\(\int (b d+2 c d x) \sqrt {a+b x+c x^2} \, dx\) [1194]

Optimal result

Integrand size = 24, antiderivative size = 19 \[ \int (b d+2 c d x) \sqrt {a+b x+c x^2} \, dx=\frac {2}{3} d \left (a+b x+c x^2\right )^{3/2} \]

[Out]

2/3*d*(c*x^2+b*x+a)^(3/2)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.042, Rules used = {643} \[ \int (b d+2 c d x) \sqrt {a+b x+c x^2} \, dx=\frac {2}{3} d \left (a+b x+c x^2\right )^{3/2} \]

[In]

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

[Out]

(2*d*(a + b*x + c*x^2)^(3/2))/3

Rule 643

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

Rubi steps \begin{align*} \text {integral}& = \frac {2}{3} d \left (a+b x+c x^2\right )^{3/2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.95 \[ \int (b d+2 c d x) \sqrt {a+b x+c x^2} \, dx=\frac {2}{3} d (a+x (b+c x))^{3/2} \]

[In]

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

[Out]

(2*d*(a + x*(b + c*x))^(3/2))/3

Maple [A] (verified)

Time = 2.49 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.84

method	result	size
gosper	\(\frac {2 d \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}{3}\)	\(16\)
default	\(\frac {2 d \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}{3}\)	\(16\)
risch	\(\frac {2 d \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}{3}\)	\(16\)
pseudoelliptic	\(\frac {2 d \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}{3}\)	\(16\)
trager	\(d \left (\frac {2}{3} c \,x^{2}+\frac {2}{3} b x +\frac {2}{3} a \right ) \sqrt {c \,x^{2}+b x +a}\)	\(29\)

[In]

int((2*c*d*x+b*d)*(c*x^2+b*x+a)^(1/2),x,method=_RETURNVERBOSE)

[Out]

2/3*d*(c*x^2+b*x+a)^(3/2)

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.47 \[ \int (b d+2 c d x) \sqrt {a+b x+c x^2} \, dx=\frac {2}{3} \, {\left (c d x^{2} + b d x + a d\right )} \sqrt {c x^{2} + b x + a} \]

[In]

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

[Out]

2/3*(c*d*x^2 + b*d*x + a*d)*sqrt(c*x^2 + b*x + a)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 65 vs. \(2 (17) = 34\).

Time = 0.08 (sec) , antiderivative size = 65, normalized size of antiderivative = 3.42 \[ \int (b d+2 c d x) \sqrt {a+b x+c x^2} \, dx=\frac {2 a d \sqrt {a + b x + c x^{2}}}{3} + \frac {2 b d x \sqrt {a + b x + c x^{2}}}{3} + \frac {2 c d x^{2} \sqrt {a + b x + c x^{2}}}{3} \]

[In]

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

[Out]

2*a*d*sqrt(a + b*x + c*x**2)/3 + 2*b*d*x*sqrt(a + b*x + c*x**2)/3 + 2*c*d*x**2*sqrt(a + b*x + c*x**2)/3

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int (b d+2 c d x) \sqrt {a+b x+c x^2} \, dx=\frac {2}{3} \, {\left (c x^{2} + b x + a\right )}^{\frac {3}{2}} d \]

[In]

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

[Out]

2/3*(c*x^2 + b*x + a)^(3/2)*d

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int (b d+2 c d x) \sqrt {a+b x+c x^2} \, dx=\frac {2}{3} \, {\left (c x^{2} + b x + a\right )}^{\frac {3}{2}} d \]

[In]

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

[Out]

2/3*(c*x^2 + b*x + a)^(3/2)*d

Mupad [B] (verification not implemented)

Time = 9.64 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int (b d+2 c d x) \sqrt {a+b x+c x^2} \, dx=\frac {2\,d\,{\left (c\,x^2+b\,x+a\right )}^{3/2}}{3} \]

[In]

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

[Out]

(2*d*(a + b*x + c*x^2)^(3/2))/3