\(\int \sqrt {f+g x} (a+b \log (c (d+e x)^n)) \, dx\) [139]

Optimal result

Integrand size = 24, antiderivative size = 132 \[ \int \sqrt {f+g x} \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx=-\frac {4 b (e f-d g) n \sqrt {f+g x}}{3 e g}-\frac {4 b n (f+g x)^{3/2}}{9 g}+\frac {4 b (e f-d g)^{3/2} n \text {arctanh}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )}{3 e^{3/2} g}+\frac {2 (f+g x)^{3/2} \left (a+b \log \left (c (d+e x)^n\right )\right )}{3 g} \] Output:

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

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.89 \[ \int \sqrt {f+g x} \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx=\frac {2 \left (6 b (e f-d g)^{3/2} n \text {arctanh}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )+\sqrt {e} \sqrt {f+g x} \left (3 a e (f+g x)-2 b n (4 e f-3 d g+e g x)+3 b e (f+g x) \log \left (c (d+e x)^n\right )\right )\right )}{9 e^{3/2} g} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {2842, 60, 60, 73, 221}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \sqrt {f+g x} \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx\)

\(\Big \downarrow \) 2842

\(\displaystyle \frac {2 (f+g x)^{3/2} \left (a+b \log \left (c (d+e x)^n\right )\right )}{3 g}-\frac {2 b e n \int \frac {(f+g x)^{3/2}}{d+e x}dx}{3 g}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {2 (f+g x)^{3/2} \left (a+b \log \left (c (d+e x)^n\right )\right )}{3 g}-\frac {2 b e n \left (\frac {(e f-d g) \int \frac {\sqrt {f+g x}}{d+e x}dx}{e}+\frac {2 (f+g x)^{3/2}}{3 e}\right )}{3 g}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {2 (f+g x)^{3/2} \left (a+b \log \left (c (d+e x)^n\right )\right )}{3 g}-\frac {2 b e n \left (\frac {(e f-d g) \left (\frac {(e f-d g) \int \frac {1}{(d+e x) \sqrt {f+g x}}dx}{e}+\frac {2 \sqrt {f+g x}}{e}\right )}{e}+\frac {2 (f+g x)^{3/2}}{3 e}\right )}{3 g}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {2 (f+g x)^{3/2} \left (a+b \log \left (c (d+e x)^n\right )\right )}{3 g}-\frac {2 b e n \left (\frac {(e f-d g) \left (\frac {2 (e f-d g) \int \frac {1}{d+\frac {e (f+g x)}{g}-\frac {e f}{g}}d\sqrt {f+g x}}{e g}+\frac {2 \sqrt {f+g x}}{e}\right )}{e}+\frac {2 (f+g x)^{3/2}}{3 e}\right )}{3 g}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {2 (f+g x)^{3/2} \left (a+b \log \left (c (d+e x)^n\right )\right )}{3 g}-\frac {2 b e n \left (\frac {(e f-d g) \left (\frac {2 \sqrt {f+g x}}{e}-\frac {2 \sqrt {e f-d g} \text {arctanh}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )}{e^{3/2}}\right )}{e}+\frac {2 (f+g x)^{3/2}}{3 e}\right )}{3 g}\)

Input:

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

Output:

(-2*b*e*n*((2*(f + g*x)^(3/2))/(3*e) + ((e*f - d*g)*((2*Sqrt[f + g*x])/e - 
 (2*Sqrt[e*f - d*g]*ArcTanh[(Sqrt[e]*Sqrt[f + g*x])/Sqrt[e*f - d*g]])/e^(3 
/2)))/e))/(3*g) + (2*(f + g*x)^(3/2)*(a + b*Log[c*(d + e*x)^n]))/(3*g)
 

Defintions of rubi rules used

rule 60

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] + Simp[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] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !Integer 
Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinear 
Q[a, b, c, d, m, n, x]
 

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[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] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 221

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 2842

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] - Simp[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]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 311, normalized size of antiderivative = 2.36 \[ \int \sqrt {f+g x} \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx=\left [-\frac {2 \, {\left (3 \, {\left (b e f - b d g\right )} n \sqrt {\frac {e f - d g}{e}} \log \left (\frac {e g x + 2 \, e f - d g - 2 \, \sqrt {g x + f} e \sqrt {\frac {e f - d g}{e}}}{e x + d}\right ) - {\left (3 \, a e f - 2 \, {\left (4 \, b e f - 3 \, b d g\right )} n - {\left (2 \, b e g n - 3 \, a e g\right )} x + 3 \, {\left (b e g n x + b e f n\right )} \log \left (e x + d\right ) + 3 \, {\left (b e g x + b e f\right )} \log \left (c\right )\right )} \sqrt {g x + f}\right )}}{9 \, e g}, \frac {2 \, {\left (6 \, {\left (b e f - b d g\right )} n \sqrt {-\frac {e f - d g}{e}} \arctan \left (-\frac {\sqrt {g x + f} e \sqrt {-\frac {e f - d g}{e}}}{e f - d g}\right ) + {\left (3 \, a e f - 2 \, {\left (4 \, b e f - 3 \, b d g\right )} n - {\left (2 \, b e g n - 3 \, a e g\right )} x + 3 \, {\left (b e g n x + b e f n\right )} \log \left (e x + d\right ) + 3 \, {\left (b e g x + b e f\right )} \log \left (c\right )\right )} \sqrt {g x + f}\right )}}{9 \, e g}\right ] \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.69 \[ \int \sqrt {f+g x} \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx=\frac {-\frac {4 \sqrt {e}\, \sqrt {d g -e f}\, \mathit {atan} \left (\frac {\sqrt {g x +f}\, e}{\sqrt {e}\, \sqrt {d g -e f}}\right ) b d g n}{3}+\frac {4 \sqrt {e}\, \sqrt {d g -e f}\, \mathit {atan} \left (\frac {\sqrt {g x +f}\, e}{\sqrt {e}\, \sqrt {d g -e f}}\right ) b e f n}{3}+\frac {2 \sqrt {g x +f}\, \mathrm {log}\left (\frac {\left (e g x +d g \right )^{n} c}{g^{n}}\right ) b \,e^{2} f}{3}+\frac {2 \sqrt {g x +f}\, \mathrm {log}\left (\frac {\left (e g x +d g \right )^{n} c}{g^{n}}\right ) b \,e^{2} g x}{3}+\frac {2 \sqrt {g x +f}\, a \,e^{2} f}{3}+\frac {2 \sqrt {g x +f}\, a \,e^{2} g x}{3}+\frac {4 \sqrt {g x +f}\, b d e g n}{3}-\frac {16 \sqrt {g x +f}\, b \,e^{2} f n}{9}-\frac {4 \sqrt {g x +f}\, b \,e^{2} g n x}{9}}{e^{2} g} \] Input:

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

Output:

(2*( - 6*sqrt(e)*sqrt(d*g - e*f)*atan((sqrt(f + g*x)*e)/(sqrt(e)*sqrt(d*g 
- e*f)))*b*d*g*n + 6*sqrt(e)*sqrt(d*g - e*f)*atan((sqrt(f + g*x)*e)/(sqrt( 
e)*sqrt(d*g - e*f)))*b*e*f*n + 3*sqrt(f + g*x)*log(((d*g + e*g*x)**n*c)/g* 
*n)*b*e**2*f + 3*sqrt(f + g*x)*log(((d*g + e*g*x)**n*c)/g**n)*b*e**2*g*x + 
 3*sqrt(f + g*x)*a*e**2*f + 3*sqrt(f + g*x)*a*e**2*g*x + 6*sqrt(f + g*x)*b 
*d*e*g*n - 8*sqrt(f + g*x)*b*e**2*f*n - 2*sqrt(f + g*x)*b*e**2*g*n*x))/(9* 
e**2*g)