\(\int F^{a+\frac {b}{(c+d x)^2}} (c+d x)^2 \, dx\) [265]

Optimal result

Integrand size = 21, antiderivative size = 102 \[ \int F^{a+\frac {b}{(c+d x)^2}} (c+d x)^2 \, dx=\frac {F^{a+\frac {b}{(c+d x)^2}} (c+d x)^3}{3 d}+\frac {2 b F^{a+\frac {b}{(c+d x)^2}} (c+d x) \log (F)}{3 d}-\frac {2 b^{3/2} F^a \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {b} \sqrt {\log (F)}}{c+d x}\right ) \log ^{\frac {3}{2}}(F)}{3 d} \] Output:

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

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.77 \[ \int F^{a+\frac {b}{(c+d x)^2}} (c+d x)^2 \, dx=\frac {F^a \left (-2 b^{3/2} \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {b} \sqrt {\log (F)}}{c+d x}\right ) \log ^{\frac {3}{2}}(F)+F^{\frac {b}{(c+d x)^2}} (c+d x) \left ((c+d x)^2+2 b \log (F)\right )\right )}{3 d} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.57 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {2643, 2635, 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.

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

\(\Big \downarrow \) 2643

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

\(\Big \downarrow \) 2635

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

\(\Big \downarrow \) 2640

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

\(\Big \downarrow \) 2633

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

Input:

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

Output:

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

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 2635

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

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 2643

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((c_.) + (d_.)*(x_))^(m_ 
.), x_Symbol] :> Simp[(c + d*x)^(m + 1)*(F^(a + b*(c + d*x)^n)/(d*(m + 1))) 
, x] - Simp[b*n*(Log[F]/(m + 1))   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[ 
-4, (m + 1)/n, 5] && IntegerQ[n] && ((GtQ[n, 0] && LtQ[m, -1]) || (GtQ[-n, 
0] && LeQ[-n, m + 1]))
 

Maple [A] (verified)

Time = 0.29 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.66

method	result	size
risch	\(\frac {F^{a} d^{2} F^{\frac {b}{\left (d x +c \right )^{2}}} x^{3}}{3}+F^{a} d \,F^{\frac {b}{\left (d x +c \right )^{2}}} c \,x^{2}+F^{a} F^{\frac {b}{\left (d x +c \right )^{2}}} c^{2} x +\frac {F^{a} F^{\frac {b}{\left (d x +c \right )^{2}}} c^{3}}{3 d}+\frac {2 F^{a} b \ln \left (F \right ) F^{\frac {b}{\left (d x +c \right )^{2}}} x}{3}+\frac {2 F^{a} b \ln \left (F \right ) F^{\frac {b}{\left (d x +c \right )^{2}}} c}{3 d}-\frac {2 F^{a} b^{2} \ln \left (F \right )^{2} \sqrt {\pi }\, \operatorname {erf}\left (\frac {\sqrt {-b \ln \left (F \right )}}{d x +c}\right )}{3 d \sqrt {-b \ln \left (F \right )}}\)	\(169\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.27 \[ \int F^{a+\frac {b}{(c+d x)^2}} (c+d x)^2 \, dx=\frac {2 \, \sqrt {\pi } F^{a} b d \sqrt {-\frac {b \log \left (F\right )}{d^{2}}} \operatorname {erf}\left (\frac {d \sqrt {-\frac {b \log \left (F\right )}{d^{2}}}}{d x + c}\right ) \log \left (F\right ) + {\left (d^{3} x^{3} + 3 \, c d^{2} x^{2} + 3 \, c^{2} d x + c^{3} + 2 \, {\left (b d x + b c\right )} \log \left (F\right )\right )} F^{\frac {a d^{2} x^{2} + 2 \, a c d x + a c^{2} + b}{d^{2} x^{2} + 2 \, c d x + c^{2}}}}{3 \, d} \] Input:

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

Output:

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

Sympy [F]

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

Maxima [F]

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

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

Output:

1/3*(F^a*d^2*x^3 + 3*F^a*c*d*x^2 + (3*F^a*c^2 + 2*F^a*b*log(F))*x)*F^(b/(d 
^2*x^2 + 2*c*d*x + c^2)) + integrate(2/3*(2*F^a*b^2*d*x*log(F)^2 - F^a*b*c 
^3*log(F))*F^(b/(d^2*x^2 + 2*c*d*x + c^2))/(d^3*x^3 + 3*c*d^2*x^2 + 3*c^2* 
d*x + c^3), x)
 

Giac [F]

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

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

Output:

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

Mupad [B] (verification not implemented)

Time = 0.50 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.95 \[ \int F^{a+\frac {b}{(c+d x)^2}} (c+d x)^2 \, dx=\frac {\left (\frac {F^a\,F^{\frac {b}{{\left (c+d\,x\right )}^2}}}{3}+\frac {2\,F^a\,F^{\frac {b}{{\left (c+d\,x\right )}^2}}\,b\,\ln \left (F\right )}{3\,{\left (c+d\,x\right )}^2}\right )\,{\left (c+d\,x\right )}^3}{d}-\frac {2\,F^a\,b^2\,\sqrt {\pi }\,\mathrm {erfi}\left (\frac {b\,\ln \left (F\right )}{\sqrt {b\,\ln \left (F\right )}\,\left (c+d\,x\right )}\right )\,{\ln \left (F\right )}^2}{3\,d\,\sqrt {b\,\ln \left (F\right )}} \] Input:

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

Output:

(((F^a*F^(b/(c + d*x)^2))/3 + (2*F^a*F^(b/(c + d*x)^2)*b*log(F))/(3*(c + d 
*x)^2))*(c + d*x)^3)/d - (2*F^a*b^2*pi^(1/2)*erfi((b*log(F))/((b*log(F))^( 
1/2)*(c + d*x)))*log(F)^2)/(3*d*(b*log(F))^(1/2))
 

Reduce [F]

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

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

Output:

(8*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)**3*b**3*d*x + 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))*log(f)**2*b**2*c**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))*log(f)**2*b**2*c**2*d*x - 2 
0*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*c*d**2*x**2 - 12*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**3*x**3 - 10*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**5 
 - 34*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**4*d*x - 36*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*d**2*x**2 - 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))*log(f)*b*c**2*d**3*x 
**3 + 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))*log(f)*b*c*d**4*x**4 + 6*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**5*x**5 + 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))*c**7 + 21*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**6*d*x + 
63*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*d**2*x**2 + 105*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**3*x**3 + 105*f**((a*c**2 + 2*a*c*d*x + a*...