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

Optimal. Leaf size=86 \[ \frac{i (c+d x)^2 \left (B \log \left (e \left (\frac{a+b x}{c+d x}\right )^n\right )+A\right )}{2 d}-\frac{B i n (b c-a d)^2 \log (a+b x)}{2 b^2 d}-\frac{B i n x (b c-a d)}{2 b} \]

[Out]

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

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Rubi [A]  time = 0.0608629, antiderivative size = 86, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.097, Rules used = {2525, 12, 43} \[ \frac{i (c+d x)^2 \left (B \log \left (e \left (\frac{a+b x}{c+d x}\right )^n\right )+A\right )}{2 d}-\frac{B i n (b c-a d)^2 \log (a+b x)}{2 b^2 d}-\frac{B i n x (b c-a d)}{2 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(B*(b*c - a*d)*i*n*x)/(2*b) - (B*(b*c - a*d)^2*i*n*Log[a + b*x])/(2*b^2*d) + (i*(c + d*x)^2*(A + B*Log[e*((a
+ b*x)/(c + d*x))^n]))/(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 (111 c+111 d x) \left (A+B \log \left (e \left (\frac{a+b x}{c+d x}\right )^n\right )\right ) \, dx &=\frac{111 (c+d x)^2 \left (A+B \log \left (e \left (\frac{a+b x}{c+d x}\right )^n\right )\right )}{2 d}-\frac{(B n) \int \frac{12321 (b c-a d) (c+d x)}{a+b x} \, dx}{222 d}\\ &=\frac{111 (c+d x)^2 \left (A+B \log \left (e \left (\frac{a+b x}{c+d x}\right )^n\right )\right )}{2 d}-\frac{(111 B (b c-a d) n) \int \frac{c+d x}{a+b x} \, dx}{2 d}\\ &=\frac{111 (c+d x)^2 \left (A+B \log \left (e \left (\frac{a+b x}{c+d x}\right )^n\right )\right )}{2 d}-\frac{(111 B (b c-a d) n) \int \left (\frac{d}{b}+\frac{b c-a d}{b (a+b x)}\right ) \, dx}{2 d}\\ &=-\frac{111 B (b c-a d) n x}{2 b}-\frac{111 B (b c-a d)^2 n \log (a+b x)}{2 b^2 d}+\frac{111 (c+d x)^2 \left (A+B \log \left (e \left (\frac{a+b x}{c+d x}\right )^n\right )\right )}{2 d}\\ \end{align*}

Mathematica [A]  time = 0.0388606, size = 74, normalized size = 0.86 \[ \frac{i \left ((c+d x)^2 \left (B \log \left (e \left (\frac{a+b x}{c+d x}\right )^n\right )+A\right )-\frac{B n (b c-a d) ((b c-a d) \log (a+b x)+b d x)}{b^2}\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(i*(-((B*(b*c - a*d)*n*(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
))^n])))/(2*d)

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Maple [F]  time = 0.371, size = 0, normalized size = 0. \begin{align*} \int \left ( dix+ci \right ) \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)),x)

[Out]

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

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Maxima [A]  time = 1.24362, size = 211, normalized size = 2.45 \begin{align*} \frac{1}{2} \, B d i x^{2} \log \left (e{\left (\frac{b x}{d x + c} + \frac{a}{d x + c}\right )}^{n}\right ) + \frac{1}{2} \, A d i x^{2} - \frac{1}{2} \, B d i 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 i n{\left (\frac{a \log \left (b x + a\right )}{b} - \frac{c \log \left (d x + c\right )}{d}\right )} + B c i x \log \left (e{\left (\frac{b x}{d x + c} + \frac{a}{d x + c}\right )}^{n}\right ) + 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))^n)),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 0.5185, size = 360, normalized size = 4.19 \begin{align*} \frac{A b^{2} d^{2} i x^{2} - B b^{2} c^{2} i n \log \left (d x + c\right ) +{\left (2 \, B a b c d - B a^{2} d^{2}\right )} i n \log \left (b x + a\right ) +{\left (2 \, A b^{2} c d i -{\left (B b^{2} c d - B a b d^{2}\right )} i n\right )} x +{\left (B b^{2} d^{2} i x^{2} + 2 \, B b^{2} c d i x\right )} \log \left (e\right ) +{\left (B b^{2} d^{2} i n x^{2} + 2 \, B b^{2} c d i n x\right )} \log \left (\frac{b x + a}{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))^n)),x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [B]  time = 1.47673, size = 339, normalized size = 3.94 \begin{align*} \frac{1}{2} \,{\left (A d i + B d i\right )} x^{2} + \frac{1}{2} \,{\left (B d i n x^{2} + 2 \, B c i n x\right )} \log \left (\frac{b x + a}{d x + c}\right ) - \frac{{\left (B b c i n - B a d i n - 2 \, A b c i - 2 \, B b c i\right )} x}{2 \, b} - \frac{{\left (B b^{2} c^{2} i n - 2 \, B a b c d i n + B a^{2} d^{2} i n\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 n + B a b^{2} c^{2} d i n - 3 \, B a^{2} b c d^{2} i n + B a^{3} d^{3} 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 )}{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))^n)),x, algorithm="giac")

[Out]

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