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

Optimal. Leaf size=158 \[ \frac{(g+h x)^4 \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{4 h}-\frac{b p q x (f g-e h)^3}{4 f^3}-\frac{b p q (g+h x)^2 (f g-e h)^2}{8 f^2 h}-\frac{b p q (f g-e h)^4 \log (e+f x)}{4 f^4 h}-\frac{b p q (g+h x)^3 (f g-e h)}{12 f h}-\frac{b p q (g+h x)^4}{16 h} \]

[Out]

-(b*(f*g - e*h)^3*p*q*x)/(4*f^3) - (b*(f*g - e*h)^2*p*q*(g + h*x)^2)/(8*f^2*h) - (b*(f*g - e*h)*p*q*(g + h*x)^
3)/(12*f*h) - (b*p*q*(g + h*x)^4)/(16*h) - (b*(f*g - e*h)^4*p*q*Log[e + f*x])/(4*f^4*h) + ((g + h*x)^4*(a + b*
Log[c*(d*(e + f*x)^p)^q]))/(4*h)

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Rubi [A]  time = 0.163642, antiderivative size = 158, 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, 43, 2445} \[ \frac{(g+h x)^4 \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{4 h}-\frac{b p q x (f g-e h)^3}{4 f^3}-\frac{b p q (g+h x)^2 (f g-e h)^2}{8 f^2 h}-\frac{b p q (f g-e h)^4 \log (e+f x)}{4 f^4 h}-\frac{b p q (g+h x)^3 (f g-e h)}{12 f h}-\frac{b p q (g+h x)^4}{16 h} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*(f*g - e*h)^3*p*q*x)/(4*f^3) - (b*(f*g - e*h)^2*p*q*(g + h*x)^2)/(8*f^2*h) - (b*(f*g - e*h)*p*q*(g + h*x)^
3)/(12*f*h) - (b*p*q*(g + h*x)^4)/(16*h) - (b*(f*g - e*h)^4*p*q*Log[e + f*x])/(4*f^4*h) + ((g + h*x)^4*(a + b*
Log[c*(d*(e + f*x)^p)^q]))/(4*h)

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 43

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[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 (g+h x)^3 \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right ) \, dx &=\operatorname{Subst}\left (\int (g+h x)^3 \left (a+b \log \left (c d^q (e+f x)^{p q}\right )\right ) \, dx,c d^q (e+f x)^{p q},c \left (d (e+f x)^p\right )^q\right )\\ &=\frac{(g+h x)^4 \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{4 h}-\operatorname{Subst}\left (\frac{(b f p q) \int \frac{(g+h x)^4}{e+f x} \, dx}{4 h},c d^q (e+f x)^{p q},c \left (d (e+f x)^p\right )^q\right )\\ &=\frac{(g+h x)^4 \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{4 h}-\operatorname{Subst}\left (\frac{(b f p q) \int \left (\frac{h (f g-e h)^3}{f^4}+\frac{(f g-e h)^4}{f^4 (e+f x)}+\frac{h (f g-e h)^2 (g+h x)}{f^3}+\frac{h (f g-e h) (g+h x)^2}{f^2}+\frac{h (g+h x)^3}{f}\right ) \, dx}{4 h},c d^q (e+f x)^{p q},c \left (d (e+f x)^p\right )^q\right )\\ &=-\frac{b (f g-e h)^3 p q x}{4 f^3}-\frac{b (f g-e h)^2 p q (g+h x)^2}{8 f^2 h}-\frac{b (f g-e h) p q (g+h x)^3}{12 f h}-\frac{b p q (g+h x)^4}{16 h}-\frac{b (f g-e h)^4 p q \log (e+f x)}{4 f^4 h}+\frac{(g+h x)^4 \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{4 h}\\ \end{align*}

Mathematica [A]  time = 0.291921, size = 232, normalized size = 1.47 \[ \frac{f x \left (12 a f^3 \left (6 g^2 h x+4 g^3+4 g h^2 x^2+h^3 x^3\right )-b p q \left (6 e^2 f h^2 (8 g+h x)-12 e^3 h^3-4 e f^2 h \left (18 g^2+6 g h x+h^2 x^2\right )+f^3 \left (36 g^2 h x+48 g^3+16 g h^2 x^2+3 h^3 x^3\right )\right )\right )+12 b f^3 \left (4 e g^3+f x \left (6 g^2 h x+4 g^3+4 g h^2 x^2+h^3 x^3\right )\right ) \log \left (c \left (d (e+f x)^p\right )^q\right )-12 b e^2 h p q \left (e^2 h^2-4 e f g h+6 f^2 g^2\right ) \log (e+f x)}{48 f^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(f*x*(12*a*f^3*(4*g^3 + 6*g^2*h*x + 4*g*h^2*x^2 + h^3*x^3) - b*p*q*(-12*e^3*h^3 + 6*e^2*f*h^2*(8*g + h*x) - 4*
e*f^2*h*(18*g^2 + 6*g*h*x + h^2*x^2) + f^3*(48*g^3 + 36*g^2*h*x + 16*g*h^2*x^2 + 3*h^3*x^3))) - 12*b*e^2*h*(6*
f^2*g^2 - 4*e*f*g*h + e^2*h^2)*p*q*Log[e + f*x] + 12*b*f^3*(4*e*g^3 + f*x*(4*g^3 + 6*g^2*h*x + 4*g*h^2*x^2 + h
^3*x^3))*Log[c*(d*(e + f*x)^p)^q])/(48*f^4)

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Maple [F]  time = 0.662, size = 0, normalized size = 0. \begin{align*} \int \left ( hx+g \right ) ^{3} \left ( a+b\ln \left ( c \left ( d \left ( fx+e \right ) ^{p} \right ) ^{q} \right ) \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.12445, size = 410, normalized size = 2.59 \begin{align*} \frac{1}{4} \, b h^{3} x^{4} \log \left (\left ({\left (f x + e\right )}^{p} d\right )^{q} c\right ) + \frac{1}{4} \, a h^{3} x^{4} - b f g^{3} p q{\left (\frac{x}{f} - \frac{e \log \left (f x + e\right )}{f^{2}}\right )} - \frac{1}{48} \, b f h^{3} p q{\left (\frac{12 \, e^{4} \log \left (f x + e\right )}{f^{5}} + \frac{3 \, f^{3} x^{4} - 4 \, e f^{2} x^{3} + 6 \, e^{2} f x^{2} - 12 \, e^{3} x}{f^{4}}\right )} + \frac{1}{6} \, b f g h^{2} p q{\left (\frac{6 \, e^{3} \log \left (f x + e\right )}{f^{4}} - \frac{2 \, f^{2} x^{3} - 3 \, e f x^{2} + 6 \, e^{2} x}{f^{3}}\right )} - \frac{3}{4} \, b f g^{2} h p q{\left (\frac{2 \, e^{2} \log \left (f x + e\right )}{f^{3}} + \frac{f x^{2} - 2 \, e x}{f^{2}}\right )} + b g h^{2} x^{3} \log \left (\left ({\left (f x + e\right )}^{p} d\right )^{q} c\right ) + a g h^{2} x^{3} + \frac{3}{2} \, b g^{2} h x^{2} \log \left (\left ({\left (f x + e\right )}^{p} d\right )^{q} c\right ) + \frac{3}{2} \, a g^{2} h x^{2} + b g^{3} x \log \left (\left ({\left (f x + e\right )}^{p} d\right )^{q} c\right ) + a g^{3} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*b*h^3*x^4*log(((f*x + e)^p*d)^q*c) + 1/4*a*h^3*x^4 - b*f*g^3*p*q*(x/f - e*log(f*x + e)/f^2) - 1/48*b*f*h^3
*p*q*(12*e^4*log(f*x + e)/f^5 + (3*f^3*x^4 - 4*e*f^2*x^3 + 6*e^2*f*x^2 - 12*e^3*x)/f^4) + 1/6*b*f*g*h^2*p*q*(6
*e^3*log(f*x + e)/f^4 - (2*f^2*x^3 - 3*e*f*x^2 + 6*e^2*x)/f^3) - 3/4*b*f*g^2*h*p*q*(2*e^2*log(f*x + e)/f^3 + (
f*x^2 - 2*e*x)/f^2) + b*g*h^2*x^3*log(((f*x + e)^p*d)^q*c) + a*g*h^2*x^3 + 3/2*b*g^2*h*x^2*log(((f*x + e)^p*d)
^q*c) + 3/2*a*g^2*h*x^2 + b*g^3*x*log(((f*x + e)^p*d)^q*c) + a*g^3*x

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Fricas [B]  time = 2.08039, size = 861, normalized size = 5.45 \begin{align*} -\frac{3 \,{\left (b f^{4} h^{3} p q - 4 \, a f^{4} h^{3}\right )} x^{4} - 4 \,{\left (12 \, a f^{4} g h^{2} -{\left (4 \, b f^{4} g h^{2} - b e f^{3} h^{3}\right )} p q\right )} x^{3} - 6 \,{\left (12 \, a f^{4} g^{2} h -{\left (6 \, b f^{4} g^{2} h - 4 \, b e f^{3} g h^{2} + b e^{2} f^{2} h^{3}\right )} p q\right )} x^{2} - 12 \,{\left (4 \, a f^{4} g^{3} -{\left (4 \, b f^{4} g^{3} - 6 \, b e f^{3} g^{2} h + 4 \, b e^{2} f^{2} g h^{2} - b e^{3} f h^{3}\right )} p q\right )} x - 12 \,{\left (b f^{4} h^{3} p q x^{4} + 4 \, b f^{4} g h^{2} p q x^{3} + 6 \, b f^{4} g^{2} h p q x^{2} + 4 \, b f^{4} g^{3} p q x +{\left (4 \, b e f^{3} g^{3} - 6 \, b e^{2} f^{2} g^{2} h + 4 \, b e^{3} f g h^{2} - b e^{4} h^{3}\right )} p q\right )} \log \left (f x + e\right ) - 12 \,{\left (b f^{4} h^{3} x^{4} + 4 \, b f^{4} g h^{2} x^{3} + 6 \, b f^{4} g^{2} h x^{2} + 4 \, b f^{4} g^{3} x\right )} \log \left (c\right ) - 12 \,{\left (b f^{4} h^{3} q x^{4} + 4 \, b f^{4} g h^{2} q x^{3} + 6 \, b f^{4} g^{2} h q x^{2} + 4 \, b f^{4} g^{3} q x\right )} \log \left (d\right )}{48 \, f^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/48*(3*(b*f^4*h^3*p*q - 4*a*f^4*h^3)*x^4 - 4*(12*a*f^4*g*h^2 - (4*b*f^4*g*h^2 - b*e*f^3*h^3)*p*q)*x^3 - 6*(1
2*a*f^4*g^2*h - (6*b*f^4*g^2*h - 4*b*e*f^3*g*h^2 + b*e^2*f^2*h^3)*p*q)*x^2 - 12*(4*a*f^4*g^3 - (4*b*f^4*g^3 -
6*b*e*f^3*g^2*h + 4*b*e^2*f^2*g*h^2 - b*e^3*f*h^3)*p*q)*x - 12*(b*f^4*h^3*p*q*x^4 + 4*b*f^4*g*h^2*p*q*x^3 + 6*
b*f^4*g^2*h*p*q*x^2 + 4*b*f^4*g^3*p*q*x + (4*b*e*f^3*g^3 - 6*b*e^2*f^2*g^2*h + 4*b*e^3*f*g*h^2 - b*e^4*h^3)*p*
q)*log(f*x + e) - 12*(b*f^4*h^3*x^4 + 4*b*f^4*g*h^2*x^3 + 6*b*f^4*g^2*h*x^2 + 4*b*f^4*g^3*x)*log(c) - 12*(b*f^
4*h^3*q*x^4 + 4*b*f^4*g*h^2*q*x^3 + 6*b*f^4*g^2*h*q*x^2 + 4*b*f^4*g^3*q*x)*log(d))/f^4

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Sympy [A]  time = 24.7956, size = 546, normalized size = 3.46 \begin{align*} \begin{cases} a g^{3} x + \frac{3 a g^{2} h x^{2}}{2} + a g h^{2} x^{3} + \frac{a h^{3} x^{4}}{4} - \frac{b e^{4} h^{3} p q \log{\left (e + f x \right )}}{4 f^{4}} + \frac{b e^{3} g h^{2} p q \log{\left (e + f x \right )}}{f^{3}} + \frac{b e^{3} h^{3} p q x}{4 f^{3}} - \frac{3 b e^{2} g^{2} h p q \log{\left (e + f x \right )}}{2 f^{2}} - \frac{b e^{2} g h^{2} p q x}{f^{2}} - \frac{b e^{2} h^{3} p q x^{2}}{8 f^{2}} + \frac{b e g^{3} p q \log{\left (e + f x \right )}}{f} + \frac{3 b e g^{2} h p q x}{2 f} + \frac{b e g h^{2} p q x^{2}}{2 f} + \frac{b e h^{3} p q x^{3}}{12 f} + b g^{3} p q x \log{\left (e + f x \right )} - b g^{3} p q x + b g^{3} q x \log{\left (d \right )} + b g^{3} x \log{\left (c \right )} + \frac{3 b g^{2} h p q x^{2} \log{\left (e + f x \right )}}{2} - \frac{3 b g^{2} h p q x^{2}}{4} + \frac{3 b g^{2} h q x^{2} \log{\left (d \right )}}{2} + \frac{3 b g^{2} h x^{2} \log{\left (c \right )}}{2} + b g h^{2} p q x^{3} \log{\left (e + f x \right )} - \frac{b g h^{2} p q x^{3}}{3} + b g h^{2} q x^{3} \log{\left (d \right )} + b g h^{2} x^{3} \log{\left (c \right )} + \frac{b h^{3} p q x^{4} \log{\left (e + f x \right )}}{4} - \frac{b h^{3} p q x^{4}}{16} + \frac{b h^{3} q x^{4} \log{\left (d \right )}}{4} + \frac{b h^{3} x^{4} \log{\left (c \right )}}{4} & \text{for}\: f \neq 0 \\\left (a + b \log{\left (c \left (d e^{p}\right )^{q} \right )}\right ) \left (g^{3} x + \frac{3 g^{2} h x^{2}}{2} + g h^{2} x^{3} + \frac{h^{3} x^{4}}{4}\right ) & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a*g**3*x + 3*a*g**2*h*x**2/2 + a*g*h**2*x**3 + a*h**3*x**4/4 - b*e**4*h**3*p*q*log(e + f*x)/(4*f**4
) + b*e**3*g*h**2*p*q*log(e + f*x)/f**3 + b*e**3*h**3*p*q*x/(4*f**3) - 3*b*e**2*g**2*h*p*q*log(e + f*x)/(2*f**
2) - b*e**2*g*h**2*p*q*x/f**2 - b*e**2*h**3*p*q*x**2/(8*f**2) + b*e*g**3*p*q*log(e + f*x)/f + 3*b*e*g**2*h*p*q
*x/(2*f) + b*e*g*h**2*p*q*x**2/(2*f) + b*e*h**3*p*q*x**3/(12*f) + b*g**3*p*q*x*log(e + f*x) - b*g**3*p*q*x + b
*g**3*q*x*log(d) + b*g**3*x*log(c) + 3*b*g**2*h*p*q*x**2*log(e + f*x)/2 - 3*b*g**2*h*p*q*x**2/4 + 3*b*g**2*h*q
*x**2*log(d)/2 + 3*b*g**2*h*x**2*log(c)/2 + b*g*h**2*p*q*x**3*log(e + f*x) - b*g*h**2*p*q*x**3/3 + b*g*h**2*q*
x**3*log(d) + b*g*h**2*x**3*log(c) + b*h**3*p*q*x**4*log(e + f*x)/4 - b*h**3*p*q*x**4/16 + b*h**3*q*x**4*log(d
)/4 + b*h**3*x**4*log(c)/4, Ne(f, 0)), ((a + b*log(c*(d*e**p)**q))*(g**3*x + 3*g**2*h*x**2/2 + g*h**2*x**3 + h
**3*x**4/4), True))

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Giac [B]  time = 1.23648, size = 1413, normalized size = 8.94 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(f*x + e)*b*g^3*p*q*log(f*x + e)/f + 3/2*(f*x + e)^2*b*g^2*h*p*q*log(f*x + e)/f^2 + (f*x + e)^3*b*g*h^2*p*q*lo
g(f*x + e)/f^3 + 1/4*(f*x + e)^4*b*h^3*p*q*log(f*x + e)/f^4 - 3*(f*x + e)*b*g^2*h*p*q*e*log(f*x + e)/f^2 - 3*(
f*x + e)^2*b*g*h^2*p*q*e*log(f*x + e)/f^3 - (f*x + e)^3*b*h^3*p*q*e*log(f*x + e)/f^4 - (f*x + e)*b*g^3*p*q/f -
3/4*(f*x + e)^2*b*g^2*h*p*q/f^2 - 1/3*(f*x + e)^3*b*g*h^2*p*q/f^3 - 1/16*(f*x + e)^4*b*h^3*p*q/f^4 + 3*(f*x +
e)*b*g^2*h*p*q*e/f^2 + 3/2*(f*x + e)^2*b*g*h^2*p*q*e/f^3 + 1/3*(f*x + e)^3*b*h^3*p*q*e/f^4 + 3*(f*x + e)*b*g*
h^2*p*q*e^2*log(f*x + e)/f^3 + 3/2*(f*x + e)^2*b*h^3*p*q*e^2*log(f*x + e)/f^4 + (f*x + e)*b*g^3*q*log(d)/f + 3
/2*(f*x + e)^2*b*g^2*h*q*log(d)/f^2 + (f*x + e)^3*b*g*h^2*q*log(d)/f^3 + 1/4*(f*x + e)^4*b*h^3*q*log(d)/f^4 -
3*(f*x + e)*b*g^2*h*q*e*log(d)/f^2 - 3*(f*x + e)^2*b*g*h^2*q*e*log(d)/f^3 - (f*x + e)^3*b*h^3*q*e*log(d)/f^4 -
3*(f*x + e)*b*g*h^2*p*q*e^2/f^3 - 3/4*(f*x + e)^2*b*h^3*p*q*e^2/f^4 - (f*x + e)*b*h^3*p*q*e^3*log(f*x + e)/f^
4 + (f*x + e)*b*g^3*log(c)/f + 3/2*(f*x + e)^2*b*g^2*h*log(c)/f^2 + (f*x + e)^3*b*g*h^2*log(c)/f^3 + 1/4*(f*x
+ e)^4*b*h^3*log(c)/f^4 - 3*(f*x + e)*b*g^2*h*e*log(c)/f^2 - 3*(f*x + e)^2*b*g*h^2*e*log(c)/f^3 - (f*x + e)^3*
b*h^3*e*log(c)/f^4 + 3*(f*x + e)*b*g*h^2*q*e^2*log(d)/f^3 + 3/2*(f*x + e)^2*b*h^3*q*e^2*log(d)/f^4 + (f*x + e)
*a*g^3/f + 3/2*(f*x + e)^2*a*g^2*h/f^2 + (f*x + e)^3*a*g*h^2/f^3 + 1/4*(f*x + e)^4*a*h^3/f^4 + (f*x + e)*b*h^3
*p*q*e^3/f^4 - 3*(f*x + e)*a*g^2*h*e/f^2 - 3*(f*x + e)^2*a*g*h^2*e/f^3 - (f*x + e)^3*a*h^3*e/f^4 + 3*(f*x + e)
*b*g*h^2*e^2*log(c)/f^3 + 3/2*(f*x + e)^2*b*h^3*e^2*log(c)/f^4 - (f*x + e)*b*h^3*q*e^3*log(d)/f^4 + 3*(f*x + e
)*a*g*h^2*e^2/f^3 + 3/2*(f*x + e)^2*a*h^3*e^2/f^4 - (f*x + e)*b*h^3*e^3*log(c)/f^4 - (f*x + e)*a*h^3*e^3/f^4