\(\int \frac {d+e x^n}{(a+b x^n+c x^{2 n})^{3/2}} \, dx\) [86]

Optimal result

Integrand size = 26, antiderivative size = 298 \[ \int \frac {d+e x^n}{\left (a+b x^n+c x^{2 n}\right )^{3/2}} \, dx=\frac {e x^{1+n} \sqrt {1+\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}} \sqrt {1+\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}} \operatorname {AppellF1}\left (1+\frac {1}{n},\frac {3}{2},\frac {3}{2},2+\frac {1}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{a (1+n) \sqrt {a+b x^n+c x^{2 n}}}+\frac {d x \sqrt {1+\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}} \sqrt {1+\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}} \operatorname {AppellF1}\left (\frac {1}{n},\frac {3}{2},\frac {3}{2},1+\frac {1}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{a \sqrt {a+b x^n+c x^{2 n}}} \] Output:

e*x^(1+n)*(1+2*c*x^n/(b-(-4*a*c+b^2)^(1/2)))^(1/2)*(1+2*c*x^n/(b+(-4*a*c+b 
^2)^(1/2)))^(1/2)*AppellF1(1+1/n,3/2,3/2,2+1/n,-2*c*x^n/(b-(-4*a*c+b^2)^(1 
/2)),-2*c*x^n/(b+(-4*a*c+b^2)^(1/2)))/a/(1+n)/(a+b*x^n+c*x^(2*n))^(1/2)+d* 
x*(1+2*c*x^n/(b-(-4*a*c+b^2)^(1/2)))^(1/2)*(1+2*c*x^n/(b+(-4*a*c+b^2)^(1/2 
)))^(1/2)*AppellF1(1/n,3/2,3/2,1+1/n,-2*c*x^n/(b-(-4*a*c+b^2)^(1/2)),-2*c* 
x^n/(b+(-4*a*c+b^2)^(1/2)))/a/(a+b*x^n+c*x^(2*n))^(1/2)
 

Mathematica [A] (warning: unable to verify)

Time = 2.25 (sec) , antiderivative size = 414, normalized size of antiderivative = 1.39 \[ \int \frac {d+e x^n}{\left (a+b x^n+c x^{2 n}\right )^{3/2}} \, dx=\frac {x \left (2 c (b d-2 a e) x^n \sqrt {\frac {b-\sqrt {b^2-4 a c}+2 c x^n}{b-\sqrt {b^2-4 a c}}} \sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x^n}{b+\sqrt {b^2-4 a c}}} \operatorname {AppellF1}\left (1+\frac {1}{n},\frac {1}{2},\frac {1}{2},2+\frac {1}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}},\frac {2 c x^n}{-b+\sqrt {b^2-4 a c}}\right )-(1+n) \left (2 \left (b^2 d+b \left (-a e+c d x^n\right )-2 a c \left (d+e x^n\right )\right )+\left (2 a b e+b^2 d (-2+n)-4 a c d (-1+n)\right ) \sqrt {\frac {b-\sqrt {b^2-4 a c}+2 c x^n}{b-\sqrt {b^2-4 a c}}} \sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x^n}{b+\sqrt {b^2-4 a c}}} \operatorname {AppellF1}\left (\frac {1}{n},\frac {1}{2},\frac {1}{2},1+\frac {1}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}},\frac {2 c x^n}{-b+\sqrt {b^2-4 a c}}\right )\right )\right )}{a \left (-b^2+4 a c\right ) n (1+n) \sqrt {a+x^n \left (b+c x^n\right )}} \] Input:

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

Output:

(x*(2*c*(b*d - 2*a*e)*x^n*Sqrt[(b - Sqrt[b^2 - 4*a*c] + 2*c*x^n)/(b - Sqrt 
[b^2 - 4*a*c])]*Sqrt[(b + Sqrt[b^2 - 4*a*c] + 2*c*x^n)/(b + Sqrt[b^2 - 4*a 
*c])]*AppellF1[1 + n^(-1), 1/2, 1/2, 2 + n^(-1), (-2*c*x^n)/(b + Sqrt[b^2 
- 4*a*c]), (2*c*x^n)/(-b + Sqrt[b^2 - 4*a*c])] - (1 + n)*(2*(b^2*d + b*(-( 
a*e) + c*d*x^n) - 2*a*c*(d + e*x^n)) + (2*a*b*e + b^2*d*(-2 + n) - 4*a*c*d 
*(-1 + n))*Sqrt[(b - Sqrt[b^2 - 4*a*c] + 2*c*x^n)/(b - Sqrt[b^2 - 4*a*c])] 
*Sqrt[(b + Sqrt[b^2 - 4*a*c] + 2*c*x^n)/(b + Sqrt[b^2 - 4*a*c])]*AppellF1[ 
n^(-1), 1/2, 1/2, 1 + n^(-1), (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c]), (2*c*x^n 
)/(-b + Sqrt[b^2 - 4*a*c])])))/(a*(-b^2 + 4*a*c)*n*(1 + n)*Sqrt[a + x^n*(b 
 + c*x^n)])
 

Rubi [A] (verified)

Time = 0.51 (sec) , antiderivative size = 298, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {1762, 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.

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

\(\Big \downarrow \) 1762

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {d x \sqrt {\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}+1} \sqrt {\frac {2 c x^n}{\sqrt {b^2-4 a c}+b}+1} \operatorname {AppellF1}\left (\frac {1}{n},\frac {3}{2},\frac {3}{2},1+\frac {1}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{a \sqrt {a+b x^n+c x^{2 n}}}+\frac {e x^{n+1} \sqrt {\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}+1} \sqrt {\frac {2 c x^n}{\sqrt {b^2-4 a c}+b}+1} \operatorname {AppellF1}\left (1+\frac {1}{n},\frac {3}{2},\frac {3}{2},2+\frac {1}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{a (n+1) \sqrt {a+b x^n+c x^{2 n}}}\)

Input:

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

Output:

(e*x^(1 + n)*Sqrt[1 + (2*c*x^n)/(b - Sqrt[b^2 - 4*a*c])]*Sqrt[1 + (2*c*x^n 
)/(b + Sqrt[b^2 - 4*a*c])]*AppellF1[1 + n^(-1), 3/2, 3/2, 2 + n^(-1), (-2* 
c*x^n)/(b - Sqrt[b^2 - 4*a*c]), (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c])])/(a*(1 
 + n)*Sqrt[a + b*x^n + c*x^(2*n)]) + (d*x*Sqrt[1 + (2*c*x^n)/(b - Sqrt[b^2 
 - 4*a*c])]*Sqrt[1 + (2*c*x^n)/(b + Sqrt[b^2 - 4*a*c])]*AppellF1[n^(-1), 3 
/2, 3/2, 1 + n^(-1), (-2*c*x^n)/(b - Sqrt[b^2 - 4*a*c]), (-2*c*x^n)/(b + S 
qrt[b^2 - 4*a*c])])/(a*Sqrt[a + b*x^n + c*x^(2*n)])
 

Defintions of rubi rules used

rule 1762

Int[((d_) + (e_.)*(x_)^(n_))*((a_) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_))^(p 
_), x_Symbol] :> Int[ExpandIntegrand[(d + e*x^n)*(a + b*x^n + c*x^(2*n))^p, 
 x], x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 
 0]
 

rule 2009

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

Maple [F]

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

Input:

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

Output:

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

Fricas [F(-2)]

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

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

Output:

Exception raised: TypeError >>  Error detected within library code:   inte 
grate: implementation incomplete (constant residues)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

int(sqrt(x**(2*n)*c + x**n*b + a)/(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 + int((x**n*sqrt(x**(2* 
n)*c + x**n*b + a))/(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)*e