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

Optimal. Leaf size=84 \[ \frac{(a+b x)^2 \left (B \log \left (e (a+b x)^n (c+d x)^{-n}\right )+A\right )}{2 b}+\frac{B n (b c-a d)^2 \log (c+d x)}{2 b d^2}-\frac{B n x (b c-a d)}{2 d} \]

[Out]

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

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Rubi [A]  time = 0.0908031, antiderivative size = 96, normalized size of antiderivative = 1.14, number of steps used = 5, number of rules used = 3, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.103, Rules used = {6742, 2492, 43} \[ \frac{A (a+b x)^2}{2 b}+\frac{B n (b c-a d)^2 \log (c+d x)}{2 b d^2}+\frac{B (a+b x)^2 \log \left (e (a+b x)^n (c+d x)^{-n}\right )}{2 b}-\frac{B n x (b c-a d)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rule 2492

Int[Log[(e_.)*((f_.)*((a_.) + (b_.)*(x_))^(p_.)*((c_.) + (d_.)*(x_))^(q_.))^(r_.)]^(s_.)*((g_.) + (h_.)*(x_))^
(m_.), x_Symbol] :> Simp[((g + h*x)^(m + 1)*Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r]^s)/(h*(m + 1)), x] - Dist[(p*
r*s*(b*c - a*d))/(h*(m + 1)), Int[((g + h*x)^(m + 1)*Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r]^(s - 1))/((a + b*x)*
(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, p, q, r, s}, x] && NeQ[b*c - a*d, 0] && EqQ[p + q, 0]
&& IGtQ[s, 0] && NeQ[m, -1]

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

Mathematica [A]  time = 0.146117, size = 126, normalized size = 1.5 \[ \frac{d \left (B d \left (2 a^2+2 a b x+b^2 x^2\right ) \log \left (e (a+b x)^n (c+d x)^{-n}\right )+b x (2 a A d+a B d n+A b d x-b B c n)\right )+B n \left (2 a^2 d^2-2 a b c d+b^2 c^2\right ) \log (c+d x)-a^2 B d^2 n \log (a+b x)}{2 b d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.51, size = 817, normalized size = 9.7 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*A*b*x^2+A*a*x+1/2*B*a^2*n/b*ln(-b*x-a)+1/2*b*B*x^2*ln((b*x+a)^n)+1/2*B*ln(e)*b*x^2+B*ln(e)*a*x-1/2*B*x*(b*
x+2*a)*ln((d*x+c)^n)+1/2*B*n*a*x+B*a*x*ln((b*x+a)^n)-1/2*I*B*Pi*a*x*csgn(I*e)*csgn(I*(b*x+a)^n/((d*x+c)^n))*cs
gn(I*e/((d*x+c)^n)*(b*x+a)^n)-1/2*I*B*Pi*a*x*csgn(I*(b*x+a)^n)*csgn(I/((d*x+c)^n))*csgn(I*(b*x+a)^n/((d*x+c)^n
))+1/4*I*b*B*Pi*x^2*csgn(I*e)*csgn(I*e/((d*x+c)^n)*(b*x+a)^n)^2+1/4*I*b*B*Pi*x^2*csgn(I*(b*x+a)^n/((d*x+c)^n))
*csgn(I*e/((d*x+c)^n)*(b*x+a)^n)^2+1/4*I*b*B*Pi*x^2*csgn(I*(b*x+a)^n)*csgn(I*(b*x+a)^n/((d*x+c)^n))^2+1/4*I*b*
B*Pi*x^2*csgn(I/((d*x+c)^n))*csgn(I*(b*x+a)^n/((d*x+c)^n))^2+1/2*I*B*Pi*a*x*csgn(I*e)*csgn(I*e/((d*x+c)^n)*(b*
x+a)^n)^2+1/2*I*B*Pi*a*x*csgn(I*(b*x+a)^n/((d*x+c)^n))*csgn(I*e/((d*x+c)^n)*(b*x+a)^n)^2+1/2*I*B*Pi*a*x*csgn(I
*(b*x+a)^n)*csgn(I*(b*x+a)^n/((d*x+c)^n))^2+1/2*I*B*Pi*a*x*csgn(I/((d*x+c)^n))*csgn(I*(b*x+a)^n/((d*x+c)^n))^2
-1/2*b/d*B*c*n*x-1/d*B*ln(d*x+c)*a*c*n+1/2*b/d^2*B*ln(d*x+c)*c^2*n-1/4*I*b*B*Pi*x^2*csgn(I*e/((d*x+c)^n)*(b*x+
a)^n)^3-1/4*I*b*B*Pi*x^2*csgn(I*(b*x+a)^n/((d*x+c)^n))^3-1/2*I*B*Pi*a*x*csgn(I*e/((d*x+c)^n)*(b*x+a)^n)^3-1/2*
I*B*Pi*a*x*csgn(I*(b*x+a)^n/((d*x+c)^n))^3-1/4*I*b*B*Pi*x^2*csgn(I*e)*csgn(I*(b*x+a)^n/((d*x+c)^n))*csgn(I*e/(
(d*x+c)^n)*(b*x+a)^n)-1/4*I*b*B*Pi*x^2*csgn(I*(b*x+a)^n)*csgn(I/((d*x+c)^n))*csgn(I*(b*x+a)^n/((d*x+c)^n))

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Maxima [A]  time = 1.15689, size = 208, normalized size = 2.48 \begin{align*} \frac{1}{2} \, B b x^{2} \log \left (\frac{{\left (b x + a\right )}^{n} e}{{\left (d x + c\right )}^{n}}\right ) + \frac{1}{2} \, A b x^{2} + B a x \log \left (\frac{{\left (b x + a\right )}^{n} e}{{\left (d x + c\right )}^{n}}\right ) + A a x + \frac{{\left (\frac{a e n \log \left (b x + a\right )}{b} - \frac{c e n \log \left (d x + c\right )}{d}\right )} B a}{e} - \frac{{\left (\frac{a^{2} e n \log \left (b x + a\right )}{b^{2}} - \frac{c^{2} e n \log \left (d x + c\right )}{d^{2}} + \frac{{\left (b c e n - a d e n\right )} x}{b d}\right )} B b}{2 \, e} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.07428, size = 354, normalized size = 4.21 \begin{align*} \frac{A b^{2} d^{2} x^{2} +{\left (2 \, A a b d^{2} -{\left (B b^{2} c d - B a b d^{2}\right )} n\right )} x +{\left (B b^{2} d^{2} n x^{2} + 2 \, B a b d^{2} n x + B a^{2} d^{2} n\right )} \log \left (b x + a\right ) -{\left (B b^{2} d^{2} n x^{2} + 2 \, B a b d^{2} n x -{\left (B b^{2} c^{2} - 2 \, B a b c d\right )} n\right )} \log \left (d x + c\right ) +{\left (B b^{2} d^{2} x^{2} + 2 \, B a b d^{2} x\right )} \log \left (e\right )}{2 \, b d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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Giac [A]  time = 1.70434, size = 171, normalized size = 2.04 \begin{align*} \frac{B a^{2} n \log \left (b x + a\right )}{2 \, b} + \frac{1}{2} \,{\left (A b + B b\right )} x^{2} + \frac{1}{2} \,{\left (B b n x^{2} + 2 \, B a n x\right )} \log \left (b x + a\right ) - \frac{1}{2} \,{\left (B b n x^{2} + 2 \, B a n x\right )} \log \left (d x + c\right ) - \frac{{\left (B b c n - B a d n - 2 \, A a d - 2 \, B a d\right )} x}{2 \, d} + \frac{{\left (B b c^{2} n - 2 \, B a c d n\right )} \log \left (d x + c\right )}{2 \, d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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