3.5.25 \(\int \frac {a+b \log (c (d (e+f x)^p)^q)}{(g+h x)^2} \, dx\) [425]

Optimal. Leaf size=80 \[ \frac {b f p q \log (e+f x)}{h (f g-e h)}-\frac {a+b \log \left (c \left (d (e+f x)^p\right )^q\right )}{h (g+h x)}-\frac {b f p q \log (g+h x)}{h (f g-e h)} \]

[Out]

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

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Rubi [A]
time = 0.06, antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2442, 36, 31, 2495} \begin {gather*} -\frac {a+b \log \left (c \left (d (e+f x)^p\right )^q\right )}{h (g+h x)}+\frac {b f p q \log (e+f x)}{h (f g-e h)}-\frac {b f p q \log (g+h x)}{h (f g-e h)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 31

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

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 2442

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

Rule 2495

Int[((a_.) + Log[(c_.)*((d_.)*((e_.) + (f_.)*(x_))^(m_.))^(n_)]*(b_.))^(p_.)*(u_.), x_Symbol] :> Subst[Int[u*(
a + b*Log[c*d^n*(e + f*x)^(m*n)])^p, x], c*d^n*(e + f*x)^(m*n), c*(d*(e + f*x)^m)^n] /; FreeQ[{a, b, c, d, e,
f, m, n, p}, x] &&  !IntegerQ[n] &&  !(EqQ[d, 1] && EqQ[m, 1]) && IntegralFreeQ[IntHide[u*(a + b*Log[c*d^n*(e
+ f*x)^(m*n)])^p, x]]

Rubi steps

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

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Mathematica [A]
time = 0.06, size = 97, normalized size = 1.21 \begin {gather*} \frac {a f g-a e h-b f p q (g+h x) \log (e+f x)+b (f g-e h) \log \left (c \left (d (e+f x)^p\right )^q\right )+b f g p q \log (g+h x)+b f h p q x \log (g+h x)}{h (-f g+e h) (g+h x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.22, size = 0, normalized size = 0.00 \[\int \frac {a +b \ln \left (c \left (d \left (f x +e \right )^{p}\right )^{q}\right )}{\left (h x +g \right )^{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.29, size = 94, normalized size = 1.18 \begin {gather*} b f p q {\left (\frac {\log \left (f x + e\right )}{f g h - h^{2} e} - \frac {\log \left (h x + g\right )}{f g h - h^{2} e}\right )} - \frac {b \log \left (\left ({\left (f x + e\right )}^{p} d\right )^{q} c\right )}{h^{2} x + g h} - \frac {a}{h^{2} x + g h} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*(d*(f*x+e)^p)^q))/(h*x+g)^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.40, size = 119, normalized size = 1.49 \begin {gather*} -\frac {a f g - a h e - {\left (b f h p q x + b h p q e\right )} \log \left (f x + e\right ) + {\left (b f h p q x + b f g p q\right )} \log \left (h x + g\right ) + {\left (b f g - b h e\right )} \log \left (c\right ) + {\left (b f g q - b h q e\right )} \log \left (d\right )}{f g h^{2} x + f g^{2} h - {\left (h^{3} x + g h^{2}\right )} e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*(d*(f*x+e)^p)^q))/(h*x+g)^2,x, algorithm="fricas")

[Out]

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

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: NotImplementedError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*ln(c*(d*(f*x+e)**p)**q))/(h*x+g)**2,x)

[Out]

Exception raised: NotImplementedError >> no valid subset found

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Giac [A]
time = 6.06, size = 129, normalized size = 1.61 \begin {gather*} \frac {b f h p q x \log \left (f x + e\right ) - b f h p q x \log \left (h x + g\right ) + b h p q e \log \left (f x + e\right ) - b f g p q \log \left (h x + g\right ) - b f g q \log \left (d\right ) + b h q e \log \left (d\right ) - b f g \log \left (c\right ) + b h e \log \left (c\right ) - a f g + a h e}{f g h^{2} x - h^{3} x e + f g^{2} h - g h^{2} e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*(d*(f*x+e)^p)^q))/(h*x+g)^2,x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 2.04, size = 89, normalized size = 1.11 \begin {gather*} -\frac {a}{x\,h^2+g\,h}-\frac {b\,\ln \left (c\,{\left (d\,{\left (e+f\,x\right )}^p\right )}^q\right )}{h\,\left (g+h\,x\right )}+\frac {b\,f\,p\,q\,\mathrm {atan}\left (\frac {f\,g\,2{}\mathrm {i}+f\,h\,x\,2{}\mathrm {i}}{e\,h-f\,g}+1{}\mathrm {i}\right )\,2{}\mathrm {i}}{h\,\left (e\,h-f\,g\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*log(c*(d*(e + f*x)^p)^q))/(g + h*x)^2,x)

[Out]

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

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