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

Optimal. Leaf size=264 \[ -\frac {(-i a-i b x+1)^{5/2}}{2 \left (a^2+1\right ) x^2 \sqrt {i a+i b x+1}}-\frac {3 (2 a+3 i) b^2 \sqrt {-i a-i b x+1}}{(1+i a)^3 (a+i) \sqrt {i a+i b x+1}}+\frac {3 (3-2 i a) b^2 \tanh ^{-1}\left (\frac {\sqrt {a+i} \sqrt {i a+i b x+1}}{\sqrt {-a+i} \sqrt {-i a-i b x+1}}\right )}{(-a+i)^{7/2} \sqrt {a+i}}+\frac {(3-2 i a) b (-i a-i b x+1)^{3/2}}{2 (-a+i)^2 (a+i) x \sqrt {i a+i b x+1}} \]

[Out]

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

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Rubi [A]  time = 0.16, antiderivative size = 264, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {5095, 96, 94, 93, 208} \[ -\frac {(-i a-i b x+1)^{5/2}}{2 \left (a^2+1\right ) x^2 \sqrt {i a+i b x+1}}-\frac {3 (2 a+3 i) b^2 \sqrt {-i a-i b x+1}}{(1+i a)^3 (a+i) \sqrt {i a+i b x+1}}+\frac {3 (3-2 i a) b^2 \tanh ^{-1}\left (\frac {\sqrt {a+i} \sqrt {i a+i b x+1}}{\sqrt {-a+i} \sqrt {-i a-i b x+1}}\right )}{(-a+i)^{7/2} \sqrt {a+i}}+\frac {(3-2 i a) b (-i a-i b x+1)^{3/2}}{2 (-a+i)^2 (a+i) x \sqrt {i a+i b x+1}} \]

Antiderivative was successfully verified.

[In]

Int[1/(E^((3*I)*ArcTan[a + b*x])*x^3),x]

[Out]

(-3*(3*I + 2*a)*b^2*Sqrt[1 - I*a - I*b*x])/((1 + I*a)^3*(I + a)*Sqrt[1 + I*a + I*b*x]) + ((3 - (2*I)*a)*b*(1 -
I*a - I*b*x)^(3/2))/(2*(I - a)^2*(I + a)*x*Sqrt[1 + I*a + I*b*x]) - (1 - I*a - I*b*x)^(5/2)/(2*(1 + a^2)*x^2*
Sqrt[1 + I*a + I*b*x]) + (3*(3 - (2*I)*a)*b^2*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)*Sqrt[I + a])

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 94

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[(n*(d*e - c*f))/((m + 1)*(b*e - a*
f)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[
m + n + p + 2, 0] && GtQ[n, 0] &&  !(SumSimplerQ[p, 1] &&  !SumSimplerQ[m, 1])

Rule 96

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(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[(a*d*f*(m + 1)
+ b*c*f*(n + 1) + b*d*e*(p + 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, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[Simplify[m + n + p + 3], 0] && (LtQ[m, -1] || Sum
SimplerQ[m, 1])

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^{-3 i \tan ^{-1}(a+b x)}}{x^3} \, dx &=\int \frac {(1-i a-i b x)^{3/2}}{x^3 (1+i a+i b x)^{3/2}} \, dx\\ &=-\frac {(1-i a-i b x)^{5/2}}{2 \left (1+a^2\right ) x^2 \sqrt {1+i a+i b x}}-\frac {((3 i+2 a) b) \int \frac {(1-i a-i b x)^{3/2}}{x^2 (1+i a+i b x)^{3/2}} \, dx}{2 \left (1+a^2\right )}\\ &=\frac {(3-2 i a) b (1-i a-i b x)^{3/2}}{2 (i-a)^2 (i+a) x \sqrt {1+i a+i b x}}-\frac {(1-i a-i b x)^{5/2}}{2 \left (1+a^2\right ) x^2 \sqrt {1+i a+i b x}}+\frac {\left (3 (3 i+2 a) b^2\right ) \int \frac {\sqrt {1-i a-i b x}}{x (1+i a+i b x)^{3/2}} \, dx}{2 (i-a)^2 (i+a)}\\ &=-\frac {3 (3-2 i a) b^2 \sqrt {1-i a-i b x}}{(i-a)^3 (i+a) \sqrt {1+i a+i b x}}+\frac {(3-2 i a) b (1-i a-i b x)^{3/2}}{2 (i-a)^2 (i+a) x \sqrt {1+i a+i b x}}-\frac {(1-i a-i b x)^{5/2}}{2 \left (1+a^2\right ) x^2 \sqrt {1+i a+i b x}}+\frac {\left (3 (3 i+2 a) b^2\right ) \int \frac {1}{x \sqrt {1-i a-i b x} \sqrt {1+i a+i b x}} \, dx}{2 (i-a)^3}\\ &=-\frac {3 (3-2 i a) b^2 \sqrt {1-i a-i b x}}{(i-a)^3 (i+a) \sqrt {1+i a+i b x}}+\frac {(3-2 i a) b (1-i a-i b x)^{3/2}}{2 (i-a)^2 (i+a) x \sqrt {1+i a+i b x}}-\frac {(1-i a-i b x)^{5/2}}{2 \left (1+a^2\right ) x^2 \sqrt {1+i a+i b x}}+\frac {\left (3 (3 i+2 a) b^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 )}{(i-a)^3}\\ &=-\frac {3 (3-2 i a) b^2 \sqrt {1-i a-i b x}}{(i-a)^3 (i+a) \sqrt {1+i a+i b x}}+\frac {(3-2 i a) b (1-i a-i b x)^{3/2}}{2 (i-a)^2 (i+a) x \sqrt {1+i a+i b x}}-\frac {(1-i a-i b x)^{5/2}}{2 \left (1+a^2\right ) x^2 \sqrt {1+i a+i b x}}+\frac {3 (3-2 i a) b^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)^{7/2} \sqrt {i+a}}\\ \end {align*}

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Mathematica [A]  time = 0.27, size = 194, normalized size = 0.73 \[ \frac {\frac {\sqrt {-i (a+b x+i)} \left (a^3-i a^2-a b^2 x^2-5 i a b x+a-14 i b^2 x^2-5 b x-i\right )}{x^2 \sqrt {i a+i b x+1}}+\frac {6 i \sqrt {-1+i a} (2 a+3 i) b^2 \tanh ^{-1}\left (\frac {\sqrt {-1-i a} \sqrt {-i (a+b x+i)}}{\sqrt {-1+i a} \sqrt {i a+i b x+1}}\right )}{\sqrt {-1-i a} (a+i)}}{2 (a-i)^3} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/(E^((3*I)*ArcTan[a + b*x])*x^3),x]

[Out]

((Sqrt[(-I)*(I + a + b*x)]*(-I + a - I*a^2 + a^3 - 5*b*x - (5*I)*a*b*x - (14*I)*b^2*x^2 - a*b^2*x^2))/(x^2*Sqr
t[1 + I*a + I*b*x]) + ((6*I)*Sqrt[-1 + I*a]*(3*I + 2*a)*b^2*ArcTanh[(Sqrt[-1 - I*a]*Sqrt[(-I)*(I + a + b*x)])/
(Sqrt[-1 + I*a]*Sqrt[1 + I*a + I*b*x])])/(Sqrt[-1 - I*a]*(I + a)))/(2*(-I + a)^3)

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fricas [B]  time = 0.49, size = 580, normalized size = 2.20 \[ \frac {{\left (i \, a - 14\right )} b^{3} x^{3} + {\left (i \, a^{2} - 13 \, a + 14 i\right )} b^{2} x^{2} - 3 \, {\left ({\left (a^{3} - 3 i \, a^{2} - 3 \, a + i\right )} b x^{3} + {\left (a^{4} - 4 i \, a^{3} - 6 \, a^{2} + 4 i \, a + 1\right )} x^{2}\right )} \sqrt {\frac {{\left (4 \, a^{2} + 12 i \, a - 9\right )} b^{4}}{a^{8} - 6 i \, a^{7} - 14 \, a^{6} + 14 i \, a^{5} + 14 i \, a^{3} + 14 \, a^{2} - 6 i \, a - 1}} \log \left (-\frac {{\left (6 \, a + 9 i\right )} b^{3} x - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} {\left (6 \, a + 9 i\right )} b^{2} + 3 \, {\left (a^{5} - 3 i \, a^{4} - 2 \, a^{3} - 2 i \, a^{2} - 3 \, a + i\right )} \sqrt {\frac {{\left (4 \, a^{2} + 12 i \, a - 9\right )} b^{4}}{a^{8} - 6 i \, a^{7} - 14 \, a^{6} + 14 i \, a^{5} + 14 i \, a^{3} + 14 \, a^{2} - 6 i \, a - 1}}}{{\left (6 \, a + 9 i\right )} b^{2}}\right ) + 3 \, {\left ({\left (a^{3} - 3 i \, a^{2} - 3 \, a + i\right )} b x^{3} + {\left (a^{4} - 4 i \, a^{3} - 6 \, a^{2} + 4 i \, a + 1\right )} x^{2}\right )} \sqrt {\frac {{\left (4 \, a^{2} + 12 i \, a - 9\right )} b^{4}}{a^{8} - 6 i \, a^{7} - 14 \, a^{6} + 14 i \, a^{5} + 14 i \, a^{3} + 14 \, a^{2} - 6 i \, a - 1}} \log \left (-\frac {{\left (6 \, a + 9 i\right )} b^{3} x - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} {\left (6 \, a + 9 i\right )} b^{2} - 3 \, {\left (a^{5} - 3 i \, a^{4} - 2 \, a^{3} - 2 i \, a^{2} - 3 \, a + i\right )} \sqrt {\frac {{\left (4 \, a^{2} + 12 i \, a - 9\right )} b^{4}}{a^{8} - 6 i \, a^{7} - 14 \, a^{6} + 14 i \, a^{5} + 14 i \, a^{3} + 14 \, a^{2} - 6 i \, a - 1}}}{{\left (6 \, a + 9 i\right )} b^{2}}\right ) + {\left ({\left (i \, a - 14\right )} b^{2} x^{2} - i \, a^{3} - {\left (5 \, a - 5 i\right )} b x - a^{2} - i \, a - 1\right )} \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{{\left (2 \, a^{3} - 6 i \, a^{2} - 6 \, a + 2 i\right )} b x^{3} + {\left (2 \, a^{4} - 8 i \, a^{3} - 12 \, a^{2} + 8 i \, a + 2\right )} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((I*a - 14)*b^3*x^3 + (I*a^2 - 13*a + 14*I)*b^2*x^2 - 3*((a^3 - 3*I*a^2 - 3*a + I)*b*x^3 + (a^4 - 4*I*a^3 - 6*
a^2 + 4*I*a + 1)*x^2)*sqrt((4*a^2 + 12*I*a - 9)*b^4/(a^8 - 6*I*a^7 - 14*a^6 + 14*I*a^5 + 14*I*a^3 + 14*a^2 - 6
*I*a - 1))*log(-((6*a + 9*I)*b^3*x - sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*(6*a + 9*I)*b^2 + 3*(a^5 - 3*I*a^4 - 2*
a^3 - 2*I*a^2 - 3*a + I)*sqrt((4*a^2 + 12*I*a - 9)*b^4/(a^8 - 6*I*a^7 - 14*a^6 + 14*I*a^5 + 14*I*a^3 + 14*a^2
- 6*I*a - 1)))/((6*a + 9*I)*b^2)) + 3*((a^3 - 3*I*a^2 - 3*a + I)*b*x^3 + (a^4 - 4*I*a^3 - 6*a^2 + 4*I*a + 1)*x
^2)*sqrt((4*a^2 + 12*I*a - 9)*b^4/(a^8 - 6*I*a^7 - 14*a^6 + 14*I*a^5 + 14*I*a^3 + 14*a^2 - 6*I*a - 1))*log(-((
6*a + 9*I)*b^3*x - sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*(6*a + 9*I)*b^2 - 3*(a^5 - 3*I*a^4 - 2*a^3 - 2*I*a^2 - 3*
a + I)*sqrt((4*a^2 + 12*I*a - 9)*b^4/(a^8 - 6*I*a^7 - 14*a^6 + 14*I*a^5 + 14*I*a^3 + 14*a^2 - 6*I*a - 1)))/((6
*a + 9*I)*b^2)) + ((I*a - 14)*b^2*x^2 - I*a^3 - (5*a - 5*I)*b*x - a^2 - I*a - 1)*sqrt(b^2*x^2 + 2*a*b*x + a^2
+ 1))/((2*a^3 - 6*I*a^2 - 6*a + 2*I)*b*x^3 + (2*a^4 - 8*I*a^3 - 12*a^2 + 8*I*a + 2)*x^2)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {undef} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

undef

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maple [B]  time = 0.20, size = 3042, normalized size = 11.52 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-9/4*I/(I-a)^3*b^2/(a^2+1)*a^2*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)+3/2*I/(I-a)^3*b^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)*a^2-9/2*I/(I-a)^4*b^3/(a^2+1)*(b^2*x^2+2*a*b*x+a^2+1)^(1/
2)*x-9/2*I/(I-a)^4*b^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)+3*I/(I-a)
^4*b/(a^2+1)/x*(b^2*x^2+2*a*b*x+a^2+1)^(5/2)-6*I/(I-a)^4*b^2*a/(a^2+1)*(b^2*x^2+2*a*b*x+a^2+1)^(3/2)-27/2*I/(I
-a)^4*b^2*a^3/(a^2+1)*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)-1/(I-a)^3/b/(x-I/b+a/b)^3*((x-(I-a)/b)^2*b^2+2*I*b*(x-(I-a
)/b))^(5/2)-3/(I-a)^3*b^3*((x-(I-a)/b)^2*b^2+2*I*b*(x-(I-a)/b))^(1/2)*x-3/(I-a)^3*b^2*((x-(I-a)/b)^2*b^2+2*I*b
*(x-(I-a)/b))^(1/2)*a+9/2*I/(I-a)^4*b^2*((x-(I-a)/b)^2*b^2+2*I*b*(x-(I-a)/b))^(1/2)*a+9/2*I/(I-a)^4*b^3*((x-(I
-a)/b)^2*b^2+2*I*b*(x-(I-a)/b))^(1/2)*x-9/2*I/(I-a)^4*b^3*a^2/(a^2+1)*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)*x-27/2*I/(
I-a)^4*b^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)-9*I/(I-a)^4*b^3*a
^4/(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)-3/4*I/(I-a)^3*b^3/(a^2+1)*a*(
b^2*x^2+2*a*b*x+a^2+1)^(1/2)*x-9/4*I/(I-a)^3*b^3/(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)-3/2*I/(I-a)^3*b^3/(a^2+1)*a^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)^3*a*b/(a^2+1)^2/x*(b^2*x^2+2*a*b*x+a^2+1)^(5/2)-3/4*I/(I-a)^3*a^3*b^3/(a^2+1)^2*(b^2*x^2+
2*a*b*x+a^2+1)^(1/2)*x-9/4*I/(I-a)^3*a^3*b^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)-3/2*I/(I-a)^3*a^5*b^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/(I-a)^3*a*b^3/(a^2+1)^2*(b^2*x^2+2*a*b*x+a^2+1)^(3/2)*x-3/4*I/(I-a)^3*a*b^3/(a^2+1)^2*(b^2*x^2+2*
a*b*x+a^2+1)^(1/2)*x-3/4*I/(I-a)^3*a*b^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)-27/2*I/(I-a)^4*b^2*a/(a^2+1)*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)+9*I/(I-a)^4*b^2*a^3/(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)+9*I/(I-a)^4*b^2*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)-3*I/(I-a)^4*b^3/(a^2+1)*(b^2*x^2+2*a*b*x+a^2+1)^(3/2)*
x-3*I/(I-a)^5*b^3*a*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)*x-9*I/(I-a)^5*b^3*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)-6*I/(I-a)^5*b^3*a^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)+6*I/(I-a)^5*b^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)*
a^2-I/(I-a)^3*a^2*b^2/(a^2+1)^2*(b^2*x^2+2*a*b*x+a^2+1)^(3/2)-9/4*I/(I-a)^3*a^4*b^2/(a^2+1)^2*(b^2*x^2+2*a*b*x
+a^2+1)^(1/2)-9/4*I/(I-a)^3*a^2*b^2/(a^2+1)^2*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)+3/2*I/(I-a)^3*a^4*b^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)+3/2*I/(I-a)^3*a^2*b^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)+9/2*I/(I-a)^4*b^3*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)-9*I/(I-a)^5*b^2*a^2*(b^2*x^2+2*a*b*x+a^2
+1)^(1/2)+6*I/(I-a)^5*b^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)+
1/2*I/(I-a)^3/(a^2+1)/x^2*(b^2*x^2+2*a*b*x+a^2+1)^(5/2)-1/2*I/(I-a)^3*b^2/(a^2+1)*(b^2*x^2+2*a*b*x+a^2+1)^(3/2
)-3/2*I/(I-a)^3*b^2/(a^2+1)*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)+3/2*I/(I-a)^3*b^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)-3/(I-a)^4/(x-I/b+a/b)^2*((x-(I-a)/b)^2*b^2+2*I*b*(x-(I-a)/b)
)^(5/2)+3/(I-a)^4*b^2*((x-(I-a)/b)^2*b^2+2*I*b*(x-(I-a)/b))^(3/2)-3/(I-a)^3*b^3*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)-2*I/(I-a)^3/(x-I/b+a/b)^2*((x-(I-a)/b)^2*b^2+2
*I*b*(x-(I-a)/b))^(5/2)+2*I/(I-a)^3*b^2*((x-(I-a)/b)^2*b^2+2*I*b*(x-(I-a)/b))^(3/2)-2*I/(I-a)^5*b^2*(b^2*x^2+2
*a*b*x+a^2+1)^(3/2)-6*I/(I-a)^5*b^2*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)-3/(I-a)^5*b^3*((x-(I-a)/b)^2*b^2+2*I*b*(x-(I
-a)/b))^(1/2)*x-3/(I-a)^5*b^2*((x-(I-a)/b)^2*b^2+2*I*b*(x-(I-a)/b))^(1/2)*a-3/(I-a)^5*b^3*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)+2*I/(I-a)^5*b^2*((x-(I-a)/b)^2*b^2+2
*I*b*(x-(I-a)/b))^(3/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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