\(\int F^{a+b (c+d x)^2} (c+d x)^6 \, dx\) [204]

Optimal result

Integrand size = 21, antiderivative size = 145 \[ \int F^{a+b (c+d x)^2} (c+d x)^6 \, dx=-\frac {15 F^a \sqrt {\pi } \text {erfi}\left (\sqrt {b} (c+d x) \sqrt {\log (F)}\right )}{16 b^{7/2} d \log ^{\frac {7}{2}}(F)}+\frac {15 F^{a+b (c+d x)^2} (c+d x)}{8 b^3 d \log ^3(F)}-\frac {5 F^{a+b (c+d x)^2} (c+d x)^3}{4 b^2 d \log ^2(F)}+\frac {F^{a+b (c+d x)^2} (c+d x)^5}{2 b d \log (F)} \] Output:

-15/16*F^a*Pi^(1/2)*erfi(b^(1/2)*(d*x+c)*ln(F)^(1/2))/b^(7/2)/d/ln(F)^(7/2 
)+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)
 

Mathematica [A] (verified)

Time = 0.50 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.87 \[ \int F^{a+b (c+d x)^2} (c+d x)^6 \, dx=\frac {F^a \left (8 F^{b (c+d x)^2} (c+d x)^5-\frac {15 \sqrt {\pi } \text {erfi}\left (\sqrt {b} (c+d x) \sqrt {\log (F)}\right )}{b^{5/2} \log ^{\frac {5}{2}}(F)}+\frac {30 F^{b (c+d x)^2} (c+d x)}{b^2 \log ^2(F)}-\frac {20 F^{b (c+d x)^2} (c+d x)^3}{b \log (F)}\right )}{16 b d \log (F)} \] Input:

Integrate[F^(a + b*(c + d*x)^2)*(c + d*x)^6,x]
 

Output:

(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])
 

Rubi [A] (verified)

Time = 0.80 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.17, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {2641, 2641, 2641, 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.

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

\(\Big \downarrow \) 2641

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

\(\Big \downarrow \) 2641

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

\(\Big \downarrow \) 2641

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

\(\Big \downarrow \) 2633

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

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)^5)/(2*b*d*Log[F]) - (5*((F^(a + b*(c + d* 
x)^2)*(c + d*x)^3)/(2*b*d*Log[F]) - (3*(-1/4*(F^a*Sqrt[Pi]*Erfi[Sqrt[b]*(c 
 + d*x)*Sqrt[Log[F]]])/(b^(3/2)*d*Log[F]^(3/2)) + (F^(a + b*(c + d*x)^2)*( 
c + d*x))/(2*b*d*Log[F])))/(2*b*Log[F])))/(2*b*Log[F])
 

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 2641

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*L 
og[F])), x] - Simp[(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])
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(575\) vs. \(2(127)=254\).

Time = 0.40 (sec) , antiderivative size = 576, normalized size of antiderivative = 3.97

method	result	size
risch	\(-\frac {5 F^{b \,c^{2}} F^{a} c^{3} F^{b \,d^{2} x^{2}} F^{2 b c d x}}{4 d \ln \left (F \right )^{2} b^{2}}-\frac {15 F^{b \,c^{2}} F^{a} c^{2} x \,F^{b \,d^{2} x^{2}} F^{2 b c d x}}{4 \ln \left (F \right )^{2} b^{2}}+\frac {15 F^{b \,c^{2}} F^{a} c \,F^{b \,d^{2} x^{2}} F^{2 b c d x}}{8 d \ln \left (F \right )^{3} b^{3}}-\frac {5 F^{b \,c^{2}} F^{a} d^{2} x^{3} F^{b \,d^{2} x^{2}} F^{2 b c d x}}{4 \ln \left (F \right )^{2} b^{2}}+\frac {F^{b \,c^{2}} F^{a} d^{4} x^{5} F^{b \,d^{2} x^{2}} F^{2 b c d x}}{2 \ln \left (F \right ) b}+\frac {5 F^{b \,c^{2}} F^{a} c^{4} x \,F^{b \,d^{2} x^{2}} F^{2 b c d x}}{2 \ln \left (F \right ) b}+\frac {F^{b \,c^{2}} F^{a} c^{5} F^{b \,d^{2} x^{2}} F^{2 b c d x}}{2 d \ln \left (F \right ) b}+\frac {5 F^{b \,c^{2}} F^{a} d^{3} c \,x^{4} F^{b \,d^{2} x^{2}} F^{2 b c d x}}{2 \ln \left (F \right ) b}+\frac {5 F^{b \,c^{2}} F^{a} d^{2} c^{2} x^{3} F^{b \,d^{2} x^{2}} F^{2 b c d x}}{\ln \left (F \right ) b}+\frac {5 F^{b \,c^{2}} F^{a} d \,c^{3} x^{2} F^{b \,d^{2} x^{2}} F^{2 b c d x}}{\ln \left (F \right ) b}-\frac {15 F^{b \,c^{2}} F^{a} d c \,x^{2} F^{b \,d^{2} x^{2}} F^{2 b c d x}}{4 \ln \left (F \right )^{2} b^{2}}+\frac {15 F^{b \,c^{2}} F^{a} \sqrt {\pi }\, F^{-b \,c^{2}} \operatorname {erf}\left (-d \sqrt {-b \ln \left (F \right )}\, x +\frac {b c \ln \left (F \right )}{\sqrt {-b \ln \left (F \right )}}\right )}{16 d \ln \left (F \right )^{3} b^{3} \sqrt {-b \ln \left (F \right )}}+\frac {15 F^{b \,c^{2}} F^{a} x \,F^{b \,d^{2} x^{2}} F^{2 b c d x}}{8 \ln \left (F \right )^{3} b^{3}}\)	\(576\)

Input:

int(F^(a+b*(d*x+c)^2)*(d*x+c)^6,x,method=_RETURNVERBOSE)
 

Output:

-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)+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)+5/2 
*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/l 
n(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)*erf(-d*(-b*ln(F))^(1/2)*x+b*c*ln(F)/(-b*ln(F) 
)^(1/2))+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)

Time = 0.08 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.50 \[ \int F^{a+b (c+d x)^2} (c+d x)^6 \, dx=\frac {15 \, \sqrt {\pi } \sqrt {-b d^{2} \log \left (F\right )} F^{a} \operatorname {erf}\left (\frac {\sqrt {-b d^{2} \log \left (F\right )} {\left (d x + c\right )}}{d}\right ) + 2 \, {\left (4 \, {\left (b^{3} d^{6} x^{5} + 5 \, b^{3} c d^{5} x^{4} + 10 \, b^{3} c^{2} d^{4} x^{3} + 10 \, b^{3} c^{3} d^{3} x^{2} + 5 \, b^{3} c^{4} d^{2} x + b^{3} c^{5} d\right )} \log \left (F\right )^{3} - 10 \, {\left (b^{2} d^{4} x^{3} + 3 \, b^{2} c d^{3} x^{2} + 3 \, b^{2} c^{2} d^{2} x + b^{2} c^{3} d\right )} \log \left (F\right )^{2} + 15 \, {\left (b d^{2} x + b c d\right )} \log \left (F\right )\right )} F^{b d^{2} x^{2} + 2 \, b c d x + b c^{2} + a}}{16 \, b^{4} d^{2} \log \left (F\right )^{4}} \] Input:

integrate(F^(a+b*(d*x+c)^2)*(d*x+c)^6,x, algorithm="fricas")
 

Output:

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]

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1922 vs. \(2 (127) = 254\).

Time = 0.72 (sec) , antiderivative size = 1922, normalized size of antiderivative = 13.26 \[ \int F^{a+b (c+d x)^2} (c+d x)^6 \, dx=\text {Too large to display} \] Input:

integrate(F^(a+b*(d*x+c)^2)*(d*x+c)^6,x, algorithm="maxima")
 

Output:

-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*(er 
f(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*ga 
mma(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*c^4*d/sqrt(b*log(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*lo 
g(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*log(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*c^3*d^2/sqrt(b*log(F)) + 15/2*(sq 
rt(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*...
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.91 \[ \int F^{a+b (c+d x)^2} (c+d x)^6 \, dx=\frac {{\left (4 \, b^{2} d^{4} {\left (x + \frac {c}{d}\right )}^{5} \log \left (F\right )^{2} - 10 \, b d^{2} {\left (x + \frac {c}{d}\right )}^{3} \log \left (F\right ) + 15 \, x + \frac {15 \, c}{d}\right )} e^{\left (b d^{2} x^{2} \log \left (F\right ) + 2 \, b c d x \log \left (F\right ) + b c^{2} \log \left (F\right ) + a \log \left (F\right )\right )}}{8 \, b^{3} \log \left (F\right )^{3}} + \frac {15 \, \sqrt {\pi } F^{a} \operatorname {erf}\left (-\sqrt {-b \log \left (F\right )} d {\left (x + \frac {c}{d}\right )}\right )}{16 \, \sqrt {-b \log \left (F\right )} b^{3} d \log \left (F\right )^{3}} \] Input:

integrate(F^(a+b*(d*x+c)^2)*(d*x+c)^6,x, algorithm="giac")
 

Output:

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))/(sq 
rt(-b*log(F))*b^3*d*log(F)^3)
 

Mupad [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 378, normalized size of antiderivative = 2.61 \[ \int F^{a+b (c+d x)^2} (c+d x)^6 \, dx=F^{b\,d^2\,x^2}\,F^a\,F^{b\,c^2}\,F^{2\,b\,c\,d\,x}\,\left (\frac {15\,c}{8\,b^3\,d\,{\ln \left (F\right )}^3}+\frac {c^5}{2\,b\,d\,\ln \left (F\right )}-\frac {5\,c^3}{4\,b^2\,d\,{\ln \left (F\right )}^2}\right )-\frac {15\,F^a\,\sqrt {\pi }\,\mathrm {erfi}\left (\frac {b\,x\,\ln \left (F\right )\,d^2+b\,c\,\ln \left (F\right )\,d}{\sqrt {b\,d^2\,\ln \left (F\right )}}\right )}{16\,b^3\,{\ln \left (F\right )}^3\,\sqrt {b\,d^2\,\ln \left (F\right )}}-\frac {5\,F^{b\,d^2\,x^2}\,F^a\,F^{b\,c^2}\,F^{2\,b\,c\,d\,x}\,x^2\,\left (3\,c\,d-4\,b\,c^3\,d\,\ln \left (F\right )\right )}{4\,b^2\,{\ln \left (F\right )}^2}+\frac {5\,F^{b\,d^2\,x^2}\,F^a\,F^{b\,c^2}\,F^{2\,b\,c\,d\,x}\,x\,\left (4\,b^2\,c^4\,{\ln \left (F\right )}^2-6\,b\,c^2\,\ln \left (F\right )+3\right )}{8\,b^3\,{\ln \left (F\right )}^3}+\frac {F^{b\,d^2\,x^2}\,F^a\,F^{b\,c^2}\,F^{2\,b\,c\,d\,x}\,d^4\,x^5}{2\,b\,\ln \left (F\right )}+\frac {5\,F^{b\,d^2\,x^2}\,F^a\,F^{b\,c^2}\,F^{2\,b\,c\,d\,x}\,c\,d^3\,x^4}{2\,b\,\ln \left (F\right )}+\frac {5\,F^{b\,d^2\,x^2}\,F^a\,F^{b\,c^2}\,F^{2\,b\,c\,d\,x}\,d^2\,x^3\,\left (4\,b\,c^2\,\ln \left (F\right )-1\right )}{4\,b^2\,{\ln \left (F\right )}^2} \] Input:

int(F^(a + b*(c + d*x)^2)*(c + d*x)^6,x)
 

Output:

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*log(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*l 
og(F)^2)
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 520, normalized size of antiderivative = 3.59 \[ \int F^{a+b (c+d x)^2} (c+d x)^6 \, dx=\frac {f^{a} \left (15 \sqrt {\pi }\, \mathrm {erf}\left (\frac {\mathrm {log}\left (f \right ) b c i +\mathrm {log}\left (f \right ) b d i x}{\sqrt {b}\, \sqrt {\mathrm {log}\left (f \right )}}\right ) i +8 f^{b \,d^{2} x^{2}+2 b c d x +b \,c^{2}} \sqrt {b}\, \sqrt {\mathrm {log}\left (f \right )}\, \mathrm {log}\left (f \right )^{2} b^{2} c^{5}+40 f^{b \,d^{2} x^{2}+2 b c d x +b \,c^{2}} \sqrt {b}\, \sqrt {\mathrm {log}\left (f \right )}\, \mathrm {log}\left (f \right )^{2} b^{2} c^{4} d x +80 f^{b \,d^{2} x^{2}+2 b c d x +b \,c^{2}} \sqrt {b}\, \sqrt {\mathrm {log}\left (f \right )}\, \mathrm {log}\left (f \right )^{2} b^{2} c^{3} d^{2} x^{2}+80 f^{b \,d^{2} x^{2}+2 b c d x +b \,c^{2}} \sqrt {b}\, \sqrt {\mathrm {log}\left (f \right )}\, \mathrm {log}\left (f \right )^{2} b^{2} c^{2} d^{3} x^{3}+40 f^{b \,d^{2} x^{2}+2 b c d x +b \,c^{2}} \sqrt {b}\, \sqrt {\mathrm {log}\left (f \right )}\, \mathrm {log}\left (f \right )^{2} b^{2} c \,d^{4} x^{4}+8 f^{b \,d^{2} x^{2}+2 b c d x +b \,c^{2}} \sqrt {b}\, \sqrt {\mathrm {log}\left (f \right )}\, \mathrm {log}\left (f \right )^{2} b^{2} d^{5} x^{5}-20 f^{b \,d^{2} x^{2}+2 b c d x +b \,c^{2}} \sqrt {b}\, \sqrt {\mathrm {log}\left (f \right )}\, \mathrm {log}\left (f \right ) b \,c^{3}-60 f^{b \,d^{2} x^{2}+2 b c d x +b \,c^{2}} \sqrt {b}\, \sqrt {\mathrm {log}\left (f \right )}\, \mathrm {log}\left (f \right ) b \,c^{2} d x -60 f^{b \,d^{2} x^{2}+2 b c d x +b \,c^{2}} \sqrt {b}\, \sqrt {\mathrm {log}\left (f \right )}\, \mathrm {log}\left (f \right ) b c \,d^{2} x^{2}-20 f^{b \,d^{2} x^{2}+2 b c d x +b \,c^{2}} \sqrt {b}\, \sqrt {\mathrm {log}\left (f \right )}\, \mathrm {log}\left (f \right ) b \,d^{3} x^{3}+30 f^{b \,d^{2} x^{2}+2 b c d x +b \,c^{2}} \sqrt {b}\, \sqrt {\mathrm {log}\left (f \right )}\, c +30 f^{b \,d^{2} x^{2}+2 b c d x +b \,c^{2}} \sqrt {b}\, \sqrt {\mathrm {log}\left (f \right )}\, d x \right )}{16 \sqrt {b}\, \sqrt {\mathrm {log}\left (f \right )}\, \mathrm {log}\left (f \right )^{3} b^{3} d} \] Input:

int(F^(a+b*(d*x+c)^2)*(d*x+c)^6,x)
 

Output:

(f**a*(15*sqrt(pi)*erf((log(f)*b*c*i + log(f)*b*d*i*x)/(sqrt(b)*sqrt(log(f 
))))*i + 8*f**(b*c**2 + 2*b*c*d*x + b*d**2*x**2)*sqrt(b)*sqrt(log(f))*log( 
f)**2*b**2*c**5 + 40*f**(b*c**2 + 2*b*c*d*x + b*d**2*x**2)*sqrt(b)*sqrt(lo 
g(f))*log(f)**2*b**2*c**4*d*x + 80*f**(b*c**2 + 2*b*c*d*x + b*d**2*x**2)*s 
qrt(b)*sqrt(log(f))*log(f)**2*b**2*c**3*d**2*x**2 + 80*f**(b*c**2 + 2*b*c* 
d*x + b*d**2*x**2)*sqrt(b)*sqrt(log(f))*log(f)**2*b**2*c**2*d**3*x**3 + 40 
*f**(b*c**2 + 2*b*c*d*x + b*d**2*x**2)*sqrt(b)*sqrt(log(f))*log(f)**2*b**2 
*c*d**4*x**4 + 8*f**(b*c**2 + 2*b*c*d*x + b*d**2*x**2)*sqrt(b)*sqrt(log(f) 
)*log(f)**2*b**2*d**5*x**5 - 20*f**(b*c**2 + 2*b*c*d*x + b*d**2*x**2)*sqrt 
(b)*sqrt(log(f))*log(f)*b*c**3 - 60*f**(b*c**2 + 2*b*c*d*x + b*d**2*x**2)* 
sqrt(b)*sqrt(log(f))*log(f)*b*c**2*d*x - 60*f**(b*c**2 + 2*b*c*d*x + b*d** 
2*x**2)*sqrt(b)*sqrt(log(f))*log(f)*b*c*d**2*x**2 - 20*f**(b*c**2 + 2*b*c* 
d*x + b*d**2*x**2)*sqrt(b)*sqrt(log(f))*log(f)*b*d**3*x**3 + 30*f**(b*c**2 
 + 2*b*c*d*x + b*d**2*x**2)*sqrt(b)*sqrt(log(f))*c + 30*f**(b*c**2 + 2*b*c 
*d*x + b*d**2*x**2)*sqrt(b)*sqrt(log(f))*d*x))/(16*sqrt(b)*sqrt(log(f))*lo 
g(f)**3*b**3*d)