3.1.6 \(\int \frac {(c i+d i x) (A+B \log (\frac {e (a+b x)}{c+d x}))}{(a g+b g x)^2} \, dx\) [6]

Optimal. Leaf size=142 \[ -\frac {B i (c+d x)}{b g^2 (a+b x)}-\frac {i (c+d x) \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{b g^2 (a+b x)}-\frac {d i \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right ) \log \left (1-\frac {b (c+d x)}{d (a+b x)}\right )}{b^2 g^2}+\frac {B d i \text {Li}_2\left (\frac {b (c+d x)}{d (a+b x)}\right )}{b^2 g^2} \]

[Out]

-B*i*(d*x+c)/b/g^2/(b*x+a)-i*(d*x+c)*(A+B*ln(e*(b*x+a)/(d*x+c)))/b/g^2/(b*x+a)-d*i*(A+B*ln(e*(b*x+a)/(d*x+c)))
*ln(1-b*(d*x+c)/d/(b*x+a))/b^2/g^2+B*d*i*polylog(2,b*(d*x+c)/d/(b*x+a))/b^2/g^2

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Rubi [A]
time = 0.13, antiderivative size = 142, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.132, Rules used = {2562, 2380, 2341, 2379, 2438} \begin {gather*} \frac {B d i \text {PolyLog}\left (2,\frac {b (c+d x)}{d (a+b x)}\right )}{b^2 g^2}-\frac {d i \log \left (1-\frac {b (c+d x)}{d (a+b x)}\right ) \left (B \log \left (\frac {e (a+b x)}{c+d x}\right )+A\right )}{b^2 g^2}-\frac {i (c+d x) \left (B \log \left (\frac {e (a+b x)}{c+d x}\right )+A\right )}{b g^2 (a+b x)}-\frac {B i (c+d x)}{b g^2 (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2341

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(d*x)^(m + 1)*((a + b*Log[c*x^
n])/(d*(m + 1))), x] - Simp[b*n*((d*x)^(m + 1)/(d*(m + 1)^2)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1
]

Rule 2379

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_)^(r_.))), x_Symbol] :> Simp[(-Log[1 +
d/(e*x^r)])*((a + b*Log[c*x^n])^p/(d*r)), x] + Dist[b*n*(p/(d*r)), Int[Log[1 + d/(e*x^r)]*((a + b*Log[c*x^n])^
(p - 1)/x), x], x] /; FreeQ[{a, b, c, d, e, n, r}, x] && IGtQ[p, 0]

Rule 2380

Int[(((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(x_)^(m_.))/((d_) + (e_.)*(x_)^(r_.)), x_Symbol] :> Dist[1/d,
Int[x^m*(a + b*Log[c*x^n])^p, x], x] - Dist[e/d, Int[(x^(m + r)*(a + b*Log[c*x^n])^p)/(d + e*x^r), x], x] /;
FreeQ[{a, b, c, d, e, m, n, r}, x] && IGtQ[p, 0] && IGtQ[r, 0] && ILtQ[m, -1]

Rule 2438

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2, (-c)*e*x^n]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 2562

Int[((A_.) + Log[(e_.)*((a_.) + (b_.)*(x_))^(n_.)*((c_.) + (d_.)*(x_))^(mn_)]*(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] && EqQ[n + mn, 0] && IGtQ[n, 0] && NeQ[b*c - a*d, 0] && EqQ[b*f - a*g, 0] && EqQ[d*h - c*i
, 0] && IntegersQ[m, q]

Rubi steps

\begin {align*} \int \frac {(6 c+6 d x) \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{(a g+b g x)^2} \, dx &=\int \left (\frac {6 (b c-a d) \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{b g^2 (a+b x)^2}+\frac {6 d \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{b g^2 (a+b x)}\right ) \, dx\\ &=\frac {(6 d) \int \frac {A+B \log \left (\frac {e (a+b x)}{c+d x}\right )}{a+b x} \, dx}{b g^2}+\frac {(6 (b c-a d)) \int \frac {A+B \log \left (\frac {e (a+b x)}{c+d x}\right )}{(a+b x)^2} \, dx}{b g^2}\\ &=-\frac {6 (b c-a d) \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{b^2 g^2 (a+b x)}+\frac {6 d \log (a+b x) \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{b^2 g^2}-\frac {(6 B d) \int \frac {(c+d x) \left (-\frac {d e (a+b x)}{(c+d x)^2}+\frac {b e}{c+d x}\right ) \log (a+b x)}{e (a+b x)} \, dx}{b^2 g^2}+\frac {(6 B (b c-a d)) \int \frac {b c-a d}{(a+b x)^2 (c+d x)} \, dx}{b^2 g^2}\\ &=-\frac {6 (b c-a d) \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{b^2 g^2 (a+b x)}+\frac {6 d \log (a+b x) \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{b^2 g^2}+\frac {\left (6 B (b c-a d)^2\right ) \int \frac {1}{(a+b x)^2 (c+d x)} \, dx}{b^2 g^2}-\frac {(6 B d) \int \frac {(c+d x) \left (-\frac {d e (a+b x)}{(c+d x)^2}+\frac {b e}{c+d x}\right ) \log (a+b x)}{a+b x} \, dx}{b^2 e g^2}\\ &=-\frac {6 (b c-a d) \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{b^2 g^2 (a+b x)}+\frac {6 d \log (a+b x) \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{b^2 g^2}+\frac {\left (6 B (b c-a d)^2\right ) \int \left (\frac {b}{(b c-a d) (a+b x)^2}-\frac {b d}{(b c-a d)^2 (a+b x)}+\frac {d^2}{(b c-a d)^2 (c+d x)}\right ) \, dx}{b^2 g^2}-\frac {(6 B d) \int \left (\frac {b e \log (a+b x)}{a+b x}-\frac {d e \log (a+b x)}{c+d x}\right ) \, dx}{b^2 e g^2}\\ &=-\frac {6 B (b c-a d)}{b^2 g^2 (a+b x)}-\frac {6 B d \log (a+b x)}{b^2 g^2}-\frac {6 (b c-a d) \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{b^2 g^2 (a+b x)}+\frac {6 d \log (a+b x) \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{b^2 g^2}+\frac {6 B d \log (c+d x)}{b^2 g^2}-\frac {(6 B d) \int \frac {\log (a+b x)}{a+b x} \, dx}{b g^2}+\frac {\left (6 B d^2\right ) \int \frac {\log (a+b x)}{c+d x} \, dx}{b^2 g^2}\\ &=-\frac {6 B (b c-a d)}{b^2 g^2 (a+b x)}-\frac {6 B d \log (a+b x)}{b^2 g^2}-\frac {6 (b c-a d) \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{b^2 g^2 (a+b x)}+\frac {6 d \log (a+b x) \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{b^2 g^2}+\frac {6 B d \log (c+d x)}{b^2 g^2}+\frac {6 B d \log (a+b x) \log \left (\frac {b (c+d x)}{b c-a d}\right )}{b^2 g^2}-\frac {(6 B d) \text {Subst}\left (\int \frac {\log (x)}{x} \, dx,x,a+b x\right )}{b^2 g^2}-\frac {(6 B d) \int \frac {\log \left (\frac {b (c+d x)}{b c-a d}\right )}{a+b x} \, dx}{b g^2}\\ &=-\frac {6 B (b c-a d)}{b^2 g^2 (a+b x)}-\frac {6 B d \log (a+b x)}{b^2 g^2}-\frac {3 B d \log ^2(a+b x)}{b^2 g^2}-\frac {6 (b c-a d) \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{b^2 g^2 (a+b x)}+\frac {6 d \log (a+b x) \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{b^2 g^2}+\frac {6 B d \log (c+d x)}{b^2 g^2}+\frac {6 B d \log (a+b x) \log \left (\frac {b (c+d x)}{b c-a d}\right )}{b^2 g^2}-\frac {(6 B d) \text {Subst}\left (\int \frac {\log \left (1+\frac {d x}{b c-a d}\right )}{x} \, dx,x,a+b x\right )}{b^2 g^2}\\ &=-\frac {6 B (b c-a d)}{b^2 g^2 (a+b x)}-\frac {6 B d \log (a+b x)}{b^2 g^2}-\frac {3 B d \log ^2(a+b x)}{b^2 g^2}-\frac {6 (b c-a d) \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{b^2 g^2 (a+b x)}+\frac {6 d \log (a+b x) \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{b^2 g^2}+\frac {6 B d \log (c+d x)}{b^2 g^2}+\frac {6 B d \log (a+b x) \log \left (\frac {b (c+d x)}{b c-a d}\right )}{b^2 g^2}+\frac {6 B d \text {Li}_2\left (-\frac {d (a+b x)}{b c-a d}\right )}{b^2 g^2}\\ \end {align*}

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Mathematica [A]
time = 0.11, size = 175, normalized size = 1.23 \begin {gather*} \frac {i \left (-2 (A+B) (b c-a d)-B d (a+b x) \log ^2(a+b x)+2 (-b B c+a B d) \log \left (\frac {e (a+b x)}{c+d x}\right )+2 B d (a+b x) \log (c+d x)+2 d (a+b x) \log (a+b x) \left (A-B+B \log \left (\frac {e (a+b x)}{c+d x}\right )+B \log \left (\frac {b (c+d x)}{b c-a d}\right )\right )+2 B d (a+b x) \text {Li}_2\left (\frac {d (a+b x)}{-b c+a d}\right )\right )}{2 b^2 g^2 (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(530\) vs. \(2(142)=284\).
time = 1.23, size = 531, normalized size = 3.74 Too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*i*x+c*i)*(A+B*ln(e*(b*x+a)/(d*x+c)))/(b*g*x+a*g)^2,x,method=_RETURNVERBOSE)

[Out]

-1/d^2*e*(a*d-b*c)*(i*d^2/(a*d-b*c)/g^2*A/b/(b*e/d+(a*d-b*c)*e/d/(d*x+c))-i*d^3/e/(a*d-b*c)/g^2*A/b^2*ln(b*e/d
+(a*d-b*c)*e/d/(d*x+c))+i*d^3/e/(a*d-b*c)/g^2*A/b^2*ln(b*e-(b*e/d+(a*d-b*c)*e/d/(d*x+c))*d)-1/2*i*d^3/e/(a*d-b
*c)/g^2*B*ln(b*e/d+(a*d-b*c)*e/d/(d*x+c))^2/b^2+i*d^2/(a*d-b*c)/g^2*B/b/(b*e/d+(a*d-b*c)*e/d/(d*x+c))*ln(b*e/d
+(a*d-b*c)*e/d/(d*x+c))+i*d^2/(a*d-b*c)/g^2*B/b/(b*e/d+(a*d-b*c)*e/d/(d*x+c))+i*d^3/e/(a*d-b*c)/g^2*B/b^2*dilo
g(-(-b*e+(b*e/d+(a*d-b*c)*e/d/(d*x+c))*d)/b/e)+i*d^3/e/(a*d-b*c)/g^2*B/b^2*ln(b*e/d+(a*d-b*c)*e/d/(d*x+c))*ln(
-(-b*e+(b*e/d+(a*d-b*c)*e/d/(d*x+c))*d)/b/e))

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-I*B*d*(((b*x + a)*log(b*x + a) + a)*log(d*x + c)/(b^3*g^2*x + a*b^2*g^2) - integrate((b^2*d*x^2 + a^2*d + (b^
2*c + a*b*d)*x + (2*b^2*d*x^2 + a^2*d + (b^2*c + 2*a*b*d)*x)*log(b*x + a))/(b^4*d*g^2*x^3 + a^2*b^2*c*g^2 + (b
^4*c*g^2 + 2*a*b^3*d*g^2)*x^2 + (2*a*b^3*c*g^2 + a^2*b^2*d*g^2)*x), x)) + I*A*d*(a/(b^3*g^2*x + a*b^2*g^2) + l
og(b*x + a)/(b^2*g^2)) - I*B*c*(log(b*x*e/(d*x + c) + a*e/(d*x + c))/(b^2*g^2*x + a*b*g^2) + 1/(b^2*g^2*x + a*
b*g^2) + d*log(b*x + a)/((b^2*c - a*b*d)*g^2) - d*log(d*x + c)/((b^2*c - a*b*d)*g^2)) - I*A*c/(b^2*g^2*x + a*b
*g^2)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((I*A*d*x + I*A*c + (I*B*d*x + I*B*c)*log((b*x + a)*e/(d*x + c)))/(b^2*g^2*x^2 + 2*a*b*g^2*x + a^2*g^2
), x)

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

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((I*d*x + I*c)*(B*log((b*x + a)*e/(d*x + c)) + A)/(b*g*x + a*g)^2, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\left (c\,i+d\,i\,x\right )\,\left (A+B\,\ln \left (\frac {e\,\left (a+b\,x\right )}{c+d\,x}\right )\right )}{{\left (a\,g+b\,g\,x\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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