Integrand size = 20, antiderivative size = 250 \[ \int \frac {x^4 (d+e x)^n}{a+c x^2} \, dx=\frac {\left (c d^2-a e^2\right ) (d+e x)^{1+n}}{c^2 e^3 (1+n)}-\frac {2 d (d+e x)^{2+n}}{c e^3 (2+n)}+\frac {(d+e x)^{3+n}}{c e^3 (3+n)}+\frac {(-a)^{3/2} (d+e x)^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt {c} (d+e x)}{\sqrt {c} d-\sqrt {-a} e}\right )}{2 c^2 \left (\sqrt {c} d-\sqrt {-a} e\right ) (1+n)}-\frac {(-a)^{3/2} (d+e x)^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt {c} (d+e x)}{\sqrt {c} d+\sqrt {-a} e}\right )}{2 c^2 \left (\sqrt {c} d+\sqrt {-a} e\right ) (1+n)} \]
(-a*e^2+c*d^2)*(e*x+d)^(1+n)/c^2/e^3/(1+n)-2*d*(e*x+d)^(2+n)/c/e^3/(2+n)+( e*x+d)^(3+n)/c/e^3/(3+n)+1/2*(-a)^(3/2)*(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)))/c^2/(1+n)/(-e*(-a)^(1/2)+d *c^(1/2))-1/2*(-a)^(3/2)*(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)))/c^2/(1+n)/(e*(-a)^(1/2)+d*c^(1/2))
Time = 0.40 (sec) , antiderivative size = 217, normalized size of antiderivative = 0.87 \[ \int \frac {x^4 (d+e x)^n}{a+c x^2} \, dx=\frac {(d+e x)^{1+n} \left (\frac {2 \left (c d^2-a e^2\right )}{e^3 (1+n)}-\frac {4 c d (d+e x)}{e^3 (2+n)}+\frac {2 c (d+e x)^2}{e^3 (3+n)}+\frac {(-a)^{3/2} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt {c} (d+e x)}{\sqrt {c} d-\sqrt {-a} e}\right )}{\left (\sqrt {c} d-\sqrt {-a} e\right ) (1+n)}+\frac {\sqrt {-a} a \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt {c} (d+e x)}{\sqrt {c} d+\sqrt {-a} e}\right )}{\left (\sqrt {c} d+\sqrt {-a} e\right ) (1+n)}\right )}{2 c^2} \]
((d + e*x)^(1 + n)*((2*(c*d^2 - a*e^2))/(e^3*(1 + n)) - (4*c*d*(d + e*x))/ (e^3*(2 + n)) + (2*c*(d + e*x)^2)/(e^3*(3 + n)) + ((-a)^(3/2)*Hypergeometr ic2F1[1, 1 + n, 2 + n, (Sqrt[c]*(d + e*x))/(Sqrt[c]*d - Sqrt[-a]*e)])/((Sq rt[c]*d - Sqrt[-a]*e)*(1 + n)) + (Sqrt[-a]*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))))/(2*c^2)
Time = 0.76 (sec) , antiderivative size = 276, normalized size of antiderivative = 1.10, 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.
\(\displaystyle \int \frac {x^4 (d+e x)^n}{a+c x^2} \, dx\) |
\(\Big \downarrow \) 604 |
\(\displaystyle \frac {\int -\frac {(d+e x)^n \left (2 c d (n+3) x^3 e^3+2 a d (n+3) x e^3+\left (c d^2+a e^2\right ) (n+3) x^2 e^2+a d^2 (n+3) e^2\right )}{c x^2+a}dx}{c e^4 (n+3)}+\frac {(d+e x)^{n+3}}{c e^3 (n+3)}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {(d+e x)^{n+3}}{c e^3 (n+3)}-\frac {\int \frac {(d+e x)^n \left (2 c d (n+3) x^3 e^3+2 a d (n+3) x e^3+\left (c d^2+a e^2\right ) (n+3) x^2 e^2+a d^2 (n+3) e^2\right )}{c x^2+a}dx}{c e^4 (n+3)}\) |
\(\Big \downarrow \) 2160 |
\(\displaystyle \frac {(d+e x)^{n+3}}{c e^3 (n+3)}-\frac {\int \left (-\frac {e^2 \left (c d^2-a e^2\right ) (n+3) (d+e x)^n}{c}+\frac {\left (-\frac {a^2 n e^4}{c}-\frac {3 a^2 e^4}{c}\right ) (d+e x)^n}{c x^2+a}+2 d e^2 (n+3) (d+e x)^{n+1}\right )dx}{c e^4 (n+3)}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {(d+e x)^{n+3}}{c e^3 (n+3)}-\frac {-\frac {e (n+3) \left (c d^2-a e^2\right ) (d+e x)^{n+1}}{c (n+1)}-\frac {(-a)^{3/2} e^4 (n+3) (d+e x)^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {\sqrt {c} (d+e x)}{\sqrt {c} d-\sqrt {-a} e}\right )}{2 c (n+1) \left (\sqrt {c} d-\sqrt {-a} e\right )}+\frac {(-a)^{3/2} e^4 (n+3) (d+e x)^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {\sqrt {c} (d+e x)}{\sqrt {c} d+\sqrt {-a} e}\right )}{2 c (n+1) \left (\sqrt {-a} e+\sqrt {c} d\right )}+\frac {2 d e (n+3) (d+e x)^{n+2}}{n+2}}{c e^4 (n+3)}\) |
(d + e*x)^(3 + n)/(c*e^3*(3 + n)) - (-((e*(c*d^2 - a*e^2)*(3 + n)*(d + e*x )^(1 + n))/(c*(1 + n))) + (2*d*e*(3 + n)*(d + e*x)^(2 + n))/(2 + n) - ((-a )^(3/2)*e^4*(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*(Sqrt[c]*d - Sqrt[-a]*e )*(1 + n)) + ((-a)^(3/2)*e^4*(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*(Sqrt[ c]*d + Sqrt[-a]*e)*(1 + n)))/(c*e^4*(3 + n))
3.4.63.3.1 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]
\[\int \frac {x^{4} \left (e x +d \right )^{n}}{c \,x^{2}+a}d x\]
\[ \int \frac {x^4 (d+e x)^n}{a+c x^2} \, dx=\int { \frac {{\left (e x + d\right )}^{n} x^{4}}{c x^{2} + a} \,d x } \]
\[ \int \frac {x^4 (d+e x)^n}{a+c x^2} \, dx=\int \frac {x^{4} \left (d + e x\right )^{n}}{a + c x^{2}}\, dx \]
\[ \int \frac {x^4 (d+e x)^n}{a+c x^2} \, dx=\int { \frac {{\left (e x + d\right )}^{n} x^{4}}{c x^{2} + a} \,d x } \]
\[ \int \frac {x^4 (d+e x)^n}{a+c x^2} \, dx=\int { \frac {{\left (e x + d\right )}^{n} x^{4}}{c x^{2} + a} \,d x } \]
Timed out. \[ \int \frac {x^4 (d+e x)^n}{a+c x^2} \, dx=\int \frac {x^4\,{\left (d+e\,x\right )}^n}{c\,x^2+a} \,d x \]