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

Optimal. Leaf size=276 \[ \frac{\sqrt{-i a-i b x+1} (i a+i b x+1)^{3/2} \left (-2 \left (-36 a^2-14 i a+13\right ) b x-96 a^3-86 i a^2+114 a+19 i\right )}{120 b^5}+\frac{\left (8 i a^4-16 a^3-24 i a^2+12 a+3 i\right ) \sqrt{-i a-i b x+1} \sqrt{i a+i b x+1}}{8 b^5}+\frac{\left (8 a^4+16 i a^3-24 a^2-12 i a+3\right ) \sinh ^{-1}(a+b x)}{8 b^5}+\frac{x^3 \sqrt{-i a-i b x+1} (i a+i b x+1)^{3/2}}{5 b^2}-\frac{(8 a+i) x^2 \sqrt{-i a-i b x+1} (i a+i b x+1)^{3/2}}{20 b^3} \]

[Out]

((3*I + 12*a - (24*I)*a^2 - 16*a^3 + (8*I)*a^4)*Sqrt[1 - I*a - I*b*x]*Sqrt[1 + I*a + I*b*x])/(8*b^5) - ((I + 8
*a)*x^2*Sqrt[1 - I*a - I*b*x]*(1 + I*a + I*b*x)^(3/2))/(20*b^3) + (x^3*Sqrt[1 - I*a - I*b*x]*(1 + I*a + I*b*x)
^(3/2))/(5*b^2) + (Sqrt[1 - I*a - I*b*x]*(1 + I*a + I*b*x)^(3/2)*(19*I + 114*a - (86*I)*a^2 - 96*a^3 - 2*(13 -
(14*I)*a - 36*a^2)*b*x))/(120*b^5) + ((3 - (12*I)*a - 24*a^2 + (16*I)*a^3 + 8*a^4)*ArcSinh[a + b*x])/(8*b^5)

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Rubi [A]  time = 0.201435, antiderivative size = 276, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {5095, 100, 153, 147, 50, 53, 619, 215} \[ \frac{\sqrt{-i a-i b x+1} (i a+i b x+1)^{3/2} \left (-2 \left (-36 a^2-14 i a+13\right ) b x-96 a^3-86 i a^2+114 a+19 i\right )}{120 b^5}+\frac{\left (8 i a^4-16 a^3-24 i a^2+12 a+3 i\right ) \sqrt{-i a-i b x+1} \sqrt{i a+i b x+1}}{8 b^5}+\frac{\left (8 a^4+16 i a^3-24 a^2-12 i a+3\right ) \sinh ^{-1}(a+b x)}{8 b^5}+\frac{x^3 \sqrt{-i a-i b x+1} (i a+i b x+1)^{3/2}}{5 b^2}-\frac{(8 a+i) x^2 \sqrt{-i a-i b x+1} (i a+i b x+1)^{3/2}}{20 b^3} \]

Antiderivative was successfully verified.

[In]

Int[E^(I*ArcTan[a + b*x])*x^4,x]

[Out]

((3*I + 12*a - (24*I)*a^2 - 16*a^3 + (8*I)*a^4)*Sqrt[1 - I*a - I*b*x]*Sqrt[1 + I*a + I*b*x])/(8*b^5) - ((I + 8
*a)*x^2*Sqrt[1 - I*a - I*b*x]*(1 + I*a + I*b*x)^(3/2))/(20*b^3) + (x^3*Sqrt[1 - I*a - I*b*x]*(1 + I*a + I*b*x)
^(3/2))/(5*b^2) + (Sqrt[1 - I*a - I*b*x]*(1 + I*a + I*b*x)^(3/2)*(19*I + 114*a - (86*I)*a^2 - 96*a^3 - 2*(13 -
(14*I)*a - 36*a^2)*b*x))/(120*b^5) + ((3 - (12*I)*a - 24*a^2 + (16*I)*a^3 + 8*a^4)*ArcSinh[a + b*x])/(8*b^5)

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 153

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[(h*(a + b*x)^m*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d*f*(m + n + p + 2)), x] + Dist[1/(d*f*(m + n
+ p + 2)), Int[(a + b*x)^(m - 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*g*(m + n + p + 2) - h*(b*c*e*m + a*(d*e*(
n + 1) + c*f*(p + 1))) + (b*d*f*g*(m + n + p + 2) + h*(a*d*f*m - b*(d*e*(m + n + 1) + c*f*(m + p + 1))))*x, x]
, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && GtQ[m, 0] && NeQ[m + n + p + 2, 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^4 \, dx &=\int \frac{x^4 \sqrt{1+i a+i b x}}{\sqrt{1-i a-i b x}} \, dx\\ &=\frac{x^3 \sqrt{1-i a-i b x} (1+i a+i b x)^{3/2}}{5 b^2}+\frac{\int \frac{x^2 \sqrt{1+i a+i b x} \left (-3 \left (1+a^2\right )-(i+8 a) b x\right )}{\sqrt{1-i a-i b x}} \, dx}{5 b^2}\\ &=-\frac{(i+8 a) x^2 \sqrt{1-i a-i b x} (1+i a+i b x)^{3/2}}{20 b^3}+\frac{x^3 \sqrt{1-i a-i b x} (1+i a+i b x)^{3/2}}{5 b^2}+\frac{\int \frac{x \sqrt{1+i a+i b x} \left (-2 (i-a) (i+a) (i+8 a) b-\left (13-14 i a-36 a^2\right ) b^2 x\right )}{\sqrt{1-i a-i b x}} \, dx}{20 b^4}\\ &=-\frac{(i+8 a) x^2 \sqrt{1-i a-i b x} (1+i a+i b x)^{3/2}}{20 b^3}+\frac{x^3 \sqrt{1-i a-i b x} (1+i a+i b x)^{3/2}}{5 b^2}+\frac{\sqrt{1-i a-i b x} (1+i a+i b x)^{3/2} \left (19 i+114 a-86 i a^2-96 a^3-2 \left (13-14 i a-36 a^2\right ) b x\right )}{120 b^5}+\frac{\left (3-12 i a-24 a^2+16 i a^3+8 a^4\right ) \int \frac{\sqrt{1+i a+i b x}}{\sqrt{1-i a-i b x}} \, dx}{8 b^4}\\ &=\frac{\left (3 i+12 a-24 i a^2-16 a^3+8 i a^4\right ) \sqrt{1-i a-i b x} \sqrt{1+i a+i b x}}{8 b^5}-\frac{(i+8 a) x^2 \sqrt{1-i a-i b x} (1+i a+i b x)^{3/2}}{20 b^3}+\frac{x^3 \sqrt{1-i a-i b x} (1+i a+i b x)^{3/2}}{5 b^2}+\frac{\sqrt{1-i a-i b x} (1+i a+i b x)^{3/2} \left (19 i+114 a-86 i a^2-96 a^3-2 \left (13-14 i a-36 a^2\right ) b x\right )}{120 b^5}+\frac{\left (3-12 i a-24 a^2+16 i a^3+8 a^4\right ) \int \frac{1}{\sqrt{1-i a-i b x} \sqrt{1+i a+i b x}} \, dx}{8 b^4}\\ &=\frac{\left (3 i+12 a-24 i a^2-16 a^3+8 i a^4\right ) \sqrt{1-i a-i b x} \sqrt{1+i a+i b x}}{8 b^5}-\frac{(i+8 a) x^2 \sqrt{1-i a-i b x} (1+i a+i b x)^{3/2}}{20 b^3}+\frac{x^3 \sqrt{1-i a-i b x} (1+i a+i b x)^{3/2}}{5 b^2}+\frac{\sqrt{1-i a-i b x} (1+i a+i b x)^{3/2} \left (19 i+114 a-86 i a^2-96 a^3-2 \left (13-14 i a-36 a^2\right ) b x\right )}{120 b^5}+\frac{\left (3-12 i a-24 a^2+16 i a^3+8 a^4\right ) \int \frac{1}{\sqrt{(1-i a) (1+i a)+2 a b x+b^2 x^2}} \, dx}{8 b^4}\\ &=\frac{\left (3 i+12 a-24 i a^2-16 a^3+8 i a^4\right ) \sqrt{1-i a-i b x} \sqrt{1+i a+i b x}}{8 b^5}-\frac{(i+8 a) x^2 \sqrt{1-i a-i b x} (1+i a+i b x)^{3/2}}{20 b^3}+\frac{x^3 \sqrt{1-i a-i b x} (1+i a+i b x)^{3/2}}{5 b^2}+\frac{\sqrt{1-i a-i b x} (1+i a+i b x)^{3/2} \left (19 i+114 a-86 i a^2-96 a^3-2 \left (13-14 i a-36 a^2\right ) b x\right )}{120 b^5}+\frac{\left (3-12 i a-24 a^2+16 i a^3+8 a^4\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^6}\\ &=\frac{\left (3 i+12 a-24 i a^2-16 a^3+8 i a^4\right ) \sqrt{1-i a-i b x} \sqrt{1+i a+i b x}}{8 b^5}-\frac{(i+8 a) x^2 \sqrt{1-i a-i b x} (1+i a+i b x)^{3/2}}{20 b^3}+\frac{x^3 \sqrt{1-i a-i b x} (1+i a+i b x)^{3/2}}{5 b^2}+\frac{\sqrt{1-i a-i b x} (1+i a+i b x)^{3/2} \left (19 i+114 a-86 i a^2-96 a^3-2 \left (13-14 i a-36 a^2\right ) b x\right )}{120 b^5}+\frac{\left (3-12 i a-24 a^2+16 i a^3+8 a^4\right ) \sinh ^{-1}(a+b x)}{8 b^5}\\ \end{align*}

Mathematica [A]  time = 0.49234, size = 217, normalized size = 0.79 \[ \frac{i \sqrt{a^2+2 a b x+b^2 x^2+1} \left (2 a^2 \left (12 b^2 x^2-65 i b x-166\right )+a^3 (-24 b x+250 i)+24 a^4+a \left (-24 b^3 x^3+70 i b^2 x^2+116 b x-275 i\right )+24 b^4 x^4-30 i b^3 x^3-32 b^2 x^2+45 i b x+64\right )}{120 b^5}+\frac{\sqrt [4]{-1} \left (8 a^4+16 i a^3-24 a^2-12 i a+3\right ) \sqrt{-i b} \sinh ^{-1}\left (\frac{\left (\frac{1}{2}+\frac{i}{2}\right ) \sqrt{b} \sqrt{-i (a+b x+i)}}{\sqrt{-i b}}\right )}{4 b^{11/2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^(I*ArcTan[a + b*x])*x^4,x]

[Out]

((I/120)*Sqrt[1 + a^2 + 2*a*b*x + b^2*x^2]*(64 + 24*a^4 + (45*I)*b*x - 32*b^2*x^2 - (30*I)*b^3*x^3 + 24*b^4*x^
4 + a^3*(250*I - 24*b*x) + 2*a^2*(-166 - (65*I)*b*x + 12*b^2*x^2) + a*(-275*I + 116*b*x + (70*I)*b^2*x^2 - 24*
b^3*x^3)))/b^5 + ((-1)^(1/4)*(3 - (12*I)*a - 24*a^2 + (16*I)*a^3 + 8*a^4)*Sqrt[(-I)*b]*ArcSinh[((1/2 + I/2)*Sq
rt[b]*Sqrt[(-I)*(I + a + b*x)])/Sqrt[(-I)*b]])/(4*b^(11/2))

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Maple [B]  time = 0.138, size = 656, normalized size = 2.4 \begin{align*}{\frac{-{\frac{4\,i}{15}}{x}^{2}}{{b}^{3}}\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}-{\frac{3\,x}{8\,{b}^{4}}\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}+{\frac{3}{8\,{b}^{4}}\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{25\,{a}^{3}}{12\,{b}^{5}}\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}+{\frac{55\,a}{24\,{b}^{5}}\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}-{\frac{{\frac{83\,i}{30}}{a}^{2}}{{b}^{5}}\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}+{\frac{2\,i{a}^{3}}{{b}^{4}}\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}{5}}{a}^{4}}{{b}^{5}}\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}+{\frac{{\frac{8\,i}{15}}}{{b}^{5}}\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}-{\frac{{\frac{3\,i}{2}}a}{{b}^{4}}\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{{x}^{3}}{4\,{b}^{2}}\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}+{\frac{{\frac{i}{5}}{x}^{4}}{b}\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}-{\frac{7\,a{x}^{2}}{12\,{b}^{3}}\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}+{\frac{13\,{a}^{2}x}{12\,{b}^{4}}\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}+{\frac{{a}^{4}}{{b}^{4}}\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}}}}}-3\,{\frac{{a}^{2}}{{b}^{4}\sqrt{{b}^{2}}}\ln \left ({\frac{{b}^{2}x+ab}{\sqrt{{b}^{2}}}}+\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1} \right ) }+{\frac{{\frac{i}{5}}{a}^{2}{x}^{2}}{{b}^{3}}\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}-{\frac{{\frac{i}{5}}a{x}^{3}}{{b}^{2}}\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}+{\frac{{\frac{29\,i}{30}}ax}{{b}^{4}}\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}-{\frac{{\frac{i}{5}}{a}^{3}x}{{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^4,x)

[Out]

-4/15*I/b^3*x^2*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)-3/8/b^4*x*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)+3/8/b^4*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)-25/12*a^3/b^5*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)+55/24*a/b^5*
(b^2*x^2+2*a*b*x+a^2+1)^(1/2)-83/30*I/b^5*a^2*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)+2*I/b^4*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)+1/5*I/b^5*a^4*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)+8/15*I/b^5*(b^2*x^
2+2*a*b*x+a^2+1)^(1/2)-3/2*I/b^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)+1/4*x
^3/b^2*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)+1/5*I/b*x^4*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)-7/12*a/b^3*x^2*(b^2*x^2+2*a*b*x
+a^2+1)^(1/2)+13/12*a^2/b^4*x*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)+a^4/b^4*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*a^2/b^4*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/
5*I/b^3*a^2*x^2*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)-1/5*I/b^2*a*x^3*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)+29/30*I/b^4*a*x*(b
^2*x^2+2*a*b*x+a^2+1)^(1/2)-1/5*I/b^4*a^3*x*(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^4,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.78394, size = 520, normalized size = 1.88 \begin{align*} \frac{186 i \, a^{5} - 1345 \, a^{4} - 1730 i \, a^{3} + 1320 \, a^{2} -{\left (960 \, a^{4} + 1920 i \, a^{3} - 2880 \, a^{2} - 1440 i \, a + 360\right )} \log \left (-b x - a + \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right ) +{\left (192 i \, b^{4} x^{4} - 48 \,{\left (4 i \, a - 5\right )} b^{3} x^{3} +{\left (192 i \, a^{2} - 560 \, a - 256 i\right )} b^{2} x^{2} + 192 i \, a^{4} - 2000 \, a^{3} +{\left (-192 i \, a^{3} + 1040 \, a^{2} + 928 i \, a - 360\right )} b x - 2656 i \, a^{2} + 2200 \, a + 512 i\right )} \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1} + 300 i \, a}{960 \, b^{5}} \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^4,x, algorithm="fricas")

[Out]

1/960*(186*I*a^5 - 1345*a^4 - 1730*I*a^3 + 1320*a^2 - (960*a^4 + 1920*I*a^3 - 2880*a^2 - 1440*I*a + 360)*log(-
b*x - a + sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)) + (192*I*b^4*x^4 - 48*(4*I*a - 5)*b^3*x^3 + (192*I*a^2 - 560*a -
256*I)*b^2*x^2 + 192*I*a^4 - 2000*a^3 + (-192*I*a^3 + 1040*a^2 + 928*I*a - 360)*b*x - 2656*I*a^2 + 2200*a + 51
2*I)*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1) + 300*I*a)/b^5

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{4} \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**4,x)

[Out]

Integral(x**4*(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 = 1.15667, size = 289, normalized size = 1.05 \begin{align*} \frac{1}{120} \, \sqrt{{\left (b x + a\right )}^{2} + 1}{\left ({\left (2 \,{\left (3 \,{\left (\frac{4 \, i x}{b} - \frac{4 \, a b^{7} i - 5 \, b^{7}}{b^{9}}\right )} x + \frac{12 \, a^{2} b^{6} i - 35 \, a b^{6} - 16 \, b^{6} i}{b^{9}}\right )} x - \frac{24 \, a^{3} b^{5} i - 130 \, a^{2} b^{5} - 116 \, a b^{5} i + 45 \, b^{5}}{b^{9}}\right )} x + \frac{24 \, a^{4} b^{4} i - 250 \, a^{3} b^{4} - 332 \, a^{2} b^{4} i + 275 \, a b^{4} + 64 \, b^{4} i}{b^{9}}\right )} - \frac{{\left (8 \, a^{4} + 16 \, a^{3} i - 24 \, a^{2} - 12 \, a i + 3\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^{4}{\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^4,x, algorithm="giac")

[Out]

1/120*sqrt((b*x + a)^2 + 1)*((2*(3*(4*i*x/b - (4*a*b^7*i - 5*b^7)/b^9)*x + (12*a^2*b^6*i - 35*a*b^6 - 16*b^6*i
)/b^9)*x - (24*a^3*b^5*i - 130*a^2*b^5 - 116*a*b^5*i + 45*b^5)/b^9)*x + (24*a^4*b^4*i - 250*a^3*b^4 - 332*a^2*
b^4*i + 275*a*b^4 + 64*b^4*i)/b^9) - 1/8*(8*a^4 + 16*a^3*i - 24*a^2 - 12*a*i + 3)*log(-a*b - (x*abs(b) - sqrt(
(b*x + a)^2 + 1))*abs(b))/(b^4*abs(b))