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

Optimal. Leaf size=210 \[ -\frac {\sqrt [4]{1-i (a+b x)} (-a-b x+i)}{(-a+i) x \sqrt [4]{1+i (a+b x)}}-\frac {i b \tan ^{-1}\left (\frac {\sqrt [4]{-a+i} \sqrt [4]{1-i (a+b x)}}{\sqrt [4]{a+i} \sqrt [4]{1+i (a+b x)}}\right )}{(-a+i)^{5/4} (a+i)^{3/4}}-\frac {i b \tanh ^{-1}\left (\frac {\sqrt [4]{-a+i} \sqrt [4]{1-i (a+b x)}}{\sqrt [4]{a+i} \sqrt [4]{1+i (a+b x)}}\right )}{(-a+i)^{5/4} (a+i)^{3/4}} \]

[Out]

-(I-a-b*x)*(1-I*(b*x+a))^(1/4)/(I-a)/x/(1+I*(b*x+a))^(1/4)-I*b*arctan((I-a)^(1/4)*(1-I*(b*x+a))^(1/4)/(I+a)^(1
/4)/(1+I*(b*x+a))^(1/4))/(I-a)^(5/4)/(I+a)^(3/4)-I*b*arctanh((I-a)^(1/4)*(1-I*(b*x+a))^(1/4)/(I+a)^(1/4)/(1+I*
(b*x+a))^(1/4))/(I-a)^(5/4)/(I+a)^(3/4)

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Rubi [A]  time = 0.10, antiderivative size = 210, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {5094, 288, 212, 208, 205} \[ -\frac {\sqrt [4]{1-i (a+b x)} (-a-b x+i)}{(-a+i) x \sqrt [4]{1+i (a+b x)}}-\frac {i b \tan ^{-1}\left (\frac {\sqrt [4]{-a+i} \sqrt [4]{1-i (a+b x)}}{\sqrt [4]{a+i} \sqrt [4]{1+i (a+b x)}}\right )}{(-a+i)^{5/4} (a+i)^{3/4}}-\frac {i b \tanh ^{-1}\left (\frac {\sqrt [4]{-a+i} \sqrt [4]{1-i (a+b x)}}{\sqrt [4]{a+i} \sqrt [4]{1+i (a+b x)}}\right )}{(-a+i)^{5/4} (a+i)^{3/4}} \]

Antiderivative was successfully verified.

[In]

Int[1/(E^((I/2)*ArcTan[a + b*x])*x^2),x]

[Out]

-(((I - a - b*x)*(1 - I*(a + b*x))^(1/4))/((I - a)*x*(1 + I*(a + b*x))^(1/4))) - (I*b*ArcTan[((I - a)^(1/4)*(1
- I*(a + b*x))^(1/4))/((I + a)^(1/4)*(1 + I*(a + b*x))^(1/4))])/((I - a)^(5/4)*(I + a)^(3/4)) - (I*b*ArcTanh[
((I - a)^(1/4)*(1 - I*(a + b*x))^(1/4))/((I + a)^(1/4)*(1 + I*(a + b*x))^(1/4))])/((I - a)^(5/4)*(I + a)^(3/4)
)

Rule 205

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

Rule 208

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

Rule 212

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

Rule 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 5094

Int[E^(ArcTan[(c_.)*((a_) + (b_.)*(x_))]*(n_))*(x_)^(m_), x_Symbol] :> Dist[4/(I^m*n*b^(m + 1)*c^(m + 1)), Sub
st[Int[(x^(2/(I*n))*(1 - I*a*c - (1 + I*a*c)*x^(2/(I*n)))^m)/(1 + x^(2/(I*n)))^(m + 2), x], x, (1 - I*c*(a + b
*x))^((I*n)/2)/(1 + I*c*(a + b*x))^((I*n)/2)], x] /; FreeQ[{a, b, c}, x] && ILtQ[m, 0] && LtQ[-1, I*n, 1]

Rubi steps

\begin {align*} \int \frac {e^{-\frac {1}{2} i \tan ^{-1}(a+b x)}}{x^2} \, dx &=-\left ((8 i b) \operatorname {Subst}\left (\int \frac {x^4}{\left (1-i a-(1+i a) x^4\right )^2} \, dx,x,\frac {\sqrt [4]{1-i (a+b x)}}{\sqrt [4]{1+i (a+b x)}}\right )\right )\\ &=-\frac {(i-a-b x) \sqrt [4]{1-i (a+b x)}}{(i-a) x \sqrt [4]{1+i (a+b x)}}-\frac {(2 b) \operatorname {Subst}\left (\int \frac {1}{1-i a+(-1-i a) x^4} \, dx,x,\frac {\sqrt [4]{1-i (a+b x)}}{\sqrt [4]{1+i (a+b x)}}\right )}{i-a}\\ &=-\frac {(i-a-b x) \sqrt [4]{1-i (a+b x)}}{(i-a) x \sqrt [4]{1+i (a+b x)}}-\frac {b \operatorname {Subst}\left (\int \frac {1}{\sqrt {i+a}-\sqrt {i-a} x^2} \, dx,x,\frac {\sqrt [4]{1-i (a+b x)}}{\sqrt [4]{1+i (a+b x)}}\right )}{(1+i a) \sqrt {i+a}}-\frac {b \operatorname {Subst}\left (\int \frac {1}{\sqrt {i+a}+\sqrt {i-a} x^2} \, dx,x,\frac {\sqrt [4]{1-i (a+b x)}}{\sqrt [4]{1+i (a+b x)}}\right )}{(1+i a) \sqrt {i+a}}\\ &=-\frac {(i-a-b x) \sqrt [4]{1-i (a+b x)}}{(i-a) x \sqrt [4]{1+i (a+b x)}}-\frac {i b \tan ^{-1}\left (\frac {\sqrt [4]{i-a} \sqrt [4]{1-i (a+b x)}}{\sqrt [4]{i+a} \sqrt [4]{1+i (a+b x)}}\right )}{(i-a)^{5/4} (i+a)^{3/4}}-\frac {i b \tanh ^{-1}\left (\frac {\sqrt [4]{i-a} \sqrt [4]{1-i (a+b x)}}{\sqrt [4]{i+a} \sqrt [4]{1+i (a+b x)}}\right )}{(i-a)^{5/4} (i+a)^{3/4}}\\ \end {align*}

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Mathematica [C]  time = 0.03, size = 107, normalized size = 0.51 \[ -\frac {\sqrt [4]{-i (a+b x+i)} \left (-2 i b x \, _2F_1\left (\frac {1}{4},1;\frac {5}{4};\frac {a^2+b x a-i b x+1}{a^2+b x a+i b x+1}\right )+a^2+a b x+i b x+1\right )}{\left (a^2+1\right ) x \sqrt [4]{i a+i b x+1}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/(E^((I/2)*ArcTan[a + b*x])*x^2),x]

[Out]

-((((-I)*(I + a + b*x))^(1/4)*(1 + a^2 + I*b*x + a*b*x - (2*I)*b*x*Hypergeometric2F1[1/4, 1, 5/4, (1 + a^2 - I
*b*x + a*b*x)/(1 + a^2 + I*b*x + a*b*x)]))/((1 + a^2)*x*(1 + I*a + I*b*x)^(1/4)))

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fricas [B]  time = 0.47, size = 728, normalized size = 3.47 \[ \frac {\left (-\frac {b^{4}}{16 \, a^{8} - 32 i \, a^{7} + 32 \, a^{6} - 96 i \, a^{5} - 96 i \, a^{3} - 32 \, a^{2} - 32 i \, a - 16}\right )^{\frac {1}{4}} {\left (-i \, a - 1\right )} x \log \left (\frac {b^{3} \sqrt {\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}} + {\left (8 \, a^{6} - 16 i \, a^{5} + 8 \, a^{4} - 32 i \, a^{3} - 8 \, a^{2} - 16 i \, a - 8\right )} \left (-\frac {b^{4}}{16 \, a^{8} - 32 i \, a^{7} + 32 \, a^{6} - 96 i \, a^{5} - 96 i \, a^{3} - 32 \, a^{2} - 32 i \, a - 16}\right )^{\frac {3}{4}}}{b^{3}}\right ) + \left (-\frac {b^{4}}{16 \, a^{8} - 32 i \, a^{7} + 32 \, a^{6} - 96 i \, a^{5} - 96 i \, a^{3} - 32 \, a^{2} - 32 i \, a - 16}\right )^{\frac {1}{4}} {\left (i \, a + 1\right )} x \log \left (\frac {b^{3} \sqrt {\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}} - {\left (8 \, a^{6} - 16 i \, a^{5} + 8 \, a^{4} - 32 i \, a^{3} - 8 \, a^{2} - 16 i \, a - 8\right )} \left (-\frac {b^{4}}{16 \, a^{8} - 32 i \, a^{7} + 32 \, a^{6} - 96 i \, a^{5} - 96 i \, a^{3} - 32 \, a^{2} - 32 i \, a - 16}\right )^{\frac {3}{4}}}{b^{3}}\right ) - \left (-\frac {b^{4}}{16 \, a^{8} - 32 i \, a^{7} + 32 \, a^{6} - 96 i \, a^{5} - 96 i \, a^{3} - 32 \, a^{2} - 32 i \, a - 16}\right )^{\frac {1}{4}} {\left (a - i\right )} x \log \left (\frac {b^{3} \sqrt {\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}} + {\left (8 i \, a^{6} + 16 \, a^{5} + 8 i \, a^{4} + 32 \, a^{3} - 8 i \, a^{2} + 16 \, a - 8 i\right )} \left (-\frac {b^{4}}{16 \, a^{8} - 32 i \, a^{7} + 32 \, a^{6} - 96 i \, a^{5} - 96 i \, a^{3} - 32 \, a^{2} - 32 i \, a - 16}\right )^{\frac {3}{4}}}{b^{3}}\right ) + \left (-\frac {b^{4}}{16 \, a^{8} - 32 i \, a^{7} + 32 \, a^{6} - 96 i \, a^{5} - 96 i \, a^{3} - 32 \, a^{2} - 32 i \, a - 16}\right )^{\frac {1}{4}} {\left (a - i\right )} x \log \left (\frac {b^{3} \sqrt {\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}} + {\left (-8 i \, a^{6} - 16 \, a^{5} - 8 i \, a^{4} - 32 \, a^{3} + 8 i \, a^{2} - 16 \, a + 8 i\right )} \left (-\frac {b^{4}}{16 \, a^{8} - 32 i \, a^{7} + 32 \, a^{6} - 96 i \, a^{5} - 96 i \, a^{3} - 32 \, a^{2} - 32 i \, a - 16}\right )^{\frac {3}{4}}}{b^{3}}\right ) + i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} \sqrt {\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}}}{{\left (a - i\right )} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((-b^4/(16*a^8 - 32*I*a^7 + 32*a^6 - 96*I*a^5 - 96*I*a^3 - 32*a^2 - 32*I*a - 16))^(1/4)*(-I*a - 1)*x*log((b^3*
sqrt(I*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)/(b*x + a + I)) + (8*a^6 - 16*I*a^5 + 8*a^4 - 32*I*a^3 - 8*a^2 - 16*I*
a - 8)*(-b^4/(16*a^8 - 32*I*a^7 + 32*a^6 - 96*I*a^5 - 96*I*a^3 - 32*a^2 - 32*I*a - 16))^(3/4))/b^3) + (-b^4/(1
6*a^8 - 32*I*a^7 + 32*a^6 - 96*I*a^5 - 96*I*a^3 - 32*a^2 - 32*I*a - 16))^(1/4)*(I*a + 1)*x*log((b^3*sqrt(I*sqr
t(b^2*x^2 + 2*a*b*x + a^2 + 1)/(b*x + a + I)) - (8*a^6 - 16*I*a^5 + 8*a^4 - 32*I*a^3 - 8*a^2 - 16*I*a - 8)*(-b
^4/(16*a^8 - 32*I*a^7 + 32*a^6 - 96*I*a^5 - 96*I*a^3 - 32*a^2 - 32*I*a - 16))^(3/4))/b^3) - (-b^4/(16*a^8 - 32
*I*a^7 + 32*a^6 - 96*I*a^5 - 96*I*a^3 - 32*a^2 - 32*I*a - 16))^(1/4)*(a - I)*x*log((b^3*sqrt(I*sqrt(b^2*x^2 +
2*a*b*x + a^2 + 1)/(b*x + a + I)) + (8*I*a^6 + 16*a^5 + 8*I*a^4 + 32*a^3 - 8*I*a^2 + 16*a - 8*I)*(-b^4/(16*a^8
- 32*I*a^7 + 32*a^6 - 96*I*a^5 - 96*I*a^3 - 32*a^2 - 32*I*a - 16))^(3/4))/b^3) + (-b^4/(16*a^8 - 32*I*a^7 + 3
2*a^6 - 96*I*a^5 - 96*I*a^3 - 32*a^2 - 32*I*a - 16))^(1/4)*(a - I)*x*log((b^3*sqrt(I*sqrt(b^2*x^2 + 2*a*b*x +
a^2 + 1)/(b*x + a + I)) + (-8*I*a^6 - 16*a^5 - 8*I*a^4 - 32*a^3 + 8*I*a^2 - 16*a + 8*I)*(-b^4/(16*a^8 - 32*I*a
^7 + 32*a^6 - 96*I*a^5 - 96*I*a^3 - 32*a^2 - 32*I*a - 16))^(3/4))/b^3) + I*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*s
qrt(I*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)/(b*x + a + I)))/((a - I)*x)

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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/((1+I*(b*x+a))/(1+(b*x+a)^2)^(1/2))^(1/2)/x^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 ValueWarning, need to choose a branch for the root of a polynomial wi
th parameters. This might be wrong.The choice was done assuming 0=[0,0,0]index.cc index_m operator + Error: Ba
d Argument ValueWarning, need to choose a branch for the root of a polynomial with parameters. This might be w
rong.The choice was done assuming 0=[0,0,0]index.cc index_m operator + Error: Bad Argument ValueWarning, need
to choose a branch for the root of a polynomial with parameters. This might be wrong.The choice was done assum
ing 0=[0,0,0]index.cc index_m operator + Error: Bad Argument ValueWarning, need to choose a branch for the roo
t 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 ValueEvaluation time: 1.74Done

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maple [F]  time = 0.17, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {\frac {1+i \left (b x +a \right )}{\sqrt {1+\left (b x +a \right )^{2}}}}\, x^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/(x^2*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 \[ \int \frac {1}{x^2\,\sqrt {\frac {1+a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}{\sqrt {{\left (a+b\,x\right )}^2+1}}}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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