3.2804 \(\int \sqrt{c (a+b x)^2} \, dx\)

Optimal. Leaf size=25 \[ \frac{(a+b x) \sqrt{c (a+b x)^2}}{2 b} \]

[Out]

((a + b*x)*Sqrt[c*(a + b*x)^2])/(2*b)

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Rubi [A]  time = 0.010434, antiderivative size = 25, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {247, 15, 30} \[ \frac{(a+b x) \sqrt{c (a+b x)^2}}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[c*(a + b*x)^2],x]

[Out]

((a + b*x)*Sqrt[c*(a + b*x)^2])/(2*b)

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]

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]

Rubi steps

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

Mathematica [A]  time = 0.0133636, size = 31, normalized size = 1.24 \[ \frac{c x (a+b x) (2 a+b x)}{2 \sqrt{c (a+b x)^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[c*(a + b*x)^2],x]

[Out]

(c*x*(a + b*x)*(2*a + b*x))/(2*Sqrt[c*(a + b*x)^2])

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Maple [A]  time = 0.001, size = 29, normalized size = 1.2 \begin{align*}{\frac{x \left ( bx+2\,a \right ) }{2\,bx+2\,a}\sqrt{c \left ( bx+a \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x*(b*x+2*a)*(c*(b*x+a)^2)^(1/2)/(b*x+a)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.23573, size = 90, normalized size = 3.6 \begin{align*} \frac{\sqrt{b^{2} c x^{2} + 2 \, a b c x + a^{2} c}{\left (b x^{2} + 2 \, a x\right )}}{2 \,{\left (b x + a\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sqrt(b^2*c*x^2 + 2*a*b*c*x + a^2*c)*(b*x^2 + 2*a*x)/(b*x + a)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{c \left (a + b x\right )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(c*(a + b*x)**2), x)

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Giac [A]  time = 1.11334, size = 49, normalized size = 1.96 \begin{align*} \frac{1}{2} \,{\left ({\left (b x^{2} + 2 \, a x\right )} \mathrm{sgn}\left (b x + a\right ) + \frac{a^{2} \mathrm{sgn}\left (b x + a\right )}{b}\right )} \sqrt{c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((b*x^2 + 2*a*x)*sgn(b*x + a) + a^2*sgn(b*x + a)/b)*sqrt(c)