\(\int F^{a+b (c+d x)^2} (c+d x)^4 \, dx\) [271]

Optimal result

Integrand size = 21, antiderivative size = 111 \[ \int F^{a+b (c+d x)^2} (c+d x)^4 \, dx=\frac {3 F^a \sqrt {\pi } \text {erfi}\left (\sqrt {b} (c+d x) \sqrt {\log (F)}\right )}{8 b^{5/2} d \log ^{\frac {5}{2}}(F)}-\frac {3 F^{a+b (c+d x)^2} (c+d x)}{4 b^2 d \log ^2(F)}+\frac {F^{a+b (c+d x)^2} (c+d x)^3}{2 b d \log (F)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2243, 2235} \[ \int F^{a+b (c+d x)^2} (c+d x)^4 \, dx=\frac {3 \sqrt {\pi } F^a \text {erfi}\left (\sqrt {b} \sqrt {\log (F)} (c+d x)\right )}{8 b^{5/2} d \log ^{\frac {5}{2}}(F)}-\frac {3 (c+d x) F^{a+b (c+d x)^2}}{4 b^2 d \log ^2(F)}+\frac {(c+d x)^3 F^{a+b (c+d x)^2}}{2 b d \log (F)} \]

[In]

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

[Out]

(3*F^a*Sqrt[Pi]*Erfi[Sqrt[b]*(c + d*x)*Sqrt[Log[F]]])/(8*b^(5/2)*d*Log[F]^(5/2)) - (3*F^(a + b*(c + d*x)^2)*(c
+ d*x))/(4*b^2*d*Log[F]^2) + (F^(a + b*(c + d*x)^2)*(c + d*x)^3)/(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)^3}{2 b d \log (F)}-\frac {3 \int F^{a+b (c+d x)^2} (c+d x)^2 \, dx}{2 b \log (F)} \\ & = -\frac {3 F^{a+b (c+d x)^2} (c+d x)}{4 b^2 d \log ^2(F)}+\frac {F^{a+b (c+d x)^2} (c+d x)^3}{2 b d \log (F)}+\frac {3 \int F^{a+b (c+d x)^2} \, dx}{4 b^2 \log ^2(F)} \\ & = \frac {3 F^a \sqrt {\pi } \text {erfi}\left (\sqrt {b} (c+d x) \sqrt {\log (F)}\right )}{8 b^{5/2} d \log ^{\frac {5}{2}}(F)}-\frac {3 F^{a+b (c+d x)^2} (c+d x)}{4 b^2 d \log ^2(F)}+\frac {F^{a+b (c+d x)^2} (c+d x)^3}{2 b d \log (F)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.81 \[ \int F^{a+b (c+d x)^2} (c+d x)^4 \, dx=\frac {F^a \left (3 \sqrt {\pi } \text {erfi}\left (\sqrt {b} (c+d x) \sqrt {\log (F)}\right )+2 \sqrt {b} F^{b (c+d x)^2} (c+d x) \sqrt {\log (F)} \left (-3+2 b (c+d x)^2 \log (F)\right )\right )}{8 b^{5/2} d \log ^{\frac {5}{2}}(F)} \]

[In]

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

[Out]

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(314\) vs. \(2(95)=190\).

Time = 0.37 (sec) , antiderivative size = 315, normalized size of antiderivative = 2.84

method	result	size
risch	\(\frac {F^{b \,c^{2}} F^{a} d^{2} x^{3} F^{b \,d^{2} x^{2}} F^{2 b c d x}}{2 \ln \left (F \right ) b}+\frac {3 F^{b \,c^{2}} F^{a} d c \,x^{2} F^{b \,d^{2} x^{2}} F^{2 b c d x}}{2 \ln \left (F \right ) b}+\frac {3 F^{b \,c^{2}} F^{a} c^{2} x \,F^{b \,d^{2} x^{2}} F^{2 b c d x}}{2 \ln \left (F \right ) b}+\frac {F^{b \,c^{2}} F^{a} c^{3} F^{b \,d^{2} x^{2}} F^{2 b c d x}}{2 d \ln \left (F \right ) b}-\frac {3 F^{b \,c^{2}} F^{a} c \,F^{b \,d^{2} x^{2}} F^{2 b c d x}}{4 d \ln \left (F \right )^{2} b^{2}}-\frac {3 F^{b \,c^{2}} F^{a} x \,F^{b \,d^{2} x^{2}} F^{2 b c d x}}{4 \ln \left (F \right )^{2} b^{2}}-\frac {3 F^{b \,c^{2}} F^{a} \sqrt {\pi }\, F^{-b \,c^{2}} \operatorname {erf}\left (-d \sqrt {-b \ln \left (F \right )}\, x +\frac {b c \ln \left (F \right )}{\sqrt {-b \ln \left (F \right )}}\right )}{8 d \ln \left (F \right )^{2} b^{2} \sqrt {-b \ln \left (F \right )}}\)	\(315\)

[In]

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

[Out]

1/2*F^(b*c^2)*F^a*d^2/ln(F)/b*x^3*F^(b*d^2*x^2)*F^(2*b*c*d*x)+3/2*F^(b*c^2)*F^a*d*c/ln(F)/b*x^2*F^(b*d^2*x^2)*
F^(2*b*c*d*x)+3/2*F^(b*c^2)*F^a*c^2/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^3/ln(F)/b*F^(b
*d^2*x^2)*F^(2*b*c*d*x)-3/4*F^(b*c^2)*F^a/d*c/ln(F)^2/b^2*F^(b*d^2*x^2)*F^(2*b*c*d*x)-3/4*F^(b*c^2)*F^a/ln(F)^
2/b^2*x*F^(b*d^2*x^2)*F^(2*b*c*d*x)-3/8*F^(b*c^2)*F^a/d/ln(F)^2/b^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))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.27 \[ \int F^{a+b (c+d x)^2} (c+d x)^4 \, dx=-\frac {3 \, \sqrt {\pi } \sqrt {-b d^{2} \log \left (F\right )} F^{a} \operatorname {erf}\left (\frac {\sqrt {-b d^{2} \log \left (F\right )} {\left (d x + c\right )}}{d}\right ) - 2 \, {\left (2 \, {\left (b^{2} d^{4} x^{3} + 3 \, b^{2} c d^{3} x^{2} + 3 \, b^{2} c^{2} d^{2} x + b^{2} c^{3} d\right )} \log \left (F\right )^{2} - 3 \, {\left (b d^{2} x + b c d\right )} \log \left (F\right )\right )} F^{b d^{2} x^{2} + 2 \, b c d x + b c^{2} + a}}{8 \, b^{3} d^{2} \log \left (F\right )^{3}} \]

[In]

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

[Out]

-1/8*(3*sqrt(pi)*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*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 - 3*(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^3*d^2*log(F)^3)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1037 vs. \(2 (95) = 190\).

Time = 0.61 (sec) , antiderivative size = 1037, normalized size of antiderivative = 9.34 \[ \int F^{a+b (c+d x)^2} (c+d x)^4 \, dx=\text {Too large to display} \]

[In]

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

[Out]

-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*c^3/sqrt(b*log(F)) + 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*c^2*d/sqrt(b*log(F
)) - 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*l
og(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*d^2/sqrt(b*log(F)) + 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*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*d^3/sqrt(b*log(F)) + 1/2*sqrt(pi)*F^(b*c^2 + a)*c^4*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 = 0.31 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.00 \[ \int F^{a+b (c+d x)^2} (c+d x)^4 \, dx=\frac {{\left (2 \, b d^{2} {\left (x + \frac {c}{d}\right )}^{3} \log \left (F\right ) - 3 \, x - \frac {3 \, c}{d}\right )} e^{\left (b d^{2} x^{2} \log \left (F\right ) + 2 \, b c d x \log \left (F\right ) + b c^{2} \log \left (F\right ) + a \log \left (F\right )\right )}}{4 \, b^{2} \log \left (F\right )^{2}} - \frac {3 \, \sqrt {\pi } F^{a} \operatorname {erf}\left (-\sqrt {-b \log \left (F\right )} d {\left (x + \frac {c}{d}\right )}\right )}{8 \, \sqrt {-b \log \left (F\right )} b^{2} d \log \left (F\right )^{2}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.36 (sec) , antiderivative size = 243, normalized size of antiderivative = 2.19 \[ \int F^{a+b (c+d x)^2} (c+d x)^4 \, dx=\frac {3\,F^a\,\sqrt {\pi }\,\mathrm {erfi}\left (\frac {b\,x\,\ln \left (F\right )\,d^2+b\,c\,\ln \left (F\right )\,d}{\sqrt {b\,d^2\,\ln \left (F\right )}}\right )}{8\,b^2\,{\ln \left (F\right )}^2\,\sqrt {b\,d^2\,\ln \left (F\right )}}-F^{b\,d^2\,x^2}\,F^a\,F^{b\,c^2}\,F^{2\,b\,c\,d\,x}\,\left (\frac {3\,c}{4\,b^2\,d\,{\ln \left (F\right )}^2}-\frac {c^3}{2\,b\,d\,\ln \left (F\right )}\right )+\frac {3\,F^{b\,d^2\,x^2}\,F^a\,F^{b\,c^2}\,F^{2\,b\,c\,d\,x}\,x\,\left (2\,b\,c^2\,\ln \left (F\right )-1\right )}{4\,b^2\,{\ln \left (F\right )}^2}+\frac {F^{b\,d^2\,x^2}\,F^a\,F^{b\,c^2}\,F^{2\,b\,c\,d\,x}\,d^2\,x^3}{2\,b\,\ln \left (F\right )}+\frac {3\,F^{b\,d^2\,x^2}\,F^a\,F^{b\,c^2}\,F^{2\,b\,c\,d\,x}\,c\,d\,x^2}{2\,b\,\ln \left (F\right )} \]

[In]

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

[Out]

(3*F^a*pi^(1/2)*erfi((b*c*d*log(F) + b*d^2*x*log(F))/(b*d^2*log(F))^(1/2)))/(8*b^2*log(F)^2*(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)*((3*c)/(4*b^2*d*log(F)^2) - c^3/(2*b*d*log(F))) + (3*F^(b*d^2*
x^2)*F^a*F^(b*c^2)*F^(2*b*c*d*x)*x*(2*b*c^2*log(F) - 1))/(4*b^2*log(F)^2) + (F^(b*d^2*x^2)*F^a*F^(b*c^2)*F^(2*
b*c*d*x)*d^2*x^3)/(2*b*log(F)) + (3*F^(b*d^2*x^2)*F^a*F^(b*c^2)*F^(2*b*c*d*x)*c*d*x^2)/(2*b*log(F))