\(\int (b d+2 c d x) (a+b x+c x^2)^{3/2} \, dx\) [1206]

Optimal result

Integrand size = 24, antiderivative size = 19 \[ \int (b d+2 c d x) \left (a+b x+c x^2\right )^{3/2} \, dx=\frac {2}{5} d \left (a+b x+c x^2\right )^{5/2} \]

[Out]

2/5*d*(c*x^2+b*x+a)^(5/2)

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} \[ \int (b d+2 c d x) \left (a+b x+c x^2\right )^{3/2} \, dx=\frac {2}{5} d \left (a+b x+c x^2\right )^{5/2} \]

[In]

Int[(b*d + 2*c*d*x)*(a + b*x + c*x^2)^(3/2),x]

[Out]

(2*d*(a + b*x + c*x^2)^(5/2))/5

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}{5} d \left (a+b x+c x^2\right )^{5/2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.95 \[ \int (b d+2 c d x) \left (a+b x+c x^2\right )^{3/2} \, dx=\frac {2}{5} d (a+x (b+c x))^{5/2} \]

[In]

Integrate[(b*d + 2*c*d*x)*(a + b*x + c*x^2)^(3/2),x]

[Out]

(2*d*(a + x*(b + c*x))^(5/2))/5

Maple [A] (verified)

Time = 2.86 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.84

method	result	size
gosper	\(\frac {2 d \left (c \,x^{2}+b x +a \right )^{\frac {5}{2}}}{5}\)	\(16\)
default	\(\frac {2 d \left (c \,x^{2}+b x +a \right )^{\frac {5}{2}}}{5}\)	\(16\)
pseudoelliptic	\(\frac {2 d \left (c \,x^{2}+b x +a \right )^{\frac {5}{2}}}{5}\)	\(16\)
risch	\(\frac {2 d \left (x^{4} c^{2}+2 b c \,x^{3}+2 a c \,x^{2}+b^{2} x^{2}+2 a b x +a^{2}\right ) \sqrt {c \,x^{2}+b x +a}}{5}\)	\(53\)
trager	\(d \left (\frac {2}{5} x^{4} c^{2}+\frac {4}{5} b c \,x^{3}+\frac {4}{5} a c \,x^{2}+\frac {2}{5} b^{2} x^{2}+\frac {4}{5} a b x +\frac {2}{5} a^{2}\right ) \sqrt {c \,x^{2}+b x +a}\)	\(56\)

[In]

int((2*c*d*x+b*d)*(c*x^2+b*x+a)^(3/2),x,method=_RETURNVERBOSE)

[Out]

2/5*d*(c*x^2+b*x+a)^(5/2)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 55 vs. \(2 (15) = 30\).

Time = 0.26 (sec) , antiderivative size = 55, normalized size of antiderivative = 2.89 \[ \int (b d+2 c d x) \left (a+b x+c x^2\right )^{3/2} \, dx=\frac {2}{5} \, {\left (c^{2} d x^{4} + 2 \, b c d x^{3} + 2 \, a b d x + {\left (b^{2} + 2 \, a c\right )} d x^{2} + a^{2} d\right )} \sqrt {c x^{2} + b x + a} \]

[In]

integrate((2*c*d*x+b*d)*(c*x^2+b*x+a)^(3/2),x, algorithm="fricas")

[Out]

2/5*(c^2*d*x^4 + 2*b*c*d*x^3 + 2*a*b*d*x + (b^2 + 2*a*c)*d*x^2 + a^2*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. 146 vs. \(2 (17) = 34\).

Time = 0.14 (sec) , antiderivative size = 146, normalized size of antiderivative = 7.68 \[ \int (b d+2 c d x) \left (a+b x+c x^2\right )^{3/2} \, dx=\frac {2 a^{2} d \sqrt {a + b x + c x^{2}}}{5} + \frac {4 a b d x \sqrt {a + b x + c x^{2}}}{5} + \frac {4 a c d x^{2} \sqrt {a + b x + c x^{2}}}{5} + \frac {2 b^{2} d x^{2} \sqrt {a + b x + c x^{2}}}{5} + \frac {4 b c d x^{3} \sqrt {a + b x + c x^{2}}}{5} + \frac {2 c^{2} d x^{4} \sqrt {a + b x + c x^{2}}}{5} \]

[In]

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

[Out]

2*a**2*d*sqrt(a + b*x + c*x**2)/5 + 4*a*b*d*x*sqrt(a + b*x + c*x**2)/5 + 4*a*c*d*x**2*sqrt(a + b*x + c*x**2)/5
+ 2*b**2*d*x**2*sqrt(a + b*x + c*x**2)/5 + 4*b*c*d*x**3*sqrt(a + b*x + c*x**2)/5 + 2*c**2*d*x**4*sqrt(a + b*x
+ c*x**2)/5

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int (b d+2 c d x) \left (a+b x+c x^2\right )^{3/2} \, dx=\frac {2}{5} \, {\left (c x^{2} + b x + a\right )}^{\frac {5}{2}} d \]

[In]

integrate((2*c*d*x+b*d)*(c*x^2+b*x+a)^(3/2),x, algorithm="maxima")

[Out]

2/5*(c*x^2 + b*x + a)^(5/2)*d

Giac [A] (verification not implemented)

none

Time = 0.46 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int (b d+2 c d x) \left (a+b x+c x^2\right )^{3/2} \, dx=\frac {2}{5} \, {\left (c x^{2} + b x + a\right )}^{\frac {5}{2}} d \]

[In]

integrate((2*c*d*x+b*d)*(c*x^2+b*x+a)^(3/2),x, algorithm="giac")

[Out]

2/5*(c*x^2 + b*x + a)^(5/2)*d

Mupad [B] (verification not implemented)

Time = 9.66 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int (b d+2 c d x) \left (a+b x+c x^2\right )^{3/2} \, dx=\frac {2\,d\,{\left (c\,x^2+b\,x+a\right )}^{5/2}}{5} \]

[In]

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

[Out]

(2*d*(a + b*x + c*x^2)^(5/2))/5