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

[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)

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Rubi [A]  time = 0.0668842, antiderivative size = 118, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, 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 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]

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])

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*}

Mathematica [A]  time = 0.0398882, 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*(-(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]))/(2*b^3) + (c + d
*x)^3*(A + B*Log[(e*(a + b*x))/(c + d*x)])))/(3*d)

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Maple [B]  time = 0.155, size = 1522, normalized size = 12.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.38073, size = 378, normalized size = 3.2 \begin{align*} \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 \end{align*}

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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Fricas [B]  time = 1.35517, size = 467, normalized size = 3.96 \begin{align*} \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} \end{align*}

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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Sympy [B]  time = 3.70134, size = 503, normalized size = 4.26 \begin{align*} \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} + \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 )} + \frac{x^{2} \left (6 A b c d i^{2} + B a d^{2} i^{2} - B b c d i^{2}\right )}{6 b} - \frac{x \left (- 3 A b^{2} c^{2} i^{2} + B a^{2} d^{2} i^{2} - 3 B a b c d i^{2} + 2 B b^{2} c^{2} i^{2}\right )}{3 b^{2}} \end{align*}

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

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Giac [A]  time = 2.04041, size = 240, normalized size = 2.03 \begin{align*} -\frac{1}{3} \,{\left (A d^{2} + B d^{2}\right )} x^{3} + \frac{B c^{3} \log \left (d x + c\right )}{3 \, d} - \frac{{\left (6 \, A b c d + 5 \, B b c d + B a d^{2}\right )} x^{2}}{6 \, b} - \frac{1}{3} \,{\left (B d^{2} x^{3} + 3 \, B c d x^{2} + 3 \, B c^{2} x\right )} \log \left (\frac{b x + a}{d x + c}\right ) - \frac{{\left (3 \, A b^{2} c^{2} + B b^{2} c^{2} + 3 \, B a b c d - B a^{2} d^{2}\right )} x}{3 \, b^{2}} - \frac{{\left (3 \, B a b^{2} c^{2} - 3 \, B a^{2} b c d + B a^{3} d^{2}\right )} \log \left (b x + a\right )}{3 \, b^{3}} \end{align*}

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]

-1/3*(A*d^2 + B*d^2)*x^3 + 1/3*B*c^3*log(d*x + c)/d - 1/6*(6*A*b*c*d + 5*B*b*c*d + B*a*d^2)*x^2/b - 1/3*(B*d^2
*x^3 + 3*B*c*d*x^2 + 3*B*c^2*x)*log((b*x + a)/(d*x + c)) - 1/3*(3*A*b^2*c^2 + B*b^2*c^2 + 3*B*a*b*c*d - B*a^2*
d^2)*x/b^2 - 1/3*(3*B*a*b^2*c^2 - 3*B*a^2*b*c*d + B*a^3*d^2)*log(b*x + a)/b^3