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

Optimal. Leaf size=157 \[ \frac{(f+g x)^3 \left (B \log \left (e \left (\frac{a+b x}{c+d x}\right )^n\right )+A\right )}{3 g}-\frac{B g n x (b c-a d) (-a d g-b c g+3 b d f)}{3 b^2 d^2}-\frac{B n (b f-a g)^3 \log (a+b x)}{3 b^3 g}-\frac{B g^2 n x^2 (b c-a d)}{6 b d}+\frac{B n (d f-c g)^3 \log (c+d x)}{3 d^3 g} \]

[Out]

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

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Rubi [A]  time = 0.180485, antiderivative size = 157, 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, 72} \[ \frac{(f+g x)^3 \left (B \log \left (e \left (\frac{a+b x}{c+d x}\right )^n\right )+A\right )}{3 g}-\frac{B g n x (b c-a d) (-a d g-b c g+3 b d f)}{3 b^2 d^2}-\frac{B n (b f-a g)^3 \log (a+b x)}{3 b^3 g}-\frac{B g^2 n x^2 (b c-a d)}{6 b d}+\frac{B n (d f-c g)^3 \log (c+d x)}{3 d^3 g} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 72

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

Rubi steps

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

Mathematica [A]  time = 0.143738, size = 146, normalized size = 0.93 \[ \frac{(f+g x)^3 \left (B \log \left (e \left (\frac{a+b x}{c+d x}\right )^n\right )+A\right )-\frac{B n \left (b^2 d^2 g^3 x^2 (b c-a d)+2 b d g^2 x (b c-a d) (-a d g-b c g+3 b d f)+2 d^3 (b f-a g)^3 \log (a+b x)-2 b^3 (d f-c g)^3 \log (c+d x)\right )}{2 b^3 d^3}}{3 g} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.41, size = 0, normalized size = 0. \begin{align*} \int \left ( gx+f \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((g*x+f)^2*(A+B*ln(e*((b*x+a)/(d*x+c))^n)),x)

[Out]

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

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Maxima [A]  time = 1.18207, size = 381, normalized size = 2.43 \begin{align*} \frac{1}{3} \, B 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 g^{2} x^{3} + B f g x^{2} \log \left (e{\left (\frac{b x}{d x + c} + \frac{a}{d x + c}\right )}^{n}\right ) + A f g x^{2} + \frac{1}{6} \, B 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 f g 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 f^{2} n{\left (\frac{a \log \left (b x + a\right )}{b} - \frac{c \log \left (d x + c\right )}{d}\right )} + B f^{2} x \log \left (e{\left (\frac{b x}{d x + c} + \frac{a}{d x + c}\right )}^{n}\right ) + A f^{2} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.20537, size = 695, normalized size = 4.43 \begin{align*} \frac{2 \, A b^{3} d^{3} g^{2} x^{3} +{\left (6 \, A b^{3} d^{3} f g -{\left (B b^{3} c d^{2} - B a b^{2} d^{3}\right )} g^{2} n\right )} x^{2} + 2 \,{\left (3 \, B a b^{2} d^{3} f^{2} - 3 \, B a^{2} b d^{3} f g + B a^{3} d^{3} g^{2}\right )} n \log \left (b x + a\right ) - 2 \,{\left (3 \, B b^{3} c d^{2} f^{2} - 3 \, B b^{3} c^{2} d f g + B b^{3} c^{3} g^{2}\right )} n \log \left (d x + c\right ) + 2 \,{\left (3 \, A b^{3} d^{3} f^{2} -{\left (3 \,{\left (B b^{3} c d^{2} - B a b^{2} d^{3}\right )} f g -{\left (B b^{3} c^{2} d - B a^{2} b d^{3}\right )} g^{2}\right )} n\right )} x + 2 \,{\left (B b^{3} d^{3} g^{2} x^{3} + 3 \, B b^{3} d^{3} f g x^{2} + 3 \, B b^{3} d^{3} f^{2} x\right )} \log \left (e\right ) + 2 \,{\left (B b^{3} d^{3} g^{2} n x^{3} + 3 \, B b^{3} d^{3} f g n x^{2} + 3 \, B b^{3} d^{3} f^{2} n x\right )} \log \left (\frac{b x + a}{d x + c}\right )}{6 \, b^{3} d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

[Out]

Timed out

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Giac [A]  time = 81.8263, size = 366, normalized size = 2.33 \begin{align*} \frac{1}{3} \,{\left (A g^{2} + B g^{2}\right )} x^{3} + \frac{1}{3} \,{\left (B g^{2} n x^{3} + 3 \, B f g n x^{2} + 3 \, B f^{2} n x\right )} \log \left (\frac{b x + a}{d x + c}\right ) - \frac{{\left (B b c g^{2} n - B a d g^{2} n - 6 \, A b d f g - 6 \, B b d f g\right )} x^{2}}{6 \, b d} + \frac{{\left (3 \, B a b^{2} f^{2} n - 3 \, B a^{2} b f g n + B a^{3} g^{2} n\right )} \log \left (b x + a\right )}{3 \, b^{3}} - \frac{{\left (3 \, B c d^{2} f^{2} n - 3 \, B c^{2} d f g n + B c^{3} g^{2} n\right )} \log \left (-d x - c\right )}{3 \, d^{3}} - \frac{{\left (3 \, B b^{2} c d f g n - 3 \, B a b d^{2} f g n - B b^{2} c^{2} g^{2} n + B a^{2} d^{2} g^{2} n - 3 \, A b^{2} d^{2} f^{2} - 3 \, B b^{2} d^{2} f^{2}\right )} x}{3 \, b^{2} d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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