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

Optimal. Leaf size=118 \[ \frac {i^2 (c+d x)^3 \left (B \log \left (\frac {e (a+b x)}{c+d x}\right )+A\right )}{3 d}-\frac {B i^2 (b c-a d)^3 \log (a+b x)}{3 b^3 d}-\frac {B i^2 x (b c-a d)^2}{3 b^2}-\frac {B i^2 (c+d x)^2 (b c-a d)}{6 b d} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {2525, 12, 43} \[ \frac {i^2 (c+d x)^3 \left (B \log \left (\frac {e (a+b x)}{c+d x}\right )+A\right )}{3 d}-\frac {B i^2 x (b c-a d)^2}{3 b^2}-\frac {B i^2 (b c-a d)^3 \log (a+b x)}{3 b^3 d}-\frac {B i^2 (c+d x)^2 (b c-a d)}{6 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(B*(b*c - a*d)^2*i^2*x)/(3*b^2) - (B*(b*c - a*d)*i^2*(c + d*x)^2)/(6*b*d) - (B*(b*c - a*d)^3*i^2*Log[a + b*x]
)/(3*b^3*d) + (i^2*(c + d*x)^3*(A + B*Log[(e*(a + b*x))/(c + d*x)]))/(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 (13 c+13 d x)^2 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right ) \, dx &=\frac {169 (c+d x)^3 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{3 d}-\frac {B \int \frac {2197 (b c-a d) (c+d x)^2}{a+b x} \, dx}{39 d}\\ &=\frac {169 (c+d x)^3 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{3 d}-\frac {(169 B (b c-a d)) \int \frac {(c+d x)^2}{a+b x} \, dx}{3 d}\\ &=\frac {169 (c+d x)^3 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{3 d}-\frac {(169 B (b c-a d)) \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}{3 d}\\ &=-\frac {169 B (b c-a d)^2 x}{3 b^2}-\frac {169 B (b c-a d) (c+d x)^2}{6 b d}-\frac {169 B (b c-a d)^3 \log (a+b x)}{3 b^3 d}+\frac {169 (c+d x)^3 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{3 d}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 97, normalized size = 0.82 \[ \frac {i^2 \left ((c+d x)^3 \left (B \log \left (\frac {e (a+b x)}{c+d x}\right )+A\right )-\frac {B (b c-a d) \left (2 b d x (b c-a d)+2 (b c-a d)^2 \log (a+b x)+b^2 (c+d x)^2\right )}{2 b^3}\right )}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(i^2*(-1/2*(B*(b*c - a*d)*(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)])))/(3*d)

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fricas [B]  time = 0.95, size = 223, normalized size = 1.89 \[ \frac {2 \, A b^{3} d^{3} i^{2} x^{3} - 2 \, B b^{3} c^{3} i^{2} \log \left (d x + c\right ) + {\left ({\left (6 \, A - B\right )} b^{3} c d^{2} + B a b^{2} d^{3}\right )} i^{2} x^{2} + 2 \, {\left ({\left (3 \, A - 2 \, B\right )} b^{3} c^{2} d + 3 \, B a b^{2} c d^{2} - B a^{2} b d^{3}\right )} i^{2} x + 2 \, {\left (3 \, B a b^{2} c^{2} d - 3 \, B a^{2} b c d^{2} + B a^{3} d^{3}\right )} i^{2} \log \left (b x + a\right ) + 2 \, {\left (B b^{3} d^{3} i^{2} x^{3} + 3 \, B b^{3} c d^{2} i^{2} x^{2} + 3 \, B b^{3} c^{2} d i^{2} x\right )} \log \left (\frac {b e x + a e}{d x + c}\right )}{6 \, b^{3} d} \]

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))),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.83, size = 2475, normalized size = 20.97 \[ \text {result too large to display} \]

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))),x, algorithm="giac")

[Out]

-1/6*(2*B*b^7*c^4*e^4*log(-b*e + (b*x*e + a*e)*d/(d*x + c)) - 8*B*a*b^6*c^3*d*e^4*log(-b*e + (b*x*e + a*e)*d/(
d*x + c)) + 12*B*a^2*b^5*c^2*d^2*e^4*log(-b*e + (b*x*e + a*e)*d/(d*x + c)) - 8*B*a^3*b^4*c*d^3*e^4*log(-b*e +
(b*x*e + a*e)*d/(d*x + c)) + 2*B*a^4*b^3*d^4*e^4*log(-b*e + (b*x*e + a*e)*d/(d*x + c)) - 6*(b*x*e + a*e)*B*b^6
*c^4*d*e^3*log(-b*e + (b*x*e + a*e)*d/(d*x + c))/(d*x + c) + 24*(b*x*e + a*e)*B*a*b^5*c^3*d^2*e^3*log(-b*e + (
b*x*e + a*e)*d/(d*x + c))/(d*x + c) - 36*(b*x*e + a*e)*B*a^2*b^4*c^2*d^3*e^3*log(-b*e + (b*x*e + a*e)*d/(d*x +
c))/(d*x + c) + 24*(b*x*e + a*e)*B*a^3*b^3*c*d^4*e^3*log(-b*e + (b*x*e + a*e)*d/(d*x + c))/(d*x + c) - 6*(b*x
*e + a*e)*B*a^4*b^2*d^5*e^3*log(-b*e + (b*x*e + a*e)*d/(d*x + c))/(d*x + c) + 6*(b*x*e + a*e)^2*B*b^5*c^4*d^2*
e^2*log(-b*e + (b*x*e + a*e)*d/(d*x + c))/(d*x + c)^2 - 24*(b*x*e + a*e)^2*B*a*b^4*c^3*d^3*e^2*log(-b*e + (b*x
*e + a*e)*d/(d*x + c))/(d*x + c)^2 + 36*(b*x*e + a*e)^2*B*a^2*b^3*c^2*d^4*e^2*log(-b*e + (b*x*e + a*e)*d/(d*x
+ c))/(d*x + c)^2 - 24*(b*x*e + a*e)^2*B*a^3*b^2*c*d^5*e^2*log(-b*e + (b*x*e + a*e)*d/(d*x + c))/(d*x + c)^2 +
6*(b*x*e + a*e)^2*B*a^4*b*d^6*e^2*log(-b*e + (b*x*e + a*e)*d/(d*x + c))/(d*x + c)^2 - 2*(b*x*e + a*e)^3*B*b^4
*c^4*d^3*e*log(-b*e + (b*x*e + a*e)*d/(d*x + c))/(d*x + c)^3 + 8*(b*x*e + a*e)^3*B*a*b^3*c^3*d^4*e*log(-b*e +
(b*x*e + a*e)*d/(d*x + c))/(d*x + c)^3 - 12*(b*x*e + a*e)^3*B*a^2*b^2*c^2*d^5*e*log(-b*e + (b*x*e + a*e)*d/(d*
x + c))/(d*x + c)^3 + 8*(b*x*e + a*e)^3*B*a^3*b*c*d^6*e*log(-b*e + (b*x*e + a*e)*d/(d*x + c))/(d*x + c)^3 - 2*
(b*x*e + a*e)^3*B*a^4*d^7*e*log(-b*e + (b*x*e + a*e)*d/(d*x + c))/(d*x + c)^3 + 6*(b*x*e + a*e)*B*b^6*c^4*d*e^
3*log((b*x*e + a*e)/(d*x + c))/(d*x + c) - 24*(b*x*e + a*e)*B*a*b^5*c^3*d^2*e^3*log((b*x*e + a*e)/(d*x + c))/(
d*x + c) + 36*(b*x*e + a*e)*B*a^2*b^4*c^2*d^3*e^3*log((b*x*e + a*e)/(d*x + c))/(d*x + c) - 24*(b*x*e + a*e)*B*
a^3*b^3*c*d^4*e^3*log((b*x*e + a*e)/(d*x + c))/(d*x + c) + 6*(b*x*e + a*e)*B*a^4*b^2*d^5*e^3*log((b*x*e + a*e)
/(d*x + c))/(d*x + c) - 6*(b*x*e + a*e)^2*B*b^5*c^4*d^2*e^2*log((b*x*e + a*e)/(d*x + c))/(d*x + c)^2 + 24*(b*x
*e + a*e)^2*B*a*b^4*c^3*d^3*e^2*log((b*x*e + a*e)/(d*x + c))/(d*x + c)^2 - 36*(b*x*e + a*e)^2*B*a^2*b^3*c^2*d^
4*e^2*log((b*x*e + a*e)/(d*x + c))/(d*x + c)^2 + 24*(b*x*e + a*e)^2*B*a^3*b^2*c*d^5*e^2*log((b*x*e + a*e)/(d*x
+ c))/(d*x + c)^2 - 6*(b*x*e + a*e)^2*B*a^4*b*d^6*e^2*log((b*x*e + a*e)/(d*x + c))/(d*x + c)^2 + 2*(b*x*e + a
*e)^3*B*b^4*c^4*d^3*e*log((b*x*e + a*e)/(d*x + c))/(d*x + c)^3 - 8*(b*x*e + a*e)^3*B*a*b^3*c^3*d^4*e*log((b*x*
e + a*e)/(d*x + c))/(d*x + c)^3 + 12*(b*x*e + a*e)^3*B*a^2*b^2*c^2*d^5*e*log((b*x*e + a*e)/(d*x + c))/(d*x + c
)^3 - 8*(b*x*e + a*e)^3*B*a^3*b*c*d^6*e*log((b*x*e + a*e)/(d*x + c))/(d*x + c)^3 + 2*(b*x*e + a*e)^3*B*a^4*d^7
*e*log((b*x*e + a*e)/(d*x + c))/(d*x + c)^3 + 2*A*b^7*c^4*e^4 - 3*B*b^7*c^4*e^4 - 8*A*a*b^6*c^3*d*e^4 + 12*B*a
*b^6*c^3*d*e^4 + 12*A*a^2*b^5*c^2*d^2*e^4 - 18*B*a^2*b^5*c^2*d^2*e^4 - 8*A*a^3*b^4*c*d^3*e^4 + 12*B*a^3*b^4*c*
d^3*e^4 + 2*A*a^4*b^3*d^4*e^4 - 3*B*a^4*b^3*d^4*e^4 + 5*(b*x*e + a*e)*B*b^6*c^4*d*e^3/(d*x + c) - 20*(b*x*e +
a*e)*B*a*b^5*c^3*d^2*e^3/(d*x + c) + 30*(b*x*e + a*e)*B*a^2*b^4*c^2*d^3*e^3/(d*x + c) - 20*(b*x*e + a*e)*B*a^3
*b^3*c*d^4*e^3/(d*x + c) + 5*(b*x*e + a*e)*B*a^4*b^2*d^5*e^3/(d*x + c) - 2*(b*x*e + a*e)^2*B*b^5*c^4*d^2*e^2/(
d*x + c)^2 + 8*(b*x*e + a*e)^2*B*a*b^4*c^3*d^3*e^2/(d*x + c)^2 - 12*(b*x*e + a*e)^2*B*a^2*b^3*c^2*d^4*e^2/(d*x
+ c)^2 + 8*(b*x*e + a*e)^2*B*a^3*b^2*c*d^5*e^2/(d*x + c)^2 - 2*(b*x*e + a*e)^2*B*a^4*b*d^6*e^2/(d*x + c)^2)*(
b*c/((b*c*e - a*d*e)*(b*c - a*d)) - a*d/((b*c*e - a*d*e)*(b*c - a*d)))/(b^6*d*e^3 - 3*(b*x*e + a*e)*b^5*d^2*e^
2/(d*x + c) + 3*(b*x*e + a*e)^2*b^4*d^3*e/(d*x + c)^2 - (b*x*e + a*e)^3*b^3*d^4/(d*x + c)^3)

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maple [B]  time = 0.13, size = 1522, normalized size = 12.90 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3/d*e^3*B*i^2*ln(b/d*e+(a*d-b*c)/(d*x+c)/d*e)*b^3/(1/(d*x+c)*a*d*e-1/(d*x+c)*b*c*e)^3*c^3+e^3*A*i^2/(1/(d*x
+c)*a*d*e-1/(d*x+c)*b*c*e)^3*a*b^2*c^2+d*B*i^2/b^2*ln(-b*e+(b/d*e+(a*d-b*c)/(d*x+c)/d*e)*d)*a^2*c+1/2*e^2*B*i^
2/(1/(d*x+c)*a*d*e-1/(d*x+c)*b*c*e)^2*a*c^2*b+1/3/d*B*i^2*ln(-b*e+(b/d*e+(a*d-b*c)/(d*x+c)/d*e)*d)*c^3+d*e*B*i
^2/b/(1/(d*x+c)*a*d*e-1/(d*x+c)*b*c*e)*a^2*c-B*i^2/b*ln(-b*e+(b/d*e+(a*d-b*c)/(d*x+c)/d*e)*d)*a*c^2-1/3*d^2*B*
i^2/b^3*ln(-b*e+(b/d*e+(a*d-b*c)/(d*x+c)/d*e)*d)*a^3+1/6*d^2*e^2*B*i^2/b/(1/(d*x+c)*a*d*e-1/(d*x+c)*b*c*e)^2*a
^3-1/6/d*e^2*B*i^2*b^2/(1/(d*x+c)*a*d*e-1/(d*x+c)*b*c*e)^2*c^3+1/3/d*e*B*i^2/(1/(d*x+c)*a*d*e-1/(d*x+c)*b*c*e)
*c^3*b-e*B*i^2/(1/(d*x+c)*a*d*e-1/(d*x+c)*b*c*e)*c^2*a+1/3*d^2*e^3*A*i^2/(1/(d*x+c)*a*d*e-1/(d*x+c)*b*c*e)^3*a
^3+e^3*B*i^2*ln(b/d*e+(a*d-b*c)/(d*x+c)/d*e)*b^2/(1/(d*x+c)*a*d*e-1/(d*x+c)*b*c*e)^3*c^2*a-d*e^3*A*i^2/(1/(d*x
+c)*a*d*e-1/(d*x+c)*b*c*e)^3*a^2*b*c-1/2*d*e^2*B*i^2/(1/(d*x+c)*a*d*e-1/(d*x+c)*b*c*e)^2*a^2*c-1/3*d^2*e*B*i^2
/b^2/(1/(d*x+c)*a*d*e-1/(d*x+c)*b*c*e)*a^3+1/3*d^2*e^3*B*i^2*ln(b/d*e+(a*d-b*c)/(d*x+c)/d*e)/(1/(d*x+c)*a*d*e-
1/(d*x+c)*b*c*e)^3*a^3-1/3/d*e^3*A*i^2/(1/(d*x+c)*a*d*e-1/(d*x+c)*b*c*e)^3*b^3*c^3+5*d^3*e^3*B*i^2*ln(b/d*e+(a
*d-b*c)/(d*x+c)/d*e)/b/(1/(d*x+c)*a*d*e-1/(d*x+c)*b*c*e)^3*a^4/(d*x+c)^3*c^2+5*d*e^3*B*i^2*ln(b/d*e+(a*d-b*c)/
(d*x+c)/d*e)*b/(1/(d*x+c)*a*d*e-1/(d*x+c)*b*c*e)^3*a^2*c^4/(d*x+c)^3-2*d^4*e^3*B*i^2*ln(b/d*e+(a*d-b*c)/(d*x+c
)/d*e)/b^2/(1/(d*x+c)*a*d*e-1/(d*x+c)*b*c*e)^3*a^5/(d*x+c)^3*c-d*e^3*B*i^2*ln(b/d*e+(a*d-b*c)/(d*x+c)/d*e)/(1/
(d*x+c)*a*d*e-1/(d*x+c)*b*c*e)^3*a^2*b*c+1/3*d^5*e^3*B*i^2*ln(b/d*e+(a*d-b*c)/(d*x+c)/d*e)/b^3/(1/(d*x+c)*a*d*
e-1/(d*x+c)*b*c*e)^3*a^6/(d*x+c)^3-2*e^3*B*i^2*ln(b/d*e+(a*d-b*c)/(d*x+c)/d*e)*b^2/(1/(d*x+c)*a*d*e-1/(d*x+c)*
b*c*e)^3*a*c^5/(d*x+c)^3+1/3/d*e^3*B*i^2*ln(b/d*e+(a*d-b*c)/(d*x+c)/d*e)*b^3/(1/(d*x+c)*a*d*e-1/(d*x+c)*b*c*e)
^3*c^6/(d*x+c)^3-20/3*d^2*e^3*B*i^2*ln(b/d*e+(a*d-b*c)/(d*x+c)/d*e)/(1/(d*x+c)*a*d*e-1/(d*x+c)*b*c*e)^3*a^3/(d
*x+c)^3*c^3

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maxima [B]  time = 1.10, size = 280, normalized size = 2.37 \[ \frac {1}{3} \, A d^{2} i^{2} x^{3} + A c d i^{2} x^{2} + {\left (x \log \left (\frac {b e x}{d x + c} + \frac {a e}{d x + c}\right ) + \frac {a \log \left (b x + a\right )}{b} - \frac {c \log \left (d x + c\right )}{d}\right )} B c^{2} i^{2} + {\left (x^{2} \log \left (\frac {b e x}{d x + c} + \frac {a e}{d x + c}\right ) - \frac {a^{2} \log \left (b x + a\right )}{b^{2}} + \frac {c^{2} \log \left (d x + c\right )}{d^{2}} - \frac {{\left (b c - a d\right )} x}{b d}\right )} B c d i^{2} + \frac {1}{6} \, {\left (2 \, x^{3} \log \left (\frac {b e x}{d x + c} + \frac {a e}{d x + c}\right ) + \frac {2 \, a^{3} \log \left (b x + a\right )}{b^{3}} - \frac {2 \, c^{3} \log \left (d x + c\right )}{d^{3}} - \frac {{\left (b^{2} c d - a b d^{2}\right )} x^{2} - 2 \, {\left (b^{2} c^{2} - a^{2} d^{2}\right )} x}{b^{2} d^{2}}\right )} B d^{2} i^{2} + A c^{2} i^{2} x \]

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))),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 4.59, size = 290, normalized size = 2.46 \[ x^2\,\left (\frac {d\,i^2\,\left (3\,A\,a\,d+9\,A\,b\,c+B\,a\,d-B\,b\,c\right )}{6\,b}-\frac {A\,d\,i^2\,\left (3\,a\,d+3\,b\,c\right )}{6\,b}\right )-x\,\left (\frac {\left (3\,a\,d+3\,b\,c\right )\,\left (\frac {d\,i^2\,\left (3\,A\,a\,d+9\,A\,b\,c+B\,a\,d-B\,b\,c\right )}{3\,b}-\frac {A\,d\,i^2\,\left (3\,a\,d+3\,b\,c\right )}{3\,b}\right )}{3\,b\,d}-\frac {c\,i^2\,\left (3\,A\,a\,d+3\,A\,b\,c+B\,a\,d-B\,b\,c\right )}{b}+\frac {A\,a\,c\,d\,i^2}{b}\right )+\ln \left (\frac {e\,\left (a+b\,x\right )}{c+d\,x}\right )\,\left (B\,c^2\,i^2\,x+B\,c\,d\,i^2\,x^2+\frac {B\,d^2\,i^2\,x^3}{3}\right )+\frac {\ln \left (a+b\,x\right )\,\left (B\,a^3\,d^2\,i^2-3\,B\,a^2\,b\,c\,d\,i^2+3\,B\,a\,b^2\,c^2\,i^2\right )}{3\,b^3}+\frac {A\,d^2\,i^2\,x^3}{3}-\frac {B\,c^3\,i^2\,\ln \left (c+d\,x\right )}{3\,d} \]

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))),x)

[Out]

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

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sympy [B]  time = 2.99, size = 491, normalized size = 4.16 \[ \frac {A d^{2} i^{2} x^{3}}{3} + \frac {B a i^{2} \left (a^{2} d^{2} - 3 a b c d + 3 b^{2} c^{2}\right ) \log {\left (x + \frac {B a^{3} c d^{2} i^{2} - 3 B a^{2} b c^{2} d i^{2} + \frac {B a^{2} d i^{2} \left (a^{2} d^{2} - 3 a b c d + 3 b^{2} c^{2}\right )}{b} + 4 B a b^{2} c^{3} i^{2} - B a c i^{2} \left (a^{2} d^{2} - 3 a b c d + 3 b^{2} c^{2}\right )}{B a^{3} d^{3} i^{2} - 3 B a^{2} b c d^{2} i^{2} + 3 B a b^{2} c^{2} d i^{2} + B b^{3} c^{3} i^{2}} \right )}}{3 b^{3}} - \frac {B c^{3} i^{2} \log {\left (x + \frac {B a^{3} c d^{2} i^{2} - 3 B a^{2} b c^{2} d i^{2} + 3 B a b^{2} c^{3} i^{2} + \frac {B b^{3} c^{4} i^{2}}{d}}{B a^{3} d^{3} i^{2} - 3 B a^{2} b c d^{2} i^{2} + 3 B a b^{2} c^{2} d i^{2} + B b^{3} c^{3} i^{2}} \right )}}{3 d} + x^{2} \left (A c d i^{2} + \frac {B a d^{2} i^{2}}{6 b} - \frac {B c d i^{2}}{6}\right ) + x \left (A c^{2} i^{2} - \frac {B a^{2} d^{2} i^{2}}{3 b^{2}} + \frac {B a c d i^{2}}{b} - \frac {2 B c^{2} i^{2}}{3}\right ) + \left (B c^{2} i^{2} x + B c d i^{2} x^{2} + \frac {B d^{2} i^{2} x^{3}}{3}\right ) \log {\left (\frac {e \left (a + b x\right )}{c + d x} \right )} \]

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))),x)

[Out]

A*d**2*i**2*x**3/3 + B*a*i**2*(a**2*d**2 - 3*a*b*c*d + 3*b**2*c**2)*log(x + (B*a**3*c*d**2*i**2 - 3*B*a**2*b*c
**2*d*i**2 + B*a**2*d*i**2*(a**2*d**2 - 3*a*b*c*d + 3*b**2*c**2)/b + 4*B*a*b**2*c**3*i**2 - B*a*c*i**2*(a**2*d
**2 - 3*a*b*c*d + 3*b**2*c**2))/(B*a**3*d**3*i**2 - 3*B*a**2*b*c*d**2*i**2 + 3*B*a*b**2*c**2*d*i**2 + B*b**3*c
**3*i**2))/(3*b**3) - B*c**3*i**2*log(x + (B*a**3*c*d**2*i**2 - 3*B*a**2*b*c**2*d*i**2 + 3*B*a*b**2*c**3*i**2
+ B*b**3*c**4*i**2/d)/(B*a**3*d**3*i**2 - 3*B*a**2*b*c*d**2*i**2 + 3*B*a*b**2*c**2*d*i**2 + B*b**3*c**3*i**2))
/(3*d) + x**2*(A*c*d*i**2 + B*a*d**2*i**2/(6*b) - B*c*d*i**2/6) + x*(A*c**2*i**2 - B*a**2*d**2*i**2/(3*b**2) +
B*a*c*d*i**2/b - 2*B*c**2*i**2/3) + (B*c**2*i**2*x + B*c*d*i**2*x**2 + B*d**2*i**2*x**3/3)*log(e*(a + b*x)/(c
+ d*x))

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