3.21 \(\int \frac {\log (c (a+b x^3)^p)}{x^3} \, dx\)

Optimal. Leaf size=139 \[ -\frac {b^{2/3} p \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{4 a^{2/3}}+\frac {b^{2/3} p \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{2 a^{2/3}}-\frac {\sqrt {3} b^{2/3} p \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{2 a^{2/3}}-\frac {\log \left (c \left (a+b x^3\right )^p\right )}{2 x^2} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 139, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {2455, 200, 31, 634, 617, 204, 628} \[ -\frac {b^{2/3} p \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{4 a^{2/3}}+\frac {b^{2/3} p \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{2 a^{2/3}}-\frac {\sqrt {3} b^{2/3} p \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{2 a^{2/3}}-\frac {\log \left (c \left (a+b x^3\right )^p\right )}{2 x^2} \]

Antiderivative was successfully verified.

[In]

Int[Log[c*(a + b*x^3)^p]/x^3,x]

[Out]

-(Sqrt[3]*b^(2/3)*p*ArcTan[(a^(1/3) - 2*b^(1/3)*x)/(Sqrt[3]*a^(1/3))])/(2*a^(2/3)) + (b^(2/3)*p*Log[a^(1/3) +
b^(1/3)*x])/(2*a^(2/3)) - (b^(2/3)*p*Log[a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2])/(4*a^(2/3)) - Log[c*(a +
b*x^3)^p]/(2*x^2)

Rule 31

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

Rule 200

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 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 617

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 628

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 634

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 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 {\log \left (c \left (a+b x^3\right )^p\right )}{x^3} \, dx &=-\frac {\log \left (c \left (a+b x^3\right )^p\right )}{2 x^2}+\frac {1}{2} (3 b p) \int \frac {1}{a+b x^3} \, dx\\ &=-\frac {\log \left (c \left (a+b x^3\right )^p\right )}{2 x^2}+\frac {(b p) \int \frac {1}{\sqrt [3]{a}+\sqrt [3]{b} x} \, dx}{2 a^{2/3}}+\frac {(b p) \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}{2 a^{2/3}}\\ &=\frac {b^{2/3} p \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{2 a^{2/3}}-\frac {\log \left (c \left (a+b x^3\right )^p\right )}{2 x^2}-\frac {\left (b^{2/3} 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}{4 a^{2/3}}+\frac {(3 b p) \int \frac {1}{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2} \, dx}{4 \sqrt [3]{a}}\\ &=\frac {b^{2/3} p \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{2 a^{2/3}}-\frac {b^{2/3} p \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{4 a^{2/3}}-\frac {\log \left (c \left (a+b x^3\right )^p\right )}{2 x^2}+\frac {\left (3 b^{2/3} p\right ) \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}\right )}{2 a^{2/3}}\\ &=-\frac {\sqrt {3} b^{2/3} p \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{2 a^{2/3}}+\frac {b^{2/3} p \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{2 a^{2/3}}-\frac {b^{2/3} p \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{4 a^{2/3}}-\frac {\log \left (c \left (a+b x^3\right )^p\right )}{2 x^2}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 134, normalized size = 0.96 \[ -\frac {b^{2/3} p x^2 \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )+2 a^{2/3} \log \left (c \left (a+b x^3\right )^p\right )-2 b^{2/3} p x^2 \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )+2 \sqrt {3} b^{2/3} p x^2 \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{4 a^{2/3} x^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Log[c*(a + b*x^3)^p]/x^3,x]

[Out]

-1/4*(2*Sqrt[3]*b^(2/3)*p*x^2*ArcTan[(1 - (2*b^(1/3)*x)/a^(1/3))/Sqrt[3]] - 2*b^(2/3)*p*x^2*Log[a^(1/3) + b^(1
/3)*x] + b^(2/3)*p*x^2*Log[a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2] + 2*a^(2/3)*Log[c*(a + b*x^3)^p])/(a^(2/
3)*x^2)

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fricas [A]  time = 0.47, size = 150, normalized size = 1.08 \[ \frac {2 \, \sqrt {3} p x^{2} \left (\frac {b^{2}}{a^{2}}\right )^{\frac {1}{3}} \arctan \left (\frac {2 \, \sqrt {3} a x \left (\frac {b^{2}}{a^{2}}\right )^{\frac {2}{3}} - \sqrt {3} b}{3 \, b}\right ) - p x^{2} \left (\frac {b^{2}}{a^{2}}\right )^{\frac {1}{3}} \log \left (b^{2} x^{2} - a b x \left (\frac {b^{2}}{a^{2}}\right )^{\frac {1}{3}} + a^{2} \left (\frac {b^{2}}{a^{2}}\right )^{\frac {2}{3}}\right ) + 2 \, p x^{2} \left (\frac {b^{2}}{a^{2}}\right )^{\frac {1}{3}} \log \left (b x + a \left (\frac {b^{2}}{a^{2}}\right )^{\frac {1}{3}}\right ) - 2 \, p \log \left (b x^{3} + a\right ) - 2 \, \log \relax (c)}{4 \, x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.18, size = 138, normalized size = 0.99 \[ -\frac {1}{4} \, b p {\left (\frac {2 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}} \log \left ({\left | x - \left (-\frac {a}{b}\right )^{\frac {1}{3}} \right |}\right )}{a} - \frac {2 \, \sqrt {3} \left (-a b^{2}\right )^{\frac {1}{3}} \arctan \left (\frac {\sqrt {3} {\left (2 \, x + \left (-\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{a b} - \frac {\left (-a b^{2}\right )^{\frac {1}{3}} \log \left (x^{2} + x \left (-\frac {a}{b}\right )^{\frac {1}{3}} + \left (-\frac {a}{b}\right )^{\frac {2}{3}}\right )}{a b}\right )} - \frac {p \log \left (b x^{3} + a\right )}{2 \, x^{2}} - \frac {\log \relax (c)}{2 \, x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.34, size = 197, normalized size = 1.42 \[ -\frac {\ln \left (\left (b \,x^{3}+a \right )^{p}\right )}{2 x^{2}}-\frac {-2 x^{2} \RootOf \left (a^{2} \textit {\_Z}^{3}-b^{2} p^{3}\right ) \ln \left (-\RootOf \left (a^{2} \textit {\_Z}^{3}-b^{2} p^{3}\right ) a b \,p^{2}+\left (-4 \RootOf \left (a^{2} \textit {\_Z}^{3}-b^{2} p^{3}\right )^{3} a^{2}+3 b^{2} p^{3}\right ) x \right )-i \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i \left (b \,x^{3}+a \right )^{p}\right ) \mathrm {csgn}\left (i c \left (b \,x^{3}+a \right )^{p}\right )+i \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (b \,x^{3}+a \right )^{p}\right )^{2}+i \pi \,\mathrm {csgn}\left (i \left (b \,x^{3}+a \right )^{p}\right ) \mathrm {csgn}\left (i c \left (b \,x^{3}+a \right )^{p}\right )^{2}-i \pi \mathrm {csgn}\left (i c \left (b \,x^{3}+a \right )^{p}\right )^{3}+2 \ln \relax (c )}{4 x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(ln(c*(b*x^3+a)^p)/x^3,x)

[Out]

-1/2/x^2*ln((b*x^3+a)^p)-1/4*(I*Pi*csgn(I*(b*x^3+a)^p)*csgn(I*c*(b*x^3+a)^p)^2-I*Pi*csgn(I*c)*csgn(I*(b*x^3+a)
^p)*csgn(I*c*(b*x^3+a)^p)-I*Pi*csgn(I*c*(b*x^3+a)^p)^3+I*Pi*csgn(I*c)*csgn(I*c*(b*x^3+a)^p)^2-2*sum(_R*ln((-4*
_R^3*a^2+3*b^2*p^3)*x-a*p^2*_R*b),_R=RootOf(_Z^3*a^2-b^2*p^3))*x^2+2*ln(c))/x^2

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maxima [A]  time = 1.45, size = 120, normalized size = 0.86 \[ \frac {1}{4} \, b p {\left (\frac {2 \, \sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (2 \, x - \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{b \left (\frac {a}{b}\right )^{\frac {2}{3}}} - \frac {\log \left (x^{2} - x \left (\frac {a}{b}\right )^{\frac {1}{3}} + \left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{b \left (\frac {a}{b}\right )^{\frac {2}{3}}} + \frac {2 \, \log \left (x + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{b \left (\frac {a}{b}\right )^{\frac {2}{3}}}\right )} - \frac {\log \left ({\left (b x^{3} + a\right )}^{p} c\right )}{2 \, x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 2.61, size = 115, normalized size = 0.83 \[ \frac {b^{2/3}\,p\,\ln \left (b^{1/3}\,x+a^{1/3}\right )}{2\,a^{2/3}}-\frac {\ln \left (c\,{\left (b\,x^3+a\right )}^p\right )}{2\,x^2}-\frac {b^{2/3}\,p\,\ln \left (2\,b^{1/3}\,x-a^{1/3}-\sqrt {3}\,a^{1/3}\,1{}\mathrm {i}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{2\,a^{2/3}}+\frac {b^{2/3}\,p\,\ln \left (2\,b^{1/3}\,x-a^{1/3}+\sqrt {3}\,a^{1/3}\,1{}\mathrm {i}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{2\,a^{2/3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log(c*(a + b*x^3)^p)/x^3,x)

[Out]

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

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

[Out]

Timed out

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