3.114 \(\int \frac{(c i+d i x) (A+B \log (e (\frac{a+b x}{c+d x})^n))}{(a g+b g x)^3} \, dx\)

Optimal. Leaf size=89 \[ -\frac{i (c+d x)^2 \left (B \log \left (e \left (\frac{a+b x}{c+d x}\right )^n\right )+A\right )}{2 g^3 (a+b x)^2 (b c-a d)}-\frac{B i n (c+d x)^2}{4 g^3 (a+b x)^2 (b c-a d)} \]

[Out]

-(B*i*n*(c + d*x)^2)/(4*(b*c - a*d)*g^3*(a + b*x)^2) - (i*(c + d*x)^2*(A + B*Log[e*((a + b*x)/(c + d*x))^n]))/
(2*(b*c - a*d)*g^3*(a + b*x)^2)

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Rubi [B]  time = 0.285895, antiderivative size = 201, normalized size of antiderivative = 2.26, number of steps used = 10, number of rules used = 4, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.098, Rules used = {2528, 2525, 12, 44} \[ -\frac{d i \left (B \log \left (e \left (\frac{a+b x}{c+d x}\right )^n\right )+A\right )}{b^2 g^3 (a+b x)}-\frac{i (b c-a d) \left (B \log \left (e \left (\frac{a+b x}{c+d x}\right )^n\right )+A\right )}{2 b^2 g^3 (a+b x)^2}-\frac{B d^2 i n \log (a+b x)}{2 b^2 g^3 (b c-a d)}+\frac{B d^2 i n \log (c+d x)}{2 b^2 g^3 (b c-a d)}-\frac{B i n (b c-a d)}{4 b^2 g^3 (a+b x)^2}-\frac{B d i n}{2 b^2 g^3 (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(B*(b*c - a*d)*i*n)/(4*b^2*g^3*(a + b*x)^2) - (B*d*i*n)/(2*b^2*g^3*(a + b*x)) - (B*d^2*i*n*Log[a + b*x])/(2*b
^2*(b*c - a*d)*g^3) - ((b*c - a*d)*i*(A + B*Log[e*((a + b*x)/(c + d*x))^n]))/(2*b^2*g^3*(a + b*x)^2) - (d*i*(A
+ B*Log[e*((a + b*x)/(c + d*x))^n]))/(b^2*g^3*(a + b*x)) + (B*d^2*i*n*Log[c + d*x])/(2*b^2*(b*c - a*d)*g^3)

Rule 2528

Int[((a_.) + Log[(c_.)*(RFx_)^(p_.)]*(b_.))^(n_.)*(RGx_), x_Symbol] :> With[{u = ExpandIntegrand[(a + b*Log[c*
RFx^p])^n, RGx, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, p}, x] && RationalFunctionQ[RFx, x] && RationalF
unctionQ[RGx, x] && IGtQ[n, 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]

Rule 12

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

Rule 44

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{(114 c+114 d x) \left (A+B \log \left (e \left (\frac{a+b x}{c+d x}\right )^n\right )\right )}{(a g+b g x)^3} \, dx &=\int \left (\frac{114 (b c-a d) \left (A+B \log \left (e \left (\frac{a+b x}{c+d x}\right )^n\right )\right )}{b g^3 (a+b x)^3}+\frac{114 d \left (A+B \log \left (e \left (\frac{a+b x}{c+d x}\right )^n\right )\right )}{b g^3 (a+b x)^2}\right ) \, dx\\ &=\frac{(114 d) \int \frac{A+B \log \left (e \left (\frac{a+b x}{c+d x}\right )^n\right )}{(a+b x)^2} \, dx}{b g^3}+\frac{(114 (b c-a d)) \int \frac{A+B \log \left (e \left (\frac{a+b x}{c+d x}\right )^n\right )}{(a+b x)^3} \, dx}{b g^3}\\ &=-\frac{57 (b c-a d) \left (A+B \log \left (e \left (\frac{a+b x}{c+d x}\right )^n\right )\right )}{b^2 g^3 (a+b x)^2}-\frac{114 d \left (A+B \log \left (e \left (\frac{a+b x}{c+d x}\right )^n\right )\right )}{b^2 g^3 (a+b x)}+\frac{(114 B d n) \int \frac{b c-a d}{(a+b x)^2 (c+d x)} \, dx}{b^2 g^3}+\frac{(57 B (b c-a d) n) \int \frac{b c-a d}{(a+b x)^3 (c+d x)} \, dx}{b^2 g^3}\\ &=-\frac{57 (b c-a d) \left (A+B \log \left (e \left (\frac{a+b x}{c+d x}\right )^n\right )\right )}{b^2 g^3 (a+b x)^2}-\frac{114 d \left (A+B \log \left (e \left (\frac{a+b x}{c+d x}\right )^n\right )\right )}{b^2 g^3 (a+b x)}+\frac{(114 B d (b c-a d) n) \int \frac{1}{(a+b x)^2 (c+d x)} \, dx}{b^2 g^3}+\frac{\left (57 B (b c-a d)^2 n\right ) \int \frac{1}{(a+b x)^3 (c+d x)} \, dx}{b^2 g^3}\\ &=-\frac{57 (b c-a d) \left (A+B \log \left (e \left (\frac{a+b x}{c+d x}\right )^n\right )\right )}{b^2 g^3 (a+b x)^2}-\frac{114 d \left (A+B \log \left (e \left (\frac{a+b x}{c+d x}\right )^n\right )\right )}{b^2 g^3 (a+b x)}+\frac{(114 B d (b c-a d) n) \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^3}+\frac{\left (57 B (b c-a d)^2 n\right ) \int \left (\frac{b}{(b c-a d) (a+b x)^3}-\frac{b d}{(b c-a d)^2 (a+b x)^2}+\frac{b d^2}{(b c-a d)^3 (a+b x)}-\frac{d^3}{(b c-a d)^3 (c+d x)}\right ) \, dx}{b^2 g^3}\\ &=-\frac{57 B (b c-a d) n}{2 b^2 g^3 (a+b x)^2}-\frac{57 B d n}{b^2 g^3 (a+b x)}-\frac{57 B d^2 n \log (a+b x)}{b^2 (b c-a d) g^3}-\frac{57 (b c-a d) \left (A+B \log \left (e \left (\frac{a+b x}{c+d x}\right )^n\right )\right )}{b^2 g^3 (a+b x)^2}-\frac{114 d \left (A+B \log \left (e \left (\frac{a+b x}{c+d x}\right )^n\right )\right )}{b^2 g^3 (a+b x)}+\frac{57 B d^2 n \log (c+d x)}{b^2 (b c-a d) g^3}\\ \end{align*}

Mathematica [B]  time = 0.161494, size = 216, normalized size = 2.43 \[ \frac{i \left (-\frac{d \left (B \log \left (e \left (\frac{a+b x}{c+d x}\right )^n\right )+A\right )}{b^2 (a+b x)}-\frac{(b c-a d) \left (B \log \left (e \left (\frac{a+b x}{c+d x}\right )^n\right )+A\right )}{2 b^2 (a+b x)^2}-\frac{B n \left (-\frac{2 d^2 \log (a+b x)}{b c-a d}+\frac{2 d^2 \log (c+d x)}{b c-a d}+\frac{b c-a d}{(a+b x)^2}-\frac{2 d}{a+b x}\right )}{4 b^2}-\frac{B d n \left (\frac{d \log (a+b x)}{b c-a d}-\frac{d \log (c+d x)}{b c-a d}+\frac{1}{a+b x}\right )}{b^2}\right )}{g^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(i*(-((b*c - a*d)*(A + B*Log[e*((a + b*x)/(c + d*x))^n]))/(2*b^2*(a + b*x)^2) - (d*(A + B*Log[e*((a + b*x)/(c
+ d*x))^n]))/(b^2*(a + b*x)) - (B*d*n*((a + b*x)^(-1) + (d*Log[a + b*x])/(b*c - a*d) - (d*Log[c + d*x])/(b*c -
a*d)))/b^2 - (B*n*((b*c - a*d)/(a + b*x)^2 - (2*d)/(a + b*x) - (2*d^2*Log[a + b*x])/(b*c - a*d) + (2*d^2*Log[
c + d*x])/(b*c - a*d)))/(4*b^2)))/g^3

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Maple [F]  time = 0.514, size = 0, normalized size = 0. \begin{align*} \int{\frac{dix+ci}{ \left ( bgx+ag \right ) ^{3}} \left ( A+B\ln \left ( e \left ({\frac{bx+a}{dx+c}} \right ) ^{n} \right ) \right ) }\, dx \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))^n))/(b*g*x+a*g)^3,x)

[Out]

int((d*i*x+c*i)*(A+B*ln(e*((b*x+a)/(d*x+c))^n))/(b*g*x+a*g)^3,x)

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Maxima [B]  time = 1.29162, size = 786, normalized size = 8.83 \begin{align*} -\frac{1}{4} \, B d i n{\left (\frac{3 \, a b c - a^{2} d + 2 \,{\left (2 \, b^{2} c - a b d\right )} x}{{\left (b^{5} c - a b^{4} d\right )} g^{3} x^{2} + 2 \,{\left (a b^{4} c - a^{2} b^{3} d\right )} g^{3} x +{\left (a^{2} b^{3} c - a^{3} b^{2} d\right )} g^{3}} + \frac{2 \,{\left (2 \, b c d - a d^{2}\right )} \log \left (b x + a\right )}{{\left (b^{4} c^{2} - 2 \, a b^{3} c d + a^{2} b^{2} d^{2}\right )} g^{3}} - \frac{2 \,{\left (2 \, b c d - a d^{2}\right )} \log \left (d x + c\right )}{{\left (b^{4} c^{2} - 2 \, a b^{3} c d + a^{2} b^{2} d^{2}\right )} g^{3}}\right )} + \frac{1}{4} \, B c i n{\left (\frac{2 \, b d x - b c + 3 \, a d}{{\left (b^{4} c - a b^{3} d\right )} g^{3} x^{2} + 2 \,{\left (a b^{3} c - a^{2} b^{2} d\right )} g^{3} x +{\left (a^{2} b^{2} c - a^{3} b d\right )} g^{3}} + \frac{2 \, d^{2} \log \left (b x + a\right )}{{\left (b^{3} c^{2} - 2 \, a b^{2} c d + a^{2} b d^{2}\right )} g^{3}} - \frac{2 \, d^{2} \log \left (d x + c\right )}{{\left (b^{3} c^{2} - 2 \, a b^{2} c d + a^{2} b d^{2}\right )} g^{3}}\right )} - \frac{{\left (2 \, b x + a\right )} B d i \log \left (e{\left (\frac{b x}{d x + c} + \frac{a}{d x + c}\right )}^{n}\right )}{2 \,{\left (b^{4} g^{3} x^{2} + 2 \, a b^{3} g^{3} x + a^{2} b^{2} g^{3}\right )}} - \frac{{\left (2 \, b x + a\right )} A d i}{2 \,{\left (b^{4} g^{3} x^{2} + 2 \, a b^{3} g^{3} x + a^{2} b^{2} g^{3}\right )}} - \frac{B c i \log \left (e{\left (\frac{b x}{d x + c} + \frac{a}{d x + c}\right )}^{n}\right )}{2 \,{\left (b^{3} g^{3} x^{2} + 2 \, a b^{2} g^{3} x + a^{2} b g^{3}\right )}} - \frac{A c i}{2 \,{\left (b^{3} g^{3} x^{2} + 2 \, a b^{2} g^{3} x + a^{2} b g^{3}\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))^n))/(b*g*x+a*g)^3,x, algorithm="maxima")

[Out]

-1/4*B*d*i*n*((3*a*b*c - a^2*d + 2*(2*b^2*c - a*b*d)*x)/((b^5*c - a*b^4*d)*g^3*x^2 + 2*(a*b^4*c - a^2*b^3*d)*g
^3*x + (a^2*b^3*c - a^3*b^2*d)*g^3) + 2*(2*b*c*d - a*d^2)*log(b*x + a)/((b^4*c^2 - 2*a*b^3*c*d + a^2*b^2*d^2)*
g^3) - 2*(2*b*c*d - a*d^2)*log(d*x + c)/((b^4*c^2 - 2*a*b^3*c*d + a^2*b^2*d^2)*g^3)) + 1/4*B*c*i*n*((2*b*d*x -
b*c + 3*a*d)/((b^4*c - a*b^3*d)*g^3*x^2 + 2*(a*b^3*c - a^2*b^2*d)*g^3*x + (a^2*b^2*c - a^3*b*d)*g^3) + 2*d^2*
log(b*x + a)/((b^3*c^2 - 2*a*b^2*c*d + a^2*b*d^2)*g^3) - 2*d^2*log(d*x + c)/((b^3*c^2 - 2*a*b^2*c*d + a^2*b*d^
2)*g^3)) - 1/2*(2*b*x + a)*B*d*i*log(e*(b*x/(d*x + c) + a/(d*x + c))^n)/(b^4*g^3*x^2 + 2*a*b^3*g^3*x + a^2*b^2
*g^3) - 1/2*(2*b*x + a)*A*d*i/(b^4*g^3*x^2 + 2*a*b^3*g^3*x + a^2*b^2*g^3) - 1/2*B*c*i*log(e*(b*x/(d*x + c) + a
/(d*x + c))^n)/(b^3*g^3*x^2 + 2*a*b^2*g^3*x + a^2*b*g^3) - 1/2*A*c*i/(b^3*g^3*x^2 + 2*a*b^2*g^3*x + a^2*b*g^3)

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Fricas [B]  time = 0.509721, size = 510, normalized size = 5.73 \begin{align*} -\frac{{\left (B b^{2} c^{2} - B a^{2} d^{2}\right )} i n + 2 \,{\left (A b^{2} c^{2} - A a^{2} d^{2}\right )} i + 2 \,{\left ({\left (B b^{2} c d - B a b d^{2}\right )} i n + 2 \,{\left (A b^{2} c d - A a b d^{2}\right )} i\right )} x + 2 \,{\left (2 \,{\left (B b^{2} c d - B a b d^{2}\right )} i x +{\left (B b^{2} c^{2} - B a^{2} d^{2}\right )} i\right )} \log \left (e\right ) + 2 \,{\left (B b^{2} d^{2} i n x^{2} + 2 \, B b^{2} c d i n x + B b^{2} c^{2} i n\right )} \log \left (\frac{b x + a}{d x + c}\right )}{4 \,{\left ({\left (b^{5} c - a b^{4} d\right )} g^{3} x^{2} + 2 \,{\left (a b^{4} c - a^{2} b^{3} d\right )} g^{3} x +{\left (a^{2} b^{3} c - a^{3} b^{2} d\right )} g^{3}\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))^n))/(b*g*x+a*g)^3,x, algorithm="fricas")

[Out]

-1/4*((B*b^2*c^2 - B*a^2*d^2)*i*n + 2*(A*b^2*c^2 - A*a^2*d^2)*i + 2*((B*b^2*c*d - B*a*b*d^2)*i*n + 2*(A*b^2*c*
d - A*a*b*d^2)*i)*x + 2*(2*(B*b^2*c*d - B*a*b*d^2)*i*x + (B*b^2*c^2 - B*a^2*d^2)*i)*log(e) + 2*(B*b^2*d^2*i*n*
x^2 + 2*B*b^2*c*d*i*n*x + B*b^2*c^2*i*n)*log((b*x + a)/(d*x + c)))/((b^5*c - a*b^4*d)*g^3*x^2 + 2*(a*b^4*c - a
^2*b^3*d)*g^3*x + (a^2*b^3*c - a^3*b^2*d)*g^3)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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))**n))/(b*g*x+a*g)**3,x)

[Out]

Timed out

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Giac [B]  time = 1.23785, size = 320, normalized size = 3.6 \begin{align*} \frac{B d^{2} n \log \left (b x + a\right )}{2 \,{\left (b^{3} c g^{3} i - a b^{2} d g^{3} i\right )}} - \frac{B d^{2} n \log \left (d x + c\right )}{2 \,{\left (b^{3} c g^{3} i - a b^{2} d g^{3} i\right )}} - \frac{{\left (2 \, B b d i n x + B b c i n + B a d i n\right )} \log \left (\frac{b x + a}{d x + c}\right )}{2 \,{\left (b^{4} g^{3} x^{2} + 2 \, a b^{3} g^{3} x + a^{2} b^{2} g^{3}\right )}} - \frac{2 \, B b d i n x + B b c i n + B a d i n + 4 \, A b d i x + 4 \, B b d i x + 2 \, A b c i + 2 \, B b c i + 2 \, A a d i + 2 \, B a d i}{4 \,{\left (b^{4} g^{3} x^{2} + 2 \, a b^{3} g^{3} x + a^{2} b^{2} g^{3}\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))^n))/(b*g*x+a*g)^3,x, algorithm="giac")

[Out]

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