3.3.33 \(\int \frac {1}{(a+b x) (c+d x) (A+B \log (e (a+b x)^n (c+d x)^{-n}))^3} \, dx\) [233]

Optimal. Leaf size=45 \[ -\frac {1}{2 B (b c-a d) n \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^2} \]

[Out]

-1/2/B/(-a*d+b*c)/n/(A+B*ln(e*(b*x+a)^n/((d*x+c)^n)))^2

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Rubi [A]
time = 0.15, antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {2573, 2561, 2339, 30} \begin {gather*} -\frac {1}{2 B n (b c-a d) \left (B \log \left (e (a+b x)^n (c+d x)^{-n}\right )+A\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((a + b*x)*(c + d*x)*(A + B*Log[(e*(a + b*x)^n)/(c + d*x)^n])^3),x]

[Out]

-1/2*1/(B*(b*c - a*d)*n*(A + B*Log[(e*(a + b*x)^n)/(c + d*x)^n])^2)

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2339

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

Rule 2561

Int[((A_.) + Log[(e_.)*(((a_.) + (b_.)*(x_))/((c_.) + (d_.)*(x_)))^(n_.)]*(B_.))^(p_.)*((f_.) + (g_.)*(x_))^(m
_.)*((h_.) + (i_.)*(x_))^(q_.), x_Symbol] :> Dist[(b*c - a*d)^(m + q + 1)*(g/b)^m*(i/d)^q, Subst[Int[x^m*((A +
B*Log[e*x^n])^p/(b - d*x)^(m + q + 2)), x], x, (a + b*x)/(c + d*x)], x] /; FreeQ[{a, b, c, d, e, f, g, h, i,
A, B, n, p}, x] && NeQ[b*c - a*d, 0] && EqQ[b*f - a*g, 0] && EqQ[d*h - c*i, 0] && IntegersQ[m, q]

Rule 2573

Int[((A_.) + Log[(e_.)*(u_)^(n_.)*(v_)^(mn_)]*(B_.))^(p_.)*(w_.), x_Symbol] :> Subst[Int[w*(A + B*Log[e*(u/v)^
n])^p, x], e*(u/v)^n, e*(u^n/v^n)] /; FreeQ[{e, A, B, n, p}, x] && EqQ[n + mn, 0] && LinearQ[{u, v}, x] &&  !I
ntegerQ[n]

Rubi steps

\begin {align*} \int \frac {1}{(a+b x) (c+d x) \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^3} \, dx &=-\frac {1}{2 B (b c-a d) n \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^2}\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 43, normalized size = 0.96 \begin {gather*} -\frac {1}{2 (b B c n-a B d n) \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b*x)*(c + d*x)*(A + B*Log[(e*(a + b*x)^n)/(c + d*x)^n])^3),x]

[Out]

-1/2*1/((b*B*c*n - a*B*d*n)*(A + B*Log[(e*(a + b*x)^n)/(c + d*x)^n])^2)

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.34, size = 366, normalized size = 8.13

method	result	size
risch	\(\frac {2}{B n \left (a d -c b \right ) \left (2 A +2 B \ln \left (e \right )+2 B \ln \left (\left (b x +a \right )^{n}\right )-2 B \ln \left (\left (d x +c \right )^{n}\right )-i B \pi \,\mathrm {csgn}\left (i \left (b x +a \right )^{n}\right ) \mathrm {csgn}\left (i \left (d x +c \right )^{-n}\right ) \mathrm {csgn}\left (i \left (b x +a \right )^{n} \left (d x +c \right )^{-n}\right )+i B \pi \,\mathrm {csgn}\left (i \left (b x +a \right )^{n}\right ) \mathrm {csgn}\left (i \left (b x +a \right )^{n} \left (d x +c \right )^{-n}\right )^{2}+i B \pi \,\mathrm {csgn}\left (i \left (d x +c \right )^{-n}\right ) \mathrm {csgn}\left (i \left (b x +a \right )^{n} \left (d x +c \right )^{-n}\right )^{2}-i B \pi \mathrm {csgn}\left (i \left (b x +a \right )^{n} \left (d x +c \right )^{-n}\right )^{3}-i B \pi \,\mathrm {csgn}\left (i e \right ) \mathrm {csgn}\left (i \left (b x +a \right )^{n} \left (d x +c \right )^{-n}\right ) \mathrm {csgn}\left (i e \left (d x +c \right )^{-n} \left (b x +a \right )^{n}\right )+i B \pi \,\mathrm {csgn}\left (i e \right ) \mathrm {csgn}\left (i e \left (d x +c \right )^{-n} \left (b x +a \right )^{n}\right )^{2}+i B \pi \,\mathrm {csgn}\left (i \left (b x +a \right )^{n} \left (d x +c \right )^{-n}\right ) \mathrm {csgn}\left (i e \left (d x +c \right )^{-n} \left (b x +a \right )^{n}\right )^{2}-i B \pi \mathrm {csgn}\left (i e \left (d x +c \right )^{-n} \left (b x +a \right )^{n}\right )^{3}\right )^{2}}\)	\(366\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*x+a)/(d*x+c)/(A+B*ln(e*(b*x+a)^n/((d*x+c)^n)))^3,x,method=_RETURNVERBOSE)

[Out]

2/B/n/(a*d-b*c)/(2*A+2*B*ln(e)+2*B*ln((b*x+a)^n)-2*B*ln((d*x+c)^n)-I*B*Pi*csgn(I*(b*x+a)^n)*csgn(I/((d*x+c)^n)
)*csgn(I*(b*x+a)^n/((d*x+c)^n))+I*B*Pi*csgn(I*(b*x+a)^n)*csgn(I*(b*x+a)^n/((d*x+c)^n))^2+I*B*Pi*csgn(I/((d*x+c
)^n))*csgn(I*(b*x+a)^n/((d*x+c)^n))^2-I*B*Pi*csgn(I*(b*x+a)^n/((d*x+c)^n))^3-I*B*Pi*csgn(I*e)*csgn(I*(b*x+a)^n
/((d*x+c)^n))*csgn(I*e/((d*x+c)^n)*(b*x+a)^n)+I*B*Pi*csgn(I*e)*csgn(I*e/((d*x+c)^n)*(b*x+a)^n)^2+I*B*Pi*csgn(I
*(b*x+a)^n/((d*x+c)^n))*csgn(I*e/((d*x+c)^n)*(b*x+a)^n)^2-I*B*Pi*csgn(I*e/((d*x+c)^n)*(b*x+a)^n)^3)^2

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 200 vs. \(2 (44) = 88\).
time = 0.43, size = 200, normalized size = 4.44 \begin {gather*} -\frac {1}{2 \, {\left ({\left (b c n - a d n\right )} B^{3} \log \left ({\left (b x + a\right )}^{n}\right )^{2} + {\left (b c n - a d n\right )} B^{3} \log \left ({\left (d x + c\right )}^{n}\right )^{2} + {\left (b c n - a d n\right )} A^{2} B + 2 \, {\left (b c n - a d n\right )} A B^{2} + {\left (b c n - a d n\right )} B^{3} + 2 \, {\left ({\left (b c n - a d n\right )} A B^{2} + {\left (b c n - a d n\right )} B^{3}\right )} \log \left ({\left (b x + a\right )}^{n}\right ) - 2 \, {\left ({\left (b c n - a d n\right )} B^{3} \log \left ({\left (b x + a\right )}^{n}\right ) + {\left (b c n - a d n\right )} A B^{2} + {\left (b c n - a d n\right )} B^{3}\right )} \log \left ({\left (d x + c\right )}^{n}\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)/(d*x+c)/(A+B*log(e*(b*x+a)^n/((d*x+c)^n)))^3,x, algorithm="maxima")

[Out]

-1/2/((b*c*n - a*d*n)*B^3*log((b*x + a)^n)^2 + (b*c*n - a*d*n)*B^3*log((d*x + c)^n)^2 + (b*c*n - a*d*n)*A^2*B
+ 2*(b*c*n - a*d*n)*A*B^2 + (b*c*n - a*d*n)*B^3 + 2*((b*c*n - a*d*n)*A*B^2 + (b*c*n - a*d*n)*B^3)*log((b*x + a
)^n) - 2*((b*c*n - a*d*n)*B^3*log((b*x + a)^n) + (b*c*n - a*d*n)*A*B^2 + (b*c*n - a*d*n)*B^3)*log((d*x + c)^n)
)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 197 vs. \(2 (44) = 88\).
time = 0.34, size = 197, normalized size = 4.38 \begin {gather*} -\frac {1}{2 \, {\left ({\left (B^{3} b c - B^{3} a d\right )} n^{3} \log \left (b x + a\right )^{2} + {\left (B^{3} b c - B^{3} a d\right )} n^{3} \log \left (d x + c\right )^{2} + 2 \, {\left ({\left (A B^{2} + B^{3}\right )} b c - {\left (A B^{2} + B^{3}\right )} a d\right )} n^{2} \log \left (b x + a\right ) + {\left ({\left (A^{2} B + 2 \, A B^{2} + B^{3}\right )} b c - {\left (A^{2} B + 2 \, A B^{2} + B^{3}\right )} a d\right )} n - 2 \, {\left ({\left (B^{3} b c - B^{3} a d\right )} n^{3} \log \left (b x + a\right ) + {\left ({\left (A B^{2} + B^{3}\right )} b c - {\left (A B^{2} + B^{3}\right )} a d\right )} n^{2}\right )} \log \left (d x + c\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)/(d*x+c)/(A+B*log(e*(b*x+a)^n/((d*x+c)^n)))^3,x, algorithm="fricas")

[Out]

-1/2/((B^3*b*c - B^3*a*d)*n^3*log(b*x + a)^2 + (B^3*b*c - B^3*a*d)*n^3*log(d*x + c)^2 + 2*((A*B^2 + B^3)*b*c -
(A*B^2 + B^3)*a*d)*n^2*log(b*x + a) + ((A^2*B + 2*A*B^2 + B^3)*b*c - (A^2*B + 2*A*B^2 + B^3)*a*d)*n - 2*((B^3
*b*c - B^3*a*d)*n^3*log(b*x + a) + ((A*B^2 + B^3)*b*c - (A*B^2 + B^3)*a*d)*n^2)*log(d*x + c))

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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(1/(b*x+a)/(d*x+c)/(A+B*ln(e*(b*x+a)**n/((d*x+c)**n)))**3,x)

[Out]

Timed out

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 301 vs. \(2 (44) = 88\).
time = 3.70, size = 301, normalized size = 6.69 \begin {gather*} -\frac {1}{2 \, {\left (B^{3} b c n^{3} \log \left (b x + a\right )^{2} - B^{3} a d n^{3} \log \left (b x + a\right )^{2} - 2 \, B^{3} b c n^{3} \log \left (b x + a\right ) \log \left (d x + c\right ) + 2 \, B^{3} a d n^{3} \log \left (b x + a\right ) \log \left (d x + c\right ) + B^{3} b c n^{3} \log \left (d x + c\right )^{2} - B^{3} a d n^{3} \log \left (d x + c\right )^{2} + 2 \, A B^{2} b c n^{2} \log \left (b x + a\right ) + 2 \, B^{3} b c n^{2} \log \left (b x + a\right ) - 2 \, A B^{2} a d n^{2} \log \left (b x + a\right ) - 2 \, B^{3} a d n^{2} \log \left (b x + a\right ) - 2 \, A B^{2} b c n^{2} \log \left (d x + c\right ) - 2 \, B^{3} b c n^{2} \log \left (d x + c\right ) + 2 \, A B^{2} a d n^{2} \log \left (d x + c\right ) + 2 \, B^{3} a d n^{2} \log \left (d x + c\right ) + A^{2} B b c n + 2 \, A B^{2} b c n + B^{3} b c n - A^{2} B a d n - 2 \, A B^{2} a d n - B^{3} a d n\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)/(d*x+c)/(A+B*log(e*(b*x+a)^n/((d*x+c)^n)))^3,x, algorithm="giac")

[Out]

-1/2/(B^3*b*c*n^3*log(b*x + a)^2 - B^3*a*d*n^3*log(b*x + a)^2 - 2*B^3*b*c*n^3*log(b*x + a)*log(d*x + c) + 2*B^
3*a*d*n^3*log(b*x + a)*log(d*x + c) + B^3*b*c*n^3*log(d*x + c)^2 - B^3*a*d*n^3*log(d*x + c)^2 + 2*A*B^2*b*c*n^
2*log(b*x + a) + 2*B^3*b*c*n^2*log(b*x + a) - 2*A*B^2*a*d*n^2*log(b*x + a) - 2*B^3*a*d*n^2*log(b*x + a) - 2*A*
B^2*b*c*n^2*log(d*x + c) - 2*B^3*b*c*n^2*log(d*x + c) + 2*A*B^2*a*d*n^2*log(d*x + c) + 2*B^3*a*d*n^2*log(d*x +
c) + A^2*B*b*c*n + 2*A*B^2*b*c*n + B^3*b*c*n - A^2*B*a*d*n - 2*A*B^2*a*d*n - B^3*a*d*n)

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Mupad [B]
time = 4.54, size = 72, normalized size = 1.60 \begin {gather*} \frac {1}{2\,B\,n\,\left (a\,d-b\,c\right )\,\left (A^2+2\,A\,B\,\ln \left (\frac {e\,{\left (a+b\,x\right )}^n}{{\left (c+d\,x\right )}^n}\right )+B^2\,{\ln \left (\frac {e\,{\left (a+b\,x\right )}^n}{{\left (c+d\,x\right )}^n}\right )}^2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((A + B*log((e*(a + b*x)^n)/(c + d*x)^n))^3*(a + b*x)*(c + d*x)),x)

[Out]

1/(2*B*n*(a*d - b*c)*(B^2*log((e*(a + b*x)^n)/(c + d*x)^n)^2 + A^2 + 2*A*B*log((e*(a + b*x)^n)/(c + d*x)^n)))

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