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

Optimal. Leaf size=201 \[ -\frac{\sqrt{-i a-i b x+1} (i a+i b x+1)^{3/2} \left (-18 a^2+2 (6 a+i) b x-10 i a+7\right )}{24 b^4}-\frac{\left (8 i a^3-12 a^2-12 i a+3\right ) \sqrt{-i a-i b x+1} \sqrt{i a+i b x+1}}{8 b^4}+\frac{\left (-8 a^3-12 i a^2+12 a+3 i\right ) \sinh ^{-1}(a+b x)}{8 b^4}+\frac{x^2 \sqrt{-i a-i b x+1} (i a+i b x+1)^{3/2}}{4 b^2} \]

[Out]

-((3 - (12*I)*a - 12*a^2 + (8*I)*a^3)*Sqrt[1 - I*a - I*b*x]*Sqrt[1 + I*a + I*b*x])/(8*b^4) + (x^2*Sqrt[1 - I*a
- I*b*x]*(1 + I*a + I*b*x)^(3/2))/(4*b^2) - (Sqrt[1 - I*a - I*b*x]*(1 + I*a + I*b*x)^(3/2)*(7 - (10*I)*a - 18
*a^2 + 2*(I + 6*a)*b*x))/(24*b^4) + ((3*I + 12*a - (12*I)*a^2 - 8*a^3)*ArcSinh[a + b*x])/(8*b^4)

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Rubi [A]  time = 0.192415, antiderivative size = 201, 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, 100, 147, 50, 53, 619, 215} \[ -\frac{\sqrt{-i a-i b x+1} (i a+i b x+1)^{3/2} \left (-18 a^2+2 (6 a+i) b x-10 i a+7\right )}{24 b^4}-\frac{\left (8 i a^3-12 a^2-12 i a+3\right ) \sqrt{-i a-i b x+1} \sqrt{i a+i b x+1}}{8 b^4}+\frac{\left (-8 a^3-12 i a^2+12 a+3 i\right ) \sinh ^{-1}(a+b x)}{8 b^4}+\frac{x^2 \sqrt{-i a-i b x+1} (i a+i b x+1)^{3/2}}{4 b^2} \]

Antiderivative was successfully verified.

[In]

Int[E^(I*ArcTan[a + b*x])*x^3,x]

[Out]

-((3 - (12*I)*a - 12*a^2 + (8*I)*a^3)*Sqrt[1 - I*a - I*b*x]*Sqrt[1 + I*a + I*b*x])/(8*b^4) + (x^2*Sqrt[1 - I*a
- I*b*x]*(1 + I*a + I*b*x)^(3/2))/(4*b^2) - (Sqrt[1 - I*a - I*b*x]*(1 + I*a + I*b*x)^(3/2)*(7 - (10*I)*a - 18
*a^2 + 2*(I + 6*a)*b*x))/(24*b^4) + ((3*I + 12*a - (12*I)*a^2 - 8*a^3)*ArcSinh[a + b*x])/(8*b^4)

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 100

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))/(d*f*(m + n + p + 1)), x] + Dist[1/(d*f*(m + n + p + 1)), I
nt[(a + b*x)^(m - 2)*(c + d*x)^n*(e + f*x)^p*Simp[a^2*d*f*(m + n + p + 1) - b*(b*c*e*(m - 1) + a*(d*e*(n + 1)
+ c*f*(p + 1))) + b*(a*d*f*(2*m + n + p) - b*(d*e*(m + n) + c*f*(m + p)))*x, x], x], x] /; FreeQ[{a, b, c, d,
e, f, n, p}, x] && GtQ[m, 1] && NeQ[m + n + p + 1, 0] && IntegerQ[m]

Rule 147

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))*((g_.) + (h_.)*(x_)), x_Symbol]
:> -Simp[((a*d*f*h*(n + 2) + b*c*f*h*(m + 2) - b*d*(f*g + e*h)*(m + n + 3) - b*d*f*h*(m + n + 2)*x)*(a + b*x)^
(m + 1)*(c + d*x)^(n + 1))/(b^2*d^2*(m + n + 2)*(m + n + 3)), x] + Dist[(a^2*d^2*f*h*(n + 1)*(n + 2) + a*b*d*(
n + 1)*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b^2*(c^2*f*h*(m + 1)*(m + 2) - c*d*(f*g + e*h)*(m + 1)*
(m + n + 3) + d^2*e*g*(m + n + 2)*(m + n + 3)))/(b^2*d^2*(m + n + 2)*(m + n + 3)), Int[(a + b*x)^m*(c + d*x)^n
, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && NeQ[m + n + 2, 0] && NeQ[m + n + 3, 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^3 \, dx &=\int \frac{x^3 \sqrt{1+i a+i b x}}{\sqrt{1-i a-i b x}} \, dx\\ &=\frac{x^2 \sqrt{1-i a-i b x} (1+i a+i b x)^{3/2}}{4 b^2}+\frac{\int \frac{x \sqrt{1+i a+i b x} \left (-2 \left (1+a^2\right )-(i+6 a) b x\right )}{\sqrt{1-i a-i b x}} \, dx}{4 b^2}\\ &=\frac{x^2 \sqrt{1-i a-i b x} (1+i a+i b x)^{3/2}}{4 b^2}-\frac{\sqrt{1-i a-i b x} (1+i a+i b x)^{3/2} \left (7-10 i a-18 a^2+2 (i+6 a) b x\right )}{24 b^4}+\frac{\left (3 i+12 a-12 i a^2-8 a^3\right ) \int \frac{\sqrt{1+i a+i b x}}{\sqrt{1-i a-i b x}} \, dx}{8 b^3}\\ &=-\frac{\left (3-12 i a-12 a^2+8 i a^3\right ) \sqrt{1-i a-i b x} \sqrt{1+i a+i b x}}{8 b^4}+\frac{x^2 \sqrt{1-i a-i b x} (1+i a+i b x)^{3/2}}{4 b^2}-\frac{\sqrt{1-i a-i b x} (1+i a+i b x)^{3/2} \left (7-10 i a-18 a^2+2 (i+6 a) b x\right )}{24 b^4}+\frac{\left (3 i+12 a-12 i a^2-8 a^3\right ) \int \frac{1}{\sqrt{1-i a-i b x} \sqrt{1+i a+i b x}} \, dx}{8 b^3}\\ &=-\frac{\left (3-12 i a-12 a^2+8 i a^3\right ) \sqrt{1-i a-i b x} \sqrt{1+i a+i b x}}{8 b^4}+\frac{x^2 \sqrt{1-i a-i b x} (1+i a+i b x)^{3/2}}{4 b^2}-\frac{\sqrt{1-i a-i b x} (1+i a+i b x)^{3/2} \left (7-10 i a-18 a^2+2 (i+6 a) b x\right )}{24 b^4}+\frac{\left (3 i+12 a-12 i a^2-8 a^3\right ) \int \frac{1}{\sqrt{(1-i a) (1+i a)+2 a b x+b^2 x^2}} \, dx}{8 b^3}\\ &=-\frac{\left (3-12 i a-12 a^2+8 i a^3\right ) \sqrt{1-i a-i b x} \sqrt{1+i a+i b x}}{8 b^4}+\frac{x^2 \sqrt{1-i a-i b x} (1+i a+i b x)^{3/2}}{4 b^2}-\frac{\sqrt{1-i a-i b x} (1+i a+i b x)^{3/2} \left (7-10 i a-18 a^2+2 (i+6 a) b x\right )}{24 b^4}+\frac{\left (3 i+12 a-12 i a^2-8 a^3\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 )}{16 b^5}\\ &=-\frac{\left (3-12 i a-12 a^2+8 i a^3\right ) \sqrt{1-i a-i b x} \sqrt{1+i a+i b x}}{8 b^4}+\frac{x^2 \sqrt{1-i a-i b x} (1+i a+i b x)^{3/2}}{4 b^2}-\frac{\sqrt{1-i a-i b x} (1+i a+i b x)^{3/2} \left (7-10 i a-18 a^2+2 (i+6 a) b x\right )}{24 b^4}+\frac{\left (3 i+12 a-12 i a^2-8 a^3\right ) \sinh ^{-1}(a+b x)}{8 b^4}\\ \end{align*}

Mathematica [A]  time = 0.272815, size = 176, normalized size = 0.88 \[ \frac{\sqrt{b} \sqrt{a^2+2 a b x+b^2 x^2+1} \left (a^2 (44+6 i b x)-6 i a^3+a \left (-6 i b^2 x^2-20 b x+39 i\right )+6 i b^3 x^3+8 b^2 x^2-9 i b x-16\right )-6 \sqrt [4]{-1} \left (8 a^3+12 i a^2-12 a-3 i\right ) \sqrt{-i b} \sinh ^{-1}\left (\frac{\left (\frac{1}{2}+\frac{i}{2}\right ) \sqrt{b} \sqrt{-i (a+b x+i)}}{\sqrt{-i b}}\right )}{24 b^{9/2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Sqrt[b]*Sqrt[1 + a^2 + 2*a*b*x + b^2*x^2]*(-16 - (6*I)*a^3 - (9*I)*b*x + 8*b^2*x^2 + (6*I)*b^3*x^3 + a^2*(44
+ (6*I)*b*x) + a*(39*I - 20*b*x - (6*I)*b^2*x^2)) - 6*(-1)^(1/4)*(-3*I - 12*a + (12*I)*a^2 + 8*a^3)*Sqrt[(-I)*
b]*ArcSinh[((1/2 + I/2)*Sqrt[b]*Sqrt[(-I)*(I + a + b*x)])/Sqrt[(-I)*b]])/(24*b^(9/2))

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Maple [B]  time = 0.117, size = 465, normalized size = 2.3 \begin{align*}{\frac{{\frac{3\,i}{8}}}{{b}^{3}}\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{i}{4}}{a}^{2}x}{{b}^{3}}\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}-{\frac{{\frac{i}{4}}a{x}^{2}}{{b}^{2}}\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}-{\frac{{\frac{3\,i}{8}}x}{{b}^{3}}\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}+{\frac{{\frac{13\,i}{8}}a}{{b}^{4}}\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}-{\frac{{\frac{i}{4}}{a}^{3}}{{b}^{4}}\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}-{\frac{{\frac{3\,i}{2}}{a}^{2}}{{b}^{3}}\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{i}{4}}{x}^{3}}{b}\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}+{\frac{{x}^{2}}{3\,{b}^{2}}\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}-{\frac{5\,ax}{6\,{b}^{3}}\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}+{\frac{11\,{a}^{2}}{6\,{b}^{4}}\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}-{\frac{{a}^{3}}{{b}^{3}}\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{3\,a}{2\,{b}^{3}}\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{2}{3\,{b}^{4}}\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^3,x)

[Out]

3/8*I/b^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)+1/4*I/b^3*a^2*x*(b^2*x^2+2*a*b
*x+a^2+1)^(1/2)-1/4*I/b^2*a*x^2*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)-3/8*I/b^3*x*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)+13/8*I
/b^4*a*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)-1/4*I/b^4*a^3*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)-3/2*I/b^3*a^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)+1/4*I/b*x^3*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)+1/3*x^2/b^2*(
b^2*x^2+2*a*b*x+a^2+1)^(1/2)-5/6*a/b^3*x*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)+11/6*a^2/b^4*(b^2*x^2+2*a*b*x+a^2+1)^(1
/2)-a^3/b^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)+3/2*a/b^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)-2/3/b^4*(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^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.71922, size = 386, normalized size = 1.92 \begin{align*} \frac{-45 i \, a^{4} + 224 \, a^{3} + 192 i \, a^{2} +{\left (192 \, a^{3} + 288 i \, a^{2} - 288 \, a - 72 i\right )} \log \left (-b x - a + \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right ) +{\left (48 i \, b^{3} x^{3} - 16 \,{\left (3 i \, a - 4\right )} b^{2} x^{2} - 48 i \, a^{3} +{\left (48 i \, a^{2} - 160 \, a - 72 i\right )} b x + 352 \, a^{2} + 312 i \, a - 128\right )} \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1} - 72 \, a}{192 \, b^{4}} \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^3,x, algorithm="fricas")

[Out]

1/192*(-45*I*a^4 + 224*a^3 + 192*I*a^2 + (192*a^3 + 288*I*a^2 - 288*a - 72*I)*log(-b*x - a + sqrt(b^2*x^2 + 2*
a*b*x + a^2 + 1)) + (48*I*b^3*x^3 - 16*(3*I*a - 4)*b^2*x^2 - 48*I*a^3 + (48*I*a^2 - 160*a - 72*I)*b*x + 352*a^
2 + 312*I*a - 128)*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1) - 72*a)/b^4

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3} \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**3,x)

[Out]

Integral(x**3*(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 = 1.17295, size = 220, normalized size = 1.09 \begin{align*} \frac{1}{24} \, \sqrt{{\left (b x + a\right )}^{2} + 1}{\left ({\left (2 \,{\left (\frac{3 \, i x}{b} - \frac{3 \, a b^{5} i - 4 \, b^{5}}{b^{7}}\right )} x + \frac{6 \, a^{2} b^{4} i - 20 \, a b^{4} - 9 \, b^{4} i}{b^{7}}\right )} x - \frac{6 \, a^{3} b^{3} i - 44 \, a^{2} b^{3} - 39 \, a b^{3} i + 16 \, b^{3}}{b^{7}}\right )} + \frac{{\left (8 \, a^{3} + 12 \, a^{2} i - 12 \, a - 3 \, i\right )} \log \left (-a b -{\left (x{\left | b \right |} - \sqrt{{\left (b x + a\right )}^{2} + 1}\right )}{\left | b \right |}\right )}{8 \, b^{3}{\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^3,x, algorithm="giac")

[Out]

1/24*sqrt((b*x + a)^2 + 1)*((2*(3*i*x/b - (3*a*b^5*i - 4*b^5)/b^7)*x + (6*a^2*b^4*i - 20*a*b^4 - 9*b^4*i)/b^7)
*x - (6*a^3*b^3*i - 44*a^2*b^3 - 39*a*b^3*i + 16*b^3)/b^7) + 1/8*(8*a^3 + 12*a^2*i - 12*a - 3*i)*log(-a*b - (x
*abs(b) - sqrt((b*x + a)^2 + 1))*abs(b))/(b^3*abs(b))