\(\int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^6} \, dx\) [269]

Optimal result

Integrand size = 21, antiderivative size = 115 \[ \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^6} \, dx=-\frac {3 F^a \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {b} \sqrt {\log (F)}}{c+d x}\right )}{8 b^{5/2} d \log ^{\frac {5}{2}}(F)}+\frac {3 F^{a+\frac {b}{(c+d x)^2}}}{4 b^2 d (c+d x) \log ^2(F)}-\frac {F^{a+\frac {b}{(c+d x)^2}}}{2 b d (c+d x)^3 \log (F)} \] Output:

-3/8*F^a*Pi^(1/2)*erfi(b^(1/2)*ln(F)^(1/2)/(d*x+c))/b^(5/2)/d/ln(F)^(5/2)+ 
3/4*F^(a+b/(d*x+c)^2)/b^2/d/(d*x+c)/ln(F)^2-1/2*F^(a+b/(d*x+c)^2)/b/d/(d*x 
+c)^3/ln(F)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.83 \[ \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^6} \, dx=\frac {F^a \left (-3 \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {b} \sqrt {\log (F)}}{c+d x}\right )-\frac {2 \sqrt {b} F^{\frac {b}{(c+d x)^2}} \sqrt {\log (F)} \left (-3 (c+d x)^2+2 b \log (F)\right )}{(c+d x)^3}\right )}{8 b^{5/2} d \log ^{\frac {5}{2}}(F)} \] Input:

Integrate[F^(a + b/(c + d*x)^2)/(c + d*x)^6,x]
 

Output:

(F^a*(-3*Sqrt[Pi]*Erfi[(Sqrt[b]*Sqrt[Log[F]])/(c + d*x)] - (2*Sqrt[b]*F^(b 
/(c + d*x)^2)*Sqrt[Log[F]]*(-3*(c + d*x)^2 + 2*b*Log[F]))/(c + d*x)^3))/(8 
*b^(5/2)*d*Log[F]^(5/2))
 

Rubi [A] (verified)

Time = 0.65 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.10, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {2641, 2641, 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.

\(\displaystyle \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^6} \, dx\)

\(\Big \downarrow \) 2641

\(\displaystyle -\frac {3 \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^4}dx}{2 b \log (F)}-\frac {F^{a+\frac {b}{(c+d x)^2}}}{2 b d \log (F) (c+d x)^3}\)

\(\Big \downarrow \) 2641

\(\displaystyle -\frac {3 \left (-\frac {\int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^2}dx}{2 b \log (F)}-\frac {F^{a+\frac {b}{(c+d x)^2}}}{2 b d \log (F) (c+d x)}\right )}{2 b \log (F)}-\frac {F^{a+\frac {b}{(c+d x)^2}}}{2 b d \log (F) (c+d x)^3}\)

\(\Big \downarrow \) 2640

\(\displaystyle -\frac {3 \left (\frac {\int F^{a+\frac {b}{(c+d x)^2}}d\frac {1}{c+d x}}{2 b d \log (F)}-\frac {F^{a+\frac {b}{(c+d x)^2}}}{2 b d \log (F) (c+d x)}\right )}{2 b \log (F)}-\frac {F^{a+\frac {b}{(c+d x)^2}}}{2 b d \log (F) (c+d x)^3}\)

\(\Big \downarrow \) 2633

\(\displaystyle -\frac {3 \left (\frac {\sqrt {\pi } F^a \text {erfi}\left (\frac {\sqrt {b} \sqrt {\log (F)}}{c+d x}\right )}{4 b^{3/2} d \log ^{\frac {3}{2}}(F)}-\frac {F^{a+\frac {b}{(c+d x)^2}}}{2 b d \log (F) (c+d x)}\right )}{2 b \log (F)}-\frac {F^{a+\frac {b}{(c+d x)^2}}}{2 b d \log (F) (c+d x)^3}\)

Input:

Int[F^(a + b/(c + d*x)^2)/(c + d*x)^6,x]
 

Output:

-1/2*F^(a + b/(c + d*x)^2)/(b*d*(c + d*x)^3*Log[F]) - (3*((F^a*Sqrt[Pi]*Er 
fi[(Sqrt[b]*Sqrt[Log[F]])/(c + d*x)])/(4*b^(3/2)*d*Log[F]^(3/2)) - F^(a + 
b/(c + d*x)^2)/(2*b*d*(c + d*x)*Log[F])))/(2*b*Log[F])
 

Defintions of rubi rules used

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 2640

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)]
 

rule 2641

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((c_.) + (d_.)*(x_))^(m_ 
.), x_Symbol] :> Simp[(c + d*x)^(m - n + 1)*(F^(a + b*(c + d*x)^n)/(b*d*n*L 
og[F])), x] - Simp[(m - n + 1)/(b*n*Log[F])   Int[(c + d*x)^(m - n)*F^(a + 
b*(c + d*x)^n), x], x] /; FreeQ[{F, a, b, c, d}, x] && IntegerQ[2*((m + 1)/ 
n)] && LtQ[0, (m + 1)/n, 5] && IntegerQ[n] && (LtQ[0, n, m + 1] || LtQ[m, n 
, 0])
 

Maple [A] (verified)

Time = 1.14 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.95

method	result	size
risch	\(-\frac {F^{a} F^{\frac {b}{\left (d x +c \right )^{2}}}}{2 d \left (d x +c \right )^{3} b \ln \left (F \right )}+\frac {3 F^{a} F^{\frac {b}{\left (d x +c \right )^{2}}}}{4 d \,b^{2} \ln \left (F \right )^{2} \left (d x +c \right )}-\frac {3 F^{a} \sqrt {\pi }\, \operatorname {erf}\left (\frac {\sqrt {-b \ln \left (F \right )}}{d x +c}\right )}{8 d \,b^{2} \ln \left (F \right )^{2} \sqrt {-b \ln \left (F \right )}}\)	\(109\)

Input:

int(F^(a+b/(d*x+c)^2)/(d*x+c)^6,x,method=_RETURNVERBOSE)
 

Output:

-1/2*F^a/d*F^(b/(d*x+c)^2)/(d*x+c)^3/b/ln(F)+3/4*F^a/d/b^2/ln(F)^2*F^(b/(d 
*x+c)^2)/(d*x+c)-3/8*F^a/d/b^2/ln(F)^2*Pi^(1/2)/(-b*ln(F))^(1/2)*erf((-b*l 
n(F))^(1/2)/(d*x+c))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 199 vs. \(2 (99) = 198\).

Time = 0.09 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.73 \[ \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^6} \, dx=\frac {3 \, \sqrt {\pi } {\left (d^{4} x^{3} + 3 \, c d^{3} x^{2} + 3 \, c^{2} d^{2} x + c^{3} d\right )} F^{a} \sqrt {-\frac {b \log \left (F\right )}{d^{2}}} \operatorname {erf}\left (\frac {d \sqrt {-\frac {b \log \left (F\right )}{d^{2}}}}{d x + c}\right ) - 2 \, {\left (2 \, b^{2} \log \left (F\right )^{2} - 3 \, {\left (b d^{2} x^{2} + 2 \, b c d x + b c^{2}\right )} \log \left (F\right )\right )} F^{\frac {a d^{2} x^{2} + 2 \, a c d x + a c^{2} + b}{d^{2} x^{2} + 2 \, c d x + c^{2}}}}{8 \, {\left (b^{3} d^{4} x^{3} + 3 \, b^{3} c d^{3} x^{2} + 3 \, b^{3} c^{2} d^{2} x + b^{3} c^{3} d\right )} \log \left (F\right )^{3}} \] Input:

integrate(F^(a+b/(d*x+c)^2)/(d*x+c)^6,x, algorithm="fricas")
 

Output:

1/8*(3*sqrt(pi)*(d^4*x^3 + 3*c*d^3*x^2 + 3*c^2*d^2*x + c^3*d)*F^a*sqrt(-b* 
log(F)/d^2)*erf(d*sqrt(-b*log(F)/d^2)/(d*x + c)) - 2*(2*b^2*log(F)^2 - 3*( 
b*d^2*x^2 + 2*b*c*d*x + b*c^2)*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^3*d^4*x^3 + 3*b^3*c*d^3*x^2 + 3*b^3*c^ 
2*d^2*x + b^3*c^3*d)*log(F)^3)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^6} \, dx=\text {Timed out} \] Input:

integrate(F**(a+b/(d*x+c)**2)/(d*x+c)**6,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^6} \, dx=\int { \frac {F^{a + \frac {b}{{\left (d x + c\right )}^{2}}}}{{\left (d x + c\right )}^{6}} \,d x } \] Input:

integrate(F^(a+b/(d*x+c)^2)/(d*x+c)^6,x, algorithm="maxima")
 

Output:

integrate(F^(a + b/(d*x + c)^2)/(d*x + c)^6, x)
 

Giac [F]

\[ \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^6} \, dx=\int { \frac {F^{a + \frac {b}{{\left (d x + c\right )}^{2}}}}{{\left (d x + c\right )}^{6}} \,d x } \] Input:

integrate(F^(a+b/(d*x+c)^2)/(d*x+c)^6,x, algorithm="giac")
 

Output:

integrate(F^(a + b/(d*x + c)^2)/(d*x + c)^6, x)
 

Mupad [B] (verification not implemented)

Time = 0.42 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.91 \[ \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^6} \, dx=-\frac {F^a\,F^{\frac {b}{{\left (c+d\,x\right )}^2}}}{2\,b\,d\,\ln \left (F\right )\,{\left (c+d\,x\right )}^3}-\frac {F^a\,\left (3\,\sqrt {\pi }\,\mathrm {erfi}\left (\frac {b\,\ln \left (F\right )}{\sqrt {b\,\ln \left (F\right )}\,\left (c+d\,x\right )}\right )-\frac {6\,F^{\frac {b}{{\left (c+d\,x\right )}^2}}\,\sqrt {b\,\ln \left (F\right )}}{c+d\,x}\right )}{8\,b^2\,d\,{\ln \left (F\right )}^2\,\sqrt {b\,\ln \left (F\right )}} \] Input:

int(F^(a + b/(c + d*x)^2)/(c + d*x)^6,x)
 

Output:

- (F^a*F^(b/(c + d*x)^2))/(2*b*d*log(F)*(c + d*x)^3) - (F^a*(3*pi^(1/2)*er 
fi((b*log(F))/((b*log(F))^(1/2)*(c + d*x))) - (6*F^(b/(c + d*x)^2)*(b*log( 
F))^(1/2))/(c + d*x)))/(8*b^2*d*log(F)^2*(b*log(F))^(1/2))
 

Reduce [F]

\[ \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^6} \, dx =\text {Too large to display} \] Input:

int(F^(a+b/(d*x+c)^2)/(d*x+c)^6,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 - 7*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 - 14*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 - 7*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**10 + 10*c**9*d*x + 45 
*c**8*d**2*x**2 + 120*c**7*d**3*x**3 + 210*c**6*d**4*x**4 + 252*c**5*d**5* 
x**5 + 210*c**4*d**6*x**6 + 120*c**3*d**7*x**7 + 45*c**2*d**8*x**8 + 10*c* 
d**9*x**9 + d**10*x**10),x)*log(f)**2*b**2*c**7*d + 28*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**10 + 10*c**9 
*d*x + 45*c**8*d**2*x**2 + 120*c**7*d**3*x**3 + 210*c**6*d**4*x**4 + 252*c 
**5*d**5*x**5 + 210*c**4*d**6*x**6 + 120*c**3*d**7*x**7 + 45*c**2*d**8*x** 
8 + 10*c*d**9*x**9 + d**10*x**10),x)*log(f)**2*b**2*c**6*d**2*x + 84*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 
**10 + 10*c**9*d*x + 45*c**8*d**2*x**2 + 120*c**7*d**3*x**3 + 210*c**6*d** 
4*x**4 + 252*c**5*d**5*x**5 + 210*c**4*d**6*x**6 + 120*c**3*d**7*x**7 + 45 
*c**2*d**8*x**8 + 10*c*d**9*x**9 + d**10*x**10),x)*log(f)**2*b**2*c**5*d** 
3*x**2 + 140*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**10 + 10*c**9*d*x + 45*c**8*d**2*x**2 + 120*c**7*d**3*x 
**3 + 210*c**6*d**4*x**4 + 252*c**5*d**5*x**5 + 210*c**4*d**6*x**6 + 12...