3.4.29 \(\int F^{a+\frac {b}{(c+d x)^2}} (c+d x)^6 \, dx\) [329]

3.4.29.1 Optimal result
3.4.29.2 Mathematica [A] (verified)
3.4.29.3 Rubi [A] (verified)
3.4.29.4 Maple [B] (verified)
3.4.29.5 Fricas [A] (verification not implemented)
3.4.29.6 Sympy [F]
3.4.29.7 Maxima [F]
3.4.29.8 Giac [F]
3.4.29.9 Mupad [B] (verification not implemented)

3.4.29.1 Optimal result

Integrand size = 21, antiderivative size = 170 \[ \int F^{a+\frac {b}{(c+d x)^2}} (c+d x)^6 \, dx=\frac {F^{a+\frac {b}{(c+d x)^2}} (c+d x)^7}{7 d}+\frac {2 b F^{a+\frac {b}{(c+d x)^2}} (c+d x)^5 \log (F)}{35 d}+\frac {4 b^2 F^{a+\frac {b}{(c+d x)^2}} (c+d x)^3 \log ^2(F)}{105 d}+\frac {8 b^3 F^{a+\frac {b}{(c+d x)^2}} (c+d x) \log ^3(F)}{105 d}-\frac {8 b^{7/2} F^a \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {b} \sqrt {\log (F)}}{c+d x}\right ) \log ^{\frac {7}{2}}(F)}{105 d} \]

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

Time = 0.09 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.66 \[ \int F^{a+\frac {b}{(c+d x)^2}} (c+d x)^6 \, dx=\frac {F^a \left (-8 b^{7/2} \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {b} \sqrt {\log (F)}}{c+d x}\right ) \log ^{\frac {7}{2}}(F)+F^{\frac {b}{(c+d x)^2}} (c+d x) \left (15 (c+d x)^6+6 b (c+d x)^4 \log (F)+4 b^2 (c+d x)^2 \log ^2(F)+8 b^3 \log ^3(F)\right )\right )}{105 d} \]

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

Time = 0.55 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.01, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {2643, 2643, 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.

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

\(\Big \downarrow \) 2643

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

\(\Big \downarrow \) 2643

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

\(\Big \downarrow \) 2643

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

\(\Big \downarrow \) 2635

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

\(\Big \downarrow \) 2640

\(\displaystyle \frac {2}{7} b \log (F) \left (\frac {2}{5} b \log (F) \left (\frac {2}{3} b \log (F) \left (\frac {(c+d x) F^{a+\frac {b}{(c+d x)^2}}}{d}-\frac {2 b \log (F) \int F^{a+\frac {b}{(c+d x)^2}}d\frac {1}{c+d x}}{d}\right )+\frac {(c+d x)^3 F^{a+\frac {b}{(c+d x)^2}}}{3 d}\right )+\frac {(c+d x)^5 F^{a+\frac {b}{(c+d x)^2}}}{5 d}\right )+\frac {(c+d x)^7 F^{a+\frac {b}{(c+d x)^2}}}{7 d}\)

\(\Big \downarrow \) 2633

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

input
Int[F^(a + b/(c + d*x)^2)*(c + d*x)^6,x]
 
output
(F^(a + b/(c + d*x)^2)*(c + d*x)^7)/(7*d) + (2*b*Log[F]*((F^(a + b/(c + d* 
x)^2)*(c + d*x)^5)/(5*d) + (2*b*Log[F]*((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))/5))/ 
7
 

3.4.29.3.1 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 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]))
 
3.4.29.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(542\) vs. \(2(150)=300\).

Time = 1.52 (sec) , antiderivative size = 543, normalized size of antiderivative = 3.19

method result size
risch \(\frac {F^{a} d^{6} F^{\frac {b}{\left (d x +c \right )^{2}}} x^{7}}{7}+F^{a} d^{5} F^{\frac {b}{\left (d x +c \right )^{2}}} c \,x^{6}+3 F^{a} d^{4} F^{\frac {b}{\left (d x +c \right )^{2}}} c^{2} x^{5}+5 F^{a} d^{3} F^{\frac {b}{\left (d x +c \right )^{2}}} c^{3} x^{4}+5 F^{a} d^{2} F^{\frac {b}{\left (d x +c \right )^{2}}} c^{4} x^{3}+3 F^{a} d \,F^{\frac {b}{\left (d x +c \right )^{2}}} c^{5} x^{2}+F^{a} F^{\frac {b}{\left (d x +c \right )^{2}}} c^{6} x +\frac {F^{a} F^{\frac {b}{\left (d x +c \right )^{2}}} c^{7}}{7 d}+\frac {2 F^{a} d^{4} b \ln \left (F \right ) F^{\frac {b}{\left (d x +c \right )^{2}}} x^{5}}{35}+\frac {2 F^{a} d^{3} b \ln \left (F \right ) F^{\frac {b}{\left (d x +c \right )^{2}}} c \,x^{4}}{7}+\frac {4 F^{a} d^{2} b \ln \left (F \right ) F^{\frac {b}{\left (d x +c \right )^{2}}} c^{2} x^{3}}{7}+\frac {4 F^{a} d b \ln \left (F \right ) F^{\frac {b}{\left (d x +c \right )^{2}}} c^{3} x^{2}}{7}+\frac {2 F^{a} b \ln \left (F \right ) F^{\frac {b}{\left (d x +c \right )^{2}}} c^{4} x}{7}+\frac {2 F^{a} b \ln \left (F \right ) F^{\frac {b}{\left (d x +c \right )^{2}}} c^{5}}{35 d}+\frac {4 F^{a} d^{2} b^{2} \ln \left (F \right )^{2} F^{\frac {b}{\left (d x +c \right )^{2}}} x^{3}}{105}+\frac {4 F^{a} d \,b^{2} \ln \left (F \right )^{2} F^{\frac {b}{\left (d x +c \right )^{2}}} c \,x^{2}}{35}+\frac {4 F^{a} b^{2} \ln \left (F \right )^{2} F^{\frac {b}{\left (d x +c \right )^{2}}} c^{2} x}{35}+\frac {4 F^{a} b^{2} \ln \left (F \right )^{2} F^{\frac {b}{\left (d x +c \right )^{2}}} c^{3}}{105 d}+\frac {8 F^{a} b^{3} \ln \left (F \right )^{3} F^{\frac {b}{\left (d x +c \right )^{2}}} x}{105}+\frac {8 F^{a} b^{3} \ln \left (F \right )^{3} F^{\frac {b}{\left (d x +c \right )^{2}}} c}{105 d}-\frac {8 F^{a} b^{4} \ln \left (F \right )^{4} \sqrt {\pi }\, \operatorname {erf}\left (\frac {\sqrt {-b \ln \left (F \right )}}{d x +c}\right )}{105 d \sqrt {-b \ln \left (F \right )}}\) \(543\)

input
int(F^(a+b/(d*x+c)^2)*(d*x+c)^6,x,method=_RETURNVERBOSE)
 
output
1/7*F^a*d^6*F^(b/(d*x+c)^2)*x^7+F^a*d^5*F^(b/(d*x+c)^2)*c*x^6+3*F^a*d^4*F^ 
(b/(d*x+c)^2)*c^2*x^5+5*F^a*d^3*F^(b/(d*x+c)^2)*c^3*x^4+5*F^a*d^2*F^(b/(d* 
x+c)^2)*c^4*x^3+3*F^a*d*F^(b/(d*x+c)^2)*c^5*x^2+F^a*F^(b/(d*x+c)^2)*c^6*x+ 
1/7*F^a/d*F^(b/(d*x+c)^2)*c^7+2/35*F^a*d^4*b*ln(F)*F^(b/(d*x+c)^2)*x^5+2/7 
*F^a*d^3*b*ln(F)*F^(b/(d*x+c)^2)*c*x^4+4/7*F^a*d^2*b*ln(F)*F^(b/(d*x+c)^2) 
*c^2*x^3+4/7*F^a*d*b*ln(F)*F^(b/(d*x+c)^2)*c^3*x^2+2/7*F^a*b*ln(F)*F^(b/(d 
*x+c)^2)*c^4*x+2/35*F^a/d*b*ln(F)*F^(b/(d*x+c)^2)*c^5+4/105*F^a*d^2*b^2*ln 
(F)^2*F^(b/(d*x+c)^2)*x^3+4/35*F^a*d*b^2*ln(F)^2*F^(b/(d*x+c)^2)*c*x^2+4/3 
5*F^a*b^2*ln(F)^2*F^(b/(d*x+c)^2)*c^2*x+4/105*F^a/d*b^2*ln(F)^2*F^(b/(d*x+ 
c)^2)*c^3+8/105*F^a*b^3*ln(F)^3*F^(b/(d*x+c)^2)*x+8/105*F^a/d*b^3*ln(F)^3* 
F^(b/(d*x+c)^2)*c-8/105*F^a/d*b^4*ln(F)^4*Pi^(1/2)/(-b*ln(F))^(1/2)*erf((- 
b*ln(F))^(1/2)/(d*x+c))
 
3.4.29.5 Fricas [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 293, normalized size of antiderivative = 1.72 \[ \int F^{a+\frac {b}{(c+d x)^2}} (c+d x)^6 \, dx=\frac {8 \, \sqrt {\pi } F^{a} b^{3} d \sqrt {-\frac {b \log \left (F\right )}{d^{2}}} \operatorname {erf}\left (\frac {d \sqrt {-\frac {b \log \left (F\right )}{d^{2}}}}{d x + c}\right ) \log \left (F\right )^{3} + {\left (15 \, d^{7} x^{7} + 105 \, c d^{6} x^{6} + 315 \, c^{2} d^{5} x^{5} + 525 \, c^{3} d^{4} x^{4} + 525 \, c^{4} d^{3} x^{3} + 315 \, c^{5} d^{2} x^{2} + 105 \, c^{6} d x + 15 \, c^{7} + 8 \, {\left (b^{3} d x + b^{3} c\right )} \log \left (F\right )^{3} + 4 \, {\left (b^{2} d^{3} x^{3} + 3 \, b^{2} c d^{2} x^{2} + 3 \, b^{2} c^{2} d x + b^{2} c^{3}\right )} \log \left (F\right )^{2} + 6 \, {\left (b d^{5} x^{5} + 5 \, b c d^{4} x^{4} + 10 \, b c^{2} d^{3} x^{3} + 10 \, b c^{3} d^{2} x^{2} + 5 \, b c^{4} d x + b c^{5}\right )} \log \left (F\right )\right )} F^{\frac {a d^{2} x^{2} + 2 \, a c d x + a c^{2} + b}{d^{2} x^{2} + 2 \, c d x + c^{2}}}}{105 \, d} \]

input
integrate(F^(a+b/(d*x+c)^2)*(d*x+c)^6,x, algorithm="fricas")
 
output
1/105*(8*sqrt(pi)*F^a*b^3*d*sqrt(-b*log(F)/d^2)*erf(d*sqrt(-b*log(F)/d^2)/ 
(d*x + c))*log(F)^3 + (15*d^7*x^7 + 105*c*d^6*x^6 + 315*c^2*d^5*x^5 + 525* 
c^3*d^4*x^4 + 525*c^4*d^3*x^3 + 315*c^5*d^2*x^2 + 105*c^6*d*x + 15*c^7 + 8 
*(b^3*d*x + b^3*c)*log(F)^3 + 4*(b^2*d^3*x^3 + 3*b^2*c*d^2*x^2 + 3*b^2*c^2 
*d*x + b^2*c^3)*log(F)^2 + 6*(b*d^5*x^5 + 5*b*c*d^4*x^4 + 10*b*c^2*d^3*x^3 
 + 10*b*c^3*d^2*x^2 + 5*b*c^4*d*x + b*c^5)*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
 
3.4.29.6 Sympy [F]

\[ \int F^{a+\frac {b}{(c+d x)^2}} (c+d x)^6 \, dx=\int F^{a + \frac {b}{\left (c + d x\right )^{2}}} \left (c + d x\right )^{6}\, dx \]

input
integrate(F**(a+b/(d*x+c)**2)*(d*x+c)**6,x)
 
output
Integral(F**(a + b/(c + d*x)**2)*(c + d*x)**6, x)
 
3.4.29.7 Maxima [F]

\[ \int F^{a+\frac {b}{(c+d x)^2}} (c+d x)^6 \, dx=\int { {\left (d x + c\right )}^{6} F^{a + \frac {b}{{\left (d x + c\right )}^{2}}} \,d x } \]

input
integrate(F^(a+b/(d*x+c)^2)*(d*x+c)^6,x, algorithm="maxima")
 
output
1/105*(15*F^a*d^6*x^7 + 105*F^a*c*d^5*x^6 + 3*(105*F^a*c^2*d^4 + 2*F^a*b*d 
^4*log(F))*x^5 + 15*(35*F^a*c^3*d^3 + 2*F^a*b*c*d^3*log(F))*x^4 + (525*F^a 
*c^4*d^2 + 60*F^a*b*c^2*d^2*log(F) + 4*F^a*b^2*d^2*log(F)^2)*x^3 + 3*(105* 
F^a*c^5*d + 20*F^a*b*c^3*d*log(F) + 4*F^a*b^2*c*d*log(F)^2)*x^2 + (105*F^a 
*c^6 + 30*F^a*b*c^4*log(F) + 12*F^a*b^2*c^2*log(F)^2 + 8*F^a*b^3*log(F)^3) 
*x)*F^(b/(d^2*x^2 + 2*c*d*x + c^2)) + integrate(2/105*(8*F^a*b^4*d*x*log(F 
)^4 - 15*F^a*b*c^7*log(F) - 6*F^a*b^2*c^5*log(F)^2 - 4*F^a*b^3*c^3*log(F)^ 
3)*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)
 
3.4.29.8 Giac [F]

\[ \int F^{a+\frac {b}{(c+d x)^2}} (c+d x)^6 \, dx=\int { {\left (d x + c\right )}^{6} F^{a + \frac {b}{{\left (d x + c\right )}^{2}}} \,d x } \]

input
integrate(F^(a+b/(d*x+c)^2)*(d*x+c)^6,x, algorithm="giac")
 
output
integrate((d*x + c)^6*F^(a + b/(d*x + c)^2), x)
 
3.4.29.9 Mupad [B] (verification not implemented)

Time = 1.10 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.17 \[ \int F^{a+\frac {b}{(c+d x)^2}} (c+d x)^6 \, dx=\frac {F^a\,F^{\frac {b}{{\left (c+d\,x\right )}^2}}\,{\left (c+d\,x\right )}^7}{7\,d}-\frac {8\,F^a\,\sqrt {\pi }\,{\left (c+d\,x\right )}^7\,{\left (-\frac {b\,\ln \left (F\right )}{{\left (c+d\,x\right )}^2}\right )}^{7/2}}{105\,d}+\frac {4\,F^a\,F^{\frac {b}{{\left (c+d\,x\right )}^2}}\,b^2\,{\ln \left (F\right )}^2\,{\left (c+d\,x\right )}^3}{105\,d}+\frac {2\,F^a\,F^{\frac {b}{{\left (c+d\,x\right )}^2}}\,b\,\ln \left (F\right )\,{\left (c+d\,x\right )}^5}{35\,d}+\frac {8\,F^a\,F^{\frac {b}{{\left (c+d\,x\right )}^2}}\,b^3\,{\ln \left (F\right )}^3\,\left (c+d\,x\right )}{105\,d}+\frac {8\,F^a\,\sqrt {\pi }\,\mathrm {erfc}\left (\sqrt {-\frac {b\,\ln \left (F\right )}{{\left (c+d\,x\right )}^2}}\right )\,{\left (c+d\,x\right )}^7\,{\left (-\frac {b\,\ln \left (F\right )}{{\left (c+d\,x\right )}^2}\right )}^{7/2}}{105\,d} \]

input
int(F^(a + b/(c + d*x)^2)*(c + d*x)^6,x)
 
output
(F^a*F^(b/(c + d*x)^2)*(c + d*x)^7)/(7*d) - (8*F^a*pi^(1/2)*(c + d*x)^7*(- 
(b*log(F))/(c + d*x)^2)^(7/2))/(105*d) + (4*F^a*F^(b/(c + d*x)^2)*b^2*log( 
F)^2*(c + d*x)^3)/(105*d) + (2*F^a*F^(b/(c + d*x)^2)*b*log(F)*(c + d*x)^5) 
/(35*d) + (8*F^a*F^(b/(c + d*x)^2)*b^3*log(F)^3*(c + d*x))/(105*d) + (8*F^ 
a*pi^(1/2)*erfc((-(b*log(F))/(c + d*x)^2)^(1/2))*(c + d*x)^7*(-(b*log(F))/ 
(c + d*x)^2)^(7/2))/(105*d)