\(\int \sqrt {(3+5 x)^2} \, dx\) [2808]

Optimal result

Integrand size = 11, antiderivative size = 20 \[ \int \sqrt {(3+5 x)^2} \, dx=\frac {1}{10} (3+5 x) \sqrt {(3+5 x)^2} \]

[Out]

1/10*(3+5*x)*((3+5*x)^2)^(1/2)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {253, 15, 30} \[ \int \sqrt {(3+5 x)^2} \, dx=\frac {1}{10} (5 x+3) \sqrt {(5 x+3)^2} \]

[In]

Int[Sqrt[(3 + 5*x)^2],x]

[Out]

((3 + 5*x)*Sqrt[(3 + 5*x)^2])/10

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 253

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.25 \[ \int \sqrt {(3+5 x)^2} \, dx=\frac {x \sqrt {(3+5 x)^2} (6+5 x)}{6+10 x} \]

[In]

Integrate[Sqrt[(3 + 5*x)^2],x]

[Out]

(x*Sqrt[(3 + 5*x)^2]*(6 + 5*x))/(6 + 10*x)

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 2.

Time = 0.07 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.80

method	result	size
default	\(\frac {\operatorname {csgn}\left (3+5 x \right ) \left (3+5 x \right )^{2}}{10}\)	\(16\)
gosper	\(\frac {x \left (5 x +6\right ) \sqrt {\left (3+5 x \right )^{2}}}{6+10 x}\)	\(25\)
risch	\(\frac {5 \sqrt {\left (3+5 x \right )^{2}}\, x^{2}}{2 \left (3+5 x \right )}+\frac {3 \sqrt {\left (3+5 x \right )^{2}}\, x}{3+5 x}\)	\(42\)

[In]

int(((3+5*x)^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/10*csgn(3+5*x)*(3+5*x)^2

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.45 \[ \int \sqrt {(3+5 x)^2} \, dx=\frac {5}{2} \, x^{2} + 3 \, x \]

[In]

integrate(((3+5*x)^2)^(1/2),x, algorithm="fricas")

[Out]

5/2*x^2 + 3*x

Sympy [A] (verification not implemented)

Time = 0.41 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95 \[ \int \sqrt {(3+5 x)^2} \, dx=\left (\frac {x}{2} + \frac {3}{10}\right ) \sqrt {25 x^{2} + 30 x + 9} \]

[In]

integrate(((3+5*x)**2)**(1/2),x)

[Out]

(x/2 + 3/10)*sqrt(25*x**2 + 30*x + 9)

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.50 \[ \int \sqrt {(3+5 x)^2} \, dx=\frac {1}{2} \, \sqrt {25 \, x^{2} + 30 \, x + 9} x + \frac {3}{10} \, \sqrt {25 \, x^{2} + 30 \, x + 9} \]

[In]

integrate(((3+5*x)^2)^(1/2),x, algorithm="maxima")

[Out]

1/2*sqrt(25*x^2 + 30*x + 9)*x + 3/10*sqrt(25*x^2 + 30*x + 9)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.30 \[ \int \sqrt {(3+5 x)^2} \, dx=\frac {1}{2} \, {\left (5 \, x^{2} + 6 \, x\right )} \mathrm {sgn}\left (5 \, x + 3\right ) + \frac {9}{10} \, \mathrm {sgn}\left (5 \, x + 3\right ) \]

[In]

integrate(((3+5*x)^2)^(1/2),x, algorithm="giac")

[Out]

1/2*(5*x^2 + 6*x)*sgn(5*x + 3) + 9/10*sgn(5*x + 3)

Mupad [B] (verification not implemented)

Time = 6.57 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.65 \[ \int \sqrt {(3+5 x)^2} \, dx=\frac {\left |5\,x+3\right |\,\left (5\,x+3\right )}{10} \]

[In]

int(((5*x + 3)^2)^(1/2),x)

[Out]

(abs(5*x + 3)*(5*x + 3))/10