3.1.12 \(\int \frac {\log (e (f (a+b x)^p (c+d x)^q)^r)}{(a+b x)^2} \, dx\) [12]

Optimal. Leaf size=95 \[ -\frac {p r}{b (a+b x)}+\frac {d q r \log (a+b x)}{b (b c-a d)}-\frac {d q r \log (c+d x)}{b (b c-a d)}-\frac {\log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{b (a+b x)} \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {2581, 32, 36, 31} \begin {gather*} -\frac {\log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{b (a+b x)}+\frac {d q r \log (a+b x)}{b (b c-a d)}-\frac {d q r \log (c+d x)}{b (b c-a d)}-\frac {p r}{b (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

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 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 2581

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

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Mathematica [A]
time = 0.04, size = 89, normalized size = 0.94 \begin {gather*} \frac {r \left (-\frac {p}{a+b x}+\frac {d q \log (a+b x)}{b c-a d}-\frac {d q \log (c+d x)}{b c-a d}\right )}{b}-\frac {\log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{b (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.14, size = 0, normalized size = 0.00 \[\int \frac {\ln \left (e \left (f \left (b x +a \right )^{p} \left (d x +c \right )^{q}\right )^{r}\right )}{\left (b x +a \right )^{2}}\, dx\]

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)^2,x)

[Out]

int(ln(e*(f*(b*x+a)^p*(d*x+c)^q)^r)/(b*x+a)^2,x)

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Maxima [A]
time = 0.31, size = 100, normalized size = 1.05 \begin {gather*} \frac {{\left (d f q {\left (\frac {\log \left (b x + a\right )}{b c - a d} - \frac {\log \left (d x + c\right )}{b c - a d}\right )} - \frac {b f p}{b^{2} x + a b}\right )} r}{b f} - \frac {\log \left (\left ({\left (b x + a\right )}^{p} {\left (d x + c\right )}^{q} f\right )^{r} e\right )}{{\left (b x + a\right )} b} \end {gather*}

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)^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.36, size = 116, normalized size = 1.22 \begin {gather*} -\frac {{\left (b c - a d\right )} p r + {\left (b c - a d\right )} r \log \left (f\right ) + b c - a d - {\left (b d q r x + {\left (a d q - {\left (b c - a d\right )} p\right )} r\right )} \log \left (b x + a\right ) + {\left (b d q r x + b c q r\right )} \log \left (d x + c\right )}{a b^{2} c - a^{2} b d + {\left (b^{3} c - a b^{2} d\right )} x} \end {gather*}

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)^2,x, algorithm="fricas")

[Out]

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

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 1112 vs. \(2 (76) = 152\).
time = 141.38, size = 1112, normalized size = 11.71 \begin {gather*} \begin {cases} - \frac {p r}{a b + b^{2} x} - \frac {\log {\left (e \left (c^{q} f \left (a + b x\right )^{p}\right )^{r} \right )}}{a b + b^{2} x} & \text {for}\: d = 0 \\\tilde {\infty } \left (\frac {2 \cdot 0^{p} \tilde {\infty }^{2 p} c q r \log {\left (c + d x \right )}}{0^{p} \tilde {\infty }^{p} d - d} - \frac {0^{p} \tilde {\infty }^{2 p} d q r x}{0^{p} \tilde {\infty }^{p} d - d} - \frac {2 \cdot 0^{p} \tilde {\infty }^{p} c q r \log {\left (c + d x \right )}}{0^{p} \tilde {\infty }^{p} d - d} - \frac {0^{p} \tilde {\infty }^{p} c \log {\left (e \left (0^{p} f \left (c + d x\right )^{q}\right )^{r} \right )}}{0^{p} \tilde {\infty }^{p} d - d} + \frac {0^{p} \tilde {\infty }^{p} d q r x}{0^{p} \tilde {\infty }^{p} d - d} + \frac {0^{p} \tilde {\infty }^{p} d x \log {\left (e \left (0^{p} f \left (c + d x\right )^{q}\right )^{r} \right )}}{0^{p} \tilde {\infty }^{p} d - d} + \frac {c \log {\left (e \left (0^{p} f \left (c + d x\right )^{q}\right )^{r} \right )}}{0^{p} \tilde {\infty }^{p} d - d} - \frac {d x \log {\left (e \left (0^{p} f \left (c + d x\right )^{q}\right )^{r} \right )}}{0^{p} \tilde {\infty }^{p} d - d}\right ) & \text {for}\: a = - b x \\\frac {\frac {c \log {\left (e \left (a^{p} f \left (c + d x\right )^{q}\right )^{r} \right )}}{d} - q r x + x \log {\left (e \left (a^{p} f \left (c + d x\right )^{q}\right )^{r} \right )}}{a^{2}} & \text {for}\: b = 0 \\- \frac {p r}{a b + b^{2} x} - \frac {q r}{a b + b^{2} x} - \frac {\log {\left (e \left (f \left (a + b x\right )^{p} \left (\frac {a d}{b} + d x\right )^{q}\right )^{r} \right )}}{a b + b^{2} x} & \text {for}\: c = \frac {a d}{b} \\- \frac {a d q r \log {\left (\frac {a}{b} + x \right )}}{a^{2} b d - a b^{2} c + a b^{2} d x - b^{3} c x} + \frac {b c \log {\left (e \left (f \left (c + d x\right )^{q}\right )^{r} \right )}}{a^{2} b d - a b^{2} c + a b^{2} d x - b^{3} c x} - \frac {b d q r x \log {\left (\frac {a}{b} + x \right )}}{a^{2} b d - a b^{2} c + a b^{2} d x - b^{3} c x} + \frac {b d x \log {\left (e \left (f \left (c + d x\right )^{q}\right )^{r} \right )}}{a^{2} b d - a b^{2} c + a b^{2} d x - b^{3} c x} & \text {for}\: p = 0 \\- \frac {a d p^{2} r}{a^{2} b d p - a b^{2} c p + a b^{2} d p x - b^{3} c p x} + \frac {a d p q r \log {\left (\frac {c}{d} + x \right )}}{a^{2} b d p - a b^{2} c p + a b^{2} d p x - b^{3} c p x} - \frac {a d p \log {\left (e \left (f \left (a + b x\right )^{p} \left (c + d x\right )^{q}\right )^{r} \right )}}{a^{2} b d p - a b^{2} c p + a b^{2} d p x - b^{3} c p x} + \frac {a d q^{2} r \log {\left (\frac {c}{d} + x \right )}}{a^{2} b d p - a b^{2} c p + a b^{2} d p x - b^{3} c p x} - \frac {a d q \log {\left (e \left (f \left (a + b x\right )^{p} \left (c + d x\right )^{q}\right )^{r} \right )}}{a^{2} b d p - a b^{2} c p + a b^{2} d p x - b^{3} c p x} + \frac {b c p^{2} r}{a^{2} b d p - a b^{2} c p + a b^{2} d p x - b^{3} c p x} + \frac {b c p \log {\left (e \left (f \left (a + b x\right )^{p} \left (c + d x\right )^{q}\right )^{r} \right )}}{a^{2} b d p - a b^{2} c p + a b^{2} d p x - b^{3} c p x} + \frac {b d p q r x \log {\left (\frac {c}{d} + x \right )}}{a^{2} b d p - a b^{2} c p + a b^{2} d p x - b^{3} c p x} + \frac {b d q^{2} r x \log {\left (\frac {c}{d} + x \right )}}{a^{2} b d p - a b^{2} c p + a b^{2} d p x - b^{3} c p x} - \frac {b d q x \log {\left (e \left (f \left (a + b x\right )^{p} \left (c + d x\right )^{q}\right )^{r} \right )}}{a^{2} b d p - a b^{2} c p + a b^{2} d p x - b^{3} c p x} & \text {otherwise} \end {cases} \end {gather*}

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)**2,x)

[Out]

Piecewise((-p*r/(a*b + b**2*x) - log(e*(c**q*f*(a + b*x)**p)**r)/(a*b + b**2*x), Eq(d, 0)), (zoo*(2*0**p*zoo**
(2*p)*c*q*r*log(c + d*x)/(0**p*zoo**p*d - d) - 0**p*zoo**(2*p)*d*q*r*x/(0**p*zoo**p*d - d) - 2*0**p*zoo**p*c*q
*r*log(c + d*x)/(0**p*zoo**p*d - d) - 0**p*zoo**p*c*log(e*(0**p*f*(c + d*x)**q)**r)/(0**p*zoo**p*d - d) + 0**p
*zoo**p*d*q*r*x/(0**p*zoo**p*d - d) + 0**p*zoo**p*d*x*log(e*(0**p*f*(c + d*x)**q)**r)/(0**p*zoo**p*d - d) + c*
log(e*(0**p*f*(c + d*x)**q)**r)/(0**p*zoo**p*d - d) - d*x*log(e*(0**p*f*(c + d*x)**q)**r)/(0**p*zoo**p*d - d))
, Eq(a, -b*x)), ((c*log(e*(a**p*f*(c + d*x)**q)**r)/d - q*r*x + x*log(e*(a**p*f*(c + d*x)**q)**r))/a**2, Eq(b,
0)), (-p*r/(a*b + b**2*x) - q*r/(a*b + b**2*x) - log(e*(f*(a + b*x)**p*(a*d/b + d*x)**q)**r)/(a*b + b**2*x),
Eq(c, a*d/b)), (-a*d*q*r*log(a/b + x)/(a**2*b*d - a*b**2*c + a*b**2*d*x - b**3*c*x) + b*c*log(e*(f*(c + d*x)**
q)**r)/(a**2*b*d - a*b**2*c + a*b**2*d*x - b**3*c*x) - b*d*q*r*x*log(a/b + x)/(a**2*b*d - a*b**2*c + a*b**2*d*
x - b**3*c*x) + b*d*x*log(e*(f*(c + d*x)**q)**r)/(a**2*b*d - a*b**2*c + a*b**2*d*x - b**3*c*x), Eq(p, 0)), (-a
*d*p**2*r/(a**2*b*d*p - a*b**2*c*p + a*b**2*d*p*x - b**3*c*p*x) + a*d*p*q*r*log(c/d + x)/(a**2*b*d*p - a*b**2*
c*p + a*b**2*d*p*x - b**3*c*p*x) - a*d*p*log(e*(f*(a + b*x)**p*(c + d*x)**q)**r)/(a**2*b*d*p - a*b**2*c*p + a*
b**2*d*p*x - b**3*c*p*x) + a*d*q**2*r*log(c/d + x)/(a**2*b*d*p - a*b**2*c*p + a*b**2*d*p*x - b**3*c*p*x) - a*d
*q*log(e*(f*(a + b*x)**p*(c + d*x)**q)**r)/(a**2*b*d*p - a*b**2*c*p + a*b**2*d*p*x - b**3*c*p*x) + b*c*p**2*r/
(a**2*b*d*p - a*b**2*c*p + a*b**2*d*p*x - b**3*c*p*x) + b*c*p*log(e*(f*(a + b*x)**p*(c + d*x)**q)**r)/(a**2*b*
d*p - a*b**2*c*p + a*b**2*d*p*x - b**3*c*p*x) + b*d*p*q*r*x*log(c/d + x)/(a**2*b*d*p - a*b**2*c*p + a*b**2*d*p
*x - b**3*c*p*x) + b*d*q**2*r*x*log(c/d + x)/(a**2*b*d*p - a*b**2*c*p + a*b**2*d*p*x - b**3*c*p*x) - b*d*q*x*l
og(e*(f*(a + b*x)**p*(c + d*x)**q)**r)/(a**2*b*d*p - a*b**2*c*p + a*b**2*d*p*x - b**3*c*p*x), True))

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Giac [A]
time = 3.96, size = 112, normalized size = 1.18 \begin {gather*} \frac {d q r \log \left (b x + a\right )}{b^{2} c - a b d} - \frac {d q r \log \left (d x + c\right )}{b^{2} c - a b d} - \frac {p r \log \left (b x + a\right )}{b^{2} x + a b} - \frac {q r \log \left (d x + c\right )}{b^{2} x + a b} - \frac {p r + r \log \left (f\right ) + 1}{b^{2} x + a b} \end {gather*}

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)^2,x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 2.16, size = 99, normalized size = 1.04 \begin {gather*} -\frac {\ln \left (e\,{\left (f\,{\left (a+b\,x\right )}^p\,{\left (c+d\,x\right )}^q\right )}^r\right )\,\left (x+\frac {a}{b}\right )}{{\left (a+b\,x\right )}^2}-\frac {p\,r}{x\,b^2+a\,b}+\frac {d\,q\,r\,\mathrm {atan}\left (\frac {b\,c\,2{}\mathrm {i}+b\,d\,x\,2{}\mathrm {i}}{a\,d-b\,c}+1{}\mathrm {i}\right )\,2{}\mathrm {i}}{b\,\left (a\,d-b\,c\right )} \end {gather*}

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)^2,x)

[Out]

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

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