3.634 \(\int \log (c (d+\frac {e}{(f+g x)^3})^q) \, dx\)

Optimal. Leaf size=165 \[ \frac {(f+g x) \log \left (c \left (d+\frac {e}{(f+g x)^3}\right )^q\right )}{g}-\frac {\sqrt [3]{e} q \log \left (d^{2/3} (f+g x)^2-\sqrt [3]{d} \sqrt [3]{e} (f+g x)+e^{2/3}\right )}{2 \sqrt [3]{d} g}+\frac {\sqrt [3]{e} q \log \left (\sqrt [3]{d} (f+g x)+\sqrt [3]{e}\right )}{\sqrt [3]{d} g}-\frac {\sqrt {3} \sqrt [3]{e} q \tan ^{-1}\left (\frac {\sqrt [3]{e}-2 \sqrt [3]{d} (f+g x)}{\sqrt {3} \sqrt [3]{e}}\right )}{\sqrt [3]{d} g} \]

[Out]

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

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Rubi [A]  time = 0.17, antiderivative size = 165, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.562, Rules used = {2483, 2448, 263, 200, 31, 634, 617, 204, 628} \[ \frac {(f+g x) \log \left (c \left (d+\frac {e}{(f+g x)^3}\right )^q\right )}{g}-\frac {\sqrt [3]{e} q \log \left (d^{2/3} (f+g x)^2-\sqrt [3]{d} \sqrt [3]{e} (f+g x)+e^{2/3}\right )}{2 \sqrt [3]{d} g}+\frac {\sqrt [3]{e} q \log \left (\sqrt [3]{d} (f+g x)+\sqrt [3]{e}\right )}{\sqrt [3]{d} g}-\frac {\sqrt {3} \sqrt [3]{e} q \tan ^{-1}\left (\frac {\sqrt [3]{e}-2 \sqrt [3]{d} (f+g x)}{\sqrt {3} \sqrt [3]{e}}\right )}{\sqrt [3]{d} g} \]

Antiderivative was successfully verified.

[In]

Int[Log[c*(d + e/(f + g*x)^3)^q],x]

[Out]

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

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 263

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[x^(m + n*p)*(b + a/x^n)^p, x] /; FreeQ[{a, b, m
, n}, x] && IntegerQ[p] && NegQ[n]

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 2448

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 2483

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

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Mathematica [C]  time = 0.33, size = 66, normalized size = 0.40 \[ \frac {(f+g x) \log \left (c \left (d+\frac {e}{(f+g x)^3}\right )^q\right )}{g}-\frac {3 e q \, _2F_1\left (\frac {2}{3},1;\frac {5}{3};-\frac {e}{d (f+g x)^3}\right )}{2 d g (f+g x)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*e*q*Hypergeometric2F1[2/3, 1, 5/3, -(e/(d*(f + g*x)^3))])/(2*d*g*(f + g*x)^2) + ((f + g*x)*Log[c*(d + e/(f
+ g*x)^3)^q])/g

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fricas [C]  time = 3.90, size = 1392, normalized size = 8.44 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(4*g*q*x*log((d*g^3*x^3 + 3*d*f*g^2*x^2 + 3*d*f^2*g*x + d*f^3 + e)/(g^3*x^3 + 3*f*g^2*x^2 + 3*f^2*g*x + f^
3)) - 4*sqrt(3)*g*sqrt((((-1/2*f^3*q^3/g^3 + 1/2*e*q^3/(d*g^3) + 1/2*(d*f^3*q^3 + e*q^3)/(d*g^3))^(1/3)*(I*sqr
t(3) + 1) - 2*f*q/g)^2*g^2 + 4*((-1/2*f^3*q^3/g^3 + 1/2*e*q^3/(d*g^3) + 1/2*(d*f^3*q^3 + e*q^3)/(d*g^3))^(1/3)
*(I*sqrt(3) + 1) - 2*f*q/g)*f*g*q + 4*f^2*q^2)/g^2)*arctan(-1/24*(2*sqrt(3)*sqrt(4*g^2*q^2*x^2 + 12*f*g*q^2*x
+ ((-1/2*f^3*q^3/g^3 + 1/2*e*q^3/(d*g^3) + 1/2*(d*f^3*q^3 + e*q^3)/(d*g^3))^(1/3)*(I*sqrt(3) + 1) - 2*f*q/g)^2
*g^2 + 12*f^2*q^2 + 2*(g^2*q*x + 3*f*g*q)*((-1/2*f^3*q^3/g^3 + 1/2*e*q^3/(d*g^3) + 1/2*(d*f^3*q^3 + e*q^3)/(d*
g^3))^(1/3)*(I*sqrt(3) + 1) - 2*f*q/g))*(((-1/2*f^3*q^3/g^3 + 1/2*e*q^3/(d*g^3) + 1/2*(d*f^3*q^3 + e*q^3)/(d*g
^3))^(1/3)*(I*sqrt(3) + 1) - 2*f*q/g)*d*g^2 + 2*d*f*g*q)*sqrt((((-1/2*f^3*q^3/g^3 + 1/2*e*q^3/(d*g^3) + 1/2*(d
*f^3*q^3 + e*q^3)/(d*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*e*q^3/(d*g^3) +
1/2*(d*f^3*q^3 + e*q^3)/(d*g^3))^(1/3)*(I*sqrt(3) + 1) - 2*f*q/g)*f*g*q + 4*f^2*q^2)/g^2) - sqrt(3)*(8*d*f*g^
2*q^2*x + ((-1/2*f^3*q^3/g^3 + 1/2*e*q^3/(d*g^3) + 1/2*(d*f^3*q^3 + e*q^3)/(d*g^3))^(1/3)*(I*sqrt(3) + 1) - 2*
f*q/g)^2*d*g^3 + 12*d*f^2*g*q^2 + 4*(d*g^3*q*x + 2*d*f*g^2*q)*((-1/2*f^3*q^3/g^3 + 1/2*e*q^3/(d*g^3) + 1/2*(d*
f^3*q^3 + e*q^3)/(d*g^3))^(1/3)*(I*sqrt(3) + 1) - 2*f*q/g))*sqrt((((-1/2*f^3*q^3/g^3 + 1/2*e*q^3/(d*g^3) + 1/2
*(d*f^3*q^3 + e*q^3)/(d*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*e*q^3/(d*g^3
) + 1/2*(d*f^3*q^3 + e*q^3)/(d*g^3))^(1/3)*(I*sqrt(3) + 1) - 2*f*q/g)*f*g*q + 4*f^2*q^2)/g^2))/(e*q^3)) - 12*f
*q*log(g*x + f) - 2*((-1/2*f^3*q^3/g^3 + 1/2*e*q^3/(d*g^3) + 1/2*(d*f^3*q^3 + e*q^3)/(d*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*e*q^3/(d*g^3) + 1/2*(d*f^3*q^3 + e*q^3)/(d*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*e*q^3/(d*g^3) + 1/2*(d*f^3*q^3 + e*q^3)/
(d*g^3))^(1/3)*(I*sqrt(3) + 1) - 2*f*q/g)*g + 6*f*q)*log(4*g^2*q^2*x^2 + 12*f*g*q^2*x + ((-1/2*f^3*q^3/g^3 + 1
/2*e*q^3/(d*g^3) + 1/2*(d*f^3*q^3 + e*q^3)/(d*g^3))^(1/3)*(I*sqrt(3) + 1) - 2*f*q/g)^2*g^2 + 12*f^2*q^2 + 2*(g
^2*q*x + 3*f*g*q)*((-1/2*f^3*q^3/g^3 + 1/2*e*q^3/(d*g^3) + 1/2*(d*f^3*q^3 + e*q^3)/(d*g^3))^(1/3)*(I*sqrt(3) +
1) - 2*f*q/g)))/g

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

[Out]

Timed out

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maple [C]  time = 0.48, size = 157, normalized size = 0.95 \[ -\frac {3 f q \ln \left (g x +f \right )}{g}+x \ln \left (c \left (\frac {d \,g^{3} x^{3}+3 d f \,g^{2} x^{2}+3 d \,f^{2} g x +d \,f^{3}+e}{\left (g x +f \right )^{3}}\right )^{q}\right )+\frac {q \left (\RootOf \left (d \,g^{3} \textit {\_Z}^{3}+3 d f \,g^{2} \textit {\_Z}^{2}+3 d \,f^{2} g \textit {\_Z} +d \,f^{3}+e \right )^{2} d f \,g^{2}+2 \RootOf \left (d \,g^{3} \textit {\_Z}^{3}+3 d f \,g^{2} \textit {\_Z}^{2}+3 d \,f^{2} g \textit {\_Z} +d \,f^{3}+e \right ) d \,f^{2} g +d \,f^{3}+e \right ) \ln \left (-\RootOf \left (d \,g^{3} \textit {\_Z}^{3}+3 d f \,g^{2} \textit {\_Z}^{2}+3 d \,f^{2} g \textit {\_Z} +d \,f^{3}+e \right )+x \right )}{d g \left (g^{2} \RootOf \left (d \,g^{3} \textit {\_Z}^{3}+3 d f \,g^{2} \textit {\_Z}^{2}+3 d \,f^{2} g \textit {\_Z} +d \,f^{3}+e \right )^{2}+2 f g \RootOf \left (d \,g^{3} \textit {\_Z}^{3}+3 d f \,g^{2} \textit {\_Z}^{2}+3 d \,f^{2} g \textit {\_Z} +d \,f^{3}+e \right )+f^{2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ 3 \, q \int \frac {d f g^{2} x^{2} + 2 \, d f^{2} g x + d f^{3} + e}{d g^{3} x^{3} + 3 \, d f g^{2} x^{2} + 3 \, d f^{2} g x + d f^{3} + e}\,{d x} - \frac {3 \, f q \log \left (g x + f\right ) - g x \log \left ({\left (d g^{3} x^{3} + 3 \, d f g^{2} x^{2} + 3 \, d f^{2} g x + d f^{3} + e\right )}^{q}\right ) + 3 \, g x \log \left ({\left (g x + f\right )}^{q}\right ) - g x \log \relax (c)}{g} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.63, size = 499, normalized size = 3.02 \[ x\,\ln \left (c\,{\left (d+\frac {e}{{\left (f+g\,x\right )}^3}\right )}^q\right )-\left (\sum _{k=1}^3\ln \left (-d^2\,e^2\,g^{11}\,\left (3\,e\,q^3\,x+\mathrm {root}\left (d\,g^3\,z^3+3\,d\,f\,g^2\,q\,z^2+3\,d\,f^2\,g\,q^2\,z+d\,f^3\,q^3+e\,q^3,z,k\right )\,e\,q^2+{\mathrm {root}\left (d\,g^3\,z^3+3\,d\,f\,g^2\,q\,z^2+3\,d\,f^2\,g\,q^2\,z+d\,f^3\,q^3+e\,q^3,z,k\right )}^3\,d\,f\,g^2\,4+\mathrm {root}\left (d\,g^3\,z^3+3\,d\,f\,g^2\,q\,z^2+3\,d\,f^2\,g\,q^2\,z+d\,f^3\,q^3+e\,q^3,z,k\right )\,d\,f^3\,q^2\,4+{\mathrm {root}\left (d\,g^3\,z^3+3\,d\,f\,g^2\,q\,z^2+3\,d\,f^2\,g\,q^2\,z+d\,f^3\,q^3+e\,q^3,z,k\right )}^3\,d\,g^3\,x\,4+{\mathrm {root}\left (d\,g^3\,z^3+3\,d\,f\,g^2\,q\,z^2+3\,d\,f^2\,g\,q^2\,z+d\,f^3\,q^3+e\,q^3,z,k\right )}^2\,d\,f^2\,g\,q\,8+\mathrm {root}\left (d\,g^3\,z^3+3\,d\,f\,g^2\,q\,z^2+3\,d\,f^2\,g\,q^2\,z+d\,f^3\,q^3+e\,q^3,z,k\right )\,d\,f^2\,g\,q^2\,x\,4+{\mathrm {root}\left (d\,g^3\,z^3+3\,d\,f\,g^2\,q\,z^2+3\,d\,f^2\,g\,q^2\,z+d\,f^3\,q^3+e\,q^3,z,k\right )}^2\,d\,f\,g^2\,q\,x\,8\right )\,9\right )\,\mathrm {root}\left (d\,g^3\,z^3+3\,d\,f\,g^2\,q\,z^2+3\,d\,f^2\,g\,q^2\,z+d\,f^3\,q^3+e\,q^3,z,k\right )\right )-\frac {3\,f\,q\,\ln \left (f+g\,x\right )}{g} \]

Verification of antiderivative is not currently implemented for this CAS.

[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^2*e^2*g^11*(3*e*q^3*x + root(d*g^3*z^3 + 3*d*f*g^2*q*z^2 + 3*
d*f^2*g*q^2*z + d*f^3*q^3 + e*q^3, z, k)*e*q^2 + 4*root(d*g^3*z^3 + 3*d*f*g^2*q*z^2 + 3*d*f^2*g*q^2*z + d*f^3*
q^3 + e*q^3, z, k)^3*d*f*g^2 + 4*root(d*g^3*z^3 + 3*d*f*g^2*q*z^2 + 3*d*f^2*g*q^2*z + d*f^3*q^3 + e*q^3, z, k)
*d*f^3*q^2 + 4*root(d*g^3*z^3 + 3*d*f*g^2*q*z^2 + 3*d*f^2*g*q^2*z + d*f^3*q^3 + e*q^3, z, k)^3*d*g^3*x + 8*roo
t(d*g^3*z^3 + 3*d*f*g^2*q*z^2 + 3*d*f^2*g*q^2*z + d*f^3*q^3 + e*q^3, z, k)^2*d*f^2*g*q + 4*root(d*g^3*z^3 + 3*
d*f*g^2*q*z^2 + 3*d*f^2*g*q^2*z + d*f^3*q^3 + e*q^3, z, k)*d*f^2*g*q^2*x + 8*root(d*g^3*z^3 + 3*d*f*g^2*q*z^2
+ 3*d*f^2*g*q^2*z + d*f^3*q^3 + e*q^3, z, k)^2*d*f*g^2*q*x))*root(d*g^3*z^3 + 3*d*f*g^2*q*z^2 + 3*d*f^2*g*q^2*
z + d*f^3*q^3 + e*q^3, z, k), k, 1, 3) - (3*f*q*log(f + g*x))/g

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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*(d+e/(g*x+f)**3)**q),x)

[Out]

Timed out

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