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

Optimal. Leaf size=119 \[ -\frac{a+b \log \left (c \left (d (e+f x)^p\right )^q\right )}{2 h (g+h x)^2}+\frac{b f^2 p q \log (e+f x)}{2 h (f g-e h)^2}-\frac{b f^2 p q \log (g+h x)}{2 h (f g-e h)^2}+\frac{b f p q}{2 h (g+h x) (f g-e h)} \]

[Out]

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

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Rubi [A]  time = 0.129398, antiderivative size = 119, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.115, Rules used = {2395, 44, 2445} \[ -\frac{a+b \log \left (c \left (d (e+f x)^p\right )^q\right )}{2 h (g+h x)^2}+\frac{b f^2 p q \log (e+f x)}{2 h (f g-e h)^2}-\frac{b f^2 p q \log (g+h x)}{2 h (f g-e h)^2}+\frac{b f p q}{2 h (g+h x) (f g-e h)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2395

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 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 2445

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

Mathematica [A]  time = 0.147197, size = 88, normalized size = 0.74 \[ -\frac{a+b \log \left (c \left (d (e+f x)^p\right )^q\right )-\frac{b f p q (g+h x) (f (g+h x) \log (e+f x)-e h-f (g+h x) \log (g+h x)+f g)}{(f g-e h)^2}}{2 h (g+h x)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.664, size = 0, normalized size = 0. \begin{align*} \int{\frac{a+b\ln \left ( c \left ( d \left ( fx+e \right ) ^{p} \right ) ^{q} \right ) }{ \left ( hx+g \right ) ^{3}}}\, dx \end{align*}

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)^3,x)

[Out]

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

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Maxima [A]  time = 1.06033, size = 232, normalized size = 1.95 \begin{align*} \frac{1}{2} \, b f p q{\left (\frac{f \log \left (f x + e\right )}{f^{2} g^{2} h - 2 \, e f g h^{2} + e^{2} h^{3}} - \frac{f \log \left (h x + g\right )}{f^{2} g^{2} h - 2 \, e f g h^{2} + e^{2} h^{3}} + \frac{1}{f g^{2} h - e g h^{2} +{\left (f g h^{2} - e h^{3}\right )} x}\right )} - \frac{b \log \left (\left ({\left (f x + e\right )}^{p} d\right )^{q} c\right )}{2 \,{\left (h^{3} x^{2} + 2 \, g h^{2} x + g^{2} h\right )}} - \frac{a}{2 \,{\left (h^{3} x^{2} + 2 \, g h^{2} x + g^{2} h\right )}} \end{align*}

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)^3,x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 2.23678, size = 667, normalized size = 5.61 \begin{align*} -\frac{a f^{2} g^{2} - 2 \, a e f g h + a e^{2} h^{2} -{\left (b f^{2} g h - b e f h^{2}\right )} p q x -{\left (b f^{2} g^{2} - b e f g h\right )} p q +{\left (b f^{2} g^{2} - 2 \, b e f g h + b e^{2} h^{2}\right )} q \log \left (d\right ) -{\left (b f^{2} h^{2} p q x^{2} + 2 \, b f^{2} g h p q x +{\left (2 \, b e f g h - b e^{2} h^{2}\right )} p q\right )} \log \left (f x + e\right ) +{\left (b f^{2} h^{2} p q x^{2} + 2 \, b f^{2} g h p q x + b f^{2} g^{2} p q\right )} \log \left (h x + g\right ) +{\left (b f^{2} g^{2} - 2 \, b e f g h + b e^{2} h^{2}\right )} \log \left (c\right )}{2 \,{\left (f^{2} g^{4} h - 2 \, e f g^{3} h^{2} + e^{2} g^{2} h^{3} +{\left (f^{2} g^{2} h^{3} - 2 \, e f g h^{4} + e^{2} h^{5}\right )} x^{2} + 2 \,{\left (f^{2} g^{3} h^{2} - 2 \, e f g^{2} h^{3} + e^{2} g h^{4}\right )} x\right )}} \end{align*}

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)^3,x, algorithm="fricas")

[Out]

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

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

[Out]

Timed out

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Giac [B]  time = 1.1509, size = 485, normalized size = 4.08 \begin{align*} \frac{b f^{2} h^{2} p q x^{2} \log \left (f x + e\right ) - b f^{2} h^{2} p q x^{2} \log \left (h x + g\right ) + 2 \, b f^{2} g h p q x \log \left (f x + e\right ) - 2 \, b f^{2} g h p q x \log \left (h x + g\right ) + b f^{2} g h p q x - b f h^{2} p q x e + 2 \, b f g h p q e \log \left (f x + e\right ) - b f^{2} g^{2} p q \log \left (h x + g\right ) + b f^{2} g^{2} p q - b f g h p q e - b h^{2} p q e^{2} \log \left (f x + e\right ) - b f^{2} g^{2} q \log \left (d\right ) + 2 \, b f g h q e \log \left (d\right ) - b f^{2} g^{2} \log \left (c\right ) + 2 \, b f g h e \log \left (c\right ) - b h^{2} q e^{2} \log \left (d\right ) - a f^{2} g^{2} + 2 \, a f g h e - b h^{2} e^{2} \log \left (c\right ) - a h^{2} e^{2}}{2 \,{\left (f^{2} g^{2} h^{3} x^{2} - 2 \, f g h^{4} x^{2} e + 2 \, f^{2} g^{3} h^{2} x + h^{5} x^{2} e^{2} - 4 \, f g^{2} h^{3} x e + f^{2} g^{4} h + 2 \, g h^{4} x e^{2} - 2 \, f g^{3} h^{2} e + g^{2} h^{3} e^{2}\right )}} \end{align*}

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)^3,x, algorithm="giac")

[Out]

1/2*(b*f^2*h^2*p*q*x^2*log(f*x + e) - b*f^2*h^2*p*q*x^2*log(h*x + g) + 2*b*f^2*g*h*p*q*x*log(f*x + e) - 2*b*f^
2*g*h*p*q*x*log(h*x + g) + b*f^2*g*h*p*q*x - b*f*h^2*p*q*x*e + 2*b*f*g*h*p*q*e*log(f*x + e) - b*f^2*g^2*p*q*lo
g(h*x + g) + b*f^2*g^2*p*q - b*f*g*h*p*q*e - b*h^2*p*q*e^2*log(f*x + e) - b*f^2*g^2*q*log(d) + 2*b*f*g*h*q*e*l
og(d) - b*f^2*g^2*log(c) + 2*b*f*g*h*e*log(c) - b*h^2*q*e^2*log(d) - a*f^2*g^2 + 2*a*f*g*h*e - b*h^2*e^2*log(c
) - a*h^2*e^2)/(f^2*g^2*h^3*x^2 - 2*f*g*h^4*x^2*e + 2*f^2*g^3*h^2*x + h^5*x^2*e^2 - 4*f*g^2*h^3*x*e + f^2*g^4*
h + 2*g*h^4*x*e^2 - 2*f*g^3*h^2*e + g^2*h^3*e^2)