3.1.24 \(\int \frac {\log (c (a+b x^3)^p)}{x^6} \, dx\) [24]

Optimal. Leaf size=151 \[ -\frac {3 b p}{10 a x^2}+\frac {\sqrt {3} b^{5/3} p \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{5 a^{5/3}}-\frac {b^{5/3} p \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{5 a^{5/3}}+\frac {b^{5/3} p \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{10 a^{5/3}}-\frac {\log \left (c \left (a+b x^3\right )^p\right )}{5 x^5} \]

[Out]

-3/10*b*p/a/x^2-1/5*b^(5/3)*p*ln(a^(1/3)+b^(1/3)*x)/a^(5/3)+1/10*b^(5/3)*p*ln(a^(2/3)-a^(1/3)*b^(1/3)*x+b^(2/3
)*x^2)/a^(5/3)-1/5*ln(c*(b*x^3+a)^p)/x^5+1/5*b^(5/3)*p*arctan(1/3*(a^(1/3)-2*b^(1/3)*x)/a^(1/3)*3^(1/2))*3^(1/
2)/a^(5/3)

________________________________________________________________________________________

Rubi [A]
time = 0.06, antiderivative size = 151, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {2505, 331, 206, 31, 648, 631, 210, 642} \begin {gather*} \frac {\sqrt {3} b^{5/3} p \text {ArcTan}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{5 a^{5/3}}+\frac {b^{5/3} p \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{10 a^{5/3}}-\frac {b^{5/3} p \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{5 a^{5/3}}-\frac {\log \left (c \left (a+b x^3\right )^p\right )}{5 x^5}-\frac {3 b p}{10 a x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Log[c*(a + b*x^3)^p]/x^6,x]

[Out]

(-3*b*p)/(10*a*x^2) + (Sqrt[3]*b^(5/3)*p*ArcTan[(a^(1/3) - 2*b^(1/3)*x)/(Sqrt[3]*a^(1/3))])/(5*a^(5/3)) - (b^(
5/3)*p*Log[a^(1/3) + b^(1/3)*x])/(5*a^(5/3)) + (b^(5/3)*p*Log[a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2])/(10*
a^(5/3)) - Log[c*(a + b*x^3)^p]/(5*x^5)

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 331

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] - Dist[b*((m + n*(p + 1) + 1)/(a*c^n*(m + 1))), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && 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 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^6} \, dx &=-\frac {\log \left (c \left (a+b x^3\right )^p\right )}{5 x^5}+\frac {1}{5} (3 b p) \int \frac {1}{x^3 \left (a+b x^3\right )} \, dx\\ &=-\frac {3 b p}{10 a x^2}-\frac {\log \left (c \left (a+b x^3\right )^p\right )}{5 x^5}-\frac {\left (3 b^2 p\right ) \int \frac {1}{a+b x^3} \, dx}{5 a}\\ &=-\frac {3 b p}{10 a x^2}-\frac {\log \left (c \left (a+b x^3\right )^p\right )}{5 x^5}-\frac {\left (b^2 p\right ) \int \frac {1}{\sqrt [3]{a}+\sqrt [3]{b} x} \, dx}{5 a^{5/3}}-\frac {\left (b^2 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}{5 a^{5/3}}\\ &=-\frac {3 b p}{10 a x^2}-\frac {b^{5/3} p \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{5 a^{5/3}}-\frac {\log \left (c \left (a+b x^3\right )^p\right )}{5 x^5}+\frac {\left (b^{5/3} 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}{10 a^{5/3}}-\frac {\left (3 b^2 p\right ) \int \frac {1}{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2} \, dx}{10 a^{4/3}}\\ &=-\frac {3 b p}{10 a x^2}-\frac {b^{5/3} p \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{5 a^{5/3}}+\frac {b^{5/3} p \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{10 a^{5/3}}-\frac {\log \left (c \left (a+b x^3\right )^p\right )}{5 x^5}-\frac {\left (3 b^{5/3} p\right ) \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}\right )}{5 a^{5/3}}\\ &=-\frac {3 b p}{10 a x^2}+\frac {\sqrt {3} b^{5/3} p \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{5 a^{5/3}}-\frac {b^{5/3} p \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{5 a^{5/3}}+\frac {b^{5/3} p \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{10 a^{5/3}}-\frac {\log \left (c \left (a+b x^3\right )^p\right )}{5 x^5}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 0.00, size = 49, normalized size = 0.32 \begin {gather*} -\frac {3 b p \, _2F_1\left (-\frac {2}{3},1;\frac {1}{3};-\frac {b x^3}{a}\right )}{10 a x^2}-\frac {\log \left (c \left (a+b x^3\right )^p\right )}{5 x^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Log[c*(a + b*x^3)^p]/x^6,x]

[Out]

(-3*b*p*Hypergeometric2F1[-2/3, 1, 1/3, -((b*x^3)/a)])/(10*a*x^2) - Log[c*(a + b*x^3)^p]/(5*x^5)

________________________________________________________________________________________

Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.31, size = 216, normalized size = 1.43

method	result	size
risch	\(-\frac {\ln \left (\left (x^{3} b +a \right )^{p}\right )}{5 x^{5}}-\frac {-2 \left (\munderset {\textit {\_R} =\RootOf \left (a^{5} \textit {\_Z}^{3}+b^{5} p^{3}\right )}{\sum }\textit {\_R} \ln \left (\left (-4 \textit {\_R}^{3} a^{5}-3 b^{5} p^{3}\right ) x -a^{2} b^{3} p^{2} \textit {\_R} \right )\right ) a \,x^{5}+i \pi a \,\mathrm {csgn}\left (i \left (x^{3} b +a \right )^{p}\right ) \mathrm {csgn}\left (i c \left (x^{3} b +a \right )^{p}\right )^{2}-i \pi a \,\mathrm {csgn}\left (i \left (x^{3} b +a \right )^{p}\right ) \mathrm {csgn}\left (i c \left (x^{3} b +a \right )^{p}\right ) \mathrm {csgn}\left (i c \right )-i \pi a \mathrm {csgn}\left (i c \left (x^{3} b +a \right )^{p}\right )^{3}+i \pi a \mathrm {csgn}\left (i c \left (x^{3} b +a \right )^{p}\right )^{2} \mathrm {csgn}\left (i c \right )+3 b p \,x^{3}+2 \ln \left (c \right ) a}{10 a \,x^{5}}\)	\(216\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(ln(c*(b*x^3+a)^p)/x^6,x,method=_RETURNVERBOSE)

[Out]

-1/5/x^5*ln((b*x^3+a)^p)-1/10*(-2*sum(_R*ln((-4*_R^3*a^5-3*b^5*p^3)*x-a^2*b^3*p^2*_R),_R=RootOf(_Z^3*a^5+b^5*p
^3))*a*x^5+I*Pi*a*csgn(I*(b*x^3+a)^p)*csgn(I*c*(b*x^3+a)^p)^2-I*Pi*a*csgn(I*(b*x^3+a)^p)*csgn(I*c*(b*x^3+a)^p)
*csgn(I*c)-I*Pi*a*csgn(I*c*(b*x^3+a)^p)^3+I*Pi*a*csgn(I*c*(b*x^3+a)^p)^2*csgn(I*c)+3*b*p*x^3+2*ln(c)*a)/a/x^5

________________________________________________________________________________________

Maxima [A]
time = 0.51, size = 128, normalized size = 0.85 \begin {gather*} -\frac {1}{10} \, 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 {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 )}{a \left (\frac {a}{b}\right )^{\frac {2}{3}}} + \frac {2 \, \log \left (x + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{a \left (\frac {a}{b}\right )^{\frac {2}{3}}} + \frac {3}{a x^{2}}\right )} - \frac {\log \left ({\left (b x^{3} + a\right )}^{p} c\right )}{5 \, x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(c*(b*x^3+a)^p)/x^6,x, algorithm="maxima")

[Out]

-1/10*b*p*(2*sqrt(3)*arctan(1/3*sqrt(3)*(2*x - (a/b)^(1/3))/(a/b)^(1/3))/(a*(a/b)^(2/3)) - log(x^2 - x*(a/b)^(
1/3) + (a/b)^(2/3))/(a*(a/b)^(2/3)) + 2*log(x + (a/b)^(1/3))/(a*(a/b)^(2/3)) + 3/(a*x^2)) - 1/5*log((b*x^3 + a
)^p*c)/x^5

________________________________________________________________________________________

Fricas [A]
time = 0.37, size = 172, normalized size = 1.14 \begin {gather*} \frac {2 \, \sqrt {3} b p x^{5} \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 ) - b p x^{5} \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 \, b p x^{5} \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 ) - 3 \, b p x^{3} - 2 \, a p \log \left (b x^{3} + a\right ) - 2 \, a \log \left (c\right )}{10 \, a x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(c*(b*x^3+a)^p)/x^6,x, algorithm="fricas")

[Out]

1/10*(2*sqrt(3)*b*p*x^5*(-b^2/a^2)^(1/3)*arctan(1/3*(2*sqrt(3)*a*x*(-b^2/a^2)^(2/3) - sqrt(3)*b)/b) - b*p*x^5*
(-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*b*p*x^5*(-b^2/a^2)^(1/3)*log
(b*x - a*(-b^2/a^2)^(1/3)) - 3*b*p*x^3 - 2*a*p*log(b*x^3 + a) - 2*a*log(c))/(a*x^5)

________________________________________________________________________________________

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**6,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [A]
time = 4.63, size = 149, normalized size = 0.99 \begin {gather*} \frac {b^{2} p \left (-\frac {a}{b}\right )^{\frac {1}{3}} \log \left ({\left | x - \left (-\frac {a}{b}\right )^{\frac {1}{3}} \right |}\right )}{5 \, a^{2}} - \frac {\sqrt {3} \left (-a b^{2}\right )^{\frac {1}{3}} b 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 )}{5 \, a^{2}} - \frac {\left (-a b^{2}\right )^{\frac {1}{3}} b p \log \left (x^{2} + x \left (-\frac {a}{b}\right )^{\frac {1}{3}} + \left (-\frac {a}{b}\right )^{\frac {2}{3}}\right )}{10 \, a^{2}} - \frac {p \log \left (b x^{3} + a\right )}{5 \, x^{5}} - \frac {3 \, b p x^{3} + 2 \, a \log \left (c\right )}{10 \, a x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(c*(b*x^3+a)^p)/x^6,x, algorithm="giac")

[Out]

1/5*b^2*p*(-a/b)^(1/3)*log(abs(x - (-a/b)^(1/3)))/a^2 - 1/5*sqrt(3)*(-a*b^2)^(1/3)*b*p*arctan(1/3*sqrt(3)*(2*x
+ (-a/b)^(1/3))/(-a/b)^(1/3))/a^2 - 1/10*(-a*b^2)^(1/3)*b*p*log(x^2 + x*(-a/b)^(1/3) + (-a/b)^(2/3))/a^2 - 1/
5*p*log(b*x^3 + a)/x^5 - 1/10*(3*b*p*x^3 + 2*a*log(c))/(a*x^5)

________________________________________________________________________________________

Mupad [B]
time = 2.45, size = 156, normalized size = 1.03 \begin {gather*} \frac {{\left (-b\right )}^{5/3}\,p\,\ln \left (a^{1/3}\,{\left (-b\right )}^{11/3}-b^4\,x\right )}{5\,a^{5/3}}-\frac {\ln \left (c\,{\left (b\,x^3+a\right )}^p\right )}{5\,x^5}-\frac {3\,b\,p}{10\,a\,x^2}+\frac {{\left (-b\right )}^{5/3}\,p\,\ln \left (225\,a^2\,b^4\,p\,x-225\,a^{7/3}\,{\left (-b\right )}^{11/3}\,p\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{5\,a^{5/3}}-\frac {{\left (-b\right )}^{5/3}\,p\,\ln \left (225\,a^2\,b^4\,p\,x+225\,a^{7/3}\,{\left (-b\right )}^{11/3}\,p\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{5\,a^{5/3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log(c*(a + b*x^3)^p)/x^6,x)

[Out]

((-b)^(5/3)*p*log(a^(1/3)*(-b)^(11/3) - b^4*x))/(5*a^(5/3)) - log(c*(a + b*x^3)^p)/(5*x^5) - (3*b*p)/(10*a*x^2
) + ((-b)^(5/3)*p*log(225*a^2*b^4*p*x - 225*a^(7/3)*(-b)^(11/3)*p*((3^(1/2)*1i)/2 - 1/2))*((3^(1/2)*1i)/2 - 1/
2))/(5*a^(5/3)) - ((-b)^(5/3)*p*log(225*a^2*b^4*p*x + 225*a^(7/3)*(-b)^(11/3)*p*((3^(1/2)*1i)/2 + 1/2))*((3^(1
/2)*1i)/2 + 1/2))/(5*a^(5/3))

________________________________________________________________________________________