\(\int \frac {F^{a+b (c+d x)^2}}{(c+d x)^{12}} \, dx\) [213]

Optimal result

Integrand size = 21, antiderivative size = 49 \[ \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^{12}} \, dx=-\frac {F^a \Gamma \left (-\frac {11}{2},-b (c+d x)^2 \log (F)\right ) \left (-b (c+d x)^2 \log (F)\right )^{11/2}}{2 d (c+d x)^{11}} \] Output:

-1/2*F^a*(64/10395*Pi^(1/2)*erfc((-b*(d*x+c)^2*ln(F))^(1/2))-64/10395/(-b* 
(d*x+c)^2*ln(F))^(1/2)*exp(b*(d*x+c)^2*ln(F))+32/10395/(-b*(d*x+c)^2*ln(F) 
)^(3/2)*exp(b*(d*x+c)^2*ln(F))-16/3465/(-b*(d*x+c)^2*ln(F))^(5/2)*exp(b*(d 
*x+c)^2*ln(F))+8/693/(-b*(d*x+c)^2*ln(F))^(7/2)*exp(b*(d*x+c)^2*ln(F))-4/9 
9/(-b*(d*x+c)^2*ln(F))^(9/2)*exp(b*(d*x+c)^2*ln(F))+2/11/(-b*(d*x+c)^2*ln( 
F))^(11/2)*exp(b*(d*x+c)^2*ln(F)))*(-b*(d*x+c)^2*ln(F))^(11/2)/d/(d*x+c)^1 
1
 

Mathematica [A] (verified)

Time = 0.57 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.00 \[ \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^{12}} \, dx=-\frac {F^a \Gamma \left (-\frac {11}{2},-b (c+d x)^2 \log (F)\right ) \left (-b (c+d x)^2 \log (F)\right )^{11/2}}{2 d (c+d x)^{11}} \] Input:

Integrate[F^(a + b*(c + d*x)^2)/(c + d*x)^12,x]
 

Output:

-1/2*(F^a*Gamma[-11/2, -(b*(c + d*x)^2*Log[F])]*(-(b*(c + d*x)^2*Log[F]))^ 
(11/2))/(d*(c + d*x)^11)
 

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {2648}

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.

\(\displaystyle \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^{12}} \, dx\)

\(\Big \downarrow \) 2648

\(\displaystyle -\frac {F^a \left (-b \log (F) (c+d x)^2\right )^{11/2} \Gamma \left (-\frac {11}{2},-b (c+d x)^2 \log (F)\right )}{2 d (c+d x)^{11}}\)

Input:

Int[F^(a + b*(c + d*x)^2)/(c + d*x)^12,x]
 

Output:

-1/2*(F^a*Gamma[-11/2, -(b*(c + d*x)^2*Log[F])]*(-(b*(c + d*x)^2*Log[F]))^ 
(11/2))/(d*(c + d*x)^11)
 

Defintions of rubi rules used

rule 2648

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_ 
.), x_Symbol] :> Simp[(-F^a)*((e + f*x)^(m + 1)/(f*n*((-b)*(c + d*x)^n*Log[ 
F])^((m + 1)/n)))*Gamma[(m + 1)/n, (-b)*(c + d*x)^n*Log[F]], x] /; FreeQ[{F 
, a, b, c, d, e, f, m, n}, x] && EqQ[d*e - c*f, 0]
 

Maple [A] (verified)

Time = 1.30 (sec) , antiderivative size = 228, normalized size of antiderivative = 4.65

method	result	size
risch	\(-\frac {F^{b \left (d x +c \right )^{2}} F^{a}}{11 d \left (d x +c \right )^{11}}-\frac {2 b \ln \left (F \right ) F^{b \left (d x +c \right )^{2}} F^{a}}{99 d \left (d x +c \right )^{9}}-\frac {4 b^{2} \ln \left (F \right )^{2} F^{b \left (d x +c \right )^{2}} F^{a}}{693 d \left (d x +c \right )^{7}}-\frac {8 b^{3} \ln \left (F \right )^{3} F^{b \left (d x +c \right )^{2}} F^{a}}{3465 d \left (d x +c \right )^{5}}-\frac {16 b^{4} \ln \left (F \right )^{4} F^{b \left (d x +c \right )^{2}} F^{a}}{10395 d \left (d x +c \right )^{3}}-\frac {32 b^{5} \ln \left (F \right )^{5} F^{b \left (d x +c \right )^{2}} F^{a}}{10395 d \left (d x +c \right )}+\frac {32 b^{6} \ln \left (F \right )^{6} \sqrt {\pi }\, F^{a} \operatorname {erf}\left (\sqrt {-b \ln \left (F \right )}\, \left (d x +c \right )\right )}{10395 d \sqrt {-b \ln \left (F \right )}}\)	\(228\)

Input:

int(F^(a+b*(d*x+c)^2)/(d*x+c)^12,x,method=_RETURNVERBOSE)
 

Output:

-1/11/d/(d*x+c)^11*F^(b*(d*x+c)^2)*F^a-2/99/d*b*ln(F)/(d*x+c)^9*F^(b*(d*x+ 
c)^2)*F^a-4/693/d*b^2*ln(F)^2/(d*x+c)^7*F^(b*(d*x+c)^2)*F^a-8/3465/d*b^3*l 
n(F)^3/(d*x+c)^5*F^(b*(d*x+c)^2)*F^a-16/10395/d*b^4*ln(F)^4/(d*x+c)^3*F^(b 
*(d*x+c)^2)*F^a-32/10395/d*b^5*ln(F)^5/(d*x+c)*F^(b*(d*x+c)^2)*F^a+32/1039 
5/d*b^6*ln(F)^6*Pi^(1/2)*F^a/(-b*ln(F))^(1/2)*erf((-b*ln(F))^(1/2)*(d*x+c) 
)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 795 vs. \(2 (212) = 424\).

Time = 0.09 (sec) , antiderivative size = 795, normalized size of antiderivative = 16.22 \[ \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^{12}} \, dx =\text {Too large to display} \] Input:

integrate(F^(a+b*(d*x+c)^2)/(d*x+c)^12,x, algorithm="fricas")
 

Output:

-1/10395*(32*sqrt(pi)*(b^5*d^11*x^11 + 11*b^5*c*d^10*x^10 + 55*b^5*c^2*d^9 
*x^9 + 165*b^5*c^3*d^8*x^8 + 330*b^5*c^4*d^7*x^7 + 462*b^5*c^5*d^6*x^6 + 4 
62*b^5*c^6*d^5*x^5 + 330*b^5*c^7*d^4*x^4 + 165*b^5*c^8*d^3*x^3 + 55*b^5*c^ 
9*d^2*x^2 + 11*b^5*c^10*d*x + b^5*c^11)*sqrt(-b*d^2*log(F))*F^a*erf(sqrt(- 
b*d^2*log(F))*(d*x + c)/d)*log(F)^5 + (32*(b^5*d^11*x^10 + 10*b^5*c*d^10*x 
^9 + 45*b^5*c^2*d^9*x^8 + 120*b^5*c^3*d^8*x^7 + 210*b^5*c^4*d^7*x^6 + 252* 
b^5*c^5*d^6*x^5 + 210*b^5*c^6*d^5*x^4 + 120*b^5*c^7*d^4*x^3 + 45*b^5*c^8*d 
^3*x^2 + 10*b^5*c^9*d^2*x + b^5*c^10*d)*log(F)^5 + 16*(b^4*d^9*x^8 + 8*b^4 
*c*d^8*x^7 + 28*b^4*c^2*d^7*x^6 + 56*b^4*c^3*d^6*x^5 + 70*b^4*c^4*d^5*x^4 
+ 56*b^4*c^5*d^4*x^3 + 28*b^4*c^6*d^3*x^2 + 8*b^4*c^7*d^2*x + b^4*c^8*d)*l 
og(F)^4 + 24*(b^3*d^7*x^6 + 6*b^3*c*d^6*x^5 + 15*b^3*c^2*d^5*x^4 + 20*b^3* 
c^3*d^4*x^3 + 15*b^3*c^4*d^3*x^2 + 6*b^3*c^5*d^2*x + b^3*c^6*d)*log(F)^3 + 
 60*(b^2*d^5*x^4 + 4*b^2*c*d^4*x^3 + 6*b^2*c^2*d^3*x^2 + 4*b^2*c^3*d^2*x + 
 b^2*c^4*d)*log(F)^2 + 210*(b*d^3*x^2 + 2*b*c*d^2*x + b*c^2*d)*log(F) + 94 
5*d)*F^(b*d^2*x^2 + 2*b*c*d*x + b*c^2 + a))/(d^13*x^11 + 11*c*d^12*x^10 + 
55*c^2*d^11*x^9 + 165*c^3*d^10*x^8 + 330*c^4*d^9*x^7 + 462*c^5*d^8*x^6 + 4 
62*c^6*d^7*x^5 + 330*c^7*d^6*x^4 + 165*c^8*d^5*x^3 + 55*c^9*d^4*x^2 + 11*c 
^10*d^3*x + c^11*d^2)
 

Sympy [F]

\[ \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^{12}} \, dx=\int \frac {F^{a + b \left (c + d x\right )^{2}}}{\left (c + d x\right )^{12}}\, dx \] Input:

integrate(F**(a+b*(d*x+c)**2)/(d*x+c)**12,x)
 

Output:

Integral(F**(a + b*(c + d*x)**2)/(c + d*x)**12, x)
                                                                                    
                                                                                    
 

Maxima [F]

\[ \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^{12}} \, dx=\int { \frac {F^{{\left (d x + c\right )}^{2} b + a}}{{\left (d x + c\right )}^{12}} \,d x } \] Input:

integrate(F^(a+b*(d*x+c)^2)/(d*x+c)^12,x, algorithm="maxima")
 

Output:

integrate(F^((d*x + c)^2*b + a)/(d*x + c)^12, x)
 

Giac [F]

\[ \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^{12}} \, dx=\int { \frac {F^{{\left (d x + c\right )}^{2} b + a}}{{\left (d x + c\right )}^{12}} \,d x } \] Input:

integrate(F^(a+b*(d*x+c)^2)/(d*x+c)^12,x, algorithm="giac")
 

Output:

integrate(F^((d*x + c)^2*b + a)/(d*x + c)^12, x)
 

Mupad [B] (verification not implemented)

Time = 0.48 (sec) , antiderivative size = 267, normalized size of antiderivative = 5.45 \[ \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^{12}} \, dx=\frac {32\,F^a\,\sqrt {\pi }\,{\left (-b\,\ln \left (F\right )\,{\left (c+d\,x\right )}^2\right )}^{11/2}}{10395\,d\,{\left (c+d\,x\right )}^{11}}-\frac {F^a\,F^{b\,{\left (c+d\,x\right )}^2}}{11\,d\,{\left (c+d\,x\right )}^{11}}-\frac {4\,F^a\,F^{b\,{\left (c+d\,x\right )}^2}\,b^2\,{\ln \left (F\right )}^2}{693\,d\,{\left (c+d\,x\right )}^7}-\frac {8\,F^a\,F^{b\,{\left (c+d\,x\right )}^2}\,b^3\,{\ln \left (F\right )}^3}{3465\,d\,{\left (c+d\,x\right )}^5}-\frac {16\,F^a\,F^{b\,{\left (c+d\,x\right )}^2}\,b^4\,{\ln \left (F\right )}^4}{10395\,d\,{\left (c+d\,x\right )}^3}-\frac {32\,F^a\,F^{b\,{\left (c+d\,x\right )}^2}\,b^5\,{\ln \left (F\right )}^5}{10395\,d\,\left (c+d\,x\right )}-\frac {2\,F^a\,F^{b\,{\left (c+d\,x\right )}^2}\,b\,\ln \left (F\right )}{99\,d\,{\left (c+d\,x\right )}^9}-\frac {32\,F^a\,\sqrt {\pi }\,\mathrm {erfc}\left (\sqrt {-b\,\ln \left (F\right )\,{\left (c+d\,x\right )}^2}\right )\,{\left (-b\,\ln \left (F\right )\,{\left (c+d\,x\right )}^2\right )}^{11/2}}{10395\,d\,{\left (c+d\,x\right )}^{11}} \] Input:

int(F^(a + b*(c + d*x)^2)/(c + d*x)^12,x)
 

Output:

(32*F^a*pi^(1/2)*(-b*log(F)*(c + d*x)^2)^(11/2))/(10395*d*(c + d*x)^11) - 
(F^a*F^(b*(c + d*x)^2))/(11*d*(c + d*x)^11) - (4*F^a*F^(b*(c + d*x)^2)*b^2 
*log(F)^2)/(693*d*(c + d*x)^7) - (8*F^a*F^(b*(c + d*x)^2)*b^3*log(F)^3)/(3 
465*d*(c + d*x)^5) - (16*F^a*F^(b*(c + d*x)^2)*b^4*log(F)^4)/(10395*d*(c + 
 d*x)^3) - (32*F^a*F^(b*(c + d*x)^2)*b^5*log(F)^5)/(10395*d*(c + d*x)) - ( 
2*F^a*F^(b*(c + d*x)^2)*b*log(F))/(99*d*(c + d*x)^9) - (32*F^a*pi^(1/2)*er 
fc((-b*log(F)*(c + d*x)^2)^(1/2))*(-b*log(F)*(c + d*x)^2)^(11/2))/(10395*d 
*(c + d*x)^11)
 

Reduce [F]

\[ \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^{12}} \, dx=\text {too large to display} \] Input:

int(F^(a+b*(d*x+c)^2)/(d*x+c)^12,x)
 

Output:

(f**(a + b*c**2)*( - f**(2*b*c*d*x + b*d**2*x**2) + 8*int(f**(2*b*c*d*x + 
b*d**2*x**2)/(2*log(f)*b*c**14 + 24*log(f)*b*c**13*d*x + 132*log(f)*b*c**1 
2*d**2*x**2 + 440*log(f)*b*c**11*d**3*x**3 + 990*log(f)*b*c**10*d**4*x**4 
+ 1584*log(f)*b*c**9*d**5*x**5 + 1848*log(f)*b*c**8*d**6*x**6 + 1584*log(f 
)*b*c**7*d**7*x**7 + 990*log(f)*b*c**6*d**8*x**8 + 440*log(f)*b*c**5*d**9* 
x**9 + 132*log(f)*b*c**4*d**10*x**10 + 24*log(f)*b*c**3*d**11*x**11 + 2*lo 
g(f)*b*c**2*d**12*x**12 - 11*c**12 - 132*c**11*d*x - 726*c**10*d**2*x**2 - 
 2420*c**9*d**3*x**3 - 5445*c**8*d**4*x**4 - 8712*c**7*d**5*x**5 - 10164*c 
**6*d**6*x**6 - 8712*c**5*d**7*x**7 - 5445*c**4*d**8*x**8 - 2420*c**3*d**9 
*x**9 - 726*c**2*d**10*x**10 - 132*c*d**11*x**11 - 11*d**12*x**12),x)*log( 
f)**2*b**2*c**15*d + 88*int(f**(2*b*c*d*x + b*d**2*x**2)/(2*log(f)*b*c**14 
 + 24*log(f)*b*c**13*d*x + 132*log(f)*b*c**12*d**2*x**2 + 440*log(f)*b*c** 
11*d**3*x**3 + 990*log(f)*b*c**10*d**4*x**4 + 1584*log(f)*b*c**9*d**5*x**5 
 + 1848*log(f)*b*c**8*d**6*x**6 + 1584*log(f)*b*c**7*d**7*x**7 + 990*log(f 
)*b*c**6*d**8*x**8 + 440*log(f)*b*c**5*d**9*x**9 + 132*log(f)*b*c**4*d**10 
*x**10 + 24*log(f)*b*c**3*d**11*x**11 + 2*log(f)*b*c**2*d**12*x**12 - 11*c 
**12 - 132*c**11*d*x - 726*c**10*d**2*x**2 - 2420*c**9*d**3*x**3 - 5445*c* 
*8*d**4*x**4 - 8712*c**7*d**5*x**5 - 10164*c**6*d**6*x**6 - 8712*c**5*d**7 
*x**7 - 5445*c**4*d**8*x**8 - 2420*c**3*d**9*x**9 - 726*c**2*d**10*x**10 - 
 132*c*d**11*x**11 - 11*d**12*x**12),x)*log(f)**2*b**2*c**14*d**2*x + 4...