3.3.77 \(\int \frac {F^{a+b (c+d x)^2}}{(c+d x)^8} \, dx\) [277]

3.3.77.1 Optimal result
3.3.77.2 Mathematica [A] (verified)
3.3.77.3 Rubi [A] (verified)
3.3.77.4 Maple [A] (verified)
3.3.77.5 Fricas [B] (verification not implemented)
3.3.77.6 Sympy [F]
3.3.77.7 Maxima [F]
3.3.77.8 Giac [F]
3.3.77.9 Mupad [B] (verification not implemented)

3.3.77.1 Optimal result

Integrand size = 21, antiderivative size = 170 \[ \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^8} \, dx=-\frac {F^{a+b (c+d x)^2}}{7 d (c+d x)^7}-\frac {2 b F^{a+b (c+d x)^2} \log (F)}{35 d (c+d x)^5}-\frac {4 b^2 F^{a+b (c+d x)^2} \log ^2(F)}{105 d (c+d x)^3}-\frac {8 b^3 F^{a+b (c+d x)^2} \log ^3(F)}{105 d (c+d x)}+\frac {8 b^{7/2} F^a \sqrt {\pi } \text {erfi}\left (\sqrt {b} (c+d x) \sqrt {\log (F)}\right ) \log ^{\frac {7}{2}}(F)}{105 d} \]

output
-1/7*F^(a+b*(d*x+c)^2)/d/(d*x+c)^7-2/35*b*F^(a+b*(d*x+c)^2)*ln(F)/d/(d*x+c 
)^5-4/105*b^2*F^(a+b*(d*x+c)^2)*ln(F)^2/d/(d*x+c)^3-8/105*b^3*F^(a+b*(d*x+ 
c)^2)*ln(F)^3/d/(d*x+c)+8/105*b^(7/2)*F^a*erfi((d*x+c)*b^(1/2)*ln(F)^(1/2) 
)*ln(F)^(7/2)*Pi^(1/2)/d
 
3.3.77.2 Mathematica [A] (verified)

Time = 0.35 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.66 \[ \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^8} \, dx=\frac {F^a \left (8 b^{7/2} \sqrt {\pi } \text {erfi}\left (\sqrt {b} (c+d x) \sqrt {\log (F)}\right ) \log ^{\frac {7}{2}}(F)+\frac {F^{b (c+d x)^2} \left (-15-6 b (c+d x)^2 \log (F)-4 b^2 (c+d x)^4 \log ^2(F)-8 b^3 (c+d x)^6 \log ^3(F)\right )}{(c+d x)^7}\right )}{105 d} \]

input
Integrate[F^(a + b*(c + d*x)^2)/(c + d*x)^8,x]
 
output
(F^a*(8*b^(7/2)*Sqrt[Pi]*Erfi[Sqrt[b]*(c + d*x)*Sqrt[Log[F]]]*Log[F]^(7/2) 
 + (F^(b*(c + d*x)^2)*(-15 - 6*b*(c + d*x)^2*Log[F] - 4*b^2*(c + d*x)^4*Lo 
g[F]^2 - 8*b^3*(c + d*x)^6*Log[F]^3))/(c + d*x)^7))/(105*d)
 
3.3.77.3 Rubi [A] (verified)

Time = 0.56 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.01, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {2643, 2643, 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.

\(\displaystyle \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^8} \, dx\)

\(\Big \downarrow \) 2643

\(\displaystyle \frac {2}{7} b \log (F) \int \frac {F^{b (c+d x)^2+a}}{(c+d x)^6}dx-\frac {F^{a+b (c+d x)^2}}{7 d (c+d x)^7}\)

\(\Big \downarrow \) 2643

\(\displaystyle \frac {2}{7} b \log (F) \left (\frac {2}{5} b \log (F) \int \frac {F^{b (c+d x)^2+a}}{(c+d x)^4}dx-\frac {F^{a+b (c+d x)^2}}{5 d (c+d x)^5}\right )-\frac {F^{a+b (c+d x)^2}}{7 d (c+d x)^7}\)

\(\Big \downarrow \) 2643

\(\displaystyle \frac {2}{7} b \log (F) \left (\frac {2}{5} b \log (F) \left (\frac {2}{3} b \log (F) \int \frac {F^{b (c+d x)^2+a}}{(c+d x)^2}dx-\frac {F^{a+b (c+d x)^2}}{3 d (c+d x)^3}\right )-\frac {F^{a+b (c+d x)^2}}{5 d (c+d x)^5}\right )-\frac {F^{a+b (c+d x)^2}}{7 d (c+d x)^7}\)

\(\Big \downarrow \) 2643

\(\displaystyle \frac {2}{7} b \log (F) \left (\frac {2}{5} b \log (F) \left (\frac {2}{3} b \log (F) \left (2 b \log (F) \int F^{b (c+d x)^2+a}dx-\frac {F^{a+b (c+d x)^2}}{d (c+d x)}\right )-\frac {F^{a+b (c+d x)^2}}{3 d (c+d x)^3}\right )-\frac {F^{a+b (c+d x)^2}}{5 d (c+d x)^5}\right )-\frac {F^{a+b (c+d x)^2}}{7 d (c+d x)^7}\)

\(\Big \downarrow \) 2633

\(\displaystyle \frac {2}{7} b \log (F) \left (\frac {2}{5} b \log (F) \left (\frac {2}{3} b \log (F) \left (\frac {\sqrt {\pi } \sqrt {b} F^a \sqrt {\log (F)} \text {erfi}\left (\sqrt {b} \sqrt {\log (F)} (c+d x)\right )}{d}-\frac {F^{a+b (c+d x)^2}}{d (c+d x)}\right )-\frac {F^{a+b (c+d x)^2}}{3 d (c+d x)^3}\right )-\frac {F^{a+b (c+d x)^2}}{5 d (c+d x)^5}\right )-\frac {F^{a+b (c+d x)^2}}{7 d (c+d x)^7}\)

input
Int[F^(a + b*(c + d*x)^2)/(c + d*x)^8,x]
 
output
-1/7*F^(a + b*(c + d*x)^2)/(d*(c + d*x)^7) + (2*b*Log[F]*(-1/5*F^(a + b*(c 
 + d*x)^2)/(d*(c + d*x)^5) + (2*b*Log[F]*(-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*S 
qrt[Pi]*Erfi[Sqrt[b]*(c + d*x)*Sqrt[Log[F]]]*Sqrt[Log[F]])/d)*Log[F])/3))/ 
5))/7
 

3.3.77.3.1 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]))
 
3.3.77.4 Maple [A] (verified)

Time = 0.57 (sec) , antiderivative size = 162, normalized size of antiderivative = 0.95

method result size
risch \(-\frac {F^{b \left (d x +c \right )^{2}} F^{a}}{7 d \left (d x +c \right )^{7}}-\frac {2 b \ln \left (F \right ) F^{b \left (d x +c \right )^{2}} F^{a}}{35 d \left (d x +c \right )^{5}}-\frac {4 b^{2} \ln \left (F \right )^{2} F^{b \left (d x +c \right )^{2}} F^{a}}{105 d \left (d x +c \right )^{3}}-\frac {8 b^{3} \ln \left (F \right )^{3} F^{b \left (d x +c \right )^{2}} F^{a}}{105 d \left (d x +c \right )}+\frac {8 b^{4} \ln \left (F \right )^{4} \sqrt {\pi }\, F^{a} \operatorname {erf}\left (\sqrt {-b \ln \left (F \right )}\, \left (d x +c \right )\right )}{105 d \sqrt {-b \ln \left (F \right )}}\) \(162\)

input
int(F^(a+b*(d*x+c)^2)/(d*x+c)^8,x,method=_RETURNVERBOSE)
 
output
-1/7/d/(d*x+c)^7*F^(b*(d*x+c)^2)*F^a-2/35/d*b*ln(F)/(d*x+c)^5*F^(b*(d*x+c) 
^2)*F^a-4/105/d*b^2*ln(F)^2/(d*x+c)^3*F^(b*(d*x+c)^2)*F^a-8/105/d*b^3*ln(F 
)^3/(d*x+c)*F^(b*(d*x+c)^2)*F^a+8/105/d*b^4*ln(F)^4*Pi^(1/2)*F^a/(-b*ln(F) 
)^(1/2)*erf((-b*ln(F))^(1/2)*(d*x+c))
 
3.3.77.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 429 vs. \(2 (150) = 300\).

Time = 0.28 (sec) , antiderivative size = 429, normalized size of antiderivative = 2.52 \[ \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^8} \, dx=-\frac {8 \, \sqrt {\pi } {\left (b^{3} d^{7} x^{7} + 7 \, b^{3} c d^{6} x^{6} + 21 \, b^{3} c^{2} d^{5} x^{5} + 35 \, b^{3} c^{3} d^{4} x^{4} + 35 \, b^{3} c^{4} d^{3} x^{3} + 21 \, b^{3} c^{5} d^{2} x^{2} + 7 \, b^{3} c^{6} d x + b^{3} c^{7}\right )} \sqrt {-b d^{2} \log \left (F\right )} F^{a} \operatorname {erf}\left (\frac {\sqrt {-b d^{2} \log \left (F\right )} {\left (d x + c\right )}}{d}\right ) \log \left (F\right )^{3} + {\left (8 \, {\left (b^{3} d^{7} x^{6} + 6 \, b^{3} c d^{6} x^{5} + 15 \, b^{3} c^{2} d^{5} x^{4} + 20 \, b^{3} c^{3} d^{4} x^{3} + 15 \, b^{3} c^{4} d^{3} x^{2} + 6 \, b^{3} c^{5} d^{2} x + b^{3} c^{6} d\right )} \log \left (F\right )^{3} + 4 \, {\left (b^{2} d^{5} x^{4} + 4 \, b^{2} c d^{4} x^{3} + 6 \, b^{2} c^{2} d^{3} x^{2} + 4 \, b^{2} c^{3} d^{2} x + b^{2} c^{4} d\right )} \log \left (F\right )^{2} + 6 \, {\left (b d^{3} x^{2} + 2 \, b c d^{2} x + b c^{2} d\right )} \log \left (F\right ) + 15 \, d\right )} F^{b d^{2} x^{2} + 2 \, b c d x + b c^{2} + a}}{105 \, {\left (d^{9} x^{7} + 7 \, c d^{8} x^{6} + 21 \, c^{2} d^{7} x^{5} + 35 \, c^{3} d^{6} x^{4} + 35 \, c^{4} d^{5} x^{3} + 21 \, c^{5} d^{4} x^{2} + 7 \, c^{6} d^{3} x + c^{7} d^{2}\right )}} \]

input
integrate(F^(a+b*(d*x+c)^2)/(d*x+c)^8,x, algorithm="fricas")
 
output
-1/105*(8*sqrt(pi)*(b^3*d^7*x^7 + 7*b^3*c*d^6*x^6 + 21*b^3*c^2*d^5*x^5 + 3 
5*b^3*c^3*d^4*x^4 + 35*b^3*c^4*d^3*x^3 + 21*b^3*c^5*d^2*x^2 + 7*b^3*c^6*d* 
x + b^3*c^7)*sqrt(-b*d^2*log(F))*F^a*erf(sqrt(-b*d^2*log(F))*(d*x + c)/d)* 
log(F)^3 + (8*(b^3*d^7*x^6 + 6*b^3*c*d^6*x^5 + 15*b^3*c^2*d^5*x^4 + 20*b^3 
*c^3*d^4*x^3 + 15*b^3*c^4*d^3*x^2 + 6*b^3*c^5*d^2*x + b^3*c^6*d)*log(F)^3 
+ 4*(b^2*d^5*x^4 + 4*b^2*c*d^4*x^3 + 6*b^2*c^2*d^3*x^2 + 4*b^2*c^3*d^2*x + 
 b^2*c^4*d)*log(F)^2 + 6*(b*d^3*x^2 + 2*b*c*d^2*x + b*c^2*d)*log(F) + 15*d 
)*F^(b*d^2*x^2 + 2*b*c*d*x + b*c^2 + a))/(d^9*x^7 + 7*c*d^8*x^6 + 21*c^2*d 
^7*x^5 + 35*c^3*d^6*x^4 + 35*c^4*d^5*x^3 + 21*c^5*d^4*x^2 + 7*c^6*d^3*x + 
c^7*d^2)
 
3.3.77.6 Sympy [F]

\[ \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^8} \, dx=\int \frac {F^{a + b \left (c + d x\right )^{2}}}{\left (c + d x\right )^{8}}\, dx \]

input
integrate(F**(a+b*(d*x+c)**2)/(d*x+c)**8,x)
 
output
Integral(F**(a + b*(c + d*x)**2)/(c + d*x)**8, x)
 
3.3.77.7 Maxima [F]

\[ \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^8} \, dx=\int { \frac {F^{{\left (d x + c\right )}^{2} b + a}}{{\left (d x + c\right )}^{8}} \,d x } \]

input
integrate(F^(a+b*(d*x+c)^2)/(d*x+c)^8,x, algorithm="maxima")
 
output
integrate(F^((d*x + c)^2*b + a)/(d*x + c)^8, x)
 
3.3.77.8 Giac [F]

\[ \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^8} \, dx=\int { \frac {F^{{\left (d x + c\right )}^{2} b + a}}{{\left (d x + c\right )}^{8}} \,d x } \]

input
integrate(F^(a+b*(d*x+c)^2)/(d*x+c)^8,x, algorithm="giac")
 
output
integrate(F^((d*x + c)^2*b + a)/(d*x + c)^8, x)
 
3.3.77.9 Mupad [B] (verification not implemented)

Time = 0.80 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.18 \[ \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^8} \, dx=\frac {8\,F^a\,\sqrt {\pi }\,{\left (-b\,\ln \left (F\right )\,{\left (c+d\,x\right )}^2\right )}^{7/2}}{105\,d\,{\left (c+d\,x\right )}^7}-\frac {F^a\,F^{b\,{\left (c+d\,x\right )}^2}}{7\,d\,{\left (c+d\,x\right )}^7}-\frac {4\,F^a\,F^{b\,{\left (c+d\,x\right )}^2}\,b^2\,{\ln \left (F\right )}^2}{105\,d\,{\left (c+d\,x\right )}^3}-\frac {8\,F^a\,F^{b\,{\left (c+d\,x\right )}^2}\,b^3\,{\ln \left (F\right )}^3}{105\,d\,\left (c+d\,x\right )}-\frac {2\,F^a\,F^{b\,{\left (c+d\,x\right )}^2}\,b\,\ln \left (F\right )}{35\,d\,{\left (c+d\,x\right )}^5}-\frac {8\,F^a\,\sqrt {\pi }\,\mathrm {erfc}\left (\sqrt {-b\,\ln \left (F\right )\,{\left (c+d\,x\right )}^2}\right )\,{\left (-b\,\ln \left (F\right )\,{\left (c+d\,x\right )}^2\right )}^{7/2}}{105\,d\,{\left (c+d\,x\right )}^7} \]

input
int(F^(a + b*(c + d*x)^2)/(c + d*x)^8,x)
 
output
(8*F^a*pi^(1/2)*(-b*log(F)*(c + d*x)^2)^(7/2))/(105*d*(c + d*x)^7) - (F^a* 
F^(b*(c + d*x)^2))/(7*d*(c + d*x)^7) - (4*F^a*F^(b*(c + d*x)^2)*b^2*log(F) 
^2)/(105*d*(c + d*x)^3) - (8*F^a*F^(b*(c + d*x)^2)*b^3*log(F)^3)/(105*d*(c 
 + d*x)) - (2*F^a*F^(b*(c + d*x)^2)*b*log(F))/(35*d*(c + d*x)^5) - (8*F^a* 
pi^(1/2)*erfc((-b*log(F)*(c + d*x)^2)^(1/2))*(-b*log(F)*(c + d*x)^2)^(7/2) 
)/(105*d*(c + d*x)^7)