\(\int F^{a+\frac {b}{(c+d x)^2}} \, dx\) [332]

Optimal result

Integrand size = 13, antiderivative size = 67 \[ \int F^{a+\frac {b}{(c+d x)^2}} \, dx=\frac {F^{a+\frac {b}{(c+d x)^2}} (c+d x)}{d}-\frac {\sqrt {b} F^a \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {b} \sqrt {\log (F)}}{c+d x}\right ) \sqrt {\log (F)}}{d} \]

[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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {2237, 2242, 2235} \[ \int F^{a+\frac {b}{(c+d x)^2}} \, dx=\frac {(c+d x) F^{a+\frac {b}{(c+d x)^2}}}{d}-\frac {\sqrt {\pi } \sqrt {b} F^a \sqrt {\log (F)} \text {erfi}\left (\frac {\sqrt {b} \sqrt {\log (F)}}{c+d x}\right )}{d} \]

[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 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 2237

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 2242

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*} \text {integral}& = \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)) \text {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*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.94 \[ \int F^{a+\frac {b}{(c+d x)^2}} \, dx=\frac {F^a \left (F^{\frac {b}{(c+d x)^2}} (c+d x)-\sqrt {b} \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {b} \sqrt {\log (F)}}{c+d x}\right ) \sqrt {\log (F)}\right )}{d} \]

[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

Maple [A] (verified)

Time = 0.11 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.10

method	result	size
risch	\(F^{a} F^{\frac {b}{\left (d x +c \right )^{2}}} x +\frac {F^{a} F^{\frac {b}{\left (d x +c \right )^{2}}} c}{d}-\frac {F^{a} b \ln \left (F \right ) \sqrt {\pi }\, \operatorname {erf}\left (\frac {\sqrt {-b \ln \left (F \right )}}{d x +c}\right )}{d \sqrt {-b \ln \left (F \right )}}\)	\(74\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.36 \[ \int F^{a+\frac {b}{(c+d x)^2}} \, dx=\frac {\sqrt {\pi } F^{a} d \sqrt {-\frac {b \log \left (F\right )}{d^{2}}} \operatorname {erf}\left (\frac {d \sqrt {-\frac {b \log \left (F\right )}{d^{2}}}}{d x + c}\right ) + {\left (d x + c\right )} F^{\frac {a d^{2} x^{2} + 2 \, a c d x + a c^{2} + b}{d^{2} x^{2} + 2 \, c d x + c^{2}}}}{d} \]

[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

Sympy [F]

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

[In]

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

[Out]

Integral(F**(a + b/(c + d*x)**2), x)

Maxima [F]

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

[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

Giac [F]

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

[In]

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

[Out]

integrate(F^(a + b/(d*x + c)^2), x)

Mupad [B] (verification not implemented)

Time = 1.33 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.93 \[ \int F^{a+\frac {b}{(c+d x)^2}} \, dx=\frac {F^a\,F^{\frac {b}{{\left (c+d\,x\right )}^2}}\,\left (c+d\,x\right )}{d}-\frac {F^a\,b\,\sqrt {\pi }\,\mathrm {erfi}\left (\frac {b\,\ln \left (F\right )}{\sqrt {b\,\ln \left (F\right )}\,\left (c+d\,x\right )}\right )\,\ln \left (F\right )}{d\,\sqrt {b\,\ln \left (F\right )}} \]

[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))