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

Optimal. Leaf size=136 \[ \frac {4 \sqrt {\pi } b^{5/2} F^a \log ^{\frac {5}{2}}(F) \text {erfi}\left (\sqrt {b} \sqrt {\log (F)} (c+d x)\right )}{15 d}-\frac {4 b^2 \log ^2(F) F^{a+b (c+d x)^2}}{15 d (c+d x)}-\frac {F^{a+b (c+d x)^2}}{5 d (c+d x)^5}-\frac {2 b \log (F) F^{a+b (c+d x)^2}}{15 d (c+d x)^3} \]

[Out]

-1/5*F^(a+b*(d*x+c)^2)/d/(d*x+c)^5-2/15*b*F^(a+b*(d*x+c)^2)*ln(F)/d/(d*x+c)^3-4/15*b^2*F^(a+b*(d*x+c)^2)*ln(F)
^2/d/(d*x+c)+4/15*b^(5/2)*F^a*erfi((d*x+c)*b^(1/2)*ln(F)^(1/2))*ln(F)^(5/2)*Pi^(1/2)/d

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Rubi [A]  time = 0.22, antiderivative size = 136, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2214, 2204} \[ \frac {4 \sqrt {\pi } b^{5/2} F^a \log ^{\frac {5}{2}}(F) \text {Erfi}\left (\sqrt {b} \sqrt {\log (F)} (c+d x)\right )}{15 d}-\frac {4 b^2 \log ^2(F) F^{a+b (c+d x)^2}}{15 d (c+d x)}-\frac {F^{a+b (c+d x)^2}}{5 d (c+d x)^5}-\frac {2 b \log (F) F^{a+b (c+d x)^2}}{15 d (c+d x)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-F^(a + b*(c + d*x)^2)/(5*d*(c + d*x)^5) - (2*b*F^(a + b*(c + d*x)^2)*Log[F])/(15*d*(c + d*x)^3) - (4*b^2*F^(a
+ b*(c + d*x)^2)*Log[F]^2)/(15*d*(c + d*x)) + (4*b^(5/2)*F^a*Sqrt[Pi]*Erfi[Sqrt[b]*(c + d*x)*Sqrt[Log[F]]]*Lo
g[F]^(5/2))/(15*d)

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 2214

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^(m
+ 1)*F^(a + b*(c + d*x)^n))/(d*(m + 1)), x] - Dist[(b*n*Log[F])/(m + 1), 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[-4, (m + 1)/n, 5] && IntegerQ[n
] && ((GtQ[n, 0] && LtQ[m, -1]) || (GtQ[-n, 0] && LeQ[-n, m + 1]))

Rubi steps

\begin {align*} \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^6} \, dx &=-\frac {F^{a+b (c+d x)^2}}{5 d (c+d x)^5}+\frac {1}{5} (2 b \log (F)) \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^4} \, dx\\ &=-\frac {F^{a+b (c+d x)^2}}{5 d (c+d x)^5}-\frac {2 b F^{a+b (c+d x)^2} \log (F)}{15 d (c+d x)^3}+\frac {1}{15} \left (4 b^2 \log ^2(F)\right ) \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^2} \, dx\\ &=-\frac {F^{a+b (c+d x)^2}}{5 d (c+d x)^5}-\frac {2 b F^{a+b (c+d x)^2} \log (F)}{15 d (c+d x)^3}-\frac {4 b^2 F^{a+b (c+d x)^2} \log ^2(F)}{15 d (c+d x)}+\frac {1}{15} \left (8 b^3 \log ^3(F)\right ) \int F^{a+b (c+d x)^2} \, dx\\ &=-\frac {F^{a+b (c+d x)^2}}{5 d (c+d x)^5}-\frac {2 b F^{a+b (c+d x)^2} \log (F)}{15 d (c+d x)^3}-\frac {4 b^2 F^{a+b (c+d x)^2} \log ^2(F)}{15 d (c+d x)}+\frac {4 b^{5/2} F^a \sqrt {\pi } \text {erfi}\left (\sqrt {b} (c+d x) \sqrt {\log (F)}\right ) \log ^{\frac {5}{2}}(F)}{15 d}\\ \end {align*}

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Mathematica [A]  time = 0.12, size = 97, normalized size = 0.71 \[ \frac {F^a \left (4 \sqrt {\pi } b^{5/2} \log ^{\frac {5}{2}}(F) \text {erfi}\left (\sqrt {b} \sqrt {\log (F)} (c+d x)\right )-\frac {F^{b (c+d x)^2} \left (4 b^2 \log ^2(F) (c+d x)^4+2 b \log (F) (c+d x)^2+3\right )}{(c+d x)^5}\right )}{15 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(F^a*(4*b^(5/2)*Sqrt[Pi]*Erfi[Sqrt[b]*(c + d*x)*Sqrt[Log[F]]]*Log[F]^(5/2) - (F^(b*(c + d*x)^2)*(3 + 2*b*(c +
d*x)^2*Log[F] + 4*b^2*(c + d*x)^4*Log[F]^2))/(c + d*x)^5))/(15*d)

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fricas [B]  time = 0.44, size = 288, normalized size = 2.12 \[ -\frac {4 \, \sqrt {\pi } {\left (b^{2} d^{5} x^{5} + 5 \, b^{2} c d^{4} x^{4} + 10 \, b^{2} c^{2} d^{3} x^{3} + 10 \, b^{2} c^{3} d^{2} x^{2} + 5 \, b^{2} c^{4} d x + b^{2} c^{5}\right )} \sqrt {-b d^{2} \log \relax (F)} F^{a} \operatorname {erf}\left (\frac {\sqrt {-b d^{2} \log \relax (F)} {\left (d x + c\right )}}{d}\right ) \log \relax (F)^{2} + {\left (4 \, {\left (b^{2} d^{5} x^{4} + 4 \, b^{2} c d^{4} x^{3} + 6 \, b^{2} c^{2} d^{3} x^{2} + 4 \, b^{2} c^{3} d^{2} x + b^{2} c^{4} d\right )} \log \relax (F)^{2} + 2 \, {\left (b d^{3} x^{2} + 2 \, b c d^{2} x + b c^{2} d\right )} \log \relax (F) + 3 \, d\right )} F^{b d^{2} x^{2} + 2 \, b c d x + b c^{2} + a}}{15 \, {\left (d^{7} x^{5} + 5 \, c d^{6} x^{4} + 10 \, c^{2} d^{5} x^{3} + 10 \, c^{3} d^{4} x^{2} + 5 \, c^{4} d^{3} x + c^{5} d^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*(4*sqrt(pi)*(b^2*d^5*x^5 + 5*b^2*c*d^4*x^4 + 10*b^2*c^2*d^3*x^3 + 10*b^2*c^3*d^2*x^2 + 5*b^2*c^4*d*x + b
^2*c^5)*sqrt(-b*d^2*log(F))*F^a*erf(sqrt(-b*d^2*log(F))*(d*x + c)/d)*log(F)^2 + (4*(b^2*d^5*x^4 + 4*b^2*c*d^4*
x^3 + 6*b^2*c^2*d^3*x^2 + 4*b^2*c^3*d^2*x + b^2*c^4*d)*log(F)^2 + 2*(b*d^3*x^2 + 2*b*c*d^2*x + b*c^2*d)*log(F)
+ 3*d)*F^(b*d^2*x^2 + 2*b*c*d*x + b*c^2 + a))/(d^7*x^5 + 5*c*d^6*x^4 + 10*c^2*d^5*x^3 + 10*c^3*d^4*x^2 + 5*c^
4*d^3*x + c^5*d^2)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {F^{{\left (d x + c\right )}^{2} b + a}}{{\left (d x + c\right )}^{6}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.09, size = 129, normalized size = 0.95 \[ \frac {4 \sqrt {\pi }\, b^{3} F^{a} \erf \left (\sqrt {-b \ln \relax (F )}\, \left (d x +c \right )\right ) \ln \relax (F )^{3}}{15 \sqrt {-b \ln \relax (F )}\, d}-\frac {4 b^{2} F^{a} F^{\left (d x +c \right )^{2} b} \ln \relax (F )^{2}}{15 \left (d x +c \right ) d}-\frac {2 b \,F^{a} F^{\left (d x +c \right )^{2} b} \ln \relax (F )}{15 \left (d x +c \right )^{3} d}-\frac {F^{a} F^{\left (d x +c \right )^{2} b}}{5 \left (d x +c \right )^{5} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/5/d/(d*x+c)^5*F^((d*x+c)^2*b)*F^a-2/15/d*b*ln(F)/(d*x+c)^3*F^((d*x+c)^2*b)*F^a-4/15/d*b^2*ln(F)^2/(d*x+c)*F
^((d*x+c)^2*b)*F^a+4/15/d*b^3*ln(F)^3*Pi^(1/2)*F^a/(-b*ln(F))^(1/2)*erf((-b*ln(F))^(1/2)*(d*x+c))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {F^{{\left (d x + c\right )}^{2} b + a}}{{\left (d x + c\right )}^{6}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 4.87, size = 168, normalized size = 1.24 \[ \frac {4\,F^a\,\sqrt {\pi }\,\mathrm {erfc}\left (\sqrt {-b\,\ln \relax (F)\,{\left (c+d\,x\right )}^2}\right )\,{\left (-b\,\ln \relax (F)\,{\left (c+d\,x\right )}^2\right )}^{5/2}}{15\,d\,{\left (c+d\,x\right )}^5}-\frac {4\,F^a\,\sqrt {\pi }\,{\left (-b\,\ln \relax (F)\,{\left (c+d\,x\right )}^2\right )}^{5/2}}{15\,d\,{\left (c+d\,x\right )}^5}-\frac {4\,F^a\,F^{b\,{\left (c+d\,x\right )}^2}\,b^2\,{\ln \relax (F)}^2}{15\,d\,\left (c+d\,x\right )}-\frac {2\,F^a\,F^{b\,{\left (c+d\,x\right )}^2}\,b\,\ln \relax (F)}{15\,d\,{\left (c+d\,x\right )}^3}-\frac {F^a\,F^{b\,{\left (c+d\,x\right )}^2}}{5\,d\,{\left (c+d\,x\right )}^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(4*F^a*pi^(1/2)*erfc((-b*log(F)*(c + d*x)^2)^(1/2))*(-b*log(F)*(c + d*x)^2)^(5/2))/(15*d*(c + d*x)^5) - (4*F^a
*pi^(1/2)*(-b*log(F)*(c + d*x)^2)^(5/2))/(15*d*(c + d*x)^5) - (4*F^a*F^(b*(c + d*x)^2)*b^2*log(F)^2)/(15*d*(c
+ d*x)) - (2*F^a*F^(b*(c + d*x)^2)*b*log(F))/(15*d*(c + d*x)^3) - (F^a*F^(b*(c + d*x)^2))/(5*d*(c + d*x)^5)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F**(a+b*(d*x+c)**2)/(d*x+c)**6,x)

[Out]

Timed out

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