3.189 \(\int f^{a+b x^n} x^{-1+\frac {5 n}{2}} \, dx\)

Optimal. Leaf size=104 \[ \frac {3 \sqrt {\pi } f^a \text {erfi}\left (\sqrt {b} \sqrt {\log (f)} x^{n/2}\right )}{4 b^{5/2} n \log ^{\frac {5}{2}}(f)}-\frac {3 x^{n/2} f^{a+b x^n}}{2 b^2 n \log ^2(f)}+\frac {x^{3 n/2} f^{a+b x^n}}{b n \log (f)} \]

[Out]

-3/2*f^(a+b*x^n)*x^(1/2*n)/b^2/n/ln(f)^2+f^(a+b*x^n)*x^(3/2*n)/b/n/ln(f)+3/4*f^a*erfi(x^(1/2*n)*b^(1/2)*ln(f)^
(1/2))*Pi^(1/2)/b^(5/2)/n/ln(f)^(5/2)

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Rubi [A]  time = 0.11, antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {2213, 2211, 2204} \[ \frac {3 \sqrt {\pi } f^a \text {Erfi}\left (\sqrt {b} \sqrt {\log (f)} x^{n/2}\right )}{4 b^{5/2} n \log ^{\frac {5}{2}}(f)}-\frac {3 x^{n/2} f^{a+b x^n}}{2 b^2 n \log ^2(f)}+\frac {x^{3 n/2} f^{a+b x^n}}{b n \log (f)} \]

Antiderivative was successfully verified.

[In]

Int[f^(a + b*x^n)*x^(-1 + (5*n)/2),x]

[Out]

(3*f^a*Sqrt[Pi]*Erfi[Sqrt[b]*x^(n/2)*Sqrt[Log[f]]])/(4*b^(5/2)*n*Log[f]^(5/2)) - (3*f^(a + b*x^n)*x^(n/2))/(2*
b^2*n*Log[f]^2) + (f^(a + b*x^n)*x^((3*n)/2))/(b*n*Log[f])

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 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
)]

Rule 2213

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)^Simplify[m
- n]*F^(a + b*(c + d*x)^n), x], x] /; FreeQ[{F, a, b, c, d, m, n}, x] && IntegerQ[2*Simplify[(m + 1)/n]] && Lt
Q[0, Simplify[(m + 1)/n], 5] &&  !RationalQ[m] && SumSimplerQ[m, -n]

Rubi steps

\begin {align*} \int f^{a+b x^n} x^{-1+\frac {5 n}{2}} \, dx &=\frac {f^{a+b x^n} x^{3 n/2}}{b n \log (f)}-\frac {3 \int f^{a+b x^n} x^{-1+\frac {3 n}{2}} \, dx}{2 b \log (f)}\\ &=-\frac {3 f^{a+b x^n} x^{n/2}}{2 b^2 n \log ^2(f)}+\frac {f^{a+b x^n} x^{3 n/2}}{b n \log (f)}+\frac {3 \int f^{a+b x^n} x^{\frac {1}{2} (-2+n)} \, dx}{4 b^2 \log ^2(f)}\\ &=-\frac {3 f^{a+b x^n} x^{n/2}}{2 b^2 n \log ^2(f)}+\frac {f^{a+b x^n} x^{3 n/2}}{b n \log (f)}+\frac {3 \operatorname {Subst}\left (\int f^{a+b x^2} \, dx,x,x^{1+\frac {1}{2} (-2+n)}\right )}{2 b^2 n \log ^2(f)}\\ &=\frac {3 f^a \sqrt {\pi } \text {erfi}\left (\sqrt {b} x^{n/2} \sqrt {\log (f)}\right )}{4 b^{5/2} n \log ^{\frac {5}{2}}(f)}-\frac {3 f^{a+b x^n} x^{n/2}}{2 b^2 n \log ^2(f)}+\frac {f^{a+b x^n} x^{3 n/2}}{b n \log (f)}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 39, normalized size = 0.38 \[ -\frac {f^a x^{5 n/2} \Gamma \left (\frac {5}{2},-b x^n \log (f)\right )}{n \left (-b \log (f) x^n\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[f^(a + b*x^n)*x^(-1 + (5*n)/2),x]

[Out]

-((f^a*x^((5*n)/2)*Gamma[5/2, -(b*x^n*Log[f])])/(n*(-(b*x^n*Log[f]))^(5/2)))

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fricas [A]  time = 0.44, size = 82, normalized size = 0.79 \[ -\frac {3 \, \sqrt {\pi } \sqrt {-b \log \relax (f)} f^{a} \operatorname {erf}\left (\sqrt {-b \log \relax (f)} x^{\frac {1}{2} \, n}\right ) - 2 \, {\left (2 \, b^{2} x^{\frac {3}{2} \, n} \log \relax (f)^{2} - 3 \, b x^{\frac {1}{2} \, n} \log \relax (f)\right )} e^{\left (b x^{n} \log \relax (f) + a \log \relax (f)\right )}}{4 \, b^{3} n \log \relax (f)^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(a+b*x^n)*x^(-1+5/2*n),x, algorithm="fricas")

[Out]

-1/4*(3*sqrt(pi)*sqrt(-b*log(f))*f^a*erf(sqrt(-b*log(f))*x^(1/2*n)) - 2*(2*b^2*x^(3/2*n)*log(f)^2 - 3*b*x^(1/2
*n)*log(f))*e^(b*x^n*log(f) + a*log(f)))/(b^3*n*log(f)^3)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int f^{b x^{n} + a} x^{\frac {5}{2} \, n - 1}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(a+b*x^n)*x^(-1+5/2*n),x, algorithm="giac")

[Out]

integrate(f^(b*x^n + a)*x^(5/2*n - 1), x)

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maple [A]  time = 0.11, size = 96, normalized size = 0.92 \[ \frac {f^{a} f^{b \,x^{n}} x^{\frac {3 n}{2}}}{b n \ln \relax (f )}-\frac {3 f^{a} f^{b \,x^{n}} x^{\frac {n}{2}}}{2 b^{2} n \ln \relax (f )^{2}}+\frac {3 \sqrt {\pi }\, f^{a} \erf \left (\sqrt {-b \ln \relax (f )}\, x^{\frac {n}{2}}\right )}{4 \sqrt {-b \ln \relax (f )}\, b^{2} n \ln \relax (f )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^(b*x^n+a)*x^(-1+5/2*n),x)

[Out]

f^a/n*f^(b*x^n)*(x^(1/2*n))^3/b/ln(f)-3/2*f^a/n/ln(f)^2/b^2*x^(1/2*n)*f^(b*x^n)+3/4*f^a/n/ln(f)^2/b^2*Pi^(1/2)
/(-b*ln(f))^(1/2)*erf((-b*ln(f))^(1/2)*x^(1/2*n))

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maxima [A]  time = 1.21, size = 33, normalized size = 0.32 \[ -\frac {f^{a} x^{\frac {5}{2} \, n} \Gamma \left (\frac {5}{2}, -b x^{n} \log \relax (f)\right )}{\left (-b x^{n} \log \relax (f)\right )^{\frac {5}{2}} n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(a+b*x^n)*x^(-1+5/2*n),x, algorithm="maxima")

[Out]

-f^a*x^(5/2*n)*gamma(5/2, -b*x^n*log(f))/((-b*x^n*log(f))^(5/2)*n)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int f^{a+b\,x^n}\,x^{\frac {5\,n}{2}-1} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^(a + b*x^n)*x^((5*n)/2 - 1),x)

[Out]

int(f^(a + b*x^n)*x^((5*n)/2 - 1), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f**(a+b*x**n)*x**(-1+5/2*n),x)

[Out]

Timed out

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