Integrand size = 16, antiderivative size = 151 \[ \int \frac {\log \left (c \left (a+b x^3\right )^p\right )}{x^5} \, dx=-\frac {3 b p}{4 a x}+\frac {\sqrt {3} b^{4/3} p \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{4 a^{4/3}}+\frac {b^{4/3} p \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{4 a^{4/3}}-\frac {b^{4/3} p \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{8 a^{4/3}}-\frac {\log \left (c \left (a+b x^3\right )^p\right )}{4 x^4} \] Output:
-3/4*b*p/a/x+1/4*3^(1/2)*b^(4/3)*p*arctan(1/3*(a^(1/3)-2*b^(1/3)*x)*3^(1/2 )/a^(1/3))/a^(4/3)+1/4*b^(4/3)*p*ln(a^(1/3)+b^(1/3)*x)/a^(4/3)-1/8*b^(4/3) *p*ln(a^(2/3)-a^(1/3)*b^(1/3)*x+b^(2/3)*x^2)/a^(4/3)-1/4*ln(c*(b*x^3+a)^p) /x^4
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.01 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.32 \[ \int \frac {\log \left (c \left (a+b x^3\right )^p\right )}{x^5} \, dx=-\frac {3 b p \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},1,\frac {2}{3},-\frac {b x^3}{a}\right )}{4 a x}-\frac {\log \left (c \left (a+b x^3\right )^p\right )}{4 x^4} \] Input:
Integrate[Log[c*(a + b*x^3)^p]/x^5,x]
Output:
(-3*b*p*Hypergeometric2F1[-1/3, 1, 2/3, -((b*x^3)/a)])/(4*a*x) - Log[c*(a + b*x^3)^p]/(4*x^4)
Time = 0.55 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.04, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.625, Rules used = {2905, 847, 821, 16, 1142, 25, 27, 1082, 217, 1103}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\log \left (c \left (a+b x^3\right )^p\right )}{x^5} \, dx\) |
\(\Big \downarrow \) 2905 |
\(\displaystyle \frac {3}{4} b p \int \frac {1}{x^2 \left (b x^3+a\right )}dx-\frac {\log \left (c \left (a+b x^3\right )^p\right )}{4 x^4}\) |
\(\Big \downarrow \) 847 |
\(\displaystyle \frac {3}{4} b p \left (-\frac {b \int \frac {x}{b x^3+a}dx}{a}-\frac {1}{a x}\right )-\frac {\log \left (c \left (a+b x^3\right )^p\right )}{4 x^4}\) |
\(\Big \downarrow \) 821 |
\(\displaystyle \frac {3}{4} b p \left (-\frac {b \left (\frac {\int \frac {\sqrt [3]{b} x+\sqrt [3]{a}}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\int \frac {1}{\sqrt [3]{b} x+\sqrt [3]{a}}dx}{3 \sqrt [3]{a} \sqrt [3]{b}}\right )}{a}-\frac {1}{a x}\right )-\frac {\log \left (c \left (a+b x^3\right )^p\right )}{4 x^4}\) |
\(\Big \downarrow \) 16 |
\(\displaystyle \frac {3}{4} b p \left (-\frac {b \left (\frac {\int \frac {\sqrt [3]{b} x+\sqrt [3]{a}}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 \sqrt [3]{a} b^{2/3}}\right )}{a}-\frac {1}{a x}\right )-\frac {\log \left (c \left (a+b x^3\right )^p\right )}{4 x^4}\) |
\(\Big \downarrow \) 1142 |
\(\displaystyle \frac {3}{4} b p \left (-\frac {b \left (\frac {\frac {3}{2} \sqrt [3]{a} \int \frac {1}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx+\frac {\int -\frac {\sqrt [3]{b} \left (\sqrt [3]{a}-2 \sqrt [3]{b} x\right )}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx}{2 \sqrt [3]{b}}}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 \sqrt [3]{a} b^{2/3}}\right )}{a}-\frac {1}{a x}\right )-\frac {\log \left (c \left (a+b x^3\right )^p\right )}{4 x^4}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {3}{4} b p \left (-\frac {b \left (\frac {\frac {3}{2} \sqrt [3]{a} \int \frac {1}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx-\frac {\int \frac {\sqrt [3]{b} \left (\sqrt [3]{a}-2 \sqrt [3]{b} x\right )}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx}{2 \sqrt [3]{b}}}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 \sqrt [3]{a} b^{2/3}}\right )}{a}-\frac {1}{a x}\right )-\frac {\log \left (c \left (a+b x^3\right )^p\right )}{4 x^4}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {3}{4} b p \left (-\frac {b \left (\frac {\frac {3}{2} \sqrt [3]{a} \int \frac {1}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx-\frac {1}{2} \int \frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 \sqrt [3]{a} b^{2/3}}\right )}{a}-\frac {1}{a x}\right )-\frac {\log \left (c \left (a+b x^3\right )^p\right )}{4 x^4}\) |
\(\Big \downarrow \) 1082 |
\(\displaystyle \frac {3}{4} b p \left (-\frac {b \left (\frac {\frac {3 \int \frac {1}{-\left (1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}\right )^2-3}d\left (1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}\right )}{\sqrt [3]{b}}-\frac {1}{2} \int \frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 \sqrt [3]{a} b^{2/3}}\right )}{a}-\frac {1}{a x}\right )-\frac {\log \left (c \left (a+b x^3\right )^p\right )}{4 x^4}\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {3}{4} b p \left (-\frac {b \left (\frac {-\frac {1}{2} \int \frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx-\frac {\sqrt {3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{\sqrt [3]{b}}}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 \sqrt [3]{a} b^{2/3}}\right )}{a}-\frac {1}{a x}\right )-\frac {\log \left (c \left (a+b x^3\right )^p\right )}{4 x^4}\) |
\(\Big \downarrow \) 1103 |
\(\displaystyle \frac {3}{4} b p \left (-\frac {b \left (\frac {\frac {\log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{2 \sqrt [3]{b}}-\frac {\sqrt {3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{\sqrt [3]{b}}}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 \sqrt [3]{a} b^{2/3}}\right )}{a}-\frac {1}{a x}\right )-\frac {\log \left (c \left (a+b x^3\right )^p\right )}{4 x^4}\) |
Input:
Int[Log[c*(a + b*x^3)^p]/x^5,x]
Output:
(3*b*p*(-(1/(a*x)) - (b*(-1/3*Log[a^(1/3) + b^(1/3)*x]/(a^(1/3)*b^(2/3)) + (-((Sqrt[3]*ArcTan[(1 - (2*b^(1/3)*x)/a^(1/3))/Sqrt[3]])/b^(1/3)) + Log[a ^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2]/(2*b^(1/3)))/(3*a^(1/3)*b^(1/3)) ))/a))/4 - Log[c*(a + b*x^3)^p]/(4*x^4)
Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[a, 0] || LtQ[b, 0])
Int[(x_)/((a_) + (b_.)*(x_)^3), x_Symbol] :> Simp[-(3*Rt[a, 3]*Rt[b, 3])^(- 1) Int[1/(Rt[a, 3] + Rt[b, 3]*x), x], x] + Simp[1/(3*Rt[a, 3]*Rt[b, 3]) Int[(Rt[a, 3] + Rt[b, 3]*x)/(Rt[a, 3]^2 - Rt[a, 3]*Rt[b, 3]*x + Rt[b, 3]^2 *x^2), x], x] /; FreeQ[{a, b}, x]
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] - Simp[b*((m + n*(p + 1) + 1)/(a*c^n*(m + 1))) Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a , b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p , x]
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S implify[a*(c/b^2)]}, Simp[-2/b Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b )], x] /; RationalQ[q] && (EqQ[q^2, 1] || !RationalQ[b^2 - 4*a*c])] /; Fre eQ[{a, b, c}, x]
Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]
Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[(2*c*d - b*e)/(2*c) Int[1/(a + b*x + c*x^2), x], x] + Simp[e/(2*c) Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x]
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] - Simp[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]
Time = 1.24 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.85
method | result | size |
parts | \(-\frac {\ln \left (c \left (b \,x^{3}+a \right )^{p}\right )}{4 x^{4}}+\frac {3 p b \left (-\frac {1}{a x}-\frac {\left (-\frac {\ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {\ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right ) b}{a}\right )}{4}\) | \(128\) |
risch | \(-\frac {\ln \left (\left (b \,x^{3}+a \right )^{p}\right )}{4 x^{4}}-\frac {i \pi a \,\operatorname {csgn}\left (i \left (b \,x^{3}+a \right )^{p}\right ) {\operatorname {csgn}\left (i c \left (b \,x^{3}+a \right )^{p}\right )}^{2}-i \pi a \,\operatorname {csgn}\left (i \left (b \,x^{3}+a \right )^{p}\right ) \operatorname {csgn}\left (i c \left (b \,x^{3}+a \right )^{p}\right ) \operatorname {csgn}\left (i c \right )-i \pi a {\operatorname {csgn}\left (i c \left (b \,x^{3}+a \right )^{p}\right )}^{3}+i \pi a {\operatorname {csgn}\left (i c \left (b \,x^{3}+a \right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )-2 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a^{4} \textit {\_Z}^{3}-b^{4} p^{3}\right )}{\sum }\textit {\_R} \ln \left (\left (-4 a^{4} \textit {\_R}^{3}+3 b^{4} p^{3}\right ) x -a^{3} b p \,\textit {\_R}^{2}\right )\right ) a \,x^{4}+6 b p \,x^{3}+2 \ln \left (c \right ) a}{8 a \,x^{4}}\) | \(215\) |
Input:
int(ln(c*(b*x^3+a)^p)/x^5,x,method=_RETURNVERBOSE)
Output:
-1/4*ln(c*(b*x^3+a)^p)/x^4+3/4*p*b*(-1/a/x-(-1/3/b/(1/b*a)^(1/3)*ln(x+(1/b *a)^(1/3))+1/6/b/(1/b*a)^(1/3)*ln(x^2-(1/b*a)^(1/3)*x+(1/b*a)^(2/3))+1/3*3 ^(1/2)/b/(1/b*a)^(1/3)*arctan(1/3*3^(1/2)*(2/(1/b*a)^(1/3)*x-1)))/a*b)
Time = 0.11 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.91 \[ \int \frac {\log \left (c \left (a+b x^3\right )^p\right )}{x^5} \, dx=-\frac {2 \, \sqrt {3} b p x^{4} \left (\frac {b}{a}\right )^{\frac {1}{3}} \arctan \left (\frac {2}{3} \, \sqrt {3} x \left (\frac {b}{a}\right )^{\frac {1}{3}} - \frac {1}{3} \, \sqrt {3}\right ) + b p x^{4} \left (\frac {b}{a}\right )^{\frac {1}{3}} \log \left (b x^{2} - a x \left (\frac {b}{a}\right )^{\frac {2}{3}} + a \left (\frac {b}{a}\right )^{\frac {1}{3}}\right ) - 2 \, b p x^{4} \left (\frac {b}{a}\right )^{\frac {1}{3}} \log \left (b x + a \left (\frac {b}{a}\right )^{\frac {2}{3}}\right ) + 6 \, b p x^{3} + 2 \, a p \log \left (b x^{3} + a\right ) + 2 \, a \log \left (c\right )}{8 \, a x^{4}} \] Input:
integrate(log(c*(b*x^3+a)^p)/x^5,x, algorithm="fricas")
Output:
-1/8*(2*sqrt(3)*b*p*x^4*(b/a)^(1/3)*arctan(2/3*sqrt(3)*x*(b/a)^(1/3) - 1/3 *sqrt(3)) + b*p*x^4*(b/a)^(1/3)*log(b*x^2 - a*x*(b/a)^(2/3) + a*(b/a)^(1/3 )) - 2*b*p*x^4*(b/a)^(1/3)*log(b*x + a*(b/a)^(2/3)) + 6*b*p*x^3 + 2*a*p*lo g(b*x^3 + a) + 2*a*log(c))/(a*x^4)
Timed out. \[ \int \frac {\log \left (c \left (a+b x^3\right )^p\right )}{x^5} \, dx=\text {Timed out} \] Input:
integrate(ln(c*(b*x**3+a)**p)/x**5,x)
Output:
Timed out
Time = 0.11 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.84 \[ \int \frac {\log \left (c \left (a+b x^3\right )^p\right )}{x^5} \, dx=-\frac {1}{8} \, b p {\left (\frac {2 \, \sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (2 \, x - \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{a \left (\frac {a}{b}\right )^{\frac {1}{3}}} + \frac {\log \left (x^{2} - x \left (\frac {a}{b}\right )^{\frac {1}{3}} + \left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{a \left (\frac {a}{b}\right )^{\frac {1}{3}}} - \frac {2 \, \log \left (x + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{a \left (\frac {a}{b}\right )^{\frac {1}{3}}} + \frac {6}{a x}\right )} - \frac {\log \left ({\left (b x^{3} + a\right )}^{p} c\right )}{4 \, x^{4}} \] Input:
integrate(log(c*(b*x^3+a)^p)/x^5,x, algorithm="maxima")
Output:
-1/8*b*p*(2*sqrt(3)*arctan(1/3*sqrt(3)*(2*x - (a/b)^(1/3))/(a/b)^(1/3))/(a *(a/b)^(1/3)) + log(x^2 - x*(a/b)^(1/3) + (a/b)^(2/3))/(a*(a/b)^(1/3)) - 2 *log(x + (a/b)^(1/3))/(a*(a/b)^(1/3)) + 6/(a*x)) - 1/4*log((b*x^3 + a)^p*c )/x^4
Time = 0.13 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.01 \[ \int \frac {\log \left (c \left (a+b x^3\right )^p\right )}{x^5} \, dx=\frac {1}{8} \, b^{2} p {\left (\frac {2 \, \left (-\frac {a}{b}\right )^{\frac {2}{3}} \log \left ({\left | x - \left (-\frac {a}{b}\right )^{\frac {1}{3}} \right |}\right )}{a^{2}} + \frac {2 \, \sqrt {3} \left (-a b^{2}\right )^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3} {\left (2 \, x + \left (-\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{a^{2} b^{2}} - \frac {\left (-a b^{2}\right )^{\frac {2}{3}} \log \left (x^{2} + x \left (-\frac {a}{b}\right )^{\frac {1}{3}} + \left (-\frac {a}{b}\right )^{\frac {2}{3}}\right )}{a^{2} b^{2}}\right )} - \frac {p \log \left (b x^{3} + a\right )}{4 \, x^{4}} - \frac {3 \, b p x^{3} + a \log \left (c\right )}{4 \, a x^{4}} \] Input:
integrate(log(c*(b*x^3+a)^p)/x^5,x, algorithm="giac")
Output:
1/8*b^2*p*(2*(-a/b)^(2/3)*log(abs(x - (-a/b)^(1/3)))/a^2 + 2*sqrt(3)*(-a*b ^2)^(2/3)*arctan(1/3*sqrt(3)*(2*x + (-a/b)^(1/3))/(-a/b)^(1/3))/(a^2*b^2) - (-a*b^2)^(2/3)*log(x^2 + x*(-a/b)^(1/3) + (-a/b)^(2/3))/(a^2*b^2)) - 1/4 *p*log(b*x^3 + a)/x^4 - 1/4*(3*b*p*x^3 + a*log(c))/(a*x^4)
Time = 28.17 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.83 \[ \int \frac {\log \left (c \left (a+b x^3\right )^p\right )}{x^5} \, dx=\frac {b^{4/3}\,p\,\ln \left (b^{1/3}\,x+a^{1/3}\right )}{4\,a^{4/3}}-\frac {\ln \left (c\,{\left (b\,x^3+a\right )}^p\right )}{4\,x^4}-\frac {3\,b\,p}{4\,a\,x}+\frac {b^{4/3}\,p\,\ln \left (4\,b^{1/3}\,x-2\,a^{1/3}-\sqrt {3}\,a^{1/3}\,2{}\mathrm {i}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{4\,a^{4/3}}-\frac {b^{4/3}\,p\,\ln \left (4\,b^{1/3}\,x-2\,a^{1/3}+\sqrt {3}\,a^{1/3}\,2{}\mathrm {i}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{4\,a^{4/3}} \] Input:
int(log(c*(a + b*x^3)^p)/x^5,x)
Output:
(b^(4/3)*p*log(b^(1/3)*x + a^(1/3)))/(4*a^(4/3)) - log(c*(a + b*x^3)^p)/(4 *x^4) - (3*b*p)/(4*a*x) + (b^(4/3)*p*log(4*b^(1/3)*x - 3^(1/2)*a^(1/3)*2i - 2*a^(1/3))*((3^(1/2)*1i)/2 - 1/2))/(4*a^(4/3)) - (b^(4/3)*p*log(3^(1/2)* a^(1/3)*2i + 4*b^(1/3)*x - 2*a^(1/3))*((3^(1/2)*1i)/2 + 1/2))/(4*a^(4/3))
Time = 0.18 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.75 \[ \int \frac {\log \left (c \left (a+b x^3\right )^p\right )}{x^5} \, dx=\frac {2 \sqrt {3}\, \mathit {atan} \left (\frac {a^{\frac {1}{3}}-2 b^{\frac {1}{3}} x}{a^{\frac {1}{3}} \sqrt {3}}\right ) b^{2} p \,x^{4}-2 b^{\frac {2}{3}} a^{\frac {4}{3}} \mathrm {log}\left (\left (b \,x^{3}+a \right )^{p} c \right )-6 b^{\frac {5}{3}} a^{\frac {1}{3}} p \,x^{3}+3 \,\mathrm {log}\left (a^{\frac {1}{3}}+b^{\frac {1}{3}} x \right ) b^{2} p \,x^{4}-\mathrm {log}\left (\left (b \,x^{3}+a \right )^{p} c \right ) b^{2} x^{4}}{8 b^{\frac {2}{3}} a^{\frac {4}{3}} x^{4}} \] Input:
int(log(c*(b*x^3+a)^p)/x^5,x)
Output:
(2*sqrt(3)*atan((a**(1/3) - 2*b**(1/3)*x)/(a**(1/3)*sqrt(3)))*b**2*p*x**4 - 2*b**(2/3)*a**(1/3)*log((a + b*x**3)**p*c)*a - 6*b**(2/3)*a**(1/3)*b*p*x **3 + 3*log(a**(1/3) + b**(1/3)*x)*b**2*p*x**4 - log((a + b*x**3)**p*c)*b* *2*x**4)/(8*b**(2/3)*a**(1/3)*a*x**4)