3.2134 \(\int (a+b \sqrt {x})^3 \, dx\)

Optimal. Leaf size=38 \[ \frac {2 \left (a+b \sqrt {x}\right )^5}{5 b^2}-\frac {a \left (a+b \sqrt {x}\right )^4}{2 b^2} \]

[Out]

-1/2*a*(a+b*x^(1/2))^4/b^2+2/5*(a+b*x^(1/2))^5/b^2

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Rubi [A]  time = 0.02, antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {190, 43} \[ \frac {2 \left (a+b \sqrt {x}\right )^5}{5 b^2}-\frac {a \left (a+b \sqrt {x}\right )^4}{2 b^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sqrt[x])^3,x]

[Out]

-(a*(a + b*Sqrt[x])^4)/(2*b^2) + (2*(a + b*Sqrt[x])^5)/(5*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])

Rule 190

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

Rubi steps

\begin {align*} \int \left (a+b \sqrt {x}\right )^3 \, dx &=2 \operatorname {Subst}\left (\int x (a+b x)^3 \, dx,x,\sqrt {x}\right )\\ &=2 \operatorname {Subst}\left (\int \left (-\frac {a (a+b x)^3}{b}+\frac {(a+b x)^4}{b}\right ) \, dx,x,\sqrt {x}\right )\\ &=-\frac {a \left (a+b \sqrt {x}\right )^4}{2 b^2}+\frac {2 \left (a+b \sqrt {x}\right )^5}{5 b^2}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 28, normalized size = 0.74 \[ -\frac {\left (a-4 b \sqrt {x}\right ) \left (a+b \sqrt {x}\right )^4}{10 b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sqrt[x])^3,x]

[Out]

-1/10*((a - 4*b*Sqrt[x])*(a + b*Sqrt[x])^4)/b^2

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fricas [A]  time = 0.68, size = 35, normalized size = 0.92 \[ \frac {3}{2} \, a b^{2} x^{2} + a^{3} x + \frac {2}{5} \, {\left (b^{3} x^{2} + 5 \, a^{2} b x\right )} \sqrt {x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*x^(1/2))^3,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.15, size = 32, normalized size = 0.84 \[ \frac {2}{5} \, b^{3} x^{\frac {5}{2}} + \frac {3}{2} \, a b^{2} x^{2} + 2 \, a^{2} b x^{\frac {3}{2}} + a^{3} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*x^(1/2))^3,x, algorithm="giac")

[Out]

2/5*b^3*x^(5/2) + 3/2*a*b^2*x^2 + 2*a^2*b*x^(3/2) + a^3*x

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maple [A]  time = 0.00, size = 33, normalized size = 0.87 \[ \frac {2 b^{3} x^{\frac {5}{2}}}{5}+\frac {3 a \,b^{2} x^{2}}{2}+2 a^{2} b \,x^{\frac {3}{2}}+a^{3} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*x^(1/2))^3,x)

[Out]

2/5*x^(5/2)*b^3+3/2*a*b^2*x^2+2*a^2*b*x^(3/2)+a^3*x

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maxima [A]  time = 0.91, size = 32, normalized size = 0.84 \[ \frac {2}{5} \, b^{3} x^{\frac {5}{2}} + \frac {3}{2} \, a b^{2} x^{2} + 2 \, a^{2} b x^{\frac {3}{2}} + a^{3} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*x^(1/2))^3,x, algorithm="maxima")

[Out]

2/5*b^3*x^(5/2) + 3/2*a*b^2*x^2 + 2*a^2*b*x^(3/2) + a^3*x

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mupad [B]  time = 0.04, size = 32, normalized size = 0.84 \[ a^3\,x+\frac {2\,b^3\,x^{5/2}}{5}+\frac {3\,a\,b^2\,x^2}{2}+2\,a^2\,b\,x^{3/2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^(1/2))^3,x)

[Out]

a^3*x + (2*b^3*x^(5/2))/5 + (3*a*b^2*x^2)/2 + 2*a^2*b*x^(3/2)

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sympy [A]  time = 1.75, size = 39, normalized size = 1.03 \[ a^{3} x + 2 a^{2} b x^{\frac {3}{2}} + \frac {3 a b^{2} x^{2}}{2} + \frac {2 b^{3} x^{\frac {5}{2}}}{5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*x**(1/2))**3,x)

[Out]

a**3*x + 2*a**2*b*x**(3/2) + 3*a*b**2*x**2/2 + 2*b**3*x**(5/2)/5

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