\(\int \frac {(a+b x)^2}{x^{3/2}} \, dx\) [440]

Optimal result

Integrand size = 13, antiderivative size = 32 \[ \int \frac {(a+b x)^2}{x^{3/2}} \, dx=-\frac {2 a^2}{\sqrt {x}}+4 a b \sqrt {x}+\frac {2}{3} b^2 x^{3/2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {45} \[ \int \frac {(a+b x)^2}{x^{3/2}} \, dx=-\frac {2 a^2}{\sqrt {x}}+4 a b \sqrt {x}+\frac {2}{3} b^2 x^{3/2} \]

[In]

Int[(a + b*x)^2/x^(3/2),x]

[Out]

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

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.88 \[ \int \frac {(a+b x)^2}{x^{3/2}} \, dx=-\frac {2 \left (3 a^2-6 a b x-b^2 x^2\right )}{3 \sqrt {x}} \]

[In]

Integrate[(a + b*x)^2/x^(3/2),x]

[Out]

(-2*(3*a^2 - 6*a*b*x - b^2*x^2))/(3*Sqrt[x])

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.78

method	result	size
gosper	\(-\frac {2 \left (-b^{2} x^{2}-6 a b x +3 a^{2}\right )}{3 \sqrt {x}}\)	\(25\)
derivativedivides	\(\frac {2 b^{2} x^{\frac {3}{2}}}{3}-\frac {2 a^{2}}{\sqrt {x}}+4 a b \sqrt {x}\)	\(25\)
default	\(\frac {2 b^{2} x^{\frac {3}{2}}}{3}-\frac {2 a^{2}}{\sqrt {x}}+4 a b \sqrt {x}\)	\(25\)
trager	\(-\frac {2 \left (-b^{2} x^{2}-6 a b x +3 a^{2}\right )}{3 \sqrt {x}}\)	\(25\)
risch	\(-\frac {2 \left (-b^{2} x^{2}-6 a b x +3 a^{2}\right )}{3 \sqrt {x}}\)	\(25\)

[In]

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

[Out]

-2/3*(-b^2*x^2-6*a*b*x+3*a^2)/x^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.72 \[ \int \frac {(a+b x)^2}{x^{3/2}} \, dx=\frac {2 \, {\left (b^{2} x^{2} + 6 \, a b x - 3 \, a^{2}\right )}}{3 \, \sqrt {x}} \]

[In]

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

[Out]

2/3*(b^2*x^2 + 6*a*b*x - 3*a^2)/sqrt(x)

Sympy [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.97 \[ \int \frac {(a+b x)^2}{x^{3/2}} \, dx=- \frac {2 a^{2}}{\sqrt {x}} + 4 a b \sqrt {x} + \frac {2 b^{2} x^{\frac {3}{2}}}{3} \]

[In]

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

[Out]

-2*a**2/sqrt(x) + 4*a*b*sqrt(x) + 2*b**2*x**(3/2)/3

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.75 \[ \int \frac {(a+b x)^2}{x^{3/2}} \, dx=\frac {2}{3} \, b^{2} x^{\frac {3}{2}} + 4 \, a b \sqrt {x} - \frac {2 \, a^{2}}{\sqrt {x}} \]

[In]

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

[Out]

2/3*b^2*x^(3/2) + 4*a*b*sqrt(x) - 2*a^2/sqrt(x)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.75 \[ \int \frac {(a+b x)^2}{x^{3/2}} \, dx=\frac {2}{3} \, b^{2} x^{\frac {3}{2}} + 4 \, a b \sqrt {x} - \frac {2 \, a^{2}}{\sqrt {x}} \]

[In]

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

[Out]

2/3*b^2*x^(3/2) + 4*a*b*sqrt(x) - 2*a^2/sqrt(x)

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.75 \[ \int \frac {(a+b x)^2}{x^{3/2}} \, dx=\frac {-6\,a^2+12\,a\,b\,x+2\,b^2\,x^2}{3\,\sqrt {x}} \]

[In]

int((a + b*x)^2/x^(3/2),x)

[Out]

(2*b^2*x^2 - 6*a^2 + 12*a*b*x)/(3*x^(1/2))