\(\int \frac {F^{a+b (c+d x)^2}}{(c+d x)^7} \, dx\) [198]

Optimal result

Integrand size = 21, antiderivative size = 121 \[ \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^7} \, dx=-\frac {F^{a+b (c+d x)^2}}{6 d (c+d x)^6}-\frac {b F^{a+b (c+d x)^2} \log (F)}{12 d (c+d x)^4}-\frac {b^2 F^{a+b (c+d x)^2} \log ^2(F)}{12 d (c+d x)^2}+\frac {b^3 F^a \operatorname {ExpIntegralEi}\left (b (c+d x)^2 \log (F)\right ) \log ^3(F)}{12 d} \] Output:

-1/6*F^(a+b*(d*x+c)^2)/d/(d*x+c)^6-1/12*b*F^(a+b*(d*x+c)^2)*ln(F)/d/(d*x+c 
)^4-1/12*b^2*F^(a+b*(d*x+c)^2)*ln(F)^2/d/(d*x+c)^2+1/12*b^3*F^a*Ei(b*(d*x+ 
c)^2*ln(F))*ln(F)^3/d
 

Mathematica [A] (verified)

Time = 0.37 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.65 \[ \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^7} \, dx=\frac {F^a \left (b^3 \operatorname {ExpIntegralEi}\left (b (c+d x)^2 \log (F)\right ) \log ^3(F)-\frac {F^{b (c+d x)^2} \left (2+b (c+d x)^2 \log (F)+b^2 (c+d x)^4 \log ^2(F)\right )}{(c+d x)^6}\right )}{12 d} \] Input:

Integrate[F^(a + b*(c + d*x)^2)/(c + d*x)^7,x]
 

Output:

(F^a*(b^3*ExpIntegralEi[b*(c + d*x)^2*Log[F]]*Log[F]^3 - (F^(b*(c + d*x)^2 
)*(2 + b*(c + d*x)^2*Log[F] + b^2*(c + d*x)^4*Log[F]^2))/(c + d*x)^6))/(12 
*d)
 

Rubi [A] (verified)

Time = 0.80 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.02, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {2643, 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.

\(\displaystyle \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^7} \, dx\)

\(\Big \downarrow \) 2643

\(\displaystyle \frac {1}{3} b \log (F) \int \frac {F^{b (c+d x)^2+a}}{(c+d x)^5}dx-\frac {F^{a+b (c+d x)^2}}{6 d (c+d x)^6}\)

\(\Big \downarrow \) 2643

\(\displaystyle \frac {1}{3} b \log (F) \left (\frac {1}{2} b \log (F) \int \frac {F^{b (c+d x)^2+a}}{(c+d x)^3}dx-\frac {F^{a+b (c+d x)^2}}{4 d (c+d x)^4}\right )-\frac {F^{a+b (c+d x)^2}}{6 d (c+d x)^6}\)

\(\Big \downarrow \) 2643

\(\displaystyle \frac {1}{3} b \log (F) \left (\frac {1}{2} b \log (F) \left (b \log (F) \int \frac {F^{b (c+d x)^2+a}}{c+d x}dx-\frac {F^{a+b (c+d x)^2}}{2 d (c+d x)^2}\right )-\frac {F^{a+b (c+d x)^2}}{4 d (c+d x)^4}\right )-\frac {F^{a+b (c+d x)^2}}{6 d (c+d x)^6}\)

\(\Big \downarrow \) 2639

\(\displaystyle \frac {1}{3} b \log (F) \left (\frac {1}{2} b \log (F) \left (\frac {b F^a \log (F) \operatorname {ExpIntegralEi}\left (b (c+d x)^2 \log (F)\right )}{2 d}-\frac {F^{a+b (c+d x)^2}}{2 d (c+d x)^2}\right )-\frac {F^{a+b (c+d x)^2}}{4 d (c+d x)^4}\right )-\frac {F^{a+b (c+d x)^2}}{6 d (c+d x)^6}\)

Input:

Int[F^(a + b*(c + d*x)^2)/(c + d*x)^7,x]
 

Output:

-1/6*F^(a + b*(c + d*x)^2)/(d*(c + d*x)^6) + (b*Log[F]*(-1/4*F^(a + b*(c + 
 d*x)^2)/(d*(c + d*x)^4) + (b*Log[F]*(-1/2*F^(a + b*(c + d*x)^2)/(d*(c + d 
*x)^2) + (b*F^a*ExpIntegralEi[b*(c + d*x)^2*Log[F]]*Log[F])/(2*d)))/2))/3
 

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 [A] (verified)

Time = 0.38 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.98

method	result	size
risch	\(-\frac {F^{b \left (d x +c \right )^{2}} F^{a}}{6 d \left (d x +c \right )^{6}}-\frac {b \ln \left (F \right ) F^{b \left (d x +c \right )^{2}} F^{a}}{12 d \left (d x +c \right )^{4}}-\frac {b^{2} \ln \left (F \right )^{2} F^{b \left (d x +c \right )^{2}} F^{a}}{12 d \left (d x +c \right )^{2}}-\frac {b^{3} \ln \left (F \right )^{3} F^{a} \operatorname {expIntegral}_{1}\left (-b \left (d x +c \right )^{2} \ln \left (F \right )\right )}{12 d}\)	\(119\)

Input:

int(F^(a+b*(d*x+c)^2)/(d*x+c)^7,x,method=_RETURNVERBOSE)
 

Output:

-1/6/d/(d*x+c)^6*F^(b*(d*x+c)^2)*F^a-1/12/d*b*ln(F)/(d*x+c)^4*F^(b*(d*x+c) 
^2)*F^a-1/12/d*b^2*ln(F)^2/(d*x+c)^2*F^(b*(d*x+c)^2)*F^a-1/12/d*b^3*ln(F)^ 
3*F^a*Ei(1,-b*(d*x+c)^2*ln(F))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 292 vs. \(2 (113) = 226\).

Time = 0.08 (sec) , antiderivative size = 292, normalized size of antiderivative = 2.41 \[ \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^7} \, dx=\frac {{\left (b^{3} d^{6} x^{6} + 6 \, b^{3} c d^{5} x^{5} + 15 \, b^{3} c^{2} d^{4} x^{4} + 20 \, b^{3} c^{3} d^{3} x^{3} + 15 \, b^{3} c^{4} d^{2} x^{2} + 6 \, b^{3} c^{5} d x + b^{3} c^{6}\right )} F^{a} {\rm Ei}\left ({\left (b d^{2} x^{2} + 2 \, b c d x + b c^{2}\right )} \log \left (F\right )\right ) \log \left (F\right )^{3} - {\left ({\left (b^{2} d^{4} x^{4} + 4 \, b^{2} c d^{3} x^{3} + 6 \, b^{2} c^{2} d^{2} x^{2} + 4 \, b^{2} c^{3} d x + b^{2} c^{4}\right )} \log \left (F\right )^{2} + {\left (b d^{2} x^{2} + 2 \, b c d x + b c^{2}\right )} \log \left (F\right ) + 2\right )} F^{b d^{2} x^{2} + 2 \, b c d x + b c^{2} + a}}{12 \, {\left (d^{7} x^{6} + 6 \, c d^{6} x^{5} + 15 \, c^{2} d^{5} x^{4} + 20 \, c^{3} d^{4} x^{3} + 15 \, c^{4} d^{3} x^{2} + 6 \, c^{5} d^{2} x + c^{6} d\right )}} \] Input:

integrate(F^(a+b*(d*x+c)^2)/(d*x+c)^7,x, algorithm="fricas")
 

Output:

1/12*((b^3*d^6*x^6 + 6*b^3*c*d^5*x^5 + 15*b^3*c^2*d^4*x^4 + 20*b^3*c^3*d^3 
*x^3 + 15*b^3*c^4*d^2*x^2 + 6*b^3*c^5*d*x + b^3*c^6)*F^a*Ei((b*d^2*x^2 + 2 
*b*c*d*x + b*c^2)*log(F))*log(F)^3 - ((b^2*d^4*x^4 + 4*b^2*c*d^3*x^3 + 6*b 
^2*c^2*d^2*x^2 + 4*b^2*c^3*d*x + b^2*c^4)*log(F)^2 + (b*d^2*x^2 + 2*b*c*d* 
x + b*c^2)*log(F) + 2)*F^(b*d^2*x^2 + 2*b*c*d*x + b*c^2 + a))/(d^7*x^6 + 6 
*c*d^6*x^5 + 15*c^2*d^5*x^4 + 20*c^3*d^4*x^3 + 15*c^4*d^3*x^2 + 6*c^5*d^2* 
x + c^6*d)
 

Sympy [F]

\[ \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^7} \, dx=\int \frac {F^{a + b \left (c + d x\right )^{2}}}{\left (c + d x\right )^{7}}\, dx \] Input:

integrate(F**(a+b*(d*x+c)**2)/(d*x+c)**7,x)
 

Output:

Integral(F**(a + b*(c + d*x)**2)/(c + d*x)**7, x)
 

Maxima [F]

\[ \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^7} \, dx=\int { \frac {F^{{\left (d x + c\right )}^{2} b + a}}{{\left (d x + c\right )}^{7}} \,d x } \] Input:

integrate(F^(a+b*(d*x+c)^2)/(d*x+c)^7,x, algorithm="maxima")
 

Output:

integrate(F^((d*x + c)^2*b + a)/(d*x + c)^7, x)
 

Giac [F]

\[ \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^7} \, dx=\int { \frac {F^{{\left (d x + c\right )}^{2} b + a}}{{\left (d x + c\right )}^{7}} \,d x } \] Input:

integrate(F^(a+b*(d*x+c)^2)/(d*x+c)^7,x, algorithm="giac")
 

Output:

integrate(F^((d*x + c)^2*b + a)/(d*x + c)^7, x)
                                                                                    
                                                                                    
 

Mupad [B] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.86 \[ \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^7} \, dx=-\frac {F^a\,b^3\,{\ln \left (F\right )}^3\,\mathrm {expint}\left (-b\,\ln \left (F\right )\,{\left (c+d\,x\right )}^2\right )}{12\,d}-\frac {F^a\,F^{b\,{\left (c+d\,x\right )}^2}\,b^3\,{\ln \left (F\right )}^3\,\left (\frac {1}{6\,b\,\ln \left (F\right )\,{\left (c+d\,x\right )}^2}+\frac {1}{6\,b^2\,{\ln \left (F\right )}^2\,{\left (c+d\,x\right )}^4}+\frac {1}{3\,b^3\,{\ln \left (F\right )}^3\,{\left (c+d\,x\right )}^6}\right )}{2\,d} \] Input:

int(F^(a + b*(c + d*x)^2)/(c + d*x)^7,x)
 

Output:

- (F^a*b^3*log(F)^3*expint(-b*log(F)*(c + d*x)^2))/(12*d) - (F^a*F^(b*(c + 
 d*x)^2)*b^3*log(F)^3*(1/(6*b*log(F)*(c + d*x)^2) + 1/(6*b^2*log(F)^2*(c + 
 d*x)^4) + 1/(3*b^3*log(F)^3*(c + d*x)^6)))/(2*d)
 

Reduce [F]

\[ \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^7} \, dx=\text {too large to display} \] Input:

int(F^(a+b*(d*x+c)^2)/(d*x+c)^7,x)
 

Output:

(f**(a + b*c**2)*( - f**(2*b*c*d*x + b*d**2*x**2) + 4*int(f**(2*b*c*d*x + 
b*d**2*x**2)/(log(f)*b*c**9 + 7*log(f)*b*c**8*d*x + 21*log(f)*b*c**7*d**2* 
x**2 + 35*log(f)*b*c**6*d**3*x**3 + 35*log(f)*b*c**5*d**4*x**4 + 21*log(f) 
*b*c**4*d**5*x**5 + 7*log(f)*b*c**3*d**6*x**6 + log(f)*b*c**2*d**7*x**7 - 
3*c**7 - 21*c**6*d*x - 63*c**5*d**2*x**2 - 105*c**4*d**3*x**3 - 105*c**3*d 
**4*x**4 - 63*c**2*d**5*x**5 - 21*c*d**6*x**6 - 3*d**7*x**7),x)*log(f)**2* 
b**2*c**10*d + 24*int(f**(2*b*c*d*x + b*d**2*x**2)/(log(f)*b*c**9 + 7*log( 
f)*b*c**8*d*x + 21*log(f)*b*c**7*d**2*x**2 + 35*log(f)*b*c**6*d**3*x**3 + 
35*log(f)*b*c**5*d**4*x**4 + 21*log(f)*b*c**4*d**5*x**5 + 7*log(f)*b*c**3* 
d**6*x**6 + log(f)*b*c**2*d**7*x**7 - 3*c**7 - 21*c**6*d*x - 63*c**5*d**2* 
x**2 - 105*c**4*d**3*x**3 - 105*c**3*d**4*x**4 - 63*c**2*d**5*x**5 - 21*c* 
d**6*x**6 - 3*d**7*x**7),x)*log(f)**2*b**2*c**9*d**2*x + 60*int(f**(2*b*c* 
d*x + b*d**2*x**2)/(log(f)*b*c**9 + 7*log(f)*b*c**8*d*x + 21*log(f)*b*c**7 
*d**2*x**2 + 35*log(f)*b*c**6*d**3*x**3 + 35*log(f)*b*c**5*d**4*x**4 + 21* 
log(f)*b*c**4*d**5*x**5 + 7*log(f)*b*c**3*d**6*x**6 + log(f)*b*c**2*d**7*x 
**7 - 3*c**7 - 21*c**6*d*x - 63*c**5*d**2*x**2 - 105*c**4*d**3*x**3 - 105* 
c**3*d**4*x**4 - 63*c**2*d**5*x**5 - 21*c*d**6*x**6 - 3*d**7*x**7),x)*log( 
f)**2*b**2*c**8*d**3*x**2 + 80*int(f**(2*b*c*d*x + b*d**2*x**2)/(log(f)*b* 
c**9 + 7*log(f)*b*c**8*d*x + 21*log(f)*b*c**7*d**2*x**2 + 35*log(f)*b*c**6 
*d**3*x**3 + 35*log(f)*b*c**5*d**4*x**4 + 21*log(f)*b*c**4*d**5*x**5 + ...