3.170 \(\int \frac{e^{i \tan ^{-1}(a+b x)}}{x^4} \, dx\)

Optimal. Leaf size=283 \[ \frac{\left (-2 a^2+9 i a+4\right ) b^2 \sqrt{-i a-i b x+1} \sqrt{i a+i b x+1}}{6 (1-i a) \left (a^2+1\right )^2 x}+\frac{\left (2 a-i \left (1-2 a^2\right )\right ) b^3 \tanh ^{-1}\left (\frac{\sqrt{a+i} \sqrt{i a+i b x+1}}{\sqrt{-a+i} \sqrt{-i a-i b x+1}}\right )}{(-a+i)^{5/2} (a+i)^{7/2}}-\frac{(-2 a+3 i) b \sqrt{-i a-i b x+1} \sqrt{i a+i b x+1}}{6 (1-i a) \left (a^2+1\right ) x^2}-\frac{\sqrt{-i a-i b x+1} \sqrt{i a+i b x+1}}{3 (1-i a) x^3} \]

[Out]

-(Sqrt[1 - I*a - I*b*x]*Sqrt[1 + I*a + I*b*x])/(3*(1 - I*a)*x^3) - ((3*I - 2*a)*b*Sqrt[1 - I*a - I*b*x]*Sqrt[1
+ I*a + I*b*x])/(6*(1 - I*a)*(1 + a^2)*x^2) + ((4 + (9*I)*a - 2*a^2)*b^2*Sqrt[1 - I*a - I*b*x]*Sqrt[1 + I*a +
I*b*x])/(6*(1 - I*a)*(1 + a^2)^2*x) + ((2*a - I*(1 - 2*a^2))*b^3*ArcTanh[(Sqrt[I + a]*Sqrt[1 + I*a + I*b*x])/
(Sqrt[I - a]*Sqrt[1 - I*a - I*b*x])])/((I - a)^(5/2)*(I + a)^(7/2))

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Rubi [A]  time = 0.183258, antiderivative size = 283, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.375, Rules used = {5095, 99, 151, 12, 93, 208} \[ \frac{\left (-2 a^2+9 i a+4\right ) b^2 \sqrt{-i a-i b x+1} \sqrt{i a+i b x+1}}{6 (1-i a) \left (a^2+1\right )^2 x}+\frac{\left (2 a-i \left (1-2 a^2\right )\right ) b^3 \tanh ^{-1}\left (\frac{\sqrt{a+i} \sqrt{i a+i b x+1}}{\sqrt{-a+i} \sqrt{-i a-i b x+1}}\right )}{(-a+i)^{5/2} (a+i)^{7/2}}-\frac{(-2 a+3 i) b \sqrt{-i a-i b x+1} \sqrt{i a+i b x+1}}{6 (1-i a) \left (a^2+1\right ) x^2}-\frac{\sqrt{-i a-i b x+1} \sqrt{i a+i b x+1}}{3 (1-i a) x^3} \]

Antiderivative was successfully verified.

[In]

Int[E^(I*ArcTan[a + b*x])/x^4,x]

[Out]

-(Sqrt[1 - I*a - I*b*x]*Sqrt[1 + I*a + I*b*x])/(3*(1 - I*a)*x^3) - ((3*I - 2*a)*b*Sqrt[1 - I*a - I*b*x]*Sqrt[1
+ I*a + I*b*x])/(6*(1 - I*a)*(1 + a^2)*x^2) + ((4 + (9*I)*a - 2*a^2)*b^2*Sqrt[1 - I*a - I*b*x]*Sqrt[1 + I*a +
I*b*x])/(6*(1 - I*a)*(1 + a^2)^2*x) + ((2*a - I*(1 - 2*a^2))*b^3*ArcTanh[(Sqrt[I + a]*Sqrt[1 + I*a + I*b*x])/
(Sqrt[I - a]*Sqrt[1 - I*a - I*b*x])])/((I - a)^(5/2)*(I + a)^(7/2))

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 99

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((a + b
*x)^(m + 1)*(c + d*x)^n*(e + f*x)^(p + 1))/((m + 1)*(b*e - a*f)), x] - Dist[1/((m + 1)*(b*e - a*f)), Int[(a +
b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p*Simp[d*e*n + c*f*(m + p + 2) + d*f*(m + n + p + 2)*x, x], x], x] /;
FreeQ[{a, b, c, d, e, f, p}, x] && LtQ[m, -1] && GtQ[n, 0] && (IntegersQ[2*m, 2*n, 2*p] || IntegersQ[m, n + p
] || IntegersQ[p, m + n])

Rule 151

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[((b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*
f)), x] + Dist[1/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[(a*d*f*
g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p
+ 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && LtQ[m, -1] && IntegerQ[m]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 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^{i \tan ^{-1}(a+b x)}}{x^4} \, dx &=\int \frac{\sqrt{1+i a+i b x}}{x^4 \sqrt{1-i a-i b x}} \, dx\\ &=-\frac{\sqrt{1-i a-i b x} \sqrt{1+i a+i b x}}{3 (1-i a) x^3}+\frac{\int \frac{(3 i-2 a) b-2 b^2 x}{x^3 \sqrt{1-i a-i b x} \sqrt{1+i a+i b x}} \, dx}{3 (1-i a)}\\ &=-\frac{\sqrt{1-i a-i b x} \sqrt{1+i a+i b x}}{3 (1-i a) x^3}-\frac{(3 i-2 a) b \sqrt{1-i a-i b x} \sqrt{1+i a+i b x}}{6 (1-i a) \left (1+a^2\right ) x^2}-\frac{\int \frac{\left (4+9 i a-2 a^2\right ) b^2+(3 i-2 a) b^3 x}{x^2 \sqrt{1-i a-i b x} \sqrt{1+i a+i b x}} \, dx}{6 (1-i a) \left (1+a^2\right )}\\ &=-\frac{\sqrt{1-i a-i b x} \sqrt{1+i a+i b x}}{3 (1-i a) x^3}-\frac{(3 i-2 a) b \sqrt{1-i a-i b x} \sqrt{1+i a+i b x}}{6 (1-i a) \left (1+a^2\right ) x^2}+\frac{\left (4+9 i a-2 a^2\right ) b^2 \sqrt{1-i a-i b x} \sqrt{1+i a+i b x}}{6 (1-i a) \left (1+a^2\right )^2 x}+\frac{\int -\frac{3 \left (i-2 a-2 i a^2\right ) b^3}{x \sqrt{1-i a-i b x} \sqrt{1+i a+i b x}} \, dx}{6 (1-i a) \left (1+a^2\right )^2}\\ &=-\frac{\sqrt{1-i a-i b x} \sqrt{1+i a+i b x}}{3 (1-i a) x^3}-\frac{(3 i-2 a) b \sqrt{1-i a-i b x} \sqrt{1+i a+i b x}}{6 (1-i a) \left (1+a^2\right ) x^2}+\frac{\left (4+9 i a-2 a^2\right ) b^2 \sqrt{1-i a-i b x} \sqrt{1+i a+i b x}}{6 (1-i a) \left (1+a^2\right )^2 x}+\frac{\left (\left (1+2 i a-2 a^2\right ) b^3\right ) \int \frac{1}{x \sqrt{1-i a-i b x} \sqrt{1+i a+i b x}} \, dx}{2 (i-a)^2 (i+a)^3}\\ &=-\frac{\sqrt{1-i a-i b x} \sqrt{1+i a+i b x}}{3 (1-i a) x^3}-\frac{(3 i-2 a) b \sqrt{1-i a-i b x} \sqrt{1+i a+i b x}}{6 (1-i a) \left (1+a^2\right ) x^2}+\frac{\left (4+9 i a-2 a^2\right ) b^2 \sqrt{1-i a-i b x} \sqrt{1+i a+i b x}}{6 (1-i a) \left (1+a^2\right )^2 x}+\frac{\left (\left (1+2 i a-2 a^2\right ) b^3\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 )}{(i-a)^2 (i+a)^3}\\ &=-\frac{\sqrt{1-i a-i b x} \sqrt{1+i a+i b x}}{3 (1-i a) x^3}-\frac{(3 i-2 a) b \sqrt{1-i a-i b x} \sqrt{1+i a+i b x}}{6 (1-i a) \left (1+a^2\right ) x^2}+\frac{\left (4+9 i a-2 a^2\right ) b^2 \sqrt{1-i a-i b x} \sqrt{1+i a+i b x}}{6 (1-i a) \left (1+a^2\right )^2 x}+\frac{\left (2 a-i \left (1-2 a^2\right )\right ) b^3 \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)^{5/2} (i+a)^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.262788, size = 242, normalized size = 0.86 \[ \frac{(1+4 i a) b x (a+b x-i) \sqrt{a^2+2 a b x+b^2 x^2+1}+2 (1-i a) (a-i) (a+b x-i) \sqrt{a^2+2 a b x+b^2 x^2+1}+\frac{3 \left (2 a^2-2 i a-1\right ) b^2 x^2 \left (\sqrt{-1+i a} \sqrt{1+i a} \sqrt{a^2+2 a b x+b^2 x^2+1}-2 i b x \tan ^{-1}\left (\frac{\sqrt{-i (a+b x+i)}}{\sqrt{\frac{a+i}{a-i}} \sqrt{i a+i b x+1}}\right )\right )}{(-1+i a)^{3/2} \sqrt{1+i a}}}{6 \left (a^2+1\right )^2 x^3} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^(I*ArcTan[a + b*x])/x^4,x]

[Out]

(2*(1 - I*a)*(-I + a)*(-I + a + b*x)*Sqrt[1 + a^2 + 2*a*b*x + b^2*x^2] + (1 + (4*I)*a)*b*x*(-I + a + b*x)*Sqrt
[1 + a^2 + 2*a*b*x + b^2*x^2] + (3*(-1 - (2*I)*a + 2*a^2)*b^2*x^2*(Sqrt[-1 + I*a]*Sqrt[1 + I*a]*Sqrt[1 + a^2 +
2*a*b*x + b^2*x^2] - (2*I)*b*x*ArcTan[Sqrt[(-I)*(I + a + b*x)]/(Sqrt[(I + a)/(-I + a)]*Sqrt[1 + I*a + I*b*x])
]))/((-1 + I*a)^(3/2)*Sqrt[1 + I*a]))/(6*(1 + a^2)^2*x^3)

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Maple [B]  time = 0.118, size = 611, normalized size = 2.2 \begin{align*}{\frac{-{\frac{i}{2}}b}{ \left ({a}^{2}+1 \right ){x}^{2}}\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}+{\frac{{\frac{5\,i}{6}}{a}^{2}b}{ \left ({a}^{2}+1 \right ) ^{2}{x}^{2}}\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}-{3\,i{b}^{3}{a}^{2}\ln \left ({\frac{1}{x} \left ( 2\,{a}^{2}+2+2\,xab+2\,\sqrt{{a}^{2}+1}\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1} \right ) } \right ) \left ({a}^{2}+1 \right ) ^{-{\frac{5}{2}}}}+{{\frac{i}{2}}{b}^{3}\ln \left ({\frac{1}{x} \left ( 2\,{a}^{2}+2+2\,xab+2\,\sqrt{{a}^{2}+1}\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1} \right ) } \right ) \left ({a}^{2}+1 \right ) ^{-{\frac{3}{2}}}}-{\frac{{\frac{5\,i}{2}}{a}^{3}{b}^{2}}{ \left ({a}^{2}+1 \right ) ^{3}x}\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}-{\frac{1}{ \left ( 3\,{a}^{2}+3 \right ){x}^{3}}\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}+{{\frac{5\,i}{2}}{a}^{4}{b}^{3}\ln \left ({\frac{1}{x} \left ( 2\,{a}^{2}+2+2\,xab+2\,\sqrt{{a}^{2}+1}\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1} \right ) } \right ) \left ({a}^{2}+1 \right ) ^{-{\frac{7}{2}}}}+{\frac{5\,ab}{6\, \left ({a}^{2}+1 \right ) ^{2}{x}^{2}}\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}-{\frac{{\frac{i}{3}}a}{ \left ({a}^{2}+1 \right ){x}^{3}}\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}-{\frac{5\,{a}^{2}{b}^{2}}{2\, \left ({a}^{2}+1 \right ) ^{3}x}\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}+{\frac{{\frac{13\,i}{6}}{b}^{2}a}{ \left ({a}^{2}+1 \right ) ^{2}x}\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}+{\frac{5\,{a}^{3}{b}^{3}}{2}\ln \left ({\frac{1}{x} \left ( 2\,{a}^{2}+2+2\,xab+2\,\sqrt{{a}^{2}+1}\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1} \right ) } \right ) \left ({a}^{2}+1 \right ) ^{-{\frac{7}{2}}}}-{\frac{3\,a{b}^{3}}{2}\ln \left ({\frac{1}{x} \left ( 2\,{a}^{2}+2+2\,xab+2\,\sqrt{{a}^{2}+1}\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1} \right ) } \right ) \left ({a}^{2}+1 \right ) ^{-{\frac{5}{2}}}}+{\frac{2\,{b}^{2}}{3\, \left ({a}^{2}+1 \right ) ^{2}x}\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1+I*(b*x+a))/(1+(b*x+a)^2)^(1/2)/x^4,x)

[Out]

-1/2*I*b/(a^2+1)/x^2*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)+5/6*I*a^2*b/(a^2+1)^2/x^2*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)-3*I
*b^3*a^2/(a^2+1)^(5/2)*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)+1/2*I*b^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)-5/2*I*a^3*b^2/(a^2+1)^3/x*(b^2*x^
2+2*a*b*x+a^2+1)^(1/2)-1/3/(a^2+1)/x^3*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)+5/2*I*a^4*b^3/(a^2+1)^(7/2)*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)+5/6*a*b/(a^2+1)^2/x^2*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)-1
/3*I/(a^2+1)/x^3*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)*a-5/2*a^2*b^2/(a^2+1)^3/x*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)+13/6*I*
b^2*a/(a^2+1)^2/x*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)+5/2*a^3*b^3/(a^2+1)^(7/2)*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)-3/2*a*b^3/(a^2+1)^(5/2)*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)+2/3*b^2/(a^2+1)^2/x*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)

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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))/(1+(b*x+a)^2)^(1/2)/x^4,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.82361, size = 1767, normalized size = 6.24 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+I*(b*x+a))/(1+(b*x+a)^2)^(1/2)/x^4,x, algorithm="fricas")

[Out]

((-2*I*a^2 - 9*a + 4*I)*b^3*x^3 - sqrt((4*a^4 - 8*I*a^3 - 8*a^2 + 4*I*a + 1)*b^6/(a^12 + 2*I*a^11 + 4*a^10 + 1
0*I*a^9 + 5*a^8 + 20*I*a^7 + 20*I*a^5 - 5*a^4 + 10*I*a^3 - 4*a^2 + 2*I*a - 1))*(3*a^5 + 3*I*a^4 + 6*a^3 + 6*I*
a^2 + 3*a + 3*I)*x^3*log(-((2*a^2 - 2*I*a - 1)*b^4*x - sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*(2*a^2 - 2*I*a - 1)*b
^3 + (a^7 + I*a^6 + 3*a^5 + 3*I*a^4 + 3*a^3 + 3*I*a^2 + a + I)*sqrt((4*a^4 - 8*I*a^3 - 8*a^2 + 4*I*a + 1)*b^6/
(a^12 + 2*I*a^11 + 4*a^10 + 10*I*a^9 + 5*a^8 + 20*I*a^7 + 20*I*a^5 - 5*a^4 + 10*I*a^3 - 4*a^2 + 2*I*a - 1)))/(
(2*a^2 - 2*I*a - 1)*b^3)) + sqrt((4*a^4 - 8*I*a^3 - 8*a^2 + 4*I*a + 1)*b^6/(a^12 + 2*I*a^11 + 4*a^10 + 10*I*a^
9 + 5*a^8 + 20*I*a^7 + 20*I*a^5 - 5*a^4 + 10*I*a^3 - 4*a^2 + 2*I*a - 1))*(3*a^5 + 3*I*a^4 + 6*a^3 + 6*I*a^2 +
3*a + 3*I)*x^3*log(-((2*a^2 - 2*I*a - 1)*b^4*x - sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*(2*a^2 - 2*I*a - 1)*b^3 - (
a^7 + I*a^6 + 3*a^5 + 3*I*a^4 + 3*a^3 + 3*I*a^2 + a + I)*sqrt((4*a^4 - 8*I*a^3 - 8*a^2 + 4*I*a + 1)*b^6/(a^12
+ 2*I*a^11 + 4*a^10 + 10*I*a^9 + 5*a^8 + 20*I*a^7 + 20*I*a^5 - 5*a^4 + 10*I*a^3 - 4*a^2 + 2*I*a - 1)))/((2*a^2
- 2*I*a - 1)*b^3)) + ((-2*I*a^2 - 9*a + 4*I)*b^2*x^2 - 2*I*a^4 + (2*I*a^3 + 3*a^2 + 2*I*a + 3)*b*x - 4*I*a^2
- 2*I)*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1))/((6*a^5 + 6*I*a^4 + 12*a^3 + 12*I*a^2 + 6*a + 6*I)*x^3)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{i a + i b x + 1}{x^{4} \sqrt{a^{2} + 2 a b x + b^{2} x^{2} + 1}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+I*(b*x+a))/(1+(b*x+a)**2)**(1/2)/x**4,x)

[Out]

Integral((I*a + I*b*x + 1)/(x**4*sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1)), 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))/(1+(b*x+a)^2)^(1/2)/x^4,x, algorithm="giac")

[Out]

Exception raised: NotImplementedError