Integrand size = 21, antiderivative size = 49 \[ \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^{14}} \, dx=\frac {F^a \Gamma \left (\frac {13}{2},-\frac {b \log (F)}{(c+d x)^2}\right )}{2 d (c+d x)^{13} \left (-\frac {b \log (F)}{(c+d x)^2}\right )^{13/2}} \] Output:
1/2*F^a*(524288/5621533568633696205238621875*GAMMA(51/2,-b*ln(F)/(d*x+c)^2 )-524288/5621533568633696205238621875*(-b*ln(F)/(d*x+c)^2)^(49/2)*exp(b*ln (F)/(d*x+c)^2)-262144/114725174870075432759971875*(-b*ln(F)/(d*x+c)^2)^(47 /2)*exp(b*ln(F)/(d*x+c)^2)-131072/2440961167448413462978125*(-b*ln(F)/(d*x +c)^2)^(45/2)*exp(b*ln(F)/(d*x+c)^2)-65536/54243581498853632510625*(-b*ln( F)/(d*x+c)^2)^(43/2)*exp(b*ln(F)/(d*x+c)^2)-32768/1261478639508224011875*( -b*ln(F)/(d*x+c)^2)^(41/2)*exp(b*ln(F)/(d*x+c)^2)-16384/307677716953225368 75*(-b*ln(F)/(d*x+c)^2)^(39/2)*exp(b*ln(F)/(d*x+c)^2)-8192/788917222956988 125*(-b*ln(F)/(d*x+c)^2)^(37/2)*exp(b*ln(F)/(d*x+c)^2)-4096/21322087106945 625*(-b*ln(F)/(d*x+c)^2)^(35/2)*exp(b*ln(F)/(d*x+c)^2)-2048/60920248876987 5*(-b*ln(F)/(d*x+c)^2)^(33/2)*exp(b*ln(F)/(d*x+c)^2)-1024/18460681477875*( -b*ln(F)/(d*x+c)^2)^(31/2)*exp(b*ln(F)/(d*x+c)^2)-512/595505854125*(-b*ln( F)/(d*x+c)^2)^(29/2)*exp(b*ln(F)/(d*x+c)^2)-256/20534684625*(-b*ln(F)/(d*x +c)^2)^(27/2)*exp(b*ln(F)/(d*x+c)^2)-128/760543875*(-b*ln(F)/(d*x+c)^2)^(2 5/2)*exp(b*ln(F)/(d*x+c)^2)-64/30421755*(-b*ln(F)/(d*x+c)^2)^(23/2)*exp(b* ln(F)/(d*x+c)^2)-32/1322685*(-b*ln(F)/(d*x+c)^2)^(21/2)*exp(b*ln(F)/(d*x+c )^2)-16/62985*(-b*ln(F)/(d*x+c)^2)^(19/2)*exp(b*ln(F)/(d*x+c)^2)-8/3315*(- b*ln(F)/(d*x+c)^2)^(17/2)*exp(b*ln(F)/(d*x+c)^2)-4/195*(-b*ln(F)/(d*x+c)^2 )^(15/2)*exp(b*ln(F)/(d*x+c)^2)-2/13*(-b*ln(F)/(d*x+c)^2)^(13/2)*exp(b*ln( F)/(d*x+c)^2))/d/(d*x+c)^13/(-b*ln(F)/(d*x+c)^2)^(13/2)
Time = 0.03 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.00 \[ \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^{14}} \, dx=\frac {F^a \Gamma \left (\frac {13}{2},-\frac {b \log (F)}{(c+d x)^2}\right )}{2 d (c+d x)^{13} \left (-\frac {b \log (F)}{(c+d x)^2}\right )^{13/2}} \] Input:
Integrate[F^(a + b/(c + d*x)^2)/(c + d*x)^14,x]
Output:
(F^a*Gamma[13/2, -((b*Log[F])/(c + d*x)^2)])/(2*d*(c + d*x)^13*(-((b*Log[F ])/(c + d*x)^2))^(13/2))
Time = 0.32 (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.
| \(\displaystyle \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^{14}} \, dx\) |
| \(\Big \downarrow \) 2648 |
| \(\displaystyle \frac {F^a \Gamma \left (\frac {13}{2},-\frac {b \log (F)}{(c+d x)^2}\right )}{2 d (c+d x)^{13} \left (-\frac {b \log (F)}{(c+d x)^2}\right )^{13/2}}\) |
Input:
Int[F^(a + b/(c + d*x)^2)/(c + d*x)^14,x]
Output:
(F^a*Gamma[13/2, -((b*Log[F])/(c + d*x)^2)])/(2*d*(c + d*x)^13*(-((b*Log[F ])/(c + d*x)^2))^(13/2))
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]
Time = 10.40 (sec) , antiderivative size = 241, normalized size of antiderivative = 4.92
| method | result | size |
| risch | \(-\frac {F^{a} F^{\frac {b}{\left (d x +c \right )^{2}}}}{2 d \left (d x +c \right )^{11} b \ln \left (F \right )}+\frac {11 F^{a} F^{\frac {b}{\left (d x +c \right )^{2}}}}{4 d \,b^{2} \ln \left (F \right )^{2} \left (d x +c \right )^{9}}-\frac {99 F^{a} F^{\frac {b}{\left (d x +c \right )^{2}}}}{8 d \,b^{3} \ln \left (F \right )^{3} \left (d x +c \right )^{7}}+\frac {693 F^{a} F^{\frac {b}{\left (d x +c \right )^{2}}}}{16 d \,b^{4} \ln \left (F \right )^{4} \left (d x +c \right )^{5}}-\frac {3465 F^{a} F^{\frac {b}{\left (d x +c \right )^{2}}}}{32 d \,b^{5} \ln \left (F \right )^{5} \left (d x +c \right )^{3}}+\frac {10395 F^{a} F^{\frac {b}{\left (d x +c \right )^{2}}}}{64 d \,b^{6} \ln \left (F \right )^{6} \left (d x +c \right )}-\frac {10395 F^{a} \sqrt {\pi }\, \operatorname {erf}\left (\frac {\sqrt {-b \ln \left (F \right )}}{d x +c}\right )}{128 d \,b^{6} \ln \left (F \right )^{6} \sqrt {-b \ln \left (F \right )}}\) | \(241\) |
Input:
int(F^(a+b/(d*x+c)^2)/(d*x+c)^14,x,method=_RETURNVERBOSE)
Output:
-1/2*F^a/d*F^(b/(d*x+c)^2)/(d*x+c)^11/b/ln(F)+11/4*F^a/d/b^2/ln(F)^2*F^(b/ (d*x+c)^2)/(d*x+c)^9-99/8*F^a/d/b^3/ln(F)^3*F^(b/(d*x+c)^2)/(d*x+c)^7+693/ 16*F^a/d/b^4/ln(F)^4*F^(b/(d*x+c)^2)/(d*x+c)^5-3465/32*F^a/d/b^5/ln(F)^5*F ^(b/(d*x+c)^2)/(d*x+c)^3+10395/64*F^a/d/b^6/ln(F)^6*F^(b/(d*x+c)^2)/(d*x+c )-10395/128*F^a/d/b^6/ln(F)^6*Pi^(1/2)/(-b*ln(F))^(1/2)*erf((-b*ln(F))^(1/ 2)/(d*x+c))
Time = 0.12 (sec) , antiderivative size = 791, normalized size of antiderivative = 16.14 \[ \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^{14}} \, dx =\text {Too large to display} \] Input:
integrate(F^(a+b/(d*x+c)^2)/(d*x+c)^14,x, algorithm="fricas")
Output:
1/128*(10395*sqrt(pi)*(d^12*x^11 + 11*c*d^11*x^10 + 55*c^2*d^10*x^9 + 165* c^3*d^9*x^8 + 330*c^4*d^8*x^7 + 462*c^5*d^7*x^6 + 462*c^6*d^6*x^5 + 330*c^ 7*d^5*x^4 + 165*c^8*d^4*x^3 + 55*c^9*d^3*x^2 + 11*c^10*d^2*x + c^11*d)*F^a *sqrt(-b*log(F)/d^2)*erf(d*sqrt(-b*log(F)/d^2)/(d*x + c)) - 2*(32*b^6*log( F)^6 - 176*(b^5*d^2*x^2 + 2*b^5*c*d*x + b^5*c^2)*log(F)^5 + 792*(b^4*d^4*x ^4 + 4*b^4*c*d^3*x^3 + 6*b^4*c^2*d^2*x^2 + 4*b^4*c^3*d*x + b^4*c^4)*log(F) ^4 - 2772*(b^3*d^6*x^6 + 6*b^3*c*d^5*x^5 + 15*b^3*c^2*d^4*x^4 + 20*b^3*c^3 *d^3*x^3 + 15*b^3*c^4*d^2*x^2 + 6*b^3*c^5*d*x + b^3*c^6)*log(F)^3 + 6930*( b^2*d^8*x^8 + 8*b^2*c*d^7*x^7 + 28*b^2*c^2*d^6*x^6 + 56*b^2*c^3*d^5*x^5 + 70*b^2*c^4*d^4*x^4 + 56*b^2*c^5*d^3*x^3 + 28*b^2*c^6*d^2*x^2 + 8*b^2*c^7*d *x + b^2*c^8)*log(F)^2 - 10395*(b*d^10*x^10 + 10*b*c*d^9*x^9 + 45*b*c^2*d^ 8*x^8 + 120*b*c^3*d^7*x^7 + 210*b*c^4*d^6*x^6 + 252*b*c^5*d^5*x^5 + 210*b* c^6*d^4*x^4 + 120*b*c^7*d^3*x^3 + 45*b*c^8*d^2*x^2 + 10*b*c^9*d*x + b*c^10 )*log(F))*F^((a*d^2*x^2 + 2*a*c*d*x + a*c^2 + b)/(d^2*x^2 + 2*c*d*x + c^2) ))/((b^7*d^12*x^11 + 11*b^7*c*d^11*x^10 + 55*b^7*c^2*d^10*x^9 + 165*b^7*c^ 3*d^9*x^8 + 330*b^7*c^4*d^8*x^7 + 462*b^7*c^5*d^7*x^6 + 462*b^7*c^6*d^6*x^ 5 + 330*b^7*c^7*d^5*x^4 + 165*b^7*c^8*d^4*x^3 + 55*b^7*c^9*d^3*x^2 + 11*b^ 7*c^10*d^2*x + b^7*c^11*d)*log(F)^7)
Timed out. \[ \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^{14}} \, dx=\text {Timed out} \] Input:
integrate(F**(a+b/(d*x+c)**2)/(d*x+c)**14,x)
Output:
Timed out
\[ \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^{14}} \, dx=\int { \frac {F^{a + \frac {b}{{\left (d x + c\right )}^{2}}}}{{\left (d x + c\right )}^{14}} \,d x } \] Input:
integrate(F^(a+b/(d*x+c)^2)/(d*x+c)^14,x, algorithm="maxima")
Output:
integrate(F^(a + b/(d*x + c)^2)/(d*x + c)^14, x)
\[ \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^{14}} \, dx=\int { \frac {F^{a + \frac {b}{{\left (d x + c\right )}^{2}}}}{{\left (d x + c\right )}^{14}} \,d x } \] Input:
integrate(F^(a+b/(d*x+c)^2)/(d*x+c)^14,x, algorithm="giac")
Output:
integrate(F^(a + b/(d*x + c)^2)/(d*x + c)^14, x)
Time = 0.97 (sec) , antiderivative size = 217, normalized size of antiderivative = 4.43 \[ \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^{14}} \, dx=-\frac {\frac {F^a\,\left (\frac {10395\,\sqrt {\pi }\,\mathrm {erfi}\left (\frac {b\,\ln \left (F\right )}{\sqrt {b\,\ln \left (F\right )}\,\left (c+d\,x\right )}\right )}{128}-\frac {10395\,F^{\frac {b}{{\left (c+d\,x\right )}^2}}\,\sqrt {b\,\ln \left (F\right )}}{64\,\left (c+d\,x\right )}\right )}{\sqrt {b\,\ln \left (F\right )}}-\frac {693\,F^{a+\frac {b}{{\left (c+d\,x\right )}^2}}\,b^2\,{\ln \left (F\right )}^2}{16\,{\left (c+d\,x\right )}^5}+\frac {99\,F^{a+\frac {b}{{\left (c+d\,x\right )}^2}}\,b^3\,{\ln \left (F\right )}^3}{8\,{\left (c+d\,x\right )}^7}-\frac {11\,F^{a+\frac {b}{{\left (c+d\,x\right )}^2}}\,b^4\,{\ln \left (F\right )}^4}{4\,{\left (c+d\,x\right )}^9}+\frac {F^{a+\frac {b}{{\left (c+d\,x\right )}^2}}\,b^5\,{\ln \left (F\right )}^5}{2\,{\left (c+d\,x\right )}^{11}}+\frac {3465\,F^{a+\frac {b}{{\left (c+d\,x\right )}^2}}\,b\,\ln \left (F\right )}{32\,{\left (c+d\,x\right )}^3}}{b^6\,d\,{\ln \left (F\right )}^6} \] Input:
int(F^(a + b/(c + d*x)^2)/(c + d*x)^14,x)
Output:
-((F^a*((10395*pi^(1/2)*erfi((b*log(F))/((b*log(F))^(1/2)*(c + d*x))))/128 - (10395*F^(b/(c + d*x)^2)*(b*log(F))^(1/2))/(64*(c + d*x))))/(b*log(F))^ (1/2) - (693*F^(a + b/(c + d*x)^2)*b^2*log(F)^2)/(16*(c + d*x)^5) + (99*F^ (a + b/(c + d*x)^2)*b^3*log(F)^3)/(8*(c + d*x)^7) - (11*F^(a + b/(c + d*x) ^2)*b^4*log(F)^4)/(4*(c + d*x)^9) + (F^(a + b/(c + d*x)^2)*b^5*log(F)^5)/( 2*(c + d*x)^11) + (3465*F^(a + b/(c + d*x)^2)*b*log(F))/(32*(c + d*x)^3))/ (b^6*d*log(F)^6)
\[ \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^{14}} \, dx=\text {too large to display} \] Input:
int(F^(a+b/(d*x+c)^2)/(d*x+c)^14,x)
Output:
(2*f**((a*c**2 + 2*a*c*d*x + a*d**2*x**2 + b)/(c**2 + 2*c*d*x + d**2*x**2) )*log(f)*b - 15*f**((a*c**2 + 2*a*c*d*x + a*d**2*x**2 + b)/(c**2 + 2*c*d*x + d**2*x**2))*c**2 - 30*f**((a*c**2 + 2*a*c*d*x + a*d**2*x**2 + b)/(c**2 + 2*c*d*x + d**2*x**2))*c*d*x - 15*f**((a*c**2 + 2*a*c*d*x + a*d**2*x**2 + b)/(c**2 + 2*c*d*x + d**2*x**2))*d**2*x**2 + 4*int(f**((a*c**2 + 2*a*c*d* x + a*d**2*x**2 + b)/(c**2 + 2*c*d*x + d**2*x**2))/(c**18 + 18*c**17*d*x + 153*c**16*d**2*x**2 + 816*c**15*d**3*x**3 + 3060*c**14*d**4*x**4 + 8568*c **13*d**5*x**5 + 18564*c**12*d**6*x**6 + 31824*c**11*d**7*x**7 + 43758*c** 10*d**8*x**8 + 48620*c**9*d**9*x**9 + 43758*c**8*d**10*x**10 + 31824*c**7* d**11*x**11 + 18564*c**6*d**12*x**12 + 8568*c**5*d**13*x**13 + 3060*c**4*d **14*x**14 + 816*c**3*d**15*x**15 + 153*c**2*d**16*x**16 + 18*c*d**17*x**1 7 + d**18*x**18),x)*log(f)**2*b**2*c**15*d + 60*int(f**((a*c**2 + 2*a*c*d* x + a*d**2*x**2 + b)/(c**2 + 2*c*d*x + d**2*x**2))/(c**18 + 18*c**17*d*x + 153*c**16*d**2*x**2 + 816*c**15*d**3*x**3 + 3060*c**14*d**4*x**4 + 8568*c **13*d**5*x**5 + 18564*c**12*d**6*x**6 + 31824*c**11*d**7*x**7 + 43758*c** 10*d**8*x**8 + 48620*c**9*d**9*x**9 + 43758*c**8*d**10*x**10 + 31824*c**7* d**11*x**11 + 18564*c**6*d**12*x**12 + 8568*c**5*d**13*x**13 + 3060*c**4*d **14*x**14 + 816*c**3*d**15*x**15 + 153*c**2*d**16*x**16 + 18*c*d**17*x**1 7 + d**18*x**18),x)*log(f)**2*b**2*c**14*d**2*x + 420*int(f**((a*c**2 + 2* a*c*d*x + a*d**2*x**2 + b)/(c**2 + 2*c*d*x + d**2*x**2))/(c**18 + 18*c*...