3.5.16 \(\int \frac {1}{(a+b \log (c (d (e+f x)^m)^n))^{5/2}} \, dx\) [416]

3.5.16.1 Optimal result

Integrand size = 22, antiderivative size = 194 \[ \int \frac {1}{\left (a+b \log \left (c \left (d (e+f x)^m\right )^n\right )\right )^{5/2}} \, dx=\frac {4 e^{-\frac {a}{b m n}} \sqrt {\pi } (e+f x) \left (c \left (d (e+f x)^m\right )^n\right )^{-\frac {1}{m n}} \text {erfi}\left (\frac {\sqrt {a+b \log \left (c \left (d (e+f x)^m\right )^n\right )}}{\sqrt {b} \sqrt {m} \sqrt {n}}\right )}{3 b^{5/2} f m^{5/2} n^{5/2}}-\frac {2 (e+f x)}{3 b f m n \left (a+b \log \left (c \left (d (e+f x)^m\right )^n\right )\right )^{3/2}}-\frac {4 (e+f x)}{3 b^2 f m^2 n^2 \sqrt {a+b \log \left (c \left (d (e+f x)^m\right )^n\right )}} \]

output

-2/3*(f*x+e)/b/f/m/n/(a+b*ln(c*(d*(f*x+e)^m)^n))^(3/2)+4/3*(f*x+e)*erfi((a 
+b*ln(c*(d*(f*x+e)^m)^n))^(1/2)/b^(1/2)/m^(1/2)/n^(1/2))*Pi^(1/2)/b^(5/2)/ 
exp(a/b/m/n)/f/m^(5/2)/n^(5/2)/((c*(d*(f*x+e)^m)^n)^(1/m/n))-4/3*(f*x+e)/b 
^2/f/m^2/n^2/(a+b*ln(c*(d*(f*x+e)^m)^n))^(1/2)
 

3.5.16.2 Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.09 \[ \int \frac {1}{\left (a+b \log \left (c \left (d (e+f x)^m\right )^n\right )\right )^{5/2}} \, dx=-\frac {2 e^{-\frac {a}{b m n}} (e+f x) \left (c \left (d (e+f x)^m\right )^n\right )^{-\frac {1}{m n}} \left (2 b m n \Gamma \left (\frac {1}{2},-\frac {a+b \log \left (c \left (d (e+f x)^m\right )^n\right )}{b m n}\right ) \left (-\frac {a+b \log \left (c \left (d (e+f x)^m\right )^n\right )}{b m n}\right )^{3/2}+e^{\frac {a}{b m n}} \left (c \left (d (e+f x)^m\right )^n\right )^{\frac {1}{m n}} \left (2 a+b m n+2 b \log \left (c \left (d (e+f x)^m\right )^n\right )\right )\right )}{3 b^2 f m^2 n^2 \left (a+b \log \left (c \left (d (e+f x)^m\right )^n\right )\right )^{3/2}} \]

input

Integrate[(a + b*Log[c*(d*(e + f*x)^m)^n])^(-5/2),x]
 

output

(-2*(e + f*x)*(2*b*m*n*Gamma[1/2, -((a + b*Log[c*(d*(e + f*x)^m)^n])/(b*m* 
n))]*(-((a + b*Log[c*(d*(e + f*x)^m)^n])/(b*m*n)))^(3/2) + E^(a/(b*m*n))*( 
c*(d*(e + f*x)^m)^n)^(1/(m*n))*(2*a + b*m*n + 2*b*Log[c*(d*(e + f*x)^m)^n] 
)))/(3*b^2*E^(a/(b*m*n))*f*m^2*n^2*(c*(d*(e + f*x)^m)^n)^(1/(m*n))*(a + b* 
Log[c*(d*(e + f*x)^m)^n])^(3/2))
 

3.5.16.3 Rubi [A] (warning: unable to verify)

Time = 0.79 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.05, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {2895, 2836, 2734, 2734, 2737, 2611, 2633}

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.

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

\(\Big \downarrow \) 2895

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

\(\Big \downarrow \) 2836

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

\(\Big \downarrow \) 2734

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

\(\Big \downarrow \) 2734

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

\(\Big \downarrow \) 2737

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

\(\Big \downarrow \) 2611

\(\displaystyle \frac {\frac {2 \left (\frac {4 (e+f x) \left (c d^n (e+f x)^{m n}\right )^{-\frac {1}{m n}} \int \exp \left (\frac {a+b \log \left (c d^n (e+f x)^{m n}\right )}{b m n}-\frac {a}{b m n}\right )d\sqrt {a+b \log \left (c d^n (e+f x)^{m n}\right )}}{b^2 m^2 n^2}-\frac {2 (e+f x)}{b m n \sqrt {a+b \log \left (c d^n (e+f x)^{m n}\right )}}\right )}{3 b m n}-\frac {2 (e+f x)}{3 b m n \left (a+b \log \left (c d^n (e+f x)^{m n}\right )\right )^{3/2}}}{f}\)

\(\Big \downarrow \) 2633

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

input

Int[(a + b*Log[c*(d*(e + f*x)^m)^n])^(-5/2),x]
 

output

((-2*(e + f*x))/(3*b*m*n*(a + b*Log[c*d^n*(e + f*x)^(m*n)])^(3/2)) + (2*(( 
2*Sqrt[Pi]*(e + f*x)*Erfi[Sqrt[a + b*Log[c*d^n*(e + f*x)^(m*n)]]/(Sqrt[b]* 
Sqrt[m]*Sqrt[n])])/(b^(3/2)*E^(a/(b*m*n))*m^(3/2)*n^(3/2)*(c*d^n*(e + f*x) 
^(m*n))^(1/(m*n))) - (2*(e + f*x))/(b*m*n*Sqrt[a + b*Log[c*d^n*(e + f*x)^( 
m*n)]])))/(3*b*m*n))/f
 

3.5.16.3.1 Defintions of rubi rules used

rule 2611

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] : 
> Simp[2/d   Subst[Int[F^(g*(e - c*(f/d)) + f*g*(x^2/d)), x], x, Sqrt[c + d 
*x]], x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !TrueQ[$UseGamma]
 

rule 2633

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[F^a*Sqrt 
[Pi]*(Erfi[(c + d*x)*Rt[b*Log[F], 2]]/(2*d*Rt[b*Log[F], 2])), x] /; FreeQ[{ 
F, a, b, c, d}, x] && PosQ[b]
 

rule 2734

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_), x_Symbol] :> Simp[x*((a + b 
*Log[c*x^n])^(p + 1)/(b*n*(p + 1))), x] - Simp[1/(b*n*(p + 1))   Int[(a + b 
*Log[c*x^n])^(p + 1), x], x] /; FreeQ[{a, b, c, n}, x] && LtQ[p, -1] && Int 
egerQ[2*p]
 

rule 2737

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_), x_Symbol] :> Simp[x/(n*(c*x 
^n)^(1/n))   Subst[Int[E^(x/n)*(a + b*x)^p, x], x, Log[c*x^n]], x] /; FreeQ 
[{a, b, c, n, p}, x]
 

rule 2836

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.), x_Symbol] : 
> Simp[1/e   Subst[Int[(a + b*Log[c*x^n])^p, x], x, d + e*x], x] /; FreeQ[{ 
a, b, c, d, e, n, p}, x]
 

rule 2895

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]]
 

3.5.16.4 Maple [F]

\[\int \frac {1}{{\left (a +b \ln \left (c \left (d \left (f x +e \right )^{m}\right )^{n}\right )\right )}^{\frac {5}{2}}}d x\]

input

int(1/(a+b*ln(c*(d*(f*x+e)^m)^n))^(5/2),x)
 

output

int(1/(a+b*ln(c*(d*(f*x+e)^m)^n))^(5/2),x)
 

3.5.16.5 Fricas [F(-2)]

Exception generated. \[ \int \frac {1}{\left (a+b \log \left (c \left (d (e+f x)^m\right )^n\right )\right )^{5/2}} \, dx=\text {Exception raised: TypeError} \]

input

integrate(1/(a+b*log(c*(d*(f*x+e)^m)^n))^(5/2),x, algorithm="fricas")
 

output

Exception raised: TypeError >>  Error detected within library code:   inte 
grate: implementation incomplete (constant residues)
 

3.5.16.6 Sympy [F]

\[ \int \frac {1}{\left (a+b \log \left (c \left (d (e+f x)^m\right )^n\right )\right )^{5/2}} \, dx=\int \frac {1}{\left (a + b \log {\left (c \left (d \left (e + f x\right )^{m}\right )^{n} \right )}\right )^{\frac {5}{2}}}\, dx \]

input

integrate(1/(a+b*ln(c*(d*(f*x+e)**m)**n))**(5/2),x)
 

output

Integral((a + b*log(c*(d*(e + f*x)**m)**n))**(-5/2), x)
 

3.5.16.7 Maxima [F]

\[ \int \frac {1}{\left (a+b \log \left (c \left (d (e+f x)^m\right )^n\right )\right )^{5/2}} \, dx=\int { \frac {1}{{\left (b \log \left (\left ({\left (f x + e\right )}^{m} d\right )^{n} c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]

input

integrate(1/(a+b*log(c*(d*(f*x+e)^m)^n))^(5/2),x, algorithm="maxima")
 

output

integrate((b*log(((f*x + e)^m*d)^n*c) + a)^(-5/2), x)
 

3.5.16.8 Giac [F]

\[ \int \frac {1}{\left (a+b \log \left (c \left (d (e+f x)^m\right )^n\right )\right )^{5/2}} \, dx=\int { \frac {1}{{\left (b \log \left (\left ({\left (f x + e\right )}^{m} d\right )^{n} c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]

input

integrate(1/(a+b*log(c*(d*(f*x+e)^m)^n))^(5/2),x, algorithm="giac")
 

output

integrate((b*log(((f*x + e)^m*d)^n*c) + a)^(-5/2), x)
 

3.5.16.9 Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\left (a+b \log \left (c \left (d (e+f x)^m\right )^n\right )\right )^{5/2}} \, dx=\int \frac {1}{{\left (a+b\,\ln \left (c\,{\left (d\,{\left (e+f\,x\right )}^m\right )}^n\right )\right )}^{5/2}} \,d x \]

input

int(1/(a + b*log(c*(d*(e + f*x)^m)^n))^(5/2),x)
 

output

int(1/(a + b*log(c*(d*(e + f*x)^m)^n))^(5/2), x)