (2 p+1)\right ) \, _2F_1\left (\frac {1}{2},-p;\frac {3}{2};-\frac {b x^2}{a}\right )}{a}-\frac {3 d^2 e \left (a+b x^2\right )^{p+1} \, _2F_1\left (1,p+1;p+2;\frac {b x^2}{a}+1\right )}{2 a (p+1)}+\frac {e^3 \left (a+b x^2\right )^{p+1}}{2 b (p+1)} \]
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Rubi [A] time = 0.19, antiderivative size = 159, 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 = {1807, 1652, 446, 80, 65, 12, 246, 245} \[ \frac {d x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (3 a e^2+b d^2 (2 p+1)\right ) \, _2F_1\left (\frac {1}{2},-p;\frac {3}{2};-\frac {b x^2}{a}\right )}{a}-\frac {3 d^2 e \left (a+b x^2\right )^{p+1} \, _2F_1\left (1,p+1;p+2;\frac {b x^2}{a}+1\right )}{2 a (p+1)}-\frac {d^3 \left (a+b x^2\right )^{p+1}}{a x}+\frac {e^3 \left (a+b x^2\right )^{p+1}}{2 b (p+1)} \]
Antiderivative was successfully verified.
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Rule 12
Rule 65
Rule 80
Rule 245
Rule 246
Rule 446
Rule 1652
Rule 1807
Rubi steps
\begin {align*} \int \frac {(d+e x)^3 \left (a+b x^2\right )^p}{x^2} \, dx &=-\frac {d^3 \left (a+b x^2\right )^{1+p}}{a x}-\frac {\int \frac {\left (a+b x^2\right )^p \left (-3 a d^2 e-d \left (3 a e^2+b d^2 (1+2 p)\right ) x-a e^3 x^2\right )}{x} \, dx}{a}\\ &=-\frac {d^3 \left (a+b x^2\right )^{1+p}}{a x}+\frac {\int d \left (3 a e^2+b d^2 (1+2 p)\right ) \left (a+b x^2\right )^p \, dx}{a}-\frac {\int \frac {\left (a+b x^2\right )^p \left (-3 a d^2 e-a e^3 x^2\right )}{x} \, dx}{a}\\ &=-\frac {d^3 \left (a+b x^2\right )^{1+p}}{a x}-\frac {\operatorname {Subst}\left (\int \frac {(a+b x)^p \left (-3 a d^2 e-a e^3 x\right )}{x} \, dx,x,x^2\right )}{2 a}+\frac {\left (d \left (3 a e^2+b d^2 (1+2 p)\right )\right ) \int \left (a+b x^2\right )^p \, dx}{a}\\ &=\frac {e^3 \left (a+b x^2\right )^{1+p}}{2 b (1+p)}-\frac {d^3 \left (a+b x^2\right )^{1+p}}{a x}+\frac {1}{2} \left (3 d^2 e\right ) \operatorname {Subst}\left (\int \frac {(a+b x)^p}{x} \, dx,x,x^2\right )+\frac {\left (d \left (3 a e^2+b d^2 (1+2 p)\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}{a}\\ &=\frac {e^3 \left (a+b x^2\right )^{1+p}}{2 b (1+p)}-\frac {d^3 \left (a+b x^2\right )^{1+p}}{a x}+\frac {d \left (3 a e^2+b d^2 (1+2 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 )}{a}-\frac {3 d^2 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 (1+p)}\\ \end {align*}
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Mathematica [A] time = 0.12, size = 154, normalized size = 0.97 \[ \frac {\left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (e x \left (\left (a+b x^2\right ) \left (\frac {b x^2}{a}+1\right )^p \left (a e^2-3 b d^2 \, _2F_1\left (1,p+1;p+2;\frac {b x^2}{a}+1\right )\right )+6 a b d e (p+1) x \, _2F_1\left (\frac {1}{2},-p;\frac {3}{2};-\frac {b x^2}{a}\right )\right )-2 a b d^3 (p+1) \, _2F_1\left (-\frac {1}{2},-p;\frac {1}{2};-\frac {b x^2}{a}\right )\right )}{2 a b (p+1) x} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.90, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (e^{3} x^{3} + 3 \, d e^{2} x^{2} + 3 \, d^{2} e x + d^{3}\right )} {\left (b x^{2} + a\right )}^{p}}{x^{2}}, x\right ) \]
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 \[ \int \frac {{\left (e x + d\right )}^{3} {\left (b x^{2} + a\right )}^{p}}{x^{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.04, size = 0, normalized size = 0.00 \[ \int \frac {\left (e x +d \right )^{3} \left (b \,x^{2}+a \right )^{p}}{x^{2}}\, dx \]
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 \[ \int \frac {{\left (e x + d\right )}^{3} {\left (b x^{2} + a\right )}^{p}}{x^{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (b\,x^2+a\right )}^p\,{\left (d+e\,x\right )}^3}{x^2} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 14.91, size = 143, normalized size = 0.90 \[ - \frac {a^{p} d^{3} {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, - p \\ \frac {1}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{x} + 3 a^{p} d e^{2} x {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, - p \\ \frac {3}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )} - \frac {3 b^{p} d^{2} e x^{2 p} \Gamma \left (- p\right ) {{}_{2}F_{1}\left (\begin {matrix} - p, - p \\ 1 - p \end {matrix}\middle | {\frac {a e^{i \pi }}{b x^{2}}} \right )}}{2 \Gamma \left (1 - p\right )} + e^{3} \left (\begin {cases} \frac {a^{p} x^{2}}{2} & \text {for}\: b = 0 \\\frac {\begin {cases} \frac {\left (a + b x^{2}\right )^{p + 1}}{p + 1} & \text {for}\: p \neq -1 \\\log {\left (a + b x^{2} \right )} & \text {otherwise} \end {cases}}{2 b} & \text {otherwise} \end {cases}\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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