\(\int (d+e x)^m \sqrt {a+b x+c x^2} \, dx\) [769]

Optimal result

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

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

Mathematica [A] (warning: unable to verify)

Time = 0.77 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.10 \[ \int (d+e x)^m \sqrt {a+b x+c x^2} \, dx=\frac {(d+e x)^{1+m} \sqrt {a+x (b+c x)} \operatorname {AppellF1}\left (1+m,-\frac {1}{2},-\frac {1}{2},2+m,\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e},\frac {2 c (d+e x)}{2 c d+\left (-b+\sqrt {b^2-4 a c}\right ) e}\right )}{e (1+m) \sqrt {\frac {e \left (-b+\sqrt {b^2-4 a c}-2 c x\right )}{2 c d+\left (-b+\sqrt {b^2-4 a c}\right ) e}} \sqrt {\frac {e \left (b+\sqrt {b^2-4 a c}+2 c x\right )}{-2 c d+\left (b+\sqrt {b^2-4 a c}\right ) e}}} \] Input:

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

Output:

((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)])
 

Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 189, 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.

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

\(\Big \downarrow \) 1179

\(\displaystyle \frac {\sqrt {a+b x+c x^2} \int (d+e x)^m \sqrt {1-\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}} \sqrt {1-\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}}d(d+e x)}{e \sqrt {1-\frac {2 c (d+e x)}{2 c d-e \left (b-\sqrt {b^2-4 a c}\right )}} \sqrt {1-\frac {2 c (d+e x)}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}}}\)

\(\Big \downarrow \) 150

\(\displaystyle \frac {\sqrt {a+b x+c x^2} (d+e x)^{m+1} \operatorname {AppellF1}\left (m+1,-\frac {1}{2},-\frac {1}{2},m+2,\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e},\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{e (m+1) \sqrt {1-\frac {2 c (d+e x)}{2 c d-e \left (b-\sqrt {b^2-4 a c}\right )}} \sqrt {1-\frac {2 c (d+e x)}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}}}\)

Input:

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

Output:

((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 + S 
qrt[b^2 - 4*a*c])*e)])
 

Defintions of rubi rules used

rule 150

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])
 

rule 1179

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]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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