3.2809 \(\int \sqrt {(6+10 x)^2} \, dx\)

Optimal. Leaf size=20 \[ \frac {1}{5} (5 x+3) \sqrt {(5 x+3)^2} \]

[Out]

1/5*(3+5*x)*((3+5*x)^2)^(1/2)

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Rubi [A]  time = 0.00, antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {247, 15, 30} \[ \frac {1}{5} (5 x+3) \sqrt {(5 x+3)^2} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[(6 + 10*x)^2],x]

[Out]

((3 + 5*x)*Sqrt[(3 + 5*x)^2])/5

Rule 15

Int[(u_.)*((a_.)*(x_)^(n_))^(m_), x_Symbol] :> Dist[(a^IntPart[m]*(a*x^n)^FracPart[m])/x^(n*FracPart[m]), Int[
u*x^(m*n), x], x] /; FreeQ[{a, m, n}, x] &&  !IntegerQ[m]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 247

Int[((a_.) + (b_.)*(v_)^(n_))^(p_), x_Symbol] :> Dist[1/Coefficient[v, x, 1], Subst[Int[(a + b*x^n)^p, x], x,
v], x] /; FreeQ[{a, b, n, p}, x] && LinearQ[v, x] && NeQ[v, x]

Rubi steps

\begin {align*} \int \sqrt {(6+10 x)^2} \, dx &=\frac {1}{10} \operatorname {Subst}\left (\int \sqrt {x^2} \, dx,x,6+10 x\right )\\ &=\frac {\sqrt {(6+10 x)^2} \operatorname {Subst}(\int x \, dx,x,6+10 x)}{10 (6+10 x)}\\ &=\frac {1}{5} (3+5 x) \sqrt {(3+5 x)^2}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 25, normalized size = 1.25 \[ \frac {x \sqrt {(5 x+3)^2} (5 x+6)}{5 x+3} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[(6 + 10*x)^2],x]

[Out]

(x*Sqrt[(3 + 5*x)^2]*(6 + 5*x))/(3 + 5*x)

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fricas [A]  time = 0.77, size = 9, normalized size = 0.45 \[ 5 \, x^{2} + 6 \, x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6+10*x)^2)^(1/2),x, algorithm="fricas")

[Out]

5*x^2 + 6*x

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giac [A]  time = 0.15, size = 25, normalized size = 1.25 \[ {\left (5 \, x^{2} + 6 \, x\right )} \mathrm {sgn}\left (5 \, x + 3\right ) + \frac {9}{5} \, \mathrm {sgn}\left (5 \, x + 3\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6+10*x)^2)^(1/2),x, algorithm="giac")

[Out]

(5*x^2 + 6*x)*sgn(5*x + 3) + 9/5*sgn(5*x + 3)

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maple [A]  time = 0.00, size = 24, normalized size = 1.20 \[ \frac {\left (5 x +6\right ) \sqrt {\left (5 x +3\right )^{2}}\, x}{5 x +3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((6+10*x)^2)^(1/2),x)

[Out]

(5*x+6)*((5*x+3)^2)^(1/2)/(5*x+3)*x

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maxima [A]  time = 1.23, size = 29, normalized size = 1.45 \[ \sqrt {25 \, x^{2} + 30 \, x + 9} x + \frac {3}{5} \, \sqrt {25 \, x^{2} + 30 \, x + 9} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6+10*x)^2)^(1/2),x, algorithm="maxima")

[Out]

sqrt(25*x^2 + 30*x + 9)*x + 3/5*sqrt(25*x^2 + 30*x + 9)

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mupad [B]  time = 0.08, size = 13, normalized size = 0.65 \[ \frac {\left |10\,x+6\right |\,\left (5\,x+3\right )}{10} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((10*x + 6)^2)^(1/2),x)

[Out]

(abs(10*x + 6)*(5*x + 3))/10

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sympy [A]  time = 0.07, size = 7, normalized size = 0.35 \[ 5 x^{2} + 6 x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6+10*x)**2)**(1/2),x)

[Out]

5*x**2 + 6*x

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