\(\int x (a+b \log (c (d+\frac {e}{\sqrt {x}})^n)) \, dx\) [423]

Optimal result

Integrand size = 20, antiderivative size = 107 \[ \int x \left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )\right ) \, dx=\frac {b e^3 n \sqrt {x}}{2 d^3}-\frac {b e^2 n x}{4 d^2}+\frac {b e n x^{3/2}}{6 d}-\frac {b e^4 n \log \left (d+\frac {e}{\sqrt {x}}\right )}{2 d^4}+\frac {1}{2} x^2 \left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )\right )-\frac {b e^4 n \log (x)}{4 d^4} \]

[Out]

-1/4*b*e^2*n*x/d^2+1/6*b*e*n*x^(3/2)/d-1/4*b*e^4*n*ln(x)/d^4-1/2*b*e^4*n*ln(d+e/x^(1/2))/d^4+1/2*x^2*(a+b*ln(c
*(d+e/x^(1/2))^n))+1/2*b*e^3*n*x^(1/2)/d^3

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {2504, 2442, 46} \[ \int x \left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )\right ) \, dx=\frac {1}{2} x^2 \left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )\right )-\frac {b e^4 n \log \left (d+\frac {e}{\sqrt {x}}\right )}{2 d^4}-\frac {b e^4 n \log (x)}{4 d^4}+\frac {b e^3 n \sqrt {x}}{2 d^3}-\frac {b e^2 n x}{4 d^2}+\frac {b e n x^{3/2}}{6 d} \]

[In]

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

[Out]

(b*e^3*n*Sqrt[x])/(2*d^3) - (b*e^2*n*x)/(4*d^2) + (b*e*n*x^(3/2))/(6*d) - (b*e^4*n*Log[d + e/Sqrt[x]])/(2*d^4)
+ (x^2*(a + b*Log[c*(d + e/Sqrt[x])^n]))/2 - (b*e^4*n*Log[x])/(4*d^4)

Rule 46

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] && Lt
Q[m + n + 2, 0])

Rule 2442

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))*((f_.) + (g_.)*(x_))^(q_.), x_Symbol] :> Simp[(f + g*
x)^(q + 1)*((a + b*Log[c*(d + e*x)^n])/(g*(q + 1))), x] - Dist[b*e*(n/(g*(q + 1))), Int[(f + g*x)^(q + 1)/(d +
e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n, q}, x] && NeQ[e*f - d*g, 0] && NeQ[q, -1]

Rule 2504

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[I
nt[x^(Simplify[(m + 1)/n] - 1)*(a + b*Log[c*(d + e*x)^p])^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, e, m, n, p,
q}, x] && IntegerQ[Simplify[(m + 1)/n]] && (GtQ[(m + 1)/n, 0] || IGtQ[q, 0]) &&  !(EqQ[q, 1] && ILtQ[n, 0] &&
IGtQ[m, 0])

Rubi steps \begin{align*} \text {integral}& = -\left (2 \text {Subst}\left (\int \frac {a+b \log \left (c (d+e x)^n\right )}{x^5} \, dx,x,\frac {1}{\sqrt {x}}\right )\right ) \\ & = \frac {1}{2} x^2 \left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )\right )-\frac {1}{2} (b e n) \text {Subst}\left (\int \frac {1}{x^4 (d+e x)} \, dx,x,\frac {1}{\sqrt {x}}\right ) \\ & = \frac {1}{2} x^2 \left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )\right )-\frac {1}{2} (b e n) \text {Subst}\left (\int \left (\frac {1}{d x^4}-\frac {e}{d^2 x^3}+\frac {e^2}{d^3 x^2}-\frac {e^3}{d^4 x}+\frac {e^4}{d^4 (d+e x)}\right ) \, dx,x,\frac {1}{\sqrt {x}}\right ) \\ & = \frac {b e^3 n \sqrt {x}}{2 d^3}-\frac {b e^2 n x}{4 d^2}+\frac {b e n x^{3/2}}{6 d}-\frac {b e^4 n \log \left (d+\frac {e}{\sqrt {x}}\right )}{2 d^4}+\frac {1}{2} x^2 \left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )\right )-\frac {b e^4 n \log (x)}{4 d^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.93 \[ \int x \left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )\right ) \, dx=\frac {a x^2}{2}+\frac {1}{2} b x^2 \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )+\frac {1}{4} b e n \left (\frac {2 e^2 \sqrt {x}}{d^3}-\frac {e x}{d^2}+\frac {2 x^{3/2}}{3 d}-\frac {2 e^3 \log \left (d+\frac {e}{\sqrt {x}}\right )}{d^4}-\frac {e^3 \log (x)}{d^4}\right ) \]

[In]

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

[Out]

(a*x^2)/2 + (b*x^2*Log[c*(d + e/Sqrt[x])^n])/2 + (b*e*n*((2*e^2*Sqrt[x])/d^3 - (e*x)/d^2 + (2*x^(3/2))/(3*d) -
(2*e^3*Log[d + e/Sqrt[x]])/d^4 - (e^3*Log[x])/d^4))/4

Maple [F]

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

[In]

int(x*(a+b*ln(c*(d+e/x^(1/2))^n)),x)

[Out]

int(x*(a+b*ln(c*(d+e/x^(1/2))^n)),x)

Fricas [A] (verification not implemented)

none

Time = 0.39 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.18 \[ \int x \left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )\right ) \, dx=\frac {6 \, b d^{4} x^{2} \log \left (c\right ) - 3 \, b d^{2} e^{2} n x + 6 \, a d^{4} x^{2} - 6 \, b d^{4} n \log \left (\sqrt {x}\right ) + 6 \, {\left (b d^{4} - b e^{4}\right )} n \log \left (d \sqrt {x} + e\right ) + 6 \, {\left (b d^{4} n x^{2} - b d^{4} n\right )} \log \left (\frac {d x + e \sqrt {x}}{x}\right ) + 2 \, {\left (b d^{3} e n x + 3 \, b d e^{3} n\right )} \sqrt {x}}{12 \, d^{4}} \]

[In]

integrate(x*(a+b*log(c*(d+e/x^(1/2))^n)),x, algorithm="fricas")

[Out]

1/12*(6*b*d^4*x^2*log(c) - 3*b*d^2*e^2*n*x + 6*a*d^4*x^2 - 6*b*d^4*n*log(sqrt(x)) + 6*(b*d^4 - b*e^4)*n*log(d*
sqrt(x) + e) + 6*(b*d^4*n*x^2 - b*d^4*n)*log((d*x + e*sqrt(x))/x) + 2*(b*d^3*e*n*x + 3*b*d*e^3*n)*sqrt(x))/d^4

Sympy [A] (verification not implemented)

Time = 6.59 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.82 \[ \int x \left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )\right ) \, dx=\frac {a x^{2}}{2} + b \left (\frac {e n \left (\frac {2 x^{\frac {3}{2}}}{3 d} - \frac {e x}{d^{2}} - \frac {2 e^{3} \left (\begin {cases} \frac {\sqrt {x}}{e} & \text {for}\: d = 0 \\\frac {\log {\left (d \sqrt {x} + e \right )}}{d} & \text {otherwise} \end {cases}\right )}{d^{3}} + \frac {2 e^{2} \sqrt {x}}{d^{3}}\right )}{4} + \frac {x^{2} \log {\left (c \left (d + \frac {e}{\sqrt {x}}\right )^{n} \right )}}{2}\right ) \]

[In]

integrate(x*(a+b*ln(c*(d+e/x**(1/2))**n)),x)

[Out]

a*x**2/2 + b*(e*n*(2*x**(3/2)/(3*d) - e*x/d**2 - 2*e**3*Piecewise((sqrt(x)/e, Eq(d, 0)), (log(d*sqrt(x) + e)/d
, True))/d**3 + 2*e**2*sqrt(x)/d**3)/4 + x**2*log(c*(d + e/sqrt(x))**n)/2)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.69 \[ \int x \left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )\right ) \, dx=-\frac {1}{12} \, b e n {\left (\frac {6 \, e^{3} \log \left (d \sqrt {x} + e\right )}{d^{4}} - \frac {2 \, d^{2} x^{\frac {3}{2}} - 3 \, d e x + 6 \, e^{2} \sqrt {x}}{d^{3}}\right )} + \frac {1}{2} \, b x^{2} \log \left (c {\left (d + \frac {e}{\sqrt {x}}\right )}^{n}\right ) + \frac {1}{2} \, a x^{2} \]

[In]

integrate(x*(a+b*log(c*(d+e/x^(1/2))^n)),x, algorithm="maxima")

[Out]

-1/12*b*e*n*(6*e^3*log(d*sqrt(x) + e)/d^4 - (2*d^2*x^(3/2) - 3*d*e*x + 6*e^2*sqrt(x))/d^3) + 1/2*b*x^2*log(c*(
d + e/sqrt(x))^n) + 1/2*a*x^2

Giac [A] (verification not implemented)

none

Time = 0.51 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.76 \[ \int x \left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )\right ) \, dx=\frac {1}{2} \, b x^{2} \log \left (c\right ) + \frac {1}{12} \, {\left (6 \, x^{2} \log \left (d + \frac {e}{\sqrt {x}}\right ) - e {\left (\frac {6 \, e^{3} \log \left ({\left | d \sqrt {x} + e \right |}\right )}{d^{4}} - \frac {2 \, d^{2} x^{\frac {3}{2}} - 3 \, d e x + 6 \, e^{2} \sqrt {x}}{d^{3}}\right )}\right )} b n + \frac {1}{2} \, a x^{2} \]

[In]

integrate(x*(a+b*log(c*(d+e/x^(1/2))^n)),x, algorithm="giac")

[Out]

1/2*b*x^2*log(c) + 1/12*(6*x^2*log(d + e/sqrt(x)) - e*(6*e^3*log(abs(d*sqrt(x) + e))/d^4 - (2*d^2*x^(3/2) - 3*
d*e*x + 6*e^2*sqrt(x))/d^3))*b*n + 1/2*a*x^2

Mupad [B] (verification not implemented)

Time = 2.10 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.80 \[ \int x \left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )\right ) \, dx=\frac {x^{3/2}\,\left (\frac {b\,e\,n}{3\,d}-\frac {b\,e^2\,n}{2\,d^2\,\sqrt {x}}+\frac {b\,e^3\,n}{d^3\,x}\right )}{2}+\frac {a\,x^2}{2}+\frac {b\,x^2\,\ln \left (c\,{\left (d+\frac {e}{\sqrt {x}}\right )}^n\right )}{2}-\frac {b\,e^4\,n\,\mathrm {atanh}\left (\frac {2\,e}{d\,\sqrt {x}}+1\right )}{d^4} \]

[In]

int(x*(a + b*log(c*(d + e/x^(1/2))^n)),x)

[Out]

(x^(3/2)*((b*e*n)/(3*d) - (b*e^2*n)/(2*d^2*x^(1/2)) + (b*e^3*n)/(d^3*x)))/2 + (a*x^2)/2 + (b*x^2*log(c*(d + e/
x^(1/2))^n))/2 - (b*e^4*n*atanh((2*e)/(d*x^(1/2)) + 1))/d^4