3.240 \(\int e^{n \tan ^{-1}(a+b x)} \, dx\)

Optimal. Leaf size=91 \[ -\frac {2^{1-\frac {i n}{2}} (-i a-i b x+1)^{1+\frac {i n}{2}} \, _2F_1\left (\frac {i n}{2}+1,\frac {i n}{2};\frac {i n}{2}+2;\frac {1}{2} (-i a-i b x+1)\right )}{b (-n+2 i)} \]

[Out]

-2^(1-1/2*I*n)*(1-I*a-I*b*x)^(1+1/2*I*n)*hypergeom([1/2*I*n, 1+1/2*I*n],[2+1/2*I*n],1/2-1/2*I*a-1/2*I*b*x)/b/(
2*I-n)

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Rubi [A]  time = 0.01, antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {5093, 69} \[ -\frac {2^{1-\frac {i n}{2}} (-i a-i b x+1)^{1+\frac {i n}{2}} \, _2F_1\left (\frac {i n}{2}+1,\frac {i n}{2};\frac {i n}{2}+2;\frac {1}{2} (-i a-i b x+1)\right )}{b (-n+2 i)} \]

Antiderivative was successfully verified.

[In]

Int[E^(n*ArcTan[a + b*x]),x]

[Out]

-((2^(1 - (I/2)*n)*(1 - I*a - I*b*x)^(1 + (I/2)*n)*Hypergeometric2F1[1 + (I/2)*n, (I/2)*n, 2 + (I/2)*n, (1 - I
*a - I*b*x)/2])/(b*(2*I - n)))

Rule 69

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*Hypergeometric2F1[
-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b*(m + 1)*(b/(b*c - a*d))^n), x] /; FreeQ[{a, b, c, d, m, n}
, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] ||  !(Ra
tionalQ[n] && GtQ[-(d/(b*c - a*d)), 0]))

Rule 5093

Int[E^(ArcTan[(c_.)*((a_) + (b_.)*(x_))]*(n_.)), x_Symbol] :> Int[(1 - I*a*c - I*b*c*x)^((I*n)/2)/(1 + I*a*c +
I*b*c*x)^((I*n)/2), x] /; FreeQ[{a, b, c, n}, x]

Rubi steps

\begin {align*} \int e^{n \tan ^{-1}(a+b x)} \, dx &=\int (1-i a-i b x)^{\frac {i n}{2}} (1+i a+i b x)^{-\frac {i n}{2}} \, dx\\ &=-\frac {2^{1-\frac {i n}{2}} (1-i a-i b x)^{1+\frac {i n}{2}} \, _2F_1\left (1+\frac {i n}{2},\frac {i n}{2};2+\frac {i n}{2};\frac {1}{2} (1-i a-i b x)\right )}{b (2 i-n)}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 60, normalized size = 0.66 \[ \frac {4 e^{(n+2 i) \tan ^{-1}(a+b x)} \, _2F_1\left (2,1-\frac {i n}{2};2-\frac {i n}{2};-e^{2 i \tan ^{-1}(a+b x)}\right )}{b (n+2 i)} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^(n*ArcTan[a + b*x]),x]

[Out]

(4*E^((2*I + n)*ArcTan[a + b*x])*Hypergeometric2F1[2, 1 - (I/2)*n, 2 - (I/2)*n, -E^((2*I)*ArcTan[a + b*x])])/(
b*(2*I + n))

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fricas [F]  time = 0.43, size = 0, normalized size = 0.00 \[ {\rm integral}\left (e^{\left (n \arctan \left (b x + a\right )\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*arctan(b*x+a)),x, algorithm="fricas")

[Out]

integral(e^(n*arctan(b*x + a)), x)

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giac [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(exp(n*arctan(b*x+a)),x, algorithm="giac")

[Out]

Timed out

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maple [F]  time = 0.07, size = 0, normalized size = 0.00 \[ \int {\mathrm e}^{n \arctan \left (b x +a \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(n*arctan(b*x+a)),x)

[Out]

int(exp(n*arctan(b*x+a)),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int e^{\left (n \arctan \left (b x + a\right )\right )}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*arctan(b*x+a)),x, algorithm="maxima")

[Out]

integrate(e^(n*arctan(b*x + a)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\mathrm {e}}^{n\,\mathrm {atan}\left (a+b\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(n*atan(a + b*x)),x)

[Out]

int(exp(n*atan(a + b*x)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int e^{n \operatorname {atan}{\left (a + b x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*atan(b*x+a)),x)

[Out]

Integral(exp(n*atan(a + b*x)), x)

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