\(\int \frac {F^{a+b (c+d x)^2}}{(c+d x)^2} \, dx\) [274]

Optimal result

Integrand size = 21, antiderivative size = 67 \[ \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^2} \, dx=-\frac {F^{a+b (c+d x)^2}}{d (c+d x)}+\frac {\sqrt {b} F^a \sqrt {\pi } \text {erfi}\left (\sqrt {b} (c+d x) \sqrt {\log (F)}\right ) \sqrt {\log (F)}}{d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2245, 2235} \[ \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^2} \, dx=\frac {\sqrt {\pi } \sqrt {b} F^a \sqrt {\log (F)} \text {erfi}\left (\sqrt {b} \sqrt {\log (F)} (c+d x)\right )}{d}-\frac {F^{a+b (c+d x)^2}}{d (c+d x)} \]

[In]

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

[Out]

-(F^(a + b*(c + d*x)^2)/(d*(c + d*x))) + (Sqrt[b]*F^a*Sqrt[Pi]*Erfi[Sqrt[b]*(c + d*x)*Sqrt[Log[F]]]*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 2245

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

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.94 \[ \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^2} \, dx=\frac {F^a \left (-\frac {F^{b (c+d x)^2}}{c+d x}+\sqrt {b} \sqrt {\pi } \text {erfi}\left (\sqrt {b} (c+d x) \sqrt {\log (F)}\right ) \sqrt {\log (F)}\right )}{d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.32 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.93

method	result	size
risch	\(-\frac {F^{b \left (d x +c \right )^{2}} F^{a}}{d \left (d x +c \right )}+\frac {b \ln \left (F \right ) \sqrt {\pi }\, F^{a} \operatorname {erf}\left (\sqrt {-b \ln \left (F \right )}\, \left (d x +c \right )\right )}{d \sqrt {-b \ln \left (F \right )}}\)	\(62\)

[In]

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

[Out]

-1/d/(d*x+c)*F^(b*(d*x+c)^2)*F^a+1/d*b*ln(F)*Pi^(1/2)*F^a/(-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 = 83, normalized size of antiderivative = 1.24 \[ \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^2} \, dx=-\frac {\sqrt {\pi } \sqrt {-b d^{2} \log \left (F\right )} {\left (d x + c\right )} F^{a} \operatorname {erf}\left (\frac {\sqrt {-b d^{2} \log \left (F\right )} {\left (d x + c\right )}}{d}\right ) + F^{b d^{2} x^{2} + 2 \, b c d x + b c^{2} + a} d}{d^{3} x + c d^{2}} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.69 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.28 \[ \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^2} \, dx=\frac {F^a\,b\,\sqrt {\pi }\,\mathrm {erfi}\left (\frac {b\,x\,\ln \left (F\right )\,d^2+b\,c\,\ln \left (F\right )\,d}{\sqrt {b\,d^2\,\ln \left (F\right )}}\right )\,\ln \left (F\right )}{\sqrt {b\,d^2\,\ln \left (F\right )}}-\frac {F^{b\,d^2\,x^2}\,F^a\,F^{b\,c^2}\,F^{2\,b\,c\,d\,x}}{d\,\left (c+d\,x\right )} \]

[In]

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

[Out]

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