3.778 \(\int \frac{(c x^2)^{5/2} (a+b x)}{x^3} \, dx\)

Optimal. Leaf size=41 \[ \frac{1}{3} a c^2 x^2 \sqrt{c x^2}+\frac{1}{4} b c^2 x^3 \sqrt{c x^2} \]

[Out]

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

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Rubi [A]  time = 0.0100404, antiderivative size = 41, 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} \[ \frac{1}{3} a c^2 x^2 \sqrt{c x^2}+\frac{1}{4} b c^2 x^3 \sqrt{c x^2} \]

Antiderivative was successfully verified.

[In]

Int[((c*x^2)^(5/2)*(a + b*x))/x^3,x]

[Out]

(a*c^2*x^2*Sqrt[c*x^2])/3 + (b*c^2*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 )^{5/2} (a+b x)}{x^3} \, dx &=\frac{\left (c^2 \sqrt{c x^2}\right ) \int x^2 (a+b x) \, dx}{x}\\ &=\frac{\left (c^2 \sqrt{c x^2}\right ) \int \left (a x^2+b x^3\right ) \, dx}{x}\\ &=\frac{1}{3} a c^2 x^2 \sqrt{c x^2}+\frac{1}{4} b c^2 x^3 \sqrt{c x^2}\\ \end{align*}

Mathematica [A]  time = 0.0026706, size = 27, normalized size = 0.66 \[ \frac{1}{12} c^2 x^2 \sqrt{c x^2} (4 a+3 b x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c^2*x^2*Sqrt[c*x^2]*(4*a + 3*b*x))/12

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Maple [A]  time = 0.003, size = 21, normalized size = 0.5 \begin{align*}{\frac{3\,bx+4\,a}{12\,{x}^{2}} \left ( c{x}^{2} \right ) ^{{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12/x^2*(3*b*x+4*a)*(c*x^2)^(5/2)

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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)^(5/2)*(b*x+a)/x^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.88457, size = 62, normalized size = 1.51 \begin{align*} \frac{1}{12} \,{\left (3 \, b c^{2} x^{3} + 4 \, a c^{2} x^{2}\right )} \sqrt{c x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2)^(5/2)*(b*x+a)/x^3,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 1.52712, size = 34, normalized size = 0.83 \begin{align*} \frac{a c^{\frac{5}{2}} \left (x^{2}\right )^{\frac{5}{2}}}{3 x^{2}} + \frac{b c^{\frac{5}{2}} \left (x^{2}\right )^{\frac{5}{2}}}{4 x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**2)**(5/2)*(b*x+a)/x**3,x)

[Out]

a*c**(5/2)*(x**2)**(5/2)/(3*x**2) + b*c**(5/2)*(x**2)**(5/2)/(4*x)

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Giac [A]  time = 1.05465, size = 38, normalized size = 0.93 \begin{align*} \frac{1}{12} \,{\left (3 \, b c^{2} x^{4} \mathrm{sgn}\left (x\right ) + 4 \, a c^{2} x^{3} \mathrm{sgn}\left (x\right )\right )} \sqrt{c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2)^(5/2)*(b*x+a)/x^3,x, algorithm="giac")

[Out]

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