\(\int F^{a+\frac {b}{(c+d x)^2}} (c+d x)^3 \, dx\) [252]

Optimal result

Integrand size = 21, antiderivative size = 87 \[ \int F^{a+\frac {b}{(c+d x)^2}} (c+d x)^3 \, dx=\frac {F^{a+\frac {b}{(c+d x)^2}} (c+d x)^4}{4 d}+\frac {b F^{a+\frac {b}{(c+d x)^2}} (c+d x)^2 \log (F)}{4 d}-\frac {b^2 F^a \operatorname {ExpIntegralEi}\left (\frac {b \log (F)}{(c+d x)^2}\right ) \log ^2(F)}{4 d} \] Output:

1/4*F^(a+b/(d*x+c)^2)*(d*x+c)^4/d+1/4*b*F^(a+b/(d*x+c)^2)*(d*x+c)^2*ln(F)/ 
d-1/4*b^2*F^a*Ei(b*ln(F)/(d*x+c)^2)*ln(F)^2/d
 

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.82 \[ \int F^{a+\frac {b}{(c+d x)^2}} (c+d x)^3 \, dx=\frac {F^a \left (F^{\frac {b}{(c+d x)^2}} (c+d x)^4+b \log (F) \left (F^{\frac {b}{(c+d x)^2}} (c+d x)^2-b \operatorname {ExpIntegralEi}\left (\frac {b \log (F)}{(c+d x)^2}\right ) \log (F)\right )\right )}{4 d} \] Input:

Integrate[F^(a + b/(c + d*x)^2)*(c + d*x)^3,x]
 

Output:

(F^a*(F^(b/(c + d*x)^2)*(c + d*x)^4 + b*Log[F]*(F^(b/(c + d*x)^2)*(c + d*x 
)^2 - b*ExpIntegralEi[(b*Log[F])/(c + d*x)^2]*Log[F])))/(4*d)
 

Rubi [A] (verified)

Time = 0.55 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.01, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2643, 2643, 2639}

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.

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

\(\Big \downarrow \) 2643

\(\displaystyle \frac {1}{2} b \log (F) \int F^{a+\frac {b}{(c+d x)^2}} (c+d x)dx+\frac {(c+d x)^4 F^{a+\frac {b}{(c+d x)^2}}}{4 d}\)

\(\Big \downarrow \) 2643

\(\displaystyle \frac {1}{2} b \log (F) \left (b \log (F) \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{c+d x}dx+\frac {(c+d x)^2 F^{a+\frac {b}{(c+d x)^2}}}{2 d}\right )+\frac {(c+d x)^4 F^{a+\frac {b}{(c+d x)^2}}}{4 d}\)

\(\Big \downarrow \) 2639

\(\displaystyle \frac {1}{2} b \log (F) \left (\frac {(c+d x)^2 F^{a+\frac {b}{(c+d x)^2}}}{2 d}-\frac {b F^a \log (F) \operatorname {ExpIntegralEi}\left (\frac {b \log (F)}{(c+d x)^2}\right )}{2 d}\right )+\frac {(c+d x)^4 F^{a+\frac {b}{(c+d x)^2}}}{4 d}\)

Input:

Int[F^(a + b/(c + d*x)^2)*(c + d*x)^3,x]
 

Output:

(F^(a + b/(c + d*x)^2)*(c + d*x)^4)/(4*d) + (b*Log[F]*((F^(a + b/(c + d*x) 
^2)*(c + d*x)^2)/(2*d) - (b*F^a*ExpIntegralEi[(b*Log[F])/(c + d*x)^2]*Log[ 
F])/(2*d)))/2
 

Defintions of rubi rules used

rule 2639

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_ 
Symbol] :> Simp[F^a*(ExpIntegralEi[b*(c + d*x)^n*Log[F]]/(f*n)), x] /; Free 
Q[{F, a, b, c, d, e, f, n}, x] && EqQ[d*e - c*f, 0]
 

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 [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(207\) vs. \(2(81)=162\).

Time = 0.39 (sec) , antiderivative size = 208, normalized size of antiderivative = 2.39

method	result	size
risch	\(\frac {F^{a} d^{3} F^{\frac {b}{\left (d x +c \right )^{2}}} x^{4}}{4}+F^{a} d^{2} F^{\frac {b}{\left (d x +c \right )^{2}}} c \,x^{3}+\frac {3 F^{a} d \,F^{\frac {b}{\left (d x +c \right )^{2}}} c^{2} x^{2}}{2}+F^{a} F^{\frac {b}{\left (d x +c \right )^{2}}} c^{3} x +\frac {F^{a} F^{\frac {b}{\left (d x +c \right )^{2}}} c^{4}}{4 d}+\frac {F^{a} d b \ln \left (F \right ) F^{\frac {b}{\left (d x +c \right )^{2}}} x^{2}}{4}+\frac {F^{a} b \ln \left (F \right ) F^{\frac {b}{\left (d x +c \right )^{2}}} c x}{2}+\frac {F^{a} b \ln \left (F \right ) F^{\frac {b}{\left (d x +c \right )^{2}}} c^{2}}{4 d}+\frac {F^{a} b^{2} \ln \left (F \right )^{2} \operatorname {expIntegral}_{1}\left (-\frac {b \ln \left (F \right )}{\left (d x +c \right )^{2}}\right )}{4 d}\)	\(208\)

Input:

int(F^(a+b/(d*x+c)^2)*(d*x+c)^3,x,method=_RETURNVERBOSE)
 

Output:

1/4*F^a*d^3*F^(b/(d*x+c)^2)*x^4+F^a*d^2*F^(b/(d*x+c)^2)*c*x^3+3/2*F^a*d*F^ 
(b/(d*x+c)^2)*c^2*x^2+F^a*F^(b/(d*x+c)^2)*c^3*x+1/4*F^a/d*F^(b/(d*x+c)^2)* 
c^4+1/4*F^a*d*b*ln(F)*F^(b/(d*x+c)^2)*x^2+1/2*F^a*b*ln(F)*F^(b/(d*x+c)^2)* 
c*x+1/4*F^a/d*b*ln(F)*F^(b/(d*x+c)^2)*c^2+1/4*F^a/d*b^2*ln(F)^2*Ei(1,-b*ln 
(F)/(d*x+c)^2)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.67 \[ \int F^{a+\frac {b}{(c+d x)^2}} (c+d x)^3 \, dx=-\frac {F^{a} b^{2} {\rm Ei}\left (\frac {b \log \left (F\right )}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right ) \log \left (F\right )^{2} - {\left (d^{4} x^{4} + 4 \, c d^{3} x^{3} + 6 \, c^{2} d^{2} x^{2} + 4 \, c^{3} d x + c^{4} + {\left (b d^{2} x^{2} + 2 \, b c d x + b c^{2}\right )} \log \left (F\right )\right )} F^{\frac {a d^{2} x^{2} + 2 \, a c d x + a c^{2} + b}{d^{2} x^{2} + 2 \, c d x + c^{2}}}}{4 \, d} \] Input:

integrate(F^(a+b/(d*x+c)^2)*(d*x+c)^3,x, algorithm="fricas")
 

Output:

-1/4*(F^a*b^2*Ei(b*log(F)/(d^2*x^2 + 2*c*d*x + c^2))*log(F)^2 - (d^4*x^4 + 
 4*c*d^3*x^3 + 6*c^2*d^2*x^2 + 4*c^3*d*x + c^4 + (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)))/d
 

Sympy [F]

\[ \int F^{a+\frac {b}{(c+d x)^2}} (c+d x)^3 \, dx=\int F^{a + \frac {b}{\left (c + d x\right )^{2}}} \left (c + d x\right )^{3}\, dx \] Input:

integrate(F**(a+b/(d*x+c)**2)*(d*x+c)**3,x)
 

Output:

Integral(F**(a + b/(c + d*x)**2)*(c + d*x)**3, x)
 

Maxima [F]

\[ \int F^{a+\frac {b}{(c+d x)^2}} (c+d x)^3 \, dx=\int { {\left (d x + c\right )}^{3} F^{a + \frac {b}{{\left (d x + c\right )}^{2}}} \,d x } \] Input:

integrate(F^(a+b/(d*x+c)^2)*(d*x+c)^3,x, algorithm="maxima")
 

Output:

1/4*(F^a*d^3*x^4 + 4*F^a*c*d^2*x^3 + (6*F^a*c^2*d + F^a*b*d*log(F))*x^2 + 
2*(2*F^a*c^3 + F^a*b*c*log(F))*x)*F^(b/(d^2*x^2 + 2*c*d*x + c^2)) + integr 
ate(1/2*(F^a*b^2*d^2*x^2*log(F)^2 + 2*F^a*b^2*c*d*x*log(F)^2 - F^a*b*c^4*l 
og(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]

\[ \int F^{a+\frac {b}{(c+d x)^2}} (c+d x)^3 \, dx=\int { {\left (d x + c\right )}^{3} F^{a + \frac {b}{{\left (d x + c\right )}^{2}}} \,d x } \] Input:

integrate(F^(a+b/(d*x+c)^2)*(d*x+c)^3,x, algorithm="giac")
 

Output:

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

Mupad [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.87 \[ \int F^{a+\frac {b}{(c+d x)^2}} (c+d x)^3 \, dx=\frac {F^a\,b^2\,{\ln \left (F\right )}^2\,\left (\frac {\mathrm {expint}\left (-\frac {b\,\ln \left (F\right )}{{\left (c+d\,x\right )}^2}\right )}{2}+F^{\frac {b}{{\left (c+d\,x\right )}^2}}\,\left (\frac {{\left (c+d\,x\right )}^2}{2\,b\,\ln \left (F\right )}+\frac {{\left (c+d\,x\right )}^4}{2\,b^2\,{\ln \left (F\right )}^2}\right )\right )}{2\,d} \] Input:

int(F^(a + b/(c + d*x)^2)*(c + d*x)^3,x)
 

Output:

(F^a*b^2*log(F)^2*(expint(-(b*log(F))/(c + d*x)^2)/2 + F^(b/(c + d*x)^2)*( 
(c + d*x)^2/(2*b*log(F)) + (c + d*x)^4/(2*b^2*log(F)^2))))/(2*d)
 

Reduce [F]

\[ \int F^{a+\frac {b}{(c+d x)^2}} (c+d x)^3 \, dx=\text {too large to display} \] Input:

int(F^(a+b/(d*x+c)^2)*(d*x+c)^3,x)
 

Output:

(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 
))*log(f)**3*b**3*c*d*x + 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)**2*b**2*c**4 - 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))*log(f)**2*b**2*c**3*d*x 
 - 40*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**2*x**2 - 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))*log(f)**2*b**2*c*d**3*x**3 - 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))*log( 
f)*b*c**6 - 64*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*d*x - 52*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**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* 
*3*d**3*x**3 + 31*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**4*x**4 + 18*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**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))*log(f 
)*b*d**6*x**6 + 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**8 + 76*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*d*x + 162*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**2*x**2 + 220*f**((a*c*...