\(\int \frac {(a^2+2 a b x^2+b^2 x^4)^p}{(d x)^{5/2}} \, dx\) [700]

Optimal result

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

-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))
 

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.84 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^p}{(d x)^{5/2}} \, dx=-\frac {2 x \left (\left (a+b x^2\right )^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-2 p} \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},-2 p,\frac {1}{4},-\frac {b x^2}{a}\right )}{3 (d x)^{5/2}} \] Input:

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

Output:

(-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))
 

Rubi [A] (verified)

Time = 0.31 (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 = {1385, 278}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 1385

\(\displaystyle \left (\frac {b x^2}{a}+1\right )^{-2 p} \left (a^2+2 a b x^2+b^2 x^4\right )^p \int \frac {\left (\frac {b x^2}{a}+1\right )^{2 p}}{(d x)^{5/2}}dx\)

\(\Big \downarrow \) 278

\(\displaystyle -\frac {2 \left (\frac {b x^2}{a}+1\right )^{-2 p} \left (a^2+2 a b x^2+b^2 x^4\right )^p \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},-2 p,\frac {1}{4},-\frac {b x^2}{a}\right )}{3 d (d x)^{3/2}}\)

Input:

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

Output:

(-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))
 

Defintions of rubi rules used

rule 278

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^p*(( 
c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/2, (m + 1)/2 + 1, ( 
-b)*(x^2/a)], x] /; FreeQ[{a, b, c, m, p}, x] &&  !IGtQ[p, 0] && (ILtQ[p, 0 
] || GtQ[a, 0])
 

rule 1385

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> S 
imp[a^IntPart[p]*((a + b*x^n + c*x^(2*n))^FracPart[p]/(1 + 2*c*(x^n/b))^(2* 
FracPart[p]))   Int[u*(1 + 2*c*(x^n/b))^(2*p), x], x] /; FreeQ[{a, b, c, n, 
 p}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[2*p] && NeQ[u, 
 x^(n - 1)] && NeQ[u, x^(2*n - 1)]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^p}{(d x)^{5/2}} \, dx=\frac {2 \sqrt {d}\, \left (\left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{p}+32 \sqrt {x}\, \left (\int \frac {\sqrt {x}\, \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{p}}{8 b p \,x^{5}-3 b \,x^{5}+8 a p \,x^{3}-3 a \,x^{3}}d x \right ) a \,p^{2} x -12 \sqrt {x}\, \left (\int \frac {\sqrt {x}\, \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{p}}{8 b p \,x^{5}-3 b \,x^{5}+8 a p \,x^{3}-3 a \,x^{3}}d x \right ) a p x \right )}{\sqrt {x}\, d^{3} x \left (8 p -3\right )} \] Input:

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

Output:

(2*sqrt(d)*((a**2 + 2*a*b*x**2 + b**2*x**4)**p + 32*sqrt(x)*int((sqrt(x)*( 
a**2 + 2*a*b*x**2 + b**2*x**4)**p)/(8*a*p*x**3 - 3*a*x**3 + 8*b*p*x**5 - 3 
*b*x**5),x)*a*p**2*x - 12*sqrt(x)*int((sqrt(x)*(a**2 + 2*a*b*x**2 + b**2*x 
**4)**p)/(8*a*p*x**3 - 3*a*x**3 + 8*b*p*x**5 - 3*b*x**5),x)*a*p*x))/(sqrt( 
x)*d**3*x*(8*p - 3))