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

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

[Out]

(2*a+I*(-2*a^2+1))*b^3*arctanh((I+a)^(1/2)*(1+I*a+I*b*x)^(1/2)/(I-a)^(1/2)/(1-I*a-I*b*x)^(1/2))/(I-a)^(7/2)/(I
+a)^(5/2)-1/3*(1-I*a-I*b*x)^(1/2)*(1+I*a+I*b*x)^(1/2)/(1+I*a)/x^3+1/6*(3-2*I*a)*b*(1-I*a-I*b*x)^(1/2)*(1+I*a+I
*b*x)^(1/2)/(I-a)^2/(I+a)/x^2+1/6*(4-9*I*a-2*a^2)*b^2*(1-I*a-I*b*x)^(1/2)*(1+I*a+I*b*x)^(1/2)/(1+I*a)/(a^2+1)^
2/x

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Rubi [A]  time = 0.18, antiderivative size = 282, normalized size of antiderivative = 1.00, 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} \[ \frac {\left (-2 a^2-9 i a+4\right ) b^2 \sqrt {-i a-i b x+1} \sqrt {i a+i b x+1}}{6 (1+i a) \left (a^2+1\right )^2 x}+\frac {\left (-2 i a^2+2 a+i\right ) b^3 \tanh ^{-1}\left (\frac {\sqrt {a+i} \sqrt {i a+i b x+1}}{\sqrt {-a+i} \sqrt {-i a-i b x+1}}\right )}{(-a+i)^{7/2} (a+i)^{5/2}}+\frac {(3-2 i a) b \sqrt {-i a-i b x+1} \sqrt {i a+i b x+1}}{6 (-a+i)^2 (a+i) x^2}-\frac {\sqrt {-i a-i b x+1} \sqrt {i a+i b x+1}}{3 (1+i a) x^3} \]

Antiderivative was successfully verified.

[In]

Int[1/(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 - (2*I)*a)*b*Sqrt[1 - I*a - I*b*x]*Sqrt
[1 + I*a + I*b*x])/(6*(I - a)^2*(I + a)*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) + ((I + 2*a - (2*I)*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)^(7/2)*(I + a)^(5/2))

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 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 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 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]

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 {(3-2 i a) b \sqrt {1-i a-i b x} \sqrt {1+i a+i b x}}{6 (i-a)^2 (i+a) 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 {(3-2 i a) b \sqrt {1-i a-i b x} \sqrt {1+i a+i b x}}{6 (i-a)^2 (i+a) 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 {(3-2 i a) b \sqrt {1-i a-i b x} \sqrt {1+i a+i b x}}{6 (i-a)^2 (i+a) 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)^3 (i+a)^2}\\ &=-\frac {\sqrt {1-i a-i b x} \sqrt {1+i a+i b x}}{3 (1+i a) x^3}+\frac {(3-2 i a) b \sqrt {1-i a-i b x} \sqrt {1+i a+i b x}}{6 (i-a)^2 (i+a) 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)^3 (i+a)^2}\\ &=-\frac {\sqrt {1-i a-i b x} \sqrt {1+i a+i b x}}{3 (1+i a) x^3}+\frac {(3-2 i a) b \sqrt {1-i a-i b x} \sqrt {1+i a+i b x}}{6 (i-a)^2 (i+a) 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 (i+2 a-2 i a^2\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)^{7/2} (i+a)^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.30, size = 247, normalized size = 0.87 \[ \frac {(1-4 i a) b x (a+b x+i) \sqrt {a^2+2 a b x+b^2 x^2+1}+2 (1+i a) (a+i) (a+b x+i) \sqrt {a^2+2 a b x+b^2 x^2+1}+\frac {3 \left (2 a^2+2 i a-1\right ) b^2 x^2 \left (\sqrt {-1-i a} \sqrt {-1+i a} \sqrt {a^2+2 a b x+b^2 x^2+1}+2 i b x \tanh ^{-1}\left (\frac {\sqrt {-1-i a} \sqrt {-i (a+b x+i)}}{\sqrt {-1+i a} \sqrt {i a+i b x+1}}\right )\right )}{(-1-i a)^{3/2} \sqrt {-1+i a}}}{6 \left (a^2+1\right )^2 x^3} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/(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*ArcTanh[(Sqrt[-1 - I*a]*Sqrt[(-I)*(I + a + b*x)])/(Sqrt[-1 + 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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fricas [B]  time = 0.52, size = 700, normalized size = 2.47 \[ \frac {{\left (2 i \, a^{2} - 9 \, a - 4 i\right )} b^{3} x^{3} - \sqrt {\frac {{\left (4 \, a^{4} + 8 i \, a^{3} - 8 \, a^{2} - 4 i \, a + 1\right )} 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}} {\left (3 \, a^{5} - 3 i \, a^{4} + 6 \, a^{3} - 6 i \, a^{2} + 3 \, a - 3 i\right )} x^{3} \log \left (-\frac {{\left (2 \, a^{2} + 2 i \, a - 1\right )} b^{4} x - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} {\left (2 \, a^{2} + 2 i \, a - 1\right )} b^{3} + {\left (a^{7} - i \, a^{6} + 3 \, a^{5} - 3 i \, a^{4} + 3 \, a^{3} - 3 i \, a^{2} + a - i\right )} \sqrt {\frac {{\left (4 \, a^{4} + 8 i \, a^{3} - 8 \, a^{2} - 4 i \, a + 1\right )} 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}}}{{\left (2 \, a^{2} + 2 i \, a - 1\right )} b^{3}}\right ) + \sqrt {\frac {{\left (4 \, a^{4} + 8 i \, a^{3} - 8 \, a^{2} - 4 i \, a + 1\right )} 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}} {\left (3 \, a^{5} - 3 i \, a^{4} + 6 \, a^{3} - 6 i \, a^{2} + 3 \, a - 3 i\right )} x^{3} \log \left (-\frac {{\left (2 \, a^{2} + 2 i \, a - 1\right )} b^{4} x - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} {\left (2 \, a^{2} + 2 i \, a - 1\right )} b^{3} - {\left (a^{7} - i \, a^{6} + 3 \, a^{5} - 3 i \, a^{4} + 3 \, a^{3} - 3 i \, a^{2} + a - i\right )} \sqrt {\frac {{\left (4 \, a^{4} + 8 i \, a^{3} - 8 \, a^{2} - 4 i \, a + 1\right )} 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}}}{{\left (2 \, a^{2} + 2 i \, a - 1\right )} b^{3}}\right ) + {\left ({\left (2 i \, a^{2} - 9 \, a - 4 i\right )} b^{2} x^{2} + 2 i \, a^{4} + {\left (-2 i \, a^{3} + 3 \, a^{2} - 2 i \, a + 3\right )} b x + 4 i \, a^{2} + 2 i\right )} \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{{\left (6 \, a^{5} - 6 i \, a^{4} + 12 \, a^{3} - 12 i \, a^{2} + 6 \, a - 6 i\right )} x^{3}} \]

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)/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 - 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)) + 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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giac [B]  time = 0.23, size = 906, normalized size = 3.20 \[ \text {result too large to display} \]

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)/x^4,x, algorithm="giac")

[Out]

1/2*(2*a^2*b^3 + 2*a*b^3*i - b^3)*log(abs(2*x*abs(b) - 2*sqrt((b*x + a)^2 + 1) - 2*sqrt(a^2 + 1))/abs(2*x*abs(
b) - 2*sqrt((b*x + a)^2 + 1) + 2*sqrt(a^2 + 1)))/((a^5 - a^4*i + 2*a^3 - 2*a^2*i + a - i)*sqrt(a^2 + 1)) - 1/3
*(8*(x*abs(b) - sqrt((b*x + a)^2 + 1))^3*a^5*b^3*i + 24*(x*abs(b) - sqrt((b*x + a)^2 + 1))*a^7*b^3*i + 24*(x*a
bs(b) - sqrt((b*x + a)^2 + 1))^2*a^6*b^2*i*abs(b) + 8*a^8*b^2*i*abs(b) - 6*(x*abs(b) - sqrt((b*x + a)^2 + 1))^
5*a^2*b^3 + 24*(x*abs(b) - sqrt((b*x + a)^2 + 1))^3*a^4*b^3 - 18*(x*abs(b) - sqrt((b*x + a)^2 + 1))*a^6*b^3 -
6*(x*abs(b) - sqrt((b*x + a)^2 + 1))^5*a*b^3*i + 32*(x*abs(b) - sqrt((b*x + a)^2 + 1))^3*a^3*b^3*i + 54*(x*abs
(b) - sqrt((b*x + a)^2 + 1))*a^5*b^3*i + 12*(x*abs(b) - sqrt((b*x + a)^2 + 1))^2*a^5*b^2*abs(b) - 12*a^7*b^2*a
bs(b) + 60*(x*abs(b) - sqrt((b*x + a)^2 + 1))^2*a^4*b^2*i*abs(b) + 20*a^6*b^2*i*abs(b) + 3*(x*abs(b) - sqrt((b
*x + a)^2 + 1))^5*b^3 + 24*(x*abs(b) - sqrt((b*x + a)^2 + 1))^3*a^2*b^3 - 39*(x*abs(b) - sqrt((b*x + a)^2 + 1)
)*a^4*b^3 + 24*(x*abs(b) - sqrt((b*x + a)^2 + 1))^3*a*b^3*i + 36*(x*abs(b) - sqrt((b*x + a)^2 + 1))*a^3*b^3*i
+ 24*(x*abs(b) - sqrt((b*x + a)^2 + 1))^2*a^3*b^2*abs(b) - 36*a^5*b^2*abs(b) + 48*(x*abs(b) - sqrt((b*x + a)^2
+ 1))^2*a^2*b^2*i*abs(b) + 12*a^4*b^2*i*abs(b) - 24*(x*abs(b) - sqrt((b*x + a)^2 + 1))*a^2*b^3 + 6*(x*abs(b)
- sqrt((b*x + a)^2 + 1))*a*b^3*i + 12*(x*abs(b) - sqrt((b*x + a)^2 + 1))^2*a*b^2*abs(b) - 36*a^3*b^2*abs(b) +
12*(x*abs(b) - sqrt((b*x + a)^2 + 1))^2*b^2*i*abs(b) - 4*a^2*b^2*i*abs(b) - 3*(x*abs(b) - sqrt((b*x + a)^2 + 1
))*b^3 - 12*a*b^2*abs(b) - 4*b^2*i*abs(b))/((a^5 - a^4*i + 2*a^3 - 2*a^2*i + a - i)*((x*abs(b) - sqrt((b*x + a
)^2 + 1))^2 - a^2 - 1)^3)

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maple [B]  time = 0.20, size = 1738, normalized size = 6.14 \[ \text {result too large to display} \]

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)/x^4,x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {{\left (b x + a\right )}^{2} + 1}}{{\left (i \, b x + i \, a + 1\right )} x^{4}}\,{d x} \]

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)/x^4,x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\sqrt {{\left (a+b\,x\right )}^2+1}}{x^4\,\left (1+a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - i \int \frac {\sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1}}{a x^{4} + b x^{5} - i x^{4}}\, dx \]

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)/x**4,x)

[Out]

-I*Integral(sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1)/(a*x**4 + b*x**5 - I*x**4), x)

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