3.457 \(\int \frac {1}{(a+b \log (c (d (e+f x)^p)^q))^3} \, dx\)

Optimal. Leaf size=169 \[ \frac {(e+f x) e^{-\frac {a}{b p q}} \left (c \left (d (e+f x)^p\right )^q\right )^{-\frac {1}{p q}} \text {Ei}\left (\frac {a+b \log \left (c \left (d (e+f x)^p\right )^q\right )}{b p q}\right )}{2 b^3 f p^3 q^3}-\frac {e+f x}{2 b^2 f p^2 q^2 \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}-\frac {e+f x}{2 b f p q \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )^2} \]

[Out]

1/2*(f*x+e)*Ei((a+b*ln(c*(d*(f*x+e)^p)^q))/b/p/q)/b^3/exp(a/b/p/q)/f/p^3/q^3/((c*(d*(f*x+e)^p)^q)^(1/p/q))+1/2
*(-f*x-e)/b/f/p/q/(a+b*ln(c*(d*(f*x+e)^p)^q))^2+1/2*(-f*x-e)/b^2/f/p^2/q^2/(a+b*ln(c*(d*(f*x+e)^p)^q))

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Rubi [A]  time = 0.21, antiderivative size = 169, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2389, 2297, 2300, 2178, 2445} \[ \frac {(e+f x) e^{-\frac {a}{b p q}} \left (c \left (d (e+f x)^p\right )^q\right )^{-\frac {1}{p q}} \text {Ei}\left (\frac {a+b \log \left (c \left (d (e+f x)^p\right )^q\right )}{b p q}\right )}{2 b^3 f p^3 q^3}-\frac {e+f x}{2 b^2 f p^2 q^2 \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}-\frac {e+f x}{2 b f p q \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Log[c*(d*(e + f*x)^p)^q])^(-3),x]

[Out]

((e + f*x)*ExpIntegralEi[(a + b*Log[c*(d*(e + f*x)^p)^q])/(b*p*q)])/(2*b^3*E^(a/(b*p*q))*f*p^3*q^3*(c*(d*(e +
f*x)^p)^q)^(1/(p*q))) - (e + f*x)/(2*b*f*p*q*(a + b*Log[c*(d*(e + f*x)^p)^q])^2) - (e + f*x)/(2*b^2*f*p^2*q^2*
(a + b*Log[c*(d*(e + f*x)^p)^q]))

Rule 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

Rule 2297

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_), x_Symbol] :> Simp[(x*(a + b*Log[c*x^n])^(p + 1))/(b*n*(p + 1))
, x] - Dist[1/(b*n*(p + 1)), Int[(a + b*Log[c*x^n])^(p + 1), x], x] /; FreeQ[{a, b, c, n}, x] && LtQ[p, -1] &&
IntegerQ[2*p]

Rule 2300

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_), x_Symbol] :> Dist[x/(n*(c*x^n)^(1/n)), Subst[Int[E^(x/n)*(a +
b*x)^p, x], x, Log[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, x]

Rule 2389

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.), x_Symbol] :> Dist[1/e, Subst[Int[(a + b*Log[c*
x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, n, p}, x]

Rule 2445

Int[((a_.) + Log[(c_.)*((d_.)*((e_.) + (f_.)*(x_))^(m_.))^(n_)]*(b_.))^(p_.)*(u_.), x_Symbol] :> Subst[Int[u*(
a + b*Log[c*d^n*(e + f*x)^(m*n)])^p, x], c*d^n*(e + f*x)^(m*n), c*(d*(e + f*x)^m)^n] /; FreeQ[{a, b, c, d, e,
f, m, n, p}, x] &&  !IntegerQ[n] &&  !(EqQ[d, 1] && EqQ[m, 1]) && IntegralFreeQ[IntHide[u*(a + b*Log[c*d^n*(e
+ f*x)^(m*n)])^p, x]]

Rubi steps

\begin {align*} \int \frac {1}{\left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )^3} \, dx &=\operatorname {Subst}\left (\int \frac {1}{\left (a+b \log \left (c d^q (e+f x)^{p q}\right )\right )^3} \, dx,c d^q (e+f x)^{p q},c \left (d (e+f x)^p\right )^q\right )\\ &=\operatorname {Subst}\left (\frac {\operatorname {Subst}\left (\int \frac {1}{\left (a+b \log \left (c d^q x^{p q}\right )\right )^3} \, dx,x,e+f x\right )}{f},c d^q (e+f x)^{p q},c \left (d (e+f x)^p\right )^q\right )\\ &=-\frac {e+f x}{2 b f p q \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )^2}+\operatorname {Subst}\left (\frac {\operatorname {Subst}\left (\int \frac {1}{\left (a+b \log \left (c d^q x^{p q}\right )\right )^2} \, dx,x,e+f x\right )}{2 b f p q},c d^q (e+f x)^{p q},c \left (d (e+f x)^p\right )^q\right )\\ &=-\frac {e+f x}{2 b f p q \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )^2}-\frac {e+f x}{2 b^2 f p^2 q^2 \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}+\operatorname {Subst}\left (\frac {\operatorname {Subst}\left (\int \frac {1}{a+b \log \left (c d^q x^{p q}\right )} \, dx,x,e+f x\right )}{2 b^2 f p^2 q^2},c d^q (e+f x)^{p q},c \left (d (e+f x)^p\right )^q\right )\\ &=-\frac {e+f x}{2 b f p q \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )^2}-\frac {e+f x}{2 b^2 f p^2 q^2 \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}+\operatorname {Subst}\left (\frac {\left ((e+f x) \left (c d^q (e+f x)^{p q}\right )^{-\frac {1}{p q}}\right ) \operatorname {Subst}\left (\int \frac {e^{\frac {x}{p q}}}{a+b x} \, dx,x,\log \left (c d^q (e+f x)^{p q}\right )\right )}{2 b^2 f p^3 q^3},c d^q (e+f x)^{p q},c \left (d (e+f x)^p\right )^q\right )\\ &=\frac {e^{-\frac {a}{b p q}} (e+f x) \left (c \left (d (e+f x)^p\right )^q\right )^{-\frac {1}{p q}} \text {Ei}\left (\frac {a+b \log \left (c \left (d (e+f x)^p\right )^q\right )}{b p q}\right )}{2 b^3 f p^3 q^3}-\frac {e+f x}{2 b f p q \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )^2}-\frac {e+f x}{2 b^2 f p^2 q^2 \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}\\ \end {align*}

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Mathematica [A]  time = 0.32, size = 189, normalized size = 1.12 \[ -\frac {(e+f x) e^{-\frac {a}{b p q}} \left (c \left (d (e+f x)^p\right )^q\right )^{-\frac {1}{p q}} \left (b p q e^{\frac {a}{b p q}} \left (c \left (d (e+f x)^p\right )^q\right )^{\frac {1}{p q}} \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )+b p q\right )-\left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )^2 \text {Ei}\left (\frac {a+b \log \left (c \left (d (e+f x)^p\right )^q\right )}{b p q}\right )\right )}{2 b^3 f p^3 q^3 \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Log[c*(d*(e + f*x)^p)^q])^(-3),x]

[Out]

-1/2*((e + f*x)*(-(ExpIntegralEi[(a + b*Log[c*(d*(e + f*x)^p)^q])/(b*p*q)]*(a + b*Log[c*(d*(e + f*x)^p)^q])^2)
+ b*E^(a/(b*p*q))*p*q*(c*(d*(e + f*x)^p)^q)^(1/(p*q))*(a + b*p*q + b*Log[c*(d*(e + f*x)^p)^q])))/(b^3*E^(a/(b
*p*q))*f*p^3*q^3*(c*(d*(e + f*x)^p)^q)^(1/(p*q))*(a + b*Log[c*(d*(e + f*x)^p)^q])^2)

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fricas [B]  time = 0.46, size = 444, normalized size = 2.63 \[ -\frac {{\left ({\left (b^{2} e p^{2} q^{2} + a b e p q + {\left (b^{2} f p^{2} q^{2} + a b f p q\right )} x + {\left (b^{2} f p^{2} q^{2} x + b^{2} e p^{2} q^{2}\right )} \log \left (f x + e\right ) + {\left (b^{2} f p q x + b^{2} e p q\right )} \log \relax (c) + {\left (b^{2} f p q^{2} x + b^{2} e p q^{2}\right )} \log \relax (d)\right )} e^{\left (\frac {b q \log \relax (d) + b \log \relax (c) + a}{b p q}\right )} - {\left (b^{2} p^{2} q^{2} \log \left (f x + e\right )^{2} + b^{2} q^{2} \log \relax (d)^{2} + b^{2} \log \relax (c)^{2} + 2 \, a b \log \relax (c) + a^{2} + 2 \, {\left (b^{2} p q^{2} \log \relax (d) + b^{2} p q \log \relax (c) + a b p q\right )} \log \left (f x + e\right ) + 2 \, {\left (b^{2} q \log \relax (c) + a b q\right )} \log \relax (d)\right )} \operatorname {log\_integral}\left ({\left (f x + e\right )} e^{\left (\frac {b q \log \relax (d) + b \log \relax (c) + a}{b p q}\right )}\right )\right )} e^{\left (-\frac {b q \log \relax (d) + b \log \relax (c) + a}{b p q}\right )}}{2 \, {\left (b^{5} f p^{5} q^{5} \log \left (f x + e\right )^{2} + b^{5} f p^{3} q^{5} \log \relax (d)^{2} + b^{5} f p^{3} q^{3} \log \relax (c)^{2} + 2 \, a b^{4} f p^{3} q^{3} \log \relax (c) + a^{2} b^{3} f p^{3} q^{3} + 2 \, {\left (b^{5} f p^{4} q^{5} \log \relax (d) + b^{5} f p^{4} q^{4} \log \relax (c) + a b^{4} f p^{4} q^{4}\right )} \log \left (f x + e\right ) + 2 \, {\left (b^{5} f p^{3} q^{4} \log \relax (c) + a b^{4} f p^{3} q^{4}\right )} \log \relax (d)\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*log(c*(d*(f*x+e)^p)^q))^3,x, algorithm="fricas")

[Out]

-1/2*((b^2*e*p^2*q^2 + a*b*e*p*q + (b^2*f*p^2*q^2 + a*b*f*p*q)*x + (b^2*f*p^2*q^2*x + b^2*e*p^2*q^2)*log(f*x +
e) + (b^2*f*p*q*x + b^2*e*p*q)*log(c) + (b^2*f*p*q^2*x + b^2*e*p*q^2)*log(d))*e^((b*q*log(d) + b*log(c) + a)/
(b*p*q)) - (b^2*p^2*q^2*log(f*x + e)^2 + b^2*q^2*log(d)^2 + b^2*log(c)^2 + 2*a*b*log(c) + a^2 + 2*(b^2*p*q^2*l
og(d) + b^2*p*q*log(c) + a*b*p*q)*log(f*x + e) + 2*(b^2*q*log(c) + a*b*q)*log(d))*log_integral((f*x + e)*e^((b
*q*log(d) + b*log(c) + a)/(b*p*q))))*e^(-(b*q*log(d) + b*log(c) + a)/(b*p*q))/(b^5*f*p^5*q^5*log(f*x + e)^2 +
b^5*f*p^3*q^5*log(d)^2 + b^5*f*p^3*q^3*log(c)^2 + 2*a*b^4*f*p^3*q^3*log(c) + a^2*b^3*f*p^3*q^3 + 2*(b^5*f*p^4*
q^5*log(d) + b^5*f*p^4*q^4*log(c) + a*b^4*f*p^4*q^4)*log(f*x + e) + 2*(b^5*f*p^3*q^4*log(c) + a*b^4*f*p^3*q^4)
*log(d))

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giac [B]  time = 0.43, size = 3481, normalized size = 20.60 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*log(c*(d*(f*x+e)^p)^q))^3,x, algorithm="giac")

[Out]

-1/2*(f*x + e)*b^2*p^2*q^2*log(f*x + e)/(b^5*f*p^5*q^5*log(f*x + e)^2 + 2*b^5*f*p^4*q^5*log(f*x + e)*log(d) +
2*b^5*f*p^4*q^4*log(f*x + e)*log(c) + b^5*f*p^3*q^5*log(d)^2 + 2*a*b^4*f*p^4*q^4*log(f*x + e) + 2*b^5*f*p^3*q^
4*log(c)*log(d) + b^5*f*p^3*q^3*log(c)^2 + 2*a*b^4*f*p^3*q^4*log(d) + 2*a*b^4*f*p^3*q^3*log(c) + a^2*b^3*f*p^3
*q^3) + 1/2*b^2*p^2*q^2*Ei(log(d)/p + log(c)/(p*q) + a/(b*p*q) + log(f*x + e))*e^(-a/(b*p*q))*log(f*x + e)^2/(
(b^5*f*p^5*q^5*log(f*x + e)^2 + 2*b^5*f*p^4*q^5*log(f*x + e)*log(d) + 2*b^5*f*p^4*q^4*log(f*x + e)*log(c) + b^
5*f*p^3*q^5*log(d)^2 + 2*a*b^4*f*p^4*q^4*log(f*x + e) + 2*b^5*f*p^3*q^4*log(c)*log(d) + b^5*f*p^3*q^3*log(c)^2
+ 2*a*b^4*f*p^3*q^4*log(d) + 2*a*b^4*f*p^3*q^3*log(c) + a^2*b^3*f*p^3*q^3)*c^(1/(p*q))*d^(1/p)) - 1/2*(f*x +
e)*b^2*p^2*q^2/(b^5*f*p^5*q^5*log(f*x + e)^2 + 2*b^5*f*p^4*q^5*log(f*x + e)*log(d) + 2*b^5*f*p^4*q^4*log(f*x +
e)*log(c) + b^5*f*p^3*q^5*log(d)^2 + 2*a*b^4*f*p^4*q^4*log(f*x + e) + 2*b^5*f*p^3*q^4*log(c)*log(d) + b^5*f*p
^3*q^3*log(c)^2 + 2*a*b^4*f*p^3*q^4*log(d) + 2*a*b^4*f*p^3*q^3*log(c) + a^2*b^3*f*p^3*q^3) - 1/2*(f*x + e)*b^2
*p*q^2*log(d)/(b^5*f*p^5*q^5*log(f*x + e)^2 + 2*b^5*f*p^4*q^5*log(f*x + e)*log(d) + 2*b^5*f*p^4*q^4*log(f*x +
e)*log(c) + b^5*f*p^3*q^5*log(d)^2 + 2*a*b^4*f*p^4*q^4*log(f*x + e) + 2*b^5*f*p^3*q^4*log(c)*log(d) + b^5*f*p^
3*q^3*log(c)^2 + 2*a*b^4*f*p^3*q^4*log(d) + 2*a*b^4*f*p^3*q^3*log(c) + a^2*b^3*f*p^3*q^3) + b^2*p*q^2*Ei(log(d
)/p + log(c)/(p*q) + a/(b*p*q) + log(f*x + e))*e^(-a/(b*p*q))*log(f*x + e)*log(d)/((b^5*f*p^5*q^5*log(f*x + e)
^2 + 2*b^5*f*p^4*q^5*log(f*x + e)*log(d) + 2*b^5*f*p^4*q^4*log(f*x + e)*log(c) + b^5*f*p^3*q^5*log(d)^2 + 2*a*
b^4*f*p^4*q^4*log(f*x + e) + 2*b^5*f*p^3*q^4*log(c)*log(d) + b^5*f*p^3*q^3*log(c)^2 + 2*a*b^4*f*p^3*q^4*log(d)
+ 2*a*b^4*f*p^3*q^3*log(c) + a^2*b^3*f*p^3*q^3)*c^(1/(p*q))*d^(1/p)) - 1/2*(f*x + e)*b^2*p*q*log(c)/(b^5*f*p^
5*q^5*log(f*x + e)^2 + 2*b^5*f*p^4*q^5*log(f*x + e)*log(d) + 2*b^5*f*p^4*q^4*log(f*x + e)*log(c) + b^5*f*p^3*q
^5*log(d)^2 + 2*a*b^4*f*p^4*q^4*log(f*x + e) + 2*b^5*f*p^3*q^4*log(c)*log(d) + b^5*f*p^3*q^3*log(c)^2 + 2*a*b^
4*f*p^3*q^4*log(d) + 2*a*b^4*f*p^3*q^3*log(c) + a^2*b^3*f*p^3*q^3) + b^2*p*q*Ei(log(d)/p + log(c)/(p*q) + a/(b
*p*q) + log(f*x + e))*e^(-a/(b*p*q))*log(f*x + e)*log(c)/((b^5*f*p^5*q^5*log(f*x + e)^2 + 2*b^5*f*p^4*q^5*log(
f*x + e)*log(d) + 2*b^5*f*p^4*q^4*log(f*x + e)*log(c) + b^5*f*p^3*q^5*log(d)^2 + 2*a*b^4*f*p^4*q^4*log(f*x + e
) + 2*b^5*f*p^3*q^4*log(c)*log(d) + b^5*f*p^3*q^3*log(c)^2 + 2*a*b^4*f*p^3*q^4*log(d) + 2*a*b^4*f*p^3*q^3*log(
c) + a^2*b^3*f*p^3*q^3)*c^(1/(p*q))*d^(1/p)) + 1/2*b^2*q^2*Ei(log(d)/p + log(c)/(p*q) + a/(b*p*q) + log(f*x +
e))*e^(-a/(b*p*q))*log(d)^2/((b^5*f*p^5*q^5*log(f*x + e)^2 + 2*b^5*f*p^4*q^5*log(f*x + e)*log(d) + 2*b^5*f*p^4
*q^4*log(f*x + e)*log(c) + b^5*f*p^3*q^5*log(d)^2 + 2*a*b^4*f*p^4*q^4*log(f*x + e) + 2*b^5*f*p^3*q^4*log(c)*lo
g(d) + b^5*f*p^3*q^3*log(c)^2 + 2*a*b^4*f*p^3*q^4*log(d) + 2*a*b^4*f*p^3*q^3*log(c) + a^2*b^3*f*p^3*q^3)*c^(1/
(p*q))*d^(1/p)) - 1/2*(f*x + e)*a*b*p*q/(b^5*f*p^5*q^5*log(f*x + e)^2 + 2*b^5*f*p^4*q^5*log(f*x + e)*log(d) +
2*b^5*f*p^4*q^4*log(f*x + e)*log(c) + b^5*f*p^3*q^5*log(d)^2 + 2*a*b^4*f*p^4*q^4*log(f*x + e) + 2*b^5*f*p^3*q^
4*log(c)*log(d) + b^5*f*p^3*q^3*log(c)^2 + 2*a*b^4*f*p^3*q^4*log(d) + 2*a*b^4*f*p^3*q^3*log(c) + a^2*b^3*f*p^3
*q^3) + a*b*p*q*Ei(log(d)/p + log(c)/(p*q) + a/(b*p*q) + log(f*x + e))*e^(-a/(b*p*q))*log(f*x + e)/((b^5*f*p^5
*q^5*log(f*x + e)^2 + 2*b^5*f*p^4*q^5*log(f*x + e)*log(d) + 2*b^5*f*p^4*q^4*log(f*x + e)*log(c) + b^5*f*p^3*q^
5*log(d)^2 + 2*a*b^4*f*p^4*q^4*log(f*x + e) + 2*b^5*f*p^3*q^4*log(c)*log(d) + b^5*f*p^3*q^3*log(c)^2 + 2*a*b^4
*f*p^3*q^4*log(d) + 2*a*b^4*f*p^3*q^3*log(c) + a^2*b^3*f*p^3*q^3)*c^(1/(p*q))*d^(1/p)) + b^2*q*Ei(log(d)/p + l
og(c)/(p*q) + a/(b*p*q) + log(f*x + e))*e^(-a/(b*p*q))*log(c)*log(d)/((b^5*f*p^5*q^5*log(f*x + e)^2 + 2*b^5*f*
p^4*q^5*log(f*x + e)*log(d) + 2*b^5*f*p^4*q^4*log(f*x + e)*log(c) + b^5*f*p^3*q^5*log(d)^2 + 2*a*b^4*f*p^4*q^4
*log(f*x + e) + 2*b^5*f*p^3*q^4*log(c)*log(d) + b^5*f*p^3*q^3*log(c)^2 + 2*a*b^4*f*p^3*q^4*log(d) + 2*a*b^4*f*
p^3*q^3*log(c) + a^2*b^3*f*p^3*q^3)*c^(1/(p*q))*d^(1/p)) + 1/2*b^2*Ei(log(d)/p + log(c)/(p*q) + a/(b*p*q) + lo
g(f*x + e))*e^(-a/(b*p*q))*log(c)^2/((b^5*f*p^5*q^5*log(f*x + e)^2 + 2*b^5*f*p^4*q^5*log(f*x + e)*log(d) + 2*b
^5*f*p^4*q^4*log(f*x + e)*log(c) + b^5*f*p^3*q^5*log(d)^2 + 2*a*b^4*f*p^4*q^4*log(f*x + e) + 2*b^5*f*p^3*q^4*l
og(c)*log(d) + b^5*f*p^3*q^3*log(c)^2 + 2*a*b^4*f*p^3*q^4*log(d) + 2*a*b^4*f*p^3*q^3*log(c) + a^2*b^3*f*p^3*q^
3)*c^(1/(p*q))*d^(1/p)) + a*b*q*Ei(log(d)/p + log(c)/(p*q) + a/(b*p*q) + log(f*x + e))*e^(-a/(b*p*q))*log(d)/(
(b^5*f*p^5*q^5*log(f*x + e)^2 + 2*b^5*f*p^4*q^5*log(f*x + e)*log(d) + 2*b^5*f*p^4*q^4*log(f*x + e)*log(c) + b^
5*f*p^3*q^5*log(d)^2 + 2*a*b^4*f*p^4*q^4*log(f*x + e) + 2*b^5*f*p^3*q^4*log(c)*log(d) + b^5*f*p^3*q^3*log(c)^2
+ 2*a*b^4*f*p^3*q^4*log(d) + 2*a*b^4*f*p^3*q^3*log(c) + a^2*b^3*f*p^3*q^3)*c^(1/(p*q))*d^(1/p)) + a*b*Ei(log(
d)/p + log(c)/(p*q) + a/(b*p*q) + log(f*x + e))*e^(-a/(b*p*q))*log(c)/((b^5*f*p^5*q^5*log(f*x + e)^2 + 2*b^5*f
*p^4*q^5*log(f*x + e)*log(d) + 2*b^5*f*p^4*q^4*log(f*x + e)*log(c) + b^5*f*p^3*q^5*log(d)^2 + 2*a*b^4*f*p^4*q^
4*log(f*x + e) + 2*b^5*f*p^3*q^4*log(c)*log(d) + b^5*f*p^3*q^3*log(c)^2 + 2*a*b^4*f*p^3*q^4*log(d) + 2*a*b^4*f
*p^3*q^3*log(c) + a^2*b^3*f*p^3*q^3)*c^(1/(p*q))*d^(1/p)) + 1/2*a^2*Ei(log(d)/p + log(c)/(p*q) + a/(b*p*q) + l
og(f*x + e))*e^(-a/(b*p*q))/((b^5*f*p^5*q^5*log(f*x + e)^2 + 2*b^5*f*p^4*q^5*log(f*x + e)*log(d) + 2*b^5*f*p^4
*q^4*log(f*x + e)*log(c) + b^5*f*p^3*q^5*log(d)^2 + 2*a*b^4*f*p^4*q^4*log(f*x + e) + 2*b^5*f*p^3*q^4*log(c)*lo
g(d) + b^5*f*p^3*q^3*log(c)^2 + 2*a*b^4*f*p^3*q^4*log(d) + 2*a*b^4*f*p^3*q^3*log(c) + a^2*b^3*f*p^3*q^3)*c^(1/
(p*q))*d^(1/p))

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maple [F]  time = 0.08, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (b \ln \left (c \left (d \left (f x +e \right )^{p}\right )^{q}\right )+a \right )^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*ln(c*(d*(f*x+e)^p)^q)+a)^3,x)

[Out]

int(1/(b*ln(c*(d*(f*x+e)^p)^q)+a)^3,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {{\left (e p q + e q \log \relax (d) + e \log \relax (c)\right )} b + a e + {\left ({\left (f p q + f q \log \relax (d) + f \log \relax (c)\right )} b + a f\right )} x + {\left (b f x + b e\right )} \log \left ({\left ({\left (f x + e\right )}^{p}\right )}^{q}\right )}{2 \, {\left (b^{4} f p^{2} q^{2} \log \left ({\left ({\left (f x + e\right )}^{p}\right )}^{q}\right )^{2} + a^{2} b^{2} f p^{2} q^{2} + 2 \, {\left (f p^{2} q^{3} \log \relax (d) + f p^{2} q^{2} \log \relax (c)\right )} a b^{3} + {\left (f p^{2} q^{4} \log \relax (d)^{2} + 2 \, f p^{2} q^{3} \log \relax (c) \log \relax (d) + f p^{2} q^{2} \log \relax (c)^{2}\right )} b^{4} + 2 \, {\left (a b^{3} f p^{2} q^{2} + {\left (f p^{2} q^{3} \log \relax (d) + f p^{2} q^{2} \log \relax (c)\right )} b^{4}\right )} \log \left ({\left ({\left (f x + e\right )}^{p}\right )}^{q}\right )\right )}} + \int \frac {1}{2 \, {\left (b^{3} p^{2} q^{2} \log \left ({\left ({\left (f x + e\right )}^{p}\right )}^{q}\right ) + a b^{2} p^{2} q^{2} + {\left (p^{2} q^{3} \log \relax (d) + p^{2} q^{2} \log \relax (c)\right )} b^{3}\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*log(c*(d*(f*x+e)^p)^q))^3,x, algorithm="maxima")

[Out]

-1/2*((e*p*q + e*q*log(d) + e*log(c))*b + a*e + ((f*p*q + f*q*log(d) + f*log(c))*b + a*f)*x + (b*f*x + b*e)*lo
g(((f*x + e)^p)^q))/(b^4*f*p^2*q^2*log(((f*x + e)^p)^q)^2 + a^2*b^2*f*p^2*q^2 + 2*(f*p^2*q^3*log(d) + f*p^2*q^
2*log(c))*a*b^3 + (f*p^2*q^4*log(d)^2 + 2*f*p^2*q^3*log(c)*log(d) + f*p^2*q^2*log(c)^2)*b^4 + 2*(a*b^3*f*p^2*q
^2 + (f*p^2*q^3*log(d) + f*p^2*q^2*log(c))*b^4)*log(((f*x + e)^p)^q)) + integrate(1/2/(b^3*p^2*q^2*log(((f*x +
e)^p)^q) + a*b^2*p^2*q^2 + (p^2*q^3*log(d) + p^2*q^2*log(c))*b^3), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\left (a+b\,\ln \left (c\,{\left (d\,{\left (e+f\,x\right )}^p\right )}^q\right )\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*log(c*(d*(e + f*x)^p)^q))^3,x)

[Out]

int(1/(a + b*log(c*(d*(e + f*x)^p)^q))^3, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a + b \log {\left (c \left (d \left (e + f x\right )^{p}\right )^{q} \right )}\right )^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*ln(c*(d*(f*x+e)**p)**q))**3,x)

[Out]

Integral((a + b*log(c*(d*(e + f*x)**p)**q))**(-3), x)

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