\(\int \log (c (a+b x^3)^p) \, dx\) [18]

Optimal result

Integrand size = 12, antiderivative size = 133 \[ \int \log \left (c \left (a+b x^3\right )^p\right ) \, dx=-3 p x-\frac {\sqrt {3} \sqrt [3]{a} p \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt [3]{b}}+\frac {\sqrt [3]{a} p \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\sqrt [3]{b}}-\frac {\sqrt [3]{a} p \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{2 \sqrt [3]{b}}+x \log \left (c \left (a+b x^3\right )^p\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {2498, 327, 206, 31, 648, 631, 210, 642} \[ \int \log \left (c \left (a+b x^3\right )^p\right ) \, dx=-\frac {\sqrt [3]{a} p \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{2 \sqrt [3]{b}}-\frac {\sqrt {3} \sqrt [3]{a} p \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt [3]{b}}+x \log \left (c \left (a+b x^3\right )^p\right )+\frac {\sqrt [3]{a} p \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\sqrt [3]{b}}-3 p x \]

[In]

Int[Log[c*(a + b*x^3)^p],x]

[Out]

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 206

Int[((a_) + (b_.)*(x_)^3)^(-1), x_Symbol] :> Dist[1/(3*Rt[a, 3]^2), Int[1/(Rt[a, 3] + Rt[b, 3]*x), x], x] + Di
st[1/(3*Rt[a, 3]^2), Int[(2*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]

Rule 210

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])

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 631

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[a*(c/b^2)]}, Dist[-2/b, Sub
st[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])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[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]

Rule 648

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.97 \[ \int \log \left (c \left (a+b x^3\right )^p\right ) \, dx=-3 p x-\frac {\sqrt {3} \sqrt [3]{a} p \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{\sqrt [3]{b}}+\frac {\sqrt [3]{a} p \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\sqrt [3]{b}}-\frac {\sqrt [3]{a} p \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{2 \sqrt [3]{b}}+x \log \left (c \left (a+b x^3\right )^p\right ) \]

[In]

Integrate[Log[c*(a + b*x^3)^p],x]

[Out]

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

Maple [A] (verified)

Time = 0.39 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.92

method	result	size
default	\(x \ln \left (c \left (b \,x^{3}+a \right )^{p}\right )-3 p b \left (\frac {x}{b}-\frac {\left (\frac {\ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}-\frac {\ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}\right ) a}{b}\right )\)	\(122\)
parts	\(x \ln \left (c \left (b \,x^{3}+a \right )^{p}\right )-3 p b \left (\frac {x}{b}-\frac {\left (\frac {\ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}-\frac {\ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}\right ) a}{b}\right )\)	\(122\)
risch	\(x \ln \left (\left (b \,x^{3}+a \right )^{p}\right )+\frac {i {\operatorname {csgn}\left (i c \left (b \,x^{3}+a \right )^{p}\right )}^{2} \operatorname {csgn}\left (i \left (b \,x^{3}+a \right )^{p}\right ) x \pi }{2}-\frac {i \pi x \,\operatorname {csgn}\left (i \left (b \,x^{3}+a \right )^{p}\right ) \operatorname {csgn}\left (i c \left (b \,x^{3}+a \right )^{p}\right ) \operatorname {csgn}\left (i c \right )}{2}-\frac {i \pi x {\operatorname {csgn}\left (i c \left (b \,x^{3}+a \right )^{p}\right )}^{3}}{2}+\frac {i \operatorname {csgn}\left (i c \right ) {\operatorname {csgn}\left (i c \left (b \,x^{3}+a \right )^{p}\right )}^{2} x \pi }{2}+\frac {a p \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (b \,\textit {\_Z}^{3}+a \right )}{\sum }\frac {\ln \left (x -\textit {\_R} \right )}{\textit {\_R}^{2}}\right )}{b}+x \ln \left (c \right )-3 p x\)	\(167\)

[In]

int(ln(c*(b*x^3+a)^p),x,method=_RETURNVERBOSE)

[Out]

x*ln(c*(b*x^3+a)^p)-3*p*b*(x/b-(1/3/b/(a/b)^(2/3)*ln(x+(a/b)^(1/3))-1/6/b/(a/b)^(2/3)*ln(x^2-(a/b)^(1/3)*x+(a/
b)^(2/3))+1/3/b/(a/b)^(2/3)*3^(1/2)*arctan(1/3*3^(1/2)*(2/(a/b)^(1/3)*x-1)))*a/b)

Fricas [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.83 \[ \int \log \left (c \left (a+b x^3\right )^p\right ) \, dx=p x \log \left (b x^{3} + a\right ) + \sqrt {3} p \left (\frac {a}{b}\right )^{\frac {1}{3}} \arctan \left (\frac {2 \, \sqrt {3} b x \left (\frac {a}{b}\right )^{\frac {2}{3}} - \sqrt {3} a}{3 \, a}\right ) - \frac {1}{2} \, p \left (\frac {a}{b}\right )^{\frac {1}{3}} \log \left (x^{2} - x \left (\frac {a}{b}\right )^{\frac {1}{3}} + \left (\frac {a}{b}\right )^{\frac {2}{3}}\right ) + p \left (\frac {a}{b}\right )^{\frac {1}{3}} \log \left (x + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right ) - 3 \, p x + x \log \left (c\right ) \]

[In]

integrate(log(c*(b*x^3+a)^p),x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 24.86 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.24 \[ \int \log \left (c \left (a+b x^3\right )^p\right ) \, dx=\begin {cases} x \log {\left (0^{p} c \right )} & \text {for}\: a = 0 \wedge b = 0 \\- 3 p x + x \log {\left (c \left (b x^{3}\right )^{p} \right )} & \text {for}\: a = 0 \\x \log {\left (a^{p} c \right )} & \text {for}\: b = 0 \\- 3 p x + x \log {\left (c \left (a + b x^{3}\right )^{p} \right )} - \frac {3 b p \left (- \frac {a}{b}\right )^{\frac {4}{3}} \log {\left (4 x^{2} + 4 x \sqrt [3]{- \frac {a}{b}} + 4 \left (- \frac {a}{b}\right )^{\frac {2}{3}} \right )}}{2 a} - \frac {\sqrt {3} b p \left (- \frac {a}{b}\right )^{\frac {4}{3}} \operatorname {atan}{\left (\frac {2 \sqrt {3} x}{3 \sqrt [3]{- \frac {a}{b}}} + \frac {\sqrt {3}}{3} \right )}}{a} + \frac {b \left (- \frac {a}{b}\right )^{\frac {4}{3}} \log {\left (c \left (a + b x^{3}\right )^{p} \right )}}{a} & \text {otherwise} \end {cases} \]

[In]

integrate(ln(c*(b*x**3+a)**p),x)

[Out]

Piecewise((x*log(0**p*c), Eq(a, 0) & Eq(b, 0)), (-3*p*x + x*log(c*(b*x**3)**p), Eq(a, 0)), (x*log(a**p*c), Eq(
b, 0)), (-3*p*x + x*log(c*(a + b*x**3)**p) - 3*b*p*(-a/b)**(4/3)*log(4*x**2 + 4*x*(-a/b)**(1/3) + 4*(-a/b)**(2
/3))/(2*a) - sqrt(3)*b*p*(-a/b)**(4/3)*atan(2*sqrt(3)*x/(3*(-a/b)**(1/3)) + sqrt(3)/3)/a + b*(-a/b)**(4/3)*log
(c*(a + b*x**3)**p)/a, True))

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.94 \[ \int \log \left (c \left (a+b x^3\right )^p\right ) \, dx=-\frac {1}{2} \, b p {\left (\frac {6 \, x}{b} - \frac {2 \, \sqrt {3} a \arctan \left (\frac {\sqrt {3} {\left (2 \, x - \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{b^{2} \left (\frac {a}{b}\right )^{\frac {2}{3}}} + \frac {a \log \left (x^{2} - x \left (\frac {a}{b}\right )^{\frac {1}{3}} + \left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{b^{2} \left (\frac {a}{b}\right )^{\frac {2}{3}}} - \frac {2 \, a \log \left (x + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{b^{2} \left (\frac {a}{b}\right )^{\frac {2}{3}}}\right )} + x \log \left ({\left (b x^{3} + a\right )}^{p} c\right ) \]

[In]

integrate(log(c*(b*x^3+a)^p),x, algorithm="maxima")

[Out]

-1/2*b*p*(6*x/b - 2*sqrt(3)*a*arctan(1/3*sqrt(3)*(2*x - (a/b)^(1/3))/(a/b)^(1/3))/(b^2*(a/b)^(2/3)) + a*log(x^
2 - x*(a/b)^(1/3) + (a/b)^(2/3))/(b^2*(a/b)^(2/3)) - 2*a*log(x + (a/b)^(1/3))/(b^2*(a/b)^(2/3))) + x*log((b*x^
3 + a)^p*c)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.08 \[ \int \log \left (c \left (a+b x^3\right )^p\right ) \, dx=-\frac {1}{2} \, a b p {\left (\frac {2 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}} \log \left ({\left | x - \left (-\frac {a}{b}\right )^{\frac {1}{3}} \right |}\right )}{a b} - \frac {2 \, \sqrt {3} \left (-a b^{2}\right )^{\frac {1}{3}} \arctan \left (\frac {\sqrt {3} {\left (2 \, x + \left (-\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{a b^{2}} - \frac {\left (-a b^{2}\right )^{\frac {1}{3}} \log \left (x^{2} + x \left (-\frac {a}{b}\right )^{\frac {1}{3}} + \left (-\frac {a}{b}\right )^{\frac {2}{3}}\right )}{a b^{2}}\right )} + p x \log \left (b x^{3} + a\right ) - {\left (3 \, p - \log \left (c\right )\right )} x \]

[In]

integrate(log(c*(b*x^3+a)^p),x, algorithm="giac")

[Out]

-1/2*a*b*p*(2*(-a/b)^(1/3)*log(abs(x - (-a/b)^(1/3)))/(a*b) - 2*sqrt(3)*(-a*b^2)^(1/3)*arctan(1/3*sqrt(3)*(2*x
+ (-a/b)^(1/3))/(-a/b)^(1/3))/(a*b^2) - (-a*b^2)^(1/3)*log(x^2 + x*(-a/b)^(1/3) + (-a/b)^(2/3))/(a*b^2)) + p*
x*log(b*x^3 + a) - (3*p - log(c))*x

Mupad [B] (verification not implemented)

Time = 1.67 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.01 \[ \int \log \left (c \left (a+b x^3\right )^p\right ) \, dx=x\,\ln \left (c\,{\left (b\,x^3+a\right )}^p\right )-3\,p\,x-\frac {{\left (-a\right )}^{1/3}\,p\,\ln \left ({\left (-a\right )}^{4/3}+a\,b^{1/3}\,x\right )}{b^{1/3}}+\frac {{\left (-a\right )}^{1/3}\,p\,\ln \left (2\,a\,b^{1/3}\,x-{\left (-a\right )}^{4/3}-\sqrt {3}\,{\left (-a\right )}^{4/3}\,1{}\mathrm {i}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{b^{1/3}}-\frac {{\left (-a\right )}^{1/3}\,p\,\ln \left (2\,a\,b^{1/3}\,x-{\left (-a\right )}^{4/3}+\sqrt {3}\,{\left (-a\right )}^{4/3}\,1{}\mathrm {i}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{b^{1/3}} \]

[In]

int(log(c*(a + b*x^3)^p),x)

[Out]

x*log(c*(a + b*x^3)^p) - 3*p*x - ((-a)^(1/3)*p*log((-a)^(4/3) + a*b^(1/3)*x))/b^(1/3) + ((-a)^(1/3)*p*log(2*a*
b^(1/3)*x - 3^(1/2)*(-a)^(4/3)*1i - (-a)^(4/3))*((3^(1/2)*1i)/2 + 1/2))/b^(1/3) - ((-a)^(1/3)*p*log(3^(1/2)*(-
a)^(4/3)*1i - (-a)^(4/3) + 2*a*b^(1/3)*x)*((3^(1/2)*1i)/2 - 1/2))/b^(1/3)