\(\int \frac {a+b \log (c (d+e x)^n)}{(f+g x)^{7/2}} \, dx\) [143]

Optimal result

Integrand size = 24, antiderivative size = 145 \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{(f+g x)^{7/2}} \, dx=\frac {4 b e n}{15 g (e f-d g) (f+g x)^{3/2}}+\frac {4 b e^2 n}{5 g (e f-d g)^2 \sqrt {f+g x}}-\frac {4 b e^{5/2} n \text {arctanh}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )}{5 g (e f-d g)^{5/2}}-\frac {2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{5 g (f+g x)^{5/2}} \] Output:

4/15*b*e*n/g/(-d*g+e*f)/(g*x+f)^(3/2)+4/5*b*e^2*n/g/(-d*g+e*f)^2/(g*x+f)^( 
1/2)-4/5*b*e^(5/2)*n*arctanh(e^(1/2)*(g*x+f)^(1/2)/(-d*g+e*f)^(1/2))/g/(-d 
*g+e*f)^(5/2)-2/5*(a+b*ln(c*(e*x+d)^n))/g/(g*x+f)^(5/2)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.05 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.54 \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{(f+g x)^{7/2}} \, dx=\frac {2 \left (\frac {2 b e n (f+g x) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},\frac {e (f+g x)}{e f-d g}\right )}{e f-d g}-3 \left (a+b \log \left (c (d+e x)^n\right )\right )\right )}{15 g (f+g x)^{5/2}} \] Input:

Integrate[(a + b*Log[c*(d + e*x)^n])/(f + g*x)^(7/2),x]
 

Output:

(2*((2*b*e*n*(f + g*x)*Hypergeometric2F1[-3/2, 1, -1/2, (e*(f + g*x))/(e*f 
 - d*g)])/(e*f - d*g) - 3*(a + b*Log[c*(d + e*x)^n])))/(15*g*(f + g*x)^(5/ 
2))
 

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.01, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {2842, 61, 61, 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.

\(\displaystyle \int \frac {a+b \log \left (c (d+e x)^n\right )}{(f+g x)^{7/2}} \, dx\)

\(\Big \downarrow \) 2842

\(\displaystyle \frac {2 b e n \int \frac {1}{(d+e x) (f+g x)^{5/2}}dx}{5 g}-\frac {2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{5 g (f+g x)^{5/2}}\)

\(\Big \downarrow \) 61

\(\displaystyle \frac {2 b e n \left (\frac {e \int \frac {1}{(d+e x) (f+g x)^{3/2}}dx}{e f-d g}+\frac {2}{3 (f+g x)^{3/2} (e f-d g)}\right )}{5 g}-\frac {2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{5 g (f+g x)^{5/2}}\)

\(\Big \downarrow \) 61

\(\displaystyle \frac {2 b e n \left (\frac {e \left (\frac {e \int \frac {1}{(d+e x) \sqrt {f+g x}}dx}{e f-d g}+\frac {2}{\sqrt {f+g x} (e f-d g)}\right )}{e f-d g}+\frac {2}{3 (f+g x)^{3/2} (e f-d g)}\right )}{5 g}-\frac {2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{5 g (f+g x)^{5/2}}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {2 b e n \left (\frac {e \left (\frac {2 e \int \frac {1}{d+\frac {e (f+g x)}{g}-\frac {e f}{g}}d\sqrt {f+g x}}{g (e f-d g)}+\frac {2}{\sqrt {f+g x} (e f-d g)}\right )}{e f-d g}+\frac {2}{3 (f+g x)^{3/2} (e f-d g)}\right )}{5 g}-\frac {2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{5 g (f+g x)^{5/2}}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {2 b e n \left (\frac {e \left (\frac {2}{\sqrt {f+g x} (e f-d g)}-\frac {2 \sqrt {e} \text {arctanh}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )}{(e f-d g)^{3/2}}\right )}{e f-d g}+\frac {2}{3 (f+g x)^{3/2} (e f-d g)}\right )}{5 g}-\frac {2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{5 g (f+g x)^{5/2}}\)

Input:

Int[(a + b*Log[c*(d + e*x)^n])/(f + g*x)^(7/2),x]
 

Output:

(2*b*e*n*(2/(3*(e*f - d*g)*(f + g*x)^(3/2)) + (e*(2/((e*f - d*g)*Sqrt[f + 
g*x]) - (2*Sqrt[e]*ArcTanh[(Sqrt[e]*Sqrt[f + g*x])/Sqrt[e*f - d*g]])/(e*f 
- d*g)^(3/2)))/(e*f - d*g)))/(5*g) - (2*(a + b*Log[c*(d + e*x)^n]))/(5*g*( 
f + g*x)^(5/2))
 

Defintions of rubi rules used

rule 61

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

\[\int \frac {a +b \ln \left (c \left (e x +d \right )^{n}\right )}{\left (g x +f \right )^{\frac {7}{2}}}d x\]

Input:

int((a+b*ln(c*(e*x+d)^n))/(g*x+f)^(7/2),x)
 

Output:

int((a+b*ln(c*(e*x+d)^n))/(g*x+f)^(7/2),x)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 370 vs. \(2 (121) = 242\).

Time = 0.16 (sec) , antiderivative size = 769, normalized size of antiderivative = 5.30 \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{(f+g x)^{7/2}} \, dx =\text {Too large to display} \] Input:

integrate((a+b*log(c*(e*x+d)^n))/(g*x+f)^(7/2),x, algorithm="fricas")
 

Output:

[2/15*(3*(b*e^2*g^3*n*x^3 + 3*b*e^2*f*g^2*n*x^2 + 3*b*e^2*f^2*g*n*x + b*e^ 
2*f^3*n)*sqrt(e/(e*f - d*g))*log((e*g*x + 2*e*f - d*g - 2*(e*f - d*g)*sqrt 
(g*x + f)*sqrt(e/(e*f - d*g)))/(e*x + d)) + (6*b*e^2*g^2*n*x^2 - 3*a*e^2*f 
^2 + 6*a*d*e*f*g - 3*a*d^2*g^2 + 2*(7*b*e^2*f*g - b*d*e*g^2)*n*x - 3*(b*e^ 
2*f^2 - 2*b*d*e*f*g + b*d^2*g^2)*n*log(e*x + d) + 2*(4*b*e^2*f^2 - b*d*e*f 
*g)*n - 3*(b*e^2*f^2 - 2*b*d*e*f*g + b*d^2*g^2)*log(c))*sqrt(g*x + f))/(e^ 
2*f^5*g - 2*d*e*f^4*g^2 + d^2*f^3*g^3 + (e^2*f^2*g^4 - 2*d*e*f*g^5 + d^2*g 
^6)*x^3 + 3*(e^2*f^3*g^3 - 2*d*e*f^2*g^4 + d^2*f*g^5)*x^2 + 3*(e^2*f^4*g^2 
 - 2*d*e*f^3*g^3 + d^2*f^2*g^4)*x), 2/15*(6*(b*e^2*g^3*n*x^3 + 3*b*e^2*f*g 
^2*n*x^2 + 3*b*e^2*f^2*g*n*x + b*e^2*f^3*n)*sqrt(-e/(e*f - d*g))*arctan(sq 
rt(g*x + f)*sqrt(-e/(e*f - d*g))) + (6*b*e^2*g^2*n*x^2 - 3*a*e^2*f^2 + 6*a 
*d*e*f*g - 3*a*d^2*g^2 + 2*(7*b*e^2*f*g - b*d*e*g^2)*n*x - 3*(b*e^2*f^2 - 
2*b*d*e*f*g + b*d^2*g^2)*n*log(e*x + d) + 2*(4*b*e^2*f^2 - b*d*e*f*g)*n - 
3*(b*e^2*f^2 - 2*b*d*e*f*g + b*d^2*g^2)*log(c))*sqrt(g*x + f))/(e^2*f^5*g 
- 2*d*e*f^4*g^2 + d^2*f^3*g^3 + (e^2*f^2*g^4 - 2*d*e*f*g^5 + d^2*g^6)*x^3 
+ 3*(e^2*f^3*g^3 - 2*d*e*f^2*g^4 + d^2*f*g^5)*x^2 + 3*(e^2*f^4*g^2 - 2*d*e 
*f^3*g^3 + d^2*f^2*g^4)*x)]
 

Sympy [F(-1)]

Timed out. \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{(f+g x)^{7/2}} \, dx=\text {Timed out} \] Input:

integrate((a+b*ln(c*(e*x+d)**n))/(g*x+f)**(7/2),x)
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{(f+g x)^{7/2}} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((a+b*log(c*(e*x+d)^n))/(g*x+f)^(7/2),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]

\[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{(f+g x)^{7/2}} \, dx=\int { \frac {b \log \left ({\left (e x + d\right )}^{n} c\right ) + a}{{\left (g x + f\right )}^{\frac {7}{2}}} \,d x } \] Input:

integrate((a+b*log(c*(e*x+d)^n))/(g*x+f)^(7/2),x, algorithm="giac")
 

Output:

integrate((b*log((e*x + d)^n*c) + a)/(g*x + f)^(7/2), x)
                                                                                    
                                                                                    
 

Mupad [F(-1)]

Timed out. \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{(f+g x)^{7/2}} \, dx=\int \frac {a+b\,\ln \left (c\,{\left (d+e\,x\right )}^n\right )}{{\left (f+g\,x\right )}^{7/2}} \,d x \] Input:

int((a + b*log(c*(d + e*x)^n))/(f + g*x)^(7/2),x)
 

Output:

int((a + b*log(c*(d + e*x)^n))/(f + g*x)^(7/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 562, normalized size of antiderivative = 3.88 \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{(f+g x)^{7/2}} \, dx=\frac {\frac {4 \sqrt {e}\, \sqrt {g x +f}\, \sqrt {d g -e f}\, \mathit {atan} \left (\frac {\sqrt {g x +f}\, e}{\sqrt {e}\, \sqrt {d g -e f}}\right ) b \,e^{2} f^{2} n}{5}+\frac {8 \sqrt {e}\, \sqrt {g x +f}\, \sqrt {d g -e f}\, \mathit {atan} \left (\frac {\sqrt {g x +f}\, e}{\sqrt {e}\, \sqrt {d g -e f}}\right ) b \,e^{2} f g n x}{5}+\frac {4 \sqrt {e}\, \sqrt {g x +f}\, \sqrt {d g -e f}\, \mathit {atan} \left (\frac {\sqrt {g x +f}\, e}{\sqrt {e}\, \sqrt {d g -e f}}\right ) b \,e^{2} g^{2} n \,x^{2}}{5}-\frac {2 \,\mathrm {log}\left (\frac {\left (e g x +d g \right )^{n} c}{g^{n}}\right ) b \,d^{3} g^{3}}{5}+\frac {6 \,\mathrm {log}\left (\frac {\left (e g x +d g \right )^{n} c}{g^{n}}\right ) b \,d^{2} e f \,g^{2}}{5}-\frac {6 \,\mathrm {log}\left (\frac {\left (e g x +d g \right )^{n} c}{g^{n}}\right ) b d \,e^{2} f^{2} g}{5}+\frac {2 \,\mathrm {log}\left (\frac {\left (e g x +d g \right )^{n} c}{g^{n}}\right ) b \,e^{3} f^{3}}{5}-\frac {2 a \,d^{3} g^{3}}{5}+\frac {6 a \,d^{2} e f \,g^{2}}{5}-\frac {6 a d \,e^{2} f^{2} g}{5}+\frac {2 a \,e^{3} f^{3}}{5}-\frac {4 b \,d^{2} e f \,g^{2} n}{15}-\frac {4 b \,d^{2} e \,g^{3} n x}{15}+\frac {4 b d \,e^{2} f^{2} g n}{3}+\frac {32 b d \,e^{2} f \,g^{2} n x}{15}+\frac {4 b d \,e^{2} g^{3} n \,x^{2}}{5}-\frac {16 b \,e^{3} f^{3} n}{15}-\frac {28 b \,e^{3} f^{2} g n x}{15}-\frac {4 b \,e^{3} f \,g^{2} n \,x^{2}}{5}}{\sqrt {g x +f}\, g \left (d^{3} g^{5} x^{2}-3 d^{2} e f \,g^{4} x^{2}+3 d \,e^{2} f^{2} g^{3} x^{2}-e^{3} f^{3} g^{2} x^{2}+2 d^{3} f \,g^{4} x -6 d^{2} e \,f^{2} g^{3} x +6 d \,e^{2} f^{3} g^{2} x -2 e^{3} f^{4} g x +d^{3} f^{2} g^{3}-3 d^{2} e \,f^{3} g^{2}+3 d \,e^{2} f^{4} g -e^{3} f^{5}\right )} \] Input:

int((a+b*log(c*(e*x+d)^n))/(g*x+f)^(7/2),x)
 

Output:

(2*(6*sqrt(e)*sqrt(f + g*x)*sqrt(d*g - e*f)*atan((sqrt(f + g*x)*e)/(sqrt(e 
)*sqrt(d*g - e*f)))*b*e**2*f**2*n + 12*sqrt(e)*sqrt(f + g*x)*sqrt(d*g - e* 
f)*atan((sqrt(f + g*x)*e)/(sqrt(e)*sqrt(d*g - e*f)))*b*e**2*f*g*n*x + 6*sq 
rt(e)*sqrt(f + g*x)*sqrt(d*g - e*f)*atan((sqrt(f + g*x)*e)/(sqrt(e)*sqrt(d 
*g - e*f)))*b*e**2*g**2*n*x**2 - 3*log(((d*g + e*g*x)**n*c)/g**n)*b*d**3*g 
**3 + 9*log(((d*g + e*g*x)**n*c)/g**n)*b*d**2*e*f*g**2 - 9*log(((d*g + e*g 
*x)**n*c)/g**n)*b*d*e**2*f**2*g + 3*log(((d*g + e*g*x)**n*c)/g**n)*b*e**3* 
f**3 - 3*a*d**3*g**3 + 9*a*d**2*e*f*g**2 - 9*a*d*e**2*f**2*g + 3*a*e**3*f* 
*3 - 2*b*d**2*e*f*g**2*n - 2*b*d**2*e*g**3*n*x + 10*b*d*e**2*f**2*g*n + 16 
*b*d*e**2*f*g**2*n*x + 6*b*d*e**2*g**3*n*x**2 - 8*b*e**3*f**3*n - 14*b*e** 
3*f**2*g*n*x - 6*b*e**3*f*g**2*n*x**2))/(15*sqrt(f + g*x)*g*(d**3*f**2*g** 
3 + 2*d**3*f*g**4*x + d**3*g**5*x**2 - 3*d**2*e*f**3*g**2 - 6*d**2*e*f**2* 
g**3*x - 3*d**2*e*f*g**4*x**2 + 3*d*e**2*f**4*g + 6*d*e**2*f**3*g**2*x + 3 
*d*e**2*f**2*g**3*x**2 - e**3*f**5 - 2*e**3*f**4*g*x - e**3*f**3*g**2*x**2 
))