\(\int \log (c (d+e (f+g x)^3)^q) \, dx\) [629]

Optimal result

Integrand size = 16, antiderivative size = 169 \[ \int \log \left (c \left (d+e (f+g x)^3\right )^q\right ) \, dx=-3 q x-\frac {\sqrt {3} \sqrt [3]{d} q \arctan \left (\frac {\sqrt [3]{d}-2 \sqrt [3]{e} (f+g x)}{\sqrt {3} \sqrt [3]{d}}\right )}{\sqrt [3]{e} g}+\frac {\sqrt [3]{d} q \log \left (\sqrt [3]{d}+\sqrt [3]{e} (f+g x)\right )}{\sqrt [3]{e} g}-\frac {\sqrt [3]{d} q \log \left (d^{2/3}-\sqrt [3]{d} \sqrt [3]{e} (f+g x)+e^{2/3} (f+g x)^2\right )}{2 \sqrt [3]{e} g}+\frac {(f+g x) \log \left (c \left (d+e (f+g x)^3\right )^q\right )}{g} \]

[Out]

-3*q*x+d^(1/3)*q*ln(d^(1/3)+e^(1/3)*(g*x+f))/e^(1/3)/g-1/2*d^(1/3)*q*ln(d^(2/3)-d^(1/3)*e^(1/3)*(g*x+f)+e^(2/3
)*(g*x+f)^2)/e^(1/3)/g+(g*x+f)*ln(c*(d+e*(g*x+f)^3)^q)/g-d^(1/3)*q*arctan(1/3*(d^(1/3)-2*e^(1/3)*(g*x+f))/d^(1
/3)*3^(1/2))*3^(1/2)/e^(1/3)/g

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.562, Rules used = {2533, 2498, 327, 206, 31, 648, 631, 210, 642} \[ \int \log \left (c \left (d+e (f+g x)^3\right )^q\right ) \, dx=-\frac {\sqrt {3} \sqrt [3]{d} q \arctan \left (\frac {\sqrt [3]{d}-2 \sqrt [3]{e} (f+g x)}{\sqrt {3} \sqrt [3]{d}}\right )}{\sqrt [3]{e} g}+\frac {(f+g x) \log \left (c \left (d+e (f+g x)^3\right )^q\right )}{g}-\frac {\sqrt [3]{d} q \log \left (d^{2/3}-\sqrt [3]{d} \sqrt [3]{e} (f+g x)+e^{2/3} (f+g x)^2\right )}{2 \sqrt [3]{e} g}+\frac {\sqrt [3]{d} q \log \left (\sqrt [3]{d}+\sqrt [3]{e} (f+g x)\right )}{\sqrt [3]{e} g}-3 q x \]

[In]

Int[Log[c*(d + e*(f + g*x)^3)^q],x]

[Out]

-3*q*x - (Sqrt[3]*d^(1/3)*q*ArcTan[(d^(1/3) - 2*e^(1/3)*(f + g*x))/(Sqrt[3]*d^(1/3))])/(e^(1/3)*g) + (d^(1/3)*
q*Log[d^(1/3) + e^(1/3)*(f + g*x)])/(e^(1/3)*g) - (d^(1/3)*q*Log[d^(2/3) - d^(1/3)*e^(1/3)*(f + g*x) + e^(2/3)
*(f + g*x)^2])/(2*e^(1/3)*g) + ((f + g*x)*Log[c*(d + e*(f + g*x)^3)^q])/g

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]

Rule 2533

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*((f_.) + (g_.)*(x_))^(n_))^(p_.)]*(b_.))^(q_.), x_Symbol] :> Dist[1/g, Su
bst[Int[(a + b*Log[c*(d + e*x^n)^p])^q, x], x, f + g*x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p}, x] && IGtQ[q
, 0] && (EqQ[q, 1] || IntegerQ[n])

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \log \left (c \left (d+e x^3\right )^q\right ) \, dx,x,f+g x\right )}{g} \\ & = \frac {(f+g x) \log \left (c \left (d+e (f+g x)^3\right )^q\right )}{g}-\frac {(3 e q) \text {Subst}\left (\int \frac {x^3}{d+e x^3} \, dx,x,f+g x\right )}{g} \\ & = -3 q x+\frac {(f+g x) \log \left (c \left (d+e (f+g x)^3\right )^q\right )}{g}+\frac {(3 d q) \text {Subst}\left (\int \frac {1}{d+e x^3} \, dx,x,f+g x\right )}{g} \\ & = -3 q x+\frac {(f+g x) \log \left (c \left (d+e (f+g x)^3\right )^q\right )}{g}+\frac {\left (\sqrt [3]{d} q\right ) \text {Subst}\left (\int \frac {1}{\sqrt [3]{d}+\sqrt [3]{e} x} \, dx,x,f+g x\right )}{g}+\frac {\left (\sqrt [3]{d} q\right ) \text {Subst}\left (\int \frac {2 \sqrt [3]{d}-\sqrt [3]{e} x}{d^{2/3}-\sqrt [3]{d} \sqrt [3]{e} x+e^{2/3} x^2} \, dx,x,f+g x\right )}{g} \\ & = -3 q x+\frac {\sqrt [3]{d} q \log \left (\sqrt [3]{d}+\sqrt [3]{e} (f+g x)\right )}{\sqrt [3]{e} g}+\frac {(f+g x) \log \left (c \left (d+e (f+g x)^3\right )^q\right )}{g}+\frac {\left (3 d^{2/3} q\right ) \text {Subst}\left (\int \frac {1}{d^{2/3}-\sqrt [3]{d} \sqrt [3]{e} x+e^{2/3} x^2} \, dx,x,f+g x\right )}{2 g}-\frac {\left (\sqrt [3]{d} q\right ) \text {Subst}\left (\int \frac {-\sqrt [3]{d} \sqrt [3]{e}+2 e^{2/3} x}{d^{2/3}-\sqrt [3]{d} \sqrt [3]{e} x+e^{2/3} x^2} \, dx,x,f+g x\right )}{2 \sqrt [3]{e} g} \\ & = -3 q x+\frac {\sqrt [3]{d} q \log \left (\sqrt [3]{d}+\sqrt [3]{e} (f+g x)\right )}{\sqrt [3]{e} g}-\frac {\sqrt [3]{d} q \log \left (d^{2/3}-\sqrt [3]{d} \sqrt [3]{e} (f+g x)+e^{2/3} (f+g x)^2\right )}{2 \sqrt [3]{e} g}+\frac {(f+g x) \log \left (c \left (d+e (f+g x)^3\right )^q\right )}{g}+\frac {\left (3 \sqrt [3]{d} q\right ) \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1-\frac {2 \sqrt [3]{e} (f+g x)}{\sqrt [3]{d}}\right )}{\sqrt [3]{e} g} \\ & = -3 q x-\frac {\sqrt {3} \sqrt [3]{d} q \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{e} (f+g x)}{\sqrt [3]{d}}}{\sqrt {3}}\right )}{\sqrt [3]{e} g}+\frac {\sqrt [3]{d} q \log \left (\sqrt [3]{d}+\sqrt [3]{e} (f+g x)\right )}{\sqrt [3]{e} g}-\frac {\sqrt [3]{d} q \log \left (d^{2/3}-\sqrt [3]{d} \sqrt [3]{e} (f+g x)+e^{2/3} (f+g x)^2\right )}{2 \sqrt [3]{e} g}+\frac {(f+g x) \log \left (c \left (d+e (f+g x)^3\right )^q\right )}{g} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.87 \[ \int \log \left (c \left (d+e (f+g x)^3\right )^q\right ) \, dx=-3 q x+\frac {\sqrt [3]{d} q \left (2 \sqrt {3} \arctan \left (\frac {-\sqrt [3]{d}+2 \sqrt [3]{e} (f+g x)}{\sqrt {3} \sqrt [3]{d}}\right )+2 \log \left (\sqrt [3]{d}+\sqrt [3]{e} (f+g x)\right )-\log \left (d^{2/3}-\sqrt [3]{d} \sqrt [3]{e} (f+g x)+e^{2/3} (f+g x)^2\right )\right )}{2 \sqrt [3]{e} g}+\frac {(f+g x) \log \left (c \left (d+e (f+g x)^3\right )^q\right )}{g} \]

[In]

Integrate[Log[c*(d + e*(f + g*x)^3)^q],x]

[Out]

-3*q*x + (d^(1/3)*q*(2*Sqrt[3]*ArcTan[(-d^(1/3) + 2*e^(1/3)*(f + g*x))/(Sqrt[3]*d^(1/3))] + 2*Log[d^(1/3) + e^
(1/3)*(f + g*x)] - Log[d^(2/3) - d^(1/3)*e^(1/3)*(f + g*x) + e^(2/3)*(f + g*x)^2]))/(2*e^(1/3)*g) + ((f + g*x)
*Log[c*(d + e*(f + g*x)^3)^q])/g

Maple [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 1.46 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.78

method	result	size
parts	\(\ln \left (c \left (d +e \left (g x +f \right )^{3}\right )^{q}\right ) x -3 g e q \left (\frac {x}{g e}+\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (g^{3} e \,\textit {\_Z}^{3}+3 e f \,g^{2} \textit {\_Z}^{2}+3 e \,f^{2} g \textit {\_Z} +e \,f^{3}+d \right )}{\sum }\frac {\left (-\textit {\_R}^{2} e f \,g^{2}-2 \textit {\_R} e \,f^{2} g -e \,f^{3}-d \right ) \ln \left (x -\textit {\_R} \right )}{g^{2} \textit {\_R}^{2}+2 f g \textit {\_R} +f^{2}}}{3 e^{2} g^{2}}\right )\)	\(132\)
default	\(\ln \left (c \left (e \,g^{3} x^{3}+3 e f \,g^{2} x^{2}+3 e \,f^{2} g x +e \,f^{3}+d \right )^{q}\right ) x -3 g e q \left (\frac {x}{g e}+\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (g^{3} e \,\textit {\_Z}^{3}+3 e f \,g^{2} \textit {\_Z}^{2}+3 e \,f^{2} g \textit {\_Z} +e \,f^{3}+d \right )}{\sum }\frac {\left (-\textit {\_R}^{2} e f \,g^{2}-2 \textit {\_R} e \,f^{2} g -e \,f^{3}-d \right ) \ln \left (x -\textit {\_R} \right )}{g^{2} \textit {\_R}^{2}+2 f g \textit {\_R} +f^{2}}}{3 e^{2} g^{2}}\right )\)	\(154\)
risch	\(x \ln \left (\left (d +e \left (g x +f \right )^{3}\right )^{q}\right )+\frac {i {\operatorname {csgn}\left (i c \left (d +e \left (g x +f \right )^{3}\right )^{q}\right )}^{2} \operatorname {csgn}\left (i \left (d +e \left (g x +f \right )^{3}\right )^{q}\right ) x \pi }{2}-\frac {i \pi x \,\operatorname {csgn}\left (i \left (d +e \left (g x +f \right )^{3}\right )^{q}\right ) \operatorname {csgn}\left (i c \left (d +e \left (g x +f \right )^{3}\right )^{q}\right ) \operatorname {csgn}\left (i c \right )}{2}-\frac {i \pi x {\operatorname {csgn}\left (i c \left (d +e \left (g x +f \right )^{3}\right )^{q}\right )}^{3}}{2}+\frac {i \operatorname {csgn}\left (i c \right ) {\operatorname {csgn}\left (i c \left (d +e \left (g x +f \right )^{3}\right )^{q}\right )}^{2} x \pi }{2}+x \ln \left (c \right )-3 q x +\frac {q \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (g^{3} e \,\textit {\_Z}^{3}+3 e f \,g^{2} \textit {\_Z}^{2}+3 e \,f^{2} g \textit {\_Z} +e \,f^{3}+d \right )}{\sum }\frac {\left (\textit {\_R}^{2} e f \,g^{2}+2 \textit {\_R} e \,f^{2} g +e \,f^{3}+d \right ) \ln \left (x -\textit {\_R} \right )}{g^{2} \textit {\_R}^{2}+2 f g \textit {\_R} +f^{2}}\right )}{e g}\)	\(262\)

[In]

int(ln(c*(d+e*(g*x+f)^3)^q),x,method=_RETURNVERBOSE)

[Out]

ln(c*(d+e*(g*x+f)^3)^q)*x-3*g*e*q*(1/g/e*x+1/3/e^2/g^2*sum((-_R^2*e*f*g^2-2*_R*e*f^2*g-e*f^3-d)/(_R^2*g^2+2*_R
*f*g+f^2)*ln(x-_R),_R=RootOf(_Z^3*e*g^3+3*_Z^2*e*f*g^2+3*_Z*e*f^2*g+e*f^3+d)))

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.07 (sec) , antiderivative size = 1134, normalized size of antiderivative = 6.71 \[ \int \log \left (c \left (d+e (f+g x)^3\right )^q\right ) \, dx=\text {Too large to display} \]

[In]

integrate(log(c*(d+e*(g*x+f)^3)^q),x, algorithm="fricas")

[Out]

1/4*(4*g*q*x*log(e*g^3*x^3 + 3*e*f*g^2*x^2 + 3*e*f^2*g*x + e*f^3 + d) - 12*g*q*x - 2*((-1/2*f^3*q^3/g^3 + 1/2*
d*q^3/(e*g^3) + 1/2*(e*f^3*q^3 + d*q^3)/(e*g^3))^(1/3)*(I*sqrt(3) + 1) - 2*f*q/g)*g*log(q*x - 1/2*(-1/2*f^3*q^
3/g^3 + 1/2*d*q^3/(e*g^3) + 1/2*(e*f^3*q^3 + d*q^3)/(e*g^3))^(1/3)*(I*sqrt(3) + 1) + f*q/g) + 4*g*x*log(c) + (
((-1/2*f^3*q^3/g^3 + 1/2*d*q^3/(e*g^3) + 1/2*(e*f^3*q^3 + d*q^3)/(e*g^3))^(1/3)*(I*sqrt(3) + 1) - 2*f*q/g)*g +
6*f*q + sqrt(3)*g*sqrt(-(((-1/2*f^3*q^3/g^3 + 1/2*d*q^3/(e*g^3) + 1/2*(e*f^3*q^3 + d*q^3)/(e*g^3))^(1/3)*(I*s
qrt(3) + 1) - 2*f*q/g)^2*g^2 + 4*((-1/2*f^3*q^3/g^3 + 1/2*d*q^3/(e*g^3) + 1/2*(e*f^3*q^3 + d*q^3)/(e*g^3))^(1/
3)*(I*sqrt(3) + 1) - 2*f*q/g)*f*g*q + 4*f^2*q^2)/g^2))*log(2*g*q*x + 1/2*((-1/2*f^3*q^3/g^3 + 1/2*d*q^3/(e*g^3
) + 1/2*(e*f^3*q^3 + d*q^3)/(e*g^3))^(1/3)*(I*sqrt(3) + 1) - 2*f*q/g)*g + 3*f*q + 1/2*sqrt(3)*g*sqrt(-(((-1/2*
f^3*q^3/g^3 + 1/2*d*q^3/(e*g^3) + 1/2*(e*f^3*q^3 + d*q^3)/(e*g^3))^(1/3)*(I*sqrt(3) + 1) - 2*f*q/g)^2*g^2 + 4*
((-1/2*f^3*q^3/g^3 + 1/2*d*q^3/(e*g^3) + 1/2*(e*f^3*q^3 + d*q^3)/(e*g^3))^(1/3)*(I*sqrt(3) + 1) - 2*f*q/g)*f*g
*q + 4*f^2*q^2)/g^2)) + (((-1/2*f^3*q^3/g^3 + 1/2*d*q^3/(e*g^3) + 1/2*(e*f^3*q^3 + d*q^3)/(e*g^3))^(1/3)*(I*sq
rt(3) + 1) - 2*f*q/g)*g + 6*f*q - sqrt(3)*g*sqrt(-(((-1/2*f^3*q^3/g^3 + 1/2*d*q^3/(e*g^3) + 1/2*(e*f^3*q^3 + d
*q^3)/(e*g^3))^(1/3)*(I*sqrt(3) + 1) - 2*f*q/g)^2*g^2 + 4*((-1/2*f^3*q^3/g^3 + 1/2*d*q^3/(e*g^3) + 1/2*(e*f^3*
q^3 + d*q^3)/(e*g^3))^(1/3)*(I*sqrt(3) + 1) - 2*f*q/g)*f*g*q + 4*f^2*q^2)/g^2))*log(2*g*q*x + 1/2*((-1/2*f^3*q
^3/g^3 + 1/2*d*q^3/(e*g^3) + 1/2*(e*f^3*q^3 + d*q^3)/(e*g^3))^(1/3)*(I*sqrt(3) + 1) - 2*f*q/g)*g + 3*f*q - 1/2
*sqrt(3)*g*sqrt(-(((-1/2*f^3*q^3/g^3 + 1/2*d*q^3/(e*g^3) + 1/2*(e*f^3*q^3 + d*q^3)/(e*g^3))^(1/3)*(I*sqrt(3) +
1) - 2*f*q/g)^2*g^2 + 4*((-1/2*f^3*q^3/g^3 + 1/2*d*q^3/(e*g^3) + 1/2*(e*f^3*q^3 + d*q^3)/(e*g^3))^(1/3)*(I*sq
rt(3) + 1) - 2*f*q/g)*f*g*q + 4*f^2*q^2)/g^2)))/g

Sympy [F(-1)]

Timed out. \[ \int \log \left (c \left (d+e (f+g x)^3\right )^q\right ) \, dx=\text {Timed out} \]

[In]

integrate(ln(c*(d+e*(g*x+f)**3)**q),x)

[Out]

Timed out

Maxima [F]

\[ \int \log \left (c \left (d+e (f+g x)^3\right )^q\right ) \, dx=\int { \log \left ({\left ({\left (g x + f\right )}^{3} e + d\right )}^{q} c\right ) \,d x } \]

[In]

integrate(log(c*(d+e*(g*x+f)^3)^q),x, algorithm="maxima")

[Out]

-(3*q - log(c))*x + 3*q*integrate((e*f*g^2*x^2 + 2*e*f^2*g*x + e*f^3 + d)/(e*g^3*x^3 + 3*e*f*g^2*x^2 + 3*e*f^2
*g*x + e*f^3 + d), x) + x*log((e*g^3*x^3 + 3*e*f*g^2*x^2 + 3*e*f^2*g*x + e*f^3 + d)^q)

Giac [A] (verification not implemented)

none

Time = 0.44 (sec) , antiderivative size = 261, normalized size of antiderivative = 1.54 \[ \int \log \left (c \left (d+e (f+g x)^3\right )^q\right ) \, dx=q x \log \left (e g^{3} x^{3} + 3 \, e f g^{2} x^{2} + 3 \, e f^{2} g x + e f^{3} + d\right ) - {\left (3 \, q - \log \left (c\right )\right )} x + \frac {f q \log \left ({\left | e g^{3} x^{3} + 3 \, e f g^{2} x^{2} + 3 \, e f^{2} g x + e f^{3} + d \right |}\right )}{g} + \frac {2 \, \sqrt {3} \left (d e^{2} g^{6} q^{3}\right )^{\frac {1}{3}} \arctan \left (-\frac {e g x + e f + \left (d e^{2}\right )^{\frac {1}{3}}}{\sqrt {3} e g x + \sqrt {3} e f - \sqrt {3} \left (d e^{2}\right )^{\frac {1}{3}}}\right ) - \left (d e^{2} g^{6} q^{3}\right )^{\frac {1}{3}} \log \left (4 \, {\left (\sqrt {3} e g x + \sqrt {3} e f - \sqrt {3} \left (d e^{2}\right )^{\frac {1}{3}}\right )}^{2} + 4 \, {\left (e g x + e f + \left (d e^{2}\right )^{\frac {1}{3}}\right )}^{2}\right ) + 2 \, \left (d e^{2} g^{6} q^{3}\right )^{\frac {1}{3}} \log \left ({\left | e g x + e f + \left (d e^{2}\right )^{\frac {1}{3}} \right |}\right )}{2 \, e g^{3}} \]

[In]

integrate(log(c*(d+e*(g*x+f)^3)^q),x, algorithm="giac")

[Out]

q*x*log(e*g^3*x^3 + 3*e*f*g^2*x^2 + 3*e*f^2*g*x + e*f^3 + d) - (3*q - log(c))*x + f*q*log(abs(e*g^3*x^3 + 3*e*
f*g^2*x^2 + 3*e*f^2*g*x + e*f^3 + d))/g + 1/2*(2*sqrt(3)*(d*e^2*g^6*q^3)^(1/3)*arctan(-(e*g*x + e*f + (d*e^2)^
(1/3))/(sqrt(3)*e*g*x + sqrt(3)*e*f - sqrt(3)*(d*e^2)^(1/3))) - (d*e^2*g^6*q^3)^(1/3)*log(4*(sqrt(3)*e*g*x + s
qrt(3)*e*f - sqrt(3)*(d*e^2)^(1/3))^2 + 4*(e*g*x + e*f + (d*e^2)^(1/3))^2) + 2*(d*e^2*g^6*q^3)^(1/3)*log(abs(e
*g*x + e*f + (d*e^2)^(1/3))))/(e*g^3)

Mupad [B] (verification not implemented)

Time = 1.62 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.14 \[ \int \log \left (c \left (d+e (f+g x)^3\right )^q\right ) \, dx=x\,\ln \left (c\,{\left (d+e\,{\left (f+g\,x\right )}^3\right )}^q\right )-\left (\sum _{k=1}^3\ln \left (d\,e^2\,g^5\,\left (\mathrm {root}\left (b^3\,e\,g^3+3\,b^2\,e\,f\,g^2\,q+3\,b\,e\,f^2\,g\,q^2+e\,f^3\,q^3+d\,q^3,b,k\right )\,g+f\,q\right )\,\left (\mathrm {root}\left (b^3\,e\,g^3+3\,b^2\,e\,f\,g^2\,q+3\,b\,e\,f^2\,g\,q^2+e\,f^3\,q^3+d\,q^3,b,k\right )-q\,x\right )\,9\right )\,\mathrm {root}\left (b^3\,e\,g^3+3\,b^2\,e\,f\,g^2\,q+3\,b\,e\,f^2\,g\,q^2+e\,f^3\,q^3+d\,q^3,b,k\right )\right )-3\,q\,x \]

[In]

int(log(c*(d + e*(f + g*x)^3)^q),x)

[Out]

x*log(c*(d + e*(f + g*x)^3)^q) - symsum(log(9*d*e^2*g^5*(root(b^3*e*g^3 + 3*b^2*e*f*g^2*q + 3*b*e*f^2*g*q^2 +
e*f^3*q^3 + d*q^3, b, k)*g + f*q)*(root(b^3*e*g^3 + 3*b^2*e*f*g^2*q + 3*b*e*f^2*g*q^2 + e*f^3*q^3 + d*q^3, b,
k) - q*x))*root(b^3*e*g^3 + 3*b^2*e*f*g^2*q + 3*b*e*f^2*g*q^2 + e*f^3*q^3 + d*q^3, b, k), k, 1, 3) - 3*q*x