3.185 \(\int \frac{e^{3 i \tan ^{-1}(a+b x)}}{x} \, dx\)

Optimal. Leaf size=134 \[ \frac{4 \sqrt{i a+i b x+1}}{(1-i a) \sqrt{-i a-i b x+1}}-i \sinh ^{-1}(a+b x)-\frac{2 (-a+i)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+i} \sqrt{i a+i b x+1}}{\sqrt{-a+i} \sqrt{-i a-i b x+1}}\right )}{(a+i)^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.0950191, antiderivative size = 134, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {5095, 98, 157, 53, 619, 215, 93, 208} \[ \frac{4 \sqrt{i a+i b x+1}}{(1-i a) \sqrt{-i a-i b x+1}}-i \sinh ^{-1}(a+b x)-\frac{2 (-a+i)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+i} \sqrt{i a+i b x+1}}{\sqrt{-a+i} \sqrt{-i a-i b x+1}}\right )}{(a+i)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[E^((3*I)*ArcTan[a + b*x])/x,x]

[Out]

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

Rule 5095

Int[E^(ArcTan[(c_.)*((a_) + (b_.)*(x_))]*(n_.))*((d_.) + (e_.)*(x_))^(m_.), x_Symbol] :> Int[((d + e*x)^m*(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, d, e, m, n}, x]

Rule 98

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((b*c -
a*d)*(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^(p + 1))/(b*(b*e - a*f)*(m + 1)), x] + Dist[1/(b*(b*e - a*
f)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 2)*(e + f*x)^p*Simp[a*d*(d*e*(n - 1) + c*f*(p + 1)) + b*c*(d
*e*(m - n + 2) - c*f*(m + p + 2)) + d*(a*d*f*(n + p) + b*(d*e*(m + 1) - c*f*(m + n + p + 1)))*x, x], x], x] /;
FreeQ[{a, b, c, d, e, f, p}, x] && LtQ[m, -1] && GtQ[n, 1] && (IntegersQ[2*m, 2*n, 2*p] || IntegersQ[m, n + p
] || IntegersQ[p, m + n])

Rule 157

Int[(((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/((a_.) + (b_.)*(x_)), x_Symbol]
:> Dist[h/b, Int[(c + d*x)^n*(e + f*x)^p, x], x] + Dist[(b*g - a*h)/b, Int[((c + d*x)^n*(e + f*x)^p)/(a + b*x
), x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x]

Rule 53

Int[1/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_.) + (d_.)*(x_)]), x_Symbol] :> Int[1/Sqrt[a*c - b*(a - c)*x - b^2*x^2]
, x] /; FreeQ[{a, b, c, d}, x] && EqQ[b + d, 0] && GtQ[a + c, 0]

Rule 619

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[1/(2*c*((-4*c)/(b^2 - 4*a*c))^p), Subst[Int[Si
mp[1 - x^2/(b^2 - 4*a*c), x]^p, x], x, b + 2*c*x], x] /; FreeQ[{a, b, c, p}, x] && GtQ[4*a - b^2/c, 0]

Rule 215

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

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 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]

Rubi steps

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

Mathematica [A]  time = 0.713648, size = 229, normalized size = 1.71 \[ \frac{2 \sqrt [4]{-1} (-i b)^{3/2} \sinh ^{-1}\left (\frac{\left (\frac{1}{2}+\frac{i}{2}\right ) \sqrt{b} \sqrt{-i (a+b x+i)}}{\sqrt{-i b}}\right )}{b^{3/2}}+\frac{2 i \left (2 \sqrt{-1+i a} \sqrt{1+i a} \sqrt{i a+i b x+1}-(a-i)^2 \sqrt{-i (a+b x+i)} \tan ^{-1}\left (\frac{\sqrt{-i (a+b x+i)}}{\sqrt{\frac{a+i}{a-i}} \sqrt{i a+i b x+1}}\right )\right )}{\sqrt{-1+i a} \sqrt{1+i a} (a+i) \sqrt{-i (a+b x+i)}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2*(-1)^(1/4)*((-I)*b)^(3/2)*ArcSinh[((1/2 + I/2)*Sqrt[b]*Sqrt[(-I)*(I + a + b*x)])/Sqrt[(-I)*b]])/b^(3/2) + (
(2*I)*(2*Sqrt[-1 + I*a]*Sqrt[1 + I*a]*Sqrt[1 + I*a + I*b*x] - (-I + a)^2*Sqrt[(-I)*(I + a + b*x)]*ArcTan[Sqrt[
(-I)*(I + a + b*x)]/(Sqrt[(I + a)/(-I + a)]*Sqrt[1 + I*a + I*b*x])]))/(Sqrt[-1 + I*a]*Sqrt[1 + I*a]*(I + a)*Sq
rt[(-I)*(I + a + b*x)])

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Maple [B]  time = 0.11, size = 818, normalized size = 6.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

I*b*x/(b^2*x^2+2*a*b*x+a^2+1)^(1/2)+3*I/(a^2+1)/(b^2*x^2+2*a*b*x+a^2+1)^(1/2)*a-4*a^2/(a^2+1)/(b^2*x^2+2*a*b*x
+a^2+1)^(1/2)+3*a^4/(a^2+1)/(b^2*x^2+2*a*b*x+a^2+1)^(1/2)+3/(a^2+1)^(3/2)*ln((2*a^2+2+2*x*a*b+2*(a^2+1)^(1/2)*
(b^2*x^2+2*a*b*x+a^2+1)^(1/2))/x)*a^2+I*a^4*b/(a^2+1)/(b^2*x^2+2*a*b*x+a^2+1)^(1/2)*x+1/(a^2+1)/(b^2*x^2+2*a*b
*x+a^2+1)^(1/2)+3*a^3*b/(a^2+1)/(b^2*x^2+2*a*b*x+a^2+1)^(1/2)*x-a*b/(a^2+1)/(b^2*x^2+2*a*b*x+a^2+1)^(1/2)*x+6*
I*(1+I*a)^2*b*(2*b^2*x+2*a*b)/(4*b^2*(a^2+1)-4*a^2*b^2)/(b^2*x^2+2*a*b*x+a^2+1)^(1/2)-4*I/(a^2+1)/(b^2*x^2+2*a
*b*x+a^2+1)^(1/2)*a^3-3*I/(a^2+1)^(3/2)*ln((2*a^2+2+2*x*a*b+2*(a^2+1)^(1/2)*(b^2*x^2+2*a*b*x+a^2+1)^(1/2))/x)*
a+3/(b^2*x^2+2*a*b*x+a^2+1)^(1/2)+2*I*a/(b^2*x^2+2*a*b*x+a^2+1)^(1/2)-3*I*a^2*b/(a^2+1)/(b^2*x^2+2*a*b*x+a^2+1
)^(1/2)*x+2*I*b*a^2/(b^2*x^2+2*a*b*x+a^2+1)^(1/2)*x+I*a^5/(a^2+1)/(b^2*x^2+2*a*b*x+a^2+1)^(1/2)-I*b*ln((b^2*x+
a*b)/(b^2)^(1/2)+(b^2*x^2+2*a*b*x+a^2+1)^(1/2))/(b^2)^(1/2)+3*b*a/(b^2*x^2+2*a*b*x+a^2+1)^(1/2)*x+I/(a^2+1)^(3
/2)*ln((2*a^2+2+2*x*a*b+2*(a^2+1)^(1/2)*(b^2*x^2+2*a*b*x+a^2+1)^(1/2))/x)*a^3+2*I*a^3/(b^2*x^2+2*a*b*x+a^2+1)^
(1/2)+3*a^2/(b^2*x^2+2*a*b*x+a^2+1)^(1/2)-1/(a^2+1)^(3/2)*ln((2*a^2+2+2*x*a*b+2*(a^2+1)^(1/2)*(b^2*x^2+2*a*b*x
+a^2+1)^(1/2))/x)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.50094, size = 983, normalized size = 7.34 \begin{align*} -\frac{{\left ({\left (a + i\right )} b x + a^{2} + 2 i \, a - 1\right )} \sqrt{-\frac{4 \, a^{3} - 12 i \, a^{2} - 12 \, a + 4 i}{a^{3} + 3 i \, a^{2} - 3 \, a - i}} \log \left (-\frac{{\left (2 \, a - 2 i\right )} b x - \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}{\left (2 \, a - 2 i\right )} -{\left (i \, a^{2} - 2 \, a - i\right )} \sqrt{-\frac{4 \, a^{3} - 12 i \, a^{2} - 12 \, a + 4 i}{a^{3} + 3 i \, a^{2} - 3 \, a - i}}}{2 \, a - 2 i}\right ) -{\left ({\left (a + i\right )} b x + a^{2} + 2 i \, a - 1\right )} \sqrt{-\frac{4 \, a^{3} - 12 i \, a^{2} - 12 \, a + 4 i}{a^{3} + 3 i \, a^{2} - 3 \, a - i}} \log \left (-\frac{{\left (2 \, a - 2 i\right )} b x - \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}{\left (2 \, a - 2 i\right )} -{\left (-i \, a^{2} + 2 \, a + i\right )} \sqrt{-\frac{4 \, a^{3} - 12 i \, a^{2} - 12 \, a + 4 i}{a^{3} + 3 i \, a^{2} - 3 \, a - i}}}{2 \, a - 2 i}\right ) + 8 \, b x +{\left (2 \,{\left (-i \, a + 1\right )} b x - 2 i \, a^{2} + 4 \, a + 2 i\right )} \log \left (-b x - a + \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right ) + 8 \, a + 8 \, \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1} + 8 i}{{\left (2 \, a + 2 i\right )} b x + 2 \, a^{2} + 4 i \, a - 2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(((a + I)*b*x + a^2 + 2*I*a - 1)*sqrt(-(4*a^3 - 12*I*a^2 - 12*a + 4*I)/(a^3 + 3*I*a^2 - 3*a - I))*log(-((2*a
- 2*I)*b*x - sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*(2*a - 2*I) - (I*a^2 - 2*a - I)*sqrt(-(4*a^3 - 12*I*a^2 - 12*a
+ 4*I)/(a^3 + 3*I*a^2 - 3*a - I)))/(2*a - 2*I)) - ((a + I)*b*x + a^2 + 2*I*a - 1)*sqrt(-(4*a^3 - 12*I*a^2 - 12
*a + 4*I)/(a^3 + 3*I*a^2 - 3*a - I))*log(-((2*a - 2*I)*b*x - sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*(2*a - 2*I) - (
-I*a^2 + 2*a + I)*sqrt(-(4*a^3 - 12*I*a^2 - 12*a + 4*I)/(a^3 + 3*I*a^2 - 3*a - I)))/(2*a - 2*I)) + 8*b*x + (2*
(-I*a + 1)*b*x - 2*I*a^2 + 4*a + 2*I)*log(-b*x - a + sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)) + 8*a + 8*sqrt(b^2*x^2
+ 2*a*b*x + a^2 + 1) + 8*I)/((2*a + 2*I)*b*x + 2*a^2 + 4*I*a - 2)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (i a + i b x + 1\right )^{3}}{x \left (a^{2} + 2 a b x + b^{2} x^{2} + 1\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: NotImplementedError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotImplementedError