3.188 \(\int x (a+b x)^{p/2} \, dx\)

Optimal. Leaf size=48 \[ \frac{2 (a+b x)^{\frac{p+4}{2}}}{b^2 (p+4)}-\frac{2 a (a+b x)^{\frac{p+2}{2}}}{b^2 (p+2)} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0142299, antiderivative size = 48, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.077, Rules used = {43} \[ \frac{2 (a+b x)^{\frac{p+4}{2}}}{b^2 (p+4)}-\frac{2 a (a+b x)^{\frac{p+2}{2}}}{b^2 (p+2)} \]

Antiderivative was successfully verified.

[In]

Int[x*(a + b*x)^(p/2),x]

[Out]

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

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)^{p/2} \, dx &=\int \left (\frac{(a+b x)^{1+\frac{p}{2}}}{b}-\frac{a (a+b x)^{p/2}}{b}\right ) \, dx\\ &=-\frac{2 a (a+b x)^{\frac{2+p}{2}}}{b^2 (2+p)}+\frac{2 (a+b x)^{\frac{4+p}{2}}}{b^2 (4+p)}\\ \end{align*}

Mathematica [A]  time = 0.0199537, size = 38, normalized size = 0.79 \[ \frac{2 (a+b x)^{\frac{p}{2}+1} (b (p+2) x-2 a)}{b^2 (p+2) (p+4)} \]

Antiderivative was successfully verified.

[In]

Integrate[x*(a + b*x)^(p/2),x]

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.002, size = 43, normalized size = 0.9 \begin{align*} -2\,{\frac{ \left ( \sqrt{bx+a} \right ) ^{p} \left ( -xpb-2\,bx+2\,a \right ) \left ( bx+a \right ) }{{b}^{2} \left ({p}^{2}+6\,p+8 \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*((b*x+a)^(1/2))^p,x)

[Out]

-2*((b*x+a)^(1/2))^p*(-b*p*x-2*b*x+2*a)*(b*x+a)/b^2/(p^2+6*p+8)

________________________________________________________________________________________

Maxima [A]  time = 0.967557, size = 61, normalized size = 1.27 \begin{align*} \frac{2 \,{\left (b^{2}{\left (p + 2\right )} x^{2} + a b p x - 2 \, a^{2}\right )}{\left (b x + a\right )}^{\frac{1}{2} \, p}}{{\left (p^{2} + 6 \, p + 8\right )} b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*((b*x+a)^(1/2))^p,x, algorithm="maxima")

[Out]

2*(b^2*(p + 2)*x^2 + a*b*p*x - 2*a^2)*(b*x + a)^(1/2*p)/((p^2 + 6*p + 8)*b^2)

________________________________________________________________________________________

Fricas [A]  time = 1.81903, size = 117, normalized size = 2.44 \begin{align*} \frac{2 \,{\left (a b p x +{\left (b^{2} p + 2 \, b^{2}\right )} x^{2} - 2 \, a^{2}\right )} \sqrt{b x + a}^{p}}{b^{2} p^{2} + 6 \, b^{2} p + 8 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*((b*x+a)^(1/2))^p,x, algorithm="fricas")

[Out]

2*(a*b*p*x + (b^2*p + 2*b^2)*x^2 - 2*a^2)*sqrt(b*x + a)^p/(b^2*p^2 + 6*b^2*p + 8*b^2)

________________________________________________________________________________________

Sympy [A]  time = 0.654637, size = 216, normalized size = 4.5 \begin{align*} \begin{cases} \frac{a^{\frac{p}{2}} x^{2}}{2} & \text{for}\: b = 0 \\\frac{a \log{\left (\frac{a}{b} + x \right )}}{a b^{2} + b^{3} x} + \frac{a}{a b^{2} + b^{3} x} + \frac{b x \log{\left (\frac{a}{b} + x \right )}}{a b^{2} + b^{3} x} & \text{for}\: p = -4 \\- \frac{a \log{\left (\frac{a}{b} + x \right )}}{b^{2}} + \frac{x}{b} & \text{for}\: p = -2 \\- \frac{4 a^{2} \left (a + b x\right )^{\frac{p}{2}}}{b^{2} p^{2} + 6 b^{2} p + 8 b^{2}} + \frac{2 a b p x \left (a + b x\right )^{\frac{p}{2}}}{b^{2} p^{2} + 6 b^{2} p + 8 b^{2}} + \frac{2 b^{2} p x^{2} \left (a + b x\right )^{\frac{p}{2}}}{b^{2} p^{2} + 6 b^{2} p + 8 b^{2}} + \frac{4 b^{2} x^{2} \left (a + b x\right )^{\frac{p}{2}}}{b^{2} p^{2} + 6 b^{2} p + 8 b^{2}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*((b*x+a)**(1/2))**p,x)

[Out]

Piecewise((a**(p/2)*x**2/2, Eq(b, 0)), (a*log(a/b + x)/(a*b**2 + b**3*x) + a/(a*b**2 + b**3*x) + b*x*log(a/b +
x)/(a*b**2 + b**3*x), Eq(p, -4)), (-a*log(a/b + x)/b**2 + x/b, Eq(p, -2)), (-4*a**2*(a + b*x)**(p/2)/(b**2*p*
*2 + 6*b**2*p + 8*b**2) + 2*a*b*p*x*(a + b*x)**(p/2)/(b**2*p**2 + 6*b**2*p + 8*b**2) + 2*b**2*p*x**2*(a + b*x)
**(p/2)/(b**2*p**2 + 6*b**2*p + 8*b**2) + 4*b**2*x**2*(a + b*x)**(p/2)/(b**2*p**2 + 6*b**2*p + 8*b**2), True))

________________________________________________________________________________________

Giac [A]  time = 1.06266, size = 116, normalized size = 2.42 \begin{align*} \frac{2 \,{\left ({\left (b x + a\right )}^{\frac{1}{2} \, p} b^{2} p x^{2} +{\left (b x + a\right )}^{\frac{1}{2} \, p} a b p x + 2 \,{\left (b x + a\right )}^{\frac{1}{2} \, p} b^{2} x^{2} - 2 \,{\left (b x + a\right )}^{\frac{1}{2} \, p} a^{2}\right )}}{b^{2} p^{2} + 6 \, b^{2} p + 8 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*((b*x+a)^(1/2))^p,x, algorithm="giac")

[Out]

2*((b*x + a)^(1/2*p)*b^2*p*x^2 + (b*x + a)^(1/2*p)*a*b*p*x + 2*(b*x + a)^(1/2*p)*b^2*x^2 - 2*(b*x + a)^(1/2*p)
*a^2)/(b^2*p^2 + 6*b^2*p + 8*b^2)