3.6.11 \(\int (a+b \log (c (d+\frac {e}{x^{2/3}})^n)) \, dx\) [511]

Optimal. Leaf size=65 \[ \frac {2 b e n \sqrt [3]{x}}{d}+a x-\frac {2 b e^{3/2} n \tan ^{-1}\left (\frac {\sqrt {d} \sqrt [3]{x}}{\sqrt {e}}\right )}{d^{3/2}}+b x \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right ) \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {2498, 269, 249, 327, 211} \begin {gather*} a x-\frac {2 b e^{3/2} n \text {ArcTan}\left (\frac {\sqrt {d} \sqrt [3]{x}}{\sqrt {e}}\right )}{d^{3/2}}+b x \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )+\frac {2 b e n \sqrt [3]{x}}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[a + b*Log[c*(d + e/x^(2/3))^n],x]

[Out]

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

Rule 211

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

Rule 249

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[n]}, Dist[k, Subst[Int[x^(k - 1)*(a + b*
x^(k*n))^p, x], x, x^(1/k)], x]] /; FreeQ[{a, b, p}, x] && FractionQ[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 327

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^n
)^(p + 1)/(b*(m + n*p + 1))), x] - Dist[a*c^n*((m - n + 1)/(b*(m + n*p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

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

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 0.01, size = 53, normalized size = 0.82 \begin {gather*} a x+\frac {2 b e n \sqrt [3]{x} \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};-\frac {e}{d x^{2/3}}\right )}{d}+b x \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

a*x + (2*b*e*n*x^(1/3)*Hypergeometric2F1[-1/2, 1, 1/2, -(e/(d*x^(2/3)))])/d + b*x*Log[c*(d + e/x^(2/3))^n]

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(167\) vs. \(2(51)=102\).
time = 0.22, size = 168, normalized size = 2.58

method	result	size
default	\(a x +x b \ln \left (c \left (\frac {e +d \,x^{\frac {2}{3}}}{x^{\frac {2}{3}}}\right )^{n}\right )+\frac {2 b \,e^{2} n \arctan \left (\frac {x \,d^{2}}{e \sqrt {e d}}\right )}{3 d \sqrt {e d}}+\frac {2 b e n \,x^{\frac {1}{3}}}{d}-\frac {4 b \,e^{2} n \arctan \left (\frac {d \,x^{\frac {1}{3}}}{\sqrt {e d}}\right )}{3 d \sqrt {e d}}+\frac {2 b \,e^{2} n \arctan \left (\frac {\sqrt {3}\, \sqrt {d}\, \sqrt {e}-2 d \,x^{\frac {1}{3}}}{\sqrt {e d}}\right )}{3 d \sqrt {e d}}-\frac {2 b \,e^{2} n \arctan \left (\frac {2 d \,x^{\frac {1}{3}}+\sqrt {3}\, \sqrt {d}\, \sqrt {e}}{\sqrt {e d}}\right )}{3 d \sqrt {e d}}\)	\(168\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.55, size = 54, normalized size = 0.83 \begin {gather*} -{\left (2 \, n {\left (\frac {\arctan \left (\sqrt {d} x^{\frac {1}{3}} e^{\left (-\frac {1}{2}\right )}\right ) e^{\frac {1}{2}}}{d^{\frac {3}{2}}} - \frac {x^{\frac {1}{3}}}{d}\right )} e - x \log \left (c {\left (d + \frac {e}{x^{\frac {2}{3}}}\right )}^{n}\right )\right )} b + a x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.40, size = 274, normalized size = 4.22 \begin {gather*} \left [\frac {b n \sqrt {-\frac {e}{d}} e \log \left (\frac {d^{3} x^{2} + 2 \, d^{2} x \sqrt {-\frac {e}{d}} e - 2 \, {\left (d^{3} x \sqrt {-\frac {e}{d}} - d e^{2}\right )} x^{\frac {2}{3}} - 2 \, {\left (d^{2} x e + d \sqrt {-\frac {e}{d}} e^{2}\right )} x^{\frac {1}{3}} - e^{3}}{d^{3} x^{2} + e^{3}}\right ) + b d n \log \left (d x^{\frac {2}{3}} + e\right ) + b d x \log \left (c\right ) - 2 \, b d n \log \left (x^{\frac {1}{3}}\right ) + 2 \, b n x^{\frac {1}{3}} e + a d x + {\left (b d n x - b d n\right )} \log \left (\frac {d x + x^{\frac {1}{3}} e}{x}\right )}{d}, \frac {b d n \log \left (d x^{\frac {2}{3}} + e\right ) + b d x \log \left (c\right ) - 2 \, b d n \log \left (x^{\frac {1}{3}}\right ) - \frac {2 \, b n \arctan \left (\sqrt {d} x^{\frac {1}{3}} e^{\left (-\frac {1}{2}\right )}\right ) e^{\frac {3}{2}}}{\sqrt {d}} + 2 \, b n x^{\frac {1}{3}} e + a d x + {\left (b d n x - b d n\right )} \log \left (\frac {d x + x^{\frac {1}{3}} e}{x}\right )}{d}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[(b*n*sqrt(-e/d)*e*log((d^3*x^2 + 2*d^2*x*sqrt(-e/d)*e - 2*(d^3*x*sqrt(-e/d) - d*e^2)*x^(2/3) - 2*(d^2*x*e + d
*sqrt(-e/d)*e^2)*x^(1/3) - e^3)/(d^3*x^2 + e^3)) + b*d*n*log(d*x^(2/3) + e) + b*d*x*log(c) - 2*b*d*n*log(x^(1/
3)) + 2*b*n*x^(1/3)*e + a*d*x + (b*d*n*x - b*d*n)*log((d*x + x^(1/3)*e)/x))/d, (b*d*n*log(d*x^(2/3) + e) + b*d
*x*log(c) - 2*b*d*n*log(x^(1/3)) - 2*b*n*arctan(sqrt(d)*x^(1/3)*e^(-1/2))*e^(3/2)/sqrt(d) + 2*b*n*x^(1/3)*e +
a*d*x + (b*d*n*x - b*d*n)*log((d*x + x^(1/3)*e)/x))/d]

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Sympy [A]
time = 18.16, size = 112, normalized size = 1.72 \begin {gather*} a x + b \left (\frac {2 e n \left (\begin {cases} \tilde {\infty } x & \text {for}\: d = 0 \wedge e = 0 \\\frac {x}{e} & \text {for}\: d = 0 \\\frac {3 \sqrt [3]{x}}{d} & \text {for}\: e = 0 \\\frac {3 \sqrt [3]{x}}{d} - \frac {3 e \log {\left (\sqrt [3]{x} - \sqrt {- \frac {e}{d}} \right )}}{2 d^{2} \sqrt {- \frac {e}{d}}} + \frac {3 e \log {\left (\sqrt [3]{x} + \sqrt {- \frac {e}{d}} \right )}}{2 d^{2} \sqrt {- \frac {e}{d}}} & \text {otherwise} \end {cases}\right )}{3} + x \log {\left (c \left (d + \frac {e}{x^{\frac {2}{3}}}\right )^{n} \right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*x + b*(2*e*n*Piecewise((zoo*x, Eq(d, 0) & Eq(e, 0)), (x/e, Eq(d, 0)), (3*x**(1/3)/d, Eq(e, 0)), (3*x**(1/3)/
d - 3*e*log(x**(1/3) - sqrt(-e/d))/(2*d**2*sqrt(-e/d)) + 3*e*log(x**(1/3) + sqrt(-e/d))/(2*d**2*sqrt(-e/d)), T
rue))/3 + x*log(c*(d + e/x**(2/3))**n))

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Giac [A]
time = 3.11, size = 57, normalized size = 0.88 \begin {gather*} -{\left ({\left (2 \, {\left (\frac {\arctan \left (\sqrt {d} x^{\frac {1}{3}} e^{\left (-\frac {1}{2}\right )}\right ) e^{\frac {1}{2}}}{d^{\frac {3}{2}}} - \frac {x^{\frac {1}{3}}}{d}\right )} e - x \log \left (d + \frac {e}{x^{\frac {2}{3}}}\right )\right )} n - x \log \left (c\right )\right )} b + a x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.43, size = 51, normalized size = 0.78 \begin {gather*} a\,x+b\,x\,\ln \left (c\,{\left (d+\frac {e}{x^{2/3}}\right )}^n\right )+\frac {2\,b\,e\,n\,x^{1/3}}{d}-\frac {2\,b\,e^{3/2}\,n\,\mathrm {atan}\left (\frac {\sqrt {d}\,x^{1/3}}{\sqrt {e}}\right )}{d^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(a + b*log(c*(d + e/x^(2/3))^n),x)

[Out]

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

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