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]

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

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Rubi [A]  time = 0.07, antiderivative size = 160, normalized size of antiderivative = 1.00, 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 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 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]

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*}

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Mathematica [A]  time = 0.21, 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]

-1/4*(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])))/(b^2*d^2)

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fricas [A]  time = 0.41, size = 242, normalized size = 1.51 \[ -\frac {{\left (b^{2} d^{2} h p + b^{2} d^{2} h q\right )} r x^{2} + 2 \, {\left ({\left (2 \, b^{2} d^{2} g - a b d^{2} h\right )} p + {\left (2 \, b^{2} d^{2} g - b^{2} c d h\right )} q\right )} r x - 2 \, {\left (b^{2} d^{2} h p r x^{2} + 2 \, b^{2} d^{2} g p r x + {\left (2 \, a b d^{2} g - a^{2} d^{2} h\right )} p r\right )} \log \left (b x + a\right ) - 2 \, {\left (b^{2} d^{2} h q r x^{2} + 2 \, b^{2} d^{2} g q r x + {\left (2 \, b^{2} c d g - b^{2} c^{2} h\right )} q r\right )} \log \left (d x + c\right ) - 2 \, {\left (b^{2} d^{2} h x^{2} + 2 \, b^{2} d^{2} g x\right )} \log \relax (e) - 2 \, {\left (b^{2} d^{2} h r x^{2} + 2 \, b^{2} d^{2} g r x\right )} \log \relax (f)}{4 \, b^{2} d^{2}} \]

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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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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]

Timed out

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maple [F]  time = 0.16, size = 0, normalized size = 0.00 \[ \int \left (h x +g \right ) \ln \left (e \left (f \left (b x +a \right )^{p} \left (d x +c \right )^{q}\right )^{r}\right )\, dx \]

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 = 0.57, size = 143, normalized size = 0.89 \[ \frac {1}{2} \, {\left (h x^{2} + 2 \, g x\right )} \log \left (\left ({\left (b x + a\right )}^{p} {\left (d x + c\right )}^{q} f\right )^{r} e\right ) + \frac {r {\left (\frac {2 \, {\left (2 \, a b f g p - a^{2} f h p\right )} \log \left (b x + a\right )}{b^{2}} + \frac {2 \, {\left (2 \, c d f g q - c^{2} f h q\right )} \log \left (d x + c\right )}{d^{2}} - \frac {b d f h {\left (p + q\right )} x^{2} - 2 \, {\left (a d f h p - {\left (2 \, d f g {\left (p + q\right )} - c f h q\right )} b\right )} x}{b d}\right )}}{4 \, f} \]

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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mupad [B]  time = 0.51, size = 153, normalized size = 0.96 \[ \ln \left (e\,{\left (f\,{\left (a+b\,x\right )}^p\,{\left (c+d\,x\right )}^q\right )}^r\right )\,\left (\frac {h\,x^2}{2}+g\,x\right )-x\,\left (\frac {r\,\left (b\,c\,h\,p+2\,b\,d\,g\,p+a\,d\,h\,q+2\,b\,d\,g\,q\right )}{2\,b\,d}-\frac {h\,r\,\left (p+q\right )\,\left (2\,a\,d+2\,b\,c\right )}{4\,b\,d}\right )-\frac {\ln \left (a+b\,x\right )\,\left (a^2\,h\,p\,r-2\,a\,b\,g\,p\,r\right )}{2\,b^2}-\frac {\ln \left (c+d\,x\right )\,\left (c^2\,h\,q\,r-2\,c\,d\,g\,q\,r\right )}{2\,d^2}-\frac {h\,r\,x^2\,\left (p+q\right )}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 75.96, size = 632, normalized size = 3.95 \[ \begin {cases} \left (g x + \frac {h x^{2}}{2}\right ) \log {\left (e \left (a^{p} c^{q} f\right )^{r} \right )} & \text {for}\: b = 0 \wedge d = 0 \\- \frac {c^{2} h q r \log {\left (c + d x \right )}}{2 d^{2}} + \frac {c g q r \log {\left (c + d x \right )}}{d} + \frac {c h q r x}{2 d} + g p r x \log {\relax (a )} + g q r x \log {\left (c + d x \right )} - g q r x + g r x \log {\relax (f )} + g x \log {\relax (e )} + \frac {h p r x^{2} \log {\relax (a )}}{2} + \frac {h q r x^{2} \log {\left (c + d x \right )}}{2} - \frac {h q r x^{2}}{4} + \frac {h r x^{2} \log {\relax (f )}}{2} + \frac {h x^{2} \log {\relax (e )}}{2} & \text {for}\: b = 0 \\- \frac {a^{2} h p r \log {\left (a + b x \right )}}{2 b^{2}} + \frac {a g p r \log {\left (a + b x \right )}}{b} + \frac {a h p r x}{2 b} + g p r x \log {\left (a + b x \right )} - g p r x + g q r x \log {\relax (c )} + g r x \log {\relax (f )} + g x \log {\relax (e )} + \frac {h p r x^{2} \log {\left (a + b x \right )}}{2} - \frac {h p r x^{2}}{4} + \frac {h q r x^{2} \log {\relax (c )}}{2} + \frac {h r x^{2} \log {\relax (f )}}{2} + \frac {h x^{2} \log {\relax (e )}}{2} & \text {for}\: d = 0 \\- \frac {a^{2} h p r \log {\left (a + b x \right )}}{2 b^{2}} - \frac {a^{2} h q r \log {\left (c + d x \right )}}{2 b^{2}} + \frac {a^{2} h q r \log {\left (\frac {c}{d} + x \right )}}{2 b^{2}} + \frac {a g p r \log {\left (a + b x \right )}}{b} + \frac {a g q r \log {\left (c + d x \right )}}{b} - \frac {a g q r \log {\left (\frac {c}{d} + x \right )}}{b} + \frac {a h p r x}{2 b} - \frac {c^{2} h q r \log {\left (\frac {c}{d} + x \right )}}{2 d^{2}} + \frac {c g q r \log {\left (\frac {c}{d} + x \right )}}{d} + \frac {c h q r x}{2 d} + g p r x \log {\left (a + b x \right )} - g p r x + g q r x \log {\left (c + d x \right )} - g q r x + g r x \log {\relax (f )} + g x \log {\relax (e )} + \frac {h p r x^{2} \log {\left (a + b x \right )}}{2} - \frac {h p r x^{2}}{4} + \frac {h q r x^{2} \log {\left (c + d x \right )}}{2} - \frac {h q r x^{2}}{4} + \frac {h r x^{2} \log {\relax (f )}}{2} + \frac {h x^{2} \log {\relax (e )}}{2} & \text {otherwise} \end {cases} \]

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]

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

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