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

Optimal. Leaf size=211 \[ -\frac {\sqrt [4]{-i a-i b x+1} (i a+i b x+1)^{3/4}}{(1-i a) x}-\frac {3 i b \tan ^{-1}\left (\frac {\sqrt [4]{a+i} \sqrt [4]{i a+i b x+1}}{\sqrt [4]{-a+i} \sqrt [4]{-i a-i b x+1}}\right )}{\sqrt [4]{-a+i} (a+i)^{7/4}}+\frac {3 i b \tanh ^{-1}\left (\frac {\sqrt [4]{a+i} \sqrt [4]{i a+i b x+1}}{\sqrt [4]{-a+i} \sqrt [4]{-i a-i b x+1}}\right )}{\sqrt [4]{-a+i} (a+i)^{7/4}} \]

[Out]

-(1-I*a-I*b*x)^(1/4)*(1+I*a+I*b*x)^(3/4)/(1-I*a)/x-3*I*b*arctan((I+a)^(1/4)*(1+I*a+I*b*x)^(1/4)/(I-a)^(1/4)/(1
-I*a-I*b*x)^(1/4))/(I-a)^(1/4)/(I+a)^(7/4)+3*I*b*arctanh((I+a)^(1/4)*(1+I*a+I*b*x)^(1/4)/(I-a)^(1/4)/(1-I*a-I*
b*x)^(1/4))/(I-a)^(1/4)/(I+a)^(7/4)

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Rubi [A]  time = 0.11, antiderivative size = 211, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {5095, 94, 93, 298, 205, 208} \[ -\frac {\sqrt [4]{-i a-i b x+1} (i a+i b x+1)^{3/4}}{(1-i a) x}-\frac {3 i b \tan ^{-1}\left (\frac {\sqrt [4]{a+i} \sqrt [4]{i a+i b x+1}}{\sqrt [4]{-a+i} \sqrt [4]{-i a-i b x+1}}\right )}{\sqrt [4]{-a+i} (a+i)^{7/4}}+\frac {3 i b \tanh ^{-1}\left (\frac {\sqrt [4]{a+i} \sqrt [4]{i a+i b x+1}}{\sqrt [4]{-a+i} \sqrt [4]{-i a-i b x+1}}\right )}{\sqrt [4]{-a+i} (a+i)^{7/4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((1 - I*a - I*b*x)^(1/4)*(1 + I*a + I*b*x)^(3/4))/((1 - I*a)*x)) - ((3*I)*b*ArcTan[((I + a)^(1/4)*(1 + I*a +
I*b*x)^(1/4))/((I - a)^(1/4)*(1 - I*a - I*b*x)^(1/4))])/((I - a)^(1/4)*(I + a)^(7/4)) + ((3*I)*b*ArcTanh[((I
+ a)^(1/4)*(1 + I*a + I*b*x)^(1/4))/((I - a)^(1/4)*(1 - I*a - I*b*x)^(1/4))])/((I - a)^(1/4)*(I + a)^(7/4))

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 205

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

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 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 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^{\frac {3}{2} i \tan ^{-1}(a+b x)}}{x^2} \, dx &=\int \frac {(1+i a+i b x)^{3/4}}{x^2 (1-i a-i b x)^{3/4}} \, dx\\ &=-\frac {\sqrt [4]{1-i a-i b x} (1+i a+i b x)^{3/4}}{(1-i a) x}-\frac {(3 b) \int \frac {1}{x (1-i a-i b x)^{3/4} \sqrt [4]{1+i a+i b x}} \, dx}{2 (i+a)}\\ &=-\frac {\sqrt [4]{1-i a-i b x} (1+i a+i b x)^{3/4}}{(1-i a) x}-\frac {(6 b) \operatorname {Subst}\left (\int \frac {x^2}{-1-i a-(-1+i a) x^4} \, dx,x,\frac {\sqrt [4]{1+i a+i b x}}{\sqrt [4]{1-i a-i b x}}\right )}{i+a}\\ &=-\frac {\sqrt [4]{1-i a-i b x} (1+i a+i b x)^{3/4}}{(1-i a) x}+\frac {(3 i b) \operatorname {Subst}\left (\int \frac {1}{\sqrt {i-a}-\sqrt {i+a} x^2} \, dx,x,\frac {\sqrt [4]{1+i a+i b x}}{\sqrt [4]{1-i a-i b x}}\right )}{(i+a)^{3/2}}-\frac {(3 i b) \operatorname {Subst}\left (\int \frac {1}{\sqrt {i-a}+\sqrt {i+a} x^2} \, dx,x,\frac {\sqrt [4]{1+i a+i b x}}{\sqrt [4]{1-i a-i b x}}\right )}{(i+a)^{3/2}}\\ &=-\frac {\sqrt [4]{1-i a-i b x} (1+i a+i b x)^{3/4}}{(1-i a) x}-\frac {3 i b \tan ^{-1}\left (\frac {\sqrt [4]{i+a} \sqrt [4]{1+i a+i b x}}{\sqrt [4]{i-a} \sqrt [4]{1-i a-i b x}}\right )}{\sqrt [4]{i-a} (i+a)^{7/4}}+\frac {3 i b \tanh ^{-1}\left (\frac {\sqrt [4]{i+a} \sqrt [4]{1+i a+i b x}}{\sqrt [4]{i-a} \sqrt [4]{1-i a-i b x}}\right )}{\sqrt [4]{i-a} (i+a)^{7/4}}\\ \end {align*}

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Mathematica [C]  time = 0.03, size = 106, normalized size = 0.50 \[ \frac {\sqrt [4]{-i (a+b x+i)} \left (6 i b x \, _2F_1\left (\frac {1}{4},1;\frac {5}{4};\frac {a^2+b x a-i b x+1}{a^2+b x a+i b x+1}\right )+a^2+a b x+i b x+1\right )}{(a+i)^2 x \sqrt [4]{i a+i b x+1}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(((-I)*(I + a + b*x))^(1/4)*(1 + a^2 + I*b*x + a*b*x + (6*I)*b*x*Hypergeometric2F1[1/4, 1, 5/4, (1 + a^2 - I*b
*x + a*b*x)/(1 + a^2 + I*b*x + a*b*x)]))/((I + a)^2*x*(1 + I*a + I*b*x)^(1/4))

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fricas [B]  time = 0.48, size = 711, normalized size = 3.37 \[ \frac {3 \, \left (-\frac {b^{4}}{16 \, a^{8} + 96 i \, a^{7} - 224 \, a^{6} - 224 i \, a^{5} - 224 i \, a^{3} + 224 \, a^{2} + 96 i \, a - 16}\right )^{\frac {1}{4}} {\left (-i \, a + 1\right )} x \log \left (\frac {b^{3} \sqrt {\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}} + {\left (8 \, a^{6} + 32 i \, a^{5} - 40 \, a^{4} - 40 \, a^{2} - 32 i \, a + 8\right )} \left (-\frac {b^{4}}{16 \, a^{8} + 96 i \, a^{7} - 224 \, a^{6} - 224 i \, a^{5} - 224 i \, a^{3} + 224 \, a^{2} + 96 i \, a - 16}\right )^{\frac {3}{4}}}{b^{3}}\right ) + 3 \, \left (-\frac {b^{4}}{16 \, a^{8} + 96 i \, a^{7} - 224 \, a^{6} - 224 i \, a^{5} - 224 i \, a^{3} + 224 \, a^{2} + 96 i \, a - 16}\right )^{\frac {1}{4}} {\left (i \, a - 1\right )} x \log \left (\frac {b^{3} \sqrt {\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}} - {\left (8 \, a^{6} + 32 i \, a^{5} - 40 \, a^{4} - 40 \, a^{2} - 32 i \, a + 8\right )} \left (-\frac {b^{4}}{16 \, a^{8} + 96 i \, a^{7} - 224 \, a^{6} - 224 i \, a^{5} - 224 i \, a^{3} + 224 \, a^{2} + 96 i \, a - 16}\right )^{\frac {3}{4}}}{b^{3}}\right ) - 3 \, \left (-\frac {b^{4}}{16 \, a^{8} + 96 i \, a^{7} - 224 \, a^{6} - 224 i \, a^{5} - 224 i \, a^{3} + 224 \, a^{2} + 96 i \, a - 16}\right )^{\frac {1}{4}} {\left (a + i\right )} x \log \left (\frac {b^{3} \sqrt {\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}} + {\left (8 i \, a^{6} - 32 \, a^{5} - 40 i \, a^{4} - 40 i \, a^{2} + 32 \, a + 8 i\right )} \left (-\frac {b^{4}}{16 \, a^{8} + 96 i \, a^{7} - 224 \, a^{6} - 224 i \, a^{5} - 224 i \, a^{3} + 224 \, a^{2} + 96 i \, a - 16}\right )^{\frac {3}{4}}}{b^{3}}\right ) + 3 \, \left (-\frac {b^{4}}{16 \, a^{8} + 96 i \, a^{7} - 224 \, a^{6} - 224 i \, a^{5} - 224 i \, a^{3} + 224 \, a^{2} + 96 i \, a - 16}\right )^{\frac {1}{4}} {\left (a + i\right )} x \log \left (\frac {b^{3} \sqrt {\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}} + {\left (-8 i \, a^{6} + 32 \, a^{5} + 40 i \, a^{4} + 40 i \, a^{2} - 32 \, a - 8 i\right )} \left (-\frac {b^{4}}{16 \, a^{8} + 96 i \, a^{7} - 224 \, a^{6} - 224 i \, a^{5} - 224 i \, a^{3} + 224 \, a^{2} + 96 i \, a - 16}\right )^{\frac {3}{4}}}{b^{3}}\right ) - i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} \sqrt {\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}}}{{\left (a + i\right )} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(3*(-b^4/(16*a^8 + 96*I*a^7 - 224*a^6 - 224*I*a^5 - 224*I*a^3 + 224*a^2 + 96*I*a - 16))^(1/4)*(-I*a + 1)*x*log
((b^3*sqrt(I*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)/(b*x + a + I)) + (8*a^6 + 32*I*a^5 - 40*a^4 - 40*a^2 - 32*I*a +
8)*(-b^4/(16*a^8 + 96*I*a^7 - 224*a^6 - 224*I*a^5 - 224*I*a^3 + 224*a^2 + 96*I*a - 16))^(3/4))/b^3) + 3*(-b^4
/(16*a^8 + 96*I*a^7 - 224*a^6 - 224*I*a^5 - 224*I*a^3 + 224*a^2 + 96*I*a - 16))^(1/4)*(I*a - 1)*x*log((b^3*sqr
t(I*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)/(b*x + a + I)) - (8*a^6 + 32*I*a^5 - 40*a^4 - 40*a^2 - 32*I*a + 8)*(-b^4
/(16*a^8 + 96*I*a^7 - 224*a^6 - 224*I*a^5 - 224*I*a^3 + 224*a^2 + 96*I*a - 16))^(3/4))/b^3) - 3*(-b^4/(16*a^8
+ 96*I*a^7 - 224*a^6 - 224*I*a^5 - 224*I*a^3 + 224*a^2 + 96*I*a - 16))^(1/4)*(a + I)*x*log((b^3*sqrt(I*sqrt(b^
2*x^2 + 2*a*b*x + a^2 + 1)/(b*x + a + I)) + (8*I*a^6 - 32*a^5 - 40*I*a^4 - 40*I*a^2 + 32*a + 8*I)*(-b^4/(16*a^
8 + 96*I*a^7 - 224*a^6 - 224*I*a^5 - 224*I*a^3 + 224*a^2 + 96*I*a - 16))^(3/4))/b^3) + 3*(-b^4/(16*a^8 + 96*I*
a^7 - 224*a^6 - 224*I*a^5 - 224*I*a^3 + 224*a^2 + 96*I*a - 16))^(1/4)*(a + I)*x*log((b^3*sqrt(I*sqrt(b^2*x^2 +
2*a*b*x + a^2 + 1)/(b*x + a + I)) + (-8*I*a^6 + 32*a^5 + 40*I*a^4 + 40*I*a^2 - 32*a - 8*I)*(-b^4/(16*a^8 + 96
*I*a^7 - 224*a^6 - 224*I*a^5 - 224*I*a^3 + 224*a^2 + 96*I*a - 16))^(3/4))/b^3) - I*sqrt(b^2*x^2 + 2*a*b*x + a^
2 + 1)*sqrt(I*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)/(b*x + a + I)))/((a + I)*x)

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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*(b*x+a))/(1+(b*x+a)^2)^(1/2))^(3/2)/x^2,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=[0,0,0]index.cc index_m operator + Error: Bad Argument ValueWarning, need to choose a branch f
or the root of a polynomial with parameters. This might be wrong.The choice was done assuming 0=[0,0,0]index.c
c index_m operator + Error: Bad Argument ValueWarning, need to choose a branch for the root of a polynomial wi
th parameters. This might be wrong.The choice was done assuming 0=[0,0,0]index.cc index_m operator + Error: Ba
d Argument ValueWarning, need to choose a branch for the root of a polynomial with parameters. This might be w
rong.The choice was done assuming 0=[0,0,0]index.cc index_m operator + Error: Bad Argument ValueEvaluation tim
e: 1.26Done

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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