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

Optimal. Leaf size=34 \[ \frac{2 (a+b x)^{7/2}}{7 b^2}-\frac{2 a (a+b x)^{5/2}}{5 b^2} \]

[Out]

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

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Rubi [A]  time = 0.0077795, antiderivative size = 34, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.091, Rules used = {43} \[ \frac{2 (a+b x)^{7/2}}{7 b^2}-\frac{2 a (a+b x)^{5/2}}{5 b^2} \]

Antiderivative was successfully verified.

[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 43

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*} \int x (a+b x)^{3/2} \, dx &=\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]  time = 0.0248003, size = 24, normalized size = 0.71 \[ \frac{2 (a+b x)^{5/2} (5 b x-2 a)}{35 b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x*(a + b*x)^(3/2),x]

[Out]

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

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Maple [A]  time = 0.003, size = 21, normalized size = 0.6 \begin{align*} -{\frac{-10\,bx+4\,a}{35\,{b}^{2}} \left ( bx+a \right ) ^{{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.0367, size = 35, normalized size = 1.03 \begin{align*} \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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Fricas [A]  time = 1.55714, size = 92, normalized size = 2.71 \begin{align*} \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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Sympy [A]  time = 0.843876, size = 80, normalized size = 2.35 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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Giac [B]  time = 1.17375, size = 92, normalized size = 2.71 \begin{align*} \frac{2 \,{\left (\frac{7 \,{\left (3 \,{\left (b x + a\right )}^{\frac{5}{2}} - 5 \,{\left (b x + a\right )}^{\frac{3}{2}} a\right )} a}{b} + \frac{15 \,{\left (b x + a\right )}^{\frac{7}{2}} - 42 \,{\left (b x + a\right )}^{\frac{5}{2}} a + 35 \,{\left (b x + a\right )}^{\frac{3}{2}} a^{2}}{b}\right )}}{105 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/105*(7*(3*(b*x + a)^(5/2) - 5*(b*x + a)^(3/2)*a)*a/b + (15*(b*x + a)^(7/2) - 42*(b*x + a)^(5/2)*a + 35*(b*x
+ a)^(3/2)*a^2)/b)/b