3.865 \(\int \frac{(c x^2)^{3/2}}{x^4 (a+b x)} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.0068632, antiderivative size = 44, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {15, 36, 29, 31} \[ \frac{c \sqrt{c x^2} \log (x)}{a x}-\frac{c \sqrt{c x^2} \log (a+b x)}{a x} \]

Antiderivative was successfully verified.

[In]

Int[(c*x^2)^(3/2)/(x^4*(a + b*x)),x]

[Out]

(c*Sqrt[c*x^2]*Log[x])/(a*x) - (c*Sqrt[c*x^2]*Log[a + b*x])/(a*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 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rubi steps

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

Mathematica [A]  time = 0.0076032, size = 27, normalized size = 0.61 \[ \frac{\left (c x^2\right )^{3/2} (\log (x)-\log (a+b x))}{a x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.004, size = 26, normalized size = 0.6 \begin{align*}{\frac{\ln \left ( x \right ) -\ln \left ( bx+a \right ) }{a{x}^{3}} \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)/x^4/(b*x+a),x)

[Out]

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

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Maxima [A]  time = 1.03084, size = 32, normalized size = 0.73 \begin{align*} -\frac{c^{\frac{3}{2}} \log \left (b x + a\right )}{a} + \frac{c^{\frac{3}{2}} \log \left (x\right )}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2)^(3/2)/x^4/(b*x+a),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.59278, size = 144, normalized size = 3.27 \begin{align*} \left [\frac{\sqrt{c x^{2}} c \log \left (\frac{x}{b x + a}\right )}{a x}, \frac{2 \, \sqrt{-c} c \arctan \left (\frac{\sqrt{c x^{2}}{\left (2 \, b x + a\right )} \sqrt{-c}}{a c x}\right )}{a}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2)^(3/2)/x^4/(b*x+a),x, algorithm="fricas")

[Out]

[sqrt(c*x^2)*c*log(x/(b*x + a))/(a*x), 2*sqrt(-c)*c*arctan(sqrt(c*x^2)*(2*b*x + a)*sqrt(-c)/(a*c*x))/a]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (c x^{2}\right )^{\frac{3}{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)**(3/2)/x**4/(b*x+a),x)

[Out]

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

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2)^(3/2)/x^4/(b*x+a),x, algorithm="giac")

[Out]

Exception raised: TypeError