\(\int (a+i b \arcsin (1-i d x^2)) \, dx\) [317]

Optimal result

Integrand size = 18, antiderivative size = 50 \[ \int \left (a+i b \arcsin \left (1-i d x^2\right )\right ) \, dx=a x-\frac {2 b \sqrt {2 i d x^2+d^2 x^4}}{d x}+i b x \arcsin \left (1-i d x^2\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {4924, 12, 1602} \[ \int \left (a+i b \arcsin \left (1-i d x^2\right )\right ) \, dx=a x+i b x \arcsin \left (1-i d x^2\right )-\frac {2 b \sqrt {d^2 x^4+2 i d x^2}}{d x} \]

[In]

Int[a + I*b*ArcSin[1 - I*d*x^2],x]

[Out]

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

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 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}& = a x+(i b) \int \arcsin \left (1-i d x^2\right ) \, dx \\ & = a x+i b x \arcsin \left (1-i d x^2\right )-(i b) \int -\frac {2 i d x^2}{\sqrt {2 i d x^2+d^2 x^4}} \, dx \\ & = a x+i b x \arcsin \left (1-i d x^2\right )-(2 b d) \int \frac {x^2}{\sqrt {2 i d x^2+d^2 x^4}} \, dx \\ & = a x-\frac {2 b \sqrt {2 i d x^2+d^2 x^4}}{d x}+i b x \arcsin \left (1-i d x^2\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.96 \[ \int \left (a+i b \arcsin \left (1-i d x^2\right )\right ) \, dx=a x-\frac {2 b \sqrt {d x^2 \left (2 i+d x^2\right )}}{d x}+i b x \arcsin \left (1-i d x^2\right ) \]

[In]

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

[Out]

a*x - (2*b*Sqrt[d*x^2*(2*I + d*x^2)])/(d*x) + I*b*x*ArcSin[1 - I*d*x^2]

Maple [A] (verified)

Time = 0.02 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.94

method	result	size
default	\(a x +b \left (x \,\operatorname {arcsinh}\left (d \,x^{2}+i\right )-\frac {2 x \left (d \,x^{2}+2 i\right )}{\sqrt {d^{2} x^{4}+2 i d \,x^{2}}}\right )\)	\(47\)
parts	\(a x +b \left (x \,\operatorname {arcsinh}\left (d \,x^{2}+i\right )-\frac {2 x \left (d \,x^{2}+2 i\right )}{\sqrt {d^{2} x^{4}+2 i d \,x^{2}}}\right )\)	\(47\)

[In]

int(a+b*arcsinh(I+d*x^2),x,method=_RETURNVERBOSE)

[Out]

a*x+b*(x*arcsinh(I+d*x^2)-2/(2*I*d*x^2+d^2*x^4)^(1/2)*x*(d*x^2+2*I))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.04 \[ \int \left (a+i b \arcsin \left (1-i d x^2\right )\right ) \, dx=\frac {b d x \log \left (d x^{2} + \sqrt {d^{2} x^{2} + 2 i \, d} x + i\right ) + a d x - 2 \, \sqrt {d^{2} x^{2} + 2 i \, d} b}{d} \]

[In]

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

[Out]

(b*d*x*log(d*x^2 + sqrt(d^2*x^2 + 2*I*d)*x + I) + a*d*x - 2*sqrt(d^2*x^2 + 2*I*d)*b)/d

Sympy [F(-2)]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.88 \[ \int \left (a+i b \arcsin \left (1-i d x^2\right )\right ) \, dx={\left (x \operatorname {arsinh}\left (d x^{2} + i\right ) - \frac {2 \, {\left (d^{\frac {3}{2}} x^{2} + 2 i \, \sqrt {d}\right )}}{\sqrt {d x^{2} + 2 i} d}\right )} b + a x \]

[In]

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

[Out]

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

Giac [F(-2)]

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

[In]

integrate(a+b*arcsinh(I+d*x^2),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 [B] (verification not implemented)

Time = 3.16 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.78 \[ \int \left (a+i b \arcsin \left (1-i d x^2\right )\right ) \, dx=a\,x+b\,x\,\mathrm {asinh}\left (d\,x^2+1{}\mathrm {i}\right )-\frac {2\,b\,\sqrt {{\left (d\,x^2+1{}\mathrm {i}\right )}^2+1}}{d\,x} \]

[In]

int(a + b*asinh(d*x^2 + 1i),x)

[Out]

a*x + b*x*asinh(d*x^2 + 1i) - (2*b*((d*x^2 + 1i)^2 + 1)^(1/2))/(d*x)