3.811 \(\int \frac{(a^2+2 a b x^2+b^2 x^4)^p}{\sqrt{d x}} \, dx\)

Optimal. Leaf size=65 \[ \frac{2 \sqrt{d x} \left (\frac{b x^2}{a}+1\right )^{-2 p} \left (a^2+2 a b x^2+b^2 x^4\right )^p \, _2F_1\left (\frac{1}{4},-2 p;\frac{5}{4};-\frac{b x^2}{a}\right )}{d} \]

[Out]

(2*Sqrt[d*x]*(a^2 + 2*a*b*x^2 + b^2*x^4)^p*Hypergeometric2F1[1/4, -2*p, 5/4, -((b*x^2)/a)])/(d*(1 + (b*x^2)/a)
^(2*p))

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Rubi [A]  time = 0.0206416, antiderivative size = 65, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.071, Rules used = {1113, 364} \[ \frac{2 \sqrt{d x} \left (\frac{b x^2}{a}+1\right )^{-2 p} \left (a^2+2 a b x^2+b^2 x^4\right )^p \, _2F_1\left (\frac{1}{4},-2 p;\frac{5}{4};-\frac{b x^2}{a}\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[d*x]*(a^2 + 2*a*b*x^2 + b^2*x^4)^p*Hypergeometric2F1[1/4, -2*p, 5/4, -((b*x^2)/a)])/(d*(1 + (b*x^2)/a)
^(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 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rubi steps

\begin{align*} \int \frac{\left (a^2+2 a b x^2+b^2 x^4\right )^p}{\sqrt{d x}} \, 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}}{\sqrt{d x}} \, dx\\ &=\frac{2 \sqrt{d x} \left (1+\frac{b x^2}{a}\right )^{-2 p} \left (a^2+2 a b x^2+b^2 x^4\right )^p \, _2F_1\left (\frac{1}{4},-2 p;\frac{5}{4};-\frac{b x^2}{a}\right )}{d}\\ \end{align*}

Mathematica [A]  time = 0.0088893, size = 54, normalized size = 0.83 \[ \frac{2 x \left (\left (a+b x^2\right )^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-2 p} \, _2F_1\left (\frac{1}{4},-2 p;\frac{5}{4};-\frac{b x^2}{a}\right )}{\sqrt{d x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x*((a + b*x^2)^2)^p*Hypergeometric2F1[1/4, -2*p, 5/4, -((b*x^2)/a)])/(Sqrt[d*x]*(1 + (b*x^2)/a)^(2*p))

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Maple [F]  time = 0.176, size = 0, normalized size = 0. \begin{align*} \int{ \left ({b}^{2}{x}^{4}+2\,ab{x}^{2}+{a}^{2} \right ) ^{p}{\frac{1}{\sqrt{dx}}}}\, 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/(d*x)^(1/2),x)

[Out]

int((b^2*x^4+2*a*b*x^2+a^2)^p/(d*x)^(1/2),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}}{\sqrt{d x}}\,{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/(d*x)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{d x}{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p}}{d x}, 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/(d*x)^(1/2),x, algorithm="fricas")

[Out]

integral(sqrt(d*x)*(b^2*x^4 + 2*a*b*x^2 + a^2)^p/(d*x), 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}}{\sqrt{d x}}\, 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/(d*x)**(1/2),x)

[Out]

Integral(((a + b*x**2)**2)**p/sqrt(d*x), 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}}{\sqrt{d x}}\,{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/(d*x)^(1/2),x, algorithm="giac")

[Out]

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