\(\int \frac {a+b \log (c x^n)}{(d+e x)^{3/2}} \, dx\) [154]

Optimal result

Integrand size = 20, antiderivative size = 53 \[ \int \frac {a+b \log \left (c x^n\right )}{(d+e x)^{3/2}} \, dx=-\frac {4 b n \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{\sqrt {d} e}-\frac {2 \left (a+b \log \left (c x^n\right )\right )}{e \sqrt {d+e x}} \]

[Out]

-4*b*n*arctanh((e*x+d)^(1/2)/d^(1/2))/e/d^(1/2)-2*(a+b*ln(c*x^n))/e/(e*x+d)^(1/2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {2356, 65, 214} \[ \int \frac {a+b \log \left (c x^n\right )}{(d+e x)^{3/2}} \, dx=-\frac {2 \left (a+b \log \left (c x^n\right )\right )}{e \sqrt {d+e x}}-\frac {4 b n \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{\sqrt {d} e} \]

[In]

Int[(a + b*Log[c*x^n])/(d + e*x)^(3/2),x]

[Out]

(-4*b*n*ArcTanh[Sqrt[d + e*x]/Sqrt[d]])/(Sqrt[d]*e) - (2*(a + b*Log[c*x^n]))/(e*Sqrt[d + e*x])

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 2356

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Simp[(d + e*x)^(q + 1)
*((a + b*Log[c*x^n])^p/(e*(q + 1))), x] - Dist[b*n*(p/(e*(q + 1))), Int[((d + e*x)^(q + 1)*(a + b*Log[c*x^n])^
(p - 1))/x, x], x] /; FreeQ[{a, b, c, d, e, n, p, q}, x] && GtQ[p, 0] && NeQ[q, -1] && (EqQ[p, 1] || (Integers
Q[2*p, 2*q] &&  !IGtQ[q, 0]) || (EqQ[p, 2] && NeQ[q, 1]))

Rubi steps \begin{align*} \text {integral}& = -\frac {2 \left (a+b \log \left (c x^n\right )\right )}{e \sqrt {d+e x}}+\frac {(2 b n) \int \frac {1}{x \sqrt {d+e x}} \, dx}{e} \\ & = -\frac {2 \left (a+b \log \left (c x^n\right )\right )}{e \sqrt {d+e x}}+\frac {(4 b n) \text {Subst}\left (\int \frac {1}{-\frac {d}{e}+\frac {x^2}{e}} \, dx,x,\sqrt {d+e x}\right )}{e^2} \\ & = -\frac {4 b n \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{\sqrt {d} e}-\frac {2 \left (a+b \log \left (c x^n\right )\right )}{e \sqrt {d+e x}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.00 \[ \int \frac {a+b \log \left (c x^n\right )}{(d+e x)^{3/2}} \, dx=-\frac {4 b n \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{\sqrt {d} e}-\frac {2 \left (a+b \log \left (c x^n\right )\right )}{e \sqrt {d+e x}} \]

[In]

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

[Out]

(-4*b*n*ArcTanh[Sqrt[d + e*x]/Sqrt[d]])/(Sqrt[d]*e) - (2*(a + b*Log[c*x^n]))/(e*Sqrt[d + e*x])

Maple [F]

\[\int \frac {a +b \ln \left (c \,x^{n}\right )}{\left (e x +d \right )^{\frac {3}{2}}}d x\]

[In]

int((a+b*ln(c*x^n))/(e*x+d)^(3/2),x)

[Out]

int((a+b*ln(c*x^n))/(e*x+d)^(3/2),x)

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 155, normalized size of antiderivative = 2.92 \[ \int \frac {a+b \log \left (c x^n\right )}{(d+e x)^{3/2}} \, dx=\left [\frac {2 \, {\left ({\left (b e n x + b d n\right )} \sqrt {d} \log \left (\frac {e x - 2 \, \sqrt {e x + d} \sqrt {d} + 2 \, d}{x}\right ) - {\left (b d n \log \left (x\right ) + b d \log \left (c\right ) + a d\right )} \sqrt {e x + d}\right )}}{d e^{2} x + d^{2} e}, \frac {2 \, {\left (2 \, {\left (b e n x + b d n\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {e x + d} \sqrt {-d}}{d}\right ) - {\left (b d n \log \left (x\right ) + b d \log \left (c\right ) + a d\right )} \sqrt {e x + d}\right )}}{d e^{2} x + d^{2} e}\right ] \]

[In]

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

[Out]

[2*((b*e*n*x + b*d*n)*sqrt(d)*log((e*x - 2*sqrt(e*x + d)*sqrt(d) + 2*d)/x) - (b*d*n*log(x) + b*d*log(c) + a*d)
*sqrt(e*x + d))/(d*e^2*x + d^2*e), 2*(2*(b*e*n*x + b*d*n)*sqrt(-d)*arctan(sqrt(e*x + d)*sqrt(-d)/d) - (b*d*n*l
og(x) + b*d*log(c) + a*d)*sqrt(e*x + d))/(d*e^2*x + d^2*e)]

Sympy [A] (verification not implemented)

Time = 3.85 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.66 \[ \int \frac {a+b \log \left (c x^n\right )}{(d+e x)^{3/2}} \, dx=a \left (\begin {cases} - \frac {2}{e \sqrt {d + e x}} & \text {for}\: e \neq 0 \\\frac {x}{d^{\frac {3}{2}}} & \text {otherwise} \end {cases}\right ) - b n \left (\begin {cases} \frac {4 \operatorname {asinh}{\left (\frac {\sqrt {d}}{\sqrt {e} \sqrt {x}} \right )}}{\sqrt {d} e} & \text {for}\: e > -\infty \wedge e < \infty \wedge e \neq 0 \\\frac {x}{d^{\frac {3}{2}}} & \text {otherwise} \end {cases}\right ) + b \left (\begin {cases} - \frac {2}{e \sqrt {d + e x}} & \text {for}\: e \neq 0 \\\frac {x}{d^{\frac {3}{2}}} & \text {otherwise} \end {cases}\right ) \log {\left (c x^{n} \right )} \]

[In]

integrate((a+b*ln(c*x**n))/(e*x+d)**(3/2),x)

[Out]

a*Piecewise((-2/(e*sqrt(d + e*x)), Ne(e, 0)), (x/d**(3/2), True)) - b*n*Piecewise((4*asinh(sqrt(d)/(sqrt(e)*sq
rt(x)))/(sqrt(d)*e), (e > -oo) & (e < oo) & Ne(e, 0)), (x/d**(3/2), True)) + b*Piecewise((-2/(e*sqrt(d + e*x))
, Ne(e, 0)), (x/d**(3/2), True))*log(c*x**n)

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.34 \[ \int \frac {a+b \log \left (c x^n\right )}{(d+e x)^{3/2}} \, dx=\frac {2 \, b n \log \left (\frac {\sqrt {e x + d} - \sqrt {d}}{\sqrt {e x + d} + \sqrt {d}}\right )}{\sqrt {d} e} - \frac {2 \, b \log \left (c x^{n}\right )}{\sqrt {e x + d} e} - \frac {2 \, a}{\sqrt {e x + d} e} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.34 \[ \int \frac {a+b \log \left (c x^n\right )}{(d+e x)^{3/2}} \, dx=\frac {4 \, b n \arctan \left (\frac {\sqrt {e x + d}}{\sqrt {-d}}\right )}{\sqrt {-d} e} - \frac {2 \, b n \log \left (e x\right )}{\sqrt {e x + d} e} + \frac {2 \, {\left (b n \log \left (e\right ) - b \log \left (c\right ) - a\right )}}{\sqrt {e x + d} e} \]

[In]

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

[Out]

4*b*n*arctan(sqrt(e*x + d)/sqrt(-d))/(sqrt(-d)*e) - 2*b*n*log(e*x)/(sqrt(e*x + d)*e) + 2*(b*n*log(e) - b*log(c
) - a)/(sqrt(e*x + d)*e)

Mupad [F(-1)]

Timed out. \[ \int \frac {a+b \log \left (c x^n\right )}{(d+e x)^{3/2}} \, dx=\int \frac {a+b\,\ln \left (c\,x^n\right )}{{\left (d+e\,x\right )}^{3/2}} \,d x \]

[In]

int((a + b*log(c*x^n))/(d + e*x)^(3/2),x)

[Out]

int((a + b*log(c*x^n))/(d + e*x)^(3/2), x)