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

Optimal. Leaf size=124 \[ \frac {g^2 (a+b x)^3 \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{3 b}-\frac {B g^2 n (b c-a d)^3 \log (c+d x)}{3 b d^3}+\frac {B g^2 n x (b c-a d)^2}{3 d^2}-\frac {B g^2 n (a+b x)^2 (b c-a d)}{6 b d} \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 124, normalized size of antiderivative = 1.00, 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} \[ \frac {g^2 (a+b x)^3 \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{3 b}+\frac {B g^2 n x (b c-a d)^2}{3 d^2}-\frac {B g^2 n (b c-a d)^3 \log (c+d x)}{3 b d^3}-\frac {B g^2 n (a+b x)^2 (b c-a d)}{6 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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])

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]

Rubi steps

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

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Mathematica [A]  time = 0.06, size = 103, normalized size = 0.83 \[ \frac {g^2 \left (\frac {B n (a d-b c) \left (d \left (a^2 d+4 a b d x+b^2 x (d x-2 c)\right )+2 (b c-a d)^2 \log (c+d x)\right )}{2 d^3}+(a+b x)^3 \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )\right )}{3 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.77, size = 296, normalized size = 2.39 \[ \frac {2 \, A b^{3} d^{3} g^{2} x^{3} + 2 \, B a^{3} d^{3} g^{2} n \log \left (b x + a\right ) - 2 \, {\left (B b^{3} c^{3} - 3 \, B a b^{2} c^{2} d + 3 \, B a^{2} b c d^{2}\right )} g^{2} n \log \left (d x + c\right ) + {\left (6 \, A a b^{2} d^{3} g^{2} - {\left (B b^{3} c d^{2} - B a b^{2} d^{3}\right )} g^{2} n\right )} x^{2} + 2 \, {\left (3 \, A a^{2} b d^{3} g^{2} + {\left (B b^{3} c^{2} d - 3 \, B a b^{2} c d^{2} + 2 \, B a^{2} b d^{3}\right )} g^{2} n\right )} x + 2 \, {\left (B b^{3} d^{3} g^{2} x^{3} + 3 \, B a b^{2} d^{3} g^{2} x^{2} + 3 \, B a^{2} b d^{3} g^{2} x\right )} \log \relax (e) + 2 \, {\left (B b^{3} d^{3} g^{2} n x^{3} + 3 \, B a b^{2} d^{3} g^{2} n x^{2} + 3 \, B a^{2} b d^{3} g^{2} n x\right )} \log \left (\frac {b x + a}{d x + c}\right )}{6 \, b d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 2.29, size = 1836, normalized size = 14.81 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(2*(B*b^6*c^4*g^2*n - 4*B*a*b^5*c^3*d*g^2*n - 3*(b*x + a)*B*b^5*c^4*d*g^2*n/(d*x + c) + 6*B*a^2*b^4*c^2*d^
2*g^2*n + 12*(b*x + a)*B*a*b^4*c^3*d^2*g^2*n/(d*x + c) + 3*(b*x + a)^2*B*b^4*c^4*d^2*g^2*n/(d*x + c)^2 - 4*B*a
^3*b^3*c*d^3*g^2*n - 18*(b*x + a)*B*a^2*b^3*c^2*d^3*g^2*n/(d*x + c) - 12*(b*x + a)^2*B*a*b^3*c^3*d^3*g^2*n/(d*
x + c)^2 + B*a^4*b^2*d^4*g^2*n + 12*(b*x + a)*B*a^3*b^2*c*d^4*g^2*n/(d*x + c) + 18*(b*x + a)^2*B*a^2*b^2*c^2*d
^4*g^2*n/(d*x + c)^2 - 3*(b*x + a)*B*a^4*b*d^5*g^2*n/(d*x + c) - 12*(b*x + a)^2*B*a^3*b*c*d^5*g^2*n/(d*x + c)^
2 + 3*(b*x + a)^2*B*a^4*d^6*g^2*n/(d*x + c)^2)*log((b*x + a)/(d*x + c))/(b^3*d^3 - 3*(b*x + a)*b^2*d^4/(d*x +
c) + 3*(b*x + a)^2*b*d^5/(d*x + c)^2 - (b*x + a)^3*d^6/(d*x + c)^3) + (3*B*b^6*c^4*g^2*n - 12*B*a*b^5*c^3*d*g^
2*n - 7*(b*x + a)*B*b^5*c^4*d*g^2*n/(d*x + c) + 18*B*a^2*b^4*c^2*d^2*g^2*n + 28*(b*x + a)*B*a*b^4*c^3*d^2*g^2*
n/(d*x + c) + 4*(b*x + a)^2*B*b^4*c^4*d^2*g^2*n/(d*x + c)^2 - 12*B*a^3*b^3*c*d^3*g^2*n - 42*(b*x + a)*B*a^2*b^
3*c^2*d^3*g^2*n/(d*x + c) - 16*(b*x + a)^2*B*a*b^3*c^3*d^3*g^2*n/(d*x + c)^2 + 3*B*a^4*b^2*d^4*g^2*n + 28*(b*x
+ a)*B*a^3*b^2*c*d^4*g^2*n/(d*x + c) + 24*(b*x + a)^2*B*a^2*b^2*c^2*d^4*g^2*n/(d*x + c)^2 - 7*(b*x + a)*B*a^4
*b*d^5*g^2*n/(d*x + c) - 16*(b*x + a)^2*B*a^3*b*c*d^5*g^2*n/(d*x + c)^2 + 4*(b*x + a)^2*B*a^4*d^6*g^2*n/(d*x +
c)^2 + 2*A*b^6*c^4*g^2 + 2*B*b^6*c^4*g^2 - 8*A*a*b^5*c^3*d*g^2 - 8*B*a*b^5*c^3*d*g^2 - 6*(b*x + a)*A*b^5*c^4*
d*g^2/(d*x + c) - 6*(b*x + a)*B*b^5*c^4*d*g^2/(d*x + c) + 12*A*a^2*b^4*c^2*d^2*g^2 + 12*B*a^2*b^4*c^2*d^2*g^2
+ 24*(b*x + a)*A*a*b^4*c^3*d^2*g^2/(d*x + c) + 24*(b*x + a)*B*a*b^4*c^3*d^2*g^2/(d*x + c) + 6*(b*x + a)^2*A*b^
4*c^4*d^2*g^2/(d*x + c)^2 + 6*(b*x + a)^2*B*b^4*c^4*d^2*g^2/(d*x + c)^2 - 8*A*a^3*b^3*c*d^3*g^2 - 8*B*a^3*b^3*
c*d^3*g^2 - 36*(b*x + a)*A*a^2*b^3*c^2*d^3*g^2/(d*x + c) - 36*(b*x + a)*B*a^2*b^3*c^2*d^3*g^2/(d*x + c) - 24*(
b*x + a)^2*A*a*b^3*c^3*d^3*g^2/(d*x + c)^2 - 24*(b*x + a)^2*B*a*b^3*c^3*d^3*g^2/(d*x + c)^2 + 2*A*a^4*b^2*d^4*
g^2 + 2*B*a^4*b^2*d^4*g^2 + 24*(b*x + a)*A*a^3*b^2*c*d^4*g^2/(d*x + c) + 24*(b*x + a)*B*a^3*b^2*c*d^4*g^2/(d*x
+ c) + 36*(b*x + a)^2*A*a^2*b^2*c^2*d^4*g^2/(d*x + c)^2 + 36*(b*x + a)^2*B*a^2*b^2*c^2*d^4*g^2/(d*x + c)^2 -
6*(b*x + a)*A*a^4*b*d^5*g^2/(d*x + c) - 6*(b*x + a)*B*a^4*b*d^5*g^2/(d*x + c) - 24*(b*x + a)^2*A*a^3*b*c*d^5*g
^2/(d*x + c)^2 - 24*(b*x + a)^2*B*a^3*b*c*d^5*g^2/(d*x + c)^2 + 6*(b*x + a)^2*A*a^4*d^6*g^2/(d*x + c)^2 + 6*(b
*x + a)^2*B*a^4*d^6*g^2/(d*x + c)^2)/(b^3*d^3 - 3*(b*x + a)*b^2*d^4/(d*x + c) + 3*(b*x + a)^2*b*d^5/(d*x + c)^
2 - (b*x + a)^3*d^6/(d*x + c)^3) + 2*(B*b^4*c^4*g^2*n - 4*B*a*b^3*c^3*d*g^2*n + 6*B*a^2*b^2*c^2*d^2*g^2*n - 4*
B*a^3*b*c*d^3*g^2*n + B*a^4*d^4*g^2*n)*log(b - (b*x + a)*d/(d*x + c))/(b*d^3) - 2*(B*b^4*c^4*g^2*n - 4*B*a*b^3
*c^3*d*g^2*n + 6*B*a^2*b^2*c^2*d^2*g^2*n - 4*B*a^3*b*c*d^3*g^2*n + B*a^4*d^4*g^2*n)*log((b*x + a)/(d*x + c))/(
b*d^3))*(b*c/(b*c - a*d)^2 - a*d/(b*c - a*d)^2)

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maple [F]  time = 0.27, size = 0, normalized size = 0.00 \[ \int \left (b g x +a g \right )^{2} \left (B \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right )+A \right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 1.39, size = 309, normalized size = 2.49 \[ \frac {1}{3} \, B b^{2} g^{2} x^{3} \log \left (e {\left (\frac {b x}{d x + c} + \frac {a}{d x + c}\right )}^{n}\right ) + \frac {1}{3} \, A b^{2} g^{2} x^{3} + B a b g^{2} x^{2} \log \left (e {\left (\frac {b x}{d x + c} + \frac {a}{d x + c}\right )}^{n}\right ) + A a b g^{2} x^{2} + \frac {1}{6} \, B b^{2} g^{2} n {\left (\frac {2 \, a^{3} \log \left (b x + a\right )}{b^{3}} - \frac {2 \, c^{3} \log \left (d x + c\right )}{d^{3}} - \frac {{\left (b^{2} c d - a b d^{2}\right )} x^{2} - 2 \, {\left (b^{2} c^{2} - a^{2} d^{2}\right )} x}{b^{2} d^{2}}\right )} - B a b g^{2} n {\left (\frac {a^{2} \log \left (b x + a\right )}{b^{2}} - \frac {c^{2} \log \left (d x + c\right )}{d^{2}} + \frac {{\left (b c - a d\right )} x}{b d}\right )} + B a^{2} g^{2} n {\left (\frac {a \log \left (b x + a\right )}{b} - \frac {c \log \left (d x + c\right )}{d}\right )} + B a^{2} g^{2} x \log \left (e {\left (\frac {b x}{d x + c} + \frac {a}{d x + c}\right )}^{n}\right ) + A a^{2} g^{2} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 4.28, size = 303, normalized size = 2.44 \[ \ln \left (e\,{\left (\frac {a+b\,x}{c+d\,x}\right )}^n\right )\,\left (B\,a^2\,g^2\,x+B\,a\,b\,g^2\,x^2+\frac {B\,b^2\,g^2\,x^3}{3}\right )-x\,\left (\frac {\left (3\,a\,d+3\,b\,c\right )\,\left (\frac {b\,g^2\,\left (9\,A\,a\,d+3\,A\,b\,c+B\,a\,d\,n-B\,b\,c\,n\right )}{3\,d}-\frac {A\,b\,g^2\,\left (3\,a\,d+3\,b\,c\right )}{3\,d}\right )}{3\,b\,d}-\frac {a\,g^2\,\left (3\,A\,a\,d+3\,A\,b\,c+B\,a\,d\,n-B\,b\,c\,n\right )}{d}+\frac {A\,a\,b\,c\,g^2}{d}\right )+x^2\,\left (\frac {b\,g^2\,\left (9\,A\,a\,d+3\,A\,b\,c+B\,a\,d\,n-B\,b\,c\,n\right )}{6\,d}-\frac {A\,b\,g^2\,\left (3\,a\,d+3\,b\,c\right )}{6\,d}\right )-\frac {\ln \left (c+d\,x\right )\,\left (3\,B\,n\,a^2\,c\,d^2\,g^2-3\,B\,n\,a\,b\,c^2\,d\,g^2+B\,n\,b^2\,c^3\,g^2\right )}{3\,d^3}+\frac {A\,b^2\,g^2\,x^3}{3}+\frac {B\,a^3\,g^2\,n\,\ln \left (a+b\,x\right )}{3\,b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 60.49, size = 673, normalized size = 5.43 \[ \begin {cases} a^{2} g^{2} x \left (A + B \log {\left (e \left (\frac {a}{c}\right )^{n} \right )}\right ) & \text {for}\: b = 0 \wedge d = 0 \\A a^{2} g^{2} x + A a b g^{2} x^{2} + \frac {A b^{2} g^{2} x^{3}}{3} + \frac {B a^{3} g^{2} n \log {\left (\frac {a}{c} + \frac {b x}{c} \right )}}{3 b} + B a^{2} g^{2} n x \log {\left (\frac {a}{c} + \frac {b x}{c} \right )} - \frac {B a^{2} g^{2} n x}{3} + B a^{2} g^{2} x \log {\relax (e )} + B a b g^{2} n x^{2} \log {\left (\frac {a}{c} + \frac {b x}{c} \right )} - \frac {B a b g^{2} n x^{2}}{3} + B a b g^{2} x^{2} \log {\relax (e )} + \frac {B b^{2} g^{2} n x^{3} \log {\left (\frac {a}{c} + \frac {b x}{c} \right )}}{3} - \frac {B b^{2} g^{2} n x^{3}}{9} + \frac {B b^{2} g^{2} x^{3} \log {\relax (e )}}{3} & \text {for}\: d = 0 \\a^{2} g^{2} \left (A x - \frac {B c n \log {\left (c + d x \right )}}{d} + B n x \log {\relax (a )} - B n x \log {\left (c + d x \right )} + B n x + B x \log {\relax (e )}\right ) & \text {for}\: b = 0 \\A a^{2} g^{2} x + A a b g^{2} x^{2} + \frac {A b^{2} g^{2} x^{3}}{3} + \frac {B a^{3} g^{2} n \log {\left (\frac {a}{c + d x} + \frac {b x}{c + d x} \right )}}{3 b} + \frac {B a^{3} g^{2} n \log {\left (\frac {c}{d} + x \right )}}{3 b} - \frac {B a^{2} c g^{2} n \log {\left (\frac {c}{d} + x \right )}}{d} + B a^{2} g^{2} n x \log {\left (\frac {a}{c + d x} + \frac {b x}{c + d x} \right )} + \frac {2 B a^{2} g^{2} n x}{3} + B a^{2} g^{2} x \log {\relax (e )} + \frac {B a b c^{2} g^{2} n \log {\left (\frac {c}{d} + x \right )}}{d^{2}} - \frac {B a b c g^{2} n x}{d} + B a b g^{2} n x^{2} \log {\left (\frac {a}{c + d x} + \frac {b x}{c + d x} \right )} + \frac {B a b g^{2} n x^{2}}{6} + B a b g^{2} x^{2} \log {\relax (e )} - \frac {B b^{2} c^{3} g^{2} n \log {\left (\frac {c}{d} + x \right )}}{3 d^{3}} + \frac {B b^{2} c^{2} g^{2} n x}{3 d^{2}} - \frac {B b^{2} c g^{2} n x^{2}}{6 d} + \frac {B b^{2} g^{2} n x^{3} \log {\left (\frac {a}{c + d x} + \frac {b x}{c + d x} \right )}}{3} + \frac {B b^{2} g^{2} x^{3} \log {\relax (e )}}{3} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a**2*g**2*x*(A + B*log(e*(a/c)**n)), Eq(b, 0) & Eq(d, 0)), (A*a**2*g**2*x + A*a*b*g**2*x**2 + A*b**
2*g**2*x**3/3 + B*a**3*g**2*n*log(a/c + b*x/c)/(3*b) + B*a**2*g**2*n*x*log(a/c + b*x/c) - B*a**2*g**2*n*x/3 +
B*a**2*g**2*x*log(e) + B*a*b*g**2*n*x**2*log(a/c + b*x/c) - B*a*b*g**2*n*x**2/3 + B*a*b*g**2*x**2*log(e) + B*b
**2*g**2*n*x**3*log(a/c + b*x/c)/3 - B*b**2*g**2*n*x**3/9 + B*b**2*g**2*x**3*log(e)/3, Eq(d, 0)), (a**2*g**2*(
A*x - B*c*n*log(c + d*x)/d + B*n*x*log(a) - B*n*x*log(c + d*x) + B*n*x + B*x*log(e)), Eq(b, 0)), (A*a**2*g**2*
x + A*a*b*g**2*x**2 + A*b**2*g**2*x**3/3 + B*a**3*g**2*n*log(a/(c + d*x) + b*x/(c + d*x))/(3*b) + B*a**3*g**2*
n*log(c/d + x)/(3*b) - B*a**2*c*g**2*n*log(c/d + x)/d + B*a**2*g**2*n*x*log(a/(c + d*x) + b*x/(c + d*x)) + 2*B
*a**2*g**2*n*x/3 + B*a**2*g**2*x*log(e) + B*a*b*c**2*g**2*n*log(c/d + x)/d**2 - B*a*b*c*g**2*n*x/d + B*a*b*g**
2*n*x**2*log(a/(c + d*x) + b*x/(c + d*x)) + B*a*b*g**2*n*x**2/6 + B*a*b*g**2*x**2*log(e) - B*b**2*c**3*g**2*n*
log(c/d + x)/(3*d**3) + B*b**2*c**2*g**2*n*x/(3*d**2) - B*b**2*c*g**2*n*x**2/(6*d) + B*b**2*g**2*n*x**3*log(a/
(c + d*x) + b*x/(c + d*x))/3 + B*b**2*g**2*x**3*log(e)/3, True))

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