3.1.21 \(\int \frac {\log (c (a+b x^3)^p)}{x^3} \, dx\) [21]

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

[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 = 0.06, 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 = {2505, 206, 31, 648, 631, 210, 642} \begin {gather*} -\frac {\sqrt {3} b^{2/3} p \text {ArcTan}\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 (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 {\log \left (c \left (a+b x^3\right )^p\right )}{2 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Log[c*(a + b*x^3)^p]/x^3,x]

[Out]

-1/2*(Sqrt[3]*b^(2/3)*p*ArcTan[(a^(1/3) - 2*b^(1/3)*x)/(Sqrt[3]*a^(1/3))])/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 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 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 2505

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 ) \text {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 = 0.02, size = 134, normalized size = 0.96 \begin {gather*} -\frac {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 )-2 b^{2/3} p x^2 \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )+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 )}{4 a^{2/3} x^2} \end {gather*}

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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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.32, size = 197, normalized size = 1.42

method	result	size
risch	\(-\frac {\ln \left (\left (x^{3} b +a \right )^{p}\right )}{2 x^{2}}-\frac {i \pi \,\mathrm {csgn}\left (i \left (x^{3} b +a \right )^{p}\right ) \mathrm {csgn}\left (i c \left (x^{3} b +a \right )^{p}\right )^{2}-i \pi \,\mathrm {csgn}\left (i \left (x^{3} b +a \right )^{p}\right ) \mathrm {csgn}\left (i c \left (x^{3} b +a \right )^{p}\right ) \mathrm {csgn}\left (i c \right )-i \pi \mathrm {csgn}\left (i c \left (x^{3} b +a \right )^{p}\right )^{3}+i \pi \mathrm {csgn}\left (i c \left (x^{3} b +a \right )^{p}\right )^{2} \mathrm {csgn}\left (i c \right )-2 \left (\munderset {\textit {\_R} =\RootOf \left (a^{2} \textit {\_Z}^{3}-b^{2} p^{3}\right )}{\sum }\textit {\_R} \ln \left (\left (-4 \textit {\_R}^{3} a^{2}+3 b^{2} p^{3}\right ) x -p^{2} a \textit {\_R} b \right )\right ) x^{2}+2 \ln \left (c \right )}{4 x^{2}}\)	\(197\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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*(b*x^3+a)^p)*csgn(I
*c*(b*x^3+a)^p)*csgn(I*c)-I*Pi*csgn(I*c*(b*x^3+a)^p)^3+I*Pi*csgn(I*c*(b*x^3+a)^p)^2*csgn(I*c)-2*sum(_R*ln((-4*
_R^3*a^2+3*b^2*p^3)*x-p^2*a*_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 = 0.51, size = 120, normalized size = 0.86 \begin {gather*} \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}} \end {gather*}

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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Fricas [A]
time = 0.40, size = 150, normalized size = 1.08 \begin {gather*} \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 \left (c\right )}{4 \, x^{2}} \end {gather*}

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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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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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Giac [A]
time = 3.90, size = 138, normalized size = 0.99 \begin {gather*} -\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 \left (c\right )}{2 \, x^{2}} \end {gather*}

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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Mupad [B]
time = 2.61, size = 115, normalized size = 0.83 \begin {gather*} \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}} \end {gather*}

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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