Integrand size = 26, antiderivative size = 298 \[ \int \frac {d+e x^n}{\left (a+b x^n+c x^{2 n}\right )^{5/2}} \, dx=\frac {e x^{1+n} \sqrt {1+\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}} \sqrt {1+\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}} \operatorname {AppellF1}\left (1+\frac {1}{n},\frac {5}{2},\frac {5}{2},2+\frac {1}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{a^2 (1+n) \sqrt {a+b x^n+c x^{2 n}}}+\frac {d x \sqrt {1+\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}} \sqrt {1+\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}} \operatorname {AppellF1}\left (\frac {1}{n},\frac {5}{2},\frac {5}{2},1+\frac {1}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{a^2 \sqrt {a+b x^n+c x^{2 n}}} \]
e*x^(1+n)*AppellF1(1+1/n,5/2,5/2,2+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)))*(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)/a^2/(1+n)/(a+b*x^n+c*x^(2*n))^(1/2)+ d*x*AppellF1(1/n,5/2,5/2,1+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)))*(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)/a^2/(a+b*x^n+c*x^(2*n))^(1/2)
Leaf count is larger than twice the leaf count of optimal. \(701\) vs. \(2(298)=596\).
Time = 5.36 (sec) , antiderivative size = 701, normalized size of antiderivative = 2.35 \[ \int \frac {d+e x^n}{\left (a+b x^n+c x^{2 n}\right )^{5/2}} \, dx=\frac {x \left (-2 c \left (2 a b^2 e+4 a b c d (2-5 n)+8 a^2 c e (-1+2 n)+b^3 d (-2+3 n)\right ) x^n \sqrt {\frac {b-\sqrt {b^2-4 a c}+2 c x^n}{b-\sqrt {b^2-4 a c}}} \sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x^n}{b+\sqrt {b^2-4 a c}}} \left (a+x^n \left (b+c x^n\right )\right ) \operatorname {AppellF1}\left (1+\frac {1}{n},\frac {1}{2},\frac {1}{2},2+\frac {1}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}},\frac {2 c x^n}{-b+\sqrt {b^2-4 a c}}\right )+(1+n) \left (2 \left (b^3 d (-2+3 n) x^n \left (b+c x^n\right )^2+4 a^3 c \left (b e (-2+3 n)+c d (-2+8 n)+2 c e (-1+3 n) x^n\right )+2 a b \left (b+c x^n\right ) \left (-2 c^2 d (-2+5 n) x^{2 n}+b c x^n \left (d (5-11 n)+e x^n\right )+b^2 \left (d (-1+2 n)+e x^n\right )\right )+a^2 \left (-b^3 e (-2+n)+8 b c^2 e (-2+3 n) x^{2 n}-2 b^2 c \left (d (-5+14 n)-3 e (-1+n) x^n\right )+8 c^3 x^{2 n} \left (d (-1+3 n)+e (-1+2 n) x^n\right )\right )\right )+\left (2 a b^3 e (-2+n)-8 a^2 b c e (-2+3 n)+b^4 d \left (4-8 n+3 n^2\right )+16 a^2 c^2 d \left (1-4 n+3 n^2\right )-4 a b^2 c d \left (5-14 n+6 n^2\right )\right ) \sqrt {\frac {b-\sqrt {b^2-4 a c}+2 c x^n}{b-\sqrt {b^2-4 a c}}} \sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x^n}{b+\sqrt {b^2-4 a c}}} \left (a+x^n \left (b+c x^n\right )\right ) \operatorname {AppellF1}\left (\frac {1}{n},\frac {1}{2},\frac {1}{2},1+\frac {1}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}},\frac {2 c x^n}{-b+\sqrt {b^2-4 a c}}\right )\right )\right )}{3 a^2 \left (b^2-4 a c\right )^2 n^2 (1+n) \left (a+x^n \left (b+c x^n\right )\right )^{3/2}} \]
(x*(-2*c*(2*a*b^2*e + 4*a*b*c*d*(2 - 5*n) + 8*a^2*c*e*(-1 + 2*n) + b^3*d*( -2 + 3*n))*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])]*(a + x^n*(b + c*x^n))*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^ 3*d*(-2 + 3*n)*x^n*(b + c*x^n)^2 + 4*a^3*c*(b*e*(-2 + 3*n) + c*d*(-2 + 8*n ) + 2*c*e*(-1 + 3*n)*x^n) + 2*a*b*(b + c*x^n)*(-2*c^2*d*(-2 + 5*n)*x^(2*n) + b*c*x^n*(d*(5 - 11*n) + e*x^n) + b^2*(d*(-1 + 2*n) + e*x^n)) + a^2*(-(b ^3*e*(-2 + n)) + 8*b*c^2*e*(-2 + 3*n)*x^(2*n) - 2*b^2*c*(d*(-5 + 14*n) - 3 *e*(-1 + n)*x^n) + 8*c^3*x^(2*n)*(d*(-1 + 3*n) + e*(-1 + 2*n)*x^n))) + (2* a*b^3*e*(-2 + n) - 8*a^2*b*c*e*(-2 + 3*n) + b^4*d*(4 - 8*n + 3*n^2) + 16*a ^2*c^2*d*(1 - 4*n + 3*n^2) - 4*a*b^2*c*d*(5 - 14*n + 6*n^2))*Sqrt[(b - Sqr t[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])]*(a + x^n*(b + c*x^n))*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])])))/(3*a^2*(b^2 - 4*a*c)^2*n^2*(1 + n)*(a + x^n*( b + c*x^n))^(3/2))
Time = 0.47 (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.
\(\displaystyle \int \frac {d+e x^n}{\left (a+b x^n+c x^{2 n}\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 1762 |
\(\displaystyle \int \left (\frac {d}{\left (a+b x^n+c x^{2 n}\right )^{5/2}}+\frac {e x^n}{\left (a+b x^n+c x^{2 n}\right )^{5/2}}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {d x \sqrt {\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}+1} \sqrt {\frac {2 c x^n}{\sqrt {b^2-4 a c}+b}+1} \operatorname {AppellF1}\left (\frac {1}{n},\frac {5}{2},\frac {5}{2},1+\frac {1}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{a^2 \sqrt {a+b x^n+c x^{2 n}}}+\frac {e x^{n+1} \sqrt {\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}+1} \sqrt {\frac {2 c x^n}{\sqrt {b^2-4 a c}+b}+1} \operatorname {AppellF1}\left (1+\frac {1}{n},\frac {5}{2},\frac {5}{2},2+\frac {1}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{a^2 (n+1) \sqrt {a+b x^n+c x^{2 n}}}\) |
(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), 5/2, 5/2, 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^2* (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), 5/2, 5/2, 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^2*Sqrt[a + b*x^n + c*x^(2*n)])
3.1.89.3.1 Defintions of rubi rules used
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]
\[\int \frac {d +e \,x^{n}}{\left (a +b \,x^{n}+c \,x^{2 n}\right )^{\frac {5}{2}}}d x\]
Exception generated. \[ \int \frac {d+e x^n}{\left (a+b x^n+c x^{2 n}\right )^{5/2}} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> Error detected within library code: inte grate: implementation incomplete (constant residues)
\[ \int \frac {d+e x^n}{\left (a+b x^n+c x^{2 n}\right )^{5/2}} \, dx=\int \frac {d + e x^{n}}{\left (a + b x^{n} + c x^{2 n}\right )^{\frac {5}{2}}}\, dx \]
\[ \int \frac {d+e x^n}{\left (a+b x^n+c x^{2 n}\right )^{5/2}} \, dx=\int { \frac {e x^{n} + d}{{\left (c x^{2 \, n} + b x^{n} + a\right )}^{\frac {5}{2}}} \,d x } \]
\[ \int \frac {d+e x^n}{\left (a+b x^n+c x^{2 n}\right )^{5/2}} \, dx=\int { \frac {e x^{n} + d}{{\left (c x^{2 \, n} + b x^{n} + a\right )}^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {d+e x^n}{\left (a+b x^n+c x^{2 n}\right )^{5/2}} \, dx=\int \frac {d+e\,x^n}{{\left (a+b\,x^n+c\,x^{2\,n}\right )}^{5/2}} \,d x \]