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

Optimal. Leaf size=171 \[ -\frac{\left (-2 i a^2+2 a+i\right ) \sqrt{-i a-i b x+1} \sqrt{i a+i b x+1}}{2 b^3}-\frac{\left (-2 a^2-2 i a+1\right ) \sinh ^{-1}(a+b x)}{2 b^3}+\frac{x \sqrt{-i a-i b x+1} (i a+i b x+1)^{3/2}}{3 b^2}-\frac{(4 a+i) \sqrt{-i a-i b x+1} (i a+i b x+1)^{3/2}}{6 b^3} \]

[Out]

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

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Rubi [A]  time = 0.125008, antiderivative size = 171, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.438, Rules used = {5095, 90, 80, 50, 53, 619, 215} \[ -\frac{\left (-2 i a^2+2 a+i\right ) \sqrt{-i a-i b x+1} \sqrt{i a+i b x+1}}{2 b^3}-\frac{\left (-2 a^2-2 i a+1\right ) \sinh ^{-1}(a+b x)}{2 b^3}+\frac{x \sqrt{-i a-i b x+1} (i a+i b x+1)^{3/2}}{3 b^2}-\frac{(4 a+i) \sqrt{-i a-i b x+1} (i a+i b x+1)^{3/2}}{6 b^3} \]

Antiderivative was successfully verified.

[In]

Int[E^(I*ArcTan[a + b*x])*x^2,x]

[Out]

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

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 90

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a + b*
x)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 3)), x] + Dist[1/(d*f*(n + p + 3)), Int[(c + d*x)^n*(e +
f*x)^p*Simp[a^2*d*f*(n + p + 3) - b*(b*c*e + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f*(n + p + 4) - b*(d*e*(
n + 2) + c*f*(p + 2)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 3, 0]

Rule 80

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

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, 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]

Rubi steps

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

Mathematica [A]  time = 0.134284, size = 135, normalized size = 0.79 \[ \frac{\sqrt{a^2+2 a b x+b^2 x^2+1} \left (2 i a^2+a (-9-2 i b x)+2 i b^2 x^2+3 b x-4 i\right )}{6 b^3}+\frac{\sqrt [4]{-1} \left (2 a^2+2 i a-1\right ) \sqrt{-i b} \sinh ^{-1}\left (\frac{\left (\frac{1}{2}+\frac{i}{2}\right ) \sqrt{b} \sqrt{-i (a+b x+i)}}{\sqrt{-i b}}\right )}{b^{7/2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [B]  time = 0.117, size = 302, normalized size = 1.8 \begin{align*}{\frac{{\frac{i}{3}}{x}^{2}}{b}\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}-{\frac{{\frac{i}{3}}ax}{{b}^{2}}\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}+{\frac{{\frac{i}{3}}{a}^{2}}{{b}^{3}}\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}+{\frac{ia}{{b}^{2}}\ln \left ({({b}^{2}x+ab){\frac{1}{\sqrt{{b}^{2}}}}}+\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1} \right ){\frac{1}{\sqrt{{b}^{2}}}}}-{\frac{{\frac{2\,i}{3}}}{{b}^{3}}\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}+{\frac{x}{2\,{b}^{2}}\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}-{\frac{3\,a}{2\,{b}^{3}}\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}+{\frac{{a}^{2}}{{b}^{2}}\ln \left ({({b}^{2}x+ab){\frac{1}{\sqrt{{b}^{2}}}}}+\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1} \right ){\frac{1}{\sqrt{{b}^{2}}}}}-{\frac{1}{2\,{b}^{2}}\ln \left ({({b}^{2}x+ab){\frac{1}{\sqrt{{b}^{2}}}}}+\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1} \right ){\frac{1}{\sqrt{{b}^{2}}}}} \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^2,x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.69255, size = 278, normalized size = 1.63 \begin{align*} \frac{7 i \, a^{3} - 21 \, a^{2} - 12 \,{\left (2 \, a^{2} + 2 i \, a - 1\right )} \log \left (-b x - a + \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right ) + \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}{\left (8 i \, b^{2} x^{2} - 4 \,{\left (2 i \, a - 3\right )} b x + 8 i \, a^{2} - 36 \, a - 16 i\right )} - 9 i \, a}{24 \, b^{3}} \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^2,x, algorithm="fricas")

[Out]

1/24*(7*I*a^3 - 21*a^2 - 12*(2*a^2 + 2*I*a - 1)*log(-b*x - a + sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)) + sqrt(b^2*x
^2 + 2*a*b*x + a^2 + 1)*(8*I*b^2*x^2 - 4*(2*I*a - 3)*b*x + 8*I*a^2 - 36*a - 16*I) - 9*I*a)/b^3

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

[Out]

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

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Giac [A]  time = 1.15091, size = 158, normalized size = 0.92 \begin{align*} \frac{1}{6} \, \sqrt{{\left (b x + a\right )}^{2} + 1}{\left ({\left (\frac{2 \, i x}{b} - \frac{2 \, a b^{3} i - 3 \, b^{3}}{b^{5}}\right )} x + \frac{2 \, a^{2} b^{2} i - 9 \, a b^{2} - 4 \, b^{2} i}{b^{5}}\right )} - \frac{{\left (2 \, a^{2} + 2 \, a i - 1\right )} \log \left (-a b -{\left (x{\left | b \right |} - \sqrt{{\left (b x + a\right )}^{2} + 1}\right )}{\left | b \right |}\right )}{2 \, b^{2}{\left | b \right |}} \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^2,x, algorithm="giac")

[Out]

1/6*sqrt((b*x + a)^2 + 1)*((2*i*x/b - (2*a*b^3*i - 3*b^3)/b^5)*x + (2*a^2*b^2*i - 9*a*b^2 - 4*b^2*i)/b^5) - 1/
2*(2*a^2 + 2*a*i - 1)*log(-a*b - (x*abs(b) - sqrt((b*x + a)^2 + 1))*abs(b))/(b^2*abs(b))