\(\int (b d+2 c d x)^m \sqrt {a+b x+c x^2} \, dx\) [327]

Optimal result

Integrand size = 26, antiderivative size = 98 \[ \int (b d+2 c d x)^m \sqrt {a+b x+c x^2} \, dx=-\frac {(b d+2 c d x)^{1+m} \left (4 a-\frac {b^2}{c}+\frac {(b+2 c x)^2}{c}\right )^{3/2} \operatorname {Hypergeometric2F1}\left (1,\frac {4+m}{2},\frac {3+m}{2},\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{4 \left (b^2-4 a c\right ) d (1+m)} \] Output:

-1/4*(2*c*d*x+b*d)^(1+m)*(4*a-b^2/c+(2*c*x+b)^2/c)^(3/2)*hypergeom([1, 2+1 
/2*m],[3/2+1/2*m],(2*c*x+b)^2/(-4*a*c+b^2))/(-4*a*c+b^2)/d/(1+m)
 

Mathematica [A] (verified)

Time = 0.29 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.07 \[ \int (b d+2 c d x)^m \sqrt {a+b x+c x^2} \, dx=\frac {(b+2 c x) (d (b+2 c x))^m \sqrt {a+x (b+c x)} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{4 c (1+m) \sqrt {\frac {c (a+x (b+c x))}{-b^2+4 a c}}} \] Input:

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

Output:

((b + 2*c*x)*(d*(b + 2*c*x))^m*Sqrt[a + x*(b + c*x)]*Hypergeometric2F1[-1/ 
2, (1 + m)/2, (3 + m)/2, (b + 2*c*x)^2/(b^2 - 4*a*c)])/(4*c*(1 + m)*Sqrt[( 
c*(a + x*(b + c*x)))/(-b^2 + 4*a*c)])
 

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.40, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {1118, 27, 279, 278}

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.

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

\(\Big \downarrow \) 1118

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 279

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

\(\Big \downarrow \) 278

\(\displaystyle \frac {\sqrt {4 a-\frac {b^2}{c}+\frac {(b d+2 c d x)^2}{c d^2}} (b d+2 c d x)^{m+1} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {m+1}{2},\frac {m+3}{2},\frac {(b d+2 c x d)^2}{\left (b^2-4 a c\right ) d^2}\right )}{4 c d (m+1) \sqrt {1-\frac {(b d+2 c d x)^2}{d^2 \left (b^2-4 a c\right )}}}\)

Input:

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

Output:

((b*d + 2*c*d*x)^(1 + m)*Sqrt[4*a - b^2/c + (b*d + 2*c*d*x)^2/(c*d^2)]*Hyp 
ergeometric2F1[-1/2, (1 + m)/2, (3 + m)/2, (b*d + 2*c*d*x)^2/((b^2 - 4*a*c 
)*d^2)])/(4*c*d*(1 + m)*Sqrt[1 - (b*d + 2*c*d*x)^2/((b^2 - 4*a*c)*d^2)])
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 278

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^p*(( 
c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/2, (m + 1)/2 + 1, ( 
-b)*(x^2/a)], x] /; FreeQ[{a, b, c, m, p}, x] &&  !IGtQ[p, 0] && (ILtQ[p, 0 
] || GtQ[a, 0])
 

rule 279

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^IntP 
art[p]*((a + b*x^2)^FracPart[p]/(1 + b*(x^2/a))^FracPart[p])   Int[(c*x)^m* 
(1 + b*(x^2/a))^p, x], x] /; FreeQ[{a, b, c, m, p}, x] &&  !IGtQ[p, 0] && 
!(ILtQ[p, 0] || GtQ[a, 0])
 

rule 1118

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_S 
ymbol] :> Simp[1/e   Subst[Int[x^m*(a - b^2/(4*c) + (c*x^2)/e^2)^p, x], x, 
d + e*x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && EqQ[2*c*d - b*e, 0]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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