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

[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)

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Rubi [A]  time = 0.07449, antiderivative size = 124, normalized size of antiderivative = 1., 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 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 (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*}

Mathematica [A]  time = 0.0471789, 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*(-(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]))/(2*b^3) + (c +
d*x)^3*(A + B*Log[e*((a + b*x)/(c + d*x))^n])))/(3*d)

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Maple [F]  time = 0.457, size = 0, normalized size = 0. \begin{align*} \int \left ( dix+ci \right ) ^{2} \left ( A+B\ln \left ( e \left ({\frac{bx+a}{dx+c}} \right ) ^{n} \right ) \right ) \, dx \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))^n)),x)

[Out]

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

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Maxima [B]  time = 1.32109, size = 417, normalized size = 3.36 \begin{align*} \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 \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))^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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Fricas [B]  time = 0.561858, size = 622, normalized size = 5.02 \begin{align*} \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 \left (e\right ) + 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} \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))^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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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))**n)),x)

[Out]

Timed out

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Giac [A]  time = 3.4415, size = 277, normalized size = 2.23 \begin{align*} \frac{B c^{3} n \log \left (d x + c\right )}{3 \, d} - \frac{1}{3} \,{\left (A d^{2} + B d^{2}\right )} x^{3} + \frac{{\left (B b c d n - B a d^{2} n - 6 \, A b c d - 6 \, B b c d\right )} x^{2}}{6 \, b} - \frac{1}{3} \,{\left (B d^{2} n x^{3} + 3 \, B c d n x^{2} + 3 \, B c^{2} n x\right )} \log \left (\frac{b x + a}{d x + c}\right ) + \frac{{\left (2 \, B b^{2} c^{2} n - 3 \, B a b c d n + B a^{2} d^{2} n - 3 \, A b^{2} c^{2} - 3 \, B b^{2} c^{2}\right )} x}{3 \, b^{2}} - \frac{{\left (3 \, B a b^{2} c^{2} n - 3 \, B a^{2} b c d n + B a^{3} d^{2} n\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))^n)),x, algorithm="giac")

[Out]

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