3.2803 \(\int (c (a+b x)^2)^{3/2} \, dx\)

Optimal. Leaf size=28 \[ \frac{c (a+b x)^3 \sqrt{c (a+b x)^2}}{4 b} \]

[Out]

(c*(a + b*x)^3*Sqrt[c*(a + b*x)^2])/(4*b)

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Rubi [A]  time = 0.0106663, antiderivative size = 28, 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} \[ \frac{c (a+b x)^3 \sqrt{c (a+b x)^2}}{4 b} \]

Antiderivative was successfully verified.

[In]

Int[(c*(a + b*x)^2)^(3/2),x]

[Out]

(c*(a + b*x)^3*Sqrt[c*(a + b*x)^2])/(4*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 \left (c (a+b x)^2\right )^{3/2} \, dx &=\frac{\operatorname{Subst}\left (\int \left (c x^2\right )^{3/2} \, dx,x,a+b x\right )}{b}\\ &=\frac{\left (c \sqrt{c (a+b x)^2}\right ) \operatorname{Subst}\left (\int x^3 \, dx,x,a+b x\right )}{b (a+b x)}\\ &=\frac{c (a+b x)^3 \sqrt{c (a+b x)^2}}{4 b}\\ \end{align*}

Mathematica [A]  time = 0.0090135, size = 25, normalized size = 0.89 \[ \frac{(a+b x) \left (c (a+b x)^2\right )^{3/2}}{4 b} \]

Antiderivative was successfully verified.

[In]

Integrate[(c*(a + b*x)^2)^(3/2),x]

[Out]

((a + b*x)*(c*(a + b*x)^2)^(3/2))/(4*b)

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Maple [B]  time = 0.003, size = 51, normalized size = 1.8 \begin{align*}{\frac{x \left ({b}^{3}{x}^{3}+4\,a{b}^{2}{x}^{2}+6\,{a}^{2}bx+4\,{a}^{3} \right ) }{4\, \left ( bx+a \right ) ^{3}} \left ( c \left ( bx+a \right ) ^{2} \right ) ^{{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*x*(b^3*x^3+4*a*b^2*x^2+6*a^2*b*x+4*a^3)*(c*(b*x+a)^2)^(3/2)/(b*x+a)^3

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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)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.22641, size = 144, normalized size = 5.14 \begin{align*} \frac{{\left (b^{3} c x^{4} + 4 \, a b^{2} c x^{3} + 6 \, a^{2} b c x^{2} + 4 \, a^{3} c x\right )} \sqrt{b^{2} c x^{2} + 2 \, a b c x + a^{2} c}}{4 \,{\left (b x + a\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(b^3*c*x^4 + 4*a*b^2*c*x^3 + 6*a^2*b*c*x^2 + 4*a^3*c*x)*sqrt(b^2*c*x^2 + 2*a*b*c*x + a^2*c)/(b*x + a)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (c \left (a + b x\right )^{2}\right )^{\frac{3}{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((c*(a + b*x)**2)**(3/2), x)

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Giac [A]  time = 1.09758, size = 28, normalized size = 1. \begin{align*} \frac{{\left (b x + a\right )}^{4} c^{\frac{3}{2}} \mathrm{sgn}\left (b x + a\right )}{4 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(b*x + a)^4*c^(3/2)*sgn(b*x + a)/b