Integrand size = 26, antiderivative size = 294 \[ \int \left (d+e x^n\right ) \left (a+b x^n+c x^{2 n}\right )^{3/2} \, dx=\frac {a e x^{1+n} \sqrt {a+b x^n+c x^{2 n}} \operatorname {AppellF1}\left (1+\frac {1}{n},-\frac {3}{2},-\frac {3}{2},2+\frac {1}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{(1+n) \sqrt {1+\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}} \sqrt {1+\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}}}+\frac {a d x \sqrt {a+b x^n+c x^{2 n}} \operatorname {AppellF1}\left (\frac {1}{n},-\frac {3}{2},-\frac {3}{2},1+\frac {1}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{\sqrt {1+\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}} \sqrt {1+\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}}} \]
a*e*x^(1+n)*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+b*x^n+c*x^(2*n))^(1/2)/(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)+ a*d*x*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+b*x^n+c*x^(2*n))^(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)
Leaf count is larger than twice the leaf count of optimal. \(690\) vs. \(2(294)=588\).
Time = 3.26 (sec) , antiderivative size = 690, normalized size of antiderivative = 2.35 \[ \int \left (d+e x^n\right ) \left (a+b x^n+c x^{2 n}\right )^{3/2} \, dx=\frac {x \left (3 n^2 \left (16 a^2 c^2 e \left (1+4 n+3 n^2\right )+b^4 e \left (4+8 n+3 n^2\right )-2 b^3 c d \left (2+9 n+4 n^2\right )-4 a b^2 c e \left (5+14 n+6 n^2\right )+8 a b c^2 d \left (2+11 n+12 n^2\right )\right ) x^n \sqrt {\frac {b-\sqrt {b^2-4 a c}+2 c x^n}{b-\sqrt {b^2-4 a c}}} \sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x^n}{b+\sqrt {b^2-4 a c}}} \operatorname {AppellF1}\left (1+\frac {1}{n},\frac {1}{2},\frac {1}{2},2+\frac {1}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}},\frac {2 c x^n}{-b+\sqrt {b^2-4 a c}}\right )+2 (1+n) \left (\left (a+x^n \left (b+c x^n\right )\right ) \left (-3 b^3 e n^2 (2+3 n)+6 b^2 c n^2 \left (d+4 d n+e (1+n) x^n\right )+8 c^3 \left (1+3 n+2 n^2\right ) x^{2 n} \left (d+4 d n+e (1+3 n) x^n\right )+4 b c^2 (1+n) x^n \left (d \left (2+15 n+28 n^2\right )+e \left (2+13 n+18 n^2\right ) x^n\right )+4 a c \left (3 b e n^2 (2+5 n)+2 c \left (d (1+2 n) (1+4 n)^2+e \left (1+9 n+23 n^2+15 n^3\right ) x^n\right )\right )\right )+3 a n^2 \left (b^3 e (2+3 n)-2 b^2 c d (1+4 n)-4 a b c e (2+5 n)+8 a c^2 d \left (1+6 n+8 n^2\right )\right ) \sqrt {\frac {b-\sqrt {b^2-4 a c}+2 c x^n}{b-\sqrt {b^2-4 a c}}} \sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x^n}{b+\sqrt {b^2-4 a c}}} \operatorname {AppellF1}\left (\frac {1}{n},\frac {1}{2},\frac {1}{2},1+\frac {1}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}},\frac {2 c x^n}{-b+\sqrt {b^2-4 a c}}\right )\right )\right )}{16 c^2 (1+n)^2 (1+2 n) (1+3 n) (1+4 n) \sqrt {a+x^n \left (b+c x^n\right )}} \]
(x*(3*n^2*(16*a^2*c^2*e*(1 + 4*n + 3*n^2) + b^4*e*(4 + 8*n + 3*n^2) - 2*b^ 3*c*d*(2 + 9*n + 4*n^2) - 4*a*b^2*c*e*(5 + 14*n + 6*n^2) + 8*a*b*c^2*d*(2 + 11*n + 12*n^2))*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])] + 2*(1 + n)*((a + x^n*(b + c*x^ n))*(-3*b^3*e*n^2*(2 + 3*n) + 6*b^2*c*n^2*(d + 4*d*n + e*(1 + n)*x^n) + 8* c^3*(1 + 3*n + 2*n^2)*x^(2*n)*(d + 4*d*n + e*(1 + 3*n)*x^n) + 4*b*c^2*(1 + n)*x^n*(d*(2 + 15*n + 28*n^2) + e*(2 + 13*n + 18*n^2)*x^n) + 4*a*c*(3*b*e *n^2*(2 + 5*n) + 2*c*(d*(1 + 2*n)*(1 + 4*n)^2 + e*(1 + 9*n + 23*n^2 + 15*n ^3)*x^n))) + 3*a*n^2*(b^3*e*(2 + 3*n) - 2*b^2*c*d*(1 + 4*n) - 4*a*b*c*e*(2 + 5*n) + 8*a*c^2*d*(1 + 6*n + 8*n^2))*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 + S qrt[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])])))/(16*c^2*(1 + n) ^2*(1 + 2*n)*(1 + 3*n)*(1 + 4*n)*Sqrt[a + x^n*(b + c*x^n)])
Time = 0.47 (sec) , antiderivative size = 294, 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.
\(\displaystyle \int \left (d+e x^n\right ) \left (a+b x^n+c x^{2 n}\right )^{3/2} \, dx\) |
\(\Big \downarrow \) 1762 |
\(\displaystyle \int \left (d \left (a+b x^n+c x^{2 n}\right )^{3/2}+e x^n \left (a+b x^n+c x^{2 n}\right )^{3/2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {a d x \sqrt {a+b x^n+c x^{2 n}} \operatorname {AppellF1}\left (\frac {1}{n},-\frac {3}{2},-\frac {3}{2},1+\frac {1}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{\sqrt {\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}+1} \sqrt {\frac {2 c x^n}{\sqrt {b^2-4 a c}+b}+1}}+\frac {a e x^{n+1} \sqrt {a+b x^n+c x^{2 n}} \operatorname {AppellF1}\left (1+\frac {1}{n},-\frac {3}{2},-\frac {3}{2},2+\frac {1}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{(n+1) \sqrt {\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}+1} \sqrt {\frac {2 c x^n}{\sqrt {b^2-4 a c}+b}+1}}\) |
(a*e*x^(1 + n)*Sqrt[a + b*x^n + c*x^(2*n)]*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])])/((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])]) + (a*d*x*Sqrt[a + b*x^n + c*x^(2*n)]* 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 + Sqrt[b^2 - 4*a*c])])/(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])])
3.1.86.3.1 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]
\[\int \left (d +e \,x^{n}\right ) \left (a +b \,x^{n}+c \,x^{2 n}\right )^{\frac {3}{2}}d x\]
Exception generated. \[ \int \left (d+e x^n\right ) \left (a+b x^n+c x^{2 n}\right )^{3/2} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> Error detected within library code: inte grate: implementation incomplete (has polynomial part)
\[ \int \left (d+e x^n\right ) \left (a+b x^n+c x^{2 n}\right )^{3/2} \, dx=\int \left (d + e x^{n}\right ) \left (a + b x^{n} + c x^{2 n}\right )^{\frac {3}{2}}\, dx \]
\[ \int \left (d+e x^n\right ) \left (a+b x^n+c x^{2 n}\right )^{3/2} \, dx=\int { {\left (c x^{2 \, n} + b x^{n} + a\right )}^{\frac {3}{2}} {\left (e x^{n} + d\right )} \,d x } \]
\[ \int \left (d+e x^n\right ) \left (a+b x^n+c x^{2 n}\right )^{3/2} \, dx=\int { {\left (c x^{2 \, n} + b x^{n} + a\right )}^{\frac {3}{2}} {\left (e x^{n} + d\right )} \,d x } \]
Timed out. \[ \int \left (d+e x^n\right ) \left (a+b x^n+c x^{2 n}\right )^{3/2} \, dx=\int \left (d+e\,x^n\right )\,{\left (a+b\,x^n+c\,x^{2\,n}\right )}^{3/2} \,d x \]