3.2815 \(\int \sqrt{c (a+b x)^3} \, dx\)

Optimal. Leaf size=25 \[ \frac{2 (a+b x) \sqrt{c (a+b x)^3}}{5 b} \]

[Out]

(2*(a + b*x)*Sqrt[c*(a + b*x)^3])/(5*b)

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Rubi [A]  time = 0.0112732, 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} \[ \frac{2 (a+b x) \sqrt{c (a+b x)^3}}{5 b} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[c*(a + b*x)^3],x]

[Out]

(2*(a + b*x)*Sqrt[c*(a + b*x)^3])/(5*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)^3} \, dx &=\frac{\operatorname{Subst}\left (\int \sqrt{c x^3} \, dx,x,a+b x\right )}{b}\\ &=\frac{\sqrt{c (a+b x)^3} \operatorname{Subst}\left (\int x^{3/2} \, dx,x,a+b x\right )}{b (a+b x)^{3/2}}\\ &=\frac{2 (a+b x) \sqrt{c (a+b x)^3}}{5 b}\\ \end{align*}

Mathematica [A]  time = 0.0072227, size = 25, normalized size = 1. \[ \frac{2 (a+b x) \sqrt{c (a+b x)^3}}{5 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[c*(a + b*x)^3],x]

[Out]

(2*(a + b*x)*Sqrt[c*(a + b*x)^3])/(5*b)

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Maple [A]  time = 0.003, size = 22, normalized size = 0.9 \begin{align*}{\frac{2\,bx+2\,a}{5\,b}\sqrt{c \left ( bx+a \right ) ^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*(b*x+a)*(c*(b*x+a)^3)^(1/2)/b

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Maxima [A]  time = 1.01085, size = 32, normalized size = 1.28 \begin{align*} \frac{2 \,{\left (b \sqrt{c} x + a \sqrt{c}\right )}{\left (b x + a\right )}^{\frac{3}{2}}}{5 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.25113, size = 96, normalized size = 3.84 \begin{align*} \frac{2 \, \sqrt{b^{3} c x^{3} + 3 \, a b^{2} c x^{2} + 3 \, a^{2} b c x + a^{3} c}{\left (b x + a\right )}}{5 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{c \left (a + b x\right )^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(c*(a + b*x)**3), x)

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Giac [B]  time = 1.13379, size = 89, normalized size = 3.56 \begin{align*} \frac{2 \,{\left (5 \,{\left (b c x + a c\right )}^{\frac{3}{2}} a \mathrm{sgn}\left (b x + a\right ) - \frac{{\left (5 \,{\left (b c x + a c\right )}^{\frac{3}{2}} a c - 3 \,{\left (b c x + a c\right )}^{\frac{5}{2}}\right )} \mathrm{sgn}\left (b x + a\right )}{c}\right )}}{15 \, b c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(5*(b*c*x + a*c)^(3/2)*a*sgn(b*x + a) - (5*(b*c*x + a*c)^(3/2)*a*c - 3*(b*c*x + a*c)^(5/2))*sgn(b*x + a)/
c)/(b*c)