3.2.46 \(\int \frac {A+B \log (e (\frac {a+b x}{c+d x})^n)}{(c i+d i x)^2} \, dx\) [146]

Optimal. Leaf size=102 \[ \frac {A (a+b x)}{(b c-a d) i^2 (c+d x)}-\frac {B n (a+b x)}{(b c-a d) i^2 (c+d x)}+\frac {B (a+b x) \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{(b c-a d) i^2 (c+d x)} \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.061, Rules used = {2551, 2332} \begin {gather*} \frac {A (a+b x)}{i^2 (c+d x) (b c-a d)}+\frac {B (a+b x) \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{i^2 (c+d x) (b c-a d)}-\frac {B n (a+b x)}{i^2 (c+d x) (b c-a d)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(A + B*Log[e*((a + b*x)/(c + d*x))^n])/(c*i + d*i*x)^2,x]

[Out]

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

Rule 2332

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rule 2551

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

Rubi steps

\begin {align*} \int \frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{(146 c+146 d x)^2} \, dx &=-\frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{21316 d (c+d x)}+\frac {(B n) \int \frac {b c-a d}{146 (a+b x) (c+d x)^2} \, dx}{146 d}\\ &=-\frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{21316 d (c+d x)}+\frac {(B (b c-a d) n) \int \frac {1}{(a+b x) (c+d x)^2} \, dx}{21316 d}\\ &=-\frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{21316 d (c+d x)}+\frac {(B (b c-a d) n) \int \left (\frac {b^2}{(b c-a d)^2 (a+b x)}-\frac {d}{(b c-a d) (c+d x)^2}-\frac {b d}{(b c-a d)^2 (c+d x)}\right ) \, dx}{21316 d}\\ &=\frac {B n}{21316 d (c+d x)}+\frac {b B n \log (a+b x)}{21316 d (b c-a d)}-\frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{21316 d (c+d x)}-\frac {b B n \log (c+d x)}{21316 d (b c-a d)}\\ \end {align*}

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Mathematica [A]
time = 0.04, size = 114, normalized size = 1.12 \begin {gather*} -\frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{d i (c i+d i x)}+\frac {B (b c-a d) n \left (\frac {1}{(b c-a d) (c+d x)}+\frac {b \log (a+b x)}{(b c-a d)^2}-\frac {b \log (c+d x)}{(b c-a d)^2}\right )}{d i^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Log[e*((a + b*x)/(c + d*x))^n])/(c*i + d*i*x)^2,x]

[Out]

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

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Maple [F]
time = 0.10, size = 0, normalized size = 0.00 \[\int \frac {A +B \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right )}{\left (d i x +c i \right )^{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*ln(e*((b*x+a)/(d*x+c))^n))/(d*i*x+c*i)^2,x)

[Out]

int((A+B*ln(e*((b*x+a)/(d*x+c))^n))/(d*i*x+c*i)^2,x)

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Maxima [A]
time = 0.26, size = 112, normalized size = 1.10 \begin {gather*} -B n {\left (\frac {b \log \left (b x + a\right )}{b c d - a d^{2}} - \frac {b \log \left (d x + c\right )}{b c d - a d^{2}} + \frac {1}{d^{2} x + c d}\right )} + \frac {B \log \left ({\left (\frac {b x}{d x + c} + \frac {a}{d x + c}\right )}^{n} e\right )}{d^{2} x + c d} + \frac {A}{d^{2} x + c d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*log(e*((b*x+a)/(d*x+c))^n))/(d*i*x+c*i)^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.39, size = 87, normalized size = 0.85 \begin {gather*} \frac {{\left (A + B\right )} b c - {\left (A + B\right )} a d - {\left (B b c - B a d\right )} n - {\left (B b d n x + B a d n\right )} \log \left (\frac {b x + a}{d x + c}\right )}{b c^{2} d - a c d^{2} + {\left (b c d^{2} - a d^{3}\right )} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*log(e*((b*x+a)/(d*x+c))^n))/(d*i*x+c*i)^2,x, algorithm="fricas")

[Out]

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

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: NotImplementedError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*ln(e*((b*x+a)/(d*x+c))**n))/(d*i*x+c*i)**2,x)

[Out]

Exception raised: NotImplementedError >> no valid subset found

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Giac [A]
time = 3.47, size = 84, normalized size = 0.82 \begin {gather*} -{\left (\frac {{\left (b x + a\right )} B n \log \left (\frac {b x + a}{d x + c}\right )}{d x + c} - \frac {{\left (B n - A - B\right )} {\left (b x + a\right )}}{d x + c}\right )} {\left (\frac {b c}{{\left (b c - a d\right )}^{2}} - \frac {a d}{{\left (b c - a d\right )}^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*log(e*((b*x+a)/(d*x+c))^n))/(d*i*x+c*i)^2,x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 4.84, size = 113, normalized size = 1.11 \begin {gather*} -\frac {A-B\,n}{x\,d^2\,i^2+c\,d\,i^2}-\frac {B\,\ln \left (e\,{\left (\frac {a+b\,x}{c+d\,x}\right )}^n\right )}{d\,\left (c\,i^2+d\,i^2\,x\right )}+\frac {B\,b\,n\,\mathrm {atan}\left (\frac {b\,c\,2{}\mathrm {i}+b\,d\,x\,2{}\mathrm {i}}{a\,d-b\,c}+1{}\mathrm {i}\right )\,2{}\mathrm {i}}{d\,i^2\,\left (a\,d-b\,c\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*log(e*((a + b*x)/(c + d*x))^n))/(c*i + d*i*x)^2,x)

[Out]

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

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