3.873 \(\int \frac{(c x^2)^{5/2}}{x^4 (a+b x)} \, dx\)

Optimal. Leaf size=44 \[ \frac{c^2 \sqrt{c x^2}}{b}-\frac{a c^2 \sqrt{c x^2} \log (a+b x)}{b^2 x} \]

[Out]

(c^2*Sqrt[c*x^2])/b - (a*c^2*Sqrt[c*x^2]*Log[a + b*x])/(b^2*x)

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Rubi [A]  time = 0.0122541, antiderivative size = 44, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {15, 43} \[ \frac{c^2 \sqrt{c x^2}}{b}-\frac{a c^2 \sqrt{c x^2} \log (a+b x)}{b^2 x} \]

Antiderivative was successfully verified.

[In]

Int[(c*x^2)^(5/2)/(x^4*(a + b*x)),x]

[Out]

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

Mathematica [A]  time = 0.0040608, size = 30, normalized size = 0.68 \[ \frac{c^3 x (b x-a \log (a+b x))}{b^2 \sqrt{c x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(c*x^2)^(5/2)/(x^4*(a + b*x)),x]

[Out]

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

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Maple [A]  time = 0.002, size = 29, normalized size = 0.7 \begin{align*} -{\frac{a\ln \left ( bx+a \right ) -bx}{{b}^{2}{x}^{5}} \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)/x^4/(b*x+a),x)

[Out]

-(c*x^2)^(5/2)*(a*ln(b*x+a)-b*x)/x^5/b^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)/x^4/(b*x+a),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.59132, size = 70, normalized size = 1.59 \begin{align*} \frac{{\left (b c^{2} x - a c^{2} \log \left (b x + a\right )\right )} \sqrt{c x^{2}}}{b^{2} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2)^(5/2)/x^4/(b*x+a),x, algorithm="fricas")

[Out]

(b*c^2*x - a*c^2*log(b*x + a))*sqrt(c*x^2)/(b^2*x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (c x^{2}\right )^{\frac{5}{2}}}{x^{4} \left (a + b x\right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.05969, size = 62, normalized size = 1.41 \begin{align*}{\left (\frac{c^{2} x \mathrm{sgn}\left (x\right )}{b} - \frac{a c^{2} \log \left ({\left | b x + a \right |}\right ) \mathrm{sgn}\left (x\right )}{b^{2}} + \frac{a c^{2} \log \left ({\left | a \right |}\right ) \mathrm{sgn}\left (x\right )}{b^{2}}\right )} \sqrt{c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2)^(5/2)/x^4/(b*x+a),x, algorithm="giac")

[Out]

(c^2*x*sgn(x)/b - a*c^2*log(abs(b*x + a))*sgn(x)/b^2 + a*c^2*log(abs(a))*sgn(x)/b^2)*sqrt(c)