\(\int (d+e x)^m \sqrt {a+b x+c x^2} \, dx\) [947]

Optimal result

Integrand size = 22, antiderivative size = 189 \[ \int (d+e x)^m \sqrt {a+b x+c x^2} \, dx=\frac {(d+e x)^{1+m} \sqrt {a+b x+c x^2} \operatorname {AppellF1}\left (1+m,-\frac {1}{2},-\frac {1}{2},2+m,\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e},\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{e (1+m) \sqrt {1-\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}} \sqrt {1-\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}}} \]

[Out]

(e*x+d)^(1+m)*AppellF1(1+m,-1/2,-1/2,2+m,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))))*(c*x^2+b*x+a)^(1/2)/e/(1+m)/(1-2*c*(e*x+d)/(2*c*d-e*(b-(-4*a*c+b^2)^(1/2))))^(1/2)/(1-2*
c*(e*x+d)/(2*c*d-e*(b+(-4*a*c+b^2)^(1/2))))^(1/2)

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 189, 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} \[ \int (d+e x)^m \sqrt {a+b x+c x^2} \, dx=\frac {\sqrt {a+b x+c x^2} (d+e x)^{m+1} \operatorname {AppellF1}\left (m+1,-\frac {1}{2},-\frac {1}{2},m+2,\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e},\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{e (m+1) \sqrt {1-\frac {2 c (d+e x)}{2 c d-e \left (b-\sqrt {b^2-4 a c}\right )}} \sqrt {1-\frac {2 c (d+e x)}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}}} \]

[In]

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

[Out]

((d + e*x)^(1 + m)*Sqrt[a + b*x + c*x^2]*AppellF1[1 + m, -1/2, -1/2, 2 + m, (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)])/(e*(1 + m)*Sqrt[1 - (2*c*(d + e*x))/(
2*c*d - (b - Sqrt[b^2 - 4*a*c])*e)]*Sqrt[1 - (2*c*(d + e*x))/(2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)])

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

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.10 \[ \int (d+e x)^m \sqrt {a+b x+c x^2} \, dx=\frac {(d+e x)^{1+m} \sqrt {a+x (b+c x)} \operatorname {AppellF1}\left (1+m,-\frac {1}{2},-\frac {1}{2},2+m,\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e},\frac {2 c (d+e x)}{2 c d+\left (-b+\sqrt {b^2-4 a c}\right ) e}\right )}{e (1+m) \sqrt {\frac {e \left (-b+\sqrt {b^2-4 a c}-2 c x\right )}{2 c d+\left (-b+\sqrt {b^2-4 a c}\right ) e}} \sqrt {\frac {e \left (b+\sqrt {b^2-4 a c}+2 c x\right )}{-2 c d+\left (b+\sqrt {b^2-4 a c}\right ) e}}} \]

[In]

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

[Out]

((d + e*x)^(1 + m)*Sqrt[a + x*(b + c*x)]*AppellF1[1 + m, -1/2, -1/2, 2 + m, (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)])/(e*(1 + m)*Sqrt[(e*(-b + Sqrt[b^2 -
4*a*c] - 2*c*x))/(2*c*d + (-b + Sqrt[b^2 - 4*a*c])*e)]*Sqrt[(e*(b + Sqrt[b^2 - 4*a*c] + 2*c*x))/(-2*c*d + (b +
Sqrt[b^2 - 4*a*c])*e)])

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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