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

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

[Out]

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

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Rubi [A]  time = 0.07, 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} \[ \frac {i^2 (c+d x)^3 \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{3 d}-\frac {B i^2 n x (b c-a d)^2}{3 b^2}-\frac {B i^2 n (b c-a d)^3 \log (a+b x)}{3 b^3 d}-\frac {B i^2 n (c+d x)^2 (b c-a d)}{6 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 (120 c+120 d x)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \, dx &=\frac {4800 (c+d x)^3 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{d}-\frac {(B n) \int \frac {1728000 (b c-a d) (c+d x)^2}{a+b x} \, dx}{360 d}\\ &=\frac {4800 (c+d x)^3 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{d}-\frac {(4800 B (b c-a d) n) \int \frac {(c+d x)^2}{a+b x} \, dx}{d}\\ &=\frac {4800 (c+d x)^3 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{d}-\frac {(4800 B (b c-a d) n) \int \left (\frac {d (b c-a d)}{b^2}+\frac {(b c-a d)^2}{b^2 (a+b x)}+\frac {d (c+d x)}{b}\right ) \, dx}{d}\\ &=-\frac {4800 B (b c-a d)^2 n x}{b^2}-\frac {2400 B (b c-a d) n (c+d x)^2}{b d}-\frac {4800 B (b c-a d)^3 n \log (a+b x)}{b^3 d}+\frac {4800 (c+d x)^3 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{d}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 101, normalized size = 0.81 \[ \frac {i^2 \left ((c+d x)^3 \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )-\frac {B n (b c-a d) \left (2 b d x (b c-a d)+2 (b c-a d)^2 \log (a+b x)+b^2 (c+d x)^2\right )}{2 b^3}\right )}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(i^2*(-1/2*(B*(b*c - a*d)*n*(2*b*d*(b*c - a*d)*x + b^2*(c + d*x)^2 + 2*(b*c - a*d)^2*Log[a + b*x]))/b^3 + (c +
d*x)^3*(A + B*Log[e*((a + b*x)/(c + d*x))^n])))/(3*d)

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fricas [B]  time = 0.89, size = 297, normalized size = 2.40 \[ \frac {2 \, A b^{3} d^{3} i^{2} x^{3} - 2 \, B b^{3} c^{3} i^{2} n \log \left (d x + c\right ) + 2 \, {\left (3 \, B a b^{2} c^{2} d - 3 \, B a^{2} b c d^{2} + B a^{3} d^{3}\right )} i^{2} n \log \left (b x + a\right ) + {\left (6 \, A b^{3} c d^{2} i^{2} - {\left (B b^{3} c d^{2} - B a b^{2} d^{3}\right )} i^{2} n\right )} x^{2} + 2 \, {\left (3 \, A b^{3} c^{2} d i^{2} - {\left (2 \, B b^{3} c^{2} d - 3 \, B a b^{2} c d^{2} + B a^{2} b d^{3}\right )} i^{2} n\right )} x + 2 \, {\left (B b^{3} d^{3} i^{2} x^{3} + 3 \, B b^{3} c d^{2} i^{2} x^{2} + 3 \, B b^{3} c^{2} d i^{2} x\right )} \log \relax (e) + 2 \, {\left (B b^{3} d^{3} i^{2} n x^{3} + 3 \, B b^{3} c d^{2} i^{2} n x^{2} + 3 \, B b^{3} c^{2} d i^{2} n x\right )} \log \left (\frac {b x + a}{d x + c}\right )}{6 \, b^{3} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 1.60, size = 860, normalized size = 6.94 \[ -\frac {1}{6} \, {\left (\frac {2 \, {\left (B b^{4} c^{4} n - 4 \, B a b^{3} c^{3} d n + 6 \, B a^{2} b^{2} c^{2} d^{2} n - 4 \, B a^{3} b c d^{3} n + B a^{4} d^{4} n\right )} \log \left (\frac {b x + a}{d x + c}\right )}{b^{3} d - \frac {3 \, {\left (b x + a\right )} b^{2} d^{2}}{d x + c} + \frac {3 \, {\left (b x + a\right )}^{2} b d^{3}}{{\left (d x + c\right )}^{2}} - \frac {{\left (b x + a\right )}^{3} d^{4}}{{\left (d x + c\right )}^{3}}} - \frac {3 \, B b^{6} c^{4} n - 12 \, B a b^{5} c^{3} d n - \frac {5 \, {\left (b x + a\right )} B b^{5} c^{4} d n}{d x + c} + 18 \, B a^{2} b^{4} c^{2} d^{2} n + \frac {20 \, {\left (b x + a\right )} B a b^{4} c^{3} d^{2} n}{d x + c} + \frac {2 \, {\left (b x + a\right )}^{2} B b^{4} c^{4} d^{2} n}{{\left (d x + c\right )}^{2}} - 12 \, B a^{3} b^{3} c d^{3} n - \frac {30 \, {\left (b x + a\right )} B a^{2} b^{3} c^{2} d^{3} n}{d x + c} - \frac {8 \, {\left (b x + a\right )}^{2} B a b^{3} c^{3} d^{3} n}{{\left (d x + c\right )}^{2}} + 3 \, B a^{4} b^{2} d^{4} n + \frac {20 \, {\left (b x + a\right )} B a^{3} b^{2} c d^{4} n}{d x + c} + \frac {12 \, {\left (b x + a\right )}^{2} B a^{2} b^{2} c^{2} d^{4} n}{{\left (d x + c\right )}^{2}} - \frac {5 \, {\left (b x + a\right )} B a^{4} b d^{5} n}{d x + c} - \frac {8 \, {\left (b x + a\right )}^{2} B a^{3} b c d^{5} n}{{\left (d x + c\right )}^{2}} + \frac {2 \, {\left (b x + a\right )}^{2} B a^{4} d^{6} n}{{\left (d x + c\right )}^{2}} - 2 \, A b^{6} c^{4} - 2 \, B b^{6} c^{4} + 8 \, A a b^{5} c^{3} d + 8 \, B a b^{5} c^{3} d - 12 \, A a^{2} b^{4} c^{2} d^{2} - 12 \, B a^{2} b^{4} c^{2} d^{2} + 8 \, A a^{3} b^{3} c d^{3} + 8 \, B a^{3} b^{3} c d^{3} - 2 \, A a^{4} b^{2} d^{4} - 2 \, B a^{4} b^{2} d^{4}}{b^{5} d - \frac {3 \, {\left (b x + a\right )} b^{4} d^{2}}{d x + c} + \frac {3 \, {\left (b x + a\right )}^{2} b^{3} d^{3}}{{\left (d x + c\right )}^{2}} - \frac {{\left (b x + a\right )}^{3} b^{2} d^{4}}{{\left (d x + c\right )}^{3}}} + \frac {2 \, {\left (B b^{4} c^{4} n - 4 \, B a b^{3} c^{3} d n + 6 \, B a^{2} b^{2} c^{2} d^{2} n - 4 \, B a^{3} b c d^{3} n + B a^{4} d^{4} n\right )} \log \left (b - \frac {{\left (b x + a\right )} d}{d x + c}\right )}{b^{3} d} - \frac {2 \, {\left (B b^{4} c^{4} n - 4 \, B a b^{3} c^{3} d n + 6 \, B a^{2} b^{2} c^{2} d^{2} n - 4 \, B a^{3} b c d^{3} n + B a^{4} d^{4} n\right )} \log \left (\frac {b x + a}{d x + c}\right )}{b^{3} d}\right )} {\left (\frac {b c}{{\left (b c - a d\right )}^{2}} - \frac {a d}{{\left (b c - a d\right )}^{2}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(2*(B*b^4*c^4*n - 4*B*a*b^3*c^3*d*n + 6*B*a^2*b^2*c^2*d^2*n - 4*B*a^3*b*c*d^3*n + B*a^4*d^4*n)*log((b*x +
a)/(d*x + c))/(b^3*d - 3*(b*x + a)*b^2*d^2/(d*x + c) + 3*(b*x + a)^2*b*d^3/(d*x + c)^2 - (b*x + a)^3*d^4/(d*x
+ c)^3) - (3*B*b^6*c^4*n - 12*B*a*b^5*c^3*d*n - 5*(b*x + a)*B*b^5*c^4*d*n/(d*x + c) + 18*B*a^2*b^4*c^2*d^2*n
+ 20*(b*x + a)*B*a*b^4*c^3*d^2*n/(d*x + c) + 2*(b*x + a)^2*B*b^4*c^4*d^2*n/(d*x + c)^2 - 12*B*a^3*b^3*c*d^3*n
- 30*(b*x + a)*B*a^2*b^3*c^2*d^3*n/(d*x + c) - 8*(b*x + a)^2*B*a*b^3*c^3*d^3*n/(d*x + c)^2 + 3*B*a^4*b^2*d^4*n
+ 20*(b*x + a)*B*a^3*b^2*c*d^4*n/(d*x + c) + 12*(b*x + a)^2*B*a^2*b^2*c^2*d^4*n/(d*x + c)^2 - 5*(b*x + a)*B*a
^4*b*d^5*n/(d*x + c) - 8*(b*x + a)^2*B*a^3*b*c*d^5*n/(d*x + c)^2 + 2*(b*x + a)^2*B*a^4*d^6*n/(d*x + c)^2 - 2*A
*b^6*c^4 - 2*B*b^6*c^4 + 8*A*a*b^5*c^3*d + 8*B*a*b^5*c^3*d - 12*A*a^2*b^4*c^2*d^2 - 12*B*a^2*b^4*c^2*d^2 + 8*A
*a^3*b^3*c*d^3 + 8*B*a^3*b^3*c*d^3 - 2*A*a^4*b^2*d^4 - 2*B*a^4*b^2*d^4)/(b^5*d - 3*(b*x + a)*b^4*d^2/(d*x + c)
+ 3*(b*x + a)^2*b^3*d^3/(d*x + c)^2 - (b*x + a)^3*b^2*d^4/(d*x + c)^3) + 2*(B*b^4*c^4*n - 4*B*a*b^3*c^3*d*n +
6*B*a^2*b^2*c^2*d^2*n - 4*B*a^3*b*c*d^3*n + B*a^4*d^4*n)*log(b - (b*x + a)*d/(d*x + c))/(b^3*d) - 2*(B*b^4*c^
4*n - 4*B*a*b^3*c^3*d*n + 6*B*a^2*b^2*c^2*d^2*n - 4*B*a^3*b*c*d^3*n + B*a^4*d^4*n)*log((b*x + a)/(d*x + c))/(b
^3*d))*(b*c/(b*c - a*d)^2 - a*d/(b*c - a*d)^2)

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maple [F]  time = 0.29, size = 0, normalized size = 0.00 \[ \int \left (d i x +c i \right )^{2} \left (B \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right )+A \right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 1.18, size = 309, normalized size = 2.49 \[ \frac {1}{3} \, B d^{2} i^{2} x^{3} \log \left (e {\left (\frac {b x}{d x + c} + \frac {a}{d x + c}\right )}^{n}\right ) + \frac {1}{3} \, A d^{2} i^{2} x^{3} + B c d i^{2} x^{2} \log \left (e {\left (\frac {b x}{d x + c} + \frac {a}{d x + c}\right )}^{n}\right ) + A c d i^{2} x^{2} + \frac {1}{6} \, B d^{2} i^{2} n {\left (\frac {2 \, a^{3} \log \left (b x + a\right )}{b^{3}} - \frac {2 \, c^{3} \log \left (d x + c\right )}{d^{3}} - \frac {{\left (b^{2} c d - a b d^{2}\right )} x^{2} - 2 \, {\left (b^{2} c^{2} - a^{2} d^{2}\right )} x}{b^{2} d^{2}}\right )} - B c d i^{2} n {\left (\frac {a^{2} \log \left (b x + a\right )}{b^{2}} - \frac {c^{2} \log \left (d x + c\right )}{d^{2}} + \frac {{\left (b c - a d\right )} x}{b d}\right )} + B c^{2} i^{2} n {\left (\frac {a \log \left (b x + a\right )}{b} - \frac {c \log \left (d x + c\right )}{d}\right )} + B c^{2} i^{2} x \log \left (e {\left (\frac {b x}{d x + c} + \frac {a}{d x + c}\right )}^{n}\right ) + A c^{2} i^{2} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 4.62, size = 303, normalized size = 2.44 \[ \ln \left (e\,{\left (\frac {a+b\,x}{c+d\,x}\right )}^n\right )\,\left (B\,c^2\,i^2\,x+B\,c\,d\,i^2\,x^2+\frac {B\,d^2\,i^2\,x^3}{3}\right )-x\,\left (\frac {\left (3\,a\,d+3\,b\,c\right )\,\left (\frac {d\,i^2\,\left (3\,A\,a\,d+9\,A\,b\,c+B\,a\,d\,n-B\,b\,c\,n\right )}{3\,b}-\frac {A\,d\,i^2\,\left (3\,a\,d+3\,b\,c\right )}{3\,b}\right )}{3\,b\,d}-\frac {c\,i^2\,\left (3\,A\,a\,d+3\,A\,b\,c+B\,a\,d\,n-B\,b\,c\,n\right )}{b}+\frac {A\,a\,c\,d\,i^2}{b}\right )+x^2\,\left (\frac {d\,i^2\,\left (3\,A\,a\,d+9\,A\,b\,c+B\,a\,d\,n-B\,b\,c\,n\right )}{6\,b}-\frac {A\,d\,i^2\,\left (3\,a\,d+3\,b\,c\right )}{6\,b}\right )+\frac {\ln \left (a+b\,x\right )\,\left (B\,n\,a^3\,d^2\,i^2-3\,B\,n\,a^2\,b\,c\,d\,i^2+3\,B\,n\,a\,b^2\,c^2\,i^2\right )}{3\,b^3}+\frac {A\,d^2\,i^2\,x^3}{3}-\frac {B\,c^3\,i^2\,n\,\ln \left (c+d\,x\right )}{3\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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sympy [A]  time = 58.61, size = 779, normalized size = 6.28 \[ \begin {cases} c^{2} i^{2} x \left (A + B \log {\left (e \left (\frac {a}{c}\right )^{n} \right )}\right ) & \text {for}\: b = 0 \wedge d = 0 \\A c^{2} i^{2} x + A c d i^{2} x^{2} + \frac {A d^{2} i^{2} x^{3}}{3} - \frac {B c^{3} i^{2} n \log {\left (c + d x \right )}}{3 d} + B c^{2} i^{2} n x \log {\relax (a )} - B c^{2} i^{2} n x \log {\left (c + d x \right )} + \frac {B c^{2} i^{2} n x}{3} + B c^{2} i^{2} x \log {\relax (e )} + B c d i^{2} n x^{2} \log {\relax (a )} - B c d i^{2} n x^{2} \log {\left (c + d x \right )} + \frac {B c d i^{2} n x^{2}}{3} + B c d i^{2} x^{2} \log {\relax (e )} + \frac {B d^{2} i^{2} n x^{3} \log {\relax (a )}}{3} - \frac {B d^{2} i^{2} n x^{3} \log {\left (c + d x \right )}}{3} + \frac {B d^{2} i^{2} n x^{3}}{9} + \frac {B d^{2} i^{2} x^{3} \log {\relax (e )}}{3} & \text {for}\: b = 0 \\c^{2} i^{2} \left (A x + \frac {B a n \log {\left (\frac {a}{c} + \frac {b x}{c} \right )}}{b} + B n x \log {\left (\frac {a}{c} + \frac {b x}{c} \right )} - B n x + B x \log {\relax (e )}\right ) & \text {for}\: d = 0 \\A c^{2} i^{2} x + A c d i^{2} x^{2} + \frac {A d^{2} i^{2} x^{3}}{3} + \frac {B a^{3} d^{2} i^{2} n \log {\left (\frac {a}{c + d x} + \frac {b x}{c + d x} \right )}}{3 b^{3}} + \frac {B a^{3} d^{2} i^{2} n \log {\left (\frac {c}{d} + x \right )}}{3 b^{3}} - \frac {B a^{2} c d i^{2} n \log {\left (\frac {a}{c + d x} + \frac {b x}{c + d x} \right )}}{b^{2}} - \frac {B a^{2} c d i^{2} n \log {\left (\frac {c}{d} + x \right )}}{b^{2}} - \frac {B a^{2} d^{2} i^{2} n x}{3 b^{2}} + \frac {B a c^{2} i^{2} n \log {\left (\frac {a}{c + d x} + \frac {b x}{c + d x} \right )}}{b} + \frac {B a c^{2} i^{2} n \log {\left (\frac {c}{d} + x \right )}}{b} + \frac {B a c d i^{2} n x}{b} + \frac {B a d^{2} i^{2} n x^{2}}{6 b} - \frac {B c^{3} i^{2} n \log {\left (\frac {c}{d} + x \right )}}{3 d} + B c^{2} i^{2} n x \log {\left (\frac {a}{c + d x} + \frac {b x}{c + d x} \right )} - \frac {2 B c^{2} i^{2} n x}{3} + B c^{2} i^{2} x \log {\relax (e )} + B c d i^{2} n x^{2} \log {\left (\frac {a}{c + d x} + \frac {b x}{c + d x} \right )} - \frac {B c d i^{2} n x^{2}}{6} + B c d i^{2} x^{2} \log {\relax (e )} + \frac {B d^{2} i^{2} n x^{3} \log {\left (\frac {a}{c + d x} + \frac {b x}{c + d x} \right )}}{3} + \frac {B d^{2} i^{2} x^{3} \log {\relax (e )}}{3} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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