\(\int \frac {F^{a+b (c+d x)^2}}{(c+d x)^4} \, dx\) [209]

Optimal result

Integrand size = 21, antiderivative size = 102 \[ \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^4} \, dx=-\frac {F^{a+b (c+d x)^2}}{3 d (c+d x)^3}-\frac {2 b F^{a+b (c+d x)^2} \log (F)}{3 d (c+d x)}+\frac {2 b^{3/2} F^a \sqrt {\pi } \text {erfi}\left (\sqrt {b} (c+d x) \sqrt {\log (F)}\right ) \log ^{\frac {3}{2}}(F)}{3 d} \] Output:

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

Mathematica [A] (verified)

Time = 0.45 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.79 \[ \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^4} \, dx=\frac {F^a \left (2 b^{3/2} \sqrt {\pi } \text {erfi}\left (\sqrt {b} (c+d x) \sqrt {\log (F)}\right ) \log ^{\frac {3}{2}}(F)-\frac {F^{b (c+d x)^2} \left (1+2 b (c+d x)^2 \log (F)\right )}{(c+d x)^3}\right )}{3 d} \] Input:

Integrate[F^(a + b*(c + d*x)^2)/(c + d*x)^4,x]
 

Output:

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

Rubi [A] (verified)

Time = 0.55 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2643, 2643, 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.

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

\(\Big \downarrow \) 2643

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

\(\Big \downarrow \) 2643

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

\(\Big \downarrow \) 2633

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

Input:

Int[F^(a + b*(c + d*x)^2)/(c + d*x)^4,x]
 

Output:

-1/3*F^(a + b*(c + d*x)^2)/(d*(c + d*x)^3) + (2*b*(-(F^(a + b*(c + d*x)^2) 
/(d*(c + d*x))) + (Sqrt[b]*F^a*Sqrt[Pi]*Erfi[Sqrt[b]*(c + d*x)*Sqrt[Log[F] 
]]*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 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 = 0.16 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.94

method	result	size
risch	\(-\frac {F^{b \left (d x +c \right )^{2}} F^{a}}{3 d \left (d x +c \right )^{3}}-\frac {2 b \ln \left (F \right ) F^{b \left (d x +c \right )^{2}} F^{a}}{3 d \left (d x +c \right )}+\frac {2 b^{2} \ln \left (F \right )^{2} \sqrt {\pi }\, F^{a} \operatorname {erf}\left (\sqrt {-b \ln \left (F \right )}\, \left (d x +c \right )\right )}{3 d \sqrt {-b \ln \left (F \right )}}\)	\(96\)

Input:

int(F^(a+b*(d*x+c)^2)/(d*x+c)^4,x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.60 \[ \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^4} \, dx=-\frac {2 \, \sqrt {\pi } {\left (b d^{3} x^{3} + 3 \, b c d^{2} x^{2} + 3 \, b c^{2} d x + b c^{3}\right )} \sqrt {-b d^{2} \log \left (F\right )} F^{a} \operatorname {erf}\left (\frac {\sqrt {-b d^{2} \log \left (F\right )} {\left (d x + c\right )}}{d}\right ) \log \left (F\right ) + {\left (2 \, {\left (b d^{3} x^{2} + 2 \, b c d^{2} x + b c^{2} d\right )} \log \left (F\right ) + d\right )} F^{b d^{2} x^{2} + 2 \, b c d x + b c^{2} + a}}{3 \, {\left (d^{5} x^{3} + 3 \, c d^{4} x^{2} + 3 \, c^{2} d^{3} x + c^{3} d^{2}\right )}} \] Input:

integrate(F^(a+b*(d*x+c)^2)/(d*x+c)^4,x, algorithm="fricas")
 

Output:

-1/3*(2*sqrt(pi)*(b*d^3*x^3 + 3*b*c*d^2*x^2 + 3*b*c^2*d*x + b*c^3)*sqrt(-b 
*d^2*log(F))*F^a*erf(sqrt(-b*d^2*log(F))*(d*x + c)/d)*log(F) + (2*(b*d^3*x 
^2 + 2*b*c*d^2*x + b*c^2*d)*log(F) + d)*F^(b*d^2*x^2 + 2*b*c*d*x + b*c^2 + 
 a))/(d^5*x^3 + 3*c*d^4*x^2 + 3*c^2*d^3*x + c^3*d^2)
 

Sympy [F]

\[ \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^4} \, dx=\int \frac {F^{a + b \left (c + d x\right )^{2}}}{\left (c + d x\right )^{4}}\, dx \] Input:

integrate(F**(a+b*(d*x+c)**2)/(d*x+c)**4,x)
 

Output:

Integral(F**(a + b*(c + d*x)**2)/(c + d*x)**4, x)
 

Maxima [F]

\[ \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^4} \, dx=\int { \frac {F^{{\left (d x + c\right )}^{2} b + a}}{{\left (d x + c\right )}^{4}} \,d x } \] Input:

integrate(F^(a+b*(d*x+c)^2)/(d*x+c)^4,x, algorithm="maxima")
 

Output:

integrate(F^((d*x + c)^2*b + a)/(d*x + c)^4, x)
 

Giac [F]

\[ \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^4} \, dx=\int { \frac {F^{{\left (d x + c\right )}^{2} b + a}}{{\left (d x + c\right )}^{4}} \,d x } \] Input:

integrate(F^(a+b*(d*x+c)^2)/(d*x+c)^4,x, algorithm="giac")
 

Output:

integrate(F^((d*x + c)^2*b + a)/(d*x + c)^4, x)
                                                                                    
                                                                                    
 

Mupad [B] (verification not implemented)

Time = 0.86 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.97 \[ \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^4} \, dx=\frac {2\,F^a\,b^2\,\sqrt {\pi }\,\mathrm {erfi}\left (\frac {b\,x\,\ln \left (F\right )\,d^2+b\,c\,\ln \left (F\right )\,d}{\sqrt {b\,d^2\,\ln \left (F\right )}}\right )\,{\ln \left (F\right )}^2}{3\,\sqrt {b\,d^2\,\ln \left (F\right )}}-\frac {F^{b\,d^2\,x^2}\,F^a\,F^{b\,c^2}\,F^{2\,b\,c\,d\,x}\,\left (\frac {1}{3\,d}+\frac {2\,b\,c^2\,\ln \left (F\right )}{3\,d}\right )+\frac {4\,F^{b\,d^2\,x^2}\,F^a\,F^{b\,c^2}\,F^{2\,b\,c\,d\,x}\,b\,c\,x\,\ln \left (F\right )}{3}+\frac {2\,F^{b\,d^2\,x^2}\,F^a\,F^{b\,c^2}\,F^{2\,b\,c\,d\,x}\,b\,d\,x^2\,\ln \left (F\right )}{3}}{c^3+3\,c^2\,d\,x+3\,c\,d^2\,x^2+d^3\,x^3} \] Input:

int(F^(a + b*(c + d*x)^2)/(c + d*x)^4,x)
 

Output:

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

Reduce [F]

\[ \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^4} \, dx=\text {too large to display} \] Input:

int(F^(a+b*(d*x+c)^2)/(d*x+c)^4,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**6 + 8*log(f)*b*c**5*d*x + 12*log(f)*b*c**4*d** 
2*x**2 + 8*log(f)*b*c**3*d**3*x**3 + 2*log(f)*b*c**2*d**4*x**4 - 3*c**4 - 
12*c**3*d*x - 18*c**2*d**2*x**2 - 12*c*d**3*x**3 - 3*d**4*x**4),x)*log(f)* 
*2*b**2*c**7*d + 24*int(f**(2*b*c*d*x + b*d**2*x**2)/(2*log(f)*b*c**6 + 8* 
log(f)*b*c**5*d*x + 12*log(f)*b*c**4*d**2*x**2 + 8*log(f)*b*c**3*d**3*x**3 
 + 2*log(f)*b*c**2*d**4*x**4 - 3*c**4 - 12*c**3*d*x - 18*c**2*d**2*x**2 - 
12*c*d**3*x**3 - 3*d**4*x**4),x)*log(f)**2*b**2*c**6*d**2*x + 24*int(f**(2 
*b*c*d*x + b*d**2*x**2)/(2*log(f)*b*c**6 + 8*log(f)*b*c**5*d*x + 12*log(f) 
*b*c**4*d**2*x**2 + 8*log(f)*b*c**3*d**3*x**3 + 2*log(f)*b*c**2*d**4*x**4 
- 3*c**4 - 12*c**3*d*x - 18*c**2*d**2*x**2 - 12*c*d**3*x**3 - 3*d**4*x**4) 
,x)*log(f)**2*b**2*c**5*d**3*x**2 + 8*int(f**(2*b*c*d*x + b*d**2*x**2)/(2* 
log(f)*b*c**6 + 8*log(f)*b*c**5*d*x + 12*log(f)*b*c**4*d**2*x**2 + 8*log(f 
)*b*c**3*d**3*x**3 + 2*log(f)*b*c**2*d**4*x**4 - 3*c**4 - 12*c**3*d*x - 18 
*c**2*d**2*x**2 - 12*c*d**3*x**3 - 3*d**4*x**4),x)*log(f)**2*b**2*c**4*d** 
4*x**3 - 24*int(f**(2*b*c*d*x + b*d**2*x**2)/(2*log(f)*b*c**6 + 8*log(f)*b 
*c**5*d*x + 12*log(f)*b*c**4*d**2*x**2 + 8*log(f)*b*c**3*d**3*x**3 + 2*log 
(f)*b*c**2*d**4*x**4 - 3*c**4 - 12*c**3*d*x - 18*c**2*d**2*x**2 - 12*c*d** 
3*x**3 - 3*d**4*x**4),x)*log(f)*b*c**5*d - 72*int(f**(2*b*c*d*x + b*d**2*x 
**2)/(2*log(f)*b*c**6 + 8*log(f)*b*c**5*d*x + 12*log(f)*b*c**4*d**2*x**...