3.809 \(\int (d x)^{3/2} (a^2+2 a b x^2+b^2 x^4)^p \, dx\)

Optimal. Leaf size=67 \[ \frac {2 (d x)^{5/2} \left (\frac {b x^2}{a}+1\right )^{-2 p} \left (a^2+2 a b x^2+b^2 x^4\right )^p \, _2F_1\left (\frac {5}{4},-2 p;\frac {9}{4};-\frac {b x^2}{a}\right )}{5 d} \]

[Out]

2/5*(d*x)^(5/2)*(b^2*x^4+2*a*b*x^2+a^2)^p*hypergeom([5/4, -2*p],[9/4],-b*x^2/a)/d/((1+b*x^2/a)^(2*p))

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Rubi [A]  time = 0.02, antiderivative size = 67, normalized size of antiderivative = 1.00, 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} \[ \frac {2 (d x)^{5/2} \left (\frac {b x^2}{a}+1\right )^{-2 p} \left (a^2+2 a b x^2+b^2 x^4\right )^p \, _2F_1\left (\frac {5}{4},-2 p;\frac {9}{4};-\frac {b x^2}{a}\right )}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(d*x)^(5/2)*(a^2 + 2*a*b*x^2 + b^2*x^4)^p*Hypergeometric2F1[5/4, -2*p, 9/4, -((b*x^2)/a)])/(5*d*(1 + (b*x^2
)/a)^(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])

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]

Rubi steps

\begin {align*} \int (d x)^{3/2} \left (a^2+2 a b x^2+b^2 x^4\right )^p \, 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 (d x)^{3/2} \left (1+\frac {b x^2}{a}\right )^{2 p} \, dx\\ &=\frac {2 (d x)^{5/2} \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 {5}{4},-2 p;\frac {9}{4};-\frac {b x^2}{a}\right )}{5 d}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 56, normalized size = 0.84 \[ \frac {2}{5} x (d x)^{3/2} \left (\left (a+b x^2\right )^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-2 p} \, _2F_1\left (\frac {5}{4},-2 p;\frac {9}{4};-\frac {b x^2}{a}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x*(d*x)^(3/2)*((a + b*x^2)^2)^p*Hypergeometric2F1[5/4, -2*p, 9/4, -((b*x^2)/a)])/(5*(1 + (b*x^2)/a)^(2*p))

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fricas [F]  time = 1.06, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {d x} {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} d x, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^(3/2)*(b^2*x^4+2*a*b*x^2+a^2)^p,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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (d x\right )^{\frac {3}{2}} {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.01, size = 0, normalized size = 0.00 \[ \int \left (d x \right )^{\frac {3}{2}} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{p}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (d x\right )^{\frac {3}{2}} {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (d\,x\right )}^{3/2}\,{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^p \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (d x\right )^{\frac {3}{2}} \left (\left (a + b x^{2}\right )^{2}\right )^{p}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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