3.9.12 \(\int \frac {(a^2+2 a b x^2+b^2 x^4)^p}{(d x)^{3/2}} \, dx\) [812]

Optimal. Leaf size=65 \[ -\frac {2 \left (1+\frac {b x^2}{a}\right )^{-2 p} \left (a^2+2 a b x^2+b^2 x^4\right )^p \, _2F_1\left (-\frac {1}{4},-2 p;\frac {3}{4};-\frac {b x^2}{a}\right )}{d \sqrt {d x}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.02, antiderivative size = 65, 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 = {1127, 371} \begin {gather*} -\frac {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 {1}{4},-2 p;\frac {3}{4};-\frac {b x^2}{a}\right )}{d \sqrt {d x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 1127

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

________________________________________________________________________________________

Mathematica [A]
time = 0.13, size = 54, normalized size = 0.83 \begin {gather*} -\frac {2 x \left (\left (a+b x^2\right )^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-2 p} \, _2F_1\left (-\frac {1}{4},-2 p;\frac {3}{4};-\frac {b x^2}{a}\right )}{(d x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

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

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)^(3/2),x)

[Out]

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

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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)^(3/2),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)^(3/2),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

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

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)**(3/2),x)

[Out]

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

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)^(3/2),x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^p}{{\left (d\,x\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________