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

Optimal. Leaf size=67 \[ \frac {(c+d x) F^{a+\frac {b}{(c+d x)^2}}}{d}-\frac {\sqrt {\pi } \sqrt {b} F^a \sqrt {\log (F)} \text {erfi}\left (\frac {\sqrt {b} \sqrt {\log (F)}}{c+d x}\right )}{d} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {2206, 2211, 2204} \[ \frac {(c+d x) F^{a+\frac {b}{(c+d x)^2}}}{d}-\frac {\sqrt {\pi } \sqrt {b} F^a \sqrt {\log (F)} \text {Erfi}\left (\frac {\sqrt {b} \sqrt {\log (F)}}{c+d x}\right )}{d} \]

Antiderivative was successfully verified.

[In]

Int[F^(a + b/(c + d*x)^2),x]

[Out]

(F^(a + b/(c + d*x)^2)*(c + d*x))/d - (Sqrt[b]*F^a*Sqrt[Pi]*Erfi[(Sqrt[b]*Sqrt[Log[F]])/(c + d*x)]*Sqrt[Log[F]
])/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 2206

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_)), x_Symbol] :> Simp[((c + d*x)*F^(a + b*(c + d*x)^n))/d, x]
- Dist[b*n*Log[F], Int[(c + d*x)^n*F^(a + b*(c + d*x)^n), x], x] /; FreeQ[{F, a, b, c, d}, x] && IntegerQ[2/n]
&& ILtQ[n, 0]

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
)]

Rubi steps

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

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Mathematica [A]  time = 0.04, size = 63, normalized size = 0.94 \[ \frac {F^a \left ((c+d x) F^{\frac {b}{(c+d x)^2}}-\sqrt {\pi } \sqrt {b} \sqrt {\log (F)} \text {erfi}\left (\frac {\sqrt {b} \sqrt {\log (F)}}{c+d x}\right )\right )}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[F^(a + b/(c + d*x)^2),x]

[Out]

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

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fricas [A]  time = 0.41, size = 91, normalized size = 1.36 \[ \frac {\sqrt {\pi } F^{a} d \sqrt {-\frac {b \log \relax (F)}{d^{2}}} \operatorname {erf}\left (\frac {d \sqrt {-\frac {b \log \relax (F)}{d^{2}}}}{d x + c}\right ) + {\left (d x + c\right )} F^{\frac {a d^{2} x^{2} + 2 \, a c d x + a c^{2} + b}{d^{2} x^{2} + 2 \, c d x + c^{2}}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(F^(a + b/(d*x + c)^2), x)

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maple [A]  time = 0.06, size = 74, normalized size = 1.10 \[ -\frac {\sqrt {\pi }\, b \,F^{a} \erf \left (\frac {\sqrt {-b \ln \relax (F )}}{d x +c}\right ) \ln \relax (F )}{\sqrt {-b \ln \relax (F )}\, d}+x \,F^{a} F^{\frac {b}{\left (d x +c \right )^{2}}}+\frac {c \,F^{a} F^{\frac {b}{\left (d x +c \right )^{2}}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

F^a*F^(1/(d*x+c)^2*b)*x+1/d*F^a*F^(1/(d*x+c)^2*b)*c-1/d*F^a*b*ln(F)*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 \[ 2 \, F^{a} b d \int \frac {F^{\frac {b}{d^{2} x^{2} + 2 \, c d x + c^{2}}} x}{d^{3} x^{3} + 3 \, c d^{2} x^{2} + 3 \, c^{2} d x + c^{3}}\,{d x} \log \relax (F) + F^{a} F^{\frac {b}{d^{2} x^{2} + 2 \, c d x + c^{2}}} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*F^a*b*d*integrate(F^(b/(d^2*x^2 + 2*c*d*x + c^2))*x/(d^3*x^3 + 3*c*d^2*x^2 + 3*c^2*d*x + c^3), x)*log(F) + F
^a*F^(b/(d^2*x^2 + 2*c*d*x + c^2))*x

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mupad [B]  time = 4.77, size = 62, normalized size = 0.93 \[ \frac {F^a\,F^{\frac {b}{{\left (c+d\,x\right )}^2}}\,\left (c+d\,x\right )}{d}-\frac {F^a\,b\,\sqrt {\pi }\,\mathrm {erfi}\left (\frac {b\,\ln \relax (F)}{\sqrt {b\,\ln \relax (F)}\,\left (c+d\,x\right )}\right )\,\ln \relax (F)}{d\,\sqrt {b\,\ln \relax (F)}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(F^(a + b/(c + d*x)^2),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(F**(a + b/(c + d*x)**2), x)

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