\(\int (a+b \sqrt {x})^p x^2 \, dx\) [2265]

Optimal result

Integrand size = 15, antiderivative size = 152 \[ \int \left (a+b \sqrt {x}\right )^p x^2 \, dx=-\frac {2 a^5 \left (a+b \sqrt {x}\right )^{1+p}}{b^6 (1+p)}+\frac {10 a^4 \left (a+b \sqrt {x}\right )^{2+p}}{b^6 (2+p)}-\frac {20 a^3 \left (a+b \sqrt {x}\right )^{3+p}}{b^6 (3+p)}+\frac {20 a^2 \left (a+b \sqrt {x}\right )^{4+p}}{b^6 (4+p)}-\frac {10 a \left (a+b \sqrt {x}\right )^{5+p}}{b^6 (5+p)}+\frac {2 \left (a+b \sqrt {x}\right )^{6+p}}{b^6 (6+p)} \]

[Out]

-2*a^5*(a+b*x^(1/2))^(p+1)/b^6/(p+1)+10*a^4*(a+b*x^(1/2))^(2+p)/b^6/(2+p)-20*a^3*(a+b*x^(1/2))^(3+p)/b^6/(3+p)
+20*a^2*(a+b*x^(1/2))^(4+p)/b^6/(4+p)-10*a*(a+b*x^(1/2))^(5+p)/b^6/(5+p)+2*(a+b*x^(1/2))^(6+p)/b^6/(6+p)

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {272, 45} \[ \int \left (a+b \sqrt {x}\right )^p x^2 \, dx=-\frac {2 a^5 \left (a+b \sqrt {x}\right )^{p+1}}{b^6 (p+1)}+\frac {10 a^4 \left (a+b \sqrt {x}\right )^{p+2}}{b^6 (p+2)}-\frac {20 a^3 \left (a+b \sqrt {x}\right )^{p+3}}{b^6 (p+3)}+\frac {20 a^2 \left (a+b \sqrt {x}\right )^{p+4}}{b^6 (p+4)}-\frac {10 a \left (a+b \sqrt {x}\right )^{p+5}}{b^6 (p+5)}+\frac {2 \left (a+b \sqrt {x}\right )^{p+6}}{b^6 (p+6)} \]

[In]

Int[(a + b*Sqrt[x])^p*x^2,x]

[Out]

(-2*a^5*(a + b*Sqrt[x])^(1 + p))/(b^6*(1 + p)) + (10*a^4*(a + b*Sqrt[x])^(2 + p))/(b^6*(2 + p)) - (20*a^3*(a +
b*Sqrt[x])^(3 + p))/(b^6*(3 + p)) + (20*a^2*(a + b*Sqrt[x])^(4 + p))/(b^6*(4 + p)) - (10*a*(a + b*Sqrt[x])^(5
+ p))/(b^6*(5 + p)) + (2*(a + b*Sqrt[x])^(6 + p))/(b^6*(6 + p))

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])

Rule 272

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]]

Rubi steps \begin{align*} \text {integral}& = 2 \text {Subst}\left (\int x^5 (a+b x)^p \, dx,x,\sqrt {x}\right ) \\ & = 2 \text {Subst}\left (\int \left (-\frac {a^5 (a+b x)^p}{b^5}+\frac {5 a^4 (a+b x)^{1+p}}{b^5}-\frac {10 a^3 (a+b x)^{2+p}}{b^5}+\frac {10 a^2 (a+b x)^{3+p}}{b^5}-\frac {5 a (a+b x)^{4+p}}{b^5}+\frac {(a+b x)^{5+p}}{b^5}\right ) \, dx,x,\sqrt {x}\right ) \\ & = -\frac {2 a^5 \left (a+b \sqrt {x}\right )^{1+p}}{b^6 (1+p)}+\frac {10 a^4 \left (a+b \sqrt {x}\right )^{2+p}}{b^6 (2+p)}-\frac {20 a^3 \left (a+b \sqrt {x}\right )^{3+p}}{b^6 (3+p)}+\frac {20 a^2 \left (a+b \sqrt {x}\right )^{4+p}}{b^6 (4+p)}-\frac {10 a \left (a+b \sqrt {x}\right )^{5+p}}{b^6 (5+p)}+\frac {2 \left (a+b \sqrt {x}\right )^{6+p}}{b^6 (6+p)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.83 \[ \int \left (a+b \sqrt {x}\right )^p x^2 \, dx=\frac {2 \left (-\frac {a^5}{1+p}+\frac {5 a^4 \left (a+b \sqrt {x}\right )}{2+p}-\frac {10 a^3 \left (a+b \sqrt {x}\right )^2}{3+p}+\frac {10 a^2 \left (a+b \sqrt {x}\right )^3}{4+p}-\frac {5 a \left (a+b \sqrt {x}\right )^4}{5+p}+\frac {\left (a+b \sqrt {x}\right )^5}{6+p}\right ) \left (a+b \sqrt {x}\right )^{1+p}}{b^6} \]

[In]

Integrate[(a + b*Sqrt[x])^p*x^2,x]

[Out]

(2*(-(a^5/(1 + p)) + (5*a^4*(a + b*Sqrt[x]))/(2 + p) - (10*a^3*(a + b*Sqrt[x])^2)/(3 + p) + (10*a^2*(a + b*Sqr
t[x])^3)/(4 + p) - (5*a*(a + b*Sqrt[x])^4)/(5 + p) + (a + b*Sqrt[x])^5/(6 + p))*(a + b*Sqrt[x])^(1 + p))/b^6

Maple [F]

\[\int x^{2} \left (a +b \sqrt {x}\right )^{p}d x\]

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 281 vs. \(2 (140) = 280\).

Time = 0.31 (sec) , antiderivative size = 281, normalized size of antiderivative = 1.85 \[ \int \left (a+b \sqrt {x}\right )^p x^2 \, dx=-\frac {2 \, {\left (120 \, a^{6} - {\left (b^{6} p^{5} + 15 \, b^{6} p^{4} + 85 \, b^{6} p^{3} + 225 \, b^{6} p^{2} + 274 \, b^{6} p + 120 \, b^{6}\right )} x^{3} + 5 \, {\left (a^{2} b^{4} p^{4} + 6 \, a^{2} b^{4} p^{3} + 11 \, a^{2} b^{4} p^{2} + 6 \, a^{2} b^{4} p\right )} x^{2} + 60 \, {\left (a^{4} b^{2} p^{2} + a^{4} b^{2} p\right )} x - {\left (120 \, a^{5} b p + {\left (a b^{5} p^{5} + 10 \, a b^{5} p^{4} + 35 \, a b^{5} p^{3} + 50 \, a b^{5} p^{2} + 24 \, a b^{5} p\right )} x^{2} + 20 \, {\left (a^{3} b^{3} p^{3} + 3 \, a^{3} b^{3} p^{2} + 2 \, a^{3} b^{3} p\right )} x\right )} \sqrt {x}\right )} {\left (b \sqrt {x} + a\right )}^{p}}{b^{6} p^{6} + 21 \, b^{6} p^{5} + 175 \, b^{6} p^{4} + 735 \, b^{6} p^{3} + 1624 \, b^{6} p^{2} + 1764 \, b^{6} p + 720 \, b^{6}} \]

[In]

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

[Out]

-2*(120*a^6 - (b^6*p^5 + 15*b^6*p^4 + 85*b^6*p^3 + 225*b^6*p^2 + 274*b^6*p + 120*b^6)*x^3 + 5*(a^2*b^4*p^4 + 6
*a^2*b^4*p^3 + 11*a^2*b^4*p^2 + 6*a^2*b^4*p)*x^2 + 60*(a^4*b^2*p^2 + a^4*b^2*p)*x - (120*a^5*b*p + (a*b^5*p^5
+ 10*a*b^5*p^4 + 35*a*b^5*p^3 + 50*a*b^5*p^2 + 24*a*b^5*p)*x^2 + 20*(a^3*b^3*p^3 + 3*a^3*b^3*p^2 + 2*a^3*b^3*p
)*x)*sqrt(x))*(b*sqrt(x) + a)^p/(b^6*p^6 + 21*b^6*p^5 + 175*b^6*p^4 + 735*b^6*p^3 + 1624*b^6*p^2 + 1764*b^6*p
+ 720*b^6)

Sympy [F(-1)]

Timed out. \[ \int \left (a+b \sqrt {x}\right )^p x^2 \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.22 \[ \int \left (a+b \sqrt {x}\right )^p x^2 \, dx=\frac {2 \, {\left ({\left (p^{5} + 15 \, p^{4} + 85 \, p^{3} + 225 \, p^{2} + 274 \, p + 120\right )} b^{6} x^{3} + {\left (p^{5} + 10 \, p^{4} + 35 \, p^{3} + 50 \, p^{2} + 24 \, p\right )} a b^{5} x^{\frac {5}{2}} - 5 \, {\left (p^{4} + 6 \, p^{3} + 11 \, p^{2} + 6 \, p\right )} a^{2} b^{4} x^{2} + 20 \, {\left (p^{3} + 3 \, p^{2} + 2 \, p\right )} a^{3} b^{3} x^{\frac {3}{2}} - 60 \, {\left (p^{2} + p\right )} a^{4} b^{2} x + 120 \, a^{5} b p \sqrt {x} - 120 \, a^{6}\right )} {\left (b \sqrt {x} + a\right )}^{p}}{{\left (p^{6} + 21 \, p^{5} + 175 \, p^{4} + 735 \, p^{3} + 1624 \, p^{2} + 1764 \, p + 720\right )} b^{6}} \]

[In]

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

[Out]

2*((p^5 + 15*p^4 + 85*p^3 + 225*p^2 + 274*p + 120)*b^6*x^3 + (p^5 + 10*p^4 + 35*p^3 + 50*p^2 + 24*p)*a*b^5*x^(
5/2) - 5*(p^4 + 6*p^3 + 11*p^2 + 6*p)*a^2*b^4*x^2 + 20*(p^3 + 3*p^2 + 2*p)*a^3*b^3*x^(3/2) - 60*(p^2 + p)*a^4*
b^2*x + 120*a^5*b*p*sqrt(x) - 120*a^6)*(b*sqrt(x) + a)^p/((p^6 + 21*p^5 + 175*p^4 + 735*p^3 + 1624*p^2 + 1764*
p + 720)*b^6)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 922 vs. \(2 (140) = 280\).

Time = 0.28 (sec) , antiderivative size = 922, normalized size of antiderivative = 6.07 \[ \int \left (a+b \sqrt {x}\right )^p x^2 \, dx=\text {Too large to display} \]

[In]

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

[Out]

2*((b*sqrt(x) + a)^6*(b*sqrt(x) + a)^p*p^5 - 5*(b*sqrt(x) + a)^5*(b*sqrt(x) + a)^p*a*p^5 + 10*(b*sqrt(x) + a)^
4*(b*sqrt(x) + a)^p*a^2*p^5 - 10*(b*sqrt(x) + a)^3*(b*sqrt(x) + a)^p*a^3*p^5 + 5*(b*sqrt(x) + a)^2*(b*sqrt(x)
+ a)^p*a^4*p^5 - (b*sqrt(x) + a)*(b*sqrt(x) + a)^p*a^5*p^5 + 15*(b*sqrt(x) + a)^6*(b*sqrt(x) + a)^p*p^4 - 80*(
b*sqrt(x) + a)^5*(b*sqrt(x) + a)^p*a*p^4 + 170*(b*sqrt(x) + a)^4*(b*sqrt(x) + a)^p*a^2*p^4 - 180*(b*sqrt(x) +
a)^3*(b*sqrt(x) + a)^p*a^3*p^4 + 95*(b*sqrt(x) + a)^2*(b*sqrt(x) + a)^p*a^4*p^4 - 20*(b*sqrt(x) + a)*(b*sqrt(x
) + a)^p*a^5*p^4 + 85*(b*sqrt(x) + a)^6*(b*sqrt(x) + a)^p*p^3 - 475*(b*sqrt(x) + a)^5*(b*sqrt(x) + a)^p*a*p^3
+ 1070*(b*sqrt(x) + a)^4*(b*sqrt(x) + a)^p*a^2*p^3 - 1210*(b*sqrt(x) + a)^3*(b*sqrt(x) + a)^p*a^3*p^3 + 685*(b
*sqrt(x) + a)^2*(b*sqrt(x) + a)^p*a^4*p^3 - 155*(b*sqrt(x) + a)*(b*sqrt(x) + a)^p*a^5*p^3 + 225*(b*sqrt(x) + a
)^6*(b*sqrt(x) + a)^p*p^2 - 1300*(b*sqrt(x) + a)^5*(b*sqrt(x) + a)^p*a*p^2 + 3070*(b*sqrt(x) + a)^4*(b*sqrt(x)
+ a)^p*a^2*p^2 - 3720*(b*sqrt(x) + a)^3*(b*sqrt(x) + a)^p*a^3*p^2 + 2305*(b*sqrt(x) + a)^2*(b*sqrt(x) + a)^p*
a^4*p^2 - 580*(b*sqrt(x) + a)*(b*sqrt(x) + a)^p*a^5*p^2 + 274*(b*sqrt(x) + a)^6*(b*sqrt(x) + a)^p*p - 1620*(b*
sqrt(x) + a)^5*(b*sqrt(x) + a)^p*a*p + 3960*(b*sqrt(x) + a)^4*(b*sqrt(x) + a)^p*a^2*p - 5080*(b*sqrt(x) + a)^3
*(b*sqrt(x) + a)^p*a^3*p + 3510*(b*sqrt(x) + a)^2*(b*sqrt(x) + a)^p*a^4*p - 1044*(b*sqrt(x) + a)*(b*sqrt(x) +
a)^p*a^5*p + 120*(b*sqrt(x) + a)^6*(b*sqrt(x) + a)^p - 720*(b*sqrt(x) + a)^5*(b*sqrt(x) + a)^p*a + 1800*(b*sqr
t(x) + a)^4*(b*sqrt(x) + a)^p*a^2 - 2400*(b*sqrt(x) + a)^3*(b*sqrt(x) + a)^p*a^3 + 1800*(b*sqrt(x) + a)^2*(b*s
qrt(x) + a)^p*a^4 - 720*(b*sqrt(x) + a)*(b*sqrt(x) + a)^p*a^5)/((b^5*p^6 + 21*b^5*p^5 + 175*b^5*p^4 + 735*b^5*
p^3 + 1624*b^5*p^2 + 1764*b^5*p + 720*b^5)*b)

Mupad [B] (verification not implemented)

Time = 6.28 (sec) , antiderivative size = 355, normalized size of antiderivative = 2.34 \[ \int \left (a+b \sqrt {x}\right )^p x^2 \, dx={\left (a+b\,\sqrt {x}\right )}^p\,\left (\frac {2\,x^3\,\left (p^5+15\,p^4+85\,p^3+225\,p^2+274\,p+120\right )}{p^6+21\,p^5+175\,p^4+735\,p^3+1624\,p^2+1764\,p+720}-\frac {240\,a^6}{b^6\,\left (p^6+21\,p^5+175\,p^4+735\,p^3+1624\,p^2+1764\,p+720\right )}+\frac {240\,a^5\,p\,\sqrt {x}}{b^5\,\left (p^6+21\,p^5+175\,p^4+735\,p^3+1624\,p^2+1764\,p+720\right )}+\frac {2\,a\,p\,x^{5/2}\,\left (p^4+10\,p^3+35\,p^2+50\,p+24\right )}{b\,\left (p^6+21\,p^5+175\,p^4+735\,p^3+1624\,p^2+1764\,p+720\right )}+\frac {40\,a^3\,p\,x^{3/2}\,\left (p^2+3\,p+2\right )}{b^3\,\left (p^6+21\,p^5+175\,p^4+735\,p^3+1624\,p^2+1764\,p+720\right )}-\frac {10\,a^2\,p\,x^2\,\left (p^3+6\,p^2+11\,p+6\right )}{b^2\,\left (p^6+21\,p^5+175\,p^4+735\,p^3+1624\,p^2+1764\,p+720\right )}-\frac {120\,a^4\,p\,x\,\left (p+1\right )}{b^4\,\left (p^6+21\,p^5+175\,p^4+735\,p^3+1624\,p^2+1764\,p+720\right )}\right ) \]

[In]

int(x^2*(a + b*x^(1/2))^p,x)

[Out]

(a + b*x^(1/2))^p*((2*x^3*(274*p + 225*p^2 + 85*p^3 + 15*p^4 + p^5 + 120))/(1764*p + 1624*p^2 + 735*p^3 + 175*
p^4 + 21*p^5 + p^6 + 720) - (240*a^6)/(b^6*(1764*p + 1624*p^2 + 735*p^3 + 175*p^4 + 21*p^5 + p^6 + 720)) + (24
0*a^5*p*x^(1/2))/(b^5*(1764*p + 1624*p^2 + 735*p^3 + 175*p^4 + 21*p^5 + p^6 + 720)) + (2*a*p*x^(5/2)*(50*p + 3
5*p^2 + 10*p^3 + p^4 + 24))/(b*(1764*p + 1624*p^2 + 735*p^3 + 175*p^4 + 21*p^5 + p^6 + 720)) + (40*a^3*p*x^(3/
2)*(3*p + p^2 + 2))/(b^3*(1764*p + 1624*p^2 + 735*p^3 + 175*p^4 + 21*p^5 + p^6 + 720)) - (10*a^2*p*x^2*(11*p +
6*p^2 + p^3 + 6))/(b^2*(1764*p + 1624*p^2 + 735*p^3 + 175*p^4 + 21*p^5 + p^6 + 720)) - (120*a^4*p*x*(p + 1))/
(b^4*(1764*p + 1624*p^2 + 735*p^3 + 175*p^4 + 21*p^5 + p^6 + 720)))