Integrand size = 20, antiderivative size = 194 \[ \int \frac {x^2 (d+e x)^n}{a+c x^2} \, dx=\frac {(d+e x)^{1+n}}{c e (1+n)}+\frac {\sqrt {-a} (d+e x)^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt {c} (d+e x)}{\sqrt {c} d-\sqrt {-a} e}\right )}{2 c \left (\sqrt {c} d-\sqrt {-a} e\right ) (1+n)}-\frac {\sqrt {-a} (d+e x)^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt {c} (d+e x)}{\sqrt {c} d+\sqrt {-a} e}\right )}{2 c \left (\sqrt {c} d+\sqrt {-a} e\right ) (1+n)} \]
(e*x+d)^(1+n)/c/e/(1+n)+1/2*(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)))*(-a)^(1/2)/c/(1+n)/(-e*(-a)^(1/2)+d*c^ (1/2))-1/2*(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)))*(-a)^(1/2)/c/(1+n)/(e*(-a)^(1/2)+d*c^(1/2))
Time = 0.15 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.88 \[ \int \frac {x^2 (d+e x)^n}{a+c x^2} \, dx=\frac {(d+e x)^{1+n} \left (2 \left (c d^2+a e^2\right )+e \left (\sqrt {-a} \sqrt {c} d-a e\right ) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt {c} (d+e x)}{\sqrt {c} d-\sqrt {-a} e}\right )-e \left (\sqrt {-a} \sqrt {c} d+a e\right ) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt {c} (d+e x)}{\sqrt {c} d+\sqrt {-a} e}\right )\right )}{2 c e \left (c d^2+a e^2\right ) (1+n)} \]
((d + e*x)^(1 + n)*(2*(c*d^2 + a*e^2) + e*(Sqrt[-a]*Sqrt[c]*d - a*e)*Hyper geometric2F1[1, 1 + n, 2 + n, (Sqrt[c]*(d + e*x))/(Sqrt[c]*d - Sqrt[-a]*e) ] - e*(Sqrt[-a]*Sqrt[c]*d + a*e)*Hypergeometric2F1[1, 1 + n, 2 + n, (Sqrt[ c]*(d + e*x))/(Sqrt[c]*d + Sqrt[-a]*e)]))/(2*c*e*(c*d^2 + a*e^2)*(1 + n))
Time = 0.34 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.01, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {604, 25, 27, 485, 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.
\(\displaystyle \int \frac {x^2 (d+e x)^n}{a+c x^2} \, dx\) |
\(\Big \downarrow \) 604 |
\(\displaystyle \frac {\int -\frac {a e^2 (n+1) (d+e x)^n}{c x^2+a}dx}{c e^2 (n+1)}+\frac {(d+e x)^{n+1}}{c e (n+1)}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {(d+e x)^{n+1}}{c e (n+1)}-\frac {\int \frac {a e^2 (n+1) (d+e x)^n}{c x^2+a}dx}{c e^2 (n+1)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {(d+e x)^{n+1}}{c e (n+1)}-\frac {a \int \frac {(d+e x)^n}{c x^2+a}dx}{c}\) |
\(\Big \downarrow \) 485 |
\(\displaystyle \frac {(d+e x)^{n+1}}{c e (n+1)}-\frac {a \int \left (\frac {\sqrt {-a} (d+e x)^n}{2 a \left (\sqrt {-a}-\sqrt {c} x\right )}+\frac {\sqrt {-a} (d+e x)^n}{2 a \left (\sqrt {c} x+\sqrt {-a}\right )}\right )dx}{c}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {(d+e x)^{n+1}}{c e (n+1)}-\frac {a \left (\frac {(d+e x)^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {\sqrt {c} (d+e x)}{\sqrt {c} d-\sqrt {-a} e}\right )}{2 \sqrt {-a} (n+1) \left (\sqrt {c} d-\sqrt {-a} e\right )}-\frac {(d+e x)^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {\sqrt {c} (d+e x)}{\sqrt {c} d+\sqrt {-a} e}\right )}{2 \sqrt {-a} (n+1) \left (\sqrt {-a} e+\sqrt {c} d\right )}\right )}{c}\) |
(d + e*x)^(1 + n)/(c*e*(1 + n)) - (a*(((d + e*x)^(1 + n)*Hypergeometric2F1 [1, 1 + n, 2 + n, (Sqrt[c]*(d + e*x))/(Sqrt[c]*d - Sqrt[-a]*e)])/(2*Sqrt[- a]*(Sqrt[c]*d - Sqrt[-a]*e)*(1 + n)) - ((d + e*x)^(1 + n)*Hypergeometric2F 1[1, 1 + n, 2 + n, (Sqrt[c]*(d + e*x))/(Sqrt[c]*d + Sqrt[-a]*e)])/(2*Sqrt[ -a]*(Sqrt[c]*d + Sqrt[-a]*e)*(1 + n))))/c
3.4.65.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((c_) + (d_.)*(x_))^(n_)/((a_) + (b_.)*(x_)^2), x_Symbol] :> Int[Expand Integrand[(c + d*x)^n, 1/(a + b*x^2), x], x] /; FreeQ[{a, b, c, d, n}, x] & & !IntegerQ[2*n]
Int[(x_)^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol ] :> Simp[(c + d*x)^(m + n - 1)*((a + b*x^2)^(p + 1)/(b*d^(m - 1)*(m + n + 2*p + 1))), x] + Simp[1/(b*d^m*(m + n + 2*p + 1)) Int[(c + d*x)^n*(a + b* x^2)^p*ExpandToSum[b*d^m*(m + n + 2*p + 1)*x^m - b*(m + n + 2*p + 1)*(c + d *x)^m - (c + d*x)^(m - 2)*(a*d^2*(m + n - 1) - b*c^2*(m + n + 2*p + 1) - 2* b*c*d*(m + n + p)*x), x], x], x] /; FreeQ[{a, b, c, d, n, p}, x] && IGtQ[m, 1] && NeQ[m + n + 2*p + 1, 0] && IntegerQ[2*p]
\[\int \frac {x^{2} \left (e x +d \right )^{n}}{c \,x^{2}+a}d x\]
\[ \int \frac {x^2 (d+e x)^n}{a+c x^2} \, dx=\int { \frac {{\left (e x + d\right )}^{n} x^{2}}{c x^{2} + a} \,d x } \]
\[ \int \frac {x^2 (d+e x)^n}{a+c x^2} \, dx=\int \frac {x^{2} \left (d + e x\right )^{n}}{a + c x^{2}}\, dx \]
\[ \int \frac {x^2 (d+e x)^n}{a+c x^2} \, dx=\int { \frac {{\left (e x + d\right )}^{n} x^{2}}{c x^{2} + a} \,d x } \]
\[ \int \frac {x^2 (d+e x)^n}{a+c x^2} \, dx=\int { \frac {{\left (e x + d\right )}^{n} x^{2}}{c x^{2} + a} \,d x } \]
Timed out. \[ \int \frac {x^2 (d+e x)^n}{a+c x^2} \, dx=\int \frac {x^2\,{\left (d+e\,x\right )}^n}{c\,x^2+a} \,d x \]