\(\int \frac {a+b \log (c (d (e+f x)^p)^q)}{\sqrt {g+h x}} \, dx\) [484]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [F]
   Maxima [F(-2)]
   Giac [A] (verification not implemented)
   Mupad [F(-1)]

Optimal result

Integrand size = 28, antiderivative size = 103 \[ \int \frac {a+b \log \left (c \left (d (e+f x)^p\right )^q\right )}{\sqrt {g+h x}} \, dx=-\frac {4 b p q \sqrt {g+h x}}{h}+\frac {4 b \sqrt {f g-e h} p q \text {arctanh}\left (\frac {\sqrt {f} \sqrt {g+h x}}{\sqrt {f g-e h}}\right )}{\sqrt {f} h}+\frac {2 \sqrt {g+h x} \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{h} \]

[Out]

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

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {2442, 52, 65, 214, 2495} \[ \int \frac {a+b \log \left (c \left (d (e+f x)^p\right )^q\right )}{\sqrt {g+h x}} \, dx=\frac {2 \sqrt {g+h x} \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{h}+\frac {4 b p q \sqrt {f g-e h} \text {arctanh}\left (\frac {\sqrt {f} \sqrt {g+h x}}{\sqrt {f g-e h}}\right )}{\sqrt {f} h}-\frac {4 b p q \sqrt {g+h x}}{h} \]

[In]

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

[Out]

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

Rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + n + 1))), x] + Dist[n*((b*c - a*d)/(b*(m + n + 1))), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 2442

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 2495

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*} \text {integral}& = \text {Subst}\left (\int \frac {a+b \log \left (c d^q (e+f x)^{p q}\right )}{\sqrt {g+h x}} \, dx,c d^q (e+f x)^{p q},c \left (d (e+f x)^p\right )^q\right ) \\ & = \frac {2 \sqrt {g+h x} \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{h}-\text {Subst}\left (\frac {(2 b f p q) \int \frac {\sqrt {g+h x}}{e+f x} \, dx}{h},c d^q (e+f x)^{p q},c \left (d (e+f x)^p\right )^q\right ) \\ & = -\frac {4 b p q \sqrt {g+h x}}{h}+\frac {2 \sqrt {g+h x} \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{h}-\text {Subst}\left (\frac {(2 b (f g-e h) p q) \int \frac {1}{(e+f x) \sqrt {g+h x}} \, dx}{h},c d^q (e+f x)^{p q},c \left (d (e+f x)^p\right )^q\right ) \\ & = -\frac {4 b p q \sqrt {g+h x}}{h}+\frac {2 \sqrt {g+h x} \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{h}-\text {Subst}\left (\frac {(4 b (f g-e h) p q) \text {Subst}\left (\int \frac {1}{e-\frac {f g}{h}+\frac {f x^2}{h}} \, dx,x,\sqrt {g+h x}\right )}{h^2},c d^q (e+f x)^{p q},c \left (d (e+f x)^p\right )^q\right ) \\ & = -\frac {4 b p q \sqrt {g+h x}}{h}+\frac {4 b \sqrt {f g-e h} p q \tanh ^{-1}\left (\frac {\sqrt {f} \sqrt {g+h x}}{\sqrt {f g-e h}}\right )}{\sqrt {f} h}+\frac {2 \sqrt {g+h x} \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{h} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.86 \[ \int \frac {a+b \log \left (c \left (d (e+f x)^p\right )^q\right )}{\sqrt {g+h x}} \, dx=\frac {2 \left (\frac {2 b \sqrt {f g-e h} p q \text {arctanh}\left (\frac {\sqrt {f} \sqrt {g+h x}}{\sqrt {f g-e h}}\right )}{\sqrt {f}}+\sqrt {g+h x} \left (a-2 b p q+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )\right )}{h} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.65 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.15

method result size
derivativedivides \(\frac {2 \sqrt {h x +g}\, a +2 b \left (\ln \left (c \left (d \left (\frac {f \left (h x +g \right )+e h -f g}{h}\right )^{p}\right )^{q}\right ) \sqrt {h x +g}-2 q p f \left (\frac {\sqrt {h x +g}}{f}+\frac {\left (-e h +f g \right ) \arctan \left (\frac {f \sqrt {h x +g}}{\sqrt {\left (e h -f g \right ) f}}\right )}{f \sqrt {\left (e h -f g \right ) f}}\right )\right )}{h}\) \(118\)
default \(\frac {2 \sqrt {h x +g}\, a +2 b \left (\ln \left (c \left (d \left (\frac {f \left (h x +g \right )+e h -f g}{h}\right )^{p}\right )^{q}\right ) \sqrt {h x +g}-2 q p f \left (\frac {\sqrt {h x +g}}{f}+\frac {\left (-e h +f g \right ) \arctan \left (\frac {f \sqrt {h x +g}}{\sqrt {\left (e h -f g \right ) f}}\right )}{f \sqrt {\left (e h -f g \right ) f}}\right )\right )}{h}\) \(118\)
parts \(\frac {2 a \sqrt {h x +g}}{h}+\frac {2 b \left (\ln \left (c \left (d \left (\frac {f \left (h x +g \right )+e h -f g}{h}\right )^{p}\right )^{q}\right ) \sqrt {h x +g}-2 q p f \left (\frac {\sqrt {h x +g}}{f}+\frac {\left (-e h +f g \right ) \arctan \left (\frac {f \sqrt {h x +g}}{\sqrt {\left (e h -f g \right ) f}}\right )}{f \sqrt {\left (e h -f g \right ) f}}\right )\right )}{h}\) \(121\)

[In]

int((a+b*ln(c*(d*(f*x+e)^p)^q))/(h*x+g)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.95 \[ \int \frac {a+b \log \left (c \left (d (e+f x)^p\right )^q\right )}{\sqrt {g+h x}} \, dx=\left [\frac {2 \, {\left (b p q \sqrt {\frac {f g - e h}{f}} \log \left (\frac {f h x + 2 \, f g - e h + 2 \, \sqrt {h x + g} f \sqrt {\frac {f g - e h}{f}}}{f x + e}\right ) + {\left (b p q \log \left (f x + e\right ) - 2 \, b p q + b q \log \left (d\right ) + b \log \left (c\right ) + a\right )} \sqrt {h x + g}\right )}}{h}, \frac {2 \, {\left (2 \, b p q \sqrt {-\frac {f g - e h}{f}} \arctan \left (-\frac {\sqrt {h x + g} f \sqrt {-\frac {f g - e h}{f}}}{f g - e h}\right ) + {\left (b p q \log \left (f x + e\right ) - 2 \, b p q + b q \log \left (d\right ) + b \log \left (c\right ) + a\right )} \sqrt {h x + g}\right )}}{h}\right ] \]

[In]

integrate((a+b*log(c*(d*(f*x+e)^p)^q))/(h*x+g)^(1/2),x, algorithm="fricas")

[Out]

[2*(b*p*q*sqrt((f*g - e*h)/f)*log((f*h*x + 2*f*g - e*h + 2*sqrt(h*x + g)*f*sqrt((f*g - e*h)/f))/(f*x + e)) + (
b*p*q*log(f*x + e) - 2*b*p*q + b*q*log(d) + b*log(c) + a)*sqrt(h*x + g))/h, 2*(2*b*p*q*sqrt(-(f*g - e*h)/f)*ar
ctan(-sqrt(h*x + g)*f*sqrt(-(f*g - e*h)/f)/(f*g - e*h)) + (b*p*q*log(f*x + e) - 2*b*p*q + b*q*log(d) + b*log(c
) + a)*sqrt(h*x + g))/h]

Sympy [F]

\[ \int \frac {a+b \log \left (c \left (d (e+f x)^p\right )^q\right )}{\sqrt {g+h x}} \, dx=\int \frac {a + b \log {\left (c \left (d \left (e + f x\right )^{p}\right )^{q} \right )}}{\sqrt {g + h x}}\, dx \]

[In]

integrate((a+b*ln(c*(d*(f*x+e)**p)**q))/(h*x+g)**(1/2),x)

[Out]

Integral((a + b*log(c*(d*(e + f*x)**p)**q))/sqrt(g + h*x), x)

Maxima [F(-2)]

Exception generated. \[ \int \frac {a+b \log \left (c \left (d (e+f x)^p\right )^q\right )}{\sqrt {g+h x}} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate((a+b*log(c*(d*(f*x+e)^p)^q))/(h*x+g)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(e*h-f*g>0)', see `assume?` for
 more detail

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.20 \[ \int \frac {a+b \log \left (c \left (d (e+f x)^p\right )^q\right )}{\sqrt {g+h x}} \, dx=-\frac {2 \, {\left ({\left (2 \, f {\left (\frac {{\left (f g - e h\right )} \arctan \left (\frac {\sqrt {h x + g} f}{\sqrt {-f^{2} g + e f h}}\right )}{\sqrt {-f^{2} g + e f h} f} + \frac {\sqrt {h x + g}}{f}\right )} - \sqrt {h x + g} \log \left (f x + e\right )\right )} b p q - \sqrt {h x + g} b q \log \left (d\right ) - \sqrt {h x + g} b \log \left (c\right ) - \sqrt {h x + g} a\right )}}{h} \]

[In]

integrate((a+b*log(c*(d*(f*x+e)^p)^q))/(h*x+g)^(1/2),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {a+b \log \left (c \left (d (e+f x)^p\right )^q\right )}{\sqrt {g+h x}} \, dx=\int \frac {a+b\,\ln \left (c\,{\left (d\,{\left (e+f\,x\right )}^p\right )}^q\right )}{\sqrt {g+h\,x}} \,d x \]

[In]

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

[Out]

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