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

Optimal. Leaf size=37 \[ \frac{1}{3} a c x^2 \sqrt{c x^2}+\frac{1}{4} b c x^3 \sqrt{c x^2} \]

[Out]

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

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Rubi [A]  time = 0.0088063, antiderivative size = 37, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {15, 43} \[ \frac{1}{3} a c x^2 \sqrt{c x^2}+\frac{1}{4} b c x^3 \sqrt{c x^2} \]

Antiderivative was successfully verified.

[In]

Int[((c*x^2)^(3/2)*(a + b*x))/x,x]

[Out]

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

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 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{\left (c x^2\right )^{3/2} (a+b x)}{x} \, dx &=\frac{\left (c \sqrt{c x^2}\right ) \int x^2 (a+b x) \, dx}{x}\\ &=\frac{\left (c \sqrt{c x^2}\right ) \int \left (a x^2+b x^3\right ) \, dx}{x}\\ &=\frac{1}{3} a c x^2 \sqrt{c x^2}+\frac{1}{4} b c x^3 \sqrt{c x^2}\\ \end{align*}

Mathematica [A]  time = 0.0021944, size = 25, normalized size = 0.68 \[ \frac{1}{12} c x^2 \sqrt{c x^2} (4 a+3 b x) \]

Antiderivative was successfully verified.

[In]

Integrate[((c*x^2)^(3/2)*(a + b*x))/x,x]

[Out]

(c*x^2*Sqrt[c*x^2]*(4*a + 3*b*x))/12

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Maple [A]  time = 0.003, size = 18, normalized size = 0.5 \begin{align*}{\frac{3\,bx+4\,a}{12} \left ( c{x}^{2} \right ) ^{{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(3*b*x+4*a)*(c*x^2)^(3/2)

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.39588, size = 57, normalized size = 1.54 \begin{align*} \frac{1}{12} \,{\left (3 \, b c x^{3} + 4 \, a c x^{2}\right )} \sqrt{c x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(3*b*c*x^3 + 4*a*c*x^2)*sqrt(c*x^2)

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Sympy [A]  time = 0.520868, size = 31, normalized size = 0.84 \begin{align*} \frac{a c^{\frac{3}{2}} \left (x^{2}\right )^{\frac{3}{2}}}{3} + \frac{b c^{\frac{3}{2}} x \left (x^{2}\right )^{\frac{3}{2}}}{4} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.05761, size = 30, normalized size = 0.81 \begin{align*} \frac{1}{12} \,{\left (3 \, b x^{4} \mathrm{sgn}\left (x\right ) + 4 \, a x^{3} \mathrm{sgn}\left (x\right )\right )} c^{\frac{3}{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(3*b*x^4*sgn(x) + 4*a*x^3*sgn(x))*c^(3/2)