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

Optimal. Leaf size=160 \[ -\frac{p r (b g-a h)^2 \log (a+b x)}{2 b^2 h}+\frac{(g+h x)^2 \log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{2 h}-\frac{p r x (b g-a h)}{2 b}-\frac{q r (d g-c h)^2 \log (c+d x)}{2 d^2 h}-\frac{q r x (d g-c h)}{2 d}-\frac{p r (g+h x)^2}{4 h}-\frac{q r (g+h x)^2}{4 h} \]

[Out]

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

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Rubi [A]  time = 0.0687969, antiderivative size = 160, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 2, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.074, Rules used = {2495, 43} \[ -\frac{p r (b g-a h)^2 \log (a+b x)}{2 b^2 h}+\frac{(g+h x)^2 \log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{2 h}-\frac{p r x (b g-a h)}{2 b}-\frac{q r (d g-c h)^2 \log (c+d x)}{2 d^2 h}-\frac{q r x (d g-c h)}{2 d}-\frac{p r (g+h x)^2}{4 h}-\frac{q r (g+h x)^2}{4 h} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2495

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]

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])

Rubi steps

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

Mathematica [A]  time = 0.196375, size = 120, normalized size = 0.75 \[ -\frac{b \left (d x \left (r (-2 a d h p-2 b c h q+b d (p+q) (4 g+h x))-2 b d (2 g+h x) \log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )\right )+2 b c q r (c h-2 d g) \log (c+d x)\right )+2 a d^2 p r (a h-2 b g) \log (a+b x)}{4 b^2 d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.174, size = 0, normalized size = 0. \begin{align*} \int \left ( hx+g \right ) \ln \left ( e \left ( f \left ( bx+a \right ) ^{p} \left ( dx+c \right ) ^{q} \right ) ^{r} \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.20219, size = 193, normalized size = 1.21 \begin{align*} \frac{1}{2} \,{\left (h x^{2} + 2 \, g x\right )} \log \left (\left ({\left (b x + a\right )}^{p}{\left (d x + c\right )}^{q} f\right )^{r} e\right ) + \frac{r{\left (\frac{2 \,{\left (2 \, a b f g p - a^{2} f h p\right )} \log \left (b x + a\right )}{b^{2}} + \frac{2 \,{\left (2 \, c d f g q - c^{2} f h q\right )} \log \left (d x + c\right )}{d^{2}} - \frac{b d f h{\left (p + q\right )} x^{2} - 2 \,{\left (a d f h p -{\left (2 \, d f g{\left (p + q\right )} - c f h q\right )} b\right )} x}{b d}\right )}}{4 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.20831, size = 524, normalized size = 3.28 \begin{align*} -\frac{{\left (b^{2} d^{2} h p + b^{2} d^{2} h q\right )} r x^{2} + 2 \,{\left ({\left (2 \, b^{2} d^{2} g - a b d^{2} h\right )} p +{\left (2 \, b^{2} d^{2} g - b^{2} c d h\right )} q\right )} r x - 2 \,{\left (b^{2} d^{2} h p r x^{2} + 2 \, b^{2} d^{2} g p r x +{\left (2 \, a b d^{2} g - a^{2} d^{2} h\right )} p r\right )} \log \left (b x + a\right ) - 2 \,{\left (b^{2} d^{2} h q r x^{2} + 2 \, b^{2} d^{2} g q r x +{\left (2 \, b^{2} c d g - b^{2} c^{2} h\right )} q r\right )} \log \left (d x + c\right ) - 2 \,{\left (b^{2} d^{2} h x^{2} + 2 \, b^{2} d^{2} g x\right )} \log \left (e\right ) - 2 \,{\left (b^{2} d^{2} h r x^{2} + 2 \, b^{2} d^{2} g r x\right )} \log \left (f\right )}{4 \, b^{2} d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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Giac [B]  time = 1.31039, size = 479, normalized size = 2.99 \begin{align*} -\frac{1}{4} \,{\left (h p r + h q r - 2 \, h r \log \left (f\right ) - 2 \, h\right )} x^{2} + \frac{1}{2} \,{\left (h p r x^{2} + 2 \, g p r x\right )} \log \left (b x + a\right ) + \frac{1}{2} \,{\left (h q r x^{2} + 2 \, g q r x\right )} \log \left (d x + c\right ) - \frac{{\left (2 \, b d g p r - a d h p r + 2 \, b d g q r - b c h q r - 2 \, b d g r \log \left (f\right ) - 2 \, b d g\right )} x}{2 \, b d} + \frac{{\left (2 \, a b d^{2} g p r - a^{2} d^{2} h p r + 2 \, b^{2} c d g q r - b^{2} c^{2} h q r\right )} \log \left ({\left | b d x^{2} + b c x + a d x + a c \right |}\right )}{4 \, b^{2} d^{2}} + \frac{{\left (2 \, a b^{2} c d^{2} g p r - 2 \, a^{2} b d^{3} g p r - a^{2} b c d^{2} h p r + a^{3} d^{3} h p r - 2 \, b^{3} c^{2} d g q r + 2 \, a b^{2} c d^{2} g q r + b^{3} c^{3} h q r - a b^{2} c^{2} d h q r\right )} \log \left ({\left | \frac{2 \, b d x + b c + a d -{\left | -b c + a d \right |}}{2 \, b d x + b c + a d +{\left | -b c + a d \right |}} \right |}\right )}{4 \, b^{2} d^{2}{\left | -b c + a d \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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