\(\int \frac {(a^2+2 a b x^2+b^2 x^4)^p}{(d x)^{5/2}} \, dx\) [813]

Optimal result

Integrand size = 28, antiderivative size = 67 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^p}{(d x)^{5/2}} \, dx=-\frac {2 \left (1+\frac {b x^2}{a}\right )^{-2 p} \left (a^2+2 a b x^2+b^2 x^4\right )^p \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},-2 p,\frac {1}{4},-\frac {b x^2}{a}\right )}{3 d (d x)^{3/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {1127, 371} \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^p}{(d x)^{5/2}} \, dx=-\frac {2 \left (\frac {b x^2}{a}+1\right )^{-2 p} \left (a^2+2 a b x^2+b^2 x^4\right )^p \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},-2 p,\frac {1}{4},-\frac {b x^2}{a}\right )}{3 d (d x)^{3/2}} \]

[In]

Int[(a^2 + 2*a*b*x^2 + b^2*x^4)^p/(d*x)^(5/2),x]

[Out]

(-2*(a^2 + 2*a*b*x^2 + b^2*x^4)^p*Hypergeometric2F1[-3/4, -2*p, 1/4, -((b*x^2)/a)])/(3*d*(d*x)^(3/2)*(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*} \text {integral}& = \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)^{5/2}} \, 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 {3}{4},-2 p;\frac {1}{4};-\frac {b x^2}{a}\right )}{3 d (d x)^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.84 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^p}{(d x)^{5/2}} \, dx=-\frac {2 x \left (\left (a+b x^2\right )^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-2 p} \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},-2 p,\frac {1}{4},-\frac {b x^2}{a}\right )}{3 (d x)^{5/2}} \]

[In]

Integrate[(a^2 + 2*a*b*x^2 + b^2*x^4)^p/(d*x)^(5/2),x]

[Out]

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

Maple [F]

\[\int \frac {\left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{p}}{\left (d x \right )^{\frac {5}{2}}}d x\]

[In]

int((b^2*x^4+2*a*b*x^2+a^2)^p/(d*x)^(5/2),x)

[Out]

int((b^2*x^4+2*a*b*x^2+a^2)^p/(d*x)^(5/2),x)

Fricas [F]

\[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^p}{(d x)^{5/2}} \, dx=\int { \frac {{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p}}{\left (d x\right )^{\frac {5}{2}}} \,d x } \]

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^p/(d*x)^(5/2),x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^p}{(d x)^{5/2}} \, dx=\int \frac {\left (\left (a + b x^{2}\right )^{2}\right )^{p}}{\left (d x\right )^{\frac {5}{2}}}\, dx \]

[In]

integrate((b**2*x**4+2*a*b*x**2+a**2)**p/(d*x)**(5/2),x)

[Out]

Integral(((a + b*x**2)**2)**p/(d*x)**(5/2), x)

Maxima [F]

\[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^p}{(d x)^{5/2}} \, dx=\int { \frac {{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p}}{\left (d x\right )^{\frac {5}{2}}} \,d x } \]

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^p/(d*x)^(5/2),x, algorithm="maxima")

[Out]

integrate((b^2*x^4 + 2*a*b*x^2 + a^2)^p/(d*x)^(5/2), x)

Giac [F]

\[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^p}{(d x)^{5/2}} \, dx=\int { \frac {{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p}}{\left (d x\right )^{\frac {5}{2}}} \,d x } \]

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^p/(d*x)^(5/2),x, algorithm="giac")

[Out]

integrate((b^2*x^4 + 2*a*b*x^2 + a^2)^p/(d*x)^(5/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^p}{(d x)^{5/2}} \, dx=\int \frac {{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^p}{{\left (d\,x\right )}^{5/2}} \,d x \]

[In]

int((a^2 + b^2*x^4 + 2*a*b*x^2)^p/(d*x)^(5/2),x)

[Out]

int((a^2 + b^2*x^4 + 2*a*b*x^2)^p/(d*x)^(5/2), x)