Integrand size = 16, antiderivative size = 133 \[ \int \frac {\log \left (c \left (a+b x^3\right )^p\right )}{x^2} \, dx=-\frac {\sqrt {3} \sqrt [3]{b} p \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt [3]{a}}-\frac {\sqrt [3]{b} p \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\sqrt [3]{a}}+\frac {\sqrt [3]{b} p \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{2 \sqrt [3]{a}}-\frac {\log \left (c \left (a+b x^3\right )^p\right )}{x} \] Output:
-3^(1/2)*b^(1/3)*p*arctan(1/3*(a^(1/3)-2*b^(1/3)*x)*3^(1/2)/a^(1/3))/a^(1/ 3)-b^(1/3)*p*ln(a^(1/3)+b^(1/3)*x)/a^(1/3)+1/2*b^(1/3)*p*ln(a^(2/3)-a^(1/3 )*b^(1/3)*x+b^(2/3)*x^2)/a^(1/3)-ln(c*(b*x^3+a)^p)/x
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.01 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.35 \[ \int \frac {\log \left (c \left (a+b x^3\right )^p\right )}{x^2} \, dx=\frac {3 b p x^2 \operatorname {Hypergeometric2F1}\left (\frac {2}{3},1,\frac {5}{3},-\frac {b x^3}{a}\right )}{2 a}-\frac {\log \left (c \left (a+b x^3\right )^p\right )}{x} \] Input:
Integrate[Log[c*(a + b*x^3)^p]/x^2,x]
Output:
(3*b*p*x^2*Hypergeometric2F1[2/3, 1, 5/3, -((b*x^3)/a)])/(2*a) - Log[c*(a + b*x^3)^p]/x
Time = 0.52 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.04, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.562, Rules used = {2905, 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.
| \(\displaystyle \int \frac {\log \left (c \left (a+b x^3\right )^p\right )}{x^2} \, dx\) |
| \(\Big \downarrow \) 2905 |
| \(\displaystyle 3 b p \int \frac {x}{b x^3+a}dx-\frac {\log \left (c \left (a+b x^3\right )^p\right )}{x}\) |
| \(\Big \downarrow \) 821 |
| \(\displaystyle 3 b p \left (\frac {\int \frac {\sqrt [3]{b} x+\sqrt [3]{a}}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\int \frac {1}{\sqrt [3]{b} x+\sqrt [3]{a}}dx}{3 \sqrt [3]{a} \sqrt [3]{b}}\right )-\frac {\log \left (c \left (a+b x^3\right )^p\right )}{x}\) |
| \(\Big \downarrow \) 16 |
| \(\displaystyle 3 b p \left (\frac {\int \frac {\sqrt [3]{b} x+\sqrt [3]{a}}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 \sqrt [3]{a} b^{2/3}}\right )-\frac {\log \left (c \left (a+b x^3\right )^p\right )}{x}\) |
| \(\Big \downarrow \) 1142 |
| \(\displaystyle 3 b p \left (\frac {\frac {3}{2} \sqrt [3]{a} \int \frac {1}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx+\frac {\int -\frac {\sqrt [3]{b} \left (\sqrt [3]{a}-2 \sqrt [3]{b} x\right )}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx}{2 \sqrt [3]{b}}}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 \sqrt [3]{a} b^{2/3}}\right )-\frac {\log \left (c \left (a+b x^3\right )^p\right )}{x}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle 3 b p \left (\frac {\frac {3}{2} \sqrt [3]{a} \int \frac {1}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx-\frac {\int \frac {\sqrt [3]{b} \left (\sqrt [3]{a}-2 \sqrt [3]{b} x\right )}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx}{2 \sqrt [3]{b}}}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 \sqrt [3]{a} b^{2/3}}\right )-\frac {\log \left (c \left (a+b x^3\right )^p\right )}{x}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle 3 b p \left (\frac {\frac {3}{2} \sqrt [3]{a} \int \frac {1}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx-\frac {1}{2} \int \frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 \sqrt [3]{a} b^{2/3}}\right )-\frac {\log \left (c \left (a+b x^3\right )^p\right )}{x}\) |
| \(\Big \downarrow \) 1082 |
| \(\displaystyle 3 b p \left (\frac {\frac {3 \int \frac {1}{-\left (1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}\right )^2-3}d\left (1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}\right )}{\sqrt [3]{b}}-\frac {1}{2} \int \frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 \sqrt [3]{a} b^{2/3}}\right )-\frac {\log \left (c \left (a+b x^3\right )^p\right )}{x}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle 3 b p \left (\frac {-\frac {1}{2} \int \frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx-\frac {\sqrt {3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{\sqrt [3]{b}}}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 \sqrt [3]{a} b^{2/3}}\right )-\frac {\log \left (c \left (a+b x^3\right )^p\right )}{x}\) |
| \(\Big \downarrow \) 1103 |
| \(\displaystyle 3 b p \left (\frac {\frac {\log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{2 \sqrt [3]{b}}-\frac {\sqrt {3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{\sqrt [3]{b}}}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 \sqrt [3]{a} b^{2/3}}\right )-\frac {\log \left (c \left (a+b x^3\right )^p\right )}{x}\) |
Input:
Int[Log[c*(a + b*x^3)^p]/x^2,x]
Output:
3*b*p*(-1/3*Log[a^(1/3) + b^(1/3)*x]/(a^(1/3)*b^(2/3)) + (-((Sqrt[3]*ArcTa n[(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))) - Log[c*(a + b*x ^3)^p]/x
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[((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 = 0.65 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.85
| method | result | size |
| parts | \(-\frac {\ln \left (c \left (b \,x^{3}+a \right )^{p}\right )}{x}+3 p b \left (-\frac {\ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {\ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )\) | \(113\) |
| risch | \(-\frac {\ln \left (\left (b \,x^{3}+a \right )^{p}\right )}{x}-\frac {i \pi \,\operatorname {csgn}\left (i \left (b \,x^{3}+a \right )^{p}\right ) {\operatorname {csgn}\left (i c \left (b \,x^{3}+a \right )^{p}\right )}^{2}-i \pi \,\operatorname {csgn}\left (i \left (b \,x^{3}+a \right )^{p}\right ) \operatorname {csgn}\left (i c \left (b \,x^{3}+a \right )^{p}\right ) \operatorname {csgn}\left (i c \right )-i \pi {\operatorname {csgn}\left (i c \left (b \,x^{3}+a \right )^{p}\right )}^{3}+i \pi {\operatorname {csgn}\left (i c \left (b \,x^{3}+a \right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )-2 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{3} a +b \,p^{3}\right )}{\sum }\textit {\_R} \ln \left (\left (-4 \textit {\_R}^{3} a -3 b \,p^{3}\right ) x +a p \,\textit {\_R}^{2}\right )\right ) x +2 \ln \left (c \right )}{2 x}\) | \(184\) |
Input:
int(ln(c*(b*x^3+a)^p)/x^2,x,method=_RETURNVERBOSE)
Output:
-ln(c*(b*x^3+a)^p)/x+3*p*b*(-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)))
Time = 0.08 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.95 \[ \int \frac {\log \left (c \left (a+b x^3\right )^p\right )}{x^2} \, dx=\frac {2 \, \sqrt {3} p x \left (-\frac {b}{a}\right )^{\frac {1}{3}} \arctan \left (\frac {2}{3} \, \sqrt {3} x \left (-\frac {b}{a}\right )^{\frac {1}{3}} + \frac {1}{3} \, \sqrt {3}\right ) - p x \left (-\frac {b}{a}\right )^{\frac {1}{3}} \log \left (b x^{2} - a x \left (-\frac {b}{a}\right )^{\frac {2}{3}} - a \left (-\frac {b}{a}\right )^{\frac {1}{3}}\right ) + 2 \, p x \left (-\frac {b}{a}\right )^{\frac {1}{3}} \log \left (b x + a \left (-\frac {b}{a}\right )^{\frac {2}{3}}\right ) - 2 \, p \log \left (b x^{3} + a\right ) - 2 \, \log \left (c\right )}{2 \, x} \] Input:
integrate(log(c*(b*x^3+a)^p)/x^2,x, algorithm="fricas")
Output:
1/2*(2*sqrt(3)*p*x*(-b/a)^(1/3)*arctan(2/3*sqrt(3)*x*(-b/a)^(1/3) + 1/3*sq rt(3)) - p*x*(-b/a)^(1/3)*log(b*x^2 - a*x*(-b/a)^(2/3) - a*(-b/a)^(1/3)) + 2*p*x*(-b/a)^(1/3)*log(b*x + a*(-b/a)^(2/3)) - 2*p*log(b*x^3 + a) - 2*log (c))/x
Time = 105.77 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.24 \[ \int \frac {\log \left (c \left (a+b x^3\right )^p\right )}{x^2} \, dx=\begin {cases} - \frac {\log {\left (0^{p} c \right )}}{x} & \text {for}\: a = 0 \wedge b = 0 \\- \frac {3 p}{x} - \frac {\log {\left (c \left (b x^{3}\right )^{p} \right )}}{x} & \text {for}\: a = 0 \\- \frac {\log {\left (a^{p} c \right )}}{x} & \text {for}\: b = 0 \\- \frac {\log {\left (c \left (a + b x^{3}\right )^{p} \right )}}{x} + \frac {3 b p \left (- \frac {a}{b}\right )^{\frac {2}{3}} \log {\left (4 x^{2} + 4 x \sqrt [3]{- \frac {a}{b}} + 4 \left (- \frac {a}{b}\right )^{\frac {2}{3}} \right )}}{2 a} - \frac {\sqrt {3} b p \left (- \frac {a}{b}\right )^{\frac {2}{3}} \operatorname {atan}{\left (\frac {2 \sqrt {3} x}{3 \sqrt [3]{- \frac {a}{b}}} + \frac {\sqrt {3}}{3} \right )}}{a} - \frac {b \left (- \frac {a}{b}\right )^{\frac {2}{3}} \log {\left (c \left (a + b x^{3}\right )^{p} \right )}}{a} & \text {otherwise} \end {cases} \] Input:
integrate(ln(c*(b*x**3+a)**p)/x**2,x)
Output:
Piecewise((-log(0**p*c)/x, Eq(a, 0) & Eq(b, 0)), (-3*p/x - log(c*(b*x**3)* *p)/x, Eq(a, 0)), (-log(a**p*c)/x, Eq(b, 0)), (-log(c*(a + b*x**3)**p)/x + 3*b*p*(-a/b)**(2/3)*log(4*x**2 + 4*x*(-a/b)**(1/3) + 4*(-a/b)**(2/3))/(2* a) - sqrt(3)*b*p*(-a/b)**(2/3)*atan(2*sqrt(3)*x/(3*(-a/b)**(1/3)) + sqrt(3 )/3)/a - b*(-a/b)**(2/3)*log(c*(a + b*x**3)**p)/a, True))
Time = 0.11 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.89 \[ \int \frac {\log \left (c \left (a+b x^3\right )^p\right )}{x^2} \, dx=\frac {1}{2} \, b p {\left (\frac {2 \, \sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (2 \, x - \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{b \left (\frac {a}{b}\right )^{\frac {1}{3}}} + \frac {\log \left (x^{2} - x \left (\frac {a}{b}\right )^{\frac {1}{3}} + \left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{b \left (\frac {a}{b}\right )^{\frac {1}{3}}} - \frac {2 \, \log \left (x + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{b \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )} - \frac {\log \left ({\left (b x^{3} + a\right )}^{p} c\right )}{x} \] Input:
integrate(log(c*(b*x^3+a)^p)/x^2,x, algorithm="maxima")
Output:
1/2*b*p*(2*sqrt(3)*arctan(1/3*sqrt(3)*(2*x - (a/b)^(1/3))/(a/b)^(1/3))/(b* (a/b)^(1/3)) + log(x^2 - x*(a/b)^(1/3) + (a/b)^(2/3))/(b*(a/b)^(1/3)) - 2* log(x + (a/b)^(1/3))/(b*(a/b)^(1/3))) - log((b*x^3 + a)^p*c)/x
Time = 0.13 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.03 \[ \int \frac {\log \left (c \left (a+b x^3\right )^p\right )}{x^2} \, dx=-\frac {b p \left (-\frac {a}{b}\right )^{\frac {2}{3}} \log \left ({\left | x - \left (-\frac {a}{b}\right )^{\frac {1}{3}} \right |}\right )}{a} - \frac {\sqrt {3} \left (-a b^{2}\right )^{\frac {2}{3}} p \arctan \left (\frac {\sqrt {3} {\left (2 \, x + \left (-\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{a b} - \frac {p \log \left (b x^{3} + a\right )}{x} + \frac {\left (-a b^{2}\right )^{\frac {2}{3}} p \log \left (x^{2} + x \left (-\frac {a}{b}\right )^{\frac {1}{3}} + \left (-\frac {a}{b}\right )^{\frac {2}{3}}\right )}{2 \, a b} - \frac {\log \left (c\right )}{x} \] Input:
integrate(log(c*(b*x^3+a)^p)/x^2,x, algorithm="giac")
Output:
-b*p*(-a/b)^(2/3)*log(abs(x - (-a/b)^(1/3)))/a - sqrt(3)*(-a*b^2)^(2/3)*p* arctan(1/3*sqrt(3)*(2*x + (-a/b)^(1/3))/(-a/b)^(1/3))/(a*b) - p*log(b*x^3 + a)/x + 1/2*(-a*b^2)^(2/3)*p*log(x^2 + x*(-a/b)^(1/3) + (-a/b)^(2/3))/(a* b) - log(c)/x
Time = 26.82 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.12 \[ \int \frac {\log \left (c \left (a+b x^3\right )^p\right )}{x^2} \, dx=\frac {{\left (-b\right )}^{1/3}\,p\,\ln \left (a^{1/3}\,{\left (-b\right )}^{8/3}+b^3\,x\right )}{a^{1/3}}-\frac {\ln \left (c\,{\left (b\,x^3+a\right )}^p\right )}{x}+\frac {{\left (-b\right )}^{1/3}\,p\,\ln \left (9\,b^3\,p^2\,x+9\,a^{1/3}\,{\left (-b\right )}^{8/3}\,p^2\,{\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}^2\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{a^{1/3}}-\frac {{\left (-b\right )}^{1/3}\,p\,\ln \left (9\,b^3\,p^2\,x+9\,a^{1/3}\,{\left (-b\right )}^{8/3}\,p^2\,{\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}^2\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{a^{1/3}} \] Input:
int(log(c*(a + b*x^3)^p)/x^2,x)
Output:
((-b)^(1/3)*p*log(a^(1/3)*(-b)^(8/3) + b^3*x))/a^(1/3) - log(c*(a + b*x^3) ^p)/x + ((-b)^(1/3)*p*log(9*b^3*p^2*x + 9*a^(1/3)*(-b)^(8/3)*p^2*((3^(1/2) *1i)/2 - 1/2)^2)*((3^(1/2)*1i)/2 - 1/2))/a^(1/3) - ((-b)^(1/3)*p*log(9*b^3 *p^2*x + 9*a^(1/3)*(-b)^(8/3)*p^2*((3^(1/2)*1i)/2 + 1/2)^2)*((3^(1/2)*1i)/ 2 + 1/2))/a^(1/3)
Time = 0.20 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.66 \[ \int \frac {\log \left (c \left (a+b x^3\right )^p\right )}{x^2} \, dx=\frac {-2 \sqrt {3}\, \mathit {atan} \left (\frac {a^{\frac {1}{3}}-2 b^{\frac {1}{3}} x}{a^{\frac {1}{3}} \sqrt {3}}\right ) b p x -2 b^{\frac {2}{3}} a^{\frac {1}{3}} \mathrm {log}\left (\left (b \,x^{3}+a \right )^{p} c \right )-3 \,\mathrm {log}\left (a^{\frac {1}{3}}+b^{\frac {1}{3}} x \right ) b p x +\mathrm {log}\left (\left (b \,x^{3}+a \right )^{p} c \right ) b x}{2 b^{\frac {2}{3}} a^{\frac {1}{3}} x} \] Input:
int(log(c*(b*x^3+a)^p)/x^2,x)
Output:
( - 2*sqrt(3)*atan((a**(1/3) - 2*b**(1/3)*x)/(a**(1/3)*sqrt(3)))*b*p*x - 2 *b**(2/3)*a**(1/3)*log((a + b*x**3)**p*c) - 3*log(a**(1/3) + b**(1/3)*x)*b *p*x + log((a + b*x**3)**p*c)*b*x)/(2*b**(2/3)*a**(1/3)*x)