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

Optimal. Leaf size=156 \[ \frac{i^3 (c+d x)^4 \left (B \log \left (e \left (\frac{a+b x}{c+d x}\right )^n\right )+A\right )}{4 d}-\frac{B i^3 n x (b c-a d)^3}{4 b^3}-\frac{B i^3 n (c+d x)^2 (b c-a d)^2}{8 b^2 d}-\frac{B i^3 n (b c-a d)^4 \log (a+b x)}{4 b^4 d}-\frac{B i^3 n (c+d x)^3 (b c-a d)}{12 b d} \]

[Out]

-(B*(b*c - a*d)^3*i^3*n*x)/(4*b^3) - (B*(b*c - a*d)^2*i^3*n*(c + d*x)^2)/(8*b^2*d) - (B*(b*c - a*d)*i^3*n*(c +
d*x)^3)/(12*b*d) - (B*(b*c - a*d)^4*i^3*n*Log[a + b*x])/(4*b^4*d) + (i^3*(c + d*x)^4*(A + B*Log[e*((a + b*x)/
(c + d*x))^n]))/(4*d)

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Rubi [A]  time = 0.0859817, antiderivative size = 156, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.091, Rules used = {2525, 12, 43} \[ \frac{i^3 (c+d x)^4 \left (B \log \left (e \left (\frac{a+b x}{c+d x}\right )^n\right )+A\right )}{4 d}-\frac{B i^3 n x (b c-a d)^3}{4 b^3}-\frac{B i^3 n (c+d x)^2 (b c-a d)^2}{8 b^2 d}-\frac{B i^3 n (b c-a d)^4 \log (a+b x)}{4 b^4 d}-\frac{B i^3 n (c+d x)^3 (b c-a d)}{12 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(B*(b*c - a*d)^3*i^3*n*x)/(4*b^3) - (B*(b*c - a*d)^2*i^3*n*(c + d*x)^2)/(8*b^2*d) - (B*(b*c - a*d)*i^3*n*(c +
d*x)^3)/(12*b*d) - (B*(b*c - a*d)^4*i^3*n*Log[a + b*x])/(4*b^4*d) + (i^3*(c + d*x)^4*(A + B*Log[e*((a + b*x)/
(c + d*x))^n]))/(4*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 (130 c+130 d x)^3 \left (A+B \log \left (e \left (\frac{a+b x}{c+d x}\right )^n\right )\right ) \, dx &=\frac{549250 (c+d x)^4 \left (A+B \log \left (e \left (\frac{a+b x}{c+d x}\right )^n\right )\right )}{d}-\frac{(B n) \int \frac{285610000 (b c-a d) (c+d x)^3}{a+b x} \, dx}{520 d}\\ &=\frac{549250 (c+d x)^4 \left (A+B \log \left (e \left (\frac{a+b x}{c+d x}\right )^n\right )\right )}{d}-\frac{(549250 B (b c-a d) n) \int \frac{(c+d x)^3}{a+b x} \, dx}{d}\\ &=\frac{549250 (c+d x)^4 \left (A+B \log \left (e \left (\frac{a+b x}{c+d x}\right )^n\right )\right )}{d}-\frac{(549250 B (b c-a d) n) \int \left (\frac{d (b c-a d)^2}{b^3}+\frac{(b c-a d)^3}{b^3 (a+b x)}+\frac{d (b c-a d) (c+d x)}{b^2}+\frac{d (c+d x)^2}{b}\right ) \, dx}{d}\\ &=-\frac{549250 B (b c-a d)^3 n x}{b^3}-\frac{274625 B (b c-a d)^2 n (c+d x)^2}{b^2 d}-\frac{549250 B (b c-a d) n (c+d x)^3}{3 b d}-\frac{549250 B (b c-a d)^4 n \log (a+b x)}{b^4 d}+\frac{549250 (c+d x)^4 \left (A+B \log \left (e \left (\frac{a+b x}{c+d x}\right )^n\right )\right )}{d}\\ \end{align*}

Mathematica [A]  time = 0.0730872, size = 124, normalized size = 0.79 \[ \frac{i^3 \left ((c+d x)^4 \left (B \log \left (e \left (\frac{a+b x}{c+d x}\right )^n\right )+A\right )-\frac{B n (b c-a d) \left (3 b^2 (c+d x)^2 (b c-a d)+6 b d x (b c-a d)^2+6 (b c-a d)^3 \log (a+b x)+2 b^3 (c+d x)^3\right )}{6 b^4}\right )}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(i^3*(-(B*(b*c - a*d)*n*(6*b*d*(b*c - a*d)^2*x + 3*b^2*(b*c - a*d)*(c + d*x)^2 + 2*b^3*(c + d*x)^3 + 6*(b*c -
a*d)^3*Log[a + b*x]))/(6*b^4) + (c + d*x)^4*(A + B*Log[e*((a + b*x)/(c + d*x))^n])))/(4*d)

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Maple [F]  time = 0.498, size = 0, normalized size = 0. \begin{align*} \int \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((d*i*x+c*i)^3*(A+B*ln(e*((b*x+a)/(d*x+c))^n)),x)

[Out]

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

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Maxima [B]  time = 1.40797, size = 647, normalized size = 4.15 \begin{align*} \frac{1}{4} \, B d^{3} i^{3} x^{4} \log \left (e{\left (\frac{b x}{d x + c} + \frac{a}{d x + c}\right )}^{n}\right ) + \frac{1}{4} \, A d^{3} i^{3} x^{4} + B c d^{2} i^{3} x^{3} \log \left (e{\left (\frac{b x}{d x + c} + \frac{a}{d x + c}\right )}^{n}\right ) + A c d^{2} i^{3} x^{3} + \frac{3}{2} \, B c^{2} d i^{3} x^{2} \log \left (e{\left (\frac{b x}{d x + c} + \frac{a}{d x + c}\right )}^{n}\right ) + \frac{3}{2} \, A c^{2} d i^{3} x^{2} - \frac{1}{24} \, B d^{3} i^{3} n{\left (\frac{6 \, a^{4} \log \left (b x + a\right )}{b^{4}} - \frac{6 \, c^{4} \log \left (d x + c\right )}{d^{4}} + \frac{2 \,{\left (b^{3} c d^{2} - a b^{2} d^{3}\right )} x^{3} - 3 \,{\left (b^{3} c^{2} d - a^{2} b d^{3}\right )} x^{2} + 6 \,{\left (b^{3} c^{3} - a^{3} d^{3}\right )} x}{b^{3} d^{3}}\right )} + \frac{1}{2} \, B c d^{2} i^{3} n{\left (\frac{2 \, a^{3} \log \left (b x + a\right )}{b^{3}} - \frac{2 \, c^{3} \log \left (d x + c\right )}{d^{3}} - \frac{{\left (b^{2} c d - a b d^{2}\right )} x^{2} - 2 \,{\left (b^{2} c^{2} - a^{2} d^{2}\right )} x}{b^{2} d^{2}}\right )} - \frac{3}{2} \, B c^{2} d i^{3} n{\left (\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 c^{3} i^{3} n{\left (\frac{a \log \left (b x + a\right )}{b} - \frac{c \log \left (d x + c\right )}{d}\right )} + B c^{3} i^{3} x \log \left (e{\left (\frac{b x}{d x + c} + \frac{a}{d x + c}\right )}^{n}\right ) + A c^{3} i^{3} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*B*d^3*i^3*x^4*log(e*(b*x/(d*x + c) + a/(d*x + c))^n) + 1/4*A*d^3*i^3*x^4 + B*c*d^2*i^3*x^3*log(e*(b*x/(d*x
+ c) + a/(d*x + c))^n) + A*c*d^2*i^3*x^3 + 3/2*B*c^2*d*i^3*x^2*log(e*(b*x/(d*x + c) + a/(d*x + c))^n) + 3/2*A
*c^2*d*i^3*x^2 - 1/24*B*d^3*i^3*n*(6*a^4*log(b*x + a)/b^4 - 6*c^4*log(d*x + c)/d^4 + (2*(b^3*c*d^2 - a*b^2*d^3
)*x^3 - 3*(b^3*c^2*d - a^2*b*d^3)*x^2 + 6*(b^3*c^3 - a^3*d^3)*x)/(b^3*d^3)) + 1/2*B*c*d^2*i^3*n*(2*a^3*log(b*x
+ a)/b^3 - 2*c^3*log(d*x + c)/d^3 - ((b^2*c*d - a*b*d^2)*x^2 - 2*(b^2*c^2 - a^2*d^2)*x)/(b^2*d^2)) - 3/2*B*c^
2*d*i^3*n*(a^2*log(b*x + a)/b^2 - c^2*log(d*x + c)/d^2 + (b*c - a*d)*x/(b*d)) + B*c^3*i^3*n*(a*log(b*x + a)/b
- c*log(d*x + c)/d) + B*c^3*i^3*x*log(e*(b*x/(d*x + c) + a/(d*x + c))^n) + A*c^3*i^3*x

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Fricas [B]  time = 0.635179, size = 883, normalized size = 5.66 \begin{align*} \frac{6 \, A b^{4} d^{4} i^{3} x^{4} - 6 \, B b^{4} c^{4} i^{3} n \log \left (d x + c\right ) + 6 \,{\left (4 \, B a b^{3} c^{3} d - 6 \, B a^{2} b^{2} c^{2} d^{2} + 4 \, B a^{3} b c d^{3} - B a^{4} d^{4}\right )} i^{3} n \log \left (b x + a\right ) + 2 \,{\left (12 \, A b^{4} c d^{3} i^{3} -{\left (B b^{4} c d^{3} - B a b^{3} d^{4}\right )} i^{3} n\right )} x^{3} + 3 \,{\left (12 \, A b^{4} c^{2} d^{2} i^{3} -{\left (3 \, B b^{4} c^{2} d^{2} - 4 \, B a b^{3} c d^{3} + B a^{2} b^{2} d^{4}\right )} i^{3} n\right )} x^{2} + 6 \,{\left (4 \, A b^{4} c^{3} d i^{3} -{\left (3 \, B b^{4} c^{3} d - 6 \, B a b^{3} c^{2} d^{2} + 4 \, B a^{2} b^{2} c d^{3} - B a^{3} b d^{4}\right )} i^{3} n\right )} x + 6 \,{\left (B b^{4} d^{4} i^{3} x^{4} + 4 \, B b^{4} c d^{3} i^{3} x^{3} + 6 \, B b^{4} c^{2} d^{2} i^{3} x^{2} + 4 \, B b^{4} c^{3} d i^{3} x\right )} \log \left (e\right ) + 6 \,{\left (B b^{4} d^{4} i^{3} n x^{4} + 4 \, B b^{4} c d^{3} i^{3} n x^{3} + 6 \, B b^{4} c^{2} d^{2} i^{3} n x^{2} + 4 \, B b^{4} c^{3} d i^{3} n x\right )} \log \left (\frac{b x + a}{d x + c}\right )}{24 \, b^{4} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(6*A*b^4*d^4*i^3*x^4 - 6*B*b^4*c^4*i^3*n*log(d*x + c) + 6*(4*B*a*b^3*c^3*d - 6*B*a^2*b^2*c^2*d^2 + 4*B*a^
3*b*c*d^3 - B*a^4*d^4)*i^3*n*log(b*x + a) + 2*(12*A*b^4*c*d^3*i^3 - (B*b^4*c*d^3 - B*a*b^3*d^4)*i^3*n)*x^3 + 3
*(12*A*b^4*c^2*d^2*i^3 - (3*B*b^4*c^2*d^2 - 4*B*a*b^3*c*d^3 + B*a^2*b^2*d^4)*i^3*n)*x^2 + 6*(4*A*b^4*c^3*d*i^3
- (3*B*b^4*c^3*d - 6*B*a*b^3*c^2*d^2 + 4*B*a^2*b^2*c*d^3 - B*a^3*b*d^4)*i^3*n)*x + 6*(B*b^4*d^4*i^3*x^4 + 4*B
*b^4*c*d^3*i^3*x^3 + 6*B*b^4*c^2*d^2*i^3*x^2 + 4*B*b^4*c^3*d*i^3*x)*log(e) + 6*(B*b^4*d^4*i^3*n*x^4 + 4*B*b^4*
c*d^3*i^3*n*x^3 + 6*B*b^4*c^2*d^2*i^3*n*x^2 + 4*B*b^4*c^3*d*i^3*n*x)*log((b*x + a)/(d*x + c)))/(b^4*d)

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

[Out]

Timed out

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Giac [B]  time = 1.36517, size = 680, normalized size = 4.36 \begin{align*} -\frac{1}{4} \,{\left (A d^{3} i + B d^{3} i\right )} x^{4} + \frac{{\left (B b c d^{2} i n - B a d^{3} i n - 12 \, A b c d^{2} i - 12 \, B b c d^{2} i\right )} x^{3}}{12 \, b} - \frac{1}{4} \,{\left (B d^{3} i n x^{4} + 4 \, B c d^{2} i n x^{3} + 6 \, B c^{2} d i n x^{2} + 4 \, B c^{3} i n x\right )} \log \left (\frac{b x + a}{d x + c}\right ) + \frac{{\left (3 \, B b^{2} c^{2} d i n - 4 \, B a b c d^{2} i n + B a^{2} d^{3} i n - 12 \, A b^{2} c^{2} d i - 12 \, B b^{2} c^{2} d i\right )} x^{2}}{8 \, b^{2}} + \frac{{\left (3 \, B b^{3} c^{3} i n - 6 \, B a b^{2} c^{2} d i n + 4 \, B a^{2} b c d^{2} i n - B a^{3} d^{3} i n - 4 \, A b^{3} c^{3} i - 4 \, B b^{3} c^{3} i\right )} x}{4 \, b^{3}} + \frac{{\left (B b^{4} c^{4} i n - 4 \, B a b^{3} c^{3} d i n + 6 \, B a^{2} b^{2} c^{2} d^{2} i n - 4 \, B a^{3} b c d^{3} i n + B a^{4} d^{4} i n\right )} \log \left ({\left | b d x^{2} + b c x + a d x + a c \right |}\right )}{8 \, b^{4} d} - \frac{{\left (B b^{5} c^{5} i n + 3 \, B a b^{4} c^{4} d i n - 10 \, B a^{2} b^{3} c^{3} d^{2} i n + 10 \, B a^{3} b^{2} c^{2} d^{3} i n - 5 \, B a^{4} b c d^{4} i n + B a^{5} d^{5} i n\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 )}{8 \, b^{4} 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)^3*(A+B*log(e*((b*x+a)/(d*x+c))^n)),x, algorithm="giac")

[Out]

-1/4*(A*d^3*i + B*d^3*i)*x^4 + 1/12*(B*b*c*d^2*i*n - B*a*d^3*i*n - 12*A*b*c*d^2*i - 12*B*b*c*d^2*i)*x^3/b - 1/
4*(B*d^3*i*n*x^4 + 4*B*c*d^2*i*n*x^3 + 6*B*c^2*d*i*n*x^2 + 4*B*c^3*i*n*x)*log((b*x + a)/(d*x + c)) + 1/8*(3*B*
b^2*c^2*d*i*n - 4*B*a*b*c*d^2*i*n + B*a^2*d^3*i*n - 12*A*b^2*c^2*d*i - 12*B*b^2*c^2*d*i)*x^2/b^2 + 1/4*(3*B*b^
3*c^3*i*n - 6*B*a*b^2*c^2*d*i*n + 4*B*a^2*b*c*d^2*i*n - B*a^3*d^3*i*n - 4*A*b^3*c^3*i - 4*B*b^3*c^3*i)*x/b^3 +
1/8*(B*b^4*c^4*i*n - 4*B*a*b^3*c^3*d*i*n + 6*B*a^2*b^2*c^2*d^2*i*n - 4*B*a^3*b*c*d^3*i*n + B*a^4*d^4*i*n)*log
(abs(b*d*x^2 + b*c*x + a*d*x + a*c))/(b^4*d) - 1/8*(B*b^5*c^5*i*n + 3*B*a*b^4*c^4*d*i*n - 10*B*a^2*b^3*c^3*d^2
*i*n + 10*B*a^3*b^2*c^2*d^3*i*n - 5*B*a^4*b*c*d^4*i*n + B*a^5*d^5*i*n)*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^4*d*abs(-b*c + a*d))