3.149 \(\int (a+b x)^2 (A+B \log (e (a+b x)^n (c+d x)^{-n})) \, dx\)

Optimal. Leaf size=113 \[ \frac {(a+b x)^3 \left (B \log \left (e (a+b x)^n (c+d x)^{-n}\right )+A\right )}{3 b}-\frac {B n (b c-a d)^3 \log (c+d x)}{3 b d^3}+\frac {B n x (b c-a d)^2}{3 d^2}-\frac {B n (a+b x)^2 (b c-a d)}{6 b d} \]

[Out]

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

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Rubi [A]  time = 0.12, antiderivative size = 125, normalized size of antiderivative = 1.11, number of steps used = 5, number of rules used = 3, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {6742, 2492, 43} \[ \frac {A (a+b x)^3}{3 b}+\frac {B n x (b c-a d)^2}{3 d^2}-\frac {B n (b c-a d)^3 \log (c+d x)}{3 b d^3}+\frac {B (a+b x)^3 \log \left (e (a+b x)^n (c+d x)^{-n}\right )}{3 b}-\frac {B n (a+b x)^2 (b c-a d)}{6 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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])

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 6742

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

Rubi steps

\begin {align*} \int (a+b x)^2 \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right ) \, dx &=\int \left (A (a+b x)^2+B (a+b x)^2 \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right ) \, dx\\ &=\frac {A (a+b x)^3}{3 b}+B \int (a+b x)^2 \log \left (e (a+b x)^n (c+d x)^{-n}\right ) \, dx\\ &=\frac {A (a+b x)^3}{3 b}+\frac {B (a+b x)^3 \log \left (e (a+b x)^n (c+d x)^{-n}\right )}{3 b}-\frac {(B (b c-a d) n) \int \frac {(a+b x)^2}{c+d x} \, dx}{3 b}\\ &=\frac {A (a+b x)^3}{3 b}+\frac {B (a+b x)^3 \log \left (e (a+b x)^n (c+d x)^{-n}\right )}{3 b}-\frac {(B (b c-a d) n) \int \left (-\frac {b (b c-a d)}{d^2}+\frac {b (a+b x)}{d}+\frac {(-b c+a d)^2}{d^2 (c+d x)}\right ) \, dx}{3 b}\\ &=\frac {B (b c-a d)^2 n x}{3 d^2}-\frac {B (b c-a d) n (a+b x)^2}{6 b d}+\frac {A (a+b x)^3}{3 b}-\frac {B (b c-a d)^3 n \log (c+d x)}{3 b d^3}+\frac {B (a+b x)^3 \log \left (e (a+b x)^n (c+d x)^{-n}\right )}{3 b}\\ \end {align*}

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Mathematica [A]  time = 0.29, size = 194, normalized size = 1.72 \[ \frac {-4 a^3 B d^3 n \log (a+b x)+b d x \left (2 a^2 d^2 (3 A+2 B n)+a b d (6 A d x-6 B c n+B d n x)+b^2 \left (2 A d^2 x^2+B c n (2 c-d x)\right )\right )-2 B n \left (-3 a^3 d^3+3 a^2 b c d^2-3 a b^2 c^2 d+b^3 c^3\right ) \log (c+d x)+2 B d^3 \left (3 a^3+3 a^2 b x+3 a b^2 x^2+b^3 x^3\right ) \log \left (e (a+b x)^n (c+d x)^{-n}\right )}{6 b d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.96, size = 282, normalized size = 2.50 \[ \frac {2 \, A b^{3} d^{3} x^{3} + {\left (6 \, A a b^{2} d^{3} - {\left (B b^{3} c d^{2} - B a b^{2} d^{3}\right )} n\right )} x^{2} + 2 \, {\left (3 \, A a^{2} b d^{3} + {\left (B b^{3} c^{2} d - 3 \, B a b^{2} c d^{2} + 2 \, B a^{2} b d^{3}\right )} n\right )} x + 2 \, {\left (B b^{3} d^{3} n x^{3} + 3 \, B a b^{2} d^{3} n x^{2} + 3 \, B a^{2} b d^{3} n x + B a^{3} d^{3} n\right )} \log \left (b x + a\right ) - 2 \, {\left (B b^{3} d^{3} n x^{3} + 3 \, B a b^{2} d^{3} n x^{2} + 3 \, B a^{2} b d^{3} n x + {\left (B b^{3} c^{3} - 3 \, B a b^{2} c^{2} d + 3 \, B a^{2} b c d^{2}\right )} n\right )} \log \left (d x + c\right ) + 2 \, {\left (B b^{3} d^{3} x^{3} + 3 \, B a b^{2} d^{3} x^{2} + 3 \, B a^{2} b d^{3} x\right )} \log \relax (e)}{6 \, b d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^2*(A+B*log(e*(b*x+a)^n/((d*x+c)^n))),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 1.38, size = 235, normalized size = 2.08 \[ \frac {B a^{3} n \log \left (b x + a\right )}{3 \, b} + \frac {1}{3} \, {\left (A b^{2} + B b^{2}\right )} x^{3} - \frac {{\left (B b^{2} c n - B a b d n - 6 \, A a b d - 6 \, B a b d\right )} x^{2}}{6 \, d} + \frac {1}{3} \, {\left (B b^{2} n x^{3} + 3 \, B a b n x^{2} + 3 \, B a^{2} n x\right )} \log \left (b x + a\right ) - \frac {1}{3} \, {\left (B b^{2} n x^{3} + 3 \, B a b n x^{2} + 3 \, B a^{2} n x\right )} \log \left (d x + c\right ) + \frac {{\left (B b^{2} c^{2} n - 3 \, B a b c d n + 2 \, B a^{2} d^{2} n + 3 \, A a^{2} d^{2} + 3 \, B a^{2} d^{2}\right )} x}{3 \, d^{2}} - \frac {{\left (B b^{2} c^{3} n - 3 \, B a b c^{2} d n + 3 \, B a^{2} c d^{2} n\right )} \log \left (-d x - c\right )}{3 \, d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^2*(A+B*log(e*(b*x+a)^n/((d*x+c)^n))),x, algorithm="giac")

[Out]

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

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maple [C]  time = 0.47, size = 1325, normalized size = 11.73 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^2*(A+B*ln(e*(b*x+a)^n/((d*x+c)^n))),x)

[Out]

-1/3*(b*x+a)^3*B/b*ln((d*x+c)^n)+1/3*b^2*A*x^3+B*ln(e)*a^2*x+1/3*b^2*B*x^3*ln((b*x+a)^n)+1/3*b^2*B*ln(e)*x^3+l
n((b*x+a)^n)*x*B*a^2+1/3*B*a^3*n/b*ln(-b*x-a)-1/2*I*b*B*Pi*a*x^2*csgn(I*(b*x+a)^n)*csgn(I/((d*x+c)^n))*csgn(I*
(b*x+a)^n/((d*x+c)^n))-1/2*I*b*B*Pi*a*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)-1/2*I*B*Pi*a^2*x*csgn(I*e)*csgn(I*(b*x+a)^n/((d*x+c)^n))*csgn(I*e/((d*x+c)^n)*(b*x+a)^n)-1/2*I*B*Pi*a^2*x*
csgn(I*(b*x+a)^n)*csgn(I/((d*x+c)^n))*csgn(I*(b*x+a)^n/((d*x+c)^n))-1/6*I*b^2*B*Pi*x^3*csgn(I*e)*csgn(I*(b*x+a
)^n/((d*x+c)^n))*csgn(I*e/((d*x+c)^n)*(b*x+a)^n)-1/6*I*b^2*B*Pi*x^3*csgn(I*(b*x+a)^n)*csgn(I/((d*x+c)^n))*csgn
(I*(b*x+a)^n/((d*x+c)^n))+1/2*I*b*B*Pi*a*x^2*csgn(I*(b*x+a)^n)*csgn(I*(b*x+a)^n/((d*x+c)^n))^2+1/2*I*b*B*Pi*a*
x^2*csgn(I/((d*x+c)^n))*csgn(I*(b*x+a)^n/((d*x+c)^n))^2+1/2*I*b*B*Pi*a*x^2*csgn(I*(b*x+a)^n/((d*x+c)^n))*csgn(
I*e/((d*x+c)^n)*(b*x+a)^n)^2+1/2*I*b*B*Pi*a*x^2*csgn(I*e)*csgn(I*e/((d*x+c)^n)*(b*x+a)^n)^2-1/6*I*b^2*B*Pi*x^3
*csgn(I*e/((d*x+c)^n)*(b*x+a)^n)^3-1/2*I*B*Pi*a^2*x*csgn(I*(b*x+a)^n/((d*x+c)^n))^3-1/2*I*B*Pi*a^2*x*csgn(I*e/
((d*x+c)^n)*(b*x+a)^n)^3+2/3*B*a^2*n*x+1/3*b^2/d^2*B*c^2*n*x-1/3*b^2/d^3*B*ln(d*x+c)*c^3*n-1/d*B*ln(d*x+c)*a^2
*c*n-1/6*I*b^2*B*Pi*x^3*csgn(I*(b*x+a)^n/((d*x+c)^n))^3+1/2*I*B*Pi*a^2*x*csgn(I*(b*x+a)^n)*csgn(I*(b*x+a)^n/((
d*x+c)^n))^2+1/2*I*B*Pi*a^2*x*csgn(I/((d*x+c)^n))*csgn(I*(b*x+a)^n/((d*x+c)^n))^2+1/2*I*B*Pi*a^2*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^2*x*csgn(I*e)*csgn(I*e/((d*x+c)^n)*(b*x+a)^n
)^2+1/6*I*b^2*B*Pi*x^3*csgn(I*(b*x+a)^n)*csgn(I*(b*x+a)^n/((d*x+c)^n))^2+1/6*I*b^2*B*Pi*x^3*csgn(I/((d*x+c)^n)
)*csgn(I*(b*x+a)^n/((d*x+c)^n))^2+1/6*I*b^2*B*Pi*x^3*csgn(I*(b*x+a)^n/((d*x+c)^n))*csgn(I*e/((d*x+c)^n)*(b*x+a
)^n)^2+1/6*I*b^2*B*Pi*x^3*csgn(I*e)*csgn(I*e/((d*x+c)^n)*(b*x+a)^n)^2-1/2*I*b*B*Pi*a*x^2*csgn(I*(b*x+a)^n/((d*
x+c)^n))^3-1/2*I*b*B*Pi*a*x^2*csgn(I*e/((d*x+c)^n)*(b*x+a)^n)^3+b*A*a*x^2+A*a^2*x+1/3/b*B*ln(d*x+c)*a^3*n+b*B*
a*x^2*ln((b*x+a)^n)+b*B*ln(e)*a*x^2-b/d*B*a*c*n*x+b/d^2*B*ln(d*x+c)*a*c^2*n+1/6*b*B*a*n*x^2-1/6*b^2/d*B*c*n*x^
2

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maxima [B]  time = 1.27, size = 294, normalized size = 2.60 \[ \frac {1}{3} \, B b^{2} x^{3} \log \left (\frac {{\left (b x + a\right )}^{n} e}{{\left (d x + c\right )}^{n}}\right ) + \frac {1}{3} \, A b^{2} x^{3} + B a b x^{2} \log \left (\frac {{\left (b x + a\right )}^{n} e}{{\left (d x + c\right )}^{n}}\right ) + A a b x^{2} + B a^{2} x \log \left (\frac {{\left (b x + a\right )}^{n} e}{{\left (d x + c\right )}^{n}}\right ) + A a^{2} x + \frac {{\left (\frac {a e n \log \left (b x + a\right )}{b} - \frac {c e n \log \left (d x + c\right )}{d}\right )} B a^{2}}{e} - \frac {{\left (\frac {a^{2} e n \log \left (b x + a\right )}{b^{2}} - \frac {c^{2} e n \log \left (d x + c\right )}{d^{2}} + \frac {{\left (b c e n - a d e n\right )} x}{b d}\right )} B a b}{e} + \frac {{\left (\frac {2 \, a^{3} e n \log \left (b x + a\right )}{b^{3}} - \frac {2 \, c^{3} e n \log \left (d x + c\right )}{d^{3}} - \frac {{\left (b^{2} c d e n - a b d^{2} e n\right )} x^{2} - 2 \, {\left (b^{2} c^{2} e n - a^{2} d^{2} e n\right )} x}{b^{2} d^{2}}\right )} B b^{2}}{6 \, e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^2*(A+B*log(e*(b*x+a)^n/((d*x+c)^n))),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 4.24, size = 262, normalized size = 2.32 \[ \ln \left (\frac {e\,{\left (a+b\,x\right )}^n}{{\left (c+d\,x\right )}^n}\right )\,\left (B\,a^2\,x+B\,a\,b\,x^2+\frac {B\,b^2\,x^3}{3}\right )+x^2\,\left (\frac {b\,\left (9\,A\,a\,d+3\,A\,b\,c+B\,a\,d\,n-B\,b\,c\,n\right )}{6\,d}-\frac {A\,b\,\left (3\,a\,d+3\,b\,c\right )}{6\,d}\right )-x\,\left (\frac {\left (\frac {b\,\left (9\,A\,a\,d+3\,A\,b\,c+B\,a\,d\,n-B\,b\,c\,n\right )}{3\,d}-\frac {A\,b\,\left (3\,a\,d+3\,b\,c\right )}{3\,d}\right )\,\left (3\,a\,d+3\,b\,c\right )}{3\,b\,d}-\frac {a\,\left (3\,A\,a\,d+3\,A\,b\,c+B\,a\,d\,n-B\,b\,c\,n\right )}{d}+\frac {A\,a\,b\,c}{d}\right )+\frac {A\,b^2\,x^3}{3}-\frac {\ln \left (c+d\,x\right )\,\left (3\,B\,n\,a^2\,c\,d^2-3\,B\,n\,a\,b\,c^2\,d+B\,n\,b^2\,c^3\right )}{3\,d^3}+\frac {B\,a^3\,n\,\ln \left (a+b\,x\right )}{3\,b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log((e*(a + b*x)^n)/(c + d*x)^n)*((B*b^2*x^3)/3 + B*a^2*x + B*a*b*x^2) + x^2*((b*(9*A*a*d + 3*A*b*c + B*a*d*n
- B*b*c*n))/(6*d) - (A*b*(3*a*d + 3*b*c))/(6*d)) - x*((((b*(9*A*a*d + 3*A*b*c + B*a*d*n - B*b*c*n))/(3*d) - (A
*b*(3*a*d + 3*b*c))/(3*d))*(3*a*d + 3*b*c))/(3*b*d) - (a*(3*A*a*d + 3*A*b*c + B*a*d*n - B*b*c*n))/d + (A*a*b*c
)/d) + (A*b^2*x^3)/3 - (log(c + d*x)*(B*b^2*c^3*n + 3*B*a^2*c*d^2*n - 3*B*a*b*c^2*d*n))/(3*d^3) + (B*a^3*n*log
(a + b*x))/(3*b)

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sympy [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: HeuristicGCDFailed} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**2*(A+B*ln(e*(b*x+a)**n/((d*x+c)**n))),x)

[Out]

Exception raised: HeuristicGCDFailed

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