3.218 \(\int e^{\frac {1}{2} i \tan ^{-1}(a+b x)} \, dx\)

Optimal. Leaf size=338 \[ \frac {i (-i a-i b x+1)^{3/4} \sqrt [4]{i a+i b x+1}}{b}+\frac {i \log \left (\frac {\sqrt {-i a-i b x+1}}{\sqrt {i a+i b x+1}}-\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}+1\right )}{2 \sqrt {2} b}-\frac {i \log \left (\frac {\sqrt {-i a-i b x+1}}{\sqrt {i a+i b x+1}}+\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}+1\right )}{2 \sqrt {2} b}-\frac {i \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}\right )}{\sqrt {2} b}+\frac {i \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}\right )}{\sqrt {2} b} \]

[Out]

I*(1-I*a-I*b*x)^(3/4)*(1+I*a+I*b*x)^(1/4)/b-1/2*I*arctan(1-(1-I*a-I*b*x)^(1/4)*2^(1/2)/(1+I*a+I*b*x)^(1/4))/b*
2^(1/2)+1/2*I*arctan(1+(1-I*a-I*b*x)^(1/4)*2^(1/2)/(1+I*a+I*b*x)^(1/4))/b*2^(1/2)+1/4*I*ln(1-(1-I*a-I*b*x)^(1/
4)*2^(1/2)/(1+I*a+I*b*x)^(1/4)+(1-I*a-I*b*x)^(1/2)/(1+I*a+I*b*x)^(1/2))/b*2^(1/2)-1/4*I*ln(1+(1-I*a-I*b*x)^(1/
4)*2^(1/2)/(1+I*a+I*b*x)^(1/4)+(1-I*a-I*b*x)^(1/2)/(1+I*a+I*b*x)^(1/2))/b*2^(1/2)

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Rubi [A]  time = 0.20, antiderivative size = 338, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 10, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.714, Rules used = {5093, 50, 63, 331, 297, 1162, 617, 204, 1165, 628} \[ \frac {i (-i a-i b x+1)^{3/4} \sqrt [4]{i a+i b x+1}}{b}+\frac {i \log \left (\frac {\sqrt {-i a-i b x+1}}{\sqrt {i a+i b x+1}}-\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}+1\right )}{2 \sqrt {2} b}-\frac {i \log \left (\frac {\sqrt {-i a-i b x+1}}{\sqrt {i a+i b x+1}}+\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}+1\right )}{2 \sqrt {2} b}-\frac {i \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}\right )}{\sqrt {2} b}+\frac {i \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}\right )}{\sqrt {2} b} \]

Antiderivative was successfully verified.

[In]

Int[E^((I/2)*ArcTan[a + b*x]),x]

[Out]

(I*(1 - I*a - I*b*x)^(3/4)*(1 + I*a + I*b*x)^(1/4))/b - (I*ArcTan[1 - (Sqrt[2]*(1 - I*a - I*b*x)^(1/4))/(1 + I
*a + I*b*x)^(1/4)])/(Sqrt[2]*b) + (I*ArcTan[1 + (Sqrt[2]*(1 - I*a - I*b*x)^(1/4))/(1 + I*a + I*b*x)^(1/4)])/(S
qrt[2]*b) + ((I/2)*Log[1 + Sqrt[1 - I*a - I*b*x]/Sqrt[1 + I*a + I*b*x] - (Sqrt[2]*(1 - I*a - I*b*x)^(1/4))/(1
+ I*a + I*b*x)^(1/4)])/(Sqrt[2]*b) - ((I/2)*Log[1 + Sqrt[1 - I*a - I*b*x]/Sqrt[1 + I*a + I*b*x] + (Sqrt[2]*(1
- I*a - I*b*x)^(1/4))/(1 + I*a + I*b*x)^(1/4)])/(Sqrt[2]*b)

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 297

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]},
Dist[1/(2*s), Int[(r + s*x^2)/(a + b*x^4), x], x] - Dist[1/(2*s), Int[(r - s*x^2)/(a + b*x^4), x], x]] /; Free
Q[{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ,
b]]))

Rule 331

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + (m + 1)/n), Subst[Int[x^m/(1 - b*x^n)^(
p + (m + 1)/n + 1), x], x, x/(a + b*x^n)^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[
p, -2^(-1)] && IntegersQ[m, p + (m + 1)/n]

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1162

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e, 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1165

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e, 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

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^{\frac {1}{2} i \tan ^{-1}(a+b x)} \, dx &=\int \frac {\sqrt [4]{1+i a+i b x}}{\sqrt [4]{1-i a-i b x}} \, dx\\ &=\frac {i (1-i a-i b x)^{3/4} \sqrt [4]{1+i a+i b x}}{b}+\frac {1}{2} \int \frac {1}{\sqrt [4]{1-i a-i b x} (1+i a+i b x)^{3/4}} \, dx\\ &=\frac {i (1-i a-i b x)^{3/4} \sqrt [4]{1+i a+i b x}}{b}+\frac {(2 i) \operatorname {Subst}\left (\int \frac {x^2}{\left (2-x^4\right )^{3/4}} \, dx,x,\sqrt [4]{1-i a-i b x}\right )}{b}\\ &=\frac {i (1-i a-i b x)^{3/4} \sqrt [4]{1+i a+i b x}}{b}+\frac {(2 i) \operatorname {Subst}\left (\int \frac {x^2}{1+x^4} \, dx,x,\frac {\sqrt [4]{1-i a-i b x}}{\sqrt [4]{1+i a+i b x}}\right )}{b}\\ &=\frac {i (1-i a-i b x)^{3/4} \sqrt [4]{1+i a+i b x}}{b}-\frac {i \operatorname {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,\frac {\sqrt [4]{1-i a-i b x}}{\sqrt [4]{1+i a+i b x}}\right )}{b}+\frac {i \operatorname {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,\frac {\sqrt [4]{1-i a-i b x}}{\sqrt [4]{1+i a+i b x}}\right )}{b}\\ &=\frac {i (1-i a-i b x)^{3/4} \sqrt [4]{1+i a+i b x}}{b}+\frac {i \operatorname {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\frac {\sqrt [4]{1-i a-i b x}}{\sqrt [4]{1+i a+i b x}}\right )}{2 b}+\frac {i \operatorname {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\frac {\sqrt [4]{1-i a-i b x}}{\sqrt [4]{1+i a+i b x}}\right )}{2 b}+\frac {i \operatorname {Subst}\left (\int \frac {\sqrt {2}+2 x}{-1-\sqrt {2} x-x^2} \, dx,x,\frac {\sqrt [4]{1-i a-i b x}}{\sqrt [4]{1+i a+i b x}}\right )}{2 \sqrt {2} b}+\frac {i \operatorname {Subst}\left (\int \frac {\sqrt {2}-2 x}{-1+\sqrt {2} x-x^2} \, dx,x,\frac {\sqrt [4]{1-i a-i b x}}{\sqrt [4]{1+i a+i b x}}\right )}{2 \sqrt {2} b}\\ &=\frac {i (1-i a-i b x)^{3/4} \sqrt [4]{1+i a+i b x}}{b}+\frac {i \log \left (1+\frac {\sqrt {1-i a-i b x}}{\sqrt {1+i a+i b x}}-\frac {\sqrt {2} \sqrt [4]{1-i a-i b x}}{\sqrt [4]{1+i a+i b x}}\right )}{2 \sqrt {2} b}-\frac {i \log \left (1+\frac {\sqrt {1-i a-i b x}}{\sqrt {1+i a+i b x}}+\frac {\sqrt {2} \sqrt [4]{1-i a-i b x}}{\sqrt [4]{1+i a+i b x}}\right )}{2 \sqrt {2} b}+\frac {i \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt [4]{1-i a-i b x}}{\sqrt [4]{1+i a+i b x}}\right )}{\sqrt {2} b}-\frac {i \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt [4]{1-i a-i b x}}{\sqrt [4]{1+i a+i b x}}\right )}{\sqrt {2} b}\\ &=\frac {i (1-i a-i b x)^{3/4} \sqrt [4]{1+i a+i b x}}{b}-\frac {i \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{1-i a-i b x}}{\sqrt [4]{1+i a+i b x}}\right )}{\sqrt {2} b}+\frac {i \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{1-i a-i b x}}{\sqrt [4]{1+i a+i b x}}\right )}{\sqrt {2} b}+\frac {i \log \left (1+\frac {\sqrt {1-i a-i b x}}{\sqrt {1+i a+i b x}}-\frac {\sqrt {2} \sqrt [4]{1-i a-i b x}}{\sqrt [4]{1+i a+i b x}}\right )}{2 \sqrt {2} b}-\frac {i \log \left (1+\frac {\sqrt {1-i a-i b x}}{\sqrt {1+i a+i b x}}+\frac {\sqrt {2} \sqrt [4]{1-i a-i b x}}{\sqrt [4]{1+i a+i b x}}\right )}{2 \sqrt {2} b}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 45, normalized size = 0.13 \[ -\frac {8 i e^{\frac {5}{2} i \tan ^{-1}(a+b x)} \, _2F_1\left (\frac {5}{4},2;\frac {9}{4};-e^{2 i \tan ^{-1}(a+b x)}\right )}{5 b} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^((I/2)*ArcTan[a + b*x]),x]

[Out]

(((-8*I)/5)*E^(((5*I)/2)*ArcTan[a + b*x])*Hypergeometric2F1[5/4, 2, 9/4, -E^((2*I)*ArcTan[a + b*x])])/b

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fricas [A]  time = 0.45, size = 257, normalized size = 0.76 \[ \frac {b \sqrt {\frac {i}{b^{2}}} \log \left (i \, b \sqrt {\frac {i}{b^{2}}} + \sqrt {\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}}\right ) - b \sqrt {\frac {i}{b^{2}}} \log \left (-i \, b \sqrt {\frac {i}{b^{2}}} + \sqrt {\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}}\right ) + b \sqrt {-\frac {i}{b^{2}}} \log \left (i \, b \sqrt {-\frac {i}{b^{2}}} + \sqrt {\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}}\right ) - b \sqrt {-\frac {i}{b^{2}}} \log \left (-i \, b \sqrt {-\frac {i}{b^{2}}} + \sqrt {\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}}\right ) + {\left (2 \, b x + 2 \, a + 2 i\right )} \sqrt {\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}}}{2 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((1+I*(b*x+a))/(1+(b*x+a)^2)^(1/2))^(1/2),x, algorithm="fricas")

[Out]

1/2*(b*sqrt(I/b^2)*log(I*b*sqrt(I/b^2) + sqrt(I*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)/(b*x + a + I))) - b*sqrt(I/b
^2)*log(-I*b*sqrt(I/b^2) + sqrt(I*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)/(b*x + a + I))) + b*sqrt(-I/b^2)*log(I*b*s
qrt(-I/b^2) + sqrt(I*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)/(b*x + a + I))) - b*sqrt(-I/b^2)*log(-I*b*sqrt(-I/b^2)
+ sqrt(I*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)/(b*x + a + I))) + (2*b*x + 2*a + 2*I)*sqrt(I*sqrt(b^2*x^2 + 2*a*b*x
+ a^2 + 1)/(b*x + a + I)))/b

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((1+I*(b*x+a))/(1+(b*x+a)^2)^(1/2))^(1/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, need to choose a branch for the root of a polynomial with parameters. This might be wrong.The choice was
done assuming 0=[0,0,0]index.cc index_m operator + Error: Bad Argument ValueWarning, need to choose a branch f
or the root of a polynomial with parameters. This might be wrong.The choice was done assuming 0=[0,0,0]index.c
c index_m operator + Error: Bad Argument ValueEvaluation time: 0.54Done

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maple [F]  time = 0.12, size = 0, normalized size = 0.00 \[ \int \sqrt {\frac {1+i \left (b x +a \right )}{\sqrt {1+\left (b x +a \right )^{2}}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((1+I*(b*x+a))/(1+(b*x+a)^2)^(1/2))^(1/2),x)

[Out]

int(((1+I*(b*x+a))/(1+(b*x+a)^2)^(1/2))^(1/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {\frac {i \, b x + i \, a + 1}{\sqrt {{\left (b x + a\right )}^{2} + 1}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((1+I*(b*x+a))/(1+(b*x+a)^2)^(1/2))^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt((I*b*x + I*a + 1)/sqrt((b*x + a)^2 + 1)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \sqrt {\frac {1+a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}{\sqrt {{\left (a+b\,x\right )}^2+1}}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a*1i + b*x*1i + 1)/((a + b*x)^2 + 1)^(1/2))^(1/2),x)

[Out]

int(((a*1i + b*x*1i + 1)/((a + b*x)^2 + 1)^(1/2))^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {\frac {i \left (a + b x\right ) + 1}{\sqrt {\left (a + b x\right )^{2} + 1}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((1+I*(b*x+a))/(1+(b*x+a)**2)**(1/2))**(1/2),x)

[Out]

Integral(sqrt((I*(a + b*x) + 1)/sqrt((a + b*x)**2 + 1)), x)

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