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

Optimal. Leaf size=98 \[ \frac{(g+h x)^2 \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{2 h}-\frac{b p q (f g-e h)^2 \log (e+f x)}{2 f^2 h}-\frac{b p q x (f g-e h)}{2 f}-\frac{b p q (g+h x)^2}{4 h} \]

[Out]

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

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Rubi [A]  time = 0.0813518, antiderivative size = 98, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {2395, 43, 2445} \[ \frac{(g+h x)^2 \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{2 h}-\frac{b p q (f g-e h)^2 \log (e+f x)}{2 f^2 h}-\frac{b p q x (f g-e h)}{2 f}-\frac{b p q (g+h x)^2}{4 h} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*(f*g - e*h)*p*q*x)/(2*f) - (b*p*q*(g + h*x)^2)/(4*h) - (b*(f*g - e*h)^2*p*q*Log[e + f*x])/(2*f^2*h) + ((g
+ h*x)^2*(a + b*Log[c*(d*(e + f*x)^p)^q]))/(2*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) \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right ) \, dx &=\operatorname{Subst}\left (\int (g+h x) \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)^2 \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{2 h}-\operatorname{Subst}\left (\frac{(b f p q) \int \frac{(g+h x)^2}{e+f x} \, dx}{2 h},c d^q (e+f x)^{p q},c \left (d (e+f x)^p\right )^q\right )\\ &=\frac{(g+h x)^2 \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{2 h}-\operatorname{Subst}\left (\frac{(b f p q) \int \left (\frac{h (f g-e h)}{f^2}+\frac{(f g-e h)^2}{f^2 (e+f x)}+\frac{h (g+h x)}{f}\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 g-e h) p q x}{2 f}-\frac{b p q (g+h x)^2}{4 h}-\frac{b (f g-e h)^2 p q \log (e+f x)}{2 f^2 h}+\frac{(g+h x)^2 \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{2 h}\\ \end{align*}

Mathematica [A]  time = 0.0591751, size = 113, normalized size = 1.15 \[ a g x+\frac{1}{2} a h x^2+\frac{b g (e+f x) \log \left (c \left (d (e+f x)^p\right )^q\right )}{f}+\frac{1}{2} b h x^2 \log \left (c \left (d (e+f x)^p\right )^q\right )-\frac{b e^2 h p q \log (e+f x)}{2 f^2}+\frac{b e h p q x}{2 f}-b g p q x-\frac{1}{4} b h p q x^2 \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.278, size = 0, normalized size = 0. \begin{align*} \int \left ( hx+g \right ) \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)*(a+b*ln(c*(d*(f*x+e)^p)^q)),x)

[Out]

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

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Maxima [A]  time = 1.04245, size = 151, normalized size = 1.54 \begin{align*} -b f g p q{\left (\frac{x}{f} - \frac{e \log \left (f x + e\right )}{f^{2}}\right )} - \frac{1}{4} \, b f 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 )} + \frac{1}{2} \, b h x^{2} \log \left (\left ({\left (f x + e\right )}^{p} d\right )^{q} c\right ) + \frac{1}{2} \, a h x^{2} + b g x \log \left (\left ({\left (f x + e\right )}^{p} d\right )^{q} c\right ) + a g x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.11382, size = 339, normalized size = 3.46 \begin{align*} -\frac{{\left (b f^{2} h p q - 2 \, a f^{2} h\right )} x^{2} - 2 \,{\left (2 \, a f^{2} g -{\left (2 \, b f^{2} g - b e f h\right )} p q\right )} x - 2 \,{\left (b f^{2} h p q x^{2} + 2 \, b f^{2} g p q x +{\left (2 \, b e f g - b e^{2} h\right )} p q\right )} \log \left (f x + e\right ) - 2 \,{\left (b f^{2} h x^{2} + 2 \, b f^{2} g x\right )} \log \left (c\right ) - 2 \,{\left (b f^{2} h q x^{2} + 2 \, b f^{2} g q x\right )} \log \left (d\right )}{4 \, f^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 3.47904, size = 187, normalized size = 1.91 \begin{align*} \begin{cases} a g x + \frac{a h x^{2}}{2} - \frac{b e^{2} h p q \log{\left (e + f x \right )}}{2 f^{2}} + \frac{b e g p q \log{\left (e + f x \right )}}{f} + \frac{b e h p q x}{2 f} + b g p q x \log{\left (e + f x \right )} - b g p q x + b g q x \log{\left (d \right )} + b g x \log{\left (c \right )} + \frac{b h p q x^{2} \log{\left (e + f x \right )}}{2} - \frac{b h p q x^{2}}{4} + \frac{b h q x^{2} \log{\left (d \right )}}{2} + \frac{b h x^{2} \log{\left (c \right )}}{2} & \text{for}\: f \neq 0 \\\left (a + b \log{\left (c \left (d e^{p}\right )^{q} \right )}\right ) \left (g x + \frac{h x^{2}}{2}\right ) & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.29359, size = 350, normalized size = 3.57 \begin{align*} \frac{{\left (f x + e\right )} b g p q \log \left (f x + e\right )}{f} + \frac{{\left (f x + e\right )}^{2} b h p q \log \left (f x + e\right )}{2 \, f^{2}} - \frac{{\left (f x + e\right )} b h p q e \log \left (f x + e\right )}{f^{2}} - \frac{{\left (f x + e\right )} b g p q}{f} - \frac{{\left (f x + e\right )}^{2} b h p q}{4 \, f^{2}} + \frac{{\left (f x + e\right )} b h p q e}{f^{2}} + \frac{{\left (f x + e\right )} b g q \log \left (d\right )}{f} + \frac{{\left (f x + e\right )}^{2} b h q \log \left (d\right )}{2 \, f^{2}} - \frac{{\left (f x + e\right )} b h q e \log \left (d\right )}{f^{2}} + \frac{{\left (f x + e\right )} b g \log \left (c\right )}{f} + \frac{{\left (f x + e\right )}^{2} b h \log \left (c\right )}{2 \, f^{2}} - \frac{{\left (f x + e\right )} b h e \log \left (c\right )}{f^{2}} + \frac{{\left (f x + e\right )} a g}{f} + \frac{{\left (f x + e\right )}^{2} a h}{2 \, f^{2}} - \frac{{\left (f x + e\right )} a h e}{f^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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