Integrand size = 21, antiderivative size = 46 \[ \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^2} \, dx=-\frac {F^a \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {b} \sqrt {\log (F)}}{c+d x}\right )}{2 \sqrt {b} d \sqrt {\log (F)}} \] Output:
-1/2*F^a*Pi^(1/2)*erfi(b^(1/2)*ln(F)^(1/2)/(d*x+c))/b^(1/2)/d/ln(F)^(1/2)
Time = 0.02 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.00 \[ \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^2} \, dx=-\frac {F^a \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {b} \sqrt {\log (F)}}{c+d x}\right )}{2 \sqrt {b} d \sqrt {\log (F)}} \] Input:
Integrate[F^(a + b/(c + d*x)^2)/(c + d*x)^2,x]
Output:
-1/2*(F^a*Sqrt[Pi]*Erfi[(Sqrt[b]*Sqrt[Log[F]])/(c + d*x)])/(Sqrt[b]*d*Sqrt [Log[F]])
Time = 0.34 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2640, 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.
\(\displaystyle \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^2} \, dx\) |
\(\Big \downarrow \) 2640 |
\(\displaystyle -\frac {\int F^{a+\frac {b}{(c+d x)^2}}d\frac {1}{c+d x}}{d}\) |
\(\Big \downarrow \) 2633 |
\(\displaystyle -\frac {\sqrt {\pi } F^a \text {erfi}\left (\frac {\sqrt {b} \sqrt {\log (F)}}{c+d x}\right )}{2 \sqrt {b} d \sqrt {\log (F)}}\) |
Input:
Int[F^(a + b/(c + d*x)^2)/(c + d*x)^2,x]
Output:
-1/2*(F^a*Sqrt[Pi]*Erfi[(Sqrt[b]*Sqrt[Log[F]])/(c + d*x)])/(Sqrt[b]*d*Sqrt [Log[F]])
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]
Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((c_.) + (d_.)*(x_))^(m_ .), x_Symbol] :> Simp[1/(d*(m + 1)) Subst[Int[F^(a + b*x^2), x], x, (c + d*x)^(m + 1)], x] /; FreeQ[{F, a, b, c, d, m, n}, x] && EqQ[n, 2*(m + 1)]
Time = 0.26 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.76
method | result | size |
risch | \(-\frac {F^{a} \sqrt {\pi }\, \operatorname {erf}\left (\frac {\sqrt {-b \ln \left (F \right )}}{d x +c}\right )}{2 d \sqrt {-b \ln \left (F \right )}}\) | \(35\) |
Input:
int(F^(a+b/(d*x+c)^2)/(d*x+c)^2,x,method=_RETURNVERBOSE)
Output:
-1/2/d*F^a*Pi^(1/2)/(-b*ln(F))^(1/2)*erf((-b*ln(F))^(1/2)/(d*x+c))
Time = 0.08 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.98 \[ \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^2} \, dx=\frac {\sqrt {\pi } F^{a} \sqrt {-\frac {b \log \left (F\right )}{d^{2}}} \operatorname {erf}\left (\frac {d \sqrt {-\frac {b \log \left (F\right )}{d^{2}}}}{d x + c}\right )}{2 \, b \log \left (F\right )} \] Input:
integrate(F^(a+b/(d*x+c)^2)/(d*x+c)^2,x, algorithm="fricas")
Output:
1/2*sqrt(pi)*F^a*sqrt(-b*log(F)/d^2)*erf(d*sqrt(-b*log(F)/d^2)/(d*x + c))/ (b*log(F))
\[ \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^2} \, dx=\int \frac {F^{a + \frac {b}{\left (c + d x\right )^{2}}}}{\left (c + d x\right )^{2}}\, dx \] Input:
integrate(F**(a+b/(d*x+c)**2)/(d*x+c)**2,x)
Output:
Integral(F**(a + b/(c + d*x)**2)/(c + d*x)**2, x)
\[ \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^2} \, dx=\int { \frac {F^{a + \frac {b}{{\left (d x + c\right )}^{2}}}}{{\left (d x + c\right )}^{2}} \,d x } \] Input:
integrate(F^(a+b/(d*x+c)^2)/(d*x+c)^2,x, algorithm="maxima")
Output:
integrate(F^(a + b/(d*x + c)^2)/(d*x + c)^2, x)
\[ \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^2} \, dx=\int { \frac {F^{a + \frac {b}{{\left (d x + c\right )}^{2}}}}{{\left (d x + c\right )}^{2}} \,d x } \] Input:
integrate(F^(a+b/(d*x+c)^2)/(d*x+c)^2,x, algorithm="giac")
Output:
integrate(F^(a + b/(d*x + c)^2)/(d*x + c)^2, x)
Time = 0.21 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.76 \[ \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^2} \, dx=-\frac {F^a\,\sqrt {\pi }\,\mathrm {erfi}\left (\frac {b\,\ln \left (F\right )}{\sqrt {b\,\ln \left (F\right )}\,\left (c+d\,x\right )}\right )}{2\,d\,\sqrt {b\,\ln \left (F\right )}} \] Input:
int(F^(a + b/(c + d*x)^2)/(c + d*x)^2,x)
Output:
-(F^a*pi^(1/2)*erfi((b*log(F))/((b*log(F))^(1/2)*(c + d*x))))/(2*d*(b*log( F))^(1/2))
\[ \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^2} \, dx =\text {Too large to display} \] Input:
int(F^(a+b/(d*x+c)^2)/(d*x+c)^2,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)**2*b**2*d*x + 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*c**3 - 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*c**2*d*x - 5*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*c*d**2* x**2 - 3*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*d**3*x**3 - 4*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**5 - 16*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**4*d*x - 24*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**3*d**2*x**2 - 16*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*d**3*x**3 - 4*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**4*x**4 + 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))*x)/(c**7 + 7*c**6*d*x + 21*c**5*d**2*x* *2 + 35*c**4*d**3*x**3 + 35*c**3*d**4*x**4 + 21*c**2*d**5*x**5 + 7*c*d**6* x**6 + d**7*x**7),x)*log(f)**3*b**3*c**4*d**2 + 16*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))*x)/(c**7 + 7*c**6*d *x + 21*c**5*d**2*x**2 + 35*c**4*d**3*x**3 + 35*c**3*d**4*x**4 + 21*c**2*d **5*x**5 + 7*c*d**6*x**6 + d**7*x**7),x)*log(f)**3*b**3*c**3*d**3*x + 24*i nt((f**((a*c**2 + 2*a*c*d*x + a*d**2*x**2 + b)/(c**2 + 2*c*d*x + d**2*x...