\(\int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^4} \, dx\) [334]

Optimal result

Integrand size = 21, antiderivative size = 81 \[ \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^4} \, dx=\frac {F^a \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {b} \sqrt {\log (F)}}{c+d x}\right )}{4 b^{3/2} d \log ^{\frac {3}{2}}(F)}-\frac {F^{a+\frac {b}{(c+d x)^2}}}{2 b d (c+d x) \log (F)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2243, 2242, 2235} \[ \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^4} \, dx=\frac {\sqrt {\pi } F^a \text {erfi}\left (\frac {\sqrt {b} \sqrt {\log (F)}}{c+d x}\right )}{4 b^{3/2} d \log ^{\frac {3}{2}}(F)}-\frac {F^{a+\frac {b}{(c+d x)^2}}}{2 b d \log (F) (c+d x)} \]

[In]

Int[F^(a + b/(c + d*x)^2)/(c + d*x)^4,x]

[Out]

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

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 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
)]

Rule 2243

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.00 \[ \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^4} \, dx=\frac {F^a \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {b} \sqrt {\log (F)}}{c+d x}\right )}{4 b^{3/2} d \log ^{\frac {3}{2}}(F)}-\frac {F^{a+\frac {b}{(c+d x)^2}}}{2 b d (c+d x) \log (F)} \]

[In]

Integrate[F^(a + b/(c + d*x)^2)/(c + d*x)^4,x]

[Out]

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

Maple [A] (verified)

Time = 0.69 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.94

method	result	size
risch	\(-\frac {F^{a} F^{\frac {b}{\left (d x +c \right )^{2}}}}{2 d \left (d x +c \right ) b \ln \left (F \right )}+\frac {F^{a} \sqrt {\pi }\, \operatorname {erf}\left (\frac {\sqrt {-b \ln \left (F \right )}}{d x +c}\right )}{4 d b \ln \left (F \right ) \sqrt {-b \ln \left (F \right )}}\)	\(76\)

[In]

int(F^(a+b/(d*x+c)^2)/(d*x+c)^4,x,method=_RETURNVERBOSE)

[Out]

-1/2/d*F^a*F^(b/(d*x+c)^2)/(d*x+c)/b/ln(F)+1/4/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 = 0.27 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.44 \[ \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^4} \, dx=-\frac {\sqrt {\pi } {\left (d^{2} x + c d\right )} F^{a} \sqrt {-\frac {b \log \left (F\right )}{d^{2}}} \operatorname {erf}\left (\frac {d \sqrt {-\frac {b \log \left (F\right )}{d^{2}}}}{d x + c}\right ) + 2 \, F^{\frac {a d^{2} x^{2} + 2 \, a c d x + a c^{2} + b}{d^{2} x^{2} + 2 \, c d x + c^{2}}} b \log \left (F\right )}{4 \, {\left (b^{2} d^{2} x + b^{2} c d\right )} \log \left (F\right )^{2}} \]

[In]

integrate(F^(a+b/(d*x+c)^2)/(d*x+c)^4,x, algorithm="fricas")

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^4} \, dx=\text {Timed out} \]

[In]

integrate(F**(a+b/(d*x+c)**2)/(d*x+c)**4,x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^4} \, dx=\int { \frac {F^{a + \frac {b}{{\left (d x + c\right )}^{2}}}}{{\left (d x + c\right )}^{4}} \,d x } \]

[In]

integrate(F^(a+b/(d*x+c)^2)/(d*x+c)^4,x, algorithm="maxima")

[Out]

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

Giac [F]

\[ \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^4} \, dx=\int { \frac {F^{a + \frac {b}{{\left (d x + c\right )}^{2}}}}{{\left (d x + c\right )}^{4}} \,d x } \]

[In]

integrate(F^(a+b/(d*x+c)^2)/(d*x+c)^4,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.62 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.94 \[ \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^4} \, dx=\frac {F^a\,\sqrt {\pi }\,\mathrm {erfi}\left (\frac {b\,\ln \left (F\right )}{\sqrt {b\,\ln \left (F\right )}\,\left (c+d\,x\right )}\right )}{4\,b\,d\,\ln \left (F\right )\,\sqrt {b\,\ln \left (F\right )}}-\frac {F^a\,F^{\frac {b}{{\left (c+d\,x\right )}^2}}}{2\,b\,d\,\ln \left (F\right )\,\left (c+d\,x\right )} \]

[In]

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

[Out]

(F^a*pi^(1/2)*erfi((b*log(F))/((b*log(F))^(1/2)*(c + d*x))))/(4*b*d*log(F)*(b*log(F))^(1/2)) - (F^a*F^(b/(c +
d*x)^2))/(2*b*d*log(F)*(c + d*x))