Integrand size = 20, antiderivative size = 209 \[ \int \frac {x^3 (d+e x)^n}{a+c x^2} \, dx=-\frac {d (d+e x)^{1+n}}{c e^2 (1+n)}+\frac {(d+e x)^{2+n}}{c e^2 (2+n)}+\frac {a (d+e x)^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt {c} (d+e x)}{\sqrt {c} d-\sqrt {-a} e}\right )}{2 c^{3/2} \left (\sqrt {c} d-\sqrt {-a} e\right ) (1+n)}+\frac {a (d+e x)^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt {c} (d+e x)}{\sqrt {c} d+\sqrt {-a} e}\right )}{2 c^{3/2} \left (\sqrt {c} d+\sqrt {-a} e\right ) (1+n)} \]
-d*(e*x+d)^(1+n)/c/e^2/(1+n)+(e*x+d)^(2+n)/c/e^2/(2+n)+1/2*a*(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)))/c^(3/ 2)/(1+n)/(-e*(-a)^(1/2)+d*c^(1/2))+1/2*a*(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)))/c^(3/2)/(1+n)/(e*(-a)^(1/2 )+d*c^(1/2))
Time = 0.21 (sec) , antiderivative size = 168, normalized size of antiderivative = 0.80 \[ \int \frac {x^3 (d+e x)^n}{a+c x^2} \, dx=\frac {(d+e x)^{1+n} \left (-\frac {2 \sqrt {c} (d-e (1+n) x)}{e^2 (2+n)}+\frac {a \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt {c} (d+e x)}{\sqrt {c} d-\sqrt {-a} e}\right )}{\sqrt {c} d-\sqrt {-a} e}+\frac {a \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt {c} (d+e x)}{\sqrt {c} d+\sqrt {-a} e}\right )}{\sqrt {c} d+\sqrt {-a} e}\right )}{2 c^{3/2} (1+n)} \]
((d + e*x)^(1 + n)*((-2*Sqrt[c]*(d - e*(1 + n)*x))/(e^2*(2 + n)) + (a*Hype rgeometric2F1[1, 1 + n, 2 + n, (Sqrt[c]*(d + e*x))/(Sqrt[c]*d - Sqrt[-a]*e )])/(Sqrt[c]*d - Sqrt[-a]*e) + (a*Hypergeometric2F1[1, 1 + n, 2 + n, (Sqrt [c]*(d + e*x))/(Sqrt[c]*d + Sqrt[-a]*e)])/(Sqrt[c]*d + Sqrt[-a]*e)))/(2*c^ (3/2)*(1 + n))
Time = 0.51 (sec) , antiderivative size = 232, normalized size of antiderivative = 1.11, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {604, 25, 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.
\(\displaystyle \int \frac {x^3 (d+e x)^n}{a+c x^2} \, dx\) |
\(\Big \downarrow \) 604 |
\(\displaystyle \frac {\int -\frac {(d+e x)^n \left (a (n+2) x e^3+c d (n+2) x^2 e^2+a d (n+2) e^2\right )}{c x^2+a}dx}{c e^3 (n+2)}+\frac {(d+e x)^{n+2}}{c e^2 (n+2)}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {(d+e x)^{n+2}}{c e^2 (n+2)}-\frac {\int \frac {(d+e x)^n \left (a (n+2) x e^3+c d (n+2) x^2 e^2+a d (n+2) e^2\right )}{c x^2+a}dx}{c e^3 (n+2)}\) |
\(\Big \downarrow \) 2160 |
\(\displaystyle \frac {(d+e x)^{n+2}}{c e^2 (n+2)}-\frac {\int \left (d e^2 (n+2) (d+e x)^n+\frac {\left (2 a e^3+a n e^3\right ) x (d+e x)^n}{c x^2+a}\right )dx}{c e^3 (n+2)}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {(d+e x)^{n+2}}{c e^2 (n+2)}-\frac {-\frac {a e^3 (n+2) (d+e x)^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {\sqrt {c} (d+e x)}{\sqrt {c} d-\sqrt {-a} e}\right )}{2 \sqrt {c} (n+1) \left (\sqrt {c} d-\sqrt {-a} e\right )}-\frac {a e^3 (n+2) (d+e x)^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {\sqrt {c} (d+e x)}{\sqrt {c} d+\sqrt {-a} e}\right )}{2 \sqrt {c} (n+1) \left (\sqrt {-a} e+\sqrt {c} d\right )}+\frac {d e (n+2) (d+e x)^{n+1}}{n+1}}{c e^3 (n+2)}\) |
(d + e*x)^(2 + n)/(c*e^2*(2 + n)) - ((d*e*(2 + n)*(d + e*x)^(1 + n))/(1 + n) - (a*e^3*(2 + n)*(d + e*x)^(1 + n)*Hypergeometric2F1[1, 1 + n, 2 + n, ( Sqrt[c]*(d + e*x))/(Sqrt[c]*d - Sqrt[-a]*e)])/(2*Sqrt[c]*(Sqrt[c]*d - Sqrt [-a]*e)*(1 + n)) - (a*e^3*(2 + n)*(d + e*x)^(1 + n)*Hypergeometric2F1[1, 1 + n, 2 + n, (Sqrt[c]*(d + e*x))/(Sqrt[c]*d + Sqrt[-a]*e)])/(2*Sqrt[c]*(Sq rt[c]*d + Sqrt[-a]*e)*(1 + n)))/(c*e^3*(2 + n))
3.4.64.3.1 Defintions of rubi rules used
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[(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]
\[\int \frac {x^{3} \left (e x +d \right )^{n}}{c \,x^{2}+a}d x\]
\[ \int \frac {x^3 (d+e x)^n}{a+c x^2} \, dx=\int { \frac {{\left (e x + d\right )}^{n} x^{3}}{c x^{2} + a} \,d x } \]
\[ \int \frac {x^3 (d+e x)^n}{a+c x^2} \, dx=\int \frac {x^{3} \left (d + e x\right )^{n}}{a + c x^{2}}\, dx \]
\[ \int \frac {x^3 (d+e x)^n}{a+c x^2} \, dx=\int { \frac {{\left (e x + d\right )}^{n} x^{3}}{c x^{2} + a} \,d x } \]
\[ \int \frac {x^3 (d+e x)^n}{a+c x^2} \, dx=\int { \frac {{\left (e x + d\right )}^{n} x^{3}}{c x^{2} + a} \,d x } \]
Timed out. \[ \int \frac {x^3 (d+e x)^n}{a+c x^2} \, dx=\int \frac {x^3\,{\left (d+e\,x\right )}^n}{c\,x^2+a} \,d x \]