3.515 \(\int \frac {a+b \log (c (d+\frac {e}{x^{2/3}})^n)}{x^4} \, dx\)

Optimal. Leaf size=132 \[ -\frac {a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )}{3 x^3}+\frac {2 b d^{9/2} n \tan ^{-1}\left (\frac {\sqrt {d} \sqrt [3]{x}}{\sqrt {e}}\right )}{3 e^{9/2}}+\frac {2 b d^4 n}{3 e^4 \sqrt [3]{x}}-\frac {2 b d^3 n}{9 e^3 x}+\frac {2 b d^2 n}{15 e^2 x^{5/3}}-\frac {2 b d n}{21 e x^{7/3}}+\frac {2 b n}{27 x^3} \]

[Out]

2/27*b*n/x^3-2/21*b*d*n/e/x^(7/3)+2/15*b*d^2*n/e^2/x^(5/3)-2/9*b*d^3*n/e^3/x+2/3*b*d^4*n/e^4/x^(1/3)+2/3*b*d^(
9/2)*n*arctan(x^(1/3)*d^(1/2)/e^(1/2))/e^(9/2)+1/3*(-a-b*ln(c*(d+e/x^(2/3))^n))/x^3

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Rubi [A]  time = 0.09, antiderivative size = 132, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {2455, 263, 341, 325, 205} \[ -\frac {a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )}{3 x^3}+\frac {2 b d^2 n}{15 e^2 x^{5/3}}+\frac {2 b d^4 n}{3 e^4 \sqrt [3]{x}}-\frac {2 b d^3 n}{9 e^3 x}+\frac {2 b d^{9/2} n \tan ^{-1}\left (\frac {\sqrt {d} \sqrt [3]{x}}{\sqrt {e}}\right )}{3 e^{9/2}}-\frac {2 b d n}{21 e x^{7/3}}+\frac {2 b n}{27 x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*b*n)/(27*x^3) - (2*b*d*n)/(21*e*x^(7/3)) + (2*b*d^2*n)/(15*e^2*x^(5/3)) - (2*b*d^3*n)/(9*e^3*x) + (2*b*d^4*
n)/(3*e^4*x^(1/3)) + (2*b*d^(9/2)*n*ArcTan[(Sqrt[d]*x^(1/3))/Sqrt[e]])/(3*e^(9/2)) - (a + b*Log[c*(d + e/x^(2/
3))^n])/(3*x^3)

Rule 205

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

Rule 263

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 325

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 341

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 2455

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+\frac {e}{x^{2/3}}\right )^n\right )}{x^4} \, dx &=-\frac {a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )}{3 x^3}-\frac {1}{9} (2 b e n) \int \frac {1}{\left (d+\frac {e}{x^{2/3}}\right ) x^{14/3}} \, dx\\ &=-\frac {a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )}{3 x^3}-\frac {1}{9} (2 b e n) \int \frac {1}{\left (e+d x^{2/3}\right ) x^4} \, dx\\ &=-\frac {a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )}{3 x^3}-\frac {1}{3} (2 b e n) \operatorname {Subst}\left (\int \frac {1}{x^{10} \left (e+d x^2\right )} \, dx,x,\sqrt [3]{x}\right )\\ &=\frac {2 b n}{27 x^3}-\frac {a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )}{3 x^3}+\frac {1}{3} (2 b d n) \operatorname {Subst}\left (\int \frac {1}{x^8 \left (e+d x^2\right )} \, dx,x,\sqrt [3]{x}\right )\\ &=\frac {2 b n}{27 x^3}-\frac {2 b d n}{21 e x^{7/3}}-\frac {a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )}{3 x^3}-\frac {\left (2 b d^2 n\right ) \operatorname {Subst}\left (\int \frac {1}{x^6 \left (e+d x^2\right )} \, dx,x,\sqrt [3]{x}\right )}{3 e}\\ &=\frac {2 b n}{27 x^3}-\frac {2 b d n}{21 e x^{7/3}}+\frac {2 b d^2 n}{15 e^2 x^{5/3}}-\frac {a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )}{3 x^3}+\frac {\left (2 b d^3 n\right ) \operatorname {Subst}\left (\int \frac {1}{x^4 \left (e+d x^2\right )} \, dx,x,\sqrt [3]{x}\right )}{3 e^2}\\ &=\frac {2 b n}{27 x^3}-\frac {2 b d n}{21 e x^{7/3}}+\frac {2 b d^2 n}{15 e^2 x^{5/3}}-\frac {2 b d^3 n}{9 e^3 x}-\frac {a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )}{3 x^3}-\frac {\left (2 b d^4 n\right ) \operatorname {Subst}\left (\int \frac {1}{x^2 \left (e+d x^2\right )} \, dx,x,\sqrt [3]{x}\right )}{3 e^3}\\ &=\frac {2 b n}{27 x^3}-\frac {2 b d n}{21 e x^{7/3}}+\frac {2 b d^2 n}{15 e^2 x^{5/3}}-\frac {2 b d^3 n}{9 e^3 x}+\frac {2 b d^4 n}{3 e^4 \sqrt [3]{x}}-\frac {a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )}{3 x^3}+\frac {\left (2 b d^5 n\right ) \operatorname {Subst}\left (\int \frac {1}{e+d x^2} \, dx,x,\sqrt [3]{x}\right )}{3 e^4}\\ &=\frac {2 b n}{27 x^3}-\frac {2 b d n}{21 e x^{7/3}}+\frac {2 b d^2 n}{15 e^2 x^{5/3}}-\frac {2 b d^3 n}{9 e^3 x}+\frac {2 b d^4 n}{3 e^4 \sqrt [3]{x}}+\frac {2 b d^{9/2} n \tan ^{-1}\left (\frac {\sqrt {d} \sqrt [3]{x}}{\sqrt {e}}\right )}{3 e^{9/2}}-\frac {a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )}{3 x^3}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 137, normalized size = 1.04 \[ -\frac {a}{3 x^3}-\frac {b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )}{3 x^3}-\frac {2 b d^{9/2} n \tan ^{-1}\left (\frac {\sqrt {e}}{\sqrt {d} \sqrt [3]{x}}\right )}{3 e^{9/2}}+\frac {2 b d^4 n}{3 e^4 \sqrt [3]{x}}-\frac {2 b d^3 n}{9 e^3 x}+\frac {2 b d^2 n}{15 e^2 x^{5/3}}-\frac {2 b d n}{21 e x^{7/3}}+\frac {2 b n}{27 x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*a/x^3 + (2*b*n)/(27*x^3) - (2*b*d*n)/(21*e*x^(7/3)) + (2*b*d^2*n)/(15*e^2*x^(5/3)) - (2*b*d^3*n)/(9*e^3*x
) + (2*b*d^4*n)/(3*e^4*x^(1/3)) - (2*b*d^(9/2)*n*ArcTan[Sqrt[e]/(Sqrt[d]*x^(1/3))])/(3*e^(9/2)) - (b*Log[c*(d
+ e/x^(2/3))^n])/(3*x^3)

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fricas [A]  time = 0.48, size = 339, normalized size = 2.57 \[ \left [\frac {315 \, b d^{4} n x^{3} \sqrt {-\frac {d}{e}} \log \left (\frac {d^{3} x^{2} - 2 \, d e^{2} x \sqrt {-\frac {d}{e}} - e^{3} + 2 \, {\left (d^{2} e x \sqrt {-\frac {d}{e}} + d e^{2}\right )} x^{\frac {2}{3}} - 2 \, {\left (d^{2} e x - e^{3} \sqrt {-\frac {d}{e}}\right )} x^{\frac {1}{3}}}{d^{3} x^{2} + e^{3}}\right ) - 210 \, b d^{3} e n x^{2} + 126 \, b d^{2} e^{2} n x^{\frac {4}{3}} - 315 \, b e^{4} n \log \left (\frac {d x + e x^{\frac {1}{3}}}{x}\right ) + 70 \, b e^{4} n - 315 \, b e^{4} \log \relax (c) - 315 \, a e^{4} + 90 \, {\left (7 \, b d^{4} n x^{2} - b d e^{3} n\right )} x^{\frac {2}{3}}}{945 \, e^{4} x^{3}}, \frac {630 \, b d^{4} n x^{3} \sqrt {\frac {d}{e}} \arctan \left (x^{\frac {1}{3}} \sqrt {\frac {d}{e}}\right ) - 210 \, b d^{3} e n x^{2} + 126 \, b d^{2} e^{2} n x^{\frac {4}{3}} - 315 \, b e^{4} n \log \left (\frac {d x + e x^{\frac {1}{3}}}{x}\right ) + 70 \, b e^{4} n - 315 \, b e^{4} \log \relax (c) - 315 \, a e^{4} + 90 \, {\left (7 \, b d^{4} n x^{2} - b d e^{3} n\right )} x^{\frac {2}{3}}}{945 \, e^{4} x^{3}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*(d+e/x^(2/3))^n))/x^4,x, algorithm="fricas")

[Out]

[1/945*(315*b*d^4*n*x^3*sqrt(-d/e)*log((d^3*x^2 - 2*d*e^2*x*sqrt(-d/e) - e^3 + 2*(d^2*e*x*sqrt(-d/e) + d*e^2)*
x^(2/3) - 2*(d^2*e*x - e^3*sqrt(-d/e))*x^(1/3))/(d^3*x^2 + e^3)) - 210*b*d^3*e*n*x^2 + 126*b*d^2*e^2*n*x^(4/3)
- 315*b*e^4*n*log((d*x + e*x^(1/3))/x) + 70*b*e^4*n - 315*b*e^4*log(c) - 315*a*e^4 + 90*(7*b*d^4*n*x^2 - b*d*
e^3*n)*x^(2/3))/(e^4*x^3), 1/945*(630*b*d^4*n*x^3*sqrt(d/e)*arctan(x^(1/3)*sqrt(d/e)) - 210*b*d^3*e*n*x^2 + 12
6*b*d^2*e^2*n*x^(4/3) - 315*b*e^4*n*log((d*x + e*x^(1/3))/x) + 70*b*e^4*n - 315*b*e^4*log(c) - 315*a*e^4 + 90*
(7*b*d^4*n*x^2 - b*d*e^3*n)*x^(2/3))/(e^4*x^3)]

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giac [A]  time = 0.46, size = 103, normalized size = 0.78 \[ \frac {1}{945} \, {\left (2 \, {\left (315 \, d^{\frac {9}{2}} \arctan \left (\sqrt {d} x^{\frac {1}{3}} e^{\left (-\frac {1}{2}\right )}\right ) e^{\left (-\frac {11}{2}\right )} + \frac {{\left (315 \, d^{4} x^{\frac {8}{3}} - 105 \, d^{3} x^{2} e + 63 \, d^{2} x^{\frac {4}{3}} e^{2} - 45 \, d x^{\frac {2}{3}} e^{3} + 35 \, e^{4}\right )} e^{\left (-5\right )}}{x^{3}}\right )} e - \frac {315 \, \log \left (d + \frac {e}{x^{\frac {2}{3}}}\right )}{x^{3}}\right )} b n - \frac {b \log \relax (c)}{3 \, x^{3}} - \frac {a}{3 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*(d+e/x^(2/3))^n))/x^4,x, algorithm="giac")

[Out]

1/945*(2*(315*d^(9/2)*arctan(sqrt(d)*x^(1/3)*e^(-1/2))*e^(-11/2) + (315*d^4*x^(8/3) - 105*d^3*x^2*e + 63*d^2*x
^(4/3)*e^2 - 45*d*x^(2/3)*e^3 + 35*e^4)*e^(-5)/x^3)*e - 315*log(d + e/x^(2/3))/x^3)*b*n - 1/3*b*log(c)/x^3 - 1
/3*a/x^3

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maple [F]  time = 0.10, size = 0, normalized size = 0.00 \[ \int \frac {b \ln \left (c \left (d +\frac {e}{x^{\frac {2}{3}}}\right )^{n}\right )+a}{x^{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*ln(c*(d+e/x^(2/3))^n)+a)/x^4,x)

[Out]

int((b*ln(c*(d+e/x^(2/3))^n)+a)/x^4,x)

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maxima [A]  time = 1.00, size = 105, normalized size = 0.80 \[ \frac {2}{945} \, b e n {\left (\frac {315 \, d^{5} \arctan \left (\frac {d x^{\frac {1}{3}}}{\sqrt {d e}}\right )}{\sqrt {d e} e^{5}} + \frac {315 \, d^{4} x^{\frac {8}{3}} - 105 \, d^{3} e x^{2} + 63 \, d^{2} e^{2} x^{\frac {4}{3}} - 45 \, d e^{3} x^{\frac {2}{3}} + 35 \, e^{4}}{e^{5} x^{3}}\right )} - \frac {b \log \left (c {\left (d + \frac {e}{x^{\frac {2}{3}}}\right )}^{n}\right )}{3 \, x^{3}} - \frac {a}{3 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*(d+e/x^(2/3))^n))/x^4,x, algorithm="maxima")

[Out]

2/945*b*e*n*(315*d^5*arctan(d*x^(1/3)/sqrt(d*e))/(sqrt(d*e)*e^5) + (315*d^4*x^(8/3) - 105*d^3*e*x^2 + 63*d^2*e
^2*x^(4/3) - 45*d*e^3*x^(2/3) + 35*e^4)/(e^5*x^3)) - 1/3*b*log(c*(d + e/x^(2/3))^n)/x^3 - 1/3*a/x^3

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {a+b\,\ln \left (c\,{\left (d+\frac {e}{x^{2/3}}\right )}^n\right )}{x^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*log(c*(d + e/x^(2/3))^n))/x^4,x)

[Out]

int((a + b*log(c*(d + e/x^(2/3))^n))/x^4, x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*ln(c*(d+e/x**(2/3))**n))/x**4,x)

[Out]

Timed out

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