Integrand size = 17, antiderivative size = 125 \[ \int \frac {\left (a+c x^2\right )^p}{d+e x} \, dx=\frac {x \left (a+c x^2\right )^p \left (1+\frac {c x^2}{a}\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2},-p,1,\frac {3}{2},-\frac {c x^2}{a},\frac {e^2 x^2}{d^2}\right )}{d}-\frac {e \left (a+c x^2\right )^{1+p} \operatorname {Hypergeometric2F1}\left (1,1+p,2+p,\frac {e^2 \left (a+c x^2\right )}{c d^2+a e^2}\right )}{2 \left (c d^2+a e^2\right ) (1+p)} \]
x*(c*x^2+a)^p*AppellF1(1/2,1,-p,3/2,e^2*x^2/d^2,-c*x^2/a)/d/((1+c*x^2/a)^p )-1/2*e*(c*x^2+a)^(p+1)*hypergeom([1, p+1],[2+p],e^2*(c*x^2+a)/(a*e^2+c*d^ 2))/(a*e^2+c*d^2)/(p+1)
Time = 0.26 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.05 \[ \int \frac {\left (a+c x^2\right )^p}{d+e x} \, dx=\frac {\left (\frac {e \left (-\sqrt {-\frac {a}{c}}+x\right )}{d+e x}\right )^{-p} \left (\frac {e \left (\sqrt {-\frac {a}{c}}+x\right )}{d+e x}\right )^{-p} \left (a+c x^2\right )^p \operatorname {AppellF1}\left (-2 p,-p,-p,1-2 p,\frac {d-\sqrt {-\frac {a}{c}} e}{d+e x},\frac {d+\sqrt {-\frac {a}{c}} e}{d+e x}\right )}{2 e p} \]
((a + c*x^2)^p*AppellF1[-2*p, -p, -p, 1 - 2*p, (d - Sqrt[-(a/c)]*e)/(d + e *x), (d + Sqrt[-(a/c)]*e)/(d + e*x)])/(2*e*p*((e*(-Sqrt[-(a/c)] + x))/(d + e*x))^p*((e*(Sqrt[-(a/c)] + x))/(d + e*x))^p)
Time = 0.26 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {504, 334, 333, 353, 78}
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.
\(\displaystyle \int \frac {\left (a+c x^2\right )^p}{d+e x} \, dx\) |
\(\Big \downarrow \) 504 |
\(\displaystyle d \int \frac {\left (c x^2+a\right )^p}{d^2-e^2 x^2}dx-e \int \frac {x \left (c x^2+a\right )^p}{d^2-e^2 x^2}dx\) |
\(\Big \downarrow \) 334 |
\(\displaystyle d \left (a+c x^2\right )^p \left (\frac {c x^2}{a}+1\right )^{-p} \int \frac {\left (\frac {c x^2}{a}+1\right )^p}{d^2-e^2 x^2}dx-e \int \frac {x \left (c x^2+a\right )^p}{d^2-e^2 x^2}dx\) |
\(\Big \downarrow \) 333 |
\(\displaystyle \frac {x \left (a+c x^2\right )^p \left (\frac {c x^2}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2},-p,1,\frac {3}{2},-\frac {c x^2}{a},\frac {e^2 x^2}{d^2}\right )}{d}-e \int \frac {x \left (c x^2+a\right )^p}{d^2-e^2 x^2}dx\) |
\(\Big \downarrow \) 353 |
\(\displaystyle \frac {x \left (a+c x^2\right )^p \left (\frac {c x^2}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2},-p,1,\frac {3}{2},-\frac {c x^2}{a},\frac {e^2 x^2}{d^2}\right )}{d}-\frac {1}{2} e \int \frac {\left (c x^2+a\right )^p}{d^2-e^2 x^2}dx^2\) |
\(\Big \downarrow \) 78 |
\(\displaystyle \frac {x \left (a+c x^2\right )^p \left (\frac {c x^2}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2},-p,1,\frac {3}{2},-\frac {c x^2}{a},\frac {e^2 x^2}{d^2}\right )}{d}-\frac {e \left (a+c x^2\right )^{p+1} \operatorname {Hypergeometric2F1}\left (1,p+1,p+2,\frac {e^2 \left (c x^2+a\right )}{c d^2+a e^2}\right )}{2 (p+1) \left (a e^2+c d^2\right )}\) |
(x*(a + c*x^2)^p*AppellF1[1/2, -p, 1, 3/2, -((c*x^2)/a), (e^2*x^2)/d^2])/( d*(1 + (c*x^2)/a)^p) - (e*(a + c*x^2)^(1 + p)*Hypergeometric2F1[1, 1 + p, 2 + p, (e^2*(a + c*x^2))/(c*d^2 + a*e^2)])/(2*(c*d^2 + a*e^2)*(1 + p))
3.8.36.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(b *c - a*d)^n*((a + b*x)^(m + 1)/(b^(n + 1)*(m + 1)))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m}, x] && !IntegerQ[m] && IntegerQ[n]
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim p[a^p*c^q*x*AppellF1[1/2, -p, -q, 3/2, (-b)*(x^2/a), (-d)*(x^2/c)], x] /; F reeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim p[a^IntPart[p]*((a + b*x^2)^FracPart[p]/(1 + b*(x^2/a))^FracPart[p]) Int[ (1 + b*(x^2/a))^p*(c + d*x^2)^q, x], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && !(IntegerQ[p] || GtQ[a, 0])
Int[(x_)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_Symbol] :> Simp[1/2 Subst[Int[(a + b*x)^p*(c + d*x)^q, x], x, x^2], x] /; FreeQ[ {a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0]
Int[((a_) + (b_.)*(x_)^2)^(p_)/((c_) + (d_.)*(x_)), x_Symbol] :> Simp[c I nt[(a + b*x^2)^p/(c^2 - d^2*x^2), x], x] - Simp[d Int[x*((a + b*x^2)^p/(c ^2 - d^2*x^2)), x], x] /; FreeQ[{a, b, c, d, p}, x]
\[\int \frac {\left (c \,x^{2}+a \right )^{p}}{e x +d}d x\]
\[ \int \frac {\left (a+c x^2\right )^p}{d+e x} \, dx=\int { \frac {{\left (c x^{2} + a\right )}^{p}}{e x + d} \,d x } \]
\[ \int \frac {\left (a+c x^2\right )^p}{d+e x} \, dx=\int \frac {\left (a + c x^{2}\right )^{p}}{d + e x}\, dx \]
\[ \int \frac {\left (a+c x^2\right )^p}{d+e x} \, dx=\int { \frac {{\left (c x^{2} + a\right )}^{p}}{e x + d} \,d x } \]
\[ \int \frac {\left (a+c x^2\right )^p}{d+e x} \, dx=\int { \frac {{\left (c x^{2} + a\right )}^{p}}{e x + d} \,d x } \]
Timed out. \[ \int \frac {\left (a+c x^2\right )^p}{d+e x} \, dx=\int \frac {{\left (c\,x^2+a\right )}^p}{d+e\,x} \,d x \]