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]

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

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Rubi [A]  time = 0.01, antiderivative size = 25, normalized size of antiderivative = 1.00, 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 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 {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*}

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Mathematica [A]  time = 0.01, 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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fricas [A]  time = 0.55, size = 41, normalized size = 1.64 \[ \frac {\sqrt {b^{2} c x^{2} + 2 \, a b c x + a^{2} c} {\left (b x^{2} + 2 \, a x\right )}}{2 \, {\left (b x + a\right )}} \]

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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giac [A]  time = 0.17, size = 36, normalized size = 1.44 \[ \frac {1}{2} \, {\left ({\left (b x^{2} + 2 \, a x\right )} \mathrm {sgn}\left (b x + a\right ) + \frac {a^{2} \mathrm {sgn}\left (b x + a\right )}{b}\right )} \sqrt {c} \]

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)

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maple [A]  time = 0.00, size = 29, normalized size = 1.16 \[ \frac {\left (b x +2 a \right ) \sqrt {\left (b x +a \right )^{2} c}\, x}{2 b x +2 a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.64, size = 54, normalized size = 2.16 \[ \frac {1}{2} \, \sqrt {b^{2} c x^{2} + 2 \, a b c x + a^{2} c} x + \frac {\sqrt {b^{2} c x^{2} + 2 \, a b c x + a^{2} c} a}{2 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.47, size = 21, normalized size = 0.84 \[ \frac {\sqrt {c\,{\left (a+b\,x\right )}^2}\,\left (a+b\,x\right )}{2\,b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*(a + b*x)^2)^(1/2),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {c \left (a + b x\right )^{2}}\, dx \]

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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