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

Optimal. Leaf size=77 \[ a x+b x \log \left (c \left (d+e \sqrt [3]{x}\right )^n\right )+\frac {b d^3 n \log \left (d+e \sqrt [3]{x}\right )}{e^3}-\frac {b d^2 n \sqrt [3]{x}}{e^2}+\frac {b d n x^{2/3}}{2 e}-\frac {b n x}{3} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2448, 266, 43} \[ a x+b x \log \left (c \left (d+e \sqrt [3]{x}\right )^n\right )-\frac {b d^2 n \sqrt [3]{x}}{e^2}+\frac {b d^3 n \log \left (d+e \sqrt [3]{x}\right )}{e^3}+\frac {b d n x^{2/3}}{2 e}-\frac {b n x}{3} \]

Antiderivative was successfully verified.

[In]

Int[a + b*Log[c*(d + e*x^(1/3))^n],x]

[Out]

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

Rule 43

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 266

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

Rule 2448

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

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Mathematica [A]  time = 0.04, size = 77, normalized size = 1.00 \[ a x+b x \log \left (c \left (d+e \sqrt [3]{x}\right )^n\right )+\frac {b d^3 n \log \left (d+e \sqrt [3]{x}\right )}{e^3}-\frac {b d^2 n \sqrt [3]{x}}{e^2}+\frac {b d n x^{2/3}}{2 e}-\frac {b n x}{3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.46, size = 77, normalized size = 1.00 \[ \frac {6 \, b e^{3} x \log \relax (c) + 3 \, b d e^{2} n x^{\frac {2}{3}} - 6 \, b d^{2} e n x^{\frac {1}{3}} - 2 \, {\left (b e^{3} n - 3 \, a e^{3}\right )} x + 6 \, {\left (b e^{3} n x + b d^{3} n\right )} \log \left (e x^{\frac {1}{3}} + d\right )}{6 \, e^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.17, size = 135, normalized size = 1.75 \[ \frac {1}{6} \, {\left (6 \, x e \log \relax (c) + {\left (6 \, {\left (x^{\frac {1}{3}} e + d\right )}^{3} e^{\left (-2\right )} \log \left (x^{\frac {1}{3}} e + d\right ) - 18 \, {\left (x^{\frac {1}{3}} e + d\right )}^{2} d e^{\left (-2\right )} \log \left (x^{\frac {1}{3}} e + d\right ) + 18 \, {\left (x^{\frac {1}{3}} e + d\right )} d^{2} e^{\left (-2\right )} \log \left (x^{\frac {1}{3}} e + d\right ) - 2 \, {\left (x^{\frac {1}{3}} e + d\right )}^{3} e^{\left (-2\right )} + 9 \, {\left (x^{\frac {1}{3}} e + d\right )}^{2} d e^{\left (-2\right )} - 18 \, {\left (x^{\frac {1}{3}} e + d\right )} d^{2} e^{\left (-2\right )}\right )} n\right )} b e^{\left (-1\right )} + a x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(6*x*e*log(c) + (6*(x^(1/3)*e + d)^3*e^(-2)*log(x^(1/3)*e + d) - 18*(x^(1/3)*e + d)^2*d*e^(-2)*log(x^(1/3)
*e + d) + 18*(x^(1/3)*e + d)*d^2*e^(-2)*log(x^(1/3)*e + d) - 2*(x^(1/3)*e + d)^3*e^(-2) + 9*(x^(1/3)*e + d)^2*
d*e^(-2) - 18*(x^(1/3)*e + d)*d^2*e^(-2))*n)*b*e^(-1) + a*x

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maple [A]  time = 0.07, size = 66, normalized size = 0.86 \[ \frac {b \,d^{3} n \ln \left (e \,x^{\frac {1}{3}}+d \right )}{e^{3}}-\frac {b n x}{3}+b x \ln \left (c \left (e \,x^{\frac {1}{3}}+d \right )^{n}\right )+\frac {b d n \,x^{\frac {2}{3}}}{2 e}-\frac {b \,d^{2} n \,x^{\frac {1}{3}}}{e^{2}}+a x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.81, size = 70, normalized size = 0.91 \[ \frac {1}{6} \, {\left (e n {\left (\frac {6 \, d^{3} \log \left (e x^{\frac {1}{3}} + d\right )}{e^{4}} - \frac {2 \, e^{2} x - 3 \, d e x^{\frac {2}{3}} + 6 \, d^{2} x^{\frac {1}{3}}}{e^{3}}\right )} + 6 \, x \log \left ({\left (e x^{\frac {1}{3}} + d\right )}^{n} c\right )\right )} b + a x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.34, size = 65, normalized size = 0.84 \[ a\,x+b\,x\,\ln \left (c\,{\left (d+e\,x^{1/3}\right )}^n\right )-\frac {b\,n\,x}{3}+\frac {b\,d\,n\,x^{2/3}}{2\,e}+\frac {b\,d^3\,n\,\ln \left (d+e\,x^{1/3}\right )}{e^3}-\frac {b\,d^2\,n\,x^{1/3}}{e^2} \]

Verification of antiderivative is not currently implemented for this CAS.

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

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sympy [A]  time = 1.71, size = 82, normalized size = 1.06 \[ a x + b \left (- \frac {e n \left (- \frac {3 d^{3} \left (\begin {cases} \frac {\sqrt [3]{x}}{d} & \text {for}\: e = 0 \\\frac {\log {\left (d + e \sqrt [3]{x} \right )}}{e} & \text {otherwise} \end {cases}\right )}{e^{3}} + \frac {3 d^{2} \sqrt [3]{x}}{e^{3}} - \frac {3 d x^{\frac {2}{3}}}{2 e^{2}} + \frac {x}{e}\right )}{3} + x \log {\left (c \left (d + e \sqrt [3]{x}\right )^{n} \right )}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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