\(\int \frac {(a+b x)^3}{x^{5/2}} \, dx\) [448]

Optimal result

Integrand size = 13, antiderivative size = 47 \[ \int \frac {(a+b x)^3}{x^{5/2}} \, dx=-\frac {2 a^3}{3 x^{3/2}}-\frac {6 a^2 b}{\sqrt {x}}+6 a b^2 \sqrt {x}+\frac {2}{3} b^3 x^{3/2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 47, 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} \[ \int \frac {(a+b x)^3}{x^{5/2}} \, dx=-\frac {2 a^3}{3 x^{3/2}}-\frac {6 a^2 b}{\sqrt {x}}+6 a b^2 \sqrt {x}+\frac {2}{3} b^3 x^{3/2} \]

[In]

Int[(a + b*x)^3/x^(5/2),x]

[Out]

(-2*a^3)/(3*x^(3/2)) - (6*a^2*b)/Sqrt[x] + 6*a*b^2*Sqrt[x] + (2*b^3*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^3}{x^{5/2}}+\frac {3 a^2 b}{x^{3/2}}+\frac {3 a b^2}{\sqrt {x}}+b^3 \sqrt {x}\right ) \, dx \\ & = -\frac {2 a^3}{3 x^{3/2}}-\frac {6 a^2 b}{\sqrt {x}}+6 a b^2 \sqrt {x}+\frac {2}{3} b^3 x^{3/2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.81 \[ \int \frac {(a+b x)^3}{x^{5/2}} \, dx=\frac {2 \left (-a^3-9 a^2 b x+9 a b^2 x^2+b^3 x^3\right )}{3 x^{3/2}} \]

[In]

Integrate[(a + b*x)^3/x^(5/2),x]

[Out]

(2*(-a^3 - 9*a^2*b*x + 9*a*b^2*x^2 + b^3*x^3))/(3*x^(3/2))

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.72

method	result	size
gosper	\(-\frac {2 \left (-b^{3} x^{3}-9 a \,b^{2} x^{2}+9 a^{2} b x +a^{3}\right )}{3 x^{\frac {3}{2}}}\)	\(34\)
trager	\(-\frac {2 \left (-b^{3} x^{3}-9 a \,b^{2} x^{2}+9 a^{2} b x +a^{3}\right )}{3 x^{\frac {3}{2}}}\)	\(34\)
risch	\(-\frac {2 \left (-b^{3} x^{3}-9 a \,b^{2} x^{2}+9 a^{2} b x +a^{3}\right )}{3 x^{\frac {3}{2}}}\)	\(34\)
derivativedivides	\(-\frac {2 a^{3}}{3 x^{\frac {3}{2}}}+\frac {2 b^{3} x^{\frac {3}{2}}}{3}-\frac {6 a^{2} b}{\sqrt {x}}+6 a \,b^{2} \sqrt {x}\)	\(36\)
default	\(-\frac {2 a^{3}}{3 x^{\frac {3}{2}}}+\frac {2 b^{3} x^{\frac {3}{2}}}{3}-\frac {6 a^{2} b}{\sqrt {x}}+6 a \,b^{2} \sqrt {x}\)	\(36\)

[In]

int((b*x+a)^3/x^(5/2),x,method=_RETURNVERBOSE)

[Out]

-2/3*(-b^3*x^3-9*a*b^2*x^2+9*a^2*b*x+a^3)/x^(3/2)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.72 \[ \int \frac {(a+b x)^3}{x^{5/2}} \, dx=\frac {2 \, {\left (b^{3} x^{3} + 9 \, a b^{2} x^{2} - 9 \, a^{2} b x - a^{3}\right )}}{3 \, x^{\frac {3}{2}}} \]

[In]

integrate((b*x+a)^3/x^(5/2),x, algorithm="fricas")

[Out]

2/3*(b^3*x^3 + 9*a*b^2*x^2 - 9*a^2*b*x - a^3)/x^(3/2)

Sympy [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.98 \[ \int \frac {(a+b x)^3}{x^{5/2}} \, dx=- \frac {2 a^{3}}{3 x^{\frac {3}{2}}} - \frac {6 a^{2} b}{\sqrt {x}} + 6 a b^{2} \sqrt {x} + \frac {2 b^{3} x^{\frac {3}{2}}}{3} \]

[In]

integrate((b*x+a)**3/x**(5/2),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.72 \[ \int \frac {(a+b x)^3}{x^{5/2}} \, dx=\frac {2}{3} \, b^{3} x^{\frac {3}{2}} + 6 \, a b^{2} \sqrt {x} - \frac {2 \, {\left (9 \, a^{2} b x + a^{3}\right )}}{3 \, x^{\frac {3}{2}}} \]

[In]

integrate((b*x+a)^3/x^(5/2),x, algorithm="maxima")

[Out]

2/3*b^3*x^(3/2) + 6*a*b^2*sqrt(x) - 2/3*(9*a^2*b*x + a^3)/x^(3/2)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.72 \[ \int \frac {(a+b x)^3}{x^{5/2}} \, dx=\frac {2}{3} \, b^{3} x^{\frac {3}{2}} + 6 \, a b^{2} \sqrt {x} - \frac {2 \, {\left (9 \, a^{2} b x + a^{3}\right )}}{3 \, x^{\frac {3}{2}}} \]

[In]

integrate((b*x+a)^3/x^(5/2),x, algorithm="giac")

[Out]

2/3*b^3*x^(3/2) + 6*a*b^2*sqrt(x) - 2/3*(9*a^2*b*x + a^3)/x^(3/2)

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.74 \[ \int \frac {(a+b x)^3}{x^{5/2}} \, dx=-\frac {2\,a^3+18\,a^2\,b\,x-18\,a\,b^2\,x^2-2\,b^3\,x^3}{3\,x^{3/2}} \]

[In]

int((a + b*x)^3/x^(5/2),x)

[Out]

-(2*a^3 - 2*b^3*x^3 - 18*a*b^2*x^2 + 18*a^2*b*x)/(3*x^(3/2))