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

Optimal. Leaf size=41 \[ \frac{1}{4} a c^2 x^3 \sqrt{c x^2}+\frac{1}{5} b c^2 x^4 \sqrt{c x^2} \]

[Out]

(a*c^2*x^3*Sqrt[c*x^2])/4 + (b*c^2*x^4*Sqrt[c*x^2])/5

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Rubi [A]  time = 0.011029, 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} \[ \frac{1}{4} a c^2 x^3 \sqrt{c x^2}+\frac{1}{5} b c^2 x^4 \sqrt{c x^2} \]

Antiderivative was successfully verified.

[In]

Int[((c*x^2)^(5/2)*(a + b*x))/x^2,x]

[Out]

(a*c^2*x^3*Sqrt[c*x^2])/4 + (b*c^2*x^4*Sqrt[c*x^2])/5

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

Mathematica [A]  time = 0.002578, size = 23, normalized size = 0.56 \[ \frac{1}{20} c x \left (c x^2\right )^{3/2} (5 a+4 b x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c*x*(c*x^2)^(3/2)*(5*a + 4*b*x))/20

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Maple [A]  time = 0.002, size = 21, normalized size = 0.5 \begin{align*}{\frac{4\,bx+5\,a}{20\,x} \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^2,x)

[Out]

1/20/x*(4*b*x+5*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^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.8442, size = 62, normalized size = 1.51 \begin{align*} \frac{1}{20} \,{\left (4 \, b c^{2} x^{4} + 5 \, a c^{2} x^{3}\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^2,x, algorithm="fricas")

[Out]

1/20*(4*b*c^2*x^4 + 5*a*c^2*x^3)*sqrt(c*x^2)

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Sympy [A]  time = 1.35926, size = 31, normalized size = 0.76 \begin{align*} \frac{a c^{\frac{5}{2}} \left (x^{2}\right )^{\frac{5}{2}}}{4 x} + \frac{b c^{\frac{5}{2}} \left (x^{2}\right )^{\frac{5}{2}}}{5} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.04246, size = 38, normalized size = 0.93 \begin{align*} \frac{1}{20} \,{\left (4 \, b c^{2} x^{5} \mathrm{sgn}\left (x\right ) + 5 \, a c^{2} x^{4} \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^2,x, algorithm="giac")

[Out]

1/20*(4*b*c^2*x^5*sgn(x) + 5*a*c^2*x^4*sgn(x))*sqrt(c)