3.4.83 \(\int F^{a+b (c+d x)^2} (e+f x)^4 \, dx\) [383]

3.4.83.1 Optimal result

Integrand size = 21, antiderivative size = 389 \[ \int F^{a+b (c+d x)^2} (e+f x)^4 \, dx=\frac {3 f^4 F^a \sqrt {\pi } \text {erfi}\left (\sqrt {b} (c+d x) \sqrt {\log (F)}\right )}{8 b^{5/2} d^5 \log ^{\frac {5}{2}}(F)}-\frac {2 f^3 (d e-c f) F^{a+b (c+d x)^2}}{b^2 d^5 \log ^2(F)}-\frac {3 f^4 F^{a+b (c+d x)^2} (c+d x)}{4 b^2 d^5 \log ^2(F)}-\frac {3 f^2 (d e-c f)^2 F^a \sqrt {\pi } \text {erfi}\left (\sqrt {b} (c+d x) \sqrt {\log (F)}\right )}{2 b^{3/2} d^5 \log ^{\frac {3}{2}}(F)}+\frac {2 f (d e-c f)^3 F^{a+b (c+d x)^2}}{b d^5 \log (F)}+\frac {3 f^2 (d e-c f)^2 F^{a+b (c+d x)^2} (c+d x)}{b d^5 \log (F)}+\frac {2 f^3 (d e-c f) F^{a+b (c+d x)^2} (c+d x)^2}{b d^5 \log (F)}+\frac {f^4 F^{a+b (c+d x)^2} (c+d x)^3}{2 b d^5 \log (F)}+\frac {(d e-c f)^4 F^a \sqrt {\pi } \text {erfi}\left (\sqrt {b} (c+d x) \sqrt {\log (F)}\right )}{2 \sqrt {b} d^5 \sqrt {\log (F)}} \]

output

-2*f^3*(-c*f+d*e)*F^(a+b*(d*x+c)^2)/b^2/d^5/ln(F)^2-3/4*f^4*F^(a+b*(d*x+c) 
^2)*(d*x+c)/b^2/d^5/ln(F)^2+2*f*(-c*f+d*e)^3*F^(a+b*(d*x+c)^2)/b/d^5/ln(F) 
+3*f^2*(-c*f+d*e)^2*F^(a+b*(d*x+c)^2)*(d*x+c)/b/d^5/ln(F)+2*f^3*(-c*f+d*e) 
*F^(a+b*(d*x+c)^2)*(d*x+c)^2/b/d^5/ln(F)+1/2*f^4*F^(a+b*(d*x+c)^2)*(d*x+c) 
^3/b/d^5/ln(F)+3/8*f^4*F^a*erfi((d*x+c)*b^(1/2)*ln(F)^(1/2))*Pi^(1/2)/b^(5 
/2)/d^5/ln(F)^(5/2)-3/2*f^2*(-c*f+d*e)^2*F^a*erfi((d*x+c)*b^(1/2)*ln(F)^(1 
/2))*Pi^(1/2)/b^(3/2)/d^5/ln(F)^(3/2)+1/2*(-c*f+d*e)^4*F^a*erfi((d*x+c)*b^ 
(1/2)*ln(F)^(1/2))*Pi^(1/2)/d^5/b^(1/2)/ln(F)^(1/2)
 

3.4.83.2 Mathematica [A] (verified)

Time = 1.21 (sec) , antiderivative size = 220, normalized size of antiderivative = 0.57 \[ \int F^{a+b (c+d x)^2} (e+f x)^4 \, dx=\frac {F^a \left (2 \sqrt {b} f F^{b (c+d x)^2} \sqrt {\log (F)} \left (f^2 (-8 d e+5 c f-3 d f x)+2 b \left (-c^3 f^3+c^2 d f^2 (4 e+f x)-c d^2 f \left (6 e^2+4 e f x+f^2 x^2\right )+d^3 \left (4 e^3+6 e^2 f x+4 e f^2 x^2+f^3 x^3\right )\right ) \log (F)\right )+\sqrt {\pi } \text {erfi}\left (\sqrt {b} (c+d x) \sqrt {\log (F)}\right ) \left (3 f^4-12 b f^2 (d e-c f)^2 \log (F)+4 b^2 (d e-c f)^4 \log ^2(F)\right )\right )}{8 b^{5/2} d^5 \log ^{\frac {5}{2}}(F)} \]

input

Integrate[F^(a + b*(c + d*x)^2)*(e + f*x)^4,x]
 

output

(F^a*(2*Sqrt[b]*f*F^(b*(c + d*x)^2)*Sqrt[Log[F]]*(f^2*(-8*d*e + 5*c*f - 3* 
d*f*x) + 2*b*(-(c^3*f^3) + c^2*d*f^2*(4*e + f*x) - c*d^2*f*(6*e^2 + 4*e*f* 
x + f^2*x^2) + d^3*(4*e^3 + 6*e^2*f*x + 4*e*f^2*x^2 + f^3*x^3))*Log[F]) + 
Sqrt[Pi]*Erfi[Sqrt[b]*(c + d*x)*Sqrt[Log[F]]]*(3*f^4 - 12*b*f^2*(d*e - c*f 
)^2*Log[F] + 4*b^2*(d*e - c*f)^4*Log[F]^2)))/(8*b^(5/2)*d^5*Log[F]^(5/2))
 

3.4.83.3 Rubi [A] (verified)

Time = 0.85 (sec) , antiderivative size = 389, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2656, 2009}

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.

\(\displaystyle \int (e+f x)^4 F^{a+b (c+d x)^2} \, dx\)

\(\Big \downarrow \) 2656

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

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {3 \sqrt {\pi } f^2 F^a (d e-c f)^2 \text {erfi}\left (\sqrt {b} \sqrt {\log (F)} (c+d x)\right )}{2 b^{3/2} d^5 \log ^{\frac {3}{2}}(F)}+\frac {3 \sqrt {\pi } f^4 F^a \text {erfi}\left (\sqrt {b} \sqrt {\log (F)} (c+d x)\right )}{8 b^{5/2} d^5 \log ^{\frac {5}{2}}(F)}-\frac {2 f^3 (d e-c f) F^{a+b (c+d x)^2}}{b^2 d^5 \log ^2(F)}-\frac {3 f^4 (c+d x) F^{a+b (c+d x)^2}}{4 b^2 d^5 \log ^2(F)}+\frac {\sqrt {\pi } F^a (d e-c f)^4 \text {erfi}\left (\sqrt {b} \sqrt {\log (F)} (c+d x)\right )}{2 \sqrt {b} d^5 \sqrt {\log (F)}}+\frac {2 f^3 (c+d x)^2 (d e-c f) F^{a+b (c+d x)^2}}{b d^5 \log (F)}+\frac {3 f^2 (c+d x) (d e-c f)^2 F^{a+b (c+d x)^2}}{b d^5 \log (F)}+\frac {2 f (d e-c f)^3 F^{a+b (c+d x)^2}}{b d^5 \log (F)}+\frac {f^4 (c+d x)^3 F^{a+b (c+d x)^2}}{2 b d^5 \log (F)}\)

input

Int[F^(a + b*(c + d*x)^2)*(e + f*x)^4,x]
 

output

(3*f^4*F^a*Sqrt[Pi]*Erfi[Sqrt[b]*(c + d*x)*Sqrt[Log[F]]])/(8*b^(5/2)*d^5*L 
og[F]^(5/2)) - (2*f^3*(d*e - c*f)*F^(a + b*(c + d*x)^2))/(b^2*d^5*Log[F]^2 
) - (3*f^4*F^(a + b*(c + d*x)^2)*(c + d*x))/(4*b^2*d^5*Log[F]^2) - (3*f^2* 
(d*e - c*f)^2*F^a*Sqrt[Pi]*Erfi[Sqrt[b]*(c + d*x)*Sqrt[Log[F]]])/(2*b^(3/2 
)*d^5*Log[F]^(3/2)) + (2*f*(d*e - c*f)^3*F^(a + b*(c + d*x)^2))/(b*d^5*Log 
[F]) + (3*f^2*(d*e - c*f)^2*F^(a + b*(c + d*x)^2)*(c + d*x))/(b*d^5*Log[F] 
) + (2*f^3*(d*e - c*f)*F^(a + b*(c + d*x)^2)*(c + d*x)^2)/(b*d^5*Log[F]) + 
 (f^4*F^(a + b*(c + d*x)^2)*(c + d*x)^3)/(2*b*d^5*Log[F]) + ((d*e - c*f)^4 
*F^a*Sqrt[Pi]*Erfi[Sqrt[b]*(c + d*x)*Sqrt[Log[F]]])/(2*Sqrt[b]*d^5*Sqrt[Lo 
g[F]])
 

3.4.83.3.1 Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2656

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*(Px_), x_Symbol] :> Int[ 
ExpandLinearProduct[F^(a + b*(c + d*x)^n), Px, c, d, x], x] /; FreeQ[{F, a, 
 b, c, d, n}, x] && PolynomialQ[Px, x]
 

3.4.83.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1197\) vs. \(2(349)=698\).

Time = 0.32 (sec) , antiderivative size = 1198, normalized size of antiderivative = 3.08

method	result	size
risch	\(\text {Expression too large to display}\)	\(1198\)

input

int(F^(a+b*(d*x+c)^2)*(f*x+e)^4,x,method=_RETURNVERBOSE)
 

output

-1/2*F^(b*c^2)*F^a*e^4*Pi^(1/2)*F^(-b*c^2)/d/(-b*ln(F))^(1/2)*erf(-d*(-b*l 
n(F))^(1/2)*x+b*c*ln(F)/(-b*ln(F))^(1/2))+1/2*f^4*F^a*F^(b*c^2)/ln(F)/b/d^ 
2*x^3*F^(b*d^2*x^2)*F^(2*b*c*d*x)-1/2*f^4*F^a*F^(b*c^2)/d^3*c/ln(F)/b*x^2* 
F^(b*d^2*x^2)*F^(2*b*c*d*x)+1/2*f^4*F^a*F^(b*c^2)/d^4*c^2/ln(F)/b*x*F^(b*d 
^2*x^2)*F^(2*b*c*d*x)-1/2*f^4*F^a*F^(b*c^2)/d^5*c^3/ln(F)/b*F^(b*d^2*x^2)* 
F^(2*b*c*d*x)-1/2*f^4*F^a*F^(b*c^2)/d^5*c^4*Pi^(1/2)*F^(-b*c^2)/(-b*ln(F)) 
^(1/2)*erf(-d*(-b*ln(F))^(1/2)*x+b*c*ln(F)/(-b*ln(F))^(1/2))+3/2*f^4*F^a*F 
^(b*c^2)/d^5*c^2/ln(F)/b*Pi^(1/2)*F^(-b*c^2)/(-b*ln(F))^(1/2)*erf(-d*(-b*l 
n(F))^(1/2)*x+b*c*ln(F)/(-b*ln(F))^(1/2))+5/4*f^4*F^a*F^(b*c^2)/d^5*c/b^2/ 
ln(F)^2*F^(b*d^2*x^2)*F^(2*b*c*d*x)-3/4*f^4*F^a*F^(b*c^2)/d^4/b^2/ln(F)^2* 
x*F^(b*d^2*x^2)*F^(2*b*c*d*x)-3/8*f^4*F^a*F^(b*c^2)/d^5/b^2/ln(F)^2*Pi^(1/ 
2)*F^(-b*c^2)/(-b*ln(F))^(1/2)*erf(-d*(-b*ln(F))^(1/2)*x+b*c*ln(F)/(-b*ln( 
F))^(1/2))+2*f^3*e*F^a*F^(b*c^2)/ln(F)/b/d^2*x^2*F^(b*d^2*x^2)*F^(2*b*c*d* 
x)-2*f^3*e*F^a*F^(b*c^2)/d^3*c/ln(F)/b*x*F^(b*d^2*x^2)*F^(2*b*c*d*x)+2*f^3 
*e*F^a*F^(b*c^2)/d^4*c^2/ln(F)/b*F^(b*d^2*x^2)*F^(2*b*c*d*x)+2*f^3*e*F^a*F 
^(b*c^2)/d^4*c^3*Pi^(1/2)*F^(-b*c^2)/(-b*ln(F))^(1/2)*erf(-d*(-b*ln(F))^(1 
/2)*x+b*c*ln(F)/(-b*ln(F))^(1/2))-3*f^3*e*F^a*F^(b*c^2)/d^4*c/ln(F)/b*Pi^( 
1/2)*F^(-b*c^2)/(-b*ln(F))^(1/2)*erf(-d*(-b*ln(F))^(1/2)*x+b*c*ln(F)/(-b*l 
n(F))^(1/2))-2*f^3*e*F^a*F^(b*c^2)/d^4/b^2/ln(F)^2*F^(b*d^2*x^2)*F^(2*b*c* 
d*x)+3*f^2*e^2*F^a*F^(b*c^2)/ln(F)/b/d^2*x*F^(b*d^2*x^2)*F^(2*b*c*d*x)-...
 

3.4.83.5 Fricas [A] (verification not implemented)

Time = 0.32 (sec) , antiderivative size = 364, normalized size of antiderivative = 0.94 \[ \int F^{a+b (c+d x)^2} (e+f x)^4 \, dx=-\frac {\sqrt {\pi } {\left (3 \, f^{4} + 4 \, {\left (b^{2} d^{4} e^{4} - 4 \, b^{2} c d^{3} e^{3} f + 6 \, b^{2} c^{2} d^{2} e^{2} f^{2} - 4 \, b^{2} c^{3} d e f^{3} + b^{2} c^{4} f^{4}\right )} \log \left (F\right )^{2} - 12 \, {\left (b d^{2} e^{2} f^{2} - 2 \, b c d e f^{3} + b c^{2} f^{4}\right )} \log \left (F\right )\right )} \sqrt {-b d^{2} \log \left (F\right )} F^{a} \operatorname {erf}\left (\frac {\sqrt {-b d^{2} \log \left (F\right )} {\left (d x + c\right )}}{d}\right ) - 2 \, {\left (2 \, {\left (b^{2} d^{4} f^{4} x^{3} + 4 \, b^{2} d^{4} e^{3} f - 6 \, b^{2} c d^{3} e^{2} f^{2} + 4 \, b^{2} c^{2} d^{2} e f^{3} - b^{2} c^{3} d f^{4} + {\left (4 \, b^{2} d^{4} e f^{3} - b^{2} c d^{3} f^{4}\right )} x^{2} + {\left (6 \, b^{2} d^{4} e^{2} f^{2} - 4 \, b^{2} c d^{3} e f^{3} + b^{2} c^{2} d^{2} f^{4}\right )} x\right )} \log \left (F\right )^{2} - {\left (3 \, b d^{2} f^{4} x + 8 \, b d^{2} e f^{3} - 5 \, b c d f^{4}\right )} \log \left (F\right )\right )} F^{b d^{2} x^{2} + 2 \, b c d x + b c^{2} + a}}{8 \, b^{3} d^{6} \log \left (F\right )^{3}} \]

input

integrate(F^(a+b*(d*x+c)^2)*(f*x+e)^4,x, algorithm="fricas")
 

output

-1/8*(sqrt(pi)*(3*f^4 + 4*(b^2*d^4*e^4 - 4*b^2*c*d^3*e^3*f + 6*b^2*c^2*d^2 
*e^2*f^2 - 4*b^2*c^3*d*e*f^3 + b^2*c^4*f^4)*log(F)^2 - 12*(b*d^2*e^2*f^2 - 
 2*b*c*d*e*f^3 + b*c^2*f^4)*log(F))*sqrt(-b*d^2*log(F))*F^a*erf(sqrt(-b*d^ 
2*log(F))*(d*x + c)/d) - 2*(2*(b^2*d^4*f^4*x^3 + 4*b^2*d^4*e^3*f - 6*b^2*c 
*d^3*e^2*f^2 + 4*b^2*c^2*d^2*e*f^3 - b^2*c^3*d*f^4 + (4*b^2*d^4*e*f^3 - b^ 
2*c*d^3*f^4)*x^2 + (6*b^2*d^4*e^2*f^2 - 4*b^2*c*d^3*e*f^3 + b^2*c^2*d^2*f^ 
4)*x)*log(F)^2 - (3*b*d^2*f^4*x + 8*b*d^2*e*f^3 - 5*b*c*d*f^4)*log(F))*F^( 
b*d^2*x^2 + 2*b*c*d*x + b*c^2 + a))/(b^3*d^6*log(F)^3)
 

3.4.83.6 Sympy [F]

\[ \int F^{a+b (c+d x)^2} (e+f x)^4 \, dx=\int F^{a + b \left (c + d x\right )^{2}} \left (e + f x\right )^{4}\, dx \]

input

integrate(F**(a+b*(d*x+c)**2)*(f*x+e)**4,x)
 

output

Integral(F**(a + b*(c + d*x)**2)*(e + f*x)**4, x)
 

3.4.83.7 Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1052 vs. \(2 (349) = 698\).

Time = 0.64 (sec) , antiderivative size = 1052, normalized size of antiderivative = 2.70 \[ \int F^{a+b (c+d x)^2} (e+f x)^4 \, dx=\text {Too large to display} \]

input

integrate(F^(a+b*(d*x+c)^2)*(f*x+e)^4,x, algorithm="maxima")
 

output

-2*(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*e^3*f/(sqrt(b*log(F))*d) + 3*(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*log(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*e^2*f^2/(sqrt(b*lo 
g(F))*d) - 2*(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*log( 
F)/(b*d^2))*log(F)^2/((b*log(F))^(7/2)*d^3))*F^a*e*f^3/(sqrt(b*log(F))*d) 
+ 1/2*(sqrt(pi)*(b*d^2*x + b*c*d)*b^4*c^4*(erf(sqrt(-(b*d^2*x + b*c*d)^2*l 
og(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...
 

3.4.83.8 Giac [A] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 418, normalized size of antiderivative = 1.07 \[ \int F^{a+b (c+d x)^2} (e+f x)^4 \, dx=-\frac {\frac {\sqrt {\pi } {\left (4 \, b^{2} d^{4} e^{4} \log \left (F\right )^{2} - 16 \, b^{2} c d^{3} e^{3} f \log \left (F\right )^{2} + 24 \, b^{2} c^{2} d^{2} e^{2} f^{2} \log \left (F\right )^{2} - 16 \, b^{2} c^{3} d e f^{3} \log \left (F\right )^{2} + 4 \, b^{2} c^{4} f^{4} \log \left (F\right )^{2} - 12 \, b d^{2} e^{2} f^{2} \log \left (F\right ) + 24 \, b c d e f^{3} \log \left (F\right ) - 12 \, b c^{2} f^{4} \log \left (F\right ) + 3 \, f^{4}\right )} F^{a} \operatorname {erf}\left (-\sqrt {-b \log \left (F\right )} d {\left (x + \frac {c}{d}\right )}\right )}{\sqrt {-b \log \left (F\right )} b^{2} d \log \left (F\right )^{2}} - \frac {2 \, {\left (2 \, b d^{3} f^{4} {\left (x + \frac {c}{d}\right )}^{3} \log \left (F\right ) + 8 \, b d^{3} e f^{3} {\left (x + \frac {c}{d}\right )}^{2} \log \left (F\right ) - 8 \, b c d^{2} f^{4} {\left (x + \frac {c}{d}\right )}^{2} \log \left (F\right ) + 12 \, b d^{3} e^{2} f^{2} {\left (x + \frac {c}{d}\right )} \log \left (F\right ) - 24 \, b c d^{2} e f^{3} {\left (x + \frac {c}{d}\right )} \log \left (F\right ) + 12 \, b c^{2} d f^{4} {\left (x + \frac {c}{d}\right )} \log \left (F\right ) + 8 \, b d^{3} e^{3} f \log \left (F\right ) - 24 \, b c d^{2} e^{2} f^{2} \log \left (F\right ) + 24 \, b c^{2} d e f^{3} \log \left (F\right ) - 8 \, b c^{3} f^{4} \log \left (F\right ) - 3 \, d f^{4} {\left (x + \frac {c}{d}\right )} - 8 \, d e f^{3} + 8 \, c f^{4}\right )} e^{\left (b d^{2} x^{2} \log \left (F\right ) + 2 \, b c d x \log \left (F\right ) + b c^{2} \log \left (F\right ) + a \log \left (F\right )\right )}}{b^{2} d \log \left (F\right )^{2}}}{8 \, d^{4}} \]

input

integrate(F^(a+b*(d*x+c)^2)*(f*x+e)^4,x, algorithm="giac")
 

output

-1/8*(sqrt(pi)*(4*b^2*d^4*e^4*log(F)^2 - 16*b^2*c*d^3*e^3*f*log(F)^2 + 24* 
b^2*c^2*d^2*e^2*f^2*log(F)^2 - 16*b^2*c^3*d*e*f^3*log(F)^2 + 4*b^2*c^4*f^4 
*log(F)^2 - 12*b*d^2*e^2*f^2*log(F) + 24*b*c*d*e*f^3*log(F) - 12*b*c^2*f^4 
*log(F) + 3*f^4)*F^a*erf(-sqrt(-b*log(F))*d*(x + c/d))/(sqrt(-b*log(F))*b^ 
2*d*log(F)^2) - 2*(2*b*d^3*f^4*(x + c/d)^3*log(F) + 8*b*d^3*e*f^3*(x + c/d 
)^2*log(F) - 8*b*c*d^2*f^4*(x + c/d)^2*log(F) + 12*b*d^3*e^2*f^2*(x + c/d) 
*log(F) - 24*b*c*d^2*e*f^3*(x + c/d)*log(F) + 12*b*c^2*d*f^4*(x + c/d)*log 
(F) + 8*b*d^3*e^3*f*log(F) - 24*b*c*d^2*e^2*f^2*log(F) + 24*b*c^2*d*e*f^3* 
log(F) - 8*b*c^3*f^4*log(F) - 3*d*f^4*(x + c/d) - 8*d*e*f^3 + 8*c*f^4)*e^( 
b*d^2*x^2*log(F) + 2*b*c*d*x*log(F) + b*c^2*log(F) + a*log(F))/(b^2*d*log( 
F)^2))/d^4
 

3.4.83.9 Mupad [B] (verification not implemented)

Time = 0.57 (sec) , antiderivative size = 517, normalized size of antiderivative = 1.33 \[ \int F^{a+b (c+d x)^2} (e+f x)^4 \, dx=\mathrm {erfi}\left (\frac {b\,x\,\ln \left (F\right )\,d^2+b\,c\,\ln \left (F\right )\,d}{\sqrt {b\,d^2\,\ln \left (F\right )}}\right )\,\left (\frac {\frac {3\,F^a\,f^4\,\sqrt {\pi }}{8\,\sqrt {b\,d^2\,\ln \left (F\right )}}-\frac {F^a\,\sqrt {\pi }\,\ln \left (F\right )\,\left (12\,b\,c^2\,f^4-24\,b\,c\,d\,e\,f^3+12\,b\,d^2\,e^2\,f^2\right )}{8\,\sqrt {b\,d^2\,\ln \left (F\right )}}}{b^2\,d^4\,{\ln \left (F\right )}^2}+\frac {F^a\,\sqrt {\pi }\,\left (4\,b^2\,c^4\,f^4-16\,b^2\,c^3\,d\,e\,f^3+24\,b^2\,c^2\,d^2\,e^2\,f^2-16\,b^2\,c\,d^3\,e^3\,f+4\,b^2\,d^4\,e^4\right )}{8\,b^2\,d^4\,\sqrt {b\,d^2\,\ln \left (F\right )}}\right )+\frac {F^{b\,d^2\,x^2}\,F^a\,F^{b\,c^2}\,F^{2\,b\,c\,d\,x}\,\left (\frac {5\,F^a\,c\,f^4-8\,F^a\,d\,e\,f^3}{4\,F^a}-\frac {b\,\left (2\,F^a\,c^3\,f^4\,\ln \left (F\right )-8\,F^a\,d^3\,e^3\,f\,\ln \left (F\right )-8\,F^a\,c^2\,d\,e\,f^3\,\ln \left (F\right )+12\,F^a\,c\,d^2\,e^2\,f^2\,\ln \left (F\right )\right )}{4\,F^a}\right )}{b^2\,d^5\,{\ln \left (F\right )}^2}+\frac {F^{b\,d^2\,x^2}\,F^a\,F^{b\,c^2}\,F^{2\,b\,c\,d\,x}\,f^4\,x^3}{2\,b\,d^2\,\ln \left (F\right )}+\frac {F^{b\,d^2\,x^2}\,F^a\,F^{b\,c^2}\,F^{2\,b\,c\,d\,x}\,x\,\left (b\,\left (\frac {\ln \left (F\right )\,c^2\,f^4}{2}-2\,\ln \left (F\right )\,c\,d\,e\,f^3+3\,\ln \left (F\right )\,d^2\,e^2\,f^2\right )-\frac {3\,f^4}{4}\right )}{b^2\,d^4\,{\ln \left (F\right )}^2}-\frac {F^{b\,d^2\,x^2}\,F^a\,F^{b\,c^2}\,F^{2\,b\,c\,d\,x}\,f^3\,x^2\,\left (c\,f-4\,d\,e\right )}{2\,b\,d^3\,\ln \left (F\right )} \]

input

int(F^(a + b*(c + d*x)^2)*(e + f*x)^4,x)
 

output

erfi((b*c*d*log(F) + b*d^2*x*log(F))/(b*d^2*log(F))^(1/2))*(((3*F^a*f^4*pi 
^(1/2))/(8*(b*d^2*log(F))^(1/2)) - (F^a*pi^(1/2)*log(F)*(12*b*c^2*f^4 + 12 
*b*d^2*e^2*f^2 - 24*b*c*d*e*f^3))/(8*(b*d^2*log(F))^(1/2)))/(b^2*d^4*log(F 
)^2) + (F^a*pi^(1/2)*(4*b^2*c^4*f^4 + 4*b^2*d^4*e^4 + 24*b^2*c^2*d^2*e^2*f 
^2 - 16*b^2*c*d^3*e^3*f - 16*b^2*c^3*d*e*f^3))/(8*b^2*d^4*(b*d^2*log(F))^( 
1/2))) + (F^(b*d^2*x^2)*F^a*F^(b*c^2)*F^(2*b*c*d*x)*((5*F^a*c*f^4 - 8*F^a* 
d*e*f^3)/(4*F^a) - (b*(2*F^a*c^3*f^4*log(F) - 8*F^a*d^3*e^3*f*log(F) - 8*F 
^a*c^2*d*e*f^3*log(F) + 12*F^a*c*d^2*e^2*f^2*log(F)))/(4*F^a)))/(b^2*d^5*l 
og(F)^2) + (F^(b*d^2*x^2)*F^a*F^(b*c^2)*F^(2*b*c*d*x)*f^4*x^3)/(2*b*d^2*lo 
g(F)) + (F^(b*d^2*x^2)*F^a*F^(b*c^2)*F^(2*b*c*d*x)*x*(b*((c^2*f^4*log(F))/ 
2 + 3*d^2*e^2*f^2*log(F) - 2*c*d*e*f^3*log(F)) - (3*f^4)/4))/(b^2*d^4*log( 
F)^2) - (F^(b*d^2*x^2)*F^a*F^(b*c^2)*F^(2*b*c*d*x)*f^3*x^2*(c*f - 4*d*e))/ 
(2*b*d^3*log(F))