3.4 \(\int (c i+d i x) (A+B \log (\frac{e (a+b x)}{c+d x})) \, dx\)

Optimal. Leaf size=81 \[ \frac{i (c+d x)^2 \left (B \log \left (\frac{e (a+b x)}{c+d x}\right )+A\right )}{2 d}-\frac{B i (b c-a d)^2 \log (a+b x)}{2 b^2 d}-\frac{B i x (b c-a d)}{2 b} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0567414, antiderivative size = 81, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.107, Rules used = {2525, 12, 43} \[ \frac{i (c+d x)^2 \left (B \log \left (\frac{e (a+b x)}{c+d x}\right )+A\right )}{2 d}-\frac{B i (b c-a d)^2 \log (a+b x)}{2 b^2 d}-\frac{B i x (b c-a d)}{2 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2525

Int[((a_.) + Log[(c_.)*(RFx_)^(p_.)]*(b_.))^(n_.)*((d_.) + (e_.)*(x_))^(m_.), x_Symbol] :> Simp[((d + e*x)^(m
+ 1)*(a + b*Log[c*RFx^p])^n)/(e*(m + 1)), x] - Dist[(b*n*p)/(e*(m + 1)), Int[SimplifyIntegrand[((d + e*x)^(m +
1)*(a + b*Log[c*RFx^p])^(n - 1)*D[RFx, x])/RFx, x], x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && RationalFunc
tionQ[RFx, x] && IGtQ[n, 0] && (EqQ[n, 1] || IntegerQ[m]) && NeQ[m, -1]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int (4 c+4 d x) \left (A+B \log \left (\frac{e (a+b x)}{c+d x}\right )\right ) \, dx &=\frac{2 (c+d x)^2 \left (A+B \log \left (\frac{e (a+b x)}{c+d x}\right )\right )}{d}-\frac{B \int \frac{16 (b c-a d) (c+d x)}{a+b x} \, dx}{8 d}\\ &=\frac{2 (c+d x)^2 \left (A+B \log \left (\frac{e (a+b x)}{c+d x}\right )\right )}{d}-\frac{(2 B (b c-a d)) \int \frac{c+d x}{a+b x} \, dx}{d}\\ &=\frac{2 (c+d x)^2 \left (A+B \log \left (\frac{e (a+b x)}{c+d x}\right )\right )}{d}-\frac{(2 B (b c-a d)) \int \left (\frac{d}{b}+\frac{b c-a d}{b (a+b x)}\right ) \, dx}{d}\\ &=-\frac{2 B (b c-a d) x}{b}-\frac{2 B (b c-a d)^2 \log (a+b x)}{b^2 d}+\frac{2 (c+d x)^2 \left (A+B \log \left (\frac{e (a+b x)}{c+d x}\right )\right )}{d}\\ \end{align*}

Mathematica [A]  time = 0.0344547, size = 70, normalized size = 0.86 \[ \frac{i \left ((c+d x)^2 \left (B \log \left (\frac{e (a+b x)}{c+d x}\right )+A\right )-\frac{B (b c-a d) ((b c-a d) \log (a+b x)+b d x)}{b^2}\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 0.164, size = 940, normalized size = 11.6 \begin{align*} \text{result too large to display} \end{align*}

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))),x)

[Out]

1/2*e^2*d*A*i/(d*e/(d*x+c)*a-e/(d*x+c)*b*c)^2*a^2-e^2*A*i/(d*e/(d*x+c)*a-e/(d*x+c)*b*c)^2*a*b*c+1/2*e^2/d*A*i/
(d*e/(d*x+c)*a-e/(d*x+c)*b*c)^2*b^2*c^2+1/2*e*d*B*i/b/(d*e/(d*x+c)*a-e/(d*x+c)*b*c)*a^2-e*B*i/(d*e/(d*x+c)*a-e
/(d*x+c)*b*c)*a*c+1/2*e/d*B*i/(d*e/(d*x+c)*a-e/(d*x+c)*b*c)*c^2*b+1/2*d*B*i/b^2*ln(d*(b*e/d+(a*d-b*c)*e/d/(d*x
+c))-b*e)*a^2-B*i/b*ln(d*(b*e/d+(a*d-b*c)*e/d/(d*x+c))-b*e)*a*c+1/2/d*B*i*ln(d*(b*e/d+(a*d-b*c)*e/d/(d*x+c))-b
*e)*c^2+1/2*e^2*d*B*i*ln(b*e/d+(a*d-b*c)*e/d/(d*x+c))/(d*e/(d*x+c)*a-e/(d*x+c)*b*c)^2*a^2-e^2*B*i*ln(b*e/d+(a*
d-b*c)*e/d/(d*x+c))/(d*e/(d*x+c)*a-e/(d*x+c)*b*c)^2*a*b*c-1/2*e^2*d^3*B*i*ln(b*e/d+(a*d-b*c)*e/d/(d*x+c))/b^2/
(d*e/(d*x+c)*a-e/(d*x+c)*b*c)^2*a^4/(d*x+c)^2+2*e^2*d^2*B*i*ln(b*e/d+(a*d-b*c)*e/d/(d*x+c))/b/(d*e/(d*x+c)*a-e
/(d*x+c)*b*c)^2*a^3/(d*x+c)^2*c-3*e^2*d*B*i*ln(b*e/d+(a*d-b*c)*e/d/(d*x+c))/(d*e/(d*x+c)*a-e/(d*x+c)*b*c)^2*a^
2/(d*x+c)^2*c^2+2*e^2*B*i*ln(b*e/d+(a*d-b*c)*e/d/(d*x+c))/(d*e/(d*x+c)*a-e/(d*x+c)*b*c)^2*a/(d*x+c)^2*c^3*b+1/
2*e^2/d*B*i*ln(b*e/d+(a*d-b*c)*e/d/(d*x+c))*b^2/(d*e/(d*x+c)*a-e/(d*x+c)*b*c)^2*c^2-1/2*e^2/d*B*i*ln(b*e/d+(a*
d-b*c)*e/d/(d*x+c))*b^2/(d*e/(d*x+c)*a-e/(d*x+c)*b*c)^2*c^4/(d*x+c)^2

________________________________________________________________________________________

Maxima [A]  time = 1.45613, size = 194, normalized size = 2.4 \begin{align*} \frac{1}{2} \, A d i x^{2} +{\left (x \log \left (\frac{b e x}{d x + c} + \frac{a e}{d x + c}\right ) + \frac{a \log \left (b x + a\right )}{b} - \frac{c \log \left (d x + c\right )}{d}\right )} B c i + \frac{1}{2} \,{\left (x^{2} \log \left (\frac{b e x}{d x + c} + \frac{a e}{d x + c}\right ) - \frac{a^{2} \log \left (b x + a\right )}{b^{2}} + \frac{c^{2} \log \left (d x + c\right )}{d^{2}} - \frac{{\left (b c - a d\right )} x}{b d}\right )} B d i + A c i x \end{align*}

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))),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 1.05063, size = 278, normalized size = 3.43 \begin{align*} \frac{A b^{2} d^{2} i x^{2} - B b^{2} c^{2} i \log \left (d x + c\right ) +{\left ({\left (2 \, A - B\right )} b^{2} c d + B a b d^{2}\right )} i x +{\left (2 \, B a b c d - B a^{2} d^{2}\right )} i \log \left (b x + a\right ) +{\left (B b^{2} d^{2} i x^{2} + 2 \, B b^{2} c d i x\right )} \log \left (\frac{b e x + a e}{d x + c}\right )}{2 \, b^{2} d} \end{align*}

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))),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

Sympy [B]  time = 2.39451, size = 257, normalized size = 3.17 \begin{align*} \frac{A d i x^{2}}{2} - \frac{B a i \left (a d - 2 b c\right ) \log{\left (x + \frac{B a^{2} c d i + \frac{B a^{2} d i \left (a d - 2 b c\right )}{b} - 3 B a b c^{2} i - B a c i \left (a d - 2 b c\right )}{B a^{2} d^{2} i - 2 B a b c d i - B b^{2} c^{2} i} \right )}}{2 b^{2}} - \frac{B c^{2} i \log{\left (x + \frac{B a^{2} c d i - 2 B a b c^{2} i - \frac{B b^{2} c^{3} i}{d}}{B a^{2} d^{2} i - 2 B a b c d i - B b^{2} c^{2} i} \right )}}{2 d} + \left (B c i x + \frac{B d i x^{2}}{2}\right ) \log{\left (\frac{e \left (a + b x\right )}{c + d x} \right )} + \frac{x \left (2 A b c i + B a d i - B b c i\right )}{2 b} \end{align*}

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))),x)

[Out]

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

________________________________________________________________________________________

Giac [B]  time = 1.41107, size = 315, normalized size = 3.89 \begin{align*} \frac{1}{2} \,{\left (A d i + B d i\right )} x^{2} + \frac{1}{2} \,{\left (B d i x^{2} + 2 \, B c i x\right )} \log \left (\frac{b x + a}{d x + c}\right ) + \frac{{\left (2 \, A b c i + B b c i + B a d i\right )} x}{2 \, b} - \frac{{\left (B b^{2} c^{2} i - 2 \, B a b c d i + B a^{2} d^{2} i\right )} \log \left ({\left | b d x^{2} + b c x + a d x + a c \right |}\right )}{4 \, b^{2} d} + \frac{{\left (B b^{3} c^{3} i + B a b^{2} c^{2} d i - 3 \, B a^{2} b c d^{2} i + B a^{3} d^{3} i\right )} \log \left ({\left | \frac{2 \, b d x + b c + a d -{\left | -b c + a d \right |}}{2 \, b d x + b c + a d +{\left | -b c + a d \right |}} \right |}\right )}{4 \, b^{2} d{\left | -b c + a d \right |}} \end{align*}

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))),x, algorithm="giac")

[Out]

1/2*(A*d*i + B*d*i)*x^2 + 1/2*(B*d*i*x^2 + 2*B*c*i*x)*log((b*x + a)/(d*x + c)) + 1/2*(2*A*b*c*i + B*b*c*i + B*
a*d*i)*x/b - 1/4*(B*b^2*c^2*i - 2*B*a*b*c*d*i + B*a^2*d^2*i)*log(abs(b*d*x^2 + b*c*x + a*d*x + a*c))/(b^2*d) +
1/4*(B*b^3*c^3*i + B*a*b^2*c^2*d*i - 3*B*a^2*b*c*d^2*i + B*a^3*d^3*i)*log(abs((2*b*d*x + b*c + a*d - abs(-b*c
+ a*d))/(2*b*d*x + b*c + a*d + abs(-b*c + a*d))))/(b^2*d*abs(-b*c + a*d))