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]

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

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Rubi [A]  time = 0.01, antiderivative size = 28, 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} \[ \frac {c (a+b x)^3 \sqrt {c (a+b x)^2}}{4 b} \]

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 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 \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*}

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Mathematica [A]  time = 0.01, 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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fricas [B]  time = 0.81, size = 67, normalized size = 2.39 \[ \frac {{\left (b^{3} c x^{4} + 4 \, a b^{2} c x^{3} + 6 \, a^{2} b c x^{2} + 4 \, a^{3} c x\right )} \sqrt {b^{2} c x^{2} + 2 \, a b c x + a^{2} c}}{4 \, {\left (b x + a\right )}} \]

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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giac [A]  time = 0.15, size = 21, normalized size = 0.75 \[ \frac {{\left (b x + a\right )}^{4} c^{\frac {3}{2}} \mathrm {sgn}\left (b x + a\right )}{4 \, b} \]

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

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maple [B]  time = 0.00, size = 51, normalized size = 1.82 \[ \frac {\left (b^{3} x^{3}+4 a \,b^{2} x^{2}+6 a^{2} b x +4 a^{3}\right ) \left (\left (b x +a \right )^{2} c \right )^{\frac {3}{2}} x}{4 \left (b x +a \right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maxima [B]  time = 0.53, size = 54, normalized size = 1.93 \[ \frac {1}{4} \, {\left (b^{2} c x^{2} + 2 \, a b c x + a^{2} c\right )}^{\frac {3}{2}} x + \frac {{\left (b^{2} c x^{2} + 2 \, a b c x + a^{2} c\right )}^{\frac {3}{2}} a}{4 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.20, size = 36, normalized size = 1.29 \[ \frac {\left (x\,b^2+a\,b\right )\,{\left (c\,a^2+2\,c\,a\,b\,x+c\,b^2\,x^2\right )}^{3/2}}{4\,b^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*(a + b*x)^2)^(3/2),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (c \left (a + b x\right )^{2}\right )^{\frac {3}{2}}\, dx \]

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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