∫ e3 _2 i tan−1(ax) ___________________

3.74 \(\int \frac {e^{\frac {3}{2} i \tan ^{-1}(a x)}}{x} \, dx\)

Optimal. Leaf size=267 \[ \frac {\log \left (\frac {\sqrt {1-i a x}}{\sqrt {1+i a x}}-\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}+1\right )}{\sqrt {2}}-\frac {\log \left (\frac {\sqrt {1-i a x}}{\sqrt {1+i a x}}+\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}+1\right )}{\sqrt {2}}+2 \tan ^{-1}\left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )+\sqrt {2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )-\sqrt {2} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )-2 \tanh ^{-1}\left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right ) \]

[Out]

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

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

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

1+(1-I*a*x)^(1/4)*2^(1/2)/(1+I*a*x)^(1/4))*2^(1/2)

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Rubi [A] time = 0.17, antiderivative size = 267, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 14, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.875, Rules used = {5062, 105, 63, 240, 211, 1165, 628, 1162, 617, 204, 93, 298, 203, 206} \[ \frac {\log \left (\frac {\sqrt {1-i a x}}{\sqrt {1+i a x}}-\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}+1\right )}{\sqrt {2}}-\frac {\log \left (\frac {\sqrt {1-i a x}}{\sqrt {1+i a x}}+\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}+1\right )}{\sqrt {2}}+2 \tan ^{-1}\left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )+\sqrt {2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )-\sqrt {2} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )-2 \tanh ^{-1}\left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right ) \]

Antiderivative was successfully veriﬁed.

[In]

Int[E^(((3*I)/2)*ArcTan[a*x])/x,x]

[Out]

2*ArcTan[(1 + I*a*x)^(1/4)/(1 - I*a*x)^(1/4)] + Sqrt[2]*ArcTan[1 - (Sqrt[2]*(1 - I*a*x)^(1/4))/(1 + I*a*x)^(1/

4)] - Sqrt[2]*ArcTan[1 + (Sqrt[2]*(1 - I*a*x)^(1/4))/(1 + I*a*x)^(1/4)] - 2*ArcTanh[(1 + I*a*x)^(1/4)/(1 - I*a

*x)^(1/4)] + Log[1 + Sqrt[1 - I*a*x]/Sqrt[1 + I*a*x] - (Sqrt[2]*(1 - I*a*x)^(1/4))/(1 + I*a*x)^(1/4)]/Sqrt[2]

- Log[1 + Sqrt[1 - I*a*x]/Sqrt[1 + I*a*x] + (Sqrt[2]*(1 - I*a*x)^(1/4))/(1 + I*a*x)^(1/4)]/Sqrt[2]

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 93

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin

ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^

(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[

a + b*x, c + d*x]

Rule 105

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> Dist[b/f, Int[(a

+ b*x)^(m - 1)*(c + d*x)^n, x], x] - Dist[(b*e - a*f)/f, Int[((a + b*x)^(m - 1)*(c + d*x)^n)/(e + f*x), x], x]

/; FreeQ[{a, b, c, d, e, f, m, n}, x] && IGtQ[Simplify[m + n + 1], 0] && (GtQ[m, 0] || ( !RationalQ[m] && (Su

mSimplerQ[m, -1] || !SumSimplerQ[n, -1])))

Rule 203

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

FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

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 206

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

/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 211

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]}, Di

st[1/(2*r), Int[(r - s*x^2)/(a + b*x^4), x], x] + Dist[1/(2*r), Int[(r + s*x^2)/(a + b*x^4), x], x]] /; FreeQ[

{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ, b

]]))

Rule 240

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + 1/n), Subst[Int[1/(1 - b*x^n)^(p + 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)] && IntegerQ[p

+ 1/n]

Rule 298

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b),

2]]}, Dist[s/(2*b), Int[1/(r + s*x^2), x], x] - Dist[s/(2*b), Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &

& !GtQ[a/b, 0]

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 5062

Int[E^(ArcTan[(a_.)*(x_)]*(n_.))*(x_)^(m_.), x_Symbol] :> Int[(x^m*(1 - I*a*x)^((I*n)/2))/(1 + I*a*x)^((I*n)/2

), x] /; FreeQ[{a, m, n}, x] && !IntegerQ[(I*n - 1)/2]

Rubi steps

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

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Mathematica [C] time = 0.03, size = 96, normalized size = 0.36 \[ -2\ 2^{3/4} \sqrt [4]{1-i a x} \, _2F_1\left (\frac {1}{4},\frac {1}{4};\frac {5}{4};\frac {1}{2} (1-i a x)\right )-\frac {4 \sqrt [4]{1-i a x} \, _2F_1\left (\frac {1}{4},1;\frac {5}{4};-\frac {1-i a x}{-i a x-1}\right )}{\sqrt [4]{1+i a x}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^(((3*I)/2)*ArcTan[a*x])/x,x]

[Out]

-2*2^(3/4)*(1 - I*a*x)^(1/4)*Hypergeometric2F1[1/4, 1/4, 5/4, (1 - I*a*x)/2] - (4*(1 - I*a*x)^(1/4)*Hypergeome

tric2F1[1/4, 1, 5/4, -((1 - I*a*x)/(-1 - I*a*x))])/(1 + I*a*x)^(1/4)

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fricas [A] time = 0.53, size = 243, normalized size = 0.91 \[ \frac {1}{2} \, \sqrt {4 i} \log \left (\frac {1}{2} i \, \sqrt {4 i} + \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}\right ) - \frac {1}{2} \, \sqrt {4 i} \log \left (-\frac {1}{2} i \, \sqrt {4 i} + \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}\right ) - \frac {1}{2} \, \sqrt {-4 i} \log \left (\frac {1}{2} i \, \sqrt {-4 i} + \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}\right ) + \frac {1}{2} \, \sqrt {-4 i} \log \left (-\frac {1}{2} i \, \sqrt {-4 i} + \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}\right ) - \log \left (\sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}} + 1\right ) + i \, \log \left (\sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}} + i\right ) - i \, \log \left (\sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}} - i\right ) + \log \left (\sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}} - 1\right ) \]

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(((1+I*a*x)/(a^2*x^2+1)^(1/2))^(3/2)/x,x, algorithm="fricas")

[Out]

1/2*sqrt(4*I)*log(1/2*I*sqrt(4*I) + sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I))) - 1/2*sqrt(4*I)*log(-1/2*I*sqrt(4*I)

+ sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I))) - 1/2*sqrt(-4*I)*log(1/2*I*sqrt(-4*I) + sqrt(I*sqrt(a^2*x^2 + 1)/(a*x +

I))) + 1/2*sqrt(-4*I)*log(-1/2*I*sqrt(-4*I) + sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I))) - log(sqrt(I*sqrt(a^2*x^2

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

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

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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*a*x)/(a^2*x^2+1)^(1/2))^(3/2)/x,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]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 assuming 0=[0,0]index.cc in

dex_m operator + Error: Bad Argument ValueWarning, need to choose a branch for the root of a polynomial with p

arameters. This might be wrong.The choice was done assuming 0=[0,0]index.cc index_m operator + Error: Bad Argu

ment ValueWarning, need to choose a branch for the root of a polynomial with parameters. This might be wrong.T

he choice was done assuming 0=[0,0]index.cc index_m operator + Error: Bad Argument ValueWarning, need to choos

e a branch for the root of a polynomial with parameters. This might be wrong.The choice was done assuming 0=[0

,0]index.cc index_m operator + Error: Bad Argument ValueWarning, need to choose a branch for the root of a pol

ynomial with parameters. This might be wrong.The choice was done assuming 0=[0,0]index.cc index_m operator + E

rror: Bad Argument ValueWarning, need to choose a branch for the root of a polynomial with parameters. This mi

ght be wrong.The choice was done assuming 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

assuming 0=[0,0]index.cc index_m operator + Error: Bad Argument ValueDone

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maple [F] time = 0.18, size = 0, normalized size = 0.00 \[ \int \frac {\left (\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )^{\frac {3}{2}}}{x}\, dx \]

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(((1+I*a*x)/(a^2*x^2+1)^(1/2))^(3/2)/x,x)

[Out]

int(((1+I*a*x)/(a^2*x^2+1)^(1/2))^(3/2)/x,x)

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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (\frac {i \, a x + 1}{\sqrt {a^{2} x^{2} + 1}}\right )^{\frac {3}{2}}}{x}\,{d x} \]

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(((1+I*a*x)/(a^2*x^2+1)^(1/2))^(3/2)/x,x, algorithm="maxima")

[Out]

integrate(((I*a*x + 1)/sqrt(a^2*x^2 + 1))^(3/2)/x, x)

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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (\frac {1+a\,x\,1{}\mathrm {i}}{\sqrt {a^2\,x^2+1}}\right )}^{3/2}}{x} \,d x \]

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(((a*x*1i + 1)/(a^2*x^2 + 1)^(1/2))^(3/2)/x,x)

[Out]

int(((a*x*1i + 1)/(a^2*x^2 + 1)^(1/2))^(3/2)/x, x)

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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (\frac {i \left (a x - i\right )}{\sqrt {a^{2} x^{2} + 1}}\right )^{\frac {3}{2}}}{x}\, dx \]

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(((1+I*a*x)/(a**2*x**2+1)**(1/2))**(3/2)/x,x)

[Out]

Integral((I*(a*x - I)/sqrt(a**2*x**2 + 1))**(3/2)/x, x)

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