3.2266 \(\int (a+b \sqrt{x})^p x \, dx\)

Optimal. Leaf size=100 \[ -\frac{2 a^3 \left (a+b \sqrt{x}\right )^{p+1}}{b^4 (p+1)}+\frac{6 a^2 \left (a+b \sqrt{x}\right )^{p+2}}{b^4 (p+2)}-\frac{6 a \left (a+b \sqrt{x}\right )^{p+3}}{b^4 (p+3)}+\frac{2 \left (a+b \sqrt{x}\right )^{p+4}}{b^4 (p+4)} \]

[Out]

(-2*a^3*(a + b*Sqrt[x])^(1 + p))/(b^4*(1 + p)) + (6*a^2*(a + b*Sqrt[x])^(2 + p))/(b^4*(2 + p)) - (6*a*(a + b*S
qrt[x])^(3 + p))/(b^4*(3 + p)) + (2*(a + b*Sqrt[x])^(4 + p))/(b^4*(4 + p))

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Rubi [A]  time = 0.0507882, antiderivative size = 100, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {266, 43} \[ -\frac{2 a^3 \left (a+b \sqrt{x}\right )^{p+1}}{b^4 (p+1)}+\frac{6 a^2 \left (a+b \sqrt{x}\right )^{p+2}}{b^4 (p+2)}-\frac{6 a \left (a+b \sqrt{x}\right )^{p+3}}{b^4 (p+3)}+\frac{2 \left (a+b \sqrt{x}\right )^{p+4}}{b^4 (p+4)} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sqrt[x])^p*x,x]

[Out]

(-2*a^3*(a + b*Sqrt[x])^(1 + p))/(b^4*(1 + p)) + (6*a^2*(a + b*Sqrt[x])^(2 + p))/(b^4*(2 + p)) - (6*a*(a + b*S
qrt[x])^(3 + p))/(b^4*(3 + p)) + (2*(a + b*Sqrt[x])^(4 + p))/(b^4*(4 + p))

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

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 \left (a+b \sqrt{x}\right )^p x \, dx &=2 \operatorname{Subst}\left (\int x^3 (a+b x)^p \, dx,x,\sqrt{x}\right )\\ &=2 \operatorname{Subst}\left (\int \left (-\frac{a^3 (a+b x)^p}{b^3}+\frac{3 a^2 (a+b x)^{1+p}}{b^3}-\frac{3 a (a+b x)^{2+p}}{b^3}+\frac{(a+b x)^{3+p}}{b^3}\right ) \, dx,x,\sqrt{x}\right )\\ &=-\frac{2 a^3 \left (a+b \sqrt{x}\right )^{1+p}}{b^4 (1+p)}+\frac{6 a^2 \left (a+b \sqrt{x}\right )^{2+p}}{b^4 (2+p)}-\frac{6 a \left (a+b \sqrt{x}\right )^{3+p}}{b^4 (3+p)}+\frac{2 \left (a+b \sqrt{x}\right )^{4+p}}{b^4 (4+p)}\\ \end{align*}

Mathematica [A]  time = 0.0542191, size = 95, normalized size = 0.95 \[ \frac{2 \left (a+b \sqrt{x}\right )^{p+1} \left (6 a^2 b (p+1) \sqrt{x}-6 a^3-3 a b^2 \left (p^2+3 p+2\right ) x+b^3 \left (p^3+6 p^2+11 p+6\right ) x^{3/2}\right )}{b^4 (p+1) (p+2) (p+3) (p+4)} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sqrt[x])^p*x,x]

[Out]

(2*(a + b*Sqrt[x])^(1 + p)*(-6*a^3 + 6*a^2*b*(1 + p)*Sqrt[x] - 3*a*b^2*(2 + 3*p + p^2)*x + b^3*(6 + 11*p + 6*p
^2 + p^3)*x^(3/2)))/(b^4*(1 + p)*(2 + p)*(3 + p)*(4 + p))

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Maple [F]  time = 0.018, size = 0, normalized size = 0. \begin{align*} \int x \left ( a+b\sqrt{x} \right ) ^{p}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.01214, size = 140, normalized size = 1.4 \begin{align*} \frac{2 \,{\left ({\left (p^{3} + 6 \, p^{2} + 11 \, p + 6\right )} b^{4} x^{2} +{\left (p^{3} + 3 \, p^{2} + 2 \, p\right )} a b^{3} x^{\frac{3}{2}} - 3 \,{\left (p^{2} + p\right )} a^{2} b^{2} x + 6 \, a^{3} b p \sqrt{x} - 6 \, a^{4}\right )}{\left (b \sqrt{x} + a\right )}^{p}}{{\left (p^{4} + 10 \, p^{3} + 35 \, p^{2} + 50 \, p + 24\right )} b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*((p^3 + 6*p^2 + 11*p + 6)*b^4*x^2 + (p^3 + 3*p^2 + 2*p)*a*b^3*x^(3/2) - 3*(p^2 + p)*a^2*b^2*x + 6*a^3*b*p*sq
rt(x) - 6*a^4)*(b*sqrt(x) + a)^p/((p^4 + 10*p^3 + 35*p^2 + 50*p + 24)*b^4)

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Fricas [A]  time = 1.48636, size = 309, normalized size = 3.09 \begin{align*} -\frac{2 \,{\left (6 \, a^{4} -{\left (b^{4} p^{3} + 6 \, b^{4} p^{2} + 11 \, b^{4} p + 6 \, b^{4}\right )} x^{2} + 3 \,{\left (a^{2} b^{2} p^{2} + a^{2} b^{2} p\right )} x -{\left (6 \, a^{3} b p +{\left (a b^{3} p^{3} + 3 \, a b^{3} p^{2} + 2 \, a b^{3} p\right )} x\right )} \sqrt{x}\right )}{\left (b \sqrt{x} + a\right )}^{p}}{b^{4} p^{4} + 10 \, b^{4} p^{3} + 35 \, b^{4} p^{2} + 50 \, b^{4} p + 24 \, b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(6*a^4 - (b^4*p^3 + 6*b^4*p^2 + 11*b^4*p + 6*b^4)*x^2 + 3*(a^2*b^2*p^2 + a^2*b^2*p)*x - (6*a^3*b*p + (a*b^3
*p^3 + 3*a*b^3*p^2 + 2*a*b^3*p)*x)*sqrt(x))*(b*sqrt(x) + a)^p/(b^4*p^4 + 10*b^4*p^3 + 35*b^4*p^2 + 50*b^4*p +
24*b^4)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [B]  time = 1.13545, size = 554, normalized size = 5.54 \begin{align*} \frac{2 \,{\left ({\left (b \sqrt{x} + a\right )}^{4}{\left (b \sqrt{x} + a\right )}^{p} p^{3} - 3 \,{\left (b \sqrt{x} + a\right )}^{3}{\left (b \sqrt{x} + a\right )}^{p} a p^{3} + 3 \,{\left (b \sqrt{x} + a\right )}^{2}{\left (b \sqrt{x} + a\right )}^{p} a^{2} p^{3} -{\left (b \sqrt{x} + a\right )}{\left (b \sqrt{x} + a\right )}^{p} a^{3} p^{3} + 6 \,{\left (b \sqrt{x} + a\right )}^{4}{\left (b \sqrt{x} + a\right )}^{p} p^{2} - 21 \,{\left (b \sqrt{x} + a\right )}^{3}{\left (b \sqrt{x} + a\right )}^{p} a p^{2} + 24 \,{\left (b \sqrt{x} + a\right )}^{2}{\left (b \sqrt{x} + a\right )}^{p} a^{2} p^{2} - 9 \,{\left (b \sqrt{x} + a\right )}{\left (b \sqrt{x} + a\right )}^{p} a^{3} p^{2} + 11 \,{\left (b \sqrt{x} + a\right )}^{4}{\left (b \sqrt{x} + a\right )}^{p} p - 42 \,{\left (b \sqrt{x} + a\right )}^{3}{\left (b \sqrt{x} + a\right )}^{p} a p + 57 \,{\left (b \sqrt{x} + a\right )}^{2}{\left (b \sqrt{x} + a\right )}^{p} a^{2} p - 26 \,{\left (b \sqrt{x} + a\right )}{\left (b \sqrt{x} + a\right )}^{p} a^{3} p + 6 \,{\left (b \sqrt{x} + a\right )}^{4}{\left (b \sqrt{x} + a\right )}^{p} - 24 \,{\left (b \sqrt{x} + a\right )}^{3}{\left (b \sqrt{x} + a\right )}^{p} a + 36 \,{\left (b \sqrt{x} + a\right )}^{2}{\left (b \sqrt{x} + a\right )}^{p} a^{2} - 24 \,{\left (b \sqrt{x} + a\right )}{\left (b \sqrt{x} + a\right )}^{p} a^{3}\right )}}{{\left (b^{2} p^{4} + 10 \, b^{2} p^{3} + 35 \, b^{2} p^{2} + 50 \, b^{2} p + 24 \, b^{2}\right )} b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*((b*sqrt(x) + a)^4*(b*sqrt(x) + a)^p*p^3 - 3*(b*sqrt(x) + a)^3*(b*sqrt(x) + a)^p*a*p^3 + 3*(b*sqrt(x) + a)^2
*(b*sqrt(x) + a)^p*a^2*p^3 - (b*sqrt(x) + a)*(b*sqrt(x) + a)^p*a^3*p^3 + 6*(b*sqrt(x) + a)^4*(b*sqrt(x) + a)^p
*p^2 - 21*(b*sqrt(x) + a)^3*(b*sqrt(x) + a)^p*a*p^2 + 24*(b*sqrt(x) + a)^2*(b*sqrt(x) + a)^p*a^2*p^2 - 9*(b*sq
rt(x) + a)*(b*sqrt(x) + a)^p*a^3*p^2 + 11*(b*sqrt(x) + a)^4*(b*sqrt(x) + a)^p*p - 42*(b*sqrt(x) + a)^3*(b*sqrt
(x) + a)^p*a*p + 57*(b*sqrt(x) + a)^2*(b*sqrt(x) + a)^p*a^2*p - 26*(b*sqrt(x) + a)*(b*sqrt(x) + a)^p*a^3*p + 6
*(b*sqrt(x) + a)^4*(b*sqrt(x) + a)^p - 24*(b*sqrt(x) + a)^3*(b*sqrt(x) + a)^p*a + 36*(b*sqrt(x) + a)^2*(b*sqrt
(x) + a)^p*a^2 - 24*(b*sqrt(x) + a)*(b*sqrt(x) + a)^p*a^3)/((b^2*p^4 + 10*b^2*p^3 + 35*b^2*p^2 + 50*b^2*p + 24
*b^2)*b^2)