Integrand size = 22, antiderivative size = 185 \[ \int \sqrt {d+e x} \left (a+b x+c x^2\right )^p \, dx=\frac {2 (d+e x)^{3/2} \left (a+b x+c x^2\right )^p \left (1-\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}\right )^{-p} \left (1-\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )^{-p} \operatorname {AppellF1}\left (\frac {3}{2},-p,-p,\frac {5}{2},\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e},\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{3 e} \]
2/3*(e*x+d)^(3/2)*(c*x^2+b*x+a)^p*AppellF1(3/2,-p,-p,5/2,2*c*(e*x+d)/(2*c* d-e*(b-(-4*a*c+b^2)^(1/2))),2*c*(e*x+d)/(2*c*d-e*(b+(-4*a*c+b^2)^(1/2))))/ e/((1-2*c*(e*x+d)/(2*c*d-e*(b-(-4*a*c+b^2)^(1/2))))^p)/((1-2*c*(e*x+d)/(2* c*d-e*(b+(-4*a*c+b^2)^(1/2))))^p)
Time = 0.71 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.14 \[ \int \sqrt {d+e x} \left (a+b x+c x^2\right )^p \, dx=\frac {2^{1-2 p} \left (\frac {e \left (-b+\sqrt {b^2-4 a c}-2 c x\right )}{8 c d+4 \left (-b+\sqrt {b^2-4 a c}\right ) e}\right )^{-p} \left (\frac {e \left (b+\sqrt {b^2-4 a c}+2 c x\right )}{-2 c d+\left (b+\sqrt {b^2-4 a c}\right ) e}\right )^{-p} (d+e x)^{3/2} (a+x (b+c x))^p \operatorname {AppellF1}\left (\frac {3}{2},-p,-p,\frac {5}{2},\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e},\frac {2 c (d+e x)}{2 c d+\left (-b+\sqrt {b^2-4 a c}\right ) e}\right )}{3 e} \]
(2^(1 - 2*p)*(d + e*x)^(3/2)*(a + x*(b + c*x))^p*AppellF1[3/2, -p, -p, 5/2 , (2*c*(d + e*x))/(2*c*d - (b + Sqrt[b^2 - 4*a*c])*e), (2*c*(d + e*x))/(2* c*d + (-b + Sqrt[b^2 - 4*a*c])*e)])/(3*e*((e*(-b + Sqrt[b^2 - 4*a*c] - 2*c *x))/(8*c*d + 4*(-b + Sqrt[b^2 - 4*a*c])*e))^p*((e*(b + Sqrt[b^2 - 4*a*c] + 2*c*x))/(-2*c*d + (b + Sqrt[b^2 - 4*a*c])*e))^p)
Time = 0.28 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {1179, 150}
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.
\(\displaystyle \int \sqrt {d+e x} \left (a+b x+c x^2\right )^p \, dx\) |
\(\Big \downarrow \) 1179 |
\(\displaystyle \frac {\left (a+b x+c x^2\right )^p \left (1-\frac {2 c (d+e x)}{2 c d-e \left (b-\sqrt {b^2-4 a c}\right )}\right )^{-p} \left (1-\frac {2 c (d+e x)}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}\right )^{-p} \int \sqrt {d+e x} \left (1-\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}\right )^p \left (1-\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )^pd(d+e x)}{e}\) |
\(\Big \downarrow \) 150 |
\(\displaystyle \frac {2 (d+e x)^{3/2} \left (a+b x+c x^2\right )^p \left (1-\frac {2 c (d+e x)}{2 c d-e \left (b-\sqrt {b^2-4 a c}\right )}\right )^{-p} \left (1-\frac {2 c (d+e x)}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}\right )^{-p} \operatorname {AppellF1}\left (\frac {3}{2},-p,-p,\frac {5}{2},\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e},\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{3 e}\) |
(2*(d + e*x)^(3/2)*(a + b*x + c*x^2)^p*AppellF1[3/2, -p, -p, 5/2, (2*c*(d + e*x))/(2*c*d - (b - Sqrt[b^2 - 4*a*c])*e), (2*c*(d + e*x))/(2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)])/(3*e*(1 - (2*c*(d + e*x))/(2*c*d - (b - Sqrt[b^2 - 4*a*c])*e))^p*(1 - (2*c*(d + e*x))/(2*c*d - (b + Sqrt[b^2 - 4*a*c])*e))^ p)
3.26.71.3.1 Defintions of rubi rules used
Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_ ] :> Simp[c^n*e^p*((b*x)^(m + 1)/(b*(m + 1)))*AppellF1[m + 1, -n, -p, m + 2 , (-d)*(x/c), (-f)*(x/e)], x] /; FreeQ[{b, c, d, e, f, m, n, p}, x] && !In tegerQ[m] && !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])
Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[(a + b*x + c*x^2)^p/(e*(1 - ( d + e*x)/(d - e*((b - q)/(2*c))))^p*(1 - (d + e*x)/(d - e*((b + q)/(2*c)))) ^p) Subst[Int[x^m*Simp[1 - x/(d - e*((b - q)/(2*c))), x]^p*Simp[1 - x/(d - e*((b + q)/(2*c))), x]^p, x], x, d + e*x], x]] /; FreeQ[{a, b, c, d, e, m , p}, x]
\[\int \sqrt {e x +d}\, \left (c \,x^{2}+b x +a \right )^{p}d x\]
\[ \int \sqrt {d+e x} \left (a+b x+c x^2\right )^p \, dx=\int { \sqrt {e x + d} {\left (c x^{2} + b x + a\right )}^{p} \,d x } \]
\[ \int \sqrt {d+e x} \left (a+b x+c x^2\right )^p \, dx=\int \sqrt {d + e x} \left (a + b x + c x^{2}\right )^{p}\, dx \]
\[ \int \sqrt {d+e x} \left (a+b x+c x^2\right )^p \, dx=\int { \sqrt {e x + d} {\left (c x^{2} + b x + a\right )}^{p} \,d x } \]
\[ \int \sqrt {d+e x} \left (a+b x+c x^2\right )^p \, dx=\int { \sqrt {e x + d} {\left (c x^{2} + b x + a\right )}^{p} \,d x } \]
Timed out. \[ \int \sqrt {d+e x} \left (a+b x+c x^2\right )^p \, dx=\int \sqrt {d+e\,x}\,{\left (c\,x^2+b\,x+a\right )}^p \,d x \]