\(\int F^{-c (a+b x)^n} (a+b x)^{-1+\frac {n}{2}} \, dx\) [381]

Optimal result

Integrand size = 26, antiderivative size = 47 \[ \int F^{-c (a+b x)^n} (a+b x)^{-1+\frac {n}{2}} \, dx=\frac {\sqrt {\pi } \text {erf}\left (\sqrt {c} (a+b x)^{n/2} \sqrt {\log (F)}\right )}{b \sqrt {c} n \sqrt {\log (F)}} \]

[Out]

erf((b*x+a)^(1/2*n)*c^(1/2)*ln(F)^(1/2))*Pi^(1/2)/b/n/c^(1/2)/ln(F)^(1/2)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {2242, 2236} \[ \int F^{-c (a+b x)^n} (a+b x)^{-1+\frac {n}{2}} \, dx=\frac {\sqrt {\pi } \text {erf}\left (\sqrt {c} \sqrt {\log (F)} (a+b x)^{n/2}\right )}{b \sqrt {c} n \sqrt {\log (F)}} \]

[In]

Int[(a + b*x)^(-1 + n/2)/F^(c*(a + b*x)^n),x]

[Out]

(Sqrt[Pi]*Erf[Sqrt[c]*(a + b*x)^(n/2)*Sqrt[Log[F]]])/(b*Sqrt[c]*n*Sqrt[Log[F]])

Rule 2236

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[F^a*Sqrt[Pi]*(Erf[(c + d*x)*Rt[(-b)*Log[F],
2]]/(2*d*Rt[(-b)*Log[F], 2])), x] /; FreeQ[{F, a, b, c, d}, x] && NegQ[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
)]

Rubi steps \begin{align*} \text {integral}& = \frac {2 \text {Subst}\left (\int F^{-c x^2} \, dx,x,(a+b x)^{n/2}\right )}{b n} \\ & = \frac {\sqrt {\pi } \text {erf}\left (\sqrt {c} (a+b x)^{n/2} \sqrt {\log (F)}\right )}{b \sqrt {c} n \sqrt {\log (F)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.00 \[ \int F^{-c (a+b x)^n} (a+b x)^{-1+\frac {n}{2}} \, dx=\frac {\sqrt {\pi } \text {erf}\left (\sqrt {c} (a+b x)^{n/2} \sqrt {\log (F)}\right )}{b \sqrt {c} n \sqrt {\log (F)}} \]

[In]

Integrate[(a + b*x)^(-1 + n/2)/F^(c*(a + b*x)^n),x]

[Out]

(Sqrt[Pi]*Erf[Sqrt[c]*(a + b*x)^(n/2)*Sqrt[Log[F]]])/(b*Sqrt[c]*n*Sqrt[Log[F]])

Maple [A] (verified)

Time = 0.41 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.72

method	result	size
risch	\(\frac {\sqrt {\pi }\, \operatorname {erf}\left (\sqrt {c \ln \left (F \right )}\, \left (b x +a \right )^{\frac {n}{2}}\right )}{n b \sqrt {c \ln \left (F \right )}}\)	\(34\)

[In]

int((b*x+a)^(-1+1/2*n)/(F^(c*(b*x+a)^n)),x,method=_RETURNVERBOSE)

[Out]

1/n/b*Pi^(1/2)/(c*ln(F))^(1/2)*erf((c*ln(F))^(1/2)*(b*x+a)^(1/2*n))

Fricas [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.00 \[ \int F^{-c (a+b x)^n} (a+b x)^{-1+\frac {n}{2}} \, dx=\frac {\sqrt {\pi } \sqrt {c \log \left (F\right )} \operatorname {erf}\left ({\left (b x + a\right )} \sqrt {c \log \left (F\right )} {\left (b x + a\right )}^{\frac {1}{2} \, n - 1}\right )}{b c n \log \left (F\right )} \]

[In]

integrate((b*x+a)^(-1+1/2*n)/(F^(c*(b*x+a)^n)),x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int F^{-c (a+b x)^n} (a+b x)^{-1+\frac {n}{2}} \, dx=\int F^{- c \left (a + b x\right )^{n}} \left (a + b x\right )^{\frac {n}{2} - 1}\, dx \]

[In]

integrate((b*x+a)**(-1+1/2*n)/(F**(c*(b*x+a)**n)),x)

[Out]

Integral((a + b*x)**(n/2 - 1)/F**(c*(a + b*x)**n), x)

Maxima [F]

\[ \int F^{-c (a+b x)^n} (a+b x)^{-1+\frac {n}{2}} \, dx=\int { \frac {{\left (b x + a\right )}^{\frac {1}{2} \, n - 1}}{F^{{\left (b x + a\right )}^{n} c}} \,d x } \]

[In]

integrate((b*x+a)^(-1+1/2*n)/(F^(c*(b*x+a)^n)),x, algorithm="maxima")

[Out]

integrate((b*x + a)^(1/2*n - 1)/F^((b*x + a)^n*c), x)

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.74 \[ \int F^{-c (a+b x)^n} (a+b x)^{-1+\frac {n}{2}} \, dx=-\frac {\sqrt {\pi } \operatorname {erf}\left (-\sqrt {c \log \left (F\right )} \sqrt {{\left (b x + a\right )}^{n}}\right )}{\sqrt {c \log \left (F\right )} b n} \]

[In]

integrate((b*x+a)^(-1+1/2*n)/(F^(c*(b*x+a)^n)),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.69 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.74 \[ \int F^{-c (a+b x)^n} (a+b x)^{-1+\frac {n}{2}} \, dx=\frac {\sqrt {\pi }\,\mathrm {erf}\left (\sqrt {c}\,\sqrt {\ln \left (F\right )}\,{\left (a+b\,x\right )}^{n/2}\right )}{b\,\sqrt {c}\,n\,\sqrt {\ln \left (F\right )}} \]

[In]

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

[Out]

(pi^(1/2)*erf(c^(1/2)*log(F)^(1/2)*(a + b*x)^(n/2)))/(b*c^(1/2)*n*log(F)^(1/2))