3.891 \(\int \frac{1}{(c x^2)^{3/2} (a+b x)} \, dx\)

Optimal. Leaf size=89 \[ \frac{b^2 x \log (x)}{a^3 c \sqrt{c x^2}}-\frac{b^2 x \log (a+b x)}{a^3 c \sqrt{c x^2}}+\frac{b}{a^2 c \sqrt{c x^2}}-\frac{1}{2 a c x \sqrt{c x^2}} \]

[Out]

b/(a^2*c*Sqrt[c*x^2]) - 1/(2*a*c*x*Sqrt[c*x^2]) + (b^2*x*Log[x])/(a^3*c*Sqrt[c*x^2]) - (b^2*x*Log[a + b*x])/(a
^3*c*Sqrt[c*x^2])

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Rubi [A]  time = 0.0222743, antiderivative size = 89, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.118, Rules used = {15, 44} \[ \frac{b^2 x \log (x)}{a^3 c \sqrt{c x^2}}-\frac{b^2 x \log (a+b x)}{a^3 c \sqrt{c x^2}}+\frac{b}{a^2 c \sqrt{c x^2}}-\frac{1}{2 a c x \sqrt{c x^2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((c*x^2)^(3/2)*(a + b*x)),x]

[Out]

b/(a^2*c*Sqrt[c*x^2]) - 1/(2*a*c*x*Sqrt[c*x^2]) + (b^2*x*Log[x])/(a^3*c*Sqrt[c*x^2]) - (b^2*x*Log[a + b*x])/(a
^3*c*Sqrt[c*x^2])

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 44

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.006718, size = 51, normalized size = 0.57 \[ \frac{x \left (-2 b^2 x^2 \log (a+b x)-a (a-2 b x)+2 b^2 x^2 \log (x)\right )}{2 a^3 \left (c x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((c*x^2)^(3/2)*(a + b*x)),x]

[Out]

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

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Maple [A]  time = 0.005, size = 49, normalized size = 0.6 \begin{align*}{\frac{x \left ( 2\,{b}^{2}\ln \left ( x \right ){x}^{2}-2\,{b}^{2}\ln \left ( bx+a \right ){x}^{2}+2\,abx-{a}^{2} \right ) }{2\,{a}^{3}} \left ( c{x}^{2} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(c*x^2)^(3/2)/(b*x+a),x)

[Out]

1/2*x*(2*b^2*ln(x)*x^2-2*b^2*ln(b*x+a)*x^2+2*a*b*x-a^2)/(c*x^2)^(3/2)/a^3

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.49573, size = 103, normalized size = 1.16 \begin{align*} \frac{{\left (2 \, b^{2} x^{2} \log \left (\frac{x}{b x + a}\right ) + 2 \, a b x - a^{2}\right )} \sqrt{c x^{2}}}{2 \, a^{3} c^{2} x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x^2)^(3/2)/(b*x+a),x, algorithm="fricas")

[Out]

1/2*(2*b^2*x^2*log(x/(b*x + a)) + 2*a*b*x - a^2)*sqrt(c*x^2)/(a^3*c^2*x^3)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (c x^{2}\right )^{\frac{3}{2}} \left (a + b x\right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x**2)**(3/2)/(b*x+a),x)

[Out]

Integral(1/((c*x**2)**(3/2)*(a + b*x)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{sage}_{0} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x^2)^(3/2)/(b*x+a),x, algorithm="giac")

[Out]

sage0*x