\(\int x^3 \log (c (a+b x^3)^p) \, dx\) [15]

Optimal result

Integrand size = 16, antiderivative size = 157 \[ \int x^3 \log \left (c \left (a+b x^3\right )^p\right ) \, dx=\frac {3 a p x}{4 b}-\frac {3 p x^4}{16}+\frac {\sqrt {3} a^{4/3} p \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{4 b^{4/3}}-\frac {a^{4/3} p \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{4 b^{4/3}}+\frac {a^{4/3} p \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{8 b^{4/3}}+\frac {1}{4} x^4 \log \left (c \left (a+b x^3\right )^p\right ) \]

[Out]

3/4*a*p*x/b-3/16*p*x^4-1/4*a^(4/3)*p*ln(a^(1/3)+b^(1/3)*x)/b^(4/3)+1/8*a^(4/3)*p*ln(a^(2/3)-a^(1/3)*b^(1/3)*x+
b^(2/3)*x^2)/b^(4/3)+1/4*x^4*ln(c*(b*x^3+a)^p)+1/4*a^(4/3)*p*arctan(1/3*(a^(1/3)-2*b^(1/3)*x)/a^(1/3)*3^(1/2))
*3^(1/2)/b^(4/3)

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {2505, 308, 206, 31, 648, 631, 210, 642} \[ \int x^3 \log \left (c \left (a+b x^3\right )^p\right ) \, dx=\frac {\sqrt {3} a^{4/3} p \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{4 b^{4/3}}+\frac {a^{4/3} p \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{8 b^{4/3}}-\frac {a^{4/3} p \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{4 b^{4/3}}+\frac {1}{4} x^4 \log \left (c \left (a+b x^3\right )^p\right )+\frac {3 a p x}{4 b}-\frac {3 p x^4}{16} \]

[In]

Int[x^3*Log[c*(a + b*x^3)^p],x]

[Out]

(3*a*p*x)/(4*b) - (3*p*x^4)/16 + (Sqrt[3]*a^(4/3)*p*ArcTan[(a^(1/3) - 2*b^(1/3)*x)/(Sqrt[3]*a^(1/3))])/(4*b^(4
/3)) - (a^(4/3)*p*Log[a^(1/3) + b^(1/3)*x])/(4*b^(4/3)) + (a^(4/3)*p*Log[a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)
*x^2])/(8*b^(4/3)) + (x^4*Log[c*(a + b*x^3)^p])/4

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 308

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

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*} \text {integral}& = \frac {1}{4} x^4 \log \left (c \left (a+b x^3\right )^p\right )-\frac {1}{4} (3 b p) \int \frac {x^6}{a+b x^3} \, dx \\ & = \frac {1}{4} x^4 \log \left (c \left (a+b x^3\right )^p\right )-\frac {1}{4} (3 b p) \int \left (-\frac {a}{b^2}+\frac {x^3}{b}+\frac {a^2}{b^2 \left (a+b x^3\right )}\right ) \, dx \\ & = \frac {3 a p x}{4 b}-\frac {3 p x^4}{16}+\frac {1}{4} x^4 \log \left (c \left (a+b x^3\right )^p\right )-\frac {\left (3 a^2 p\right ) \int \frac {1}{a+b x^3} \, dx}{4 b} \\ & = \frac {3 a p x}{4 b}-\frac {3 p x^4}{16}+\frac {1}{4} x^4 \log \left (c \left (a+b x^3\right )^p\right )-\frac {\left (a^{4/3} p\right ) \int \frac {1}{\sqrt [3]{a}+\sqrt [3]{b} x} \, dx}{4 b}-\frac {\left (a^{4/3} 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}{4 b} \\ & = \frac {3 a p x}{4 b}-\frac {3 p x^4}{16}-\frac {a^{4/3} p \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{4 b^{4/3}}+\frac {1}{4} x^4 \log \left (c \left (a+b x^3\right )^p\right )+\frac {\left (a^{4/3} 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}{8 b^{4/3}}-\frac {\left (3 a^{5/3} p\right ) \int \frac {1}{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2} \, dx}{8 b} \\ & = \frac {3 a p x}{4 b}-\frac {3 p x^4}{16}-\frac {a^{4/3} p \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{4 b^{4/3}}+\frac {a^{4/3} p \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{8 b^{4/3}}+\frac {1}{4} x^4 \log \left (c \left (a+b x^3\right )^p\right )-\frac {\left (3 a^{4/3} p\right ) \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}\right )}{4 b^{4/3}} \\ & = \frac {3 a p x}{4 b}-\frac {3 p x^4}{16}+\frac {\sqrt {3} a^{4/3} p \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{4 b^{4/3}}-\frac {a^{4/3} p \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{4 b^{4/3}}+\frac {a^{4/3} p \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{8 b^{4/3}}+\frac {1}{4} x^4 \log \left (c \left (a+b x^3\right )^p\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.94 \[ \int x^3 \log \left (c \left (a+b x^3\right )^p\right ) \, dx=\frac {12 a \sqrt [3]{b} p x-3 b^{4/3} p x^4+4 \sqrt {3} a^{4/3} p \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )-4 a^{4/3} p \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )+2 a^{4/3} p \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )+4 b^{4/3} x^4 \log \left (c \left (a+b x^3\right )^p\right )}{16 b^{4/3}} \]

[In]

Integrate[x^3*Log[c*(a + b*x^3)^p],x]

[Out]

(12*a*b^(1/3)*p*x - 3*b^(4/3)*p*x^4 + 4*Sqrt[3]*a^(4/3)*p*ArcTan[(1 - (2*b^(1/3)*x)/a^(1/3))/Sqrt[3]] - 4*a^(4
/3)*p*Log[a^(1/3) + b^(1/3)*x] + 2*a^(4/3)*p*Log[a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2] + 4*b^(4/3)*x^4*Lo
g[c*(a + b*x^3)^p])/(16*b^(4/3))

Maple [A] (verified)

Time = 0.66 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.87

method	result	size
parts	\(\frac {x^{4} \ln \left (c \left (b \,x^{3}+a \right )^{p}\right )}{4}-\frac {3 p b \left (-\frac {-\frac {1}{4} b \,x^{4}+a x}{b^{2}}+\frac {\left (\frac {\ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}-\frac {\ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}\right ) a^{2}}{b^{2}}\right )}{4}\)	\(136\)
risch	\(\frac {x^{4} \ln \left (\left (b \,x^{3}+a \right )^{p}\right )}{4}+\frac {i \pi \,x^{4} \operatorname {csgn}\left (i \left (b \,x^{3}+a \right )^{p}\right ) {\operatorname {csgn}\left (i c \left (b \,x^{3}+a \right )^{p}\right )}^{2}}{8}-\frac {i \pi \,x^{4} \operatorname {csgn}\left (i \left (b \,x^{3}+a \right )^{p}\right ) \operatorname {csgn}\left (i c \left (b \,x^{3}+a \right )^{p}\right ) \operatorname {csgn}\left (i c \right )}{8}-\frac {i \pi \,x^{4} {\operatorname {csgn}\left (i c \left (b \,x^{3}+a \right )^{p}\right )}^{3}}{8}+\frac {i \pi \,x^{4} {\operatorname {csgn}\left (i c \left (b \,x^{3}+a \right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )}{8}+\frac {\ln \left (c \right ) x^{4}}{4}-\frac {3 p \,x^{4}}{16}-\frac {a^{2} p \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (b \,\textit {\_Z}^{3}+a \right )}{\sum }\frac {\ln \left (x -\textit {\_R} \right )}{\textit {\_R}^{2}}\right )}{4 b^{2}}+\frac {3 a p x}{4 b}\)	\(194\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.92 \[ \int x^3 \log \left (c \left (a+b x^3\right )^p\right ) \, dx=\frac {4 \, b p x^{4} \log \left (b x^{3} + a\right ) - 3 \, b p x^{4} + 4 \, b x^{4} \log \left (c\right ) + 4 \, \sqrt {3} a p \left (-\frac {a}{b}\right )^{\frac {1}{3}} \arctan \left (\frac {2 \, \sqrt {3} b x \left (-\frac {a}{b}\right )^{\frac {2}{3}} - \sqrt {3} a}{3 \, a}\right ) - 2 \, a p \left (-\frac {a}{b}\right )^{\frac {1}{3}} \log \left (x^{2} + x \left (-\frac {a}{b}\right )^{\frac {1}{3}} + \left (-\frac {a}{b}\right )^{\frac {2}{3}}\right ) + 4 \, a p \left (-\frac {a}{b}\right )^{\frac {1}{3}} \log \left (x - \left (-\frac {a}{b}\right )^{\frac {1}{3}}\right ) + 12 \, a p x}{16 \, b} \]

[In]

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

[Out]

1/16*(4*b*p*x^4*log(b*x^3 + a) - 3*b*p*x^4 + 4*b*x^4*log(c) + 4*sqrt(3)*a*p*(-a/b)^(1/3)*arctan(1/3*(2*sqrt(3)
*b*x*(-a/b)^(2/3) - sqrt(3)*a)/a) - 2*a*p*(-a/b)^(1/3)*log(x^2 + x*(-a/b)^(1/3) + (-a/b)^(2/3)) + 4*a*p*(-a/b)
^(1/3)*log(x - (-a/b)^(1/3)) + 12*a*p*x)/b

Sympy [F(-1)]

Timed out. \[ \int x^3 \log \left (c \left (a+b x^3\right )^p\right ) \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.92 \[ \int x^3 \log \left (c \left (a+b x^3\right )^p\right ) \, dx=\frac {1}{4} \, x^{4} \log \left ({\left (b x^{3} + a\right )}^{p} c\right ) - \frac {1}{16} \, b p {\left (\frac {4 \, \sqrt {3} a^{2} \arctan \left (\frac {\sqrt {3} {\left (2 \, x - \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{b^{3} \left (\frac {a}{b}\right )^{\frac {2}{3}}} - \frac {2 \, a^{2} \log \left (x^{2} - x \left (\frac {a}{b}\right )^{\frac {1}{3}} + \left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{b^{3} \left (\frac {a}{b}\right )^{\frac {2}{3}}} + \frac {4 \, a^{2} \log \left (x + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{b^{3} \left (\frac {a}{b}\right )^{\frac {2}{3}}} + \frac {3 \, {\left (b x^{4} - 4 \, a x\right )}}{b^{2}}\right )} \]

[In]

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

[Out]

1/4*x^4*log((b*x^3 + a)^p*c) - 1/16*b*p*(4*sqrt(3)*a^2*arctan(1/3*sqrt(3)*(2*x - (a/b)^(1/3))/(a/b)^(1/3))/(b^
3*(a/b)^(2/3)) - 2*a^2*log(x^2 - x*(a/b)^(1/3) + (a/b)^(2/3))/(b^3*(a/b)^(2/3)) + 4*a^2*log(x + (a/b)^(1/3))/(
b^3*(a/b)^(2/3)) + 3*(b*x^4 - 4*a*x)/b^2)

Giac [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.02 \[ \int x^3 \log \left (c \left (a+b x^3\right )^p\right ) \, dx=\frac {1}{8} \, a^{2} b^{3} p {\left (\frac {2 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}} \log \left ({\left | x - \left (-\frac {a}{b}\right )^{\frac {1}{3}} \right |}\right )}{a b^{4}} - \frac {2 \, \sqrt {3} \left (-a b^{2}\right )^{\frac {1}{3}} \arctan \left (\frac {\sqrt {3} {\left (2 \, x + \left (-\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{a b^{5}} - \frac {\left (-a b^{2}\right )^{\frac {1}{3}} \log \left (x^{2} + x \left (-\frac {a}{b}\right )^{\frac {1}{3}} + \left (-\frac {a}{b}\right )^{\frac {2}{3}}\right )}{a b^{5}}\right )} + \frac {1}{4} \, p x^{4} \log \left (b x^{3} + a\right ) - \frac {1}{16} \, {\left (3 \, p - 4 \, \log \left (c\right )\right )} x^{4} + \frac {3 \, a p x}{4 \, b} \]

[In]

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

[Out]

1/8*a^2*b^3*p*(2*(-a/b)^(1/3)*log(abs(x - (-a/b)^(1/3)))/(a*b^4) - 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^5) - (-a*b^2)^(1/3)*log(x^2 + x*(-a/b)^(1/3) + (-a/b)^(2/3))/(a*b^5))
+ 1/4*p*x^4*log(b*x^3 + a) - 1/16*(3*p - 4*log(c))*x^4 + 3/4*a*p*x/b

Mupad [B] (verification not implemented)

Time = 3.84 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.82 \[ \int x^3 \log \left (c \left (a+b x^3\right )^p\right ) \, dx=\frac {x^4\,\ln \left (c\,{\left (b\,x^3+a\right )}^p\right )}{4}-\frac {3\,p\,x^4}{16}+\frac {3\,a\,p\,x}{4\,b}-\frac {a^{4/3}\,p\,\ln \left (b^{1/3}\,x+a^{1/3}\right )}{4\,b^{4/3}}+\frac {a^{4/3}\,p\,\ln \left (2\,b^{1/3}\,x-a^{1/3}-\sqrt {3}\,a^{1/3}\,1{}\mathrm {i}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{4\,b^{4/3}}-\frac {a^{4/3}\,p\,\ln \left (2\,b^{1/3}\,x-a^{1/3}+\sqrt {3}\,a^{1/3}\,1{}\mathrm {i}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{4\,b^{4/3}} \]

[In]

int(x^3*log(c*(a + b*x^3)^p),x)

[Out]

(x^4*log(c*(a + b*x^3)^p))/4 - (3*p*x^4)/16 + (3*a*p*x)/(4*b) - (a^(4/3)*p*log(b^(1/3)*x + a^(1/3)))/(4*b^(4/3
)) + (a^(4/3)*p*log(2*b^(1/3)*x - 3^(1/2)*a^(1/3)*1i - a^(1/3))*((3^(1/2)*1i)/2 + 1/2))/(4*b^(4/3)) - (a^(4/3)
*p*log(3^(1/2)*a^(1/3)*1i + 2*b^(1/3)*x - a^(1/3))*((3^(1/2)*1i)/2 - 1/2))/(4*b^(4/3))