\(\int \frac {a+b x+c x^2}{(b d+2 c d x)^{3/2}} \, dx\) [1265]

Optimal result

Integrand size = 24, antiderivative size = 55 \[ \int \frac {a+b x+c x^2}{(b d+2 c d x)^{3/2}} \, dx=\frac {b^2-4 a c}{4 c^2 d \sqrt {b d+2 c d x}}+\frac {(b d+2 c d x)^{3/2}}{12 c^2 d^3} \]

[Out]

1/12*(2*c*d*x+b*d)^(3/2)/c^2/d^3+1/4*(-4*a*c+b^2)/c^2/d/(2*c*d*x+b*d)^(1/2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.042, Rules used = {697} \[ \int \frac {a+b x+c x^2}{(b d+2 c d x)^{3/2}} \, dx=\frac {b^2-4 a c}{4 c^2 d \sqrt {b d+2 c d x}}+\frac {(b d+2 c d x)^{3/2}}{12 c^2 d^3} \]

[In]

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

[Out]

(b^2 - 4*a*c)/(4*c^2*d*Sqrt[b*d + 2*c*d*x]) + (b*d + 2*c*d*x)^(3/2)/(12*c^2*d^3)

Rule 697

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d +
e*x)^m*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[2*c*d - b*e,
0] && IGtQ[p, 0] &&  !(EqQ[m, 3] && NeQ[p, 1])

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {-b^2+4 a c}{4 c (b d+2 c d x)^{3/2}}+\frac {\sqrt {b d+2 c d x}}{4 c d^2}\right ) \, dx \\ & = \frac {b^2-4 a c}{4 c^2 d \sqrt {b d+2 c d x}}+\frac {(b d+2 c d x)^{3/2}}{12 c^2 d^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.78 \[ \int \frac {a+b x+c x^2}{(b d+2 c d x)^{3/2}} \, dx=\frac {(b+2 c x) \left (3 b^2-12 a c+(b+2 c x)^2\right )}{12 c^2 (d (b+2 c x))^{3/2}} \]

[In]

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

[Out]

((b + 2*c*x)*(3*b^2 - 12*a*c + (b + 2*c*x)^2))/(12*c^2*(d*(b + 2*c*x))^(3/2))

Maple [A] (verified)

Time = 2.01 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.71

method	result	size
pseudoelliptic	\(\frac {c^{2} x^{2}+\left (b x -3 a \right ) c +b^{2}}{3 \sqrt {d \left (2 c x +b \right )}\, d \,c^{2}}\)	\(39\)
gosper	\(-\frac {\left (2 c x +b \right ) \left (-c^{2} x^{2}-b c x +3 a c -b^{2}\right )}{3 c^{2} \left (2 c d x +b d \right )^{\frac {3}{2}}}\)	\(46\)
derivativedivides	\(\frac {\frac {\left (2 c d x +b d \right )^{\frac {3}{2}}}{3}-\frac {d^{2} \left (4 a c -b^{2}\right )}{\sqrt {2 c d x +b d}}}{4 c^{2} d^{3}}\)	\(49\)
default	\(\frac {\frac {\left (2 c d x +b d \right )^{\frac {3}{2}}}{3}-\frac {d^{2} \left (4 a c -b^{2}\right )}{\sqrt {2 c d x +b d}}}{4 c^{2} d^{3}}\)	\(49\)
trager	\(-\frac {\left (-c^{2} x^{2}-b c x +3 a c -b^{2}\right ) \sqrt {2 c d x +b d}}{3 d^{2} c^{2} \left (2 c x +b \right )}\)	\(51\)

[In]

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

[Out]

1/3*(c^2*x^2+(b*x-3*a)*c+b^2)/(d*(2*c*x+b))^(1/2)/d/c^2

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.95 \[ \int \frac {a+b x+c x^2}{(b d+2 c d x)^{3/2}} \, dx=\frac {{\left (c^{2} x^{2} + b c x + b^{2} - 3 \, a c\right )} \sqrt {2 \, c d x + b d}}{3 \, {\left (2 \, c^{3} d^{2} x + b c^{2} d^{2}\right )}} \]

[In]

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

[Out]

1/3*(c^2*x^2 + b*c*x + b^2 - 3*a*c)*sqrt(2*c*d*x + b*d)/(2*c^3*d^2*x + b*c^2*d^2)

Sympy [A] (verification not implemented)

Time = 0.79 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.33 \[ \int \frac {a+b x+c x^2}{(b d+2 c d x)^{3/2}} \, dx=\begin {cases} \frac {- \frac {4 a c - b^{2}}{4 c \sqrt {b d + 2 c d x}} + \frac {\left (b d + 2 c d x\right )^{\frac {3}{2}}}{12 c d^{2}}}{c d} & \text {for}\: c d \neq 0 \\\frac {a x + \frac {b x^{2}}{2} + \frac {c x^{3}}{3}}{\left (b d\right )^{\frac {3}{2}}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise(((-(4*a*c - b**2)/(4*c*sqrt(b*d + 2*c*d*x)) + (b*d + 2*c*d*x)**(3/2)/(12*c*d**2))/(c*d), Ne(c*d, 0))
, ((a*x + b*x**2/2 + c*x**3/3)/(b*d)**(3/2), True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.93 \[ \int \frac {a+b x+c x^2}{(b d+2 c d x)^{3/2}} \, dx=\frac {\frac {3 \, {\left (b^{2} - 4 \, a c\right )}}{\sqrt {2 \, c d x + b d} c} + \frac {{\left (2 \, c d x + b d\right )}^{\frac {3}{2}}}{c d^{2}}}{12 \, c d} \]

[In]

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

[Out]

1/12*(3*(b^2 - 4*a*c)/(sqrt(2*c*d*x + b*d)*c) + (2*c*d*x + b*d)^(3/2)/(c*d^2))/(c*d)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.85 \[ \int \frac {a+b x+c x^2}{(b d+2 c d x)^{3/2}} \, dx=\frac {b^{2} - 4 \, a c}{4 \, \sqrt {2 \, c d x + b d} c^{2} d} + \frac {{\left (2 \, c d x + b d\right )}^{\frac {3}{2}}}{12 \, c^{2} d^{3}} \]

[In]

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

[Out]

1/4*(b^2 - 4*a*c)/(sqrt(2*c*d*x + b*d)*c^2*d) + 1/12*(2*c*d*x + b*d)^(3/2)/(c^2*d^3)

Mupad [B] (verification not implemented)

Time = 9.24 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.67 \[ \int \frac {a+b x+c x^2}{(b d+2 c d x)^{3/2}} \, dx=\frac {{\left (b+2\,c\,x\right )}^2-12\,a\,c+3\,b^2}{12\,c^2\,d\,\sqrt {b\,d+2\,c\,d\,x}} \]

[In]

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

[Out]

((b + 2*c*x)^2 - 12*a*c + 3*b^2)/(12*c^2*d*(b*d + 2*c*d*x)^(1/2))