3.803 \(\int \frac{(a^2+2 a b x^2+b^2 x^4)^p}{x^3} \, dx\)

Optimal. Leaf size=64 \[ \frac{b \left (a+b x^2\right ) \left (a^2+2 a b x^2+b^2 x^4\right )^p \, _2F_1\left (2,2 p+1;2 (p+1);\frac{b x^2}{a}+1\right )}{2 a^2 (2 p+1)} \]

[Out]

(b*(a + b*x^2)*(a^2 + 2*a*b*x^2 + b^2*x^4)^p*Hypergeometric2F1[2, 1 + 2*p, 2*(1 + p), 1 + (b*x^2)/a])/(2*a^2*(
1 + 2*p))

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Rubi [A]  time = 0.03992, antiderivative size = 64, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {1113, 266, 65} \[ \frac{b \left (a+b x^2\right ) \left (a^2+2 a b x^2+b^2 x^4\right )^p \, _2F_1\left (2,2 p+1;2 (p+1);\frac{b x^2}{a}+1\right )}{2 a^2 (2 p+1)} \]

Antiderivative was successfully verified.

[In]

Int[(a^2 + 2*a*b*x^2 + b^2*x^4)^p/x^3,x]

[Out]

(b*(a + b*x^2)*(a^2 + 2*a*b*x^2 + b^2*x^4)^p*Hypergeometric2F1[2, 1 + 2*p, 2*(1 + p), 1 + (b*x^2)/a])/(2*a^2*(
1 + 2*p))

Rule 1113

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^2 +
c*x^4)^FracPart[p])/(1 + (2*c*x^2)/b)^(2*FracPart[p]), Int[(d*x)^m*(1 + (2*c*x^2)/b)^(2*p), x], x] /; FreeQ[{
a, b, c, d, m, p}, x] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[2*p]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 65

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((c + d*x)^(n + 1)*Hypergeometric2F1[-m, n +
1, n + 2, 1 + (d*x)/c])/(d*(n + 1)*(-(d/(b*c)))^m), x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[n] && (Inte
gerQ[m] || GtQ[-(d/(b*c)), 0])

Rubi steps

\begin{align*} \int \frac{\left (a^2+2 a b x^2+b^2 x^4\right )^p}{x^3} \, dx &=\left (\left (1+\frac{b x^2}{a}\right )^{-2 p} \left (a^2+2 a b x^2+b^2 x^4\right )^p\right ) \int \frac{\left (1+\frac{b x^2}{a}\right )^{2 p}}{x^3} \, dx\\ &=\frac{1}{2} \left (\left (1+\frac{b x^2}{a}\right )^{-2 p} \left (a^2+2 a b x^2+b^2 x^4\right )^p\right ) \operatorname{Subst}\left (\int \frac{\left (1+\frac{b x}{a}\right )^{2 p}}{x^2} \, dx,x,x^2\right )\\ &=\frac{b \left (a+b x^2\right ) \left (a^2+2 a b x^2+b^2 x^4\right )^p \, _2F_1\left (2,1+2 p;2 (1+p);1+\frac{b x^2}{a}\right )}{2 a^2 (1+2 p)}\\ \end{align*}

Mathematica [A]  time = 0.0116051, size = 55, normalized size = 0.86 \[ \frac{b \left (a+b x^2\right ) \left (\left (a+b x^2\right )^2\right )^p \, _2F_1\left (2,2 p+1;2 p+2;\frac{b x^2}{a}+1\right )}{2 a^2 (2 p+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[(a^2 + 2*a*b*x^2 + b^2*x^4)^p/x^3,x]

[Out]

(b*(a + b*x^2)*((a + b*x^2)^2)^p*Hypergeometric2F1[2, 1 + 2*p, 2 + 2*p, 1 + (b*x^2)/a])/(2*a^2*(1 + 2*p))

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Maple [F]  time = 0.197, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ({b}^{2}{x}^{4}+2\,ab{x}^{2}+{a}^{2} \right ) ^{p}}{{x}^{3}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b^2*x^4+2*a*b*x^2+a^2)^p/x^3,x)

[Out]

int((b^2*x^4+2*a*b*x^2+a^2)^p/x^3,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p}}{x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^p/x^3,x, algorithm="maxima")

[Out]

integrate((b^2*x^4 + 2*a*b*x^2 + a^2)^p/x^3, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p}}{x^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^p/x^3,x, algorithm="fricas")

[Out]

integral((b^2*x^4 + 2*a*b*x^2 + a^2)^p/x^3, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (\left (a + b x^{2}\right )^{2}\right )^{p}}{x^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b**2*x**4+2*a*b*x**2+a**2)**p/x**3,x)

[Out]

Integral(((a + b*x**2)**2)**p/x**3, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p}}{x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^p/x^3,x, algorithm="giac")

[Out]

integrate((b^2*x^4 + 2*a*b*x^2 + a^2)^p/x^3, x)