\(\int (a+b \log (c (d+\frac {e}{\sqrt [3]{x}})^n)) \, dx\) [492]

Optimal result

Integrand size = 18, antiderivative size = 70 \[ \int \left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )\right ) \, dx=-\frac {b e^2 n \sqrt [3]{x}}{d^2}+\frac {b e n x^{2/3}}{2 d}+a x+b x \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )+\frac {b e^3 n \log \left (e+d \sqrt [3]{x}\right )}{d^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {2498, 269, 196, 45} \[ \int \left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )\right ) \, dx=a x+b x \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )+\frac {b e^3 n \log \left (d \sqrt [3]{x}+e\right )}{d^3}-\frac {b e^2 n \sqrt [3]{x}}{d^2}+\frac {b e n x^{2/3}}{2 d} \]

[In]

Int[a + b*Log[c*(d + e/x^(1/3))^n],x]

[Out]

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

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])

Rule 196

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(1/n - 1)*(a + b*x)^p, x], x, x^n], x] /
; FreeQ[{a, b, p}, x] && FractionQ[n] && IntegerQ[1/n]

Rule 269

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[x^(m + n*p)*(b + a/x^n)^p, x] /; FreeQ[{a, b, m
, n}, x] && IntegerQ[p] && NegQ[n]

Rule 2498

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)], x_Symbol] :> Simp[x*Log[c*(d + e*x^n)^p], x] - Dist[e*n*p, Int[
x^n/(d + e*x^n), x], x] /; FreeQ[{c, d, e, n, p}, x]

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.13 \[ \int \left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )\right ) \, dx=a x+b x \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )+\frac {1}{3} b e n \left (-\frac {3 e \sqrt [3]{x}}{d^2}+\frac {3 x^{2/3}}{2 d}+\frac {3 e^2 \log \left (d+\frac {e}{\sqrt [3]{x}}\right )}{d^3}+\frac {e^2 \log (x)}{d^3}\right ) \]

[In]

Integrate[a + b*Log[c*(d + e/x^(1/3))^n],x]

[Out]

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

Maple [A] (verified)

Time = 0.43 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.54

method	result	size
default	\(a x +b \left (x \ln \left (c \left (\frac {e +d \,x^{\frac {1}{3}}}{x^{\frac {1}{3}}}\right )^{n}\right )+\frac {e n \left (\frac {e^{2} \ln \left (d^{3} x +e^{3}\right )}{d^{3}}+\frac {3 x^{\frac {2}{3}}}{2 d}+\frac {2 e^{2} \ln \left (e +d \,x^{\frac {1}{3}}\right )}{d^{3}}-\frac {e^{2} \ln \left (d^{2} x^{\frac {2}{3}}-e d \,x^{\frac {1}{3}}+e^{2}\right )}{d^{3}}-\frac {3 e \,x^{\frac {1}{3}}}{d^{2}}\right )}{3}\right )\)	\(108\)
parts	\(a x +b \left (x \ln \left (c \left (\frac {e +d \,x^{\frac {1}{3}}}{x^{\frac {1}{3}}}\right )^{n}\right )+\frac {e n \left (\frac {e^{2} \ln \left (d^{3} x +e^{3}\right )}{d^{3}}+\frac {3 x^{\frac {2}{3}}}{2 d}+\frac {2 e^{2} \ln \left (e +d \,x^{\frac {1}{3}}\right )}{d^{3}}-\frac {e^{2} \ln \left (d^{2} x^{\frac {2}{3}}-e d \,x^{\frac {1}{3}}+e^{2}\right )}{d^{3}}-\frac {3 e \,x^{\frac {1}{3}}}{d^{2}}\right )}{3}\right )\)	\(108\)

[In]

int(a+b*ln(c*(d+e/x^(1/3))^n),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.53 \[ \int \left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )\right ) \, dx=\frac {2 \, b d^{3} x \log \left (c\right ) - 2 \, b d^{3} n \log \left (x^{\frac {1}{3}}\right ) + b d^{2} e n x^{\frac {2}{3}} - 2 \, b d e^{2} n x^{\frac {1}{3}} + 2 \, a d^{3} x + 2 \, {\left (b d^{3} + b e^{3}\right )} n \log \left (d x^{\frac {1}{3}} + e\right ) + 2 \, {\left (b d^{3} n x - b d^{3} n\right )} \log \left (\frac {d x + e x^{\frac {2}{3}}}{x}\right )}{2 \, d^{3}} \]

[In]

integrate(a+b*log(c*(d+e/x^(1/3))^n),x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 2.29 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.04 \[ \int \left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )\right ) \, dx=a x + b \left (\frac {e n \left (\frac {3 x^{\frac {2}{3}}}{2 d} + \frac {3 e^{2} \left (\begin {cases} \frac {\sqrt [3]{x}}{e} & \text {for}\: d = 0 \\\frac {\log {\left (d \sqrt [3]{x} + e \right )}}{d} & \text {otherwise} \end {cases}\right )}{d^{2}} - \frac {3 e \sqrt [3]{x}}{d^{2}}\right )}{3} + x \log {\left (c \left (d + \frac {e}{\sqrt [3]{x}}\right )^{n} \right )}\right ) \]

[In]

integrate(a+b*ln(c*(d+e/x**(1/3))**n),x)

[Out]

a*x + b*(e*n*(3*x**(2/3)/(2*d) + 3*e**2*Piecewise((x**(1/3)/e, Eq(d, 0)), (log(d*x**(1/3) + e)/d, True))/d**2
- 3*e*x**(1/3)/d**2)/3 + x*log(c*(d + e/x**(1/3))**n))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.84 \[ \int \left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )\right ) \, dx=\frac {1}{2} \, {\left (e n {\left (\frac {2 \, e^{2} \log \left (d x^{\frac {1}{3}} + e\right )}{d^{3}} + \frac {d x^{\frac {2}{3}} - 2 \, e x^{\frac {1}{3}}}{d^{2}}\right )} + 2 \, x \log \left (c {\left (d + \frac {e}{x^{\frac {1}{3}}}\right )}^{n}\right )\right )} b + a x \]

[In]

integrate(a+b*log(c*(d+e/x^(1/3))^n),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.90 \[ \int \left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )\right ) \, dx=\frac {1}{2} \, {\left ({\left (e {\left (\frac {2 \, e^{2} \log \left ({\left | d x^{\frac {1}{3}} + e \right |}\right )}{d^{3}} + \frac {d x^{\frac {2}{3}} - 2 \, e x^{\frac {1}{3}}}{d^{2}}\right )} + 2 \, x \log \left (d + \frac {e}{x^{\frac {1}{3}}}\right )\right )} n + 2 \, x \log \left (c\right )\right )} b + a x \]

[In]

integrate(a+b*log(c*(d+e/x^(1/3))^n),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 1.71 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.84 \[ \int \left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )\right ) \, dx=a\,x+b\,x\,\ln \left (c\,{\left (d+\frac {e}{x^{1/3}}\right )}^n\right )+\frac {b\,\left (2\,e^3\,n\,\ln \left (e+d\,x^{1/3}\right )-2\,d\,e^2\,n\,x^{1/3}+d^2\,e\,n\,x^{2/3}\right )}{2\,d^3} \]

[In]

int(a + b*log(c*(d + e/x^(1/3))^n),x)

[Out]

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