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

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

[Out]

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

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Rubi [B]  time = 0.316962, 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} \[ -\frac{b g \left (B \log \left (e \left (\frac{a+b x}{c+d x}\right )^n\right )+A\right )}{d^2 i^3 (c+d x)}+\frac{g (b c-a d) \left (B \log \left (e \left (\frac{a+b x}{c+d x}\right )^n\right )+A\right )}{2 d^2 i^3 (c+d x)^2}+\frac{b^2 B g n \log (a+b x)}{2 d^2 i^3 (b c-a d)}-\frac{b^2 B g n \log (c+d x)}{2 d^2 i^3 (b c-a d)}-\frac{B g n (b c-a d)}{4 d^2 i^3 (c+d x)^2}+\frac{b B g n}{2 d^2 i^3 (c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(B*(b*c - a*d)*g*n)/(4*d^2*i^3*(c + d*x)^2) + (b*B*g*n)/(2*d^2*i^3*(c + d*x)) + (b^2*B*g*n*Log[a + b*x])/(2*d
^2*(b*c - a*d)*i^3) + ((b*c - a*d)*g*(A + B*Log[e*((a + b*x)/(c + d*x))^n]))/(2*d^2*i^3*(c + d*x)^2) - (b*g*(A
+ B*Log[e*((a + b*x)/(c + d*x))^n]))/(d^2*i^3*(c + d*x)) - (b^2*B*g*n*Log[c + d*x])/(2*d^2*(b*c - a*d)*i^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{(a g+b g x) \left (A+B \log \left (e \left (\frac{a+b x}{c+d x}\right )^n\right )\right )}{(153 c+153 d x)^3} \, dx &=\int \left (\frac{(-b c+a d) g \left (A+B \log \left (e \left (\frac{a+b x}{c+d x}\right )^n\right )\right )}{3581577 d (c+d x)^3}+\frac{b g \left (A+B \log \left (e \left (\frac{a+b x}{c+d x}\right )^n\right )\right )}{3581577 d (c+d x)^2}\right ) \, dx\\ &=\frac{(b g) \int \frac{A+B \log \left (e \left (\frac{a+b x}{c+d x}\right )^n\right )}{(c+d x)^2} \, dx}{3581577 d}-\frac{((b c-a d) g) \int \frac{A+B \log \left (e \left (\frac{a+b x}{c+d x}\right )^n\right )}{(c+d x)^3} \, dx}{3581577 d}\\ &=\frac{(b c-a d) g \left (A+B \log \left (e \left (\frac{a+b x}{c+d x}\right )^n\right )\right )}{7163154 d^2 (c+d x)^2}-\frac{b g \left (A+B \log \left (e \left (\frac{a+b x}{c+d x}\right )^n\right )\right )}{3581577 d^2 (c+d x)}+\frac{(b B g n) \int \frac{b c-a d}{(a+b x) (c+d x)^2} \, dx}{3581577 d^2}-\frac{(B (b c-a d) g n) \int \frac{b c-a d}{(a+b x) (c+d x)^3} \, dx}{7163154 d^2}\\ &=\frac{(b c-a d) g \left (A+B \log \left (e \left (\frac{a+b x}{c+d x}\right )^n\right )\right )}{7163154 d^2 (c+d x)^2}-\frac{b g \left (A+B \log \left (e \left (\frac{a+b x}{c+d x}\right )^n\right )\right )}{3581577 d^2 (c+d x)}+\frac{(b B (b c-a d) g n) \int \frac{1}{(a+b x) (c+d x)^2} \, dx}{3581577 d^2}-\frac{\left (B (b c-a d)^2 g n\right ) \int \frac{1}{(a+b x) (c+d x)^3} \, dx}{7163154 d^2}\\ &=\frac{(b c-a d) g \left (A+B \log \left (e \left (\frac{a+b x}{c+d x}\right )^n\right )\right )}{7163154 d^2 (c+d x)^2}-\frac{b g \left (A+B \log \left (e \left (\frac{a+b x}{c+d x}\right )^n\right )\right )}{3581577 d^2 (c+d x)}+\frac{(b B (b c-a d) g 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}{3581577 d^2}-\frac{\left (B (b c-a d)^2 g n\right ) \int \left (\frac{b^3}{(b c-a d)^3 (a+b x)}-\frac{d}{(b c-a d) (c+d x)^3}-\frac{b d}{(b c-a d)^2 (c+d x)^2}-\frac{b^2 d}{(b c-a d)^3 (c+d x)}\right ) \, dx}{7163154 d^2}\\ &=-\frac{B (b c-a d) g n}{14326308 d^2 (c+d x)^2}+\frac{b B g n}{7163154 d^2 (c+d x)}+\frac{b^2 B g n \log (a+b x)}{7163154 d^2 (b c-a d)}+\frac{(b c-a d) g \left (A+B \log \left (e \left (\frac{a+b x}{c+d x}\right )^n\right )\right )}{7163154 d^2 (c+d x)^2}-\frac{b g \left (A+B \log \left (e \left (\frac{a+b x}{c+d x}\right )^n\right )\right )}{3581577 d^2 (c+d x)}-\frac{b^2 B g n \log (c+d x)}{7163154 d^2 (b c-a d)}\\ \end{align*}

Mathematica [B]  time = 0.156385, size = 215, normalized size = 2.42 \[ \frac{g \left (-\frac{b \left (B \log \left (e \left (\frac{a+b x}{c+d x}\right )^n\right )+A\right )}{d^2 (c+d x)}+\frac{(b c-a d) \left (B \log \left (e \left (\frac{a+b x}{c+d x}\right )^n\right )+A\right )}{2 d^2 (c+d x)^2}-\frac{B n \left (\frac{2 b^2 \log (a+b x)}{b c-a d}-\frac{2 b^2 \log (c+d x)}{b c-a d}+\frac{b c-a d}{(c+d x)^2}+\frac{2 b}{c+d x}\right )}{4 d^2}+\frac{b B n \left (\frac{b \log (a+b x)}{b c-a d}-\frac{b \log (c+d x)}{b c-a d}+\frac{1}{c+d x}\right )}{d^2}\right )}{i^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.521, size = 0, normalized size = 0. \begin{align*} \int{\frac{bgx+ag}{ \left ( dix+ci \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((b*g*x+a*g)*(A+B*ln(e*((b*x+a)/(d*x+c))^n))/(d*i*x+c*i)^3,x)

[Out]

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

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Maxima [B]  time = 1.25808, size = 780, normalized size = 8.76 \begin{align*} \frac{1}{4} \, B b g n{\left (\frac{b c^{2} - 3 \, a c d + 2 \,{\left (b c d - 2 \, a d^{2}\right )} x}{{\left (b c d^{4} - a d^{5}\right )} i^{3} x^{2} + 2 \,{\left (b c^{2} d^{3} - a c d^{4}\right )} i^{3} x +{\left (b c^{3} d^{2} - a c^{2} d^{3}\right )} i^{3}} + \frac{2 \,{\left (b^{2} c - 2 \, a b d\right )} \log \left (b x + a\right )}{{\left (b^{2} c^{2} d^{2} - 2 \, a b c d^{3} + a^{2} d^{4}\right )} i^{3}} - \frac{2 \,{\left (b^{2} c - 2 \, a b d\right )} \log \left (d x + c\right )}{{\left (b^{2} c^{2} d^{2} - 2 \, a b c d^{3} + a^{2} d^{4}\right )} i^{3}}\right )} + \frac{1}{4} \, B a g n{\left (\frac{2 \, b d x + 3 \, b c - a d}{{\left (b c d^{3} - a d^{4}\right )} i^{3} x^{2} + 2 \,{\left (b c^{2} d^{2} - a c d^{3}\right )} i^{3} x +{\left (b c^{3} d - a c^{2} d^{2}\right )} i^{3}} + \frac{2 \, b^{2} \log \left (b x + a\right )}{{\left (b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} d^{3}\right )} i^{3}} - \frac{2 \, b^{2} \log \left (d x + c\right )}{{\left (b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} d^{3}\right )} i^{3}}\right )} - \frac{{\left (2 \, d x + c\right )} B b g \log \left (e{\left (\frac{b x}{d x + c} + \frac{a}{d x + c}\right )}^{n}\right )}{2 \,{\left (d^{4} i^{3} x^{2} + 2 \, c d^{3} i^{3} x + c^{2} d^{2} i^{3}\right )}} - \frac{{\left (2 \, d x + c\right )} A b g}{2 \,{\left (d^{4} i^{3} x^{2} + 2 \, c d^{3} i^{3} x + c^{2} d^{2} i^{3}\right )}} - \frac{B a g \log \left (e{\left (\frac{b x}{d x + c} + \frac{a}{d x + c}\right )}^{n}\right )}{2 \,{\left (d^{3} i^{3} x^{2} + 2 \, c d^{2} i^{3} x + c^{2} d i^{3}\right )}} - \frac{A a g}{2 \,{\left (d^{3} i^{3} x^{2} + 2 \, c d^{2} i^{3} x + c^{2} d i^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*g*x+a*g)*(A+B*log(e*((b*x+a)/(d*x+c))^n))/(d*i*x+c*i)^3,x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 0.534184, size = 509, normalized size = 5.72 \begin{align*} \frac{{\left (B b^{2} c^{2} - B a^{2} d^{2}\right )} g n - 2 \,{\left (A b^{2} c^{2} - A a^{2} d^{2}\right )} g + 2 \,{\left ({\left (B b^{2} c d - B a b d^{2}\right )} g n - 2 \,{\left (A b^{2} c d - A a b d^{2}\right )} g\right )} x - 2 \,{\left (2 \,{\left (B b^{2} c d - B a b d^{2}\right )} g x +{\left (B b^{2} c^{2} - B a^{2} d^{2}\right )} g\right )} \log \left (e\right ) + 2 \,{\left (B b^{2} d^{2} g n x^{2} + 2 \, B a b d^{2} g n x + B a^{2} d^{2} g n\right )} \log \left (\frac{b x + a}{d x + c}\right )}{4 \,{\left ({\left (b c d^{4} - a d^{5}\right )} i^{3} x^{2} + 2 \,{\left (b c^{2} d^{3} - a c d^{4}\right )} i^{3} x +{\left (b c^{3} d^{2} - a c^{2} d^{3}\right )} i^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*g*x+a*g)*(A+B*log(e*((b*x+a)/(d*x+c))^n))/(d*i*x+c*i)^3,x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [B]  time = 1.25183, size = 298, normalized size = 3.35 \begin{align*} -\frac{B b^{2} g n \log \left (b x + a\right )}{2 \,{\left (b c d^{2} i - a d^{3} i\right )}} + \frac{B b^{2} g n \log \left (d x + c\right )}{2 \,{\left (b c d^{2} i - a d^{3} i\right )}} - \frac{{\left (2 \, B b d g i n x + B b c g i n + B a d g i n\right )} \log \left (\frac{b x + a}{d x + c}\right )}{2 \,{\left (d^{4} x^{2} + 2 \, c d^{3} x + c^{2} d^{2}\right )}} + \frac{2 \, B b d g i n x + B b c g i n + B a d g i n - 4 \, A b d g i x - 4 \, B b d g i x - 2 \, A b c g i - 2 \, B b c g i - 2 \, A a d g i - 2 \, B a d g i}{4 \,{\left (d^{4} x^{2} + 2 \, c d^{3} x + c^{2} d^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*g*x+a*g)*(A+B*log(e*((b*x+a)/(d*x+c))^n))/(d*i*x+c*i)^3,x, algorithm="giac")

[Out]

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