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]

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

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Rubi [A]  time = 0.06, antiderivative size = 81, normalized size of antiderivative = 1.00, 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 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])

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]

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*}

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Mathematica [A]  time = 0.03, 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)

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fricas [A]  time = 0.91, size = 127, normalized size = 1.57 \[ \frac {A b^{2} d^{2} i x^{2} - B b^{2} c^{2} i \log \left (d x + c\right ) + {\left ({\left (2 \, A - B\right )} b^{2} c d + B a b d^{2}\right )} i x + {\left (2 \, B a b c d - B a^{2} d^{2}\right )} i \log \left (b x + a\right ) + {\left (B b^{2} d^{2} i x^{2} + 2 \, B b^{2} c d i x\right )} \log \left (\frac {b e x + a e}{d x + c}\right )}{2 \, b^{2} d} \]

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)

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giac [B]  time = 0.67, size = 1395, normalized size = 17.22 \[ \text {result too large to display} \]

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*(B*b^5*c^3*i*e^3*log(-b*e + (b*x*e + a*e)*d/(d*x + c)) - 3*B*a*b^4*c^2*d*i*e^3*log(-b*e + (b*x*e + a*e)*d/
(d*x + c)) + 3*B*a^2*b^3*c*d^2*i*e^3*log(-b*e + (b*x*e + a*e)*d/(d*x + c)) - B*a^3*b^2*d^3*i*e^3*log(-b*e + (b
*x*e + a*e)*d/(d*x + c)) - 2*(b*x*e + a*e)*B*b^4*c^3*d*i*e^2*log(-b*e + (b*x*e + a*e)*d/(d*x + c))/(d*x + c) +
6*(b*x*e + a*e)*B*a*b^3*c^2*d^2*i*e^2*log(-b*e + (b*x*e + a*e)*d/(d*x + c))/(d*x + c) - 6*(b*x*e + a*e)*B*a^2
*b^2*c*d^3*i*e^2*log(-b*e + (b*x*e + a*e)*d/(d*x + c))/(d*x + c) + 2*(b*x*e + a*e)*B*a^3*b*d^4*i*e^2*log(-b*e
+ (b*x*e + a*e)*d/(d*x + c))/(d*x + c) + (b*x*e + a*e)^2*B*b^3*c^3*d^2*i*e*log(-b*e + (b*x*e + a*e)*d/(d*x + c
))/(d*x + c)^2 - 3*(b*x*e + a*e)^2*B*a*b^2*c^2*d^3*i*e*log(-b*e + (b*x*e + a*e)*d/(d*x + c))/(d*x + c)^2 + 3*(
b*x*e + a*e)^2*B*a^2*b*c*d^4*i*e*log(-b*e + (b*x*e + a*e)*d/(d*x + c))/(d*x + c)^2 - (b*x*e + a*e)^2*B*a^3*d^5
*i*e*log(-b*e + (b*x*e + a*e)*d/(d*x + c))/(d*x + c)^2 + 2*(b*x*e + a*e)*B*b^4*c^3*d*i*e^2*log((b*x*e + a*e)/(
d*x + c))/(d*x + c) - 6*(b*x*e + a*e)*B*a*b^3*c^2*d^2*i*e^2*log((b*x*e + a*e)/(d*x + c))/(d*x + c) + 6*(b*x*e
+ a*e)*B*a^2*b^2*c*d^3*i*e^2*log((b*x*e + a*e)/(d*x + c))/(d*x + c) - 2*(b*x*e + a*e)*B*a^3*b*d^4*i*e^2*log((b
*x*e + a*e)/(d*x + c))/(d*x + c) - (b*x*e + a*e)^2*B*b^3*c^3*d^2*i*e*log((b*x*e + a*e)/(d*x + c))/(d*x + c)^2
+ 3*(b*x*e + a*e)^2*B*a*b^2*c^2*d^3*i*e*log((b*x*e + a*e)/(d*x + c))/(d*x + c)^2 - 3*(b*x*e + a*e)^2*B*a^2*b*c
*d^4*i*e*log((b*x*e + a*e)/(d*x + c))/(d*x + c)^2 + (b*x*e + a*e)^2*B*a^3*d^5*i*e*log((b*x*e + a*e)/(d*x + c))
/(d*x + c)^2 + A*b^5*c^3*i*e^3 - B*b^5*c^3*i*e^3 - 3*A*a*b^4*c^2*d*i*e^3 + 3*B*a*b^4*c^2*d*i*e^3 + 3*A*a^2*b^3
*c*d^2*i*e^3 - 3*B*a^2*b^3*c*d^2*i*e^3 - A*a^3*b^2*d^3*i*e^3 + B*a^3*b^2*d^3*i*e^3 + (b*x*e + a*e)*B*b^4*c^3*d
*i*e^2/(d*x + c) - 3*(b*x*e + a*e)*B*a*b^3*c^2*d^2*i*e^2/(d*x + c) + 3*(b*x*e + a*e)*B*a^2*b^2*c*d^3*i*e^2/(d*
x + c) - (b*x*e + a*e)*B*a^3*b*d^4*i*e^2/(d*x + c))*(b*c/((b*c*e - a*d*e)*(b*c - a*d)) - a*d/((b*c*e - a*d*e)*
(b*c - a*d)))/(b^4*d*e^2 - 2*(b*x*e + a*e)*b^3*d^2*e/(d*x + c) + (b*x*e + a*e)^2*b^2*d^3/(d*x + c)^2)

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maple [B]  time = 0.13, size = 940, normalized size = 11.60 \[ -\frac {B \,a^{4} d^{3} e^{2} i \ln \left (\frac {b e}{d}+\frac {\left (a d -b c \right ) e}{\left (d x +c \right ) d}\right )}{2 \left (\frac {a d e}{d x +c}-\frac {b c e}{d x +c}\right )^{2} \left (d x +c \right )^{2} b^{2}}+\frac {2 B \,a^{3} c \,d^{2} e^{2} i \ln \left (\frac {b e}{d}+\frac {\left (a d -b c \right ) e}{\left (d x +c \right ) d}\right )}{\left (\frac {a d e}{d x +c}-\frac {b c e}{d x +c}\right )^{2} \left (d x +c \right )^{2} b}-\frac {3 B \,a^{2} c^{2} d \,e^{2} i \ln \left (\frac {b e}{d}+\frac {\left (a d -b c \right ) e}{\left (d x +c \right ) d}\right )}{\left (\frac {a d e}{d x +c}-\frac {b c e}{d x +c}\right )^{2} \left (d x +c \right )^{2}}+\frac {2 B a b \,c^{3} e^{2} i \ln \left (\frac {b e}{d}+\frac {\left (a d -b c \right ) e}{\left (d x +c \right ) d}\right )}{\left (\frac {a d e}{d x +c}-\frac {b c e}{d x +c}\right )^{2} \left (d x +c \right )^{2}}-\frac {B \,b^{2} c^{4} e^{2} i \ln \left (\frac {b e}{d}+\frac {\left (a d -b c \right ) e}{\left (d x +c \right ) d}\right )}{2 \left (\frac {a d e}{d x +c}-\frac {b c e}{d x +c}\right )^{2} \left (d x +c \right )^{2} d}+\frac {B \,a^{2} d \,e^{2} i \ln \left (\frac {b e}{d}+\frac {\left (a d -b c \right ) e}{\left (d x +c \right ) d}\right )}{2 \left (\frac {a d e}{d x +c}-\frac {b c e}{d x +c}\right )^{2}}-\frac {B a b c \,e^{2} i \ln \left (\frac {b e}{d}+\frac {\left (a d -b c \right ) e}{\left (d x +c \right ) d}\right )}{\left (\frac {a d e}{d x +c}-\frac {b c e}{d x +c}\right )^{2}}+\frac {B \,b^{2} c^{2} e^{2} i \ln \left (\frac {b e}{d}+\frac {\left (a d -b c \right ) e}{\left (d x +c \right ) d}\right )}{2 \left (\frac {a d e}{d x +c}-\frac {b c e}{d x +c}\right )^{2} d}+\frac {A \,a^{2} d \,e^{2} i}{2 \left (\frac {a d e}{d x +c}-\frac {b c e}{d x +c}\right )^{2}}-\frac {A a b c \,e^{2} i}{\left (\frac {a d e}{d x +c}-\frac {b c e}{d x +c}\right )^{2}}+\frac {A \,b^{2} c^{2} e^{2} i}{2 \left (\frac {a d e}{d x +c}-\frac {b c e}{d x +c}\right )^{2} d}+\frac {B \,a^{2} d e i}{2 \left (\frac {a d e}{d x +c}-\frac {b c e}{d x +c}\right ) b}-\frac {B a c e i}{\frac {a d e}{d x +c}-\frac {b c e}{d x +c}}+\frac {B b \,c^{2} e i}{2 \left (\frac {a d e}{d x +c}-\frac {b c e}{d x +c}\right ) d}+\frac {B \,a^{2} d i \ln \left (-b e +\left (\frac {b e}{d}+\frac {\left (a d -b c \right ) e}{\left (d x +c \right ) d}\right ) d \right )}{2 b^{2}}-\frac {B a c i \ln \left (-b e +\left (\frac {b e}{d}+\frac {\left (a d -b c \right ) e}{\left (d x +c \right ) d}\right ) d \right )}{b}+\frac {B \,c^{2} i \ln \left (-b e +\left (\frac {b e}{d}+\frac {\left (a d -b c \right ) e}{\left (d x +c \right ) d}\right ) d \right )}{2 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.10, size = 144, normalized size = 1.78 \[ \frac {1}{2} \, A d i x^{2} + {\left (x \log \left (\frac {b e x}{d x + c} + \frac {a e}{d x + c}\right ) + \frac {a \log \left (b x + a\right )}{b} - \frac {c \log \left (d x + c\right )}{d}\right )} B c i + \frac {1}{2} \, {\left (x^{2} \log \left (\frac {b e x}{d x + c} + \frac {a e}{d x + c}\right ) - \frac {a^{2} \log \left (b x + a\right )}{b^{2}} + \frac {c^{2} \log \left (d x + c\right )}{d^{2}} - \frac {{\left (b c - a d\right )} x}{b d}\right )} B d i + A c i x \]

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

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mupad [B]  time = 4.64, size = 126, normalized size = 1.56 \[ x\,\left (\frac {i\,\left (2\,A\,a\,d+4\,A\,b\,c+B\,a\,d-B\,b\,c\right )}{2\,b}-\frac {A\,i\,\left (2\,a\,d+2\,b\,c\right )}{2\,b}\right )+\ln \left (\frac {e\,\left (a+b\,x\right )}{c+d\,x}\right )\,\left (\frac {B\,d\,i\,x^2}{2}+B\,c\,i\,x\right )-\frac {\ln \left (a+b\,x\right )\,\left (B\,a^2\,d\,i-2\,B\,a\,b\,c\,i\right )}{2\,b^2}+\frac {A\,d\,i\,x^2}{2}-\frac {B\,c^2\,i\,\ln \left (c+d\,x\right )}{2\,d} \]

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))),x)

[Out]

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

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sympy [B]  time = 1.99, size = 253, normalized size = 3.12 \[ \frac {A d i x^{2}}{2} - \frac {B a i \left (a d - 2 b c\right ) \log {\left (x + \frac {B a^{2} c d i + \frac {B a^{2} d i \left (a d - 2 b c\right )}{b} - 3 B a b c^{2} i - B a c i \left (a d - 2 b c\right )}{B a^{2} d^{2} i - 2 B a b c d i - B b^{2} c^{2} i} \right )}}{2 b^{2}} - \frac {B c^{2} i \log {\left (x + \frac {B a^{2} c d i - 2 B a b c^{2} i - \frac {B b^{2} c^{3} i}{d}}{B a^{2} d^{2} i - 2 B a b c d i - B b^{2} c^{2} i} \right )}}{2 d} + x \left (A c i + \frac {B a d i}{2 b} - \frac {B c i}{2}\right ) + \left (B c i x + \frac {B d i x^{2}}{2}\right ) \log {\left (\frac {e \left (a + b x\right )}{c + d x} \right )} \]

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

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