\(\int \sqrt {d+e x} (a+b x+c x^2)^p \, dx\) [2571]

Optimal result

Integrand size = 22, antiderivative size = 185 \[ \int \sqrt {d+e x} \left (a+b x+c x^2\right )^p \, dx=\frac {2 (d+e x)^{3/2} \left (a+b x+c x^2\right )^p \left (1-\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}\right )^{-p} \left (1-\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )^{-p} \operatorname {AppellF1}\left (\frac {3}{2},-p,-p,\frac {5}{2},\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e},\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{3 e} \]

[Out]

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)

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {773, 138} \[ \int \sqrt {d+e x} \left (a+b x+c x^2\right )^p \, dx=\frac {2 (d+e x)^{3/2} \left (a+b x+c x^2\right )^p \left (1-\frac {2 c (d+e x)}{2 c d-e \left (b-\sqrt {b^2-4 a c}\right )}\right )^{-p} \left (1-\frac {2 c (d+e x)}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}\right )^{-p} \operatorname {AppellF1}\left (\frac {3}{2},-p,-p,\frac {5}{2},\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e},\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{3 e} \]

[In]

Int[Sqrt[d + e*x]*(a + b*x + c*x^2)^p,x]

[Out]

(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)

Rule 138

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_Symbol] :> 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] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])

Rule 773

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c,
2]}, Dist[(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] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] &
& NeQ[2*c*d - b*e, 0] &&  !IntegerQ[p]

Rubi steps \begin{align*} \text {integral}& = \frac {\left (\left (a+b x+c x^2\right )^p \left (1-\frac {d+e x}{d-\frac {\left (b-\sqrt {b^2-4 a c}\right ) e}{2 c}}\right )^{-p} \left (1-\frac {d+e x}{d-\frac {\left (b+\sqrt {b^2-4 a c}\right ) e}{2 c}}\right )^{-p}\right ) \text {Subst}\left (\int \sqrt {x} \left (1-\frac {2 c x}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}\right )^p \left (1-\frac {2 c x}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )^p \, dx,x,d+e x\right )}{e} \\ & = \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} F_1\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} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.71 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.14 \[ \int \sqrt {d+e x} \left (a+b x+c x^2\right )^p \, dx=\frac {2^{1-2 p} \left (\frac {e \left (-b+\sqrt {b^2-4 a c}-2 c x\right )}{8 c d+4 \left (-b+\sqrt {b^2-4 a c}\right ) e}\right )^{-p} \left (\frac {e \left (b+\sqrt {b^2-4 a c}+2 c x\right )}{-2 c d+\left (b+\sqrt {b^2-4 a c}\right ) e}\right )^{-p} (d+e x)^{3/2} (a+x (b+c x))^p \operatorname {AppellF1}\left (\frac {3}{2},-p,-p,\frac {5}{2},\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e},\frac {2 c (d+e x)}{2 c d+\left (-b+\sqrt {b^2-4 a c}\right ) e}\right )}{3 e} \]

[In]

Integrate[Sqrt[d + e*x]*(a + b*x + c*x^2)^p,x]

[Out]

(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)

Maple [F]

\[\int \sqrt {e x +d}\, \left (c \,x^{2}+b x +a \right )^{p}d x\]

[In]

int((e*x+d)^(1/2)*(c*x^2+b*x+a)^p,x)

[Out]

int((e*x+d)^(1/2)*(c*x^2+b*x+a)^p,x)

Fricas [F]

\[ \int \sqrt {d+e x} \left (a+b x+c x^2\right )^p \, dx=\int { \sqrt {e x + d} {\left (c x^{2} + b x + a\right )}^{p} \,d x } \]

[In]

integrate((e*x+d)^(1/2)*(c*x^2+b*x+a)^p,x, algorithm="fricas")

[Out]

integral(sqrt(e*x + d)*(c*x^2 + b*x + a)^p, x)

Sympy [F]

\[ \int \sqrt {d+e x} \left (a+b x+c x^2\right )^p \, dx=\int \sqrt {d + e x} \left (a + b x + c x^{2}\right )^{p}\, dx \]

[In]

integrate((e*x+d)**(1/2)*(c*x**2+b*x+a)**p,x)

[Out]

Integral(sqrt(d + e*x)*(a + b*x + c*x**2)**p, x)

Maxima [F]

\[ \int \sqrt {d+e x} \left (a+b x+c x^2\right )^p \, dx=\int { \sqrt {e x + d} {\left (c x^{2} + b x + a\right )}^{p} \,d x } \]

[In]

integrate((e*x+d)^(1/2)*(c*x^2+b*x+a)^p,x, algorithm="maxima")

[Out]

integrate(sqrt(e*x + d)*(c*x^2 + b*x + a)^p, x)

Giac [F]

\[ \int \sqrt {d+e x} \left (a+b x+c x^2\right )^p \, dx=\int { \sqrt {e x + d} {\left (c x^{2} + b x + a\right )}^{p} \,d x } \]

[In]

integrate((e*x+d)^(1/2)*(c*x^2+b*x+a)^p,x, algorithm="giac")

[Out]

integrate(sqrt(e*x + d)*(c*x^2 + b*x + a)^p, x)

Mupad [F(-1)]

Timed out. \[ \int \sqrt {d+e x} \left (a+b x+c x^2\right )^p \, dx=\int \sqrt {d+e\,x}\,{\left (c\,x^2+b\,x+a\right )}^p \,d x \]

[In]

int((d + e*x)^(1/2)*(a + b*x + c*x^2)^p,x)

[Out]

int((d + e*x)^(1/2)*(a + b*x + c*x^2)^p, x)