\(\int F^{a+b (c+d x)^2} (c+d x)^6 \, dx\) [270]
Optimal result
Integrand size = 21, antiderivative size = 145 \[
\int F^{a+b (c+d x)^2} (c+d x)^6 \, dx=-\frac {15 F^a \sqrt {\pi } \text {erfi}\left (\sqrt {b} (c+d x) \sqrt {\log (F)}\right )}{16 b^{7/2} d \log ^{\frac {7}{2}}(F)}+\frac {15 F^{a+b (c+d x)^2} (c+d x)}{8 b^3 d \log ^3(F)}-\frac {5 F^{a+b (c+d x)^2} (c+d x)^3}{4 b^2 d \log ^2(F)}+\frac {F^{a+b (c+d x)^2} (c+d x)^5}{2 b d \log (F)}
\]
[Out]
15/8*F^(a+b*(d*x+c)^2)*(d*x+c)/b^3/d/ln(F)^3-5/4*F^(a+b*(d*x+c)^2)*(d*x+c)^3/b^2/d/ln(F)^2+1/2*F^(a+b*(d*x+c)^
2)*(d*x+c)^5/b/d/ln(F)-15/16*F^a*erfi((d*x+c)*b^(1/2)*ln(F)^(1/2))*Pi^(1/2)/b^(7/2)/d/ln(F)^(7/2)
Rubi [A] (verified)
Time = 0.14 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.00, number
of steps used = 4, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2243, 2235}
\[
\int F^{a+b (c+d x)^2} (c+d x)^6 \, dx=-\frac {15 \sqrt {\pi } F^a \text {erfi}\left (\sqrt {b} \sqrt {\log (F)} (c+d x)\right )}{16 b^{7/2} d \log ^{\frac {7}{2}}(F)}+\frac {15 (c+d x) F^{a+b (c+d x)^2}}{8 b^3 d \log ^3(F)}-\frac {5 (c+d x)^3 F^{a+b (c+d x)^2}}{4 b^2 d \log ^2(F)}+\frac {(c+d x)^5 F^{a+b (c+d x)^2}}{2 b d \log (F)}
\]
[In]
Int[F^(a + b*(c + d*x)^2)*(c + d*x)^6,x]
[Out]
(-15*F^a*Sqrt[Pi]*Erfi[Sqrt[b]*(c + d*x)*Sqrt[Log[F]]])/(16*b^(7/2)*d*Log[F]^(7/2)) + (15*F^(a + b*(c + d*x)^2
)*(c + d*x))/(8*b^3*d*Log[F]^3) - (5*F^(a + b*(c + d*x)^2)*(c + d*x)^3)/(4*b^2*d*Log[F]^2) + (F^(a + b*(c + d*
x)^2)*(c + d*x)^5)/(2*b*d*Log[F])
Rule 2235
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 2243
Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(c + d*x)^(m
- n + 1)*(F^(a + b*(c + d*x)^n)/(b*d*n*Log[F])), x] - Dist[(m - n + 1)/(b*n*Log[F]), 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[0, (m + 1)/n, 5] &&
IntegerQ[n] && (LtQ[0, n, m + 1] || LtQ[m, n, 0])
Rubi steps \begin{align*}
\text {integral}& = \frac {F^{a+b (c+d x)^2} (c+d x)^5}{2 b d \log (F)}-\frac {5 \int F^{a+b (c+d x)^2} (c+d x)^4 \, dx}{2 b \log (F)} \\ & = -\frac {5 F^{a+b (c+d x)^2} (c+d x)^3}{4 b^2 d \log ^2(F)}+\frac {F^{a+b (c+d x)^2} (c+d x)^5}{2 b d \log (F)}+\frac {15 \int F^{a+b (c+d x)^2} (c+d x)^2 \, dx}{4 b^2 \log ^2(F)} \\ & = \frac {15 F^{a+b (c+d x)^2} (c+d x)}{8 b^3 d \log ^3(F)}-\frac {5 F^{a+b (c+d x)^2} (c+d x)^3}{4 b^2 d \log ^2(F)}+\frac {F^{a+b (c+d x)^2} (c+d x)^5}{2 b d \log (F)}-\frac {15 \int F^{a+b (c+d x)^2} \, dx}{8 b^3 \log ^3(F)} \\ & = -\frac {15 F^a \sqrt {\pi } \text {erfi}\left (\sqrt {b} (c+d x) \sqrt {\log (F)}\right )}{16 b^{7/2} d \log ^{\frac {7}{2}}(F)}+\frac {15 F^{a+b (c+d x)^2} (c+d x)}{8 b^3 d \log ^3(F)}-\frac {5 F^{a+b (c+d x)^2} (c+d x)^3}{4 b^2 d \log ^2(F)}+\frac {F^{a+b (c+d x)^2} (c+d x)^5}{2 b d \log (F)} \\
\end{align*}
Mathematica [A] (verified)
Time = 0.30 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.87
\[
\int F^{a+b (c+d x)^2} (c+d x)^6 \, dx=\frac {F^a \left (8 F^{b (c+d x)^2} (c+d x)^5-\frac {15 \sqrt {\pi } \text {erfi}\left (\sqrt {b} (c+d x) \sqrt {\log (F)}\right )}{b^{5/2} \log ^{\frac {5}{2}}(F)}+\frac {30 F^{b (c+d x)^2} (c+d x)}{b^2 \log ^2(F)}-\frac {20 F^{b (c+d x)^2} (c+d x)^3}{b \log (F)}\right )}{16 b d \log (F)}
\]
[In]
Integrate[F^(a + b*(c + d*x)^2)*(c + d*x)^6,x]
[Out]
(F^a*(8*F^(b*(c + d*x)^2)*(c + d*x)^5 - (15*Sqrt[Pi]*Erfi[Sqrt[b]*(c + d*x)*Sqrt[Log[F]]])/(b^(5/2)*Log[F]^(5/
2)) + (30*F^(b*(c + d*x)^2)*(c + d*x))/(b^2*Log[F]^2) - (20*F^(b*(c + d*x)^2)*(c + d*x)^3)/(b*Log[F])))/(16*b*
d*Log[F])
Maple [B] (verified)
Leaf count of result is larger than twice the leaf count of optimal. \(575\) vs. \(2(127)=254\).
Time = 0.47 (sec) , antiderivative size = 576, normalized size of antiderivative = 3.97
| | |
method | result | size |
| | |
risch |
\(\frac {5 F^{b \,c^{2}} F^{a} d^{3} c \,x^{4} F^{b \,d^{2} x^{2}} F^{2 b c d x}}{2 \ln \left (F \right ) b}+\frac {5 F^{b \,c^{2}} F^{a} d^{2} c^{2} x^{3} F^{b \,d^{2} x^{2}} F^{2 b c d x}}{\ln \left (F \right ) b}+\frac {5 F^{b \,c^{2}} F^{a} d \,c^{3} x^{2} F^{b \,d^{2} x^{2}} F^{2 b c d x}}{\ln \left (F \right ) b}-\frac {15 F^{b \,c^{2}} F^{a} d c \,x^{2} F^{b \,d^{2} x^{2}} F^{2 b c d x}}{4 \ln \left (F \right )^{2} b^{2}}+\frac {15 F^{b \,c^{2}} F^{a} \sqrt {\pi }\, F^{-b \,c^{2}} \operatorname {erf}\left (-d \sqrt {-b \ln \left (F \right )}\, x +\frac {b c \ln \left (F \right )}{\sqrt {-b \ln \left (F \right )}}\right )}{16 d \ln \left (F \right )^{3} b^{3} \sqrt {-b \ln \left (F \right )}}+\frac {F^{b \,c^{2}} F^{a} d^{4} x^{5} F^{b \,d^{2} x^{2}} F^{2 b c d x}}{2 \ln \left (F \right ) b}+\frac {5 F^{b \,c^{2}} F^{a} c^{4} x \,F^{b \,d^{2} x^{2}} F^{2 b c d x}}{2 \ln \left (F \right ) b}+\frac {F^{b \,c^{2}} F^{a} c^{5} F^{b \,d^{2} x^{2}} F^{2 b c d x}}{2 d \ln \left (F \right ) b}-\frac {5 F^{b \,c^{2}} F^{a} c^{3} F^{b \,d^{2} x^{2}} F^{2 b c d x}}{4 d \ln \left (F \right )^{2} b^{2}}-\frac {15 F^{b \,c^{2}} F^{a} c^{2} x \,F^{b \,d^{2} x^{2}} F^{2 b c d x}}{4 \ln \left (F \right )^{2} b^{2}}+\frac {15 F^{b \,c^{2}} F^{a} c \,F^{b \,d^{2} x^{2}} F^{2 b c d x}}{8 d \ln \left (F \right )^{3} b^{3}}-\frac {5 F^{b \,c^{2}} F^{a} d^{2} x^{3} F^{b \,d^{2} x^{2}} F^{2 b c d x}}{4 \ln \left (F \right )^{2} b^{2}}+\frac {15 F^{b \,c^{2}} F^{a} x \,F^{b \,d^{2} x^{2}} F^{2 b c d x}}{8 \ln \left (F \right )^{3} b^{3}}\) |
\(576\) |
| | |
|
|
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[In]
int(F^(a+b*(d*x+c)^2)*(d*x+c)^6,x,method=_RETURNVERBOSE)
[Out]
5/2*F^(b*c^2)*F^a*d^3*c/ln(F)/b*x^4*F^(b*d^2*x^2)*F^(2*b*c*d*x)+5*F^(b*c^2)*F^a*d^2*c^2/ln(F)/b*x^3*F^(b*d^2*x
^2)*F^(2*b*c*d*x)+5*F^(b*c^2)*F^a*d*c^3/ln(F)/b*x^2*F^(b*d^2*x^2)*F^(2*b*c*d*x)-15/4*F^(b*c^2)*F^a*d*c/ln(F)^2
/b^2*x^2*F^(b*d^2*x^2)*F^(2*b*c*d*x)+15/16*F^(b*c^2)*F^a/d/ln(F)^3/b^3*Pi^(1/2)*F^(-b*c^2)/(-b*ln(F))^(1/2)*er
f(-d*(-b*ln(F))^(1/2)*x+b*c*ln(F)/(-b*ln(F))^(1/2))+1/2*F^(b*c^2)*F^a*d^4/ln(F)/b*x^5*F^(b*d^2*x^2)*F^(2*b*c*d
*x)+5/2*F^(b*c^2)*F^a*c^4/ln(F)/b*x*F^(b*d^2*x^2)*F^(2*b*c*d*x)+1/2*F^(b*c^2)*F^a/d*c^5/ln(F)/b*F^(b*d^2*x^2)*
F^(2*b*c*d*x)-5/4*F^(b*c^2)*F^a/d*c^3/ln(F)^2/b^2*F^(b*d^2*x^2)*F^(2*b*c*d*x)-15/4*F^(b*c^2)*F^a*c^2/ln(F)^2/b
^2*x*F^(b*d^2*x^2)*F^(2*b*c*d*x)+15/8*F^(b*c^2)*F^a/d*c/ln(F)^3/b^3*F^(b*d^2*x^2)*F^(2*b*c*d*x)-5/4*F^(b*c^2)*
F^a*d^2/ln(F)^2/b^2*x^3*F^(b*d^2*x^2)*F^(2*b*c*d*x)+15/8*F^(b*c^2)*F^a/ln(F)^3/b^3*x*F^(b*d^2*x^2)*F^(2*b*c*d*
x)
Fricas [A] (verification not implemented)
none
Time = 0.28 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.50
\[
\int F^{a+b (c+d x)^2} (c+d x)^6 \, dx=\frac {15 \, \sqrt {\pi } \sqrt {-b d^{2} \log \left (F\right )} F^{a} \operatorname {erf}\left (\frac {\sqrt {-b d^{2} \log \left (F\right )} {\left (d x + c\right )}}{d}\right ) + 2 \, {\left (4 \, {\left (b^{3} d^{6} x^{5} + 5 \, b^{3} c d^{5} x^{4} + 10 \, b^{3} c^{2} d^{4} x^{3} + 10 \, b^{3} c^{3} d^{3} x^{2} + 5 \, b^{3} c^{4} d^{2} x + b^{3} c^{5} d\right )} \log \left (F\right )^{3} - 10 \, {\left (b^{2} d^{4} x^{3} + 3 \, b^{2} c d^{3} x^{2} + 3 \, b^{2} c^{2} d^{2} x + b^{2} c^{3} d\right )} \log \left (F\right )^{2} + 15 \, {\left (b d^{2} x + b c d\right )} \log \left (F\right )\right )} F^{b d^{2} x^{2} + 2 \, b c d x + b c^{2} + a}}{16 \, b^{4} d^{2} \log \left (F\right )^{4}}
\]
[In]
integrate(F^(a+b*(d*x+c)^2)*(d*x+c)^6,x, algorithm="fricas")
[Out]
1/16*(15*sqrt(pi)*sqrt(-b*d^2*log(F))*F^a*erf(sqrt(-b*d^2*log(F))*(d*x + c)/d) + 2*(4*(b^3*d^6*x^5 + 5*b^3*c*d
^5*x^4 + 10*b^3*c^2*d^4*x^3 + 10*b^3*c^3*d^3*x^2 + 5*b^3*c^4*d^2*x + b^3*c^5*d)*log(F)^3 - 10*(b^2*d^4*x^3 + 3
*b^2*c*d^3*x^2 + 3*b^2*c^2*d^2*x + b^2*c^3*d)*log(F)^2 + 15*(b*d^2*x + b*c*d)*log(F))*F^(b*d^2*x^2 + 2*b*c*d*x
+ b*c^2 + a))/(b^4*d^2*log(F)^4)
Sympy [F]
\[
\int F^{a+b (c+d x)^2} (c+d x)^6 \, dx=\int F^{a + b \left (c + d x\right )^{2}} \left (c + d x\right )^{6}\, dx
\]
[In]
integrate(F**(a+b*(d*x+c)**2)*(d*x+c)**6,x)
[Out]
Integral(F**(a + b*(c + d*x)**2)*(c + d*x)**6, x)
Maxima [B] (verification not implemented)
Leaf count of result is larger than twice the leaf count of optimal. 1922 vs. \(2 (127) = 254\).
Time = 0.85 (sec) , antiderivative size = 1922, normalized size of antiderivative = 13.26
\[
\int F^{a+b (c+d x)^2} (c+d x)^6 \, dx=\text {Too large to display}
\]
[In]
integrate(F^(a+b*(d*x+c)^2)*(d*x+c)^6,x, algorithm="maxima")
[Out]
-3*(sqrt(pi)*(b*d^2*x + b*c*d)*b*c*(erf(sqrt(-(b*d^2*x + b*c*d)^2*log(F)/(b*d^2))) - 1)*log(F)^2/((b*log(F))^(
3/2)*d^2*sqrt(-(b*d^2*x + b*c*d)^2*log(F)/(b*d^2))) - F^((b*d^2*x + b*c*d)^2/(b*d^2))*b*log(F)/((b*log(F))^(3/
2)*d))*F^a*c^5/sqrt(b*log(F)) + 15/2*(sqrt(pi)*(b*d^2*x + b*c*d)*b^2*c^2*(erf(sqrt(-(b*d^2*x + b*c*d)^2*log(F)
/(b*d^2))) - 1)*log(F)^3/((b*log(F))^(5/2)*d^3*sqrt(-(b*d^2*x + b*c*d)^2*log(F)/(b*d^2))) - 2*F^((b*d^2*x + b*
c*d)^2/(b*d^2))*b^2*c*log(F)^2/((b*log(F))^(5/2)*d^2) - (b*d^2*x + b*c*d)^3*gamma(3/2, -(b*d^2*x + b*c*d)^2*lo
g(F)/(b*d^2))*log(F)^3/((b*log(F))^(5/2)*d^5*(-(b*d^2*x + b*c*d)^2*log(F)/(b*d^2))^(3/2)))*F^a*c^4*d/sqrt(b*lo
g(F)) - 10*(sqrt(pi)*(b*d^2*x + b*c*d)*b^3*c^3*(erf(sqrt(-(b*d^2*x + b*c*d)^2*log(F)/(b*d^2))) - 1)*log(F)^4/(
(b*log(F))^(7/2)*d^4*sqrt(-(b*d^2*x + b*c*d)^2*log(F)/(b*d^2))) - 3*F^((b*d^2*x + b*c*d)^2/(b*d^2))*b^3*c^2*lo
g(F)^3/((b*log(F))^(7/2)*d^3) - 3*(b*d^2*x + b*c*d)^3*b*c*gamma(3/2, -(b*d^2*x + b*c*d)^2*log(F)/(b*d^2))*log(
F)^4/((b*log(F))^(7/2)*d^6*(-(b*d^2*x + b*c*d)^2*log(F)/(b*d^2))^(3/2)) + b^2*gamma(2, -(b*d^2*x + b*c*d)^2*lo
g(F)/(b*d^2))*log(F)^2/((b*log(F))^(7/2)*d^3))*F^a*c^3*d^2/sqrt(b*log(F)) + 15/2*(sqrt(pi)*(b*d^2*x + b*c*d)*b
^4*c^4*(erf(sqrt(-(b*d^2*x + b*c*d)^2*log(F)/(b*d^2))) - 1)*log(F)^5/((b*log(F))^(9/2)*d^5*sqrt(-(b*d^2*x + b*
c*d)^2*log(F)/(b*d^2))) - 4*F^((b*d^2*x + b*c*d)^2/(b*d^2))*b^4*c^3*log(F)^4/((b*log(F))^(9/2)*d^4) - 6*(b*d^2
*x + b*c*d)^3*b^2*c^2*gamma(3/2, -(b*d^2*x + b*c*d)^2*log(F)/(b*d^2))*log(F)^5/((b*log(F))^(9/2)*d^7*(-(b*d^2*
x + b*c*d)^2*log(F)/(b*d^2))^(3/2)) + 4*b^3*c*gamma(2, -(b*d^2*x + b*c*d)^2*log(F)/(b*d^2))*log(F)^3/((b*log(F
))^(9/2)*d^4) - (b*d^2*x + b*c*d)^5*gamma(5/2, -(b*d^2*x + b*c*d)^2*log(F)/(b*d^2))*log(F)^5/((b*log(F))^(9/2)
*d^9*(-(b*d^2*x + b*c*d)^2*log(F)/(b*d^2))^(5/2)))*F^a*c^2*d^3/sqrt(b*log(F)) - 3*(sqrt(pi)*(b*d^2*x + b*c*d)*
b^5*c^5*(erf(sqrt(-(b*d^2*x + b*c*d)^2*log(F)/(b*d^2))) - 1)*log(F)^6/((b*log(F))^(11/2)*d^6*sqrt(-(b*d^2*x +
b*c*d)^2*log(F)/(b*d^2))) - 5*F^((b*d^2*x + b*c*d)^2/(b*d^2))*b^5*c^4*log(F)^5/((b*log(F))^(11/2)*d^5) - 10*(b
*d^2*x + b*c*d)^3*b^3*c^3*gamma(3/2, -(b*d^2*x + b*c*d)^2*log(F)/(b*d^2))*log(F)^6/((b*log(F))^(11/2)*d^8*(-(b
*d^2*x + b*c*d)^2*log(F)/(b*d^2))^(3/2)) + 10*b^4*c^2*gamma(2, -(b*d^2*x + b*c*d)^2*log(F)/(b*d^2))*log(F)^4/(
(b*log(F))^(11/2)*d^5) - b^3*gamma(3, -(b*d^2*x + b*c*d)^2*log(F)/(b*d^2))*log(F)^3/((b*log(F))^(11/2)*d^5) -
5*(b*d^2*x + b*c*d)^5*b*c*gamma(5/2, -(b*d^2*x + b*c*d)^2*log(F)/(b*d^2))*log(F)^6/((b*log(F))^(11/2)*d^10*(-(
b*d^2*x + b*c*d)^2*log(F)/(b*d^2))^(5/2)))*F^a*c*d^4/sqrt(b*log(F)) + 1/2*(sqrt(pi)*(b*d^2*x + b*c*d)*b^6*c^6*
(erf(sqrt(-(b*d^2*x + b*c*d)^2*log(F)/(b*d^2))) - 1)*log(F)^7/((b*log(F))^(13/2)*d^7*sqrt(-(b*d^2*x + b*c*d)^2
*log(F)/(b*d^2))) - 6*F^((b*d^2*x + b*c*d)^2/(b*d^2))*b^6*c^5*log(F)^6/((b*log(F))^(13/2)*d^6) - 15*(b*d^2*x +
b*c*d)^3*b^4*c^4*gamma(3/2, -(b*d^2*x + b*c*d)^2*log(F)/(b*d^2))*log(F)^7/((b*log(F))^(13/2)*d^9*(-(b*d^2*x +
b*c*d)^2*log(F)/(b*d^2))^(3/2)) + 20*b^5*c^3*gamma(2, -(b*d^2*x + b*c*d)^2*log(F)/(b*d^2))*log(F)^5/((b*log(F
))^(13/2)*d^6) - 6*b^4*c*gamma(3, -(b*d^2*x + b*c*d)^2*log(F)/(b*d^2))*log(F)^4/((b*log(F))^(13/2)*d^6) - 15*(
b*d^2*x + b*c*d)^5*b^2*c^2*gamma(5/2, -(b*d^2*x + b*c*d)^2*log(F)/(b*d^2))*log(F)^7/((b*log(F))^(13/2)*d^11*(-
(b*d^2*x + b*c*d)^2*log(F)/(b*d^2))^(5/2)) - (b*d^2*x + b*c*d)^7*gamma(7/2, -(b*d^2*x + b*c*d)^2*log(F)/(b*d^2
))*log(F)^7/((b*log(F))^(13/2)*d^13*(-(b*d^2*x + b*c*d)^2*log(F)/(b*d^2))^(7/2)))*F^a*d^5/sqrt(b*log(F)) + 1/2
*sqrt(pi)*F^(b*c^2 + a)*c^6*erf(sqrt(-b*log(F))*d*x - b*c*log(F)/sqrt(-b*log(F)))/(sqrt(-b*log(F))*F^(b*c^2)*d
)
Giac [A] (verification not implemented)
none
Time = 0.32 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.91
\[
\int F^{a+b (c+d x)^2} (c+d x)^6 \, dx=\frac {{\left (4 \, b^{2} d^{4} {\left (x + \frac {c}{d}\right )}^{5} \log \left (F\right )^{2} - 10 \, b d^{2} {\left (x + \frac {c}{d}\right )}^{3} \log \left (F\right ) + 15 \, x + \frac {15 \, c}{d}\right )} e^{\left (b d^{2} x^{2} \log \left (F\right ) + 2 \, b c d x \log \left (F\right ) + b c^{2} \log \left (F\right ) + a \log \left (F\right )\right )}}{8 \, b^{3} \log \left (F\right )^{3}} + \frac {15 \, \sqrt {\pi } F^{a} \operatorname {erf}\left (-\sqrt {-b \log \left (F\right )} d {\left (x + \frac {c}{d}\right )}\right )}{16 \, \sqrt {-b \log \left (F\right )} b^{3} d \log \left (F\right )^{3}}
\]
[In]
integrate(F^(a+b*(d*x+c)^2)*(d*x+c)^6,x, algorithm="giac")
[Out]
1/8*(4*b^2*d^4*(x + c/d)^5*log(F)^2 - 10*b*d^2*(x + c/d)^3*log(F) + 15*x + 15*c/d)*e^(b*d^2*x^2*log(F) + 2*b*c
*d*x*log(F) + b*c^2*log(F) + a*log(F))/(b^3*log(F)^3) + 15/16*sqrt(pi)*F^a*erf(-sqrt(-b*log(F))*d*(x + c/d))/(
sqrt(-b*log(F))*b^3*d*log(F)^3)
Mupad [B] (verification not implemented)
Time = 0.46 (sec) , antiderivative size = 378, normalized size of antiderivative = 2.61
\[
\int F^{a+b (c+d x)^2} (c+d x)^6 \, dx=F^{b\,d^2\,x^2}\,F^a\,F^{b\,c^2}\,F^{2\,b\,c\,d\,x}\,\left (\frac {15\,c}{8\,b^3\,d\,{\ln \left (F\right )}^3}+\frac {c^5}{2\,b\,d\,\ln \left (F\right )}-\frac {5\,c^3}{4\,b^2\,d\,{\ln \left (F\right )}^2}\right )-\frac {15\,F^a\,\sqrt {\pi }\,\mathrm {erfi}\left (\frac {b\,x\,\ln \left (F\right )\,d^2+b\,c\,\ln \left (F\right )\,d}{\sqrt {b\,d^2\,\ln \left (F\right )}}\right )}{16\,b^3\,{\ln \left (F\right )}^3\,\sqrt {b\,d^2\,\ln \left (F\right )}}-\frac {5\,F^{b\,d^2\,x^2}\,F^a\,F^{b\,c^2}\,F^{2\,b\,c\,d\,x}\,x^2\,\left (3\,c\,d-4\,b\,c^3\,d\,\ln \left (F\right )\right )}{4\,b^2\,{\ln \left (F\right )}^2}+\frac {5\,F^{b\,d^2\,x^2}\,F^a\,F^{b\,c^2}\,F^{2\,b\,c\,d\,x}\,x\,\left (4\,b^2\,c^4\,{\ln \left (F\right )}^2-6\,b\,c^2\,\ln \left (F\right )+3\right )}{8\,b^3\,{\ln \left (F\right )}^3}+\frac {F^{b\,d^2\,x^2}\,F^a\,F^{b\,c^2}\,F^{2\,b\,c\,d\,x}\,d^4\,x^5}{2\,b\,\ln \left (F\right )}+\frac {5\,F^{b\,d^2\,x^2}\,F^a\,F^{b\,c^2}\,F^{2\,b\,c\,d\,x}\,c\,d^3\,x^4}{2\,b\,\ln \left (F\right )}+\frac {5\,F^{b\,d^2\,x^2}\,F^a\,F^{b\,c^2}\,F^{2\,b\,c\,d\,x}\,d^2\,x^3\,\left (4\,b\,c^2\,\ln \left (F\right )-1\right )}{4\,b^2\,{\ln \left (F\right )}^2}
\]
[In]
int(F^(a + b*(c + d*x)^2)*(c + d*x)^6,x)
[Out]
F^(b*d^2*x^2)*F^a*F^(b*c^2)*F^(2*b*c*d*x)*((15*c)/(8*b^3*d*log(F)^3) + c^5/(2*b*d*log(F)) - (5*c^3)/(4*b^2*d*l
og(F)^2)) - (15*F^a*pi^(1/2)*erfi((b*c*d*log(F) + b*d^2*x*log(F))/(b*d^2*log(F))^(1/2)))/(16*b^3*log(F)^3*(b*d
^2*log(F))^(1/2)) - (5*F^(b*d^2*x^2)*F^a*F^(b*c^2)*F^(2*b*c*d*x)*x^2*(3*c*d - 4*b*c^3*d*log(F)))/(4*b^2*log(F)
^2) + (5*F^(b*d^2*x^2)*F^a*F^(b*c^2)*F^(2*b*c*d*x)*x*(4*b^2*c^4*log(F)^2 - 6*b*c^2*log(F) + 3))/(8*b^3*log(F)^
3) + (F^(b*d^2*x^2)*F^a*F^(b*c^2)*F^(2*b*c*d*x)*d^4*x^5)/(2*b*log(F)) + (5*F^(b*d^2*x^2)*F^a*F^(b*c^2)*F^(2*b*
c*d*x)*c*d^3*x^4)/(2*b*log(F)) + (5*F^(b*d^2*x^2)*F^a*F^(b*c^2)*F^(2*b*c*d*x)*d^2*x^3*(4*b*c^2*log(F) - 1))/(4
*b^2*log(F)^2)