\(\int \frac {\sqrt {d x}}{(a+b x^n+c x^{2 n})^2} \, dx\) [208]

Optimal result

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

(d*x)^(3/2)*(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))-1/3 
*c*(4*a*c*(3-4*n)-b^2*(3-2*n)-b*(-4*a*c+b^2)^(1/2)*(3-2*n))*(d*x)^(3/2)*hy 
pergeom([1, 3/2/n],[1+3/2/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-1/3*c*(4*a*c*(3-4*n)-b^2*(3-2*n)+b 
*(-4*a*c+b^2)^(1/2)*(3-2*n))*(d*x)^(3/2)*hypergeom([1, 3/2/n],[1+3/2/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 = 4.55 (sec) , antiderivative size = 507, normalized size of antiderivative = 1.54 \[ \int \frac {\sqrt {d x}}{\left (a+b x^n+c x^{2 n}\right )^2} \, dx=-\frac {c x \sqrt {d x} \left (\frac {2^{\left .-\frac {3}{2}\right /n} \left (4 a c \sqrt {b^2-4 a c} (3-4 n)+4 a b c (3-2 n)+b^3 (-3+2 n)+b^2 \sqrt {b^2-4 a c} (-3+2 n)\right ) \left (\frac {c x^n}{b-\sqrt {b^2-4 a c}+2 c x^n}\right )^{\left .-\frac {3}{2}\right /n} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2 n},-\frac {3}{2 n},1-\frac {3}{2 n},\frac {b-\sqrt {b^2-4 a c}}{b-\sqrt {b^2-4 a c}+2 c x^n}\right )}{-b^2+4 a c+b \sqrt {b^2-4 a c}}+\frac {\frac {4 \left (b^2-4 a c\right ) \left (-8 a^2 c n+b^2 (-3+2 n) x^n \left (b+c x^n\right )+a \left (2 b^2 n-b c (-9+8 n) x^n-2 c^2 (-3+4 n) x^{2 n}\right )\right )}{\left (b^2-4 a c-b \sqrt {b^2-4 a c}\right ) \left (a+x^n \left (b+c x^n\right )\right )}+2^{\left .-\frac {3}{2}\right /n} \left (b^2 (3-2 n)+b \sqrt {b^2-4 a c} (-3+2 n)+4 a c (-3+4 n)\right ) \left (\frac {c x^n}{b+\sqrt {b^2-4 a c}+2 c x^n}\right )^{\left .-\frac {3}{2}\right /n} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2 n},-\frac {3}{2 n},1-\frac {3}{2 n},\frac {b+\sqrt {b^2-4 a c}}{b+\sqrt {b^2-4 a c}+2 c x^n}\right )}{b+\sqrt {b^2-4 a c}}\right )}{3 a \left (b^2-4 a c\right )^{3/2} n} \] Input:

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

Output:

-1/3*(c*x*Sqrt[d*x]*(((4*a*c*Sqrt[b^2 - 4*a*c]*(3 - 4*n) + 4*a*b*c*(3 - 2* 
n) + b^3*(-3 + 2*n) + b^2*Sqrt[b^2 - 4*a*c]*(-3 + 2*n))*Hypergeometric2F1[ 
-3/(2*n), -3/(2*n), 1 - 3/(2*n), (b - Sqrt[b^2 - 4*a*c])/(b - Sqrt[b^2 - 4 
*a*c] + 2*c*x^n)])/(2^(3/(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))^(3/(2*n))) + ((4*(b^2 - 4*a*c)*(-8* 
a^2*c*n + b^2*(-3 + 2*n)*x^n*(b + c*x^n) + a*(2*b^2*n - b*c*(-9 + 8*n)*x^n 
 - 2*c^2*(-3 + 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*(3 - 2*n) + b*Sqrt[b^2 - 4*a*c]*(-3 + 2*n) + 4*a* 
c*(-3 + 4*n))*Hypergeometric2F1[-3/(2*n), -3/(2*n), 1 - 3/(2*n), (b + Sqrt 
[b^2 - 4*a*c])/(b + Sqrt[b^2 - 4*a*c] + 2*c*x^n)])/(2^(3/(2*n))*((c*x^n)/( 
b + Sqrt[b^2 - 4*a*c] + 2*c*x^n))^(3/(2*n))))/(b + Sqrt[b^2 - 4*a*c])))/(a 
*(b^2 - 4*a*c)^(3/2)*n)
 

Rubi [A] (verified)

Time = 0.63 (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.

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

\(\Big \downarrow \) 1720

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1884

\(\displaystyle \frac {\int \left (\frac {\sqrt {d x} \left (-b c (3-2 n)-\frac {c \left (2 n b^2-3 b^2+12 a c-16 a c n\right )}{\sqrt {b^2-4 a c}}\right )}{2 c x^n+b+\sqrt {b^2-4 a c}}+\frac {\left (\frac {c \left (2 n b^2-3 b^2+12 a c-16 a c n\right )}{\sqrt {b^2-4 a c}}-b c (3-2 n)\right ) \sqrt {d x}}{2 c x^n+b-\sqrt {b^2-4 a c}}\right )dx}{2 a n \left (b^2-4 a c\right )}+\frac {(d x)^{3/2} \left (-2 a c+b^2+b c x^n\right )}{a d n \left (b^2-4 a c\right ) \left (a+b x^n+c x^{2 n}\right )}\)

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

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

Input:

int((d*x)^(1/2)/(a+b*x^n+c*x^(2*n))^2,x)
 

Output:

int((d*x)^(1/2)/(a+b*x^n+c*x^(2*n))^2,x)
 

Fricas [F]

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

integrate((d*x)^(1/2)/(a+b*x^n+c*x^(2*n))^2,x, algorithm="fricas")
 

Output:

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

integrate((d*x)**(1/2)/(a+b*x**n+c*x**(2*n))**2,x)
 

Output:

Timed out
 

Maxima [F]

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

integrate((d*x)^(1/2)/(a+b*x^n+c*x^(2*n))^2,x, algorithm="maxima")
 

Output:

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

Giac [F]

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

integrate((d*x)^(1/2)/(a+b*x^n+c*x^(2*n))^2,x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

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

int((d*x)^(1/2)/(a + b*x^n + c*x^(2*n))^2,x)
 

Output:

int((d*x)^(1/2)/(a + b*x^n + c*x^(2*n))^2, x)
 

Reduce [F]

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

int((d*x)^(1/2)/(a+b*x^n+c*x^(2*n))^2,x)
 

Output:

sqrt(d)*int(sqrt(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)