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

Optimal. Leaf size=135 \[ -\frac {d^2 q r \log (a+b x)}{2 b (b c-a d)^2}+\frac {d^2 q r \log (c+d x)}{2 b (b c-a d)^2}-\frac {\log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{2 b (a+b x)^2}-\frac {d q r}{2 b (a+b x) (b c-a d)}-\frac {p r}{4 b (a+b x)^2} \]

[Out]

-1/4*p*r/b/(b*x+a)^2-1/2*d*q*r/b/(-a*d+b*c)/(b*x+a)-1/2*d^2*q*r*ln(b*x+a)/b/(-a*d+b*c)^2+1/2*d^2*q*r*ln(d*x+c)
/b/(-a*d+b*c)^2-1/2*ln(e*(f*(b*x+a)^p*(d*x+c)^q)^r)/b/(b*x+a)^2

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Rubi [A]  time = 0.06, antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {2495, 32, 44} \[ -\frac {d^2 q r \log (a+b x)}{2 b (b c-a d)^2}+\frac {d^2 q r \log (c+d x)}{2 b (b c-a d)^2}-\frac {\log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{2 b (a+b x)^2}-\frac {d q r}{2 b (a+b x) (b c-a d)}-\frac {p r}{4 b (a+b x)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(p*r)/(4*b*(a + b*x)^2) - (d*q*r)/(2*b*(b*c - a*d)*(a + b*x)) - (d^2*q*r*Log[a + b*x])/(2*b*(b*c - a*d)^2) +
(d^2*q*r*Log[c + d*x])/(2*b*(b*c - a*d)^2) - Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r]/(2*b*(a + b*x)^2)

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

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 )}{(a+b x)^3} \, dx &=-\frac {\log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{2 b (a+b x)^2}+\frac {1}{2} (p r) \int \frac {1}{(a+b x)^3} \, dx+\frac {(d q r) \int \frac {1}{(a+b x)^2 (c+d x)} \, dx}{2 b}\\ &=-\frac {p r}{4 b (a+b x)^2}-\frac {\log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{2 b (a+b x)^2}+\frac {(d q r) \int \left (\frac {b}{(b c-a d) (a+b x)^2}-\frac {b d}{(b c-a d)^2 (a+b x)}+\frac {d^2}{(b c-a d)^2 (c+d x)}\right ) \, dx}{2 b}\\ &=-\frac {p r}{4 b (a+b x)^2}-\frac {d q r}{2 b (b c-a d) (a+b x)}-\frac {d^2 q r \log (a+b x)}{2 b (b c-a d)^2}+\frac {d^2 q r \log (c+d x)}{2 b (b c-a d)^2}-\frac {\log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{2 b (a+b x)^2}\\ \end {align*}

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Mathematica [A]  time = 0.24, size = 116, normalized size = 0.86 \[ \frac {r \left (-\frac {d^2 q \log (a+b x)}{(b c-a d)^2}+\frac {d^2 q \log (c+d x)}{(b c-a d)^2}-\frac {p-\frac {2 d q (a+b x)}{a d-b c}}{2 (a+b x)^2}\right )-\frac {\log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{(a+b x)^2}}{2 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(r*(-1/2*(p - (2*d*q*(a + b*x))/(-(b*c) + a*d))/(a + b*x)^2 - (d^2*q*Log[a + b*x])/(b*c - a*d)^2 + (d^2*q*Log[
c + d*x])/(b*c - a*d)^2) - Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r]/(a + b*x)^2)/(2*b)

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fricas [B]  time = 0.44, size = 323, normalized size = 2.39 \[ -\frac {2 \, {\left (b^{2} c d - a b d^{2}\right )} q r x + 2 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} r \log \relax (f) + {\left ({\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} p + 2 \, {\left (a b c d - a^{2} d^{2}\right )} q\right )} r + 2 \, {\left (b^{2} d^{2} q r x^{2} + 2 \, a b d^{2} q r x + {\left (a^{2} d^{2} q + {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} p\right )} r\right )} \log \left (b x + a\right ) - 2 \, {\left (b^{2} d^{2} q r x^{2} + 2 \, a b d^{2} q r x - {\left (b^{2} c^{2} - 2 \, a b c d\right )} q r\right )} \log \left (d x + c\right ) + 2 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \log \relax (e)}{4 \, {\left (a^{2} b^{3} c^{2} - 2 \, a^{3} b^{2} c d + a^{4} b d^{2} + {\left (b^{5} c^{2} - 2 \, a b^{4} c d + a^{2} b^{3} d^{2}\right )} x^{2} + 2 \, {\left (a b^{4} c^{2} - 2 \, a^{2} b^{3} c d + a^{3} b^{2} d^{2}\right )} x\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.22, size = 246, normalized size = 1.82 \[ -\frac {d^{2} q r \log \left (b x + a\right )}{2 \, {\left (b^{3} c^{2} - 2 \, a b^{2} c d + a^{2} b d^{2}\right )}} + \frac {d^{2} q r \log \left (d x + c\right )}{2 \, {\left (b^{3} c^{2} - 2 \, a b^{2} c d + a^{2} b d^{2}\right )}} - \frac {p r \log \left (b x + a\right )}{2 \, {\left (b^{3} x^{2} + 2 \, a b^{2} x + a^{2} b\right )}} - \frac {q r \log \left (d x + c\right )}{2 \, {\left (b^{3} x^{2} + 2 \, a b^{2} x + a^{2} b\right )}} - \frac {2 \, b d q r x + b c p r - a d p r + 2 \, a d q r + 2 \, b c r \log \relax (f) - 2 \, a d r \log \relax (f) + 2 \, b c - 2 \, a d}{4 \, {\left (b^{4} c x^{2} - a b^{3} d x^{2} + 2 \, a b^{3} c x - 2 \, a^{2} b^{2} d x + a^{2} b^{2} c - a^{3} b d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.75, size = 165, normalized size = 1.22 \[ -\frac {{\left (2 \, d f q {\left (\frac {d \log \left (b x + a\right )}{b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}} - \frac {d \log \left (d x + c\right )}{b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}} + \frac {1}{a b c - a^{2} d + {\left (b^{2} c - a b d\right )} x}\right )} + \frac {b f p}{b^{3} x^{2} + 2 \, a b^{2} x + a^{2} b}\right )} r}{4 \, b f} - \frac {\log \left (\left ({\left (b x + a\right )}^{p} {\left (d x + c\right )}^{q} f\right )^{r} e\right )}{2 \, {\left (b x + a\right )}^{2} b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 2.39, size = 182, normalized size = 1.35 \[ \frac {\frac {b\,c\,p\,r-a\,d\,p\,r+2\,a\,d\,q\,r}{2\,\left (a\,d-b\,c\right )}+\frac {b\,d\,q\,r\,x}{a\,d-b\,c}}{2\,a^2\,b+4\,a\,b^2\,x+2\,b^3\,x^2}-\frac {\ln \left (e\,{\left (f\,{\left (a+b\,x\right )}^p\,{\left (c+d\,x\right )}^q\right )}^r\right )\,\left (\frac {x}{2}+\frac {a}{2\,b}\right )}{{\left (a+b\,x\right )}^3}+\frac {d^2\,q\,r\,\mathrm {atanh}\left (\frac {2\,b^3\,c^2-2\,a^2\,b\,d^2}{2\,b\,{\left (a\,d-b\,c\right )}^2}-\frac {2\,b\,d\,x}{a\,d-b\,c}\right )}{b\,{\left (a\,d-b\,c\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((b*c*p*r - a*d*p*r + 2*a*d*q*r)/(2*(a*d - b*c)) + (b*d*q*r*x)/(a*d - b*c))/(2*a^2*b + 2*b^3*x^2 + 4*a*b^2*x)
- (log(e*(f*(a + b*x)^p*(c + d*x)^q)^r)*(x/2 + a/(2*b)))/(a + b*x)^3 + (d^2*q*r*atanh((2*b^3*c^2 - 2*a^2*b*d^2
)/(2*b*(a*d - b*c)^2) - (2*b*d*x)/(a*d - b*c)))/(b*(a*d - b*c)^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)/(b*x+a)**3,x)

[Out]

Timed out

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