\(\int x (a+b x)^{3/2} \, dx\) [294]

Optimal result

Integrand size = 11, antiderivative size = 34 \[ \int x (a+b x)^{3/2} \, dx=-\frac {2 a (a+b x)^{5/2}}{5 b^2}+\frac {2 (a+b x)^{7/2}}{7 b^2} \]

[Out]

-2/5*a*(b*x+a)^(5/2)/b^2+2/7*(b*x+a)^(7/2)/b^2

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {45} \[ \int x (a+b x)^{3/2} \, dx=\frac {2 (a+b x)^{7/2}}{7 b^2}-\frac {2 a (a+b x)^{5/2}}{5 b^2} \]

[In]

Int[x*(a + b*x)^(3/2),x]

[Out]

(-2*a*(a + b*x)^(5/2))/(5*b^2) + (2*(a + b*x)^(7/2))/(7*b^2)

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 (a+b x)^{3/2}}{b}+\frac {(a+b x)^{5/2}}{b}\right ) \, dx \\ & = -\frac {2 a (a+b x)^{5/2}}{5 b^2}+\frac {2 (a+b x)^{7/2}}{7 b^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.71 \[ \int x (a+b x)^{3/2} \, dx=\frac {2 (a+b x)^{5/2} (-2 a+5 b x)}{35 b^2} \]

[In]

Integrate[x*(a + b*x)^(3/2),x]

[Out]

(2*(a + b*x)^(5/2)*(-2*a + 5*b*x))/(35*b^2)

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.62

method	result	size
gosper	\(-\frac {2 \left (b x +a \right )^{\frac {5}{2}} \left (-5 b x +2 a \right )}{35 b^{2}}\)	\(21\)
pseudoelliptic	\(-\frac {2 \left (b x +a \right )^{\frac {5}{2}} \left (-5 b x +2 a \right )}{35 b^{2}}\)	\(21\)
derivativedivides	\(\frac {\frac {2 \left (b x +a \right )^{\frac {7}{2}}}{7}-\frac {2 a \left (b x +a \right )^{\frac {5}{2}}}{5}}{b^{2}}\)	\(26\)
default	\(\frac {\frac {2 \left (b x +a \right )^{\frac {7}{2}}}{7}-\frac {2 a \left (b x +a \right )^{\frac {5}{2}}}{5}}{b^{2}}\)	\(26\)
trager	\(-\frac {2 \left (-5 b^{3} x^{3}-8 a \,b^{2} x^{2}-a^{2} b x +2 a^{3}\right ) \sqrt {b x +a}}{35 b^{2}}\)	\(43\)
risch	\(-\frac {2 \left (-5 b^{3} x^{3}-8 a \,b^{2} x^{2}-a^{2} b x +2 a^{3}\right ) \sqrt {b x +a}}{35 b^{2}}\)	\(43\)

[In]

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

[Out]

-2/35*(b*x+a)^(5/2)*(-5*b*x+2*a)/b^2

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.21 \[ \int x (a+b x)^{3/2} \, dx=\frac {2 \, {\left (5 \, b^{3} x^{3} + 8 \, a b^{2} x^{2} + a^{2} b x - 2 \, a^{3}\right )} \sqrt {b x + a}}{35 \, b^{2}} \]

[In]

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

[Out]

2/35*(5*b^3*x^3 + 8*a*b^2*x^2 + a^2*b*x - 2*a^3)*sqrt(b*x + a)/b^2

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 80 vs. \(2 (31) = 62\).

Time = 0.20 (sec) , antiderivative size = 80, normalized size of antiderivative = 2.35 \[ \int x (a+b x)^{3/2} \, dx=\begin {cases} - \frac {4 a^{3} \sqrt {a + b x}}{35 b^{2}} + \frac {2 a^{2} x \sqrt {a + b x}}{35 b} + \frac {16 a x^{2} \sqrt {a + b x}}{35} + \frac {2 b x^{3} \sqrt {a + b x}}{7} & \text {for}\: b \neq 0 \\\frac {a^{\frac {3}{2}} x^{2}}{2} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((-4*a**3*sqrt(a + b*x)/(35*b**2) + 2*a**2*x*sqrt(a + b*x)/(35*b) + 16*a*x**2*sqrt(a + b*x)/35 + 2*b*
x**3*sqrt(a + b*x)/7, Ne(b, 0)), (a**(3/2)*x**2/2, True))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.76 \[ \int x (a+b x)^{3/2} \, dx=\frac {2 \, {\left (b x + a\right )}^{\frac {7}{2}}}{7 \, b^{2}} - \frac {2 \, {\left (b x + a\right )}^{\frac {5}{2}} a}{5 \, b^{2}} \]

[In]

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

[Out]

2/7*(b*x + a)^(7/2)/b^2 - 2/5*(b*x + a)^(5/2)*a/b^2

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 119 vs. \(2 (26) = 52\).

Time = 0.29 (sec) , antiderivative size = 119, normalized size of antiderivative = 3.50 \[ \int x (a+b x)^{3/2} \, dx=\frac {2 \, {\left (\frac {35 \, {\left ({\left (b x + a\right )}^{\frac {3}{2}} - 3 \, \sqrt {b x + a} a\right )} a^{2}}{b} + \frac {14 \, {\left (3 \, {\left (b x + a\right )}^{\frac {5}{2}} - 10 \, {\left (b x + a\right )}^{\frac {3}{2}} a + 15 \, \sqrt {b x + a} a^{2}\right )} a}{b} + \frac {3 \, {\left (5 \, {\left (b x + a\right )}^{\frac {7}{2}} - 21 \, {\left (b x + a\right )}^{\frac {5}{2}} a + 35 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{2} - 35 \, \sqrt {b x + a} a^{3}\right )}}{b}\right )}}{105 \, b} \]

[In]

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

[Out]

2/105*(35*((b*x + a)^(3/2) - 3*sqrt(b*x + a)*a)*a^2/b + 14*(3*(b*x + a)^(5/2) - 10*(b*x + a)^(3/2)*a + 15*sqrt
(b*x + a)*a^2)*a/b + 3*(5*(b*x + a)^(7/2) - 21*(b*x + a)^(5/2)*a + 35*(b*x + a)^(3/2)*a^2 - 35*sqrt(b*x + a)*a
^3)/b)/b

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.74 \[ \int x (a+b x)^{3/2} \, dx=-\frac {14\,a\,{\left (a+b\,x\right )}^{5/2}-10\,{\left (a+b\,x\right )}^{7/2}}{35\,b^2} \]

[In]

int(x*(a + b*x)^(3/2),x)

[Out]

-(14*a*(a + b*x)^(5/2) - 10*(a + b*x)^(7/2))/(35*b^2)