\(\int (a-i b \arcsin (1+i d x^2))^3 \, dx\) [322]

Optimal result

Integrand size = 20, antiderivative size = 129 \[ \int \left (a-i b \arcsin \left (1+i d x^2\right )\right )^3 \, dx=24 a b^2 x-\frac {48 b^3 \sqrt {-2 i d x^2+d^2 x^4}}{d x}-24 i b^3 x \arcsin \left (1+i d x^2\right )-\frac {6 b \sqrt {-2 i d x^2+d^2 x^4} \left (a-i b \arcsin \left (1+i d x^2\right )\right )^2}{d x}+x \left (a-i b \arcsin \left (1+i d x^2\right )\right )^3 \]

[Out]

24*a*b^2*x-24*I*b^3*x*arcsin(1+I*d*x^2)+x*(a-I*b*arcsin(1+I*d*x^2))^3-48*b^3*(-2*I*d*x^2+d^2*x^4)^(1/2)/d/x-6*
b*(a-I*b*arcsin(1+I*d*x^2))^2*(-2*I*d*x^2+d^2*x^4)^(1/2)/d/x

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4898, 4924, 12, 1602} \[ \int \left (a-i b \arcsin \left (1+i d x^2\right )\right )^3 \, dx=-\frac {6 b \sqrt {d^2 x^4-2 i d x^2} \left (a-i b \arcsin \left (1+i d x^2\right )\right )^2}{d x}+x \left (a-i b \arcsin \left (1+i d x^2\right )\right )^3+24 a b^2 x-24 i b^3 x \arcsin \left (1+i d x^2\right )-\frac {48 b^3 \sqrt {d^2 x^4-2 i d x^2}}{d x} \]

[In]

Int[(a - I*b*ArcSin[1 + I*d*x^2])^3,x]

[Out]

24*a*b^2*x - (48*b^3*Sqrt[(-2*I)*d*x^2 + d^2*x^4])/(d*x) - (24*I)*b^3*x*ArcSin[1 + I*d*x^2] - (6*b*Sqrt[(-2*I)
*d*x^2 + d^2*x^4]*(a - I*b*ArcSin[1 + I*d*x^2])^2)/(d*x) + x*(a - I*b*ArcSin[1 + I*d*x^2])^3

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 1602

Int[(Pp_)*(Qq_)^(m_.), x_Symbol] :> With[{p = Expon[Pp, x], q = Expon[Qq, x]}, Simp[Coeff[Pp, x, p]*x^(p - q +
1)*(Qq^(m + 1)/((p + m*q + 1)*Coeff[Qq, x, q])), x] /; NeQ[p + m*q + 1, 0] && EqQ[(p + m*q + 1)*Coeff[Qq, x,
q]*Pp, Coeff[Pp, x, p]*x^(p - q)*((p - q + 1)*Qq + (m + 1)*x*D[Qq, x])]] /; FreeQ[m, x] && PolyQ[Pp, x] && Pol
yQ[Qq, x] && NeQ[m, -1]

Rule 4898

Int[((a_.) + ArcSin[(c_) + (d_.)*(x_)^2]*(b_.))^(n_), x_Symbol] :> Simp[x*(a + b*ArcSin[c + d*x^2])^n, x] + (-
Dist[4*b^2*n*(n - 1), Int[(a + b*ArcSin[c + d*x^2])^(n - 2), x], x] + Simp[2*b*n*Sqrt[-2*c*d*x^2 - d^2*x^4]*((
a + b*ArcSin[c + d*x^2])^(n - 1)/(d*x)), x]) /; FreeQ[{a, b, c, d}, x] && EqQ[c^2, 1] && GtQ[n, 1]

Rule 4924

Int[ArcSin[u_], x_Symbol] :> Simp[x*ArcSin[u], x] - Int[SimplifyIntegrand[x*(D[u, x]/Sqrt[1 - u^2]), x], x] /;
InverseFunctionFreeQ[u, x] &&  !FunctionOfExponentialQ[u, x]

Rubi steps \begin{align*} \text {integral}& = -\frac {6 b \sqrt {-2 i d x^2+d^2 x^4} \left (a-i b \arcsin \left (1+i d x^2\right )\right )^2}{d x}+x \left (a-i b \arcsin \left (1+i d x^2\right )\right )^3+\left (24 b^2\right ) \int \left (a-i b \arcsin \left (1+i d x^2\right )\right ) \, dx \\ & = 24 a b^2 x-\frac {6 b \sqrt {-2 i d x^2+d^2 x^4} \left (a-i b \arcsin \left (1+i d x^2\right )\right )^2}{d x}+x \left (a-i b \arcsin \left (1+i d x^2\right )\right )^3-\left (24 i b^3\right ) \int \arcsin \left (1+i d x^2\right ) \, dx \\ & = 24 a b^2 x-24 i b^3 x \arcsin \left (1+i d x^2\right )-\frac {6 b \sqrt {-2 i d x^2+d^2 x^4} \left (a-i b \arcsin \left (1+i d x^2\right )\right )^2}{d x}+x \left (a-i b \arcsin \left (1+i d x^2\right )\right )^3+\left (24 i b^3\right ) \int \frac {2 i d x^2}{\sqrt {-2 i d x^2+d^2 x^4}} \, dx \\ & = 24 a b^2 x-24 i b^3 x \arcsin \left (1+i d x^2\right )-\frac {6 b \sqrt {-2 i d x^2+d^2 x^4} \left (a-i b \arcsin \left (1+i d x^2\right )\right )^2}{d x}+x \left (a-i b \arcsin \left (1+i d x^2\right )\right )^3-\left (48 b^3 d\right ) \int \frac {x^2}{\sqrt {-2 i d x^2+d^2 x^4}} \, dx \\ & = 24 a b^2 x-\frac {48 b^3 \sqrt {-2 i d x^2+d^2 x^4}}{d x}-24 i b^3 x \arcsin \left (1+i d x^2\right )-\frac {6 b \sqrt {-2 i d x^2+d^2 x^4} \left (a-i b \arcsin \left (1+i d x^2\right )\right )^2}{d x}+x \left (a-i b \arcsin \left (1+i d x^2\right )\right )^3 \\ \end{align*}

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.40 \[ \int \left (a-i b \arcsin \left (1+i d x^2\right )\right )^3 \, dx=\frac {a \left (a^2+24 b^2\right ) d x^2-6 b \left (a^2+8 b^2\right ) \sqrt {d x^2 \left (-2 i+d x^2\right )}-3 i b \left (a^2 d x^2+8 b^2 d x^2-4 a b \sqrt {d x^2 \left (-2 i+d x^2\right )}\right ) \arcsin \left (1+i d x^2\right )+3 b^2 \left (-a d x^2+2 b \sqrt {d x^2 \left (-2 i+d x^2\right )}\right ) \arcsin \left (1+i d x^2\right )^2+i b^3 d x^2 \arcsin \left (1+i d x^2\right )^3}{d x} \]

[In]

Integrate[(a - I*b*ArcSin[1 + I*d*x^2])^3,x]

[Out]

(a*(a^2 + 24*b^2)*d*x^2 - 6*b*(a^2 + 8*b^2)*Sqrt[d*x^2*(-2*I + d*x^2)] - (3*I)*b*(a^2*d*x^2 + 8*b^2*d*x^2 - 4*
a*b*Sqrt[d*x^2*(-2*I + d*x^2)])*ArcSin[1 + I*d*x^2] + 3*b^2*(-(a*d*x^2) + 2*b*Sqrt[d*x^2*(-2*I + d*x^2)])*ArcS
in[1 + I*d*x^2]^2 + I*b^3*d*x^2*ArcSin[1 + I*d*x^2]^3)/(d*x)

Maple [F]

\[\int {\left (a +b \,\operatorname {arcsinh}\left (d \,x^{2}-i\right )\right )}^{3}d x\]

[In]

int((a+b*arcsinh(-I+d*x^2))^3,x)

[Out]

int((a+b*arcsinh(-I+d*x^2))^3,x)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.46 \[ \int \left (a-i b \arcsin \left (1+i d x^2\right )\right )^3 \, dx=\frac {b^{3} d x \log \left (d x^{2} + \sqrt {d^{2} x^{2} - 2 i \, d} x - i\right )^{3} + {\left (a^{3} + 24 \, a b^{2}\right )} d x + 3 \, {\left (a b^{2} d x - 2 \, \sqrt {d^{2} x^{2} - 2 i \, d} b^{3}\right )} \log \left (d x^{2} + \sqrt {d^{2} x^{2} - 2 i \, d} x - i\right )^{2} - 3 \, {\left (4 \, \sqrt {d^{2} x^{2} - 2 i \, d} a b^{2} - {\left (a^{2} b + 8 \, b^{3}\right )} d x\right )} \log \left (d x^{2} + \sqrt {d^{2} x^{2} - 2 i \, d} x - i\right ) - 6 \, \sqrt {d^{2} x^{2} - 2 i \, d} {\left (a^{2} b + 8 \, b^{3}\right )}}{d} \]

[In]

integrate((a+b*arcsinh(-I+d*x^2))^3,x, algorithm="fricas")

[Out]

(b^3*d*x*log(d*x^2 + sqrt(d^2*x^2 - 2*I*d)*x - I)^3 + (a^3 + 24*a*b^2)*d*x + 3*(a*b^2*d*x - 2*sqrt(d^2*x^2 - 2
*I*d)*b^3)*log(d*x^2 + sqrt(d^2*x^2 - 2*I*d)*x - I)^2 - 3*(4*sqrt(d^2*x^2 - 2*I*d)*a*b^2 - (a^2*b + 8*b^3)*d*x
)*log(d*x^2 + sqrt(d^2*x^2 - 2*I*d)*x - I) - 6*sqrt(d^2*x^2 - 2*I*d)*(a^2*b + 8*b^3))/d

Sympy [F(-2)]

Exception generated. \[ \int \left (a-i b \arcsin \left (1+i d x^2\right )\right )^3 \, dx=\text {Exception raised: TypeError} \]

[In]

integrate((a+b*asinh(-I+d*x**2))**3,x)

[Out]

Exception raised: TypeError >> Invalid comparison of non-real -I

Maxima [F]

\[ \int \left (a-i b \arcsin \left (1+i d x^2\right )\right )^3 \, dx=\int { {\left (b \operatorname {arsinh}\left (d x^{2} - i\right ) + a\right )}^{3} \,d x } \]

[In]

integrate((a+b*arcsinh(-I+d*x^2))^3,x, algorithm="maxima")

[Out]

b^3*x*log(d*x^2 + sqrt(d*x^2 - 2*I)*sqrt(d)*x - I)^3 + 3*(x*arcsinh(d*x^2 - I) - 2*(d^(3/2)*x^2 - 2*I*sqrt(d))
/(sqrt(d*x^2 - 2*I)*d))*a^2*b + a^3*x + integrate(3*((a*b^2*d^2 - 2*b^3*d^2)*x^4 - 2*a*b^2 - (3*I*a*b^2*d - 4*
I*b^3*d)*x^2 + ((a*b^2*d^(3/2) - 2*b^3*d^(3/2))*x^3 - 2*(I*a*b^2*sqrt(d) - I*b^3*sqrt(d))*x)*sqrt(d*x^2 - 2*I)
)*log(d*x^2 + sqrt(d*x^2 - 2*I)*sqrt(d)*x - I)^2/(d^2*x^4 - 3*I*d*x^2 + (d^(3/2)*x^3 - 2*I*sqrt(d)*x)*sqrt(d*x
^2 - 2*I) - 2), x)

Giac [F(-2)]

Exception generated. \[ \int \left (a-i b \arcsin \left (1+i d x^2\right )\right )^3 \, dx=\text {Exception raised: TypeError} \]

[In]

integrate((a+b*arcsinh(-I+d*x^2))^3,x, algorithm="giac")

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

Mupad [F(-1)]

Timed out. \[ \int \left (a-i b \arcsin \left (1+i d x^2\right )\right )^3 \, dx=\int {\left (a+b\,\mathrm {asinh}\left (d\,x^2-\mathrm {i}\right )\right )}^3 \,d x \]

[In]

int((a + b*asinh(d*x^2 - 1i))^3,x)

[Out]

int((a + b*asinh(d*x^2 - 1i))^3, x)