\(\int \frac {x (a+b x)}{(c x^2)^{3/2}} \, dx\) [790]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 16, antiderivative size = 33 \[ \int \frac {x (a+b x)}{\left (c x^2\right )^{3/2}} \, dx=-\frac {a}{c \sqrt {c x^2}}+\frac {b x \log (x)}{c \sqrt {c x^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {15, 45} \[ \int \frac {x (a+b x)}{\left (c x^2\right )^{3/2}} \, dx=\frac {b x \log (x)}{c \sqrt {c x^2}}-\frac {a}{c \sqrt {c x^2}} \]

[In]

Int[(x*(a + b*x))/(c*x^2)^(3/2),x]

[Out]

-(a/(c*Sqrt[c*x^2])) + (b*x*Log[x])/(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 45

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.73 \[ \int \frac {x (a+b x)}{\left (c x^2\right )^{3/2}} \, dx=\frac {-a x^2+b x^3 \log (x)}{\left (c x^2\right )^{3/2}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.64

method result size
default \(\frac {x^{2} \left (b \ln \left (x \right ) x -a \right )}{\left (c \,x^{2}\right )^{\frac {3}{2}}}\) \(21\)
risch \(-\frac {a}{c \sqrt {c \,x^{2}}}+\frac {b x \ln \left (x \right )}{c \sqrt {c \,x^{2}}}\) \(30\)

[In]

int(x*(b*x+a)/(c*x^2)^(3/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.70 \[ \int \frac {x (a+b x)}{\left (c x^2\right )^{3/2}} \, dx=\frac {\sqrt {c x^{2}} {\left (b x \log \left (x\right ) - a\right )}}{c^{2} x^{2}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.81 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.09 \[ \int \frac {x (a+b x)}{\left (c x^2\right )^{3/2}} \, dx=a \left (\begin {cases} \tilde {\infty } x^{2} & \text {for}\: c = 0 \\- \frac {1}{c \sqrt {c x^{2}}} & \text {otherwise} \end {cases}\right ) + \frac {b x^{3} \log {\left (x \right )}}{\left (c x^{2}\right )^{\frac {3}{2}}} \]

[In]

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

[Out]

a*Piecewise((zoo*x**2, Eq(c, 0)), (-1/(c*sqrt(c*x**2)), True)) + b*x**3*log(x)/(c*x**2)**(3/2)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.64 \[ \int \frac {x (a+b x)}{\left (c x^2\right )^{3/2}} \, dx=\frac {b \log \left (x\right )}{c^{\frac {3}{2}}} - \frac {a}{\sqrt {c x^{2}} c} \]

[In]

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

[Out]

b*log(x)/c^(3/2) - a/(sqrt(c*x^2)*c)

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.91 \[ \int \frac {x (a+b x)}{\left (c x^2\right )^{3/2}} \, dx=\frac {\frac {b \log \left ({\left | x \right |}\right )}{\sqrt {c} \mathrm {sgn}\left (x\right )} - \frac {a}{\sqrt {c} x \mathrm {sgn}\left (x\right )}}{c} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.85 \[ \int \frac {x (a+b x)}{\left (c x^2\right )^{3/2}} \, dx=-\frac {a+b\,x-b\,\ln \left (x+\left |x\right |\right )\,\sqrt {x^2}}{c^{3/2}\,\sqrt {x^2}} \]

[In]

int((x*(a + b*x))/(c*x^2)^(3/2),x)

[Out]

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