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

Optimal. Leaf size=111 \[ \frac {3 \sqrt {\pi } F^a \text {erfi}\left (\sqrt {b} \sqrt {\log (F)} (c+d x)\right )}{8 b^{5/2} d \log ^{\frac {5}{2}}(F)}-\frac {3 (c+d x) F^{a+b (c+d x)^2}}{4 b^2 d \log ^2(F)}+\frac {(c+d x)^3 F^{a+b (c+d x)^2}}{2 b d \log (F)} \]

[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)

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Rubi [A]  time = 0.15, antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2212, 2204} \[ \frac {3 \sqrt {\pi } F^a \text {Erfi}\left (\sqrt {b} \sqrt {\log (F)} (c+d x)\right )}{8 b^{5/2} d \log ^{\frac {5}{2}}(F)}-\frac {3 (c+d x) F^{a+b (c+d x)^2}}{4 b^2 d \log ^2(F)}+\frac {(c+d x)^3 F^{a+b (c+d x)^2}}{2 b d \log (F)} \]

Antiderivative was successfully verified.

[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 2204

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 2212

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*} \int F^{a+b (c+d x)^2} (c+d x)^4 \, dx &=\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*}

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Mathematica [A]  time = 0.10, size = 90, normalized size = 0.81 \[ \frac {F^a \left (3 \sqrt {\pi } \text {erfi}\left (\sqrt {b} \sqrt {\log (F)} (c+d x)\right )+2 \sqrt {b} \sqrt {\log (F)} (c+d x) F^{b (c+d x)^2} \left (2 b \log (F) (c+d x)^2-3\right )\right )}{8 b^{5/2} d \log ^{\frac {5}{2}}(F)} \]

Antiderivative was successfully verified.

[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))

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fricas [A]  time = 0.68, size = 141, normalized size = 1.27 \[ -\frac {3 \, \sqrt {\pi } \sqrt {-b d^{2} \log \relax (F)} F^{a} \operatorname {erf}\left (\frac {\sqrt {-b d^{2} \log \relax (F)} {\left (d x + c\right )}}{d}\right ) - 2 \, {\left (2 \, {\left (b^{2} d^{4} x^{3} + 3 \, b^{2} c d^{3} x^{2} + 3 \, b^{2} c^{2} d^{2} x + b^{2} c^{3} d\right )} \log \relax (F)^{2} - 3 \, {\left (b d^{2} x + b c d\right )} \log \relax (F)\right )} F^{b d^{2} x^{2} + 2 \, b c d x + b c^{2} + a}}{8 \, b^{3} d^{2} \log \relax (F)^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [A]  time = 0.29, size = 111, normalized size = 1.00 \[ \frac {{\left (2 \, b d^{2} {\left (x + \frac {c}{d}\right )}^{3} \log \relax (F) - 3 \, x - \frac {3 \, c}{d}\right )} e^{\left (b d^{2} x^{2} \log \relax (F) + 2 \, b c d x \log \relax (F) + b c^{2} \log \relax (F) + a \log \relax (F)\right )}}{4 \, b^{2} \log \relax (F)^{2}} - \frac {3 \, \sqrt {\pi } F^{a} \operatorname {erf}\left (-\sqrt {-b \log \relax (F)} d {\left (x + \frac {c}{d}\right )}\right )}{8 \, \sqrt {-b \log \relax (F)} b^{2} d \log \relax (F)^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [B]  time = 0.09, size = 300, normalized size = 2.70 \[ \frac {d^{2} x^{3} F^{a} F^{b \,c^{2}} F^{b \,d^{2} x^{2}} F^{2 b c d x}}{2 b \ln \relax (F )}+\frac {3 c d \,x^{2} F^{a} F^{b \,c^{2}} F^{b \,d^{2} x^{2}} F^{2 b c d x}}{2 b \ln \relax (F )}+\frac {3 c^{2} x \,F^{a} F^{b \,c^{2}} F^{b \,d^{2} x^{2}} F^{2 b c d x}}{2 b \ln \relax (F )}+\frac {c^{3} F^{a} F^{b \,c^{2}} F^{b \,d^{2} x^{2}} F^{2 b c d x}}{2 b d \ln \relax (F )}-\frac {3 x \,F^{a} F^{b \,c^{2}} F^{b \,d^{2} x^{2}} F^{2 b c d x}}{4 b^{2} \ln \relax (F )^{2}}-\frac {3 c \,F^{a} F^{b \,c^{2}} F^{b \,d^{2} x^{2}} F^{2 b c d x}}{4 b^{2} d \ln \relax (F )^{2}}-\frac {3 \sqrt {\pi }\, F^{a} \erf \left (\frac {b c \ln \relax (F )}{\sqrt {-b \ln \relax (F )}}-\sqrt {-b \ln \relax (F )}\, d x \right )}{8 \sqrt {-b \ln \relax (F )}\, b^{2} d \ln \relax (F )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 5.27, size = 1037, normalized size = 9.34 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 3.60, size = 243, normalized size = 2.19 \[ \frac {3\,F^a\,\sqrt {\pi }\,\mathrm {erfi}\left (\frac {b\,x\,\ln \relax (F)\,d^2+b\,c\,\ln \relax (F)\,d}{\sqrt {b\,d^2\,\ln \relax (F)}}\right )}{8\,b^2\,{\ln \relax (F)}^2\,\sqrt {b\,d^2\,\ln \relax (F)}}-F^{b\,d^2\,x^2}\,F^a\,F^{b\,c^2}\,F^{2\,b\,c\,d\,x}\,\left (\frac {3\,c}{4\,b^2\,d\,{\ln \relax (F)}^2}-\frac {c^3}{2\,b\,d\,\ln \relax (F)}\right )+\frac {3\,F^{b\,d^2\,x^2}\,F^a\,F^{b\,c^2}\,F^{2\,b\,c\,d\,x}\,x\,\left (2\,b\,c^2\,\ln \relax (F)-1\right )}{4\,b^2\,{\ln \relax (F)}^2}+\frac {F^{b\,d^2\,x^2}\,F^a\,F^{b\,c^2}\,F^{2\,b\,c\,d\,x}\,d^2\,x^3}{2\,b\,\ln \relax (F)}+\frac {3\,F^{b\,d^2\,x^2}\,F^a\,F^{b\,c^2}\,F^{2\,b\,c\,d\,x}\,c\,d\,x^2}{2\,b\,\ln \relax (F)} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int F^{a + b \left (c + d x\right )^{2}} \left (c + d x\right )^{4}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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