3.207 \(\int e^{-3 i \tan ^{-1}(a+b x)} x^4 \, dx\)

Optimal. Leaf size=324 \[ \frac {i (-i a-i b x+1)^{3/2} \sqrt {i a+i b x+1} \left (-112 i a^3-2 \left (-52 i a^2-118 a+61 i\right ) b x-422 a^2+458 i a+163\right )}{40 b^5}+\frac {3 \left (8 i a^4+48 a^3-88 i a^2-68 a+19 i\right ) \sqrt {-i a-i b x+1} \sqrt {i a+i b x+1}}{8 b^5}-\frac {3 \left (8 a^4-48 i a^3-88 a^2+68 i a+19\right ) \sinh ^{-1}(a+b x)}{8 b^5}-\frac {3 (-16 a+17 i) x^2 (-i a-i b x+1)^{3/2} \sqrt {i a+i b x+1}}{20 b^3}-\frac {11 x^3 (-i a-i b x+1)^{3/2} \sqrt {i a+i b x+1}}{5 b^2}+\frac {2 i x^4 (-i a-i b x+1)^{3/2}}{b \sqrt {i a+i b x+1}} \]

[Out]

-3/8*(19+68*I*a-88*a^2-48*I*a^3+8*a^4)*arcsinh(b*x+a)/b^5+2*I*x^4*(1-I*a-I*b*x)^(3/2)/b/(1+I*a+I*b*x)^(1/2)-3/
20*(17*I-16*a)*x^2*(1-I*a-I*b*x)^(3/2)*(1+I*a+I*b*x)^(1/2)/b^3-11/5*x^3*(1-I*a-I*b*x)^(3/2)*(1+I*a+I*b*x)^(1/2
)/b^2+1/40*I*(1-I*a-I*b*x)^(3/2)*(163+458*I*a-422*a^2-112*I*a^3-2*(61*I-118*a-52*I*a^2)*b*x)*(1+I*a+I*b*x)^(1/
2)/b^5+3/8*(19*I-68*a-88*I*a^2+48*a^3+8*I*a^4)*(1-I*a-I*b*x)^(1/2)*(1+I*a+I*b*x)^(1/2)/b^5

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Rubi [A]  time = 0.28, antiderivative size = 324, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5095, 97, 153, 147, 50, 53, 619, 215} \[ \frac {i (-i a-i b x+1)^{3/2} \sqrt {i a+i b x+1} \left (-2 \left (-52 i a^2-118 a+61 i\right ) b x-112 i a^3-422 a^2+458 i a+163\right )}{40 b^5}+\frac {3 \left (8 i a^4+48 a^3-88 i a^2-68 a+19 i\right ) \sqrt {-i a-i b x+1} \sqrt {i a+i b x+1}}{8 b^5}-\frac {3 \left (8 a^4-48 i a^3-88 a^2+68 i a+19\right ) \sinh ^{-1}(a+b x)}{8 b^5}-\frac {11 x^3 (-i a-i b x+1)^{3/2} \sqrt {i a+i b x+1}}{5 b^2}-\frac {3 (-16 a+17 i) x^2 (-i a-i b x+1)^{3/2} \sqrt {i a+i b x+1}}{20 b^3}+\frac {2 i x^4 (-i a-i b x+1)^{3/2}}{b \sqrt {i a+i b x+1}} \]

Antiderivative was successfully verified.

[In]

Int[x^4/E^((3*I)*ArcTan[a + b*x]),x]

[Out]

((2*I)*x^4*(1 - I*a - I*b*x)^(3/2))/(b*Sqrt[1 + I*a + I*b*x]) + (3*(19*I - 68*a - (88*I)*a^2 + 48*a^3 + (8*I)*
a^4)*Sqrt[1 - I*a - I*b*x]*Sqrt[1 + I*a + I*b*x])/(8*b^5) - (3*(17*I - 16*a)*x^2*(1 - I*a - I*b*x)^(3/2)*Sqrt[
1 + I*a + I*b*x])/(20*b^3) - (11*x^3*(1 - I*a - I*b*x)^(3/2)*Sqrt[1 + I*a + I*b*x])/(5*b^2) + ((I/40)*(1 - I*a
- I*b*x)^(3/2)*Sqrt[1 + I*a + I*b*x]*(163 + (458*I)*a - 422*a^2 - (112*I)*a^3 - 2*(61*I - 118*a - (52*I)*a^2)
*b*x))/b^5 - (3*(19 + (68*I)*a - 88*a^2 - (48*I)*a^3 + 8*a^4)*ArcSinh[a + b*x])/(8*b^5)

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 97

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)/(b*(m + 1)), x] - Dist[1/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n
- 1)*(e + f*x)^(p - 1)*Simp[d*e*n + c*f*p + d*f*(n + p)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && LtQ[m
, -1] && GtQ[n, 0] && GtQ[p, 0] && (IntegersQ[2*m, 2*n, 2*p] || IntegersQ[m, n + p] || IntegersQ[p, m + n])

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 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 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]

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 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 e^{-3 i \tan ^{-1}(a+b x)} x^4 \, dx &=\int \frac {x^4 (1-i a-i b x)^{3/2}}{(1+i a+i b x)^{3/2}} \, dx\\ &=\frac {2 i x^4 (1-i a-i b x)^{3/2}}{b \sqrt {1+i a+i b x}}-\frac {(2 i) \int \frac {x^3 \sqrt {1-i a-i b x} \left (4 (1-i a)-\frac {11 i b x}{2}\right )}{\sqrt {1+i a+i b x}} \, dx}{b}\\ &=\frac {2 i x^4 (1-i a-i b x)^{3/2}}{b \sqrt {1+i a+i b x}}-\frac {11 x^3 (1-i a-i b x)^{3/2} \sqrt {1+i a+i b x}}{5 b^2}-\frac {(2 i) \int \frac {x^2 \sqrt {1-i a-i b x} \left (\frac {33}{2} (1+i a) (i+a) b+\frac {3}{2} (17+16 i a) b^2 x\right )}{\sqrt {1+i a+i b x}} \, dx}{5 b^3}\\ &=\frac {2 i x^4 (1-i a-i b x)^{3/2}}{b \sqrt {1+i a+i b x}}-\frac {3 (17 i-16 a) x^2 (1-i a-i b x)^{3/2} \sqrt {1+i a+i b x}}{20 b^3}-\frac {11 x^3 (1-i a-i b x)^{3/2} \sqrt {1+i a+i b x}}{5 b^2}-\frac {i \int \frac {x \sqrt {1-i a-i b x} \left (3 (17 i-16 a) (i-a) (1-i a) b^2-\frac {3}{2} \left (118 a-i \left (61-52 a^2\right )\right ) b^3 x\right )}{\sqrt {1+i a+i b x}} \, dx}{10 b^5}\\ &=\frac {2 i x^4 (1-i a-i b x)^{3/2}}{b \sqrt {1+i a+i b x}}-\frac {3 (17 i-16 a) x^2 (1-i a-i b x)^{3/2} \sqrt {1+i a+i b x}}{20 b^3}-\frac {11 x^3 (1-i a-i b x)^{3/2} \sqrt {1+i a+i b x}}{5 b^2}+\frac {i (1-i a-i b x)^{3/2} \sqrt {1+i a+i b x} \left (163+458 i a-422 a^2-112 i a^3-2 \left (61 i-118 a-52 i a^2\right ) b x\right )}{40 b^5}-\frac {\left (3 \left (19+68 i a-88 a^2-48 i a^3+8 a^4\right )\right ) \int \frac {\sqrt {1-i a-i b x}}{\sqrt {1+i a+i b x}} \, dx}{8 b^4}\\ &=\frac {2 i x^4 (1-i a-i b x)^{3/2}}{b \sqrt {1+i a+i b x}}+\frac {3 \left (19 i-68 a-88 i a^2+48 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 {3 (17 i-16 a) x^2 (1-i a-i b x)^{3/2} \sqrt {1+i a+i b x}}{20 b^3}-\frac {11 x^3 (1-i a-i b x)^{3/2} \sqrt {1+i a+i b x}}{5 b^2}+\frac {i (1-i a-i b x)^{3/2} \sqrt {1+i a+i b x} \left (163+458 i a-422 a^2-112 i a^3-2 \left (61 i-118 a-52 i a^2\right ) b x\right )}{40 b^5}-\frac {\left (3 \left (19+68 i a-88 a^2-48 i a^3+8 a^4\right )\right ) \int \frac {1}{\sqrt {1-i a-i b x} \sqrt {1+i a+i b x}} \, dx}{8 b^4}\\ &=\frac {2 i x^4 (1-i a-i b x)^{3/2}}{b \sqrt {1+i a+i b x}}+\frac {3 \left (19 i-68 a-88 i a^2+48 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 {3 (17 i-16 a) x^2 (1-i a-i b x)^{3/2} \sqrt {1+i a+i b x}}{20 b^3}-\frac {11 x^3 (1-i a-i b x)^{3/2} \sqrt {1+i a+i b x}}{5 b^2}+\frac {i (1-i a-i b x)^{3/2} \sqrt {1+i a+i b x} \left (163+458 i a-422 a^2-112 i a^3-2 \left (61 i-118 a-52 i a^2\right ) b x\right )}{40 b^5}-\frac {\left (3 \left (19+68 i a-88 a^2-48 i a^3+8 a^4\right )\right ) \int \frac {1}{\sqrt {(1-i a) (1+i a)+2 a b x+b^2 x^2}} \, dx}{8 b^4}\\ &=\frac {2 i x^4 (1-i a-i b x)^{3/2}}{b \sqrt {1+i a+i b x}}+\frac {3 \left (19 i-68 a-88 i a^2+48 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 {3 (17 i-16 a) x^2 (1-i a-i b x)^{3/2} \sqrt {1+i a+i b x}}{20 b^3}-\frac {11 x^3 (1-i a-i b x)^{3/2} \sqrt {1+i a+i b x}}{5 b^2}+\frac {i (1-i a-i b x)^{3/2} \sqrt {1+i a+i b x} \left (163+458 i a-422 a^2-112 i a^3-2 \left (61 i-118 a-52 i a^2\right ) b x\right )}{40 b^5}-\frac {\left (3 \left (19+68 i a-88 a^2-48 i a^3+8 a^4\right )\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 {2 i x^4 (1-i a-i b x)^{3/2}}{b \sqrt {1+i a+i b x}}+\frac {3 \left (19 i-68 a-88 i a^2+48 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 {3 (17 i-16 a) x^2 (1-i a-i b x)^{3/2} \sqrt {1+i a+i b x}}{20 b^3}-\frac {11 x^3 (1-i a-i b x)^{3/2} \sqrt {1+i a+i b x}}{5 b^2}+\frac {i (1-i a-i b x)^{3/2} \sqrt {1+i a+i b x} \left (163+458 i a-422 a^2-112 i a^3-2 \left (61 i-118 a-52 i a^2\right ) b x\right )}{40 b^5}-\frac {3 \left (19+68 i a-88 a^2-48 i a^3+8 a^4\right ) \sinh ^{-1}(a+b x)}{8 b^5}\\ \end {align*}

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Mathematica [A]  time = 0.52, size = 299, normalized size = 0.92 \[ \frac {\frac {30 \sqrt [4]{-1} \left (8 i a^4+48 a^3-88 i a^2-68 a+19 i\right ) \sqrt {b} \sinh ^{-1}\left (\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {b} \sqrt {-i (a+b x+i)}}{\sqrt {-i b}}\right )}{\sqrt {-i b}}+\frac {8 i a^6+a^5 (410+8 i b x)+2 a^4 (265 b x-638 i)+a^3 \left (60 b^2 x^2-2004 i b x-905\right )-a^2 \left (20 b^3 x^3+356 i b^2 x^2+2635 b x+836 i\right )+a \left (8 i b^5 x^5+10 b^4 x^4+116 i b^3 x^3-515 b^2 x^2+1468 i b x-1315\right )+8 i b^6 x^6-30 b^5 x^5-56 i b^4 x^4+95 b^3 x^3+224 i b^2 x^2+285 b x+448 i}{\sqrt {a^2+2 a b x+b^2 x^2+1}}}{40 b^5} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[x^4/E^((3*I)*ArcTan[a + b*x]),x]

[Out]

((448*I + (8*I)*a^6 + 285*b*x + (224*I)*b^2*x^2 + 95*b^3*x^3 - (56*I)*b^4*x^4 - 30*b^5*x^5 + (8*I)*b^6*x^6 + a
^5*(410 + (8*I)*b*x) + 2*a^4*(-638*I + 265*b*x) + a^3*(-905 - (2004*I)*b*x + 60*b^2*x^2) - a^2*(836*I + 2635*b
*x + (356*I)*b^2*x^2 + 20*b^3*x^3) + a*(-1315 + (1468*I)*b*x - 515*b^2*x^2 + (116*I)*b^3*x^3 + 10*b^4*x^4 + (8
*I)*b^5*x^5))/Sqrt[1 + a^2 + 2*a*b*x + b^2*x^2] + (30*(-1)^(1/4)*(19*I - 68*a - (88*I)*a^2 + 48*a^3 + (8*I)*a^
4)*Sqrt[b]*ArcSinh[((1/2 + I/2)*Sqrt[b]*Sqrt[(-I)*(I + a + b*x)])/Sqrt[(-I)*b]])/Sqrt[(-I)*b])/(40*b^5)

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fricas [A]  time = 0.50, size = 263, normalized size = 0.81 \[ \frac {62 i \, a^{6} + 2687 \, a^{5} - 11575 i \, a^{4} - 20350 \, a^{3} + {\left (62 i \, a^{5} + 2625 \, a^{4} - 8950 i \, a^{3} - 11400 \, a^{2} + 6340 i \, a + 1280\right )} b x + 17740 i \, a^{2} + {\left (960 \, a^{5} - 6720 i \, a^{4} - 16320 \, a^{3} + {\left (960 \, a^{4} - 5760 i \, a^{3} - 10560 \, a^{2} + 8160 i \, a + 2280\right )} b x + 18720 i \, a^{2} + 10440 \, a - 2280 i\right )} \log \left (-b x - a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right ) + {\left (64 i \, b^{5} x^{5} - 176 \, b^{4} x^{4} + {\left (256 \, a - 272 i\right )} b^{3} x^{3} + 64 i \, a^{5} - 8 \, {\left (52 \, a^{2} - 118 i \, a - 61\right )} b^{2} x^{2} + 3344 \, a^{4} - 13552 i \, a^{3} + {\left (896 \, a^{3} - 3376 i \, a^{2} - 3664 \, a + 1304 i\right )} b x - 20792 \, a^{2} + 14104 i \, a + 3584\right )} \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} + 7620 \, a - 1280 i}{320 \, b^{6} x + {\left (320 \, a - 320 i\right )} b^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(1+I*(b*x+a))^3*(1+(b*x+a)^2)^(3/2),x, algorithm="fricas")

[Out]

(62*I*a^6 + 2687*a^5 - 11575*I*a^4 - 20350*a^3 + (62*I*a^5 + 2625*a^4 - 8950*I*a^3 - 11400*a^2 + 6340*I*a + 12
80)*b*x + 17740*I*a^2 + (960*a^5 - 6720*I*a^4 - 16320*a^3 + (960*a^4 - 5760*I*a^3 - 10560*a^2 + 8160*I*a + 228
0)*b*x + 18720*I*a^2 + 10440*a - 2280*I)*log(-b*x - a + sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)) + (64*I*b^5*x^5 - 1
76*b^4*x^4 + (256*a - 272*I)*b^3*x^3 + 64*I*a^5 - 8*(52*a^2 - 118*I*a - 61)*b^2*x^2 + 3344*a^4 - 13552*I*a^3 +
(896*a^3 - 3376*I*a^2 - 3664*a + 1304*I)*b*x - 20792*a^2 + 14104*I*a + 3584)*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1
) + 7620*a - 1280*I)/(320*b^6*x + (320*a - 320*I)*b^5)

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giac [A]  time = 0.20, size = 353, normalized size = 1.09 \[ \frac {1}{40} \, \sqrt {{\left (b x + a\right )}^{2} + 1} {\left ({\left (2 \, {\left ({\left (\frac {4 \, i x}{b} - \frac {4 \, a b^{17} i + 15 \, b^{17}}{b^{19}}\right )} x + \frac {4 \, a^{2} b^{16} i + 35 \, a b^{16} - 32 \, b^{16} i}{b^{19}}\right )} x - \frac {8 \, a^{3} b^{15} i + 130 \, a^{2} b^{15} - 252 \, a b^{15} i - 125 \, b^{15}}{b^{19}}\right )} x + \frac {8 \, a^{4} b^{14} i + 250 \, a^{3} b^{14} - 804 \, a^{2} b^{14} i - 835 \, a b^{14} + 288 \, b^{14} i}{b^{19}}\right )} + \frac {{\left (8 \, a^{4} i + 48 \, a^{3} - 88 \, a^{2} i - 68 \, a + 19 \, i\right )} \log \left (3 \, {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )}^{2} a b i + a^{3} b i + {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )}^{3} i {\left | b \right |} + 3 \, {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )} a^{2} i {\left | b \right |} + 2 \, {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )}^{2} b + 2 \, a^{2} b - a b i + 4 \, {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )} a {\left | b \right |} - {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )} i {\left | b \right |}\right )}{8 \, b^{4} i {\left | b \right |}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(1+I*(b*x+a))^3*(1+(b*x+a)^2)^(3/2),x, algorithm="giac")

[Out]

1/40*sqrt((b*x + a)^2 + 1)*((2*((4*i*x/b - (4*a*b^17*i + 15*b^17)/b^19)*x + (4*a^2*b^16*i + 35*a*b^16 - 32*b^1
6*i)/b^19)*x - (8*a^3*b^15*i + 130*a^2*b^15 - 252*a*b^15*i - 125*b^15)/b^19)*x + (8*a^4*b^14*i + 250*a^3*b^14
- 804*a^2*b^14*i - 835*a*b^14 + 288*b^14*i)/b^19) + 1/8*(8*a^4*i + 48*a^3 - 88*a^2*i - 68*a + 19*i)*log(3*(x*a
bs(b) - sqrt((b*x + a)^2 + 1))^2*a*b*i + a^3*b*i + (x*abs(b) - sqrt((b*x + a)^2 + 1))^3*i*abs(b) + 3*(x*abs(b)
- sqrt((b*x + a)^2 + 1))*a^2*i*abs(b) + 2*(x*abs(b) - sqrt((b*x + a)^2 + 1))^2*b + 2*a^2*b - a*b*i + 4*(x*abs
(b) - sqrt((b*x + a)^2 + 1))*a*abs(b) - (x*abs(b) - sqrt((b*x + a)^2 + 1))*i*abs(b))/(b^4*i*abs(b))

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maple [B]  time = 0.20, size = 2058, normalized size = 6.35 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/(1+I*(b*x+a))^3*(1+(b*x+a)^2)^(3/2),x)

[Out]

2*I/b^5*((x-(I-a)/b)^2*b^2+2*I*b*(x-(I-a)/b))^(3/2)*a^4+18*I/b^5*((x-(I-a)/b)^2*b^2+2*I*b*(x-(I-a)/b))^(1/2)*a
^4-24*I/b^5*((x-(I-a)/b)^2*b^2+2*I*b*(x-(I-a)/b))^(1/2)*a^2-6*I/b^7/(x-I/b+a/b)^2*((x-(I-a)/b)^2*b^2+2*I*b*(x-
(I-a)/b))^(5/2)-22*I/b^5*((x-(I-a)/b)^2*b^2+2*I*b*(x-(I-a)/b))^(3/2)*a^2-I/b^5*a^2*(b^2*x^2+2*a*b*x+a^2+1)^(3/
2)-3/2*I/b^5*a^2*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)-3/b^4*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)*a^4+33/b^4*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)*a^2-1/b^8/(x-I/b+a/b)^3*((x-(I-a)/b)^2*b^2+2*I*b*(x-(I-a)/b))^(5/2)*a^4+6/b^8/
(x-I/b+a/b)^3*((x-(I-a)/b)^2*b^2+2*I*b*(x-(I-a)/b))^(5/2)*a^2-3/b^4*((x-(I-a)/b)^2*b^2+2*I*b*(x-(I-a)/b))^(1/2
)*x*a^4+33/b^4*((x-(I-a)/b)^2*b^2+2*I*b*(x-(I-a)/b))^(1/2)*x*a^2-12/b^7/(x-I/b+a/b)^2*((x-(I-a)/b)^2*b^2+2*I*b
*(x-(I-a)/b))^(5/2)*a^3+20/b^7/(x-I/b+a/b)^2*((x-(I-a)/b)^2*b^2+2*I*b*(x-(I-a)/b))^(5/2)*a-I/b^4*a*x*(b^2*x^2+
2*a*b*x+a^2+1)^(3/2)-3/2*I/b^4*a*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)*x-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)+18*I/b^4*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)*a^3+18*I/b^4*((x-(I-a)/b)^2*b^2+2*I*b*(x-(I-a)/b))^(1/2)*x*a^3+24*I/b^7/(x-I/b+a
/b)^2*((x-(I-a)/b)^2*b^2+2*I*b*(x-(I-a)/b))^(5/2)*a^2-24*I/b^4*((x-(I-a)/b)^2*b^2+2*I*b*(x-(I-a)/b))^(1/2)*x*a
-24*I/b^4*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)*a-2*I/
b^7/(x-I/b+a/b)^2*((x-(I-a)/b)^2*b^2+2*I*b*(x-(I-a)/b))^(5/2)*a^4+4*I/b^8/(x-I/b+a/b)^3*((x-(I-a)/b)^2*b^2+2*I
*b*(x-(I-a)/b))^(5/2)*a^3-4*I/b^8/(x-I/b+a/b)^3*((x-(I-a)/b)^2*b^2+2*I*b*(x-(I-a)/b))^(5/2)*a-9/8/b^4*x*(b^2*x
^2+2*a*b*x+a^2+1)^(1/2)-9/8/b^5*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)*a-9/8/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)+4*I/b^5*((x-(I-a)/b)^2*b^2+2*I*b*(x-(I-a)/b))^(3/2)-16/b^5*((x-(I-a)/b)^2*b^2+
2*I*b*(x-(I-a)/b))^(3/2)*a+12/b^5*((x-(I-a)/b)^2*b^2+2*I*b*(x-(I-a)/b))^(3/2)*a^3-1/b^8/(x-I/b+a/b)^3*((x-(I-a
)/b)^2*b^2+2*I*b*(x-(I-a)/b))^(5/2)-6/b^4*((x-(I-a)/b)^2*b^2+2*I*b*(x-(I-a)/b))^(1/2)*x-3/b^5*((x-(I-a)/b)^2*b
^2+2*I*b*(x-(I-a)/b))^(1/2)*a^5+1/5*I/b^5*(b^2*x^2+2*a*b*x+a^2+1)^(5/2)+33/b^5*((x-(I-a)/b)^2*b^2+2*I*b*(x-(I-
a)/b))^(1/2)*a^3-6/b^5*((x-(I-a)/b)^2*b^2+2*I*b*(x-(I-a)/b))^(1/2)*a-6/b^4*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)-3/4/b^4*x*(b^2*x^2+2*a*b*x+a^2+1)^(3/2)-3/4/b^5*a*(
b^2*x^2+2*a*b*x+a^2+1)^(3/2)

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maxima [B]  time = 0.48, size = 1368, normalized size = 4.22 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(1+I*(b*x+a))^3*(1+(b*x+a)^2)^(3/2),x, algorithm="maxima")

[Out]

I*(b^2*x^2 + 2*a*b*x + a^2 + 1)^(3/2)*a^4/(b^7*x^2 + 2*a*b^6*x + a^2*b^5 - 2*I*b^6*x - 2*I*a*b^5 - b^5) + 4*(b
^2*x^2 + 2*a*b*x + a^2 + 1)^(3/2)*a^3/(b^7*x^2 + 2*a*b^6*x + a^2*b^5 - 2*I*b^6*x - 2*I*a*b^5 - b^5) + 4*(b^2*x
^2 + 2*a*b*x + a^2 + 1)^(3/2)*a^3/(2*I*b^6*x + 2*I*a*b^5 + 2*b^5) + 6*I*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*a^4/
(I*b^6*x + I*a*b^5 + b^5) - 6*I*(b^2*x^2 + 2*a*b*x + a^2 + 1)^(3/2)*a^2/(b^7*x^2 + 2*a*b^6*x + a^2*b^5 - 2*I*b
^6*x - 2*I*a*b^5 - b^5) - 12*I*(b^2*x^2 + 2*a*b*x + a^2 + 1)^(3/2)*a^2/(2*I*b^6*x + 2*I*a*b^5 + 2*b^5) + 24*sq
rt(b^2*x^2 + 2*a*b*x + a^2 + 1)*a^3/(I*b^6*x + I*a*b^5 + b^5) - 4*(b^2*x^2 + 2*a*b*x + a^2 + 1)^(3/2)*a/(b^7*x
^2 + 2*a*b^6*x + a^2*b^5 - 2*I*b^6*x - 2*I*a*b^5 - b^5) - 12*(b^2*x^2 + 2*a*b*x + a^2 + 1)^(3/2)*a/(2*I*b^6*x
+ 2*I*a*b^5 + 2*b^5) - 36*I*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*a^2/(I*b^6*x + I*a*b^5 + b^5) + I*(b^2*x^2 + 2*a
*b*x + a^2 + 1)^(3/2)/(b^7*x^2 + 2*a*b^6*x + a^2*b^5 - 2*I*b^6*x - 2*I*a*b^5 - b^5) + 4*I*(b^2*x^2 + 2*a*b*x +
a^2 + 1)^(3/2)/(2*I*b^6*x + 2*I*a*b^5 + 2*b^5) - 24*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*a/(I*b^6*x + I*a*b^5 +
b^5) - 3*a^4*arcsinh(b*x + a)/b^5 + 6*I*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)/(I*b^6*x + I*a*b^5 + b^5) - I*(b^2*x
^2 + 2*a*b*x + a^2 + 1)^(3/2)*a*x/b^4 - 3*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 4*I*b*x + 4*I*a + 3)*a^2*x/b^4 + 18*
I*a^3*arcsinh(b*x + a)/b^5 + I*(b^2*x^2 + 2*a*b*x + a^2 + 1)^(3/2)*a^2/b^5 + 6*sqrt(b^2*x^2 + 2*a*b*x + a^2 +
1)*a^3/b^5 - 3*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 4*I*b*x + 4*I*a + 3)*a^3/b^5 - 3/4*(b^2*x^2 + 2*a*b*x + a^2 + 1
)^(3/2)*x/b^4 - 3/2*I*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*a*x/b^4 + 6*I*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 4*I*b*x
+ 4*I*a + 3)*a*x/b^4 + 3*a^2*arcsin(I*b*x + I*a + 2)/b^5 + 36*a^2*arcsinh(b*x + a)/b^5 + 1/5*I*(b^2*x^2 + 2*a*
b*x + a^2 + 1)^(5/2)/b^5 + 13/4*(b^2*x^2 + 2*a*b*x + a^2 + 1)^(3/2)*a/b^5 - 39/2*I*sqrt(b^2*x^2 + 2*a*b*x + a^
2 + 1)*a^2/b^5 + 12*I*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 4*I*b*x + 4*I*a + 3)*a^2/b^5 - 9/8*sqrt(b^2*x^2 + 2*a*b*
x + a^2 + 1)*x/b^4 + 3*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 4*I*b*x + 4*I*a + 3)*x/b^4 - 6*I*a*arcsin(I*b*x + I*a +
2)/b^5 - 63/2*I*a*arcsinh(b*x + a)/b^5 - 2*I*(b^2*x^2 + 2*a*b*x + a^2 + 1)^(3/2)/b^5 - 153/8*sqrt(b^2*x^2 + 2
*a*b*x + a^2 + 1)*a/b^5 + 15*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 4*I*b*x + 4*I*a + 3)*a/b^5 - 3*arcsin(I*b*x + I*a
+ 2)/b^5 - 81/8*arcsinh(b*x + a)/b^5 + 6*I*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)/b^5 - 6*I*sqrt(-b^2*x^2 - 2*a*b*
x - a^2 + 4*I*b*x + 4*I*a + 3)/b^5

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^4\,{\left ({\left (a+b\,x\right )}^2+1\right )}^{3/2}}{{\left (1+a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4*((a + b*x)^2 + 1)^(3/2))/(a*1i + b*x*1i + 1)^3,x)

[Out]

int((x^4*((a + b*x)^2 + 1)^(3/2))/(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(x**4/(1+I*(b*x+a))**3*(1+(b*x+a)**2)**(3/2),x)

[Out]

Timed out

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