3.32 \(\int \frac {\log (e (f (a+b x)^p (c+d x)^q)^r)}{(g+h x)^3} \, dx\)

Optimal. Leaf size=202 \[ \frac {b^2 p r \log (a+b x)}{2 h (b g-a h)^2}-\frac {b^2 p r \log (g+h x)}{2 h (b g-a h)^2}-\frac {\log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{2 h (g+h x)^2}+\frac {b p r}{2 h (g+h x) (b g-a h)}+\frac {d^2 q r \log (c+d x)}{2 h (d g-c h)^2}-\frac {d^2 q r \log (g+h x)}{2 h (d g-c h)^2}+\frac {d q r}{2 h (g+h x) (d g-c h)} \]

[Out]

1/2*b*p*r/h/(-a*h+b*g)/(h*x+g)+1/2*d*q*r/h/(-c*h+d*g)/(h*x+g)+1/2*b^2*p*r*ln(b*x+a)/h/(-a*h+b*g)^2+1/2*d^2*q*r
*ln(d*x+c)/h/(-c*h+d*g)^2-1/2*ln(e*(f*(b*x+a)^p*(d*x+c)^q)^r)/h/(h*x+g)^2-1/2*b^2*p*r*ln(h*x+g)/h/(-a*h+b*g)^2
-1/2*d^2*q*r*ln(h*x+g)/h/(-c*h+d*g)^2

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Rubi [A]  time = 0.11, antiderivative size = 202, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 2, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {2495, 44} \[ \frac {b^2 p r \log (a+b x)}{2 h (b g-a h)^2}-\frac {b^2 p r \log (g+h x)}{2 h (b g-a h)^2}-\frac {\log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{2 h (g+h x)^2}+\frac {b p r}{2 h (g+h x) (b g-a h)}+\frac {d^2 q r \log (c+d x)}{2 h (d g-c h)^2}-\frac {d^2 q r \log (g+h x)}{2 h (d g-c h)^2}+\frac {d q r}{2 h (g+h x) (d g-c h)} \]

Antiderivative was successfully verified.

[In]

Int[Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r]/(g + h*x)^3,x]

[Out]

(b*p*r)/(2*h*(b*g - a*h)*(g + h*x)) + (d*q*r)/(2*h*(d*g - c*h)*(g + h*x)) + (b^2*p*r*Log[a + b*x])/(2*h*(b*g -
a*h)^2) + (d^2*q*r*Log[c + d*x])/(2*h*(d*g - c*h)^2) - Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r]/(2*h*(g + h*x)^2)
- (b^2*p*r*Log[g + h*x])/(2*h*(b*g - a*h)^2) - (d^2*q*r*Log[g + h*x])/(2*h*(d*g - c*h)^2)

Rule 44

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rule 2495

Int[Log[(e_.)*((f_.)*((a_.) + (b_.)*(x_))^(p_.)*((c_.) + (d_.)*(x_))^(q_.))^(r_.)]*((g_.) + (h_.)*(x_))^(m_.),
x_Symbol] :> Simp[((g + h*x)^(m + 1)*Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r])/(h*(m + 1)), x] + (-Dist[(b*p*r)/(
h*(m + 1)), Int[(g + h*x)^(m + 1)/(a + b*x), x], x] - Dist[(d*q*r)/(h*(m + 1)), Int[(g + h*x)^(m + 1)/(c + d*x
), x], x]) /; FreeQ[{a, b, c, d, e, f, g, h, m, p, q, r}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1]

Rubi steps

\begin {align*} \int \frac {\log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{(g+h x)^3} \, dx &=-\frac {\log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{2 h (g+h x)^2}+\frac {(b p r) \int \frac {1}{(a+b x) (g+h x)^2} \, dx}{2 h}+\frac {(d q r) \int \frac {1}{(c+d x) (g+h x)^2} \, dx}{2 h}\\ &=-\frac {\log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{2 h (g+h x)^2}+\frac {(b p r) \int \left (\frac {b^2}{(b g-a h)^2 (a+b x)}-\frac {h}{(b g-a h) (g+h x)^2}-\frac {b h}{(b g-a h)^2 (g+h x)}\right ) \, dx}{2 h}+\frac {(d q r) \int \left (\frac {d^2}{(d g-c h)^2 (c+d x)}-\frac {h}{(d g-c h) (g+h x)^2}-\frac {d h}{(d g-c h)^2 (g+h x)}\right ) \, dx}{2 h}\\ &=\frac {b p r}{2 h (b g-a h) (g+h x)}+\frac {d q r}{2 h (d g-c h) (g+h x)}+\frac {b^2 p r \log (a+b x)}{2 h (b g-a h)^2}+\frac {d^2 q r \log (c+d x)}{2 h (d g-c h)^2}-\frac {\log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{2 h (g+h x)^2}-\frac {b^2 p r \log (g+h x)}{2 h (b g-a h)^2}-\frac {d^2 q r \log (g+h x)}{2 h (d g-c h)^2}\\ \end {align*}

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Mathematica [A]  time = 0.53, size = 206, normalized size = 1.02 \[ \frac {\frac {r (g+h x) \left ((b c-a d) (b g-a h) (d g-c h) (b d g (p+q)-h (a d q+b c p))-(g+h x) \left (d^2 q (a d-b c) (b g-a h)^2 (\log (c+d x)-\log (g+h x))-b^2 p (b c-a d) (d g-c h)^2 (\log (a+b x)-\log (g+h x))\right )\right )}{(b c-a d) (b g-a h)^2 (d g-c h)^2}-\log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{2 h (g+h x)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r]/(g + h*x)^3,x]

[Out]

(-Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r] + (r*(g + h*x)*((b*c - a*d)*(b*g - a*h)*(d*g - c*h)*(b*d*g*(p + q) - h*
(b*c*p + a*d*q)) - (g + h*x)*(-(b^2*(b*c - a*d)*(d*g - c*h)^2*p*(Log[a + b*x] - Log[g + h*x])) + d^2*(-(b*c) +
a*d)*(b*g - a*h)^2*q*(Log[c + d*x] - Log[g + h*x]))))/((b*c - a*d)*(b*g - a*h)^2*(d*g - c*h)^2))/(2*h*(g + h*
x)^2)

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(e*(f*(b*x+a)^p*(d*x+c)^q)^r)/(h*x+g)^3,x, algorithm="fricas")

[Out]

Timed out

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giac [B]  time = 0.57, size = 595, normalized size = 2.95 \[ \frac {b^{3} p r \log \left ({\left | b x + a \right |}\right )}{2 \, {\left (b^{3} g^{2} h - 2 \, a b^{2} g h^{2} + a^{2} b h^{3}\right )}} + \frac {d^{3} q r \log \left ({\left | d x + c \right |}\right )}{2 \, {\left (d^{3} g^{2} h - 2 \, c d^{2} g h^{2} + c^{2} d h^{3}\right )}} - \frac {p r \log \left (b x + a\right )}{2 \, {\left (h^{3} x^{2} + 2 \, g h^{2} x + g^{2} h\right )}} - \frac {q r \log \left (d x + c\right )}{2 \, {\left (h^{3} x^{2} + 2 \, g h^{2} x + g^{2} h\right )}} - \frac {{\left (b^{2} d^{2} g^{2} p r - 2 \, b^{2} c d g h p r + b^{2} c^{2} h^{2} p r + b^{2} d^{2} g^{2} q r - 2 \, a b d^{2} g h q r + a^{2} d^{2} h^{2} q r\right )} \log \left (h x + g\right )}{2 \, {\left (b^{2} d^{2} g^{4} h - 2 \, b^{2} c d g^{3} h^{2} - 2 \, a b d^{2} g^{3} h^{2} + b^{2} c^{2} g^{2} h^{3} + 4 \, a b c d g^{2} h^{3} + a^{2} d^{2} g^{2} h^{3} - 2 \, a b c^{2} g h^{4} - 2 \, a^{2} c d g h^{4} + a^{2} c^{2} h^{5}\right )}} + \frac {b d g h p r x - b c h^{2} p r x + b d g h q r x - a d h^{2} q r x + b d g^{2} p r - b c g h p r + b d g^{2} q r - a d g h q r - b d g^{2} r \log \relax (f) + b c g h r \log \relax (f) + a d g h r \log \relax (f) - a c h^{2} r \log \relax (f) - b d g^{2} + b c g h + a d g h - a c h^{2}}{2 \, {\left (b d g^{2} h^{3} x^{2} - b c g h^{4} x^{2} - a d g h^{4} x^{2} + a c h^{5} x^{2} + 2 \, b d g^{3} h^{2} x - 2 \, b c g^{2} h^{3} x - 2 \, a d g^{2} h^{3} x + 2 \, a c g h^{4} x + b d g^{4} h - b c g^{3} h^{2} - a d g^{3} h^{2} + a c g^{2} h^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(e*(f*(b*x+a)^p*(d*x+c)^q)^r)/(h*x+g)^3,x, algorithm="giac")

[Out]

1/2*b^3*p*r*log(abs(b*x + a))/(b^3*g^2*h - 2*a*b^2*g*h^2 + a^2*b*h^3) + 1/2*d^3*q*r*log(abs(d*x + c))/(d^3*g^2
*h - 2*c*d^2*g*h^2 + c^2*d*h^3) - 1/2*p*r*log(b*x + a)/(h^3*x^2 + 2*g*h^2*x + g^2*h) - 1/2*q*r*log(d*x + c)/(h
^3*x^2 + 2*g*h^2*x + g^2*h) - 1/2*(b^2*d^2*g^2*p*r - 2*b^2*c*d*g*h*p*r + b^2*c^2*h^2*p*r + b^2*d^2*g^2*q*r - 2
*a*b*d^2*g*h*q*r + a^2*d^2*h^2*q*r)*log(h*x + g)/(b^2*d^2*g^4*h - 2*b^2*c*d*g^3*h^2 - 2*a*b*d^2*g^3*h^2 + b^2*
c^2*g^2*h^3 + 4*a*b*c*d*g^2*h^3 + a^2*d^2*g^2*h^3 - 2*a*b*c^2*g*h^4 - 2*a^2*c*d*g*h^4 + a^2*c^2*h^5) + 1/2*(b*
d*g*h*p*r*x - b*c*h^2*p*r*x + b*d*g*h*q*r*x - a*d*h^2*q*r*x + b*d*g^2*p*r - b*c*g*h*p*r + b*d*g^2*q*r - a*d*g*
h*q*r - b*d*g^2*r*log(f) + b*c*g*h*r*log(f) + a*d*g*h*r*log(f) - a*c*h^2*r*log(f) - b*d*g^2 + b*c*g*h + a*d*g*
h - a*c*h^2)/(b*d*g^2*h^3*x^2 - b*c*g*h^4*x^2 - a*d*g*h^4*x^2 + a*c*h^5*x^2 + 2*b*d*g^3*h^2*x - 2*b*c*g^2*h^3*
x - 2*a*d*g^2*h^3*x + 2*a*c*g*h^4*x + b*d*g^4*h - b*c*g^3*h^2 - a*d*g^3*h^2 + a*c*g^2*h^3)

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maple [F]  time = 0.31, size = 0, normalized size = 0.00 \[ \int \frac {\ln \left (e \left (f \left (b x +a \right )^{p} \left (d x +c \right )^{q}\right )^{r}\right )}{\left (h x +g \right )^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(ln(e*(f*(b*x+a)^p*(d*x+c)^q)^r)/(h*x+g)^3,x)

[Out]

int(ln(e*(f*(b*x+a)^p*(d*x+c)^q)^r)/(h*x+g)^3,x)

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maxima [A]  time = 0.78, size = 232, normalized size = 1.15 \[ \frac {{\left (b f p {\left (\frac {b \log \left (b x + a\right )}{b^{2} g^{2} - 2 \, a b g h + a^{2} h^{2}} - \frac {b \log \left (h x + g\right )}{b^{2} g^{2} - 2 \, a b g h + a^{2} h^{2}} + \frac {1}{b g^{2} - a g h + {\left (b g h - a h^{2}\right )} x}\right )} + d f q {\left (\frac {d \log \left (d x + c\right )}{d^{2} g^{2} - 2 \, c d g h + c^{2} h^{2}} - \frac {d \log \left (h x + g\right )}{d^{2} g^{2} - 2 \, c d g h + c^{2} h^{2}} + \frac {1}{d g^{2} - c g h + {\left (d g h - c h^{2}\right )} x}\right )}\right )} r}{2 \, f h} - \frac {\log \left (\left ({\left (b x + a\right )}^{p} {\left (d x + c\right )}^{q} f\right )^{r} e\right )}{2 \, {\left (h x + g\right )}^{2} h} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(e*(f*(b*x+a)^p*(d*x+c)^q)^r)/(h*x+g)^3,x, algorithm="maxima")

[Out]

1/2*(b*f*p*(b*log(b*x + a)/(b^2*g^2 - 2*a*b*g*h + a^2*h^2) - b*log(h*x + g)/(b^2*g^2 - 2*a*b*g*h + a^2*h^2) +
1/(b*g^2 - a*g*h + (b*g*h - a*h^2)*x)) + d*f*q*(d*log(d*x + c)/(d^2*g^2 - 2*c*d*g*h + c^2*h^2) - d*log(h*x + g
)/(d^2*g^2 - 2*c*d*g*h + c^2*h^2) + 1/(d*g^2 - c*g*h + (d*g*h - c*h^2)*x)))*r/(f*h) - 1/2*log(((b*x + a)^p*(d*
x + c)^q*f)^r*e)/((h*x + g)^2*h)

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mupad [B]  time = 2.71, size = 384, normalized size = 1.90 \[ \frac {b^2\,p\,r\,\ln \left (a+b\,x\right )}{2\,a^2\,h^3-4\,a\,b\,g\,h^2+2\,b^2\,g^2\,h}-\frac {\ln \left (g+h\,x\right )\,\left (h^2\,\left (q\,r\,a^2\,d^2+p\,r\,b^2\,c^2\right )-h\,\left (2\,c\,g\,p\,r\,b^2\,d+2\,a\,g\,q\,r\,b\,d^2\right )+b^2\,d^2\,g^2\,p\,r+b^2\,d^2\,g^2\,q\,r\right )}{2\,a^2\,c^2\,h^5-4\,a^2\,c\,d\,g\,h^4+2\,a^2\,d^2\,g^2\,h^3-4\,a\,b\,c^2\,g\,h^4+8\,a\,b\,c\,d\,g^2\,h^3-4\,a\,b\,d^2\,g^3\,h^2+2\,b^2\,c^2\,g^2\,h^3-4\,b^2\,c\,d\,g^3\,h^2+2\,b^2\,d^2\,g^4\,h}-\frac {\ln \left (e\,{\left (f\,{\left (a+b\,x\right )}^p\,{\left (c+d\,x\right )}^q\right )}^r\right )\,\left (\frac {x}{2}+\frac {g}{2\,h}\right )}{{\left (g+h\,x\right )}^3}-\frac {b\,c\,h\,p\,r-b\,d\,g\,p\,r+a\,d\,h\,q\,r-b\,d\,g\,q\,r}{\left (2\,x\,h^2+2\,g\,h\right )\,\left (a\,c\,h^2+b\,d\,g^2-a\,d\,g\,h-b\,c\,g\,h\right )}+\frac {d^2\,q\,r\,\ln \left (c+d\,x\right )}{2\,c^2\,h^3-4\,c\,d\,g\,h^2+2\,d^2\,g^2\,h} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log(e*(f*(a + b*x)^p*(c + d*x)^q)^r)/(g + h*x)^3,x)

[Out]

(b^2*p*r*log(a + b*x))/(2*a^2*h^3 + 2*b^2*g^2*h - 4*a*b*g*h^2) - (log(g + h*x)*(h^2*(b^2*c^2*p*r + a^2*d^2*q*r
) - h*(2*a*b*d^2*g*q*r + 2*b^2*c*d*g*p*r) + b^2*d^2*g^2*p*r + b^2*d^2*g^2*q*r))/(2*a^2*c^2*h^5 + 2*b^2*d^2*g^4
*h + 2*a^2*d^2*g^2*h^3 + 2*b^2*c^2*g^2*h^3 - 4*a*b*c^2*g*h^4 - 4*a^2*c*d*g*h^4 - 4*a*b*d^2*g^3*h^2 - 4*b^2*c*d
*g^3*h^2 + 8*a*b*c*d*g^2*h^3) - (log(e*(f*(a + b*x)^p*(c + d*x)^q)^r)*(x/2 + g/(2*h)))/(g + h*x)^3 - (b*c*h*p*
r - b*d*g*p*r + a*d*h*q*r - b*d*g*q*r)/((2*g*h + 2*h^2*x)*(a*c*h^2 + b*d*g^2 - a*d*g*h - b*c*g*h)) + (d^2*q*r*
log(c + d*x))/(2*c^2*h^3 + 2*d^2*g^2*h - 4*c*d*g*h^2)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(ln(e*(f*(b*x+a)**p*(d*x+c)**q)**r)/(h*x+g)**3,x)

[Out]

Timed out

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