\(\int \frac {(d x)^{3/2}}{(a+b x^n+c x^{2 n})^2} \, dx\) [207]

Optimal result

Integrand size = 24, antiderivative size = 330 \[ \int \frac {(d x)^{3/2}}{\left (a+b x^n+c x^{2 n}\right )^2} \, dx=\frac {(d x)^{5/2} \left (b^2-2 a c+b c x^n\right )}{a \left (b^2-4 a c\right ) d n \left (a+b x^n+c x^{2 n}\right )}-\frac {c \left (4 a c (5-4 n)-b^2 (5-2 n)-b \sqrt {b^2-4 a c} (5-2 n)\right ) (d x)^{5/2} \operatorname {Hypergeometric2F1}\left (1,\frac {5}{2 n},1+\frac {5}{2 n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{5 a \left (b^2-4 a c\right ) \left (b^2-4 a c-b \sqrt {b^2-4 a c}\right ) d n}-\frac {c \left (4 a c (5-4 n)-b^2 (5-2 n)+b \sqrt {b^2-4 a c} (5-2 n)\right ) (d x)^{5/2} \operatorname {Hypergeometric2F1}\left (1,\frac {5}{2 n},1+\frac {5}{2 n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{5 a \left (b^2-4 a c\right ) \left (b^2-4 a c+b \sqrt {b^2-4 a c}\right ) d n} \] Output:

(d*x)^(5/2)*(b^2-2*a*c+b*c*x^n)/a/(-4*a*c+b^2)/d/n/(a+b*x^n+c*x^(2*n))-1/5 
*c*(4*a*c*(5-4*n)-b^2*(5-2*n)-b*(-4*a*c+b^2)^(1/2)*(5-2*n))*(d*x)^(5/2)*hy 
pergeom([1, 5/2/n],[1+5/2/n],-2*c*x^n/(b-(-4*a*c+b^2)^(1/2)))/a/(-4*a*c+b^ 
2)/(b^2-4*a*c-b*(-4*a*c+b^2)^(1/2))/d/n-1/5*c*(4*a*c*(5-4*n)-b^2*(5-2*n)+b 
*(-4*a*c+b^2)^(1/2)*(5-2*n))*(d*x)^(5/2)*hypergeom([1, 5/2/n],[1+5/2/n],-2 
*c*x^n/(b+(-4*a*c+b^2)^(1/2)))/a/(-4*a*c+b^2)/(b*(-4*a*c+b^2)^(1/2)-4*a*c+ 
b^2)/d/n
 

Mathematica [A] (warning: unable to verify)

Time = 4.54 (sec) , antiderivative size = 507, normalized size of antiderivative = 1.54 \[ \int \frac {(d x)^{3/2}}{\left (a+b x^n+c x^{2 n}\right )^2} \, dx=-\frac {c x (d x)^{3/2} \left (\frac {2^{\left .-\frac {5}{2}\right /n} \left (4 a c \sqrt {b^2-4 a c} (5-4 n)+4 a b c (5-2 n)+b^3 (-5+2 n)+b^2 \sqrt {b^2-4 a c} (-5+2 n)\right ) \left (\frac {c x^n}{b-\sqrt {b^2-4 a c}+2 c x^n}\right )^{\left .-\frac {5}{2}\right /n} \operatorname {Hypergeometric2F1}\left (-\frac {5}{2 n},-\frac {5}{2 n},1-\frac {5}{2 n},\frac {b-\sqrt {b^2-4 a c}}{b-\sqrt {b^2-4 a c}+2 c x^n}\right )}{-b^2+4 a c+b \sqrt {b^2-4 a c}}+\frac {\frac {4 \left (b^2-4 a c\right ) \left (-8 a^2 c n+b^2 (-5+2 n) x^n \left (b+c x^n\right )+a \left (2 b^2 n-b c (-15+8 n) x^n-2 c^2 (-5+4 n) x^{2 n}\right )\right )}{\left (b^2-4 a c-b \sqrt {b^2-4 a c}\right ) \left (a+x^n \left (b+c x^n\right )\right )}+2^{\left .-\frac {5}{2}\right /n} \left (b^2 (5-2 n)+b \sqrt {b^2-4 a c} (-5+2 n)+4 a c (-5+4 n)\right ) \left (\frac {c x^n}{b+\sqrt {b^2-4 a c}+2 c x^n}\right )^{\left .-\frac {5}{2}\right /n} \operatorname {Hypergeometric2F1}\left (-\frac {5}{2 n},-\frac {5}{2 n},1-\frac {5}{2 n},\frac {b+\sqrt {b^2-4 a c}}{b+\sqrt {b^2-4 a c}+2 c x^n}\right )}{b+\sqrt {b^2-4 a c}}\right )}{5 a \left (b^2-4 a c\right )^{3/2} n} \] Input:

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

Output:

-1/5*(c*x*(d*x)^(3/2)*(((4*a*c*Sqrt[b^2 - 4*a*c]*(5 - 4*n) + 4*a*b*c*(5 - 
2*n) + b^3*(-5 + 2*n) + b^2*Sqrt[b^2 - 4*a*c]*(-5 + 2*n))*Hypergeometric2F 
1[-5/(2*n), -5/(2*n), 1 - 5/(2*n), (b - Sqrt[b^2 - 4*a*c])/(b - Sqrt[b^2 - 
 4*a*c] + 2*c*x^n)])/(2^(5/(2*n))*(-b^2 + 4*a*c + b*Sqrt[b^2 - 4*a*c])*((c 
*x^n)/(b - Sqrt[b^2 - 4*a*c] + 2*c*x^n))^(5/(2*n))) + ((4*(b^2 - 4*a*c)*(- 
8*a^2*c*n + b^2*(-5 + 2*n)*x^n*(b + c*x^n) + a*(2*b^2*n - b*c*(-15 + 8*n)* 
x^n - 2*c^2*(-5 + 4*n)*x^(2*n))))/((b^2 - 4*a*c - b*Sqrt[b^2 - 4*a*c])*(a 
+ x^n*(b + c*x^n))) + ((b^2*(5 - 2*n) + b*Sqrt[b^2 - 4*a*c]*(-5 + 2*n) + 4 
*a*c*(-5 + 4*n))*Hypergeometric2F1[-5/(2*n), -5/(2*n), 1 - 5/(2*n), (b + S 
qrt[b^2 - 4*a*c])/(b + Sqrt[b^2 - 4*a*c] + 2*c*x^n)])/(2^(5/(2*n))*((c*x^n 
)/(b + Sqrt[b^2 - 4*a*c] + 2*c*x^n))^(5/(2*n))))/(b + Sqrt[b^2 - 4*a*c]))) 
/(a*(b^2 - 4*a*c)^(3/2)*n)
 

Rubi [A] (verified)

Time = 0.65 (sec) , antiderivative size = 313, normalized size of antiderivative = 0.95, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1720, 27, 1884, 2009}

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.

\(\displaystyle \int \frac {(d x)^{3/2}}{\left (a+b x^n+c x^{2 n}\right )^2} \, dx\)

\(\Big \downarrow \) 1720

\(\displaystyle \frac {(d x)^{5/2} \left (-2 a c+b^2+b c x^n\right )}{a d n \left (b^2-4 a c\right ) \left (a+b x^n+c x^{2 n}\right )}-\frac {\int -\frac {(d x)^{3/2} \left (-b c (5-2 n) x^n+2 a c (5-4 n)-2 b^2 \left (\frac {5}{2}-n\right )\right )}{2 \left (b x^n+c x^{2 n}+a\right )}dx}{a n \left (b^2-4 a c\right )}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {(d x)^{3/2} \left (-b c (5-2 n) x^n+2 a c (5-4 n)-b^2 (5-2 n)\right )}{b x^n+c x^{2 n}+a}dx}{2 a n \left (b^2-4 a c\right )}+\frac {(d x)^{5/2} \left (-2 a c+b^2+b c x^n\right )}{a d n \left (b^2-4 a c\right ) \left (a+b x^n+c x^{2 n}\right )}\)

\(\Big \downarrow \) 1884

\(\displaystyle \frac {\int \left (\frac {\left (\frac {c \left (2 n b^2-5 b^2+20 a c-16 a c n\right )}{\sqrt {b^2-4 a c}}-b c (5-2 n)\right ) (d x)^{3/2}}{2 c x^n+b-\sqrt {b^2-4 a c}}+\frac {\left (-b c (5-2 n)-\frac {c \left (2 n b^2-5 b^2+20 a c-16 a c n\right )}{\sqrt {b^2-4 a c}}\right ) (d x)^{3/2}}{2 c x^n+b+\sqrt {b^2-4 a c}}\right )dx}{2 a n \left (b^2-4 a c\right )}+\frac {(d x)^{5/2} \left (-2 a c+b^2+b c x^n\right )}{a d n \left (b^2-4 a c\right ) \left (a+b x^n+c x^{2 n}\right )}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {-\frac {2 c (d x)^{5/2} \left (-b (5-2 n) \sqrt {b^2-4 a c}+4 a c (5-4 n)-\left (b^2 (5-2 n)\right )\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {5}{2 n},1+\frac {5}{2 n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{5 d \left (-b \sqrt {b^2-4 a c}-4 a c+b^2\right )}-\frac {2 c (d x)^{5/2} \left (\frac {4 a c (5-4 n)-b^2 (5-2 n)}{\sqrt {b^2-4 a c}}+b (5-2 n)\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {5}{2 n},1+\frac {5}{2 n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{5 d \left (\sqrt {b^2-4 a c}+b\right )}}{2 a n \left (b^2-4 a c\right )}+\frac {(d x)^{5/2} \left (-2 a c+b^2+b c x^n\right )}{a d n \left (b^2-4 a c\right ) \left (a+b x^n+c x^{2 n}\right )}\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 1720

Int[((d_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x 
_Symbol] :> Simp[(-(d*x)^(m + 1))*(b^2 - 2*a*c + b*c*x^n)*((a + b*x^n + c*x 
^(2*n))^(p + 1)/(a*d*n*(p + 1)*(b^2 - 4*a*c))), x] + Simp[1/(a*n*(p + 1)*(b 
^2 - 4*a*c))   Int[(d*x)^m*(a + b*x^n + c*x^(2*n))^(p + 1)*Simp[b^2*(n*(p + 
 1) + m + 1) - 2*a*c*(m + 2*n*(p + 1) + 1) + b*c*(2*n*p + 3*n + m + 1)*x^n, 
 x], x], x] /; FreeQ[{a, b, c, d, m, n}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4* 
a*c, 0] && ILtQ[p + 1, 0]
 

rule 1884

Int[((f_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.)*( 
(d_) + (e_.)*(x_)^(n_))^(q_.), x_Symbol] :> Int[ExpandIntegrand[(f*x)^m*(d 
+ e*x^n)^q*(a + b*x^n + c*x^(2*n))^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, 
 n, p, q}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] &&  !RationalQ[n] && ( 
IGtQ[p, 0] || IGtQ[q, 0])
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

Maple [F]

\[\int \frac {\left (d x \right )^{\frac {3}{2}}}{\left (a +b \,x^{n}+c \,x^{2 n}\right )^{2}}d x\]

Input:

int((d*x)^(3/2)/(a+b*x^n+c*x^(2*n))^2,x)
 

Output:

int((d*x)^(3/2)/(a+b*x^n+c*x^(2*n))^2,x)
 

Fricas [F]

\[ \int \frac {(d x)^{3/2}}{\left (a+b x^n+c x^{2 n}\right )^2} \, dx=\int { \frac {\left (d x\right )^{\frac {3}{2}}}{{\left (c x^{2 \, n} + b x^{n} + a\right )}^{2}} \,d x } \] Input:

integrate((d*x)^(3/2)/(a+b*x^n+c*x^(2*n))^2,x, algorithm="fricas")
 

Output:

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

Sympy [F(-1)]

Timed out. \[ \int \frac {(d x)^{3/2}}{\left (a+b x^n+c x^{2 n}\right )^2} \, dx=\text {Timed out} \] Input:

integrate((d*x)**(3/2)/(a+b*x**n+c*x**(2*n))**2,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {(d x)^{3/2}}{\left (a+b x^n+c x^{2 n}\right )^2} \, dx=\int { \frac {\left (d x\right )^{\frac {3}{2}}}{{\left (c x^{2 \, n} + b x^{n} + a\right )}^{2}} \,d x } \] Input:

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

Output:

-2*d^(3/2)*x^(5/2)/(c^2*(8*n - 5)*x^(4*n) + 2*b*c*(8*n - 5)*x^(3*n) + 2*a* 
b*(8*n - 5)*x^n + a^2*(8*n - 5) + (b^2*(8*n - 5) + 2*a*c*(8*n - 5))*x^(2*n 
)) + integrate(4*(b*d^(3/2)*n*x^n + 2*a*d^(3/2)*n)*x^(3/2)/(c^3*(8*n - 5)* 
x^(6*n) + 3*b*c^2*(8*n - 5)*x^(5*n) + 3*a^2*b*(8*n - 5)*x^n + a^3*(8*n - 5 
) + 3*(b^2*c*(8*n - 5) + a*c^2*(8*n - 5))*x^(4*n) + (b^3*(8*n - 5) + 6*a*b 
*c*(8*n - 5))*x^(3*n) + 3*(a*b^2*(8*n - 5) + a^2*c*(8*n - 5))*x^(2*n)), x)
 

Giac [F]

\[ \int \frac {(d x)^{3/2}}{\left (a+b x^n+c x^{2 n}\right )^2} \, dx=\int { \frac {\left (d x\right )^{\frac {3}{2}}}{{\left (c x^{2 \, n} + b x^{n} + a\right )}^{2}} \,d x } \] Input:

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

Output:

integrate((d*x)^(3/2)/(c*x^(2*n) + b*x^n + a)^2, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(d x)^{3/2}}{\left (a+b x^n+c x^{2 n}\right )^2} \, dx=\int \frac {{\left (d\,x\right )}^{3/2}}{{\left (a+b\,x^n+c\,x^{2\,n}\right )}^2} \,d x \] Input:

int((d*x)^(3/2)/(a + b*x^n + c*x^(2*n))^2,x)
 

Output:

int((d*x)^(3/2)/(a + b*x^n + c*x^(2*n))^2, x)
 

Reduce [F]

\[ \int \frac {(d x)^{3/2}}{\left (a+b x^n+c x^{2 n}\right )^2} \, dx=\sqrt {d}\, \left (\int \frac {\sqrt {x}\, x}{x^{4 n} c^{2}+2 x^{3 n} b c +2 x^{2 n} a c +x^{2 n} b^{2}+2 x^{n} a b +a^{2}}d x \right ) d \] Input:

int((d*x)^(3/2)/(a+b*x^n+c*x^(2*n))^2,x)
 

Output:

sqrt(d)*int((sqrt(x)*x)/(x**(4*n)*c**2 + 2*x**(3*n)*b*c + 2*x**(2*n)*a*c + 
 x**(2*n)*b**2 + 2*x**n*a*b + a**2),x)*d