\(\int \frac {x^3}{(a+b x)^{5/2}} \, dx\) [352]

Optimal result

Integrand size = 13, antiderivative size = 68 \[ \int \frac {x^3}{(a+b x)^{5/2}} \, dx=\frac {2 a^3}{3 b^4 (a+b x)^{3/2}}-\frac {6 a^2}{b^4 \sqrt {a+b x}}-\frac {6 a \sqrt {a+b x}}{b^4}+\frac {2 (a+b x)^{3/2}}{3 b^4} \]

[Out]

2/3*a^3/b^4/(b*x+a)^(3/2)+2/3*(b*x+a)^(3/2)/b^4-6*a^2/b^4/(b*x+a)^(1/2)-6*a*(b*x+a)^(1/2)/b^4

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 68, 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} \[ \int \frac {x^3}{(a+b x)^{5/2}} \, dx=\frac {2 a^3}{3 b^4 (a+b x)^{3/2}}-\frac {6 a^2}{b^4 \sqrt {a+b x}}-\frac {6 a \sqrt {a+b x}}{b^4}+\frac {2 (a+b x)^{3/2}}{3 b^4} \]

[In]

Int[x^3/(a + b*x)^(5/2),x]

[Out]

(2*a^3)/(3*b^4*(a + b*x)^(3/2)) - (6*a^2)/(b^4*Sqrt[a + b*x]) - (6*a*Sqrt[a + b*x])/b^4 + (2*(a + b*x)^(3/2))/
(3*b^4)

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}{b^3 (a+b x)^{5/2}}+\frac {3 a^2}{b^3 (a+b x)^{3/2}}-\frac {3 a}{b^3 \sqrt {a+b x}}+\frac {\sqrt {a+b x}}{b^3}\right ) \, dx \\ & = \frac {2 a^3}{3 b^4 (a+b x)^{3/2}}-\frac {6 a^2}{b^4 \sqrt {a+b x}}-\frac {6 a \sqrt {a+b x}}{b^4}+\frac {2 (a+b x)^{3/2}}{3 b^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.69 \[ \int \frac {x^3}{(a+b x)^{5/2}} \, dx=\frac {2 \left (a^3-9 a^2 (a+b x)-9 a (a+b x)^2+(a+b x)^3\right )}{3 b^4 (a+b x)^{3/2}} \]

[In]

Integrate[x^3/(a + b*x)^(5/2),x]

[Out]

(2*(a^3 - 9*a^2*(a + b*x) - 9*a*(a + b*x)^2 + (a + b*x)^3))/(3*b^4*(a + b*x)^(3/2))

Maple [A] (verified)

Time = 0.15 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.57

method	result	size
trager	\(-\frac {2 \left (b x +2 a \right ) \left (-b^{2} x^{2}+8 a b x +8 a^{2}\right )}{3 b^{4} \left (b x +a \right )^{\frac {3}{2}}}\)	\(39\)
pseudoelliptic	\(\frac {\frac {2}{3} b^{3} x^{3}-4 a \,b^{2} x^{2}-16 a^{2} b x -\frac {32}{3} a^{3}}{\left (b x +a \right )^{\frac {3}{2}} b^{4}}\)	\(42\)
gosper	\(-\frac {2 \left (-b^{3} x^{3}+6 a \,b^{2} x^{2}+24 a^{2} b x +16 a^{3}\right )}{3 \left (b x +a \right )^{\frac {3}{2}} b^{4}}\)	\(43\)
risch	\(-\frac {2 \left (-b x +8 a \right ) \sqrt {b x +a}}{3 b^{4}}-\frac {2 a^{2} \left (9 b x +8 a \right )}{3 b^{4} \left (b x +a \right )^{\frac {3}{2}}}\)	\(45\)
derivativedivides	\(\frac {\frac {2 \left (b x +a \right )^{\frac {3}{2}}}{3}-6 a \sqrt {b x +a}-\frac {6 a^{2}}{\sqrt {b x +a}}+\frac {2 a^{3}}{3 \left (b x +a \right )^{\frac {3}{2}}}}{b^{4}}\)	\(50\)
default	\(\frac {\frac {2 \left (b x +a \right )^{\frac {3}{2}}}{3}-6 a \sqrt {b x +a}-\frac {6 a^{2}}{\sqrt {b x +a}}+\frac {2 a^{3}}{3 \left (b x +a \right )^{\frac {3}{2}}}}{b^{4}}\)	\(50\)

[In]

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

[Out]

-2/3*(b*x+2*a)*(-b^2*x^2+8*a*b*x+8*a^2)/b^4/(b*x+a)^(3/2)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.91 \[ \int \frac {x^3}{(a+b x)^{5/2}} \, dx=\frac {2 \, {\left (b^{3} x^{3} - 6 \, a b^{2} x^{2} - 24 \, a^{2} b x - 16 \, a^{3}\right )} \sqrt {b x + a}}{3 \, {\left (b^{6} x^{2} + 2 \, a b^{5} x + a^{2} b^{4}\right )}} \]

[In]

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

[Out]

2/3*(b^3*x^3 - 6*a*b^2*x^2 - 24*a^2*b*x - 16*a^3)*sqrt(b*x + a)/(b^6*x^2 + 2*a*b^5*x + a^2*b^4)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 163 vs. \(2 (65) = 130\).

Time = 0.65 (sec) , antiderivative size = 163, normalized size of antiderivative = 2.40 \[ \int \frac {x^3}{(a+b x)^{5/2}} \, dx=\begin {cases} - \frac {32 a^{3}}{3 a b^{4} \sqrt {a + b x} + 3 b^{5} x \sqrt {a + b x}} - \frac {48 a^{2} b x}{3 a b^{4} \sqrt {a + b x} + 3 b^{5} x \sqrt {a + b x}} - \frac {12 a b^{2} x^{2}}{3 a b^{4} \sqrt {a + b x} + 3 b^{5} x \sqrt {a + b x}} + \frac {2 b^{3} x^{3}}{3 a b^{4} \sqrt {a + b x} + 3 b^{5} x \sqrt {a + b x}} & \text {for}\: b \neq 0 \\\frac {x^{4}}{4 a^{\frac {5}{2}}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((-32*a**3/(3*a*b**4*sqrt(a + b*x) + 3*b**5*x*sqrt(a + b*x)) - 48*a**2*b*x/(3*a*b**4*sqrt(a + b*x) +
3*b**5*x*sqrt(a + b*x)) - 12*a*b**2*x**2/(3*a*b**4*sqrt(a + b*x) + 3*b**5*x*sqrt(a + b*x)) + 2*b**3*x**3/(3*a*
b**4*sqrt(a + b*x) + 3*b**5*x*sqrt(a + b*x)), Ne(b, 0)), (x**4/(4*a**(5/2)), True))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.82 \[ \int \frac {x^3}{(a+b x)^{5/2}} \, dx=\frac {2 \, {\left (b x + a\right )}^{\frac {3}{2}}}{3 \, b^{4}} - \frac {6 \, \sqrt {b x + a} a}{b^{4}} - \frac {6 \, a^{2}}{\sqrt {b x + a} b^{4}} + \frac {2 \, a^{3}}{3 \, {\left (b x + a\right )}^{\frac {3}{2}} b^{4}} \]

[In]

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

[Out]

2/3*(b*x + a)^(3/2)/b^4 - 6*sqrt(b*x + a)*a/b^4 - 6*a^2/(sqrt(b*x + a)*b^4) + 2/3*a^3/((b*x + a)^(3/2)*b^4)

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.87 \[ \int \frac {x^3}{(a+b x)^{5/2}} \, dx=-\frac {2 \, {\left (9 \, {\left (b x + a\right )} a^{2} - a^{3}\right )}}{3 \, {\left (b x + a\right )}^{\frac {3}{2}} b^{4}} + \frac {2 \, {\left ({\left (b x + a\right )}^{\frac {3}{2}} b^{8} - 9 \, \sqrt {b x + a} a b^{8}\right )}}{3 \, b^{12}} \]

[In]

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

[Out]

-2/3*(9*(b*x + a)*a^2 - a^3)/((b*x + a)^(3/2)*b^4) + 2/3*((b*x + a)^(3/2)*b^8 - 9*sqrt(b*x + a)*a*b^8)/b^12

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.69 \[ \int \frac {x^3}{(a+b x)^{5/2}} \, dx=-\frac {18\,a\,{\left (a+b\,x\right )}^2+18\,a^2\,\left (a+b\,x\right )-2\,{\left (a+b\,x\right )}^3-2\,a^3}{3\,b^4\,{\left (a+b\,x\right )}^{3/2}} \]

[In]

int(x^3/(a + b*x)^(5/2),x)

[Out]

-(18*a*(a + b*x)^2 + 18*a^2*(a + b*x) - 2*(a + b*x)^3 - 2*a^3)/(3*b^4*(a + b*x)^(3/2))