3.5.68 \(\int \frac {a+b \log (c (d+e x^{2/3})^n)}{x^2} \, dx\) [468]

Optimal. Leaf size=68 \[ -\frac {2 b e n}{d \sqrt [3]{x}}-\frac {2 b e^{3/2} n \tan ^{-1}\left (\frac {\sqrt {e} \sqrt [3]{x}}{\sqrt {d}}\right )}{d^{3/2}}-\frac {a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )}{x} \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {2505, 348, 331, 211} \begin {gather*} -\frac {a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )}{x}-\frac {2 b e^{3/2} n \text {ArcTan}\left (\frac {\sqrt {e} \sqrt [3]{x}}{\sqrt {d}}\right )}{d^{3/2}}-\frac {2 b e n}{d \sqrt [3]{x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 331

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

Rule 348

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

Rule 2505

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

Rubi steps

\begin {align*} \int \frac {a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )}{x^2} \, dx &=-\frac {a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )}{x}+\frac {1}{3} (2 b e n) \int \frac {1}{\left (d+e x^{2/3}\right ) x^{4/3}} \, dx\\ &=-\frac {a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )}{x}+(2 b e n) \text {Subst}\left (\int \frac {1}{x^2 \left (d+e x^2\right )} \, dx,x,\sqrt [3]{x}\right )\\ &=-\frac {2 b e n}{d \sqrt [3]{x}}-\frac {a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )}{x}-\frac {\left (2 b e^2 n\right ) \text {Subst}\left (\int \frac {1}{d+e x^2} \, dx,x,\sqrt [3]{x}\right )}{d}\\ &=-\frac {2 b e n}{d \sqrt [3]{x}}-\frac {2 b e^{3/2} n \tan ^{-1}\left (\frac {\sqrt {e} \sqrt [3]{x}}{\sqrt {d}}\right )}{d^{3/2}}-\frac {a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )}{x}\\ \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 = 59, normalized size = 0.87 \begin {gather*} -\frac {a}{x}-\frac {2 b e n \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};-\frac {e x^{2/3}}{d}\right )}{d \sqrt [3]{x}}-\frac {b \log \left (c \left (d+e x^{2/3}\right )^n\right )}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {a +b \ln \left (c \left (d +e \,x^{\frac {2}{3}}\right )^{n}\right )}{x^{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.50, size = 56, normalized size = 0.82 \begin {gather*} -2 \, b n {\left (\frac {\arctan \left (\frac {x^{\frac {1}{3}} e^{\frac {1}{2}}}{\sqrt {d}}\right ) e^{\frac {1}{2}}}{d^{\frac {3}{2}}} + \frac {1}{d x^{\frac {1}{3}}}\right )} e - \frac {b \log \left ({\left (x^{\frac {2}{3}} e + d\right )}^{n} c\right )}{x} - \frac {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^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.38, size = 210, normalized size = 3.09 \begin {gather*} \left [\frac {b n x \sqrt {-\frac {e}{d}} e \log \left (\frac {2 \, d^{2} x \sqrt {-\frac {e}{d}} e - d^{3} + x^{2} e^{3} - 2 \, {\left (d x \sqrt {-\frac {e}{d}} e^{2} - d^{2} e\right )} x^{\frac {2}{3}} - 2 \, {\left (d^{3} \sqrt {-\frac {e}{d}} + d x e^{2}\right )} x^{\frac {1}{3}}}{d^{3} + x^{2} e^{3}}\right ) - b d n \log \left (x^{\frac {2}{3}} e + d\right ) - 2 \, b n x^{\frac {2}{3}} e - b d \log \left (c\right ) - a d}{d x}, -\frac {\frac {2 \, b n x \arctan \left (\frac {x^{\frac {1}{3}} e^{\frac {1}{2}}}{\sqrt {d}}\right ) e^{\frac {3}{2}}}{\sqrt {d}} + b d n \log \left (x^{\frac {2}{3}} e + d\right ) + 2 \, b n x^{\frac {2}{3}} e + b d \log \left (c\right ) + a d}{d x}\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^2,x, algorithm="fricas")

[Out]

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

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \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**2,x)

[Out]

Timed out

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Giac [A]
time = 4.90, size = 61, normalized size = 0.90 \begin {gather*} -{\left (2 \, {\left (\frac {\arctan \left (\frac {x^{\frac {1}{3}} e^{\frac {1}{2}}}{\sqrt {d}}\right ) e^{\frac {1}{2}}}{d^{\frac {3}{2}}} + \frac {1}{d x^{\frac {1}{3}}}\right )} e + \frac {\log \left (x^{\frac {2}{3}} e + d\right )}{x}\right )} b n - \frac {b \log \left (c\right )}{x} - \frac {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^2,x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {a+b\,\ln \left (c\,{\left (d+e\,x^{2/3}\right )}^n\right )}{x^2} \,d x \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^2,x)

[Out]

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

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