Integrand size = 26, antiderivative size = 308 \[ \int \frac {\left (d+e x^n\right )^3}{a+b x^n+c x^{2 n}} \, dx=\frac {e^2 (3 c d-b e) x}{c^2}+\frac {e^3 x^{1+n}}{c (1+n)}+\frac {\left (3 c^2 d^2 e-3 b c d e^2+b^2 e^3-a c e^3+\frac {(2 c d-b e) \left (c^2 d^2+b^2 e^2-c e (b d+3 a e)\right )}{\sqrt {b^2-4 a c}}\right ) x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{c^2 \left (b-\sqrt {b^2-4 a c}\right )}+\frac {\left (3 c^2 d^2 e-3 b c d e^2+b^2 e^3-a c e^3-\frac {(2 c d-b e) \left (c^2 d^2+b^2 e^2-c e (b d+3 a e)\right )}{\sqrt {b^2-4 a c}}\right ) x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{c^2 \left (b+\sqrt {b^2-4 a c}\right )} \]
e^2*(-b*e+3*c*d)*x/c^2+e^3*x^(1+n)/c/(1+n)+x*hypergeom([1, 1/n],[1+1/n],-2 *c*x^n/(b-(-4*a*c+b^2)^(1/2)))*(3*c^2*d^2*e-3*b*c*d*e^2+b^2*e^3-a*c*e^3+(- b*e+2*c*d)*(c^2*d^2+b^2*e^2-c*e*(3*a*e+b*d))/(-4*a*c+b^2)^(1/2))/c^2/(b-(- 4*a*c+b^2)^(1/2))+x*hypergeom([1, 1/n],[1+1/n],-2*c*x^n/(b+(-4*a*c+b^2)^(1 /2)))*(3*c^2*d^2*e-3*b*c*d*e^2+b^2*e^3-a*c*e^3-(-b*e+2*c*d)*(c^2*d^2+b^2*e ^2-c*e*(3*a*e+b*d))/(-4*a*c+b^2)^(1/2))/c^2/(b+(-4*a*c+b^2)^(1/2))
Time = 1.88 (sec) , antiderivative size = 295, normalized size of antiderivative = 0.96 \[ \int \frac {\left (d+e x^n\right )^3}{a+b x^n+c x^{2 n}} \, dx=\frac {x \left (e^2 (3 c d-b e)+\frac {c e^3 x^n}{1+n}+\frac {\left (3 c^2 d^2 e-3 b c d e^2+b^2 e^3-a c e^3+\frac {(2 c d-b e) \left (c^2 d^2+b^2 e^2-c e (b d+3 a e)\right )}{\sqrt {b^2-4 a c}}\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},\frac {2 c x^n}{-b+\sqrt {b^2-4 a c}}\right )}{b-\sqrt {b^2-4 a c}}+\frac {\left (3 c^2 d^2 e-3 b c d e^2+b^2 e^3-a c e^3+\frac {(-2 c d+b e) \left (c^2 d^2+b^2 e^2-c e (b d+3 a e)\right )}{\sqrt {b^2-4 a c}}\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{b+\sqrt {b^2-4 a c}}\right )}{c^2} \]
(x*(e^2*(3*c*d - b*e) + (c*e^3*x^n)/(1 + n) + ((3*c^2*d^2*e - 3*b*c*d*e^2 + b^2*e^3 - a*c*e^3 + ((2*c*d - b*e)*(c^2*d^2 + b^2*e^2 - c*e*(b*d + 3*a*e )))/Sqrt[b^2 - 4*a*c])*Hypergeometric2F1[1, n^(-1), 1 + n^(-1), (2*c*x^n)/ (-b + Sqrt[b^2 - 4*a*c])])/(b - Sqrt[b^2 - 4*a*c]) + ((3*c^2*d^2*e - 3*b*c *d*e^2 + b^2*e^3 - a*c*e^3 + ((-2*c*d + b*e)*(c^2*d^2 + b^2*e^2 - c*e*(b*d + 3*a*e)))/Sqrt[b^2 - 4*a*c])*Hypergeometric2F1[1, n^(-1), 1 + n^(-1), (- 2*c*x^n)/(b + Sqrt[b^2 - 4*a*c])])/(b + Sqrt[b^2 - 4*a*c])))/c^2
Time = 0.78 (sec) , antiderivative size = 308, 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 = {1754, 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.
\(\displaystyle \int \frac {\left (d+e x^n\right )^3}{a+b x^n+c x^{2 n}} \, dx\) |
\(\Big \downarrow \) 1754 |
\(\displaystyle \int \left (\frac {x^n \left (-a c e^3+b^2 e^3-3 b c d e^2+3 c^2 d^2 e\right )+a b e^3-3 a c d e^2+c^2 d^3}{c^2 \left (a+b x^n+c x^{2 n}\right )}+\frac {e^2 (3 c d-b e)}{c^2}+\frac {e^3 x^n}{c}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {x \left (\frac {(2 c d-b e) \left (-c e (3 a e+b d)+b^2 e^2+c^2 d^2\right )}{\sqrt {b^2-4 a c}}-a c e^3+b^2 e^3-3 b c d e^2+3 c^2 d^2 e\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{c^2 \left (b-\sqrt {b^2-4 a c}\right )}+\frac {x \left (-\frac {(2 c d-b e) \left (-c e (3 a e+b d)+b^2 e^2+c^2 d^2\right )}{\sqrt {b^2-4 a c}}-a c e^3+b^2 e^3-3 b c d e^2+3 c^2 d^2 e\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{c^2 \left (\sqrt {b^2-4 a c}+b\right )}+\frac {e^2 x (3 c d-b e)}{c^2}+\frac {e^3 x^{n+1}}{c (n+1)}\) |
(e^2*(3*c*d - b*e)*x)/c^2 + (e^3*x^(1 + n))/(c*(1 + n)) + ((3*c^2*d^2*e - 3*b*c*d*e^2 + b^2*e^3 - a*c*e^3 + ((2*c*d - b*e)*(c^2*d^2 + b^2*e^2 - c*e* (b*d + 3*a*e)))/Sqrt[b^2 - 4*a*c])*x*Hypergeometric2F1[1, n^(-1), 1 + n^(- 1), (-2*c*x^n)/(b - Sqrt[b^2 - 4*a*c])])/(c^2*(b - Sqrt[b^2 - 4*a*c])) + ( (3*c^2*d^2*e - 3*b*c*d*e^2 + b^2*e^3 - a*c*e^3 - ((2*c*d - b*e)*(c^2*d^2 + b^2*e^2 - c*e*(b*d + 3*a*e)))/Sqrt[b^2 - 4*a*c])*x*Hypergeometric2F1[1, n ^(-1), 1 + n^(-1), (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c])])/(c^2*(b + Sqrt[b^2 - 4*a*c]))
3.1.69.3.1 Defintions of rubi rules used
Int[((d_) + (e_.)*(x_)^(n_))^(q_)/((a_) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_ )), x_Symbol] :> Int[ExpandIntegrand[(d + e*x^n)^q/(a + b*x^n + c*x^(2*n)), x], x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[q]
\[\int \frac {\left (d +e \,x^{n}\right )^{3}}{a +b \,x^{n}+c \,x^{2 n}}d x\]
\[ \int \frac {\left (d+e x^n\right )^3}{a+b x^n+c x^{2 n}} \, dx=\int { \frac {{\left (e x^{n} + d\right )}^{3}}{c x^{2 \, n} + b x^{n} + a} \,d x } \]
Timed out. \[ \int \frac {\left (d+e x^n\right )^3}{a+b x^n+c x^{2 n}} \, dx=\text {Timed out} \]
\[ \int \frac {\left (d+e x^n\right )^3}{a+b x^n+c x^{2 n}} \, dx=\int { \frac {{\left (e x^{n} + d\right )}^{3}}{c x^{2 \, n} + b x^{n} + a} \,d x } \]
(c*e^3*x*x^n + (3*c*d*e^2*(n + 1) - b*e^3*(n + 1))*x)/(c^2*(n + 1)) - inte grate(-(c^2*d^3 - (3*c*d*e^2 - b*e^3)*a + (3*c^2*d^2*e - 3*b*c*d*e^2 + b^2 *e^3 - a*c*e^3)*x^n)/(c^3*x^(2*n) + b*c^2*x^n + a*c^2), x)
\[ \int \frac {\left (d+e x^n\right )^3}{a+b x^n+c x^{2 n}} \, dx=\int { \frac {{\left (e x^{n} + d\right )}^{3}}{c x^{2 \, n} + b x^{n} + a} \,d x } \]
Timed out. \[ \int \frac {\left (d+e x^n\right )^3}{a+b x^n+c x^{2 n}} \, dx=\int \frac {{\left (d+e\,x^n\right )}^3}{a+b\,x^n+c\,x^{2\,n}} \,d x \]