\(\int \sqrt {g+h x} (a+b \log (c (d (e+f x)^p)^q)) \, dx\) [483]

Optimal result

Integrand size = 28, antiderivative size = 139 \[ \int \sqrt {g+h x} \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right ) \, dx=-\frac {4 b (f g-e h) p q \sqrt {g+h x}}{3 f h}-\frac {4 b p q (g+h x)^{3/2}}{9 h}+\frac {4 b (f g-e h)^{3/2} p q \text {arctanh}\left (\frac {\sqrt {f} \sqrt {g+h x}}{\sqrt {f g-e h}}\right )}{3 f^{3/2} h}+\frac {2 (g+h x)^{3/2} \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{3 h} \]

[Out]

-4/9*b*p*q*(h*x+g)^(3/2)/h+4/3*b*(-e*h+f*g)^(3/2)*p*q*arctanh(f^(1/2)*(h*x+g)^(1/2)/(-e*h+f*g)^(1/2))/f^(3/2)/
h+2/3*(h*x+g)^(3/2)*(a+b*ln(c*(d*(f*x+e)^p)^q))/h-4/3*b*(-e*h+f*g)*p*q*(h*x+g)^(1/2)/f/h

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {2442, 52, 65, 214, 2495} \[ \int \sqrt {g+h x} \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right ) \, dx=\frac {2 (g+h x)^{3/2} \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{3 h}+\frac {4 b p q (f g-e h)^{3/2} \text {arctanh}\left (\frac {\sqrt {f} \sqrt {g+h x}}{\sqrt {f g-e h}}\right )}{3 f^{3/2} h}-\frac {4 b p q \sqrt {g+h x} (f g-e h)}{3 f h}-\frac {4 b p q (g+h x)^{3/2}}{9 h} \]

[In]

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

[Out]

(-4*b*(f*g - e*h)*p*q*Sqrt[g + h*x])/(3*f*h) - (4*b*p*q*(g + h*x)^(3/2))/(9*h) + (4*b*(f*g - e*h)^(3/2)*p*q*Ar
cTanh[(Sqrt[f]*Sqrt[g + h*x])/Sqrt[f*g - e*h]])/(3*f^(3/2)*h) + (2*(g + h*x)^(3/2)*(a + b*Log[c*(d*(e + f*x)^p
)^q]))/(3*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 \sqrt {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 {2 (g+h x)^{3/2} \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{3 h}-\text {Subst}\left (\frac {(2 b f p q) \int \frac {(g+h x)^{3/2}}{e+f x} \, dx}{3 h},c d^q (e+f x)^{p q},c \left (d (e+f x)^p\right )^q\right ) \\ & = -\frac {4 b p q (g+h x)^{3/2}}{9 h}+\frac {2 (g+h x)^{3/2} \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{3 h}-\text {Subst}\left (\frac {(2 b (f g-e h) p q) \int \frac {\sqrt {g+h x}}{e+f x} \, dx}{3 h},c d^q (e+f x)^{p q},c \left (d (e+f x)^p\right )^q\right ) \\ & = -\frac {4 b (f g-e h) p q \sqrt {g+h x}}{3 f h}-\frac {4 b p q (g+h x)^{3/2}}{9 h}+\frac {2 (g+h x)^{3/2} \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{3 h}-\text {Subst}\left (\frac {\left (2 b (f g-e h)^2 p q\right ) \int \frac {1}{(e+f x) \sqrt {g+h x}} \, dx}{3 f h},c d^q (e+f x)^{p q},c \left (d (e+f x)^p\right )^q\right ) \\ & = -\frac {4 b (f g-e h) p q \sqrt {g+h x}}{3 f h}-\frac {4 b p q (g+h x)^{3/2}}{9 h}+\frac {2 (g+h x)^{3/2} \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{3 h}-\text {Subst}\left (\frac {\left (4 b (f g-e h)^2 p q\right ) \text {Subst}\left (\int \frac {1}{e-\frac {f g}{h}+\frac {f x^2}{h}} \, dx,x,\sqrt {g+h x}\right )}{3 f h^2},c d^q (e+f x)^{p q},c \left (d (e+f x)^p\right )^q\right ) \\ & = -\frac {4 b (f g-e h) p q \sqrt {g+h x}}{3 f h}-\frac {4 b p q (g+h x)^{3/2}}{9 h}+\frac {4 b (f g-e h)^{3/2} p q \tanh ^{-1}\left (\frac {\sqrt {f} \sqrt {g+h x}}{\sqrt {f g-e h}}\right )}{3 f^{3/2} h}+\frac {2 (g+h x)^{3/2} \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{3 h} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.89 \[ \int \sqrt {g+h x} \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right ) \, dx=\frac {2 \left (6 b (f g-e h)^{3/2} p q \text {arctanh}\left (\frac {\sqrt {f} \sqrt {g+h x}}{\sqrt {f g-e h}}\right )+\sqrt {f} \sqrt {g+h x} \left (3 a f (g+h x)-2 b p q (4 f g-3 e h+f h x)+3 b f (g+h x) \log \left (c \left (d (e+f x)^p\right )^q\right )\right )\right )}{9 f^{3/2} h} \]

[In]

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

[Out]

(2*(6*b*(f*g - e*h)^(3/2)*p*q*ArcTanh[(Sqrt[f]*Sqrt[g + h*x])/Sqrt[f*g - e*h]] + Sqrt[f]*Sqrt[g + h*x]*(3*a*f*
(g + h*x) - 2*b*p*q*(4*f*g - 3*e*h + f*h*x) + 3*b*f*(g + h*x)*Log[c*(d*(e + f*x)^p)^q])))/(9*f^(3/2)*h)

Maple [F]

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 353, normalized size of antiderivative = 2.54 \[ \int \sqrt {g+h x} \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right ) \, dx=\left [-\frac {2 \, {\left (3 \, {\left (b f g - b e h\right )} p q \sqrt {\frac {f g - e h}{f}} \log \left (\frac {f h x + 2 \, f g - e h - 2 \, \sqrt {h x + g} f \sqrt {\frac {f g - e h}{f}}}{f x + e}\right ) - {\left (3 \, a f g - 2 \, {\left (4 \, b f g - 3 \, b e h\right )} p q - {\left (2 \, b f h p q - 3 \, a f h\right )} x + 3 \, {\left (b f h p q x + b f g p q\right )} \log \left (f x + e\right ) + 3 \, {\left (b f h x + b f g\right )} \log \left (c\right ) + 3 \, {\left (b f h q x + b f g q\right )} \log \left (d\right )\right )} \sqrt {h x + g}\right )}}{9 \, f h}, \frac {2 \, {\left (6 \, {\left (b f g - b e h\right )} p q \sqrt {-\frac {f g - e h}{f}} \arctan \left (-\frac {\sqrt {h x + g} f \sqrt {-\frac {f g - e h}{f}}}{f g - e h}\right ) + {\left (3 \, a f g - 2 \, {\left (4 \, b f g - 3 \, b e h\right )} p q - {\left (2 \, b f h p q - 3 \, a f h\right )} x + 3 \, {\left (b f h p q x + b f g p q\right )} \log \left (f x + e\right ) + 3 \, {\left (b f h x + b f g\right )} \log \left (c\right ) + 3 \, {\left (b f h q x + b f g q\right )} \log \left (d\right )\right )} \sqrt {h x + g}\right )}}{9 \, f h}\right ] \]

[In]

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

[Out]

[-2/9*(3*(b*f*g - b*e*h)*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)) - (3*a*f*g - 2*(4*b*f*g - 3*b*e*h)*p*q - (2*b*f*h*p*q - 3*a*f*h)*x + 3*(b*f*h*p*q*x + b*f*g*p*
q)*log(f*x + e) + 3*(b*f*h*x + b*f*g)*log(c) + 3*(b*f*h*q*x + b*f*g*q)*log(d))*sqrt(h*x + g))/(f*h), 2/9*(6*(b
*f*g - b*e*h)*p*q*sqrt(-(f*g - e*h)/f)*arctan(-sqrt(h*x + g)*f*sqrt(-(f*g - e*h)/f)/(f*g - e*h)) + (3*a*f*g -
2*(4*b*f*g - 3*b*e*h)*p*q - (2*b*f*h*p*q - 3*a*f*h)*x + 3*(b*f*h*p*q*x + b*f*g*p*q)*log(f*x + e) + 3*(b*f*h*x
+ b*f*g)*log(c) + 3*(b*f*h*q*x + b*f*g*q)*log(d))*sqrt(h*x + g))/(f*h)]

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F(-2)]

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

[In]

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

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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