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

Optimal. Leaf size=115 \[ -\frac {3 \sqrt {\pi } F^a \text {erfi}\left (\frac {\sqrt {b} \sqrt {\log (F)}}{c+d x}\right )}{8 b^{5/2} d \log ^{\frac {5}{2}}(F)}+\frac {3 F^{a+\frac {b}{(c+d x)^2}}}{4 b^2 d \log ^2(F) (c+d x)}-\frac {F^{a+\frac {b}{(c+d x)^2}}}{2 b d \log (F) (c+d x)^3} \]

[Out]

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

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Rubi [A]  time = 0.15, antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2212, 2211, 2204} \[ -\frac {3 \sqrt {\pi } F^a \text {Erfi}\left (\frac {\sqrt {b} \sqrt {\log (F)}}{c+d x}\right )}{8 b^{5/2} d \log ^{\frac {5}{2}}(F)}+\frac {3 F^{a+\frac {b}{(c+d x)^2}}}{4 b^2 d \log ^2(F) (c+d x)}-\frac {F^{a+\frac {b}{(c+d x)^2}}}{2 b d \log (F) (c+d x)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*F^a*Sqrt[Pi]*Erfi[(Sqrt[b]*Sqrt[Log[F]])/(c + d*x)])/(8*b^(5/2)*d*Log[F]^(5/2)) + (3*F^(a + b/(c + d*x)^2)
)/(4*b^2*d*(c + d*x)*Log[F]^2) - F^(a + b/(c + d*x)^2)/(2*b*d*(c + d*x)^3*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 2211

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Dist[1/(d*(m + 1))
, Subst[Int[F^(a + b*x^2), x], x, (c + d*x)^(m + 1)], x] /; FreeQ[{F, a, b, c, d, m, n}, x] && EqQ[n, 2*(m + 1
)]

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

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Mathematica [A]  time = 0.10, size = 95, normalized size = 0.83 \[ \frac {F^a \left (-3 \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {b} \sqrt {\log (F)}}{c+d x}\right )-\frac {2 \sqrt {b} \sqrt {\log (F)} F^{\frac {b}{(c+d x)^2}} \left (2 b \log (F)-3 (c+d x)^2\right )}{(c+d x)^3}\right )}{8 b^{5/2} d \log ^{\frac {5}{2}}(F)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.43, size = 199, normalized size = 1.73 \[ \frac {3 \, \sqrt {\pi } {\left (d^{4} x^{3} + 3 \, c d^{3} x^{2} + 3 \, c^{2} d^{2} x + c^{3} d\right )} F^{a} \sqrt {-\frac {b \log \relax (F)}{d^{2}}} \operatorname {erf}\left (\frac {d \sqrt {-\frac {b \log \relax (F)}{d^{2}}}}{d x + c}\right ) - 2 \, {\left (2 \, b^{2} \log \relax (F)^{2} - 3 \, {\left (b d^{2} x^{2} + 2 \, b c d x + b c^{2}\right )} \log \relax (F)\right )} F^{\frac {a d^{2} x^{2} + 2 \, a c d x + a c^{2} + b}{d^{2} x^{2} + 2 \, c d x + c^{2}}}}{8 \, {\left (b^{3} d^{4} x^{3} + 3 \, b^{3} c d^{3} x^{2} + 3 \, b^{3} c^{2} d^{2} x + b^{3} c^{3} d\right )} \log \relax (F)^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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^(a + b/(d*x + c)^2)/(d*x + c)^6, x)

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maple [A]  time = 0.12, size = 109, normalized size = 0.95 \[ -\frac {F^{a} F^{\frac {b}{\left (d x +c \right )^{2}}}}{2 \left (d x +c \right )^{3} b d \ln \relax (F )}+\frac {3 F^{a} F^{\frac {b}{\left (d x +c \right )^{2}}}}{4 \left (d x +c \right ) b^{2} d \ln \relax (F )^{2}}-\frac {3 \sqrt {\pi }\, F^{a} \erf \left (\frac {\sqrt {-b \ln \relax (F )}}{d x +c}\right )}{8 \sqrt {-b \ln \relax (F )}\, b^{2} d \ln \relax (F )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(F^(a+1/(d*x+c)^2*b)/(d*x+c)^6,x)

[Out]

-1/2*F^a/d*F^(1/(d*x+c)^2*b)/(d*x+c)^3/b/ln(F)+3/4*F^a/d/b^2/ln(F)^2*F^(1/(d*x+c)^2*b)/(d*x+c)-3/8*F^a/d/b^2/l
n(F)^2*Pi^(1/2)/(-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 \[ \int \frac {F^{a + \frac {b}{{\left (d x + c\right )}^{2}}}}{{\left (d x + c\right )}^{6}}\,{d x} \]

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^(a + b/(d*x + c)^2)/(d*x + c)^6, x)

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mupad [B]  time = 3.90, size = 105, normalized size = 0.91 \[ -\frac {F^a\,F^{\frac {b}{{\left (c+d\,x\right )}^2}}}{2\,b\,d\,\ln \relax (F)\,{\left (c+d\,x\right )}^3}-\frac {F^a\,\left (3\,\sqrt {\pi }\,\mathrm {erfi}\left (\frac {b\,\ln \relax (F)}{\sqrt {b\,\ln \relax (F)}\,\left (c+d\,x\right )}\right )-\frac {6\,F^{\frac {b}{{\left (c+d\,x\right )}^2}}\,\sqrt {b\,\ln \relax (F)}}{c+d\,x}\right )}{8\,b^2\,d\,{\ln \relax (F)}^2\,\sqrt {b\,\ln \relax (F)}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- (F^a*F^(b/(c + d*x)^2))/(2*b*d*log(F)*(c + d*x)^3) - (F^a*(3*pi^(1/2)*erfi((b*log(F))/((b*log(F))^(1/2)*(c +
d*x))) - (6*F^(b/(c + d*x)^2)*(b*log(F))^(1/2))/(c + d*x)))/(8*b^2*d*log(F)^2*(b*log(F))^(1/2))

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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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