Integrand size = 20, antiderivative size = 332 \[ \int \frac {x^4 (d+e x)^n}{\left (a+c x^2\right )^2} \, dx=\frac {(d+e x)^{1+n}}{c^2 e (1+n)}+\frac {a (a e+c d x) (d+e x)^{1+n}}{2 c^2 \left (c d^2+a e^2\right ) \left (a+c x^2\right )}+\frac {\left (3 \sqrt {-a} c d^2+a \sqrt {c} d e n+\sqrt {-a} a e^2 (3+n)\right ) (d+e x)^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt {c} (d+e x)}{\sqrt {c} d-\sqrt {-a} e}\right )}{4 c^2 \left (\sqrt {c} d-\sqrt {-a} e\right ) \left (c d^2+a e^2\right ) (1+n)}-\frac {\left (3 \sqrt {-a} c d^2-a \sqrt {c} d e n+\sqrt {-a} a e^2 (3+n)\right ) (d+e x)^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt {c} (d+e x)}{\sqrt {c} d+\sqrt {-a} e}\right )}{4 c^2 \left (\sqrt {c} d+\sqrt {-a} e\right ) \left (c d^2+a e^2\right ) (1+n)} \]
(e*x+d)^(1+n)/c^2/e/(1+n)+1/2*a*(c*d*x+a*e)*(e*x+d)^(1+n)/c^2/(a*e^2+c*d^2 )/(c*x^2+a)-1/4*(e*x+d)^(1+n)*hypergeom([1, 1+n],[2+n],(e*x+d)*c^(1/2)/(e* (-a)^(1/2)+d*c^(1/2)))*(3*c*d^2*(-a)^(1/2)+a*e^2*(3+n)*(-a)^(1/2)-a*d*e*n* c^(1/2))/c^2/(a*e^2+c*d^2)/(1+n)/(e*(-a)^(1/2)+d*c^(1/2))+1/4*(e*x+d)^(1+n )*hypergeom([1, 1+n],[2+n],(e*x+d)*c^(1/2)/(-e*(-a)^(1/2)+d*c^(1/2)))*(3*c *d^2*(-a)^(1/2)+a*e^2*(3+n)*(-a)^(1/2)+a*d*e*n*c^(1/2))/c^2/(a*e^2+c*d^2)/ (1+n)/(-e*(-a)^(1/2)+d*c^(1/2))
Time = 0.55 (sec) , antiderivative size = 413, normalized size of antiderivative = 1.24 \[ \int \frac {x^4 (d+e x)^n}{\left (a+c x^2\right )^2} \, dx=\frac {(d+e x)^{1+n} \left (\frac {4}{e+e n}+\frac {2 a (a e+c d x)}{\left (c d^2+a e^2\right ) \left (a+c x^2\right )}+\frac {4 \sqrt {-a} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt {c} (d+e x)}{\sqrt {c} d-\sqrt {-a} e}\right )}{\left (\sqrt {c} d-\sqrt {-a} e\right ) (1+n)}-\frac {4 \sqrt {-a} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt {c} (d+e x)}{\sqrt {c} d+\sqrt {-a} e}\right )}{\left (\sqrt {c} d+\sqrt {-a} e\right ) (1+n)}+\frac {a \left (\frac {\left (c d^2-a e^2 (-1+n)+\sqrt {-a} \sqrt {c} d e n\right ) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt {c} (d+e x)}{\sqrt {c} d-\sqrt {-a} e}\right )}{\sqrt {c} d-\sqrt {-a} e}-\frac {\left (c d^2-a e^2 (-1+n)-\sqrt {-a} \sqrt {c} d e n\right ) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt {c} (d+e x)}{\sqrt {c} d+\sqrt {-a} e}\right )}{\sqrt {c} d+\sqrt {-a} e}\right )}{\sqrt {-a} \left (c d^2+a e^2\right ) (1+n)}\right )}{4 c^2} \]
((d + e*x)^(1 + n)*(4/(e + e*n) + (2*a*(a*e + c*d*x))/((c*d^2 + a*e^2)*(a + c*x^2)) + (4*Sqrt[-a]*Hypergeometric2F1[1, 1 + n, 2 + n, (Sqrt[c]*(d + e *x))/(Sqrt[c]*d - Sqrt[-a]*e)])/((Sqrt[c]*d - Sqrt[-a]*e)*(1 + n)) - (4*Sq rt[-a]*Hypergeometric2F1[1, 1 + n, 2 + n, (Sqrt[c]*(d + e*x))/(Sqrt[c]*d + Sqrt[-a]*e)])/((Sqrt[c]*d + Sqrt[-a]*e)*(1 + n)) + (a*(((c*d^2 - a*e^2*(- 1 + n) + Sqrt[-a]*Sqrt[c]*d*e*n)*Hypergeometric2F1[1, 1 + n, 2 + n, (Sqrt[ c]*(d + e*x))/(Sqrt[c]*d - Sqrt[-a]*e)])/(Sqrt[c]*d - Sqrt[-a]*e) - ((c*d^ 2 - a*e^2*(-1 + n) - Sqrt[-a]*Sqrt[c]*d*e*n)*Hypergeometric2F1[1, 1 + n, 2 + n, (Sqrt[c]*(d + e*x))/(Sqrt[c]*d + Sqrt[-a]*e)])/(Sqrt[c]*d + Sqrt[-a] *e)))/(Sqrt[-a]*(c*d^2 + a*e^2)*(1 + n))))/(4*c^2)
Time = 0.66 (sec) , antiderivative size = 342, normalized size of antiderivative = 1.03, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {602, 2160, 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.
| \(\displaystyle \int \frac {x^4 (d+e x)^n}{\left (a+c x^2\right )^2} \, dx\) |
| \(\Big \downarrow \) 602 |
| \(\displaystyle \frac {a (d+e x)^{n+1} (a e+c d x)}{2 c^2 \left (a+c x^2\right ) \left (a e^2+c d^2\right )}-\frac {\int \frac {(d+e x)^n \left (\frac {\left (c d^2+a e^2 (n+1)\right ) a^2}{c^2}+\frac {d e n x a^2}{c}-2 \left (d^2+\frac {a e^2}{c}\right ) x^2 a\right )}{c x^2+a}dx}{2 a \left (a e^2+c d^2\right )}\) |
| \(\Big \downarrow \) 2160 |
| \(\displaystyle \frac {a (d+e x)^{n+1} (a e+c d x)}{2 c^2 \left (a+c x^2\right ) \left (a e^2+c d^2\right )}-\frac {\int \left (-\frac {2 a \left (c d^2+a e^2\right ) (d+e x)^n}{c^2}+\frac {\left (\sqrt {-a} \left (\frac {3 e^2 a^3}{c^2}+\frac {e^2 n a^3}{c^2}+\frac {3 d^2 a^2}{c}\right )-\frac {a^3 d e n}{c^{3/2}}\right ) (d+e x)^n}{2 a \left (\sqrt {-a}-\sqrt {c} x\right )}+\frac {\left (\frac {d e n a^3}{c^{3/2}}+\sqrt {-a} \left (\frac {3 e^2 a^3}{c^2}+\frac {e^2 n a^3}{c^2}+\frac {3 d^2 a^2}{c}\right )\right ) (d+e x)^n}{2 a \left (\sqrt {c} x+\sqrt {-a}\right )}\right )dx}{2 a \left (a e^2+c d^2\right )}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {a (d+e x)^{n+1} (a e+c d x)}{2 c^2 \left (a+c x^2\right ) \left (a e^2+c d^2\right )}-\frac {-\frac {a (d+e x)^{n+1} \left (3 \sqrt {-a} c d^2+a \sqrt {c} d e n+\sqrt {-a} a e^2 (n+3)\right ) \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {\sqrt {c} (d+e x)}{\sqrt {c} d-\sqrt {-a} e}\right )}{2 c^2 (n+1) \left (\sqrt {c} d-\sqrt {-a} e\right )}+\frac {a (d+e x)^{n+1} \left (3 \sqrt {-a} c d^2-a \sqrt {c} d e n+\sqrt {-a} a e^2 (n+3)\right ) \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {\sqrt {c} (d+e x)}{\sqrt {c} d+\sqrt {-a} e}\right )}{2 c^2 (n+1) \left (\sqrt {-a} e+\sqrt {c} d\right )}-\frac {2 a \left (a e^2+c d^2\right ) (d+e x)^{n+1}}{c^2 e (n+1)}}{2 a \left (a e^2+c d^2\right )}\) |
(a*(a*e + c*d*x)*(d + e*x)^(1 + n))/(2*c^2*(c*d^2 + a*e^2)*(a + c*x^2)) - ((-2*a*(c*d^2 + a*e^2)*(d + e*x)^(1 + n))/(c^2*e*(1 + n)) - (a*(3*Sqrt[-a] *c*d^2 + a*Sqrt[c]*d*e*n + Sqrt[-a]*a*e^2*(3 + n))*(d + e*x)^(1 + n)*Hyper geometric2F1[1, 1 + n, 2 + n, (Sqrt[c]*(d + e*x))/(Sqrt[c]*d - Sqrt[-a]*e) ])/(2*c^2*(Sqrt[c]*d - Sqrt[-a]*e)*(1 + n)) + (a*(3*Sqrt[-a]*c*d^2 - a*Sqr t[c]*d*e*n + Sqrt[-a]*a*e^2*(3 + n))*(d + e*x)^(1 + n)*Hypergeometric2F1[1 , 1 + n, 2 + n, (Sqrt[c]*(d + e*x))/(Sqrt[c]*d + Sqrt[-a]*e)])/(2*c^2*(Sqr t[c]*d + Sqrt[-a]*e)*(1 + n)))/(2*a*(c*d^2 + a*e^2))
3.4.70.3.1 Defintions of rubi rules used
Int[(x_)^(m_)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo l] :> With[{Qx = PolynomialQuotient[x^m, a + b*x^2, x], e = Coeff[Polynomia lRemainder[x^m, a + b*x^2, x], x, 0], f = Coeff[PolynomialRemainder[x^m, a + b*x^2, x], x, 1]}, Simp[(-(c + d*x)^(n + 1))*(a + b*x^2)^(p + 1)*((a*(d*e - c*f) + (b*c*e + a*d*f)*x)/(2*a*(p + 1)*(b*c^2 + a*d^2))), x] + Simp[1/(2 *a*(p + 1)*(b*c^2 + a*d^2)) Int[(c + d*x)^n*(a + b*x^2)^(p + 1)*ExpandToS um[2*a*(p + 1)*(b*c^2 + a*d^2)*Qx + e*(b*c^2*(2*p + 3) + a*d^2*(n + 2*p + 3 )) - a*c*d*f*n + d*(b*c*e + a*d*f)*(n + 2*p + 4)*x, x], x], x]] /; FreeQ[{a , b, c, d, n}, x] && IGtQ[m, 1] && LtQ[p, -1] && NeQ[b*c^2 + a*d^2, 0]
Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*Pq*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]
\[\int \frac {x^{4} \left (e x +d \right )^{n}}{\left (c \,x^{2}+a \right )^{2}}d x\]
\[ \int \frac {x^4 (d+e x)^n}{\left (a+c x^2\right )^2} \, dx=\int { \frac {{\left (e x + d\right )}^{n} x^{4}}{{\left (c x^{2} + a\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {x^4 (d+e x)^n}{\left (a+c x^2\right )^2} \, dx=\text {Timed out} \]
\[ \int \frac {x^4 (d+e x)^n}{\left (a+c x^2\right )^2} \, dx=\int { \frac {{\left (e x + d\right )}^{n} x^{4}}{{\left (c x^{2} + a\right )}^{2}} \,d x } \]
\[ \int \frac {x^4 (d+e x)^n}{\left (a+c x^2\right )^2} \, dx=\int { \frac {{\left (e x + d\right )}^{n} x^{4}}{{\left (c x^{2} + a\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {x^4 (d+e x)^n}{\left (a+c x^2\right )^2} \, dx=\int \frac {x^4\,{\left (d+e\,x\right )}^n}{{\left (c\,x^2+a\right )}^2} \,d x \]