3.320 \(\int \frac{1}{(a+i b \sin ^{-1}(1-i d x^2))^3} \, dx\)
Optimal. Leaf size=275 \[ \frac{x \left (-\sinh \left (\frac{a}{2 b}\right )+i \cosh \left (\frac{a}{2 b}\right )\right ) \text{CosIntegral}\left (-\frac{i \left (a+i b \sin ^{-1}\left (1-i d x^2\right )\right )}{2 b}\right )}{16 b^3 \left (\cos \left (\frac{1}{2} \sin ^{-1}\left (1-i d x^2\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (1-i d x^2\right )\right )\right )}-\frac{x \left (\sinh \left (\frac{a}{2 b}\right )+i \cosh \left (\frac{a}{2 b}\right )\right ) \text{Si}\left (\frac{i a}{2 b}-\frac{1}{2} \sin ^{-1}\left (1-i d x^2\right )\right )}{16 b^3 \left (\cos \left (\frac{1}{2} \sin ^{-1}\left (1-i d x^2\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (1-i d x^2\right )\right )\right )}-\frac{x}{8 b^2 \left (a+i b \sin ^{-1}\left (1-i d x^2\right )\right )}-\frac{\sqrt{d^2 x^4+2 i d x^2}}{4 b d x \left (a+i b \sin ^{-1}\left (1-i d x^2\right )\right )^2} \]
[Out]
-Sqrt[(2*I)*d*x^2 + d^2*x^4]/(4*b*d*x*(a + I*b*ArcSin[1 - I*d*x^2])^2) - x/(8*b^2*(a + I*b*ArcSin[1 - I*d*x^2]
)) + (x*CosIntegral[((-I/2)*(a + I*b*ArcSin[1 - I*d*x^2]))/b]*(I*Cosh[a/(2*b)] - Sinh[a/(2*b)]))/(16*b^3*(Cos[
ArcSin[1 - I*d*x^2]/2] - Sin[ArcSin[1 - I*d*x^2]/2])) - (x*(I*Cosh[a/(2*b)] + Sinh[a/(2*b)])*SinIntegral[((I/2
)*a)/b - ArcSin[1 - I*d*x^2]/2])/(16*b^3*(Cos[ArcSin[1 - I*d*x^2]/2] - Sin[ArcSin[1 - I*d*x^2]/2]))
________________________________________________________________________________________
Rubi [A] time = 0.0541483, antiderivative size = 275, normalized size of antiderivative = 1.,
number of steps used = 2, number of rules used = 2, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used =
{4828, 4816} \[ \frac{x \left (-\sinh \left (\frac{a}{2 b}\right )+i \cosh \left (\frac{a}{2 b}\right )\right ) \text{CosIntegral}\left (-\frac{i \left (a+i b \sin ^{-1}\left (1-i d x^2\right )\right )}{2 b}\right )}{16 b^3 \left (\cos \left (\frac{1}{2} \sin ^{-1}\left (1-i d x^2\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (1-i d x^2\right )\right )\right )}-\frac{x \left (\sinh \left (\frac{a}{2 b}\right )+i \cosh \left (\frac{a}{2 b}\right )\right ) \text{Si}\left (\frac{i a}{2 b}-\frac{1}{2} \sin ^{-1}\left (1-i d x^2\right )\right )}{16 b^3 \left (\cos \left (\frac{1}{2} \sin ^{-1}\left (1-i d x^2\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (1-i d x^2\right )\right )\right )}-\frac{x}{8 b^2 \left (a+i b \sin ^{-1}\left (1-i d x^2\right )\right )}-\frac{\sqrt{d^2 x^4+2 i d x^2}}{4 b d x \left (a+i b \sin ^{-1}\left (1-i d x^2\right )\right )^2} \]
Antiderivative was successfully verified.
[In]
Int[(a + I*b*ArcSin[1 - I*d*x^2])^(-3),x]
[Out]
-Sqrt[(2*I)*d*x^2 + d^2*x^4]/(4*b*d*x*(a + I*b*ArcSin[1 - I*d*x^2])^2) - x/(8*b^2*(a + I*b*ArcSin[1 - I*d*x^2]
)) + (x*CosIntegral[((-I/2)*(a + I*b*ArcSin[1 - I*d*x^2]))/b]*(I*Cosh[a/(2*b)] - Sinh[a/(2*b)]))/(16*b^3*(Cos[
ArcSin[1 - I*d*x^2]/2] - Sin[ArcSin[1 - I*d*x^2]/2])) - (x*(I*Cosh[a/(2*b)] + Sinh[a/(2*b)])*SinIntegral[((I/2
)*a)/b - ArcSin[1 - I*d*x^2]/2])/(16*b^3*(Cos[ArcSin[1 - I*d*x^2]/2] - Sin[ArcSin[1 - I*d*x^2]/2]))
Rule 4828
Int[((a_.) + ArcSin[(c_) + (d_.)*(x_)^2]*(b_.))^(n_), x_Symbol] :> Simp[(x*(a + b*ArcSin[c + d*x^2])^(n + 2))/
(4*b^2*(n + 1)*(n + 2)), x] + (-Dist[1/(4*b^2*(n + 1)*(n + 2)), Int[(a + b*ArcSin[c + d*x^2])^(n + 2), x], x]
+ Simp[(Sqrt[-2*c*d*x^2 - d^2*x^4]*(a + b*ArcSin[c + d*x^2])^(n + 1))/(2*b*d*(n + 1)*x), x]) /; FreeQ[{a, b, c
, d}, x] && EqQ[c^2, 1] && LtQ[n, -1] && NeQ[n, -2]
Rule 4816
Int[((a_.) + ArcSin[(c_) + (d_.)*(x_)^2]*(b_.))^(-1), x_Symbol] :> -Simp[(x*(c*Cos[a/(2*b)] - Sin[a/(2*b)])*Co
sIntegral[(c/(2*b))*(a + b*ArcSin[c + d*x^2])])/(2*b*(Cos[ArcSin[c + d*x^2]/2] - c*Sin[ArcSin[c + d*x^2]/2])),
x] - Simp[(x*(c*Cos[a/(2*b)] + Sin[a/(2*b)])*SinIntegral[(c/(2*b))*(a + b*ArcSin[c + d*x^2])])/(2*b*(Cos[ArcS
in[c + d*x^2]/2] - c*Sin[ArcSin[c + d*x^2]/2])), x] /; FreeQ[{a, b, c, d}, x] && EqQ[c^2, 1]
Rubi steps
\begin{align*} \int \frac{1}{\left (a+i b \sin ^{-1}\left (1-i d x^2\right )\right )^3} \, dx &=-\frac{\sqrt{2 i d x^2+d^2 x^4}}{4 b d x \left (a+i b \sin ^{-1}\left (1-i d x^2\right )\right )^2}-\frac{x}{8 b^2 \left (a+i b \sin ^{-1}\left (1-i d x^2\right )\right )}+\frac{\int \frac{1}{a+i b \sin ^{-1}\left (1-i d x^2\right )} \, dx}{8 b^2}\\ &=-\frac{\sqrt{2 i d x^2+d^2 x^4}}{4 b d x \left (a+i b \sin ^{-1}\left (1-i d x^2\right )\right )^2}-\frac{x}{8 b^2 \left (a+i b \sin ^{-1}\left (1-i d x^2\right )\right )}+\frac{x \text{Ci}\left (-\frac{i \left (a+i b \sin ^{-1}\left (1-i d x^2\right )\right )}{2 b}\right ) \left (i \cosh \left (\frac{a}{2 b}\right )-\sinh \left (\frac{a}{2 b}\right )\right )}{16 b^3 \left (\cos \left (\frac{1}{2} \sin ^{-1}\left (1-i d x^2\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (1-i d x^2\right )\right )\right )}-\frac{x \left (i \cosh \left (\frac{a}{2 b}\right )+\sinh \left (\frac{a}{2 b}\right )\right ) \text{Si}\left (\frac{i a}{2 b}-\frac{1}{2} \sin ^{-1}\left (1-i d x^2\right )\right )}{16 b^3 \left (\cos \left (\frac{1}{2} \sin ^{-1}\left (1-i d x^2\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (1-i d x^2\right )\right )\right )}\\ \end{align*}
Mathematica [A] time = 0.552229, size = 229, normalized size = 0.83 \[ \frac{-\frac{8 b^2 \sqrt{d x^2 \left (d x^2+2 i\right )}}{d \left (a+i b \sin ^{-1}\left (1-i d x^2\right )\right )^2}+\frac{2 i x^2 \left (\left (\cosh \left (\frac{a}{2 b}\right )+i \sinh \left (\frac{a}{2 b}\right )\right ) \text{CosIntegral}\left (\frac{1}{2} \left (\sin ^{-1}\left (1-i d x^2\right )-\frac{i a}{b}\right )\right )-\left (\cosh \left (\frac{a}{2 b}\right )-i \sinh \left (\frac{a}{2 b}\right )\right ) \text{Si}\left (\frac{i a}{2 b}-\frac{1}{2} \sin ^{-1}\left (1-i d x^2\right )\right )\right )}{\cos \left (\frac{1}{2} \sin ^{-1}\left (1-i d x^2\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (1-i d x^2\right )\right )}-\frac{4 b x^2}{a+i b \sin ^{-1}\left (1-i d x^2\right )}}{32 b^3 x} \]
Antiderivative was successfully verified.
[In]
Integrate[(a + I*b*ArcSin[1 - I*d*x^2])^(-3),x]
[Out]
((-8*b^2*Sqrt[d*x^2*(2*I + d*x^2)])/(d*(a + I*b*ArcSin[1 - I*d*x^2])^2) - (4*b*x^2)/(a + I*b*ArcSin[1 - I*d*x^
2]) + ((2*I)*x^2*(CosIntegral[(((-I)*a)/b + ArcSin[1 - I*d*x^2])/2]*(Cosh[a/(2*b)] + I*Sinh[a/(2*b)]) - (Cosh[
a/(2*b)] - I*Sinh[a/(2*b)])*SinIntegral[((I/2)*a)/b - ArcSin[1 - I*d*x^2]/2]))/(Cos[ArcSin[1 - I*d*x^2]/2] - S
in[ArcSin[1 - I*d*x^2]/2]))/(32*b^3*x)
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Maple [F] time = 0.066, size = 0, normalized size = 0. \begin{align*} \int \left ( a+b{\it Arcsinh} \left ( i+d{x}^{2} \right ) \right ) ^{-3}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(1/(a+b*arcsinh(I+d*x^2))^3,x)
[Out]
int(1/(a+b*arcsinh(I+d*x^2))^3,x)
________________________________________________________________________________________
Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(a+b*arcsinh(I+d*x^2))^3,x, algorithm="maxima")
[Out]
-(4*(a*d^(11/2) + 2*b*d^(11/2))*x^10 + (24*I*a*d^(9/2) + 56*I*b*d^(9/2))*x^8 - 4*(11*a*d^(7/2) + 36*b*d^(7/2))
*x^6 + (-8*I*a*d^(5/2) - 160*I*b*d^(5/2))*x^4 - 16*(3*a*d^(3/2) - 4*b*d^(3/2))*x^2 + (4*(a*d^4 + 2*b*d^4)*x^7
+ (12*I*a*d^3 + 32*I*b*d^3)*x^5 - 8*(2*a*d^2 + 5*b*d^2)*x^3 + (-16*I*a*d - 16*I*b*d)*x)*(d*x^2 + 2*I)^(3/2) +
(12*(a*d^(9/2) + 2*b*d^(9/2))*x^8 + (48*I*a*d^(7/2) + 120*I*b*d^(7/2))*x^6 - 8*(8*a*d^(5/2) + 25*b*d^(5/2))*x^
4 + (-40*I*a*d^(3/2) - 120*I*b*d^(3/2))*x^2 + 16*a*sqrt(d) + 16*b*sqrt(d))*(d*x^2 + 2*I) + (4*b*d^(11/2)*x^10
+ 24*I*b*d^(9/2)*x^8 - 44*b*d^(7/2)*x^6 - 8*I*b*d^(5/2)*x^4 - 48*b*d^(3/2)*x^2 + (4*b*d^4*x^7 + 12*I*b*d^3*x^5
- 16*b*d^2*x^3 - 16*I*b*d*x)*(d*x^2 + 2*I)^(3/2) + (12*b*d^(9/2)*x^8 + 48*I*b*d^(7/2)*x^6 - 64*b*d^(5/2)*x^4
- 40*I*b*d^(3/2)*x^2 + 16*b*sqrt(d))*(d*x^2 + 2*I) + (12*b*d^5*x^9 + 60*I*b*d^4*x^7 - 92*b*d^3*x^5 - 28*I*b*d^
2*x^3 - 24*b*d*x)*sqrt(d*x^2 + 2*I) - 32*I*b*sqrt(d))*log(d*x^2 + sqrt(d*x^2 + 2*I)*sqrt(d)*x + I) + (12*(a*d^
5 + 2*b*d^5)*x^9 + (60*I*a*d^4 + 144*I*b*d^4)*x^7 - 4*(23*a*d^3 + 76*b*d^3)*x^5 + (-28*I*a*d^2 - 256*I*b*d^2)*
x^3 - 8*(3*a*d - 8*b*d)*x)*sqrt(d*x^2 + 2*I) - 32*I*a*sqrt(d))/(32*a^2*b^2*d^(11/2)*x^9 + 192*I*a^2*b^2*d^(9/2
)*x^7 - 384*a^2*b^2*d^(7/2)*x^5 - 256*I*a^2*b^2*d^(5/2)*x^3 + (32*b^4*d^(11/2)*x^9 + 192*I*b^4*d^(9/2)*x^7 - 3
84*b^4*d^(7/2)*x^5 - 256*I*b^4*d^(5/2)*x^3 + (32*b^4*d^4*x^6 + 96*I*b^4*d^3*x^4 - 96*b^4*d^2*x^2 - 32*I*b^4*d)
*(d*x^2 + 2*I)^(3/2) + (96*b^4*d^(9/2)*x^7 + 384*I*b^4*d^(7/2)*x^5 - 480*b^4*d^(5/2)*x^3 - 192*I*b^4*d^(3/2)*x
)*(d*x^2 + 2*I) + (96*b^4*d^5*x^8 + 480*I*b^4*d^4*x^6 - 768*b^4*d^3*x^4 - 384*I*b^4*d^2*x^2)*sqrt(d*x^2 + 2*I)
)*log(d*x^2 + sqrt(d*x^2 + 2*I)*sqrt(d)*x + I)^2 + (32*a^2*b^2*d^4*x^6 + 96*I*a^2*b^2*d^3*x^4 - 96*a^2*b^2*d^2
*x^2 - 32*I*a^2*b^2*d)*(d*x^2 + 2*I)^(3/2) + (96*a^2*b^2*d^(9/2)*x^7 + 384*I*a^2*b^2*d^(7/2)*x^5 - 480*a^2*b^2
*d^(5/2)*x^3 - 192*I*a^2*b^2*d^(3/2)*x)*(d*x^2 + 2*I) + (64*a*b^3*d^(11/2)*x^9 + 384*I*a*b^3*d^(9/2)*x^7 - 768
*a*b^3*d^(7/2)*x^5 - 512*I*a*b^3*d^(5/2)*x^3 + (64*a*b^3*d^4*x^6 + 192*I*a*b^3*d^3*x^4 - 192*a*b^3*d^2*x^2 - 6
4*I*a*b^3*d)*(d*x^2 + 2*I)^(3/2) + (192*a*b^3*d^(9/2)*x^7 + 768*I*a*b^3*d^(7/2)*x^5 - 960*a*b^3*d^(5/2)*x^3 -
384*I*a*b^3*d^(3/2)*x)*(d*x^2 + 2*I) + (192*a*b^3*d^5*x^8 + 960*I*a*b^3*d^4*x^6 - 1536*a*b^3*d^3*x^4 - 768*I*a
*b^3*d^2*x^2)*sqrt(d*x^2 + 2*I))*log(d*x^2 + sqrt(d*x^2 + 2*I)*sqrt(d)*x + I) + (96*a^2*b^2*d^5*x^8 + 480*I*a^
2*b^2*d^4*x^6 - 768*a^2*b^2*d^3*x^4 - 384*I*a^2*b^2*d^2*x^2)*sqrt(d*x^2 + 2*I)) + integrate((d^6*x^12 + 8*I*d^
5*x^10 - 27*d^4*x^8 - 56*I*d^3*x^6 + 88*d^2*x^4 + (d^4*x^8 + 4*I*d^3*x^6 - 3*d^2*x^4 + 8*I*d*x^2 + 4)*(d*x^2 +
2*I)^2 + 96*I*d*x^2 + (4*d^(9/2)*x^9 + 20*I*d^(7/2)*x^7 - 30*d^(5/2)*x^5 + 2*I*d^(3/2)*x^3 - 22*sqrt(d)*x)*(d
*x^2 + 2*I)^(3/2) + (6*d^5*x^10 + 36*I*d^4*x^8 - 78*d^3*x^6 - 72*I*d^2*x^4 + 9*d*x^2 - 30*I)*(d*x^2 + 2*I) + (
4*d^(11/2)*x^11 + 28*I*d^(9/2)*x^9 - 78*d^(7/2)*x^7 - 122*I*d^(5/2)*x^5 + 122*d^(3/2)*x^3 + 60*I*sqrt(d)*x)*sq
rt(d*x^2 + 2*I) - 48)/(8*a*b^2*d^6*x^12 + 64*I*a*b^2*d^5*x^10 - 192*a*b^2*d^4*x^8 - 256*I*a*b^2*d^3*x^6 + 128*
a*b^2*d^2*x^4 + (8*a*b^2*d^4*x^8 + 32*I*a*b^2*d^3*x^6 - 48*a*b^2*d^2*x^4 - 32*I*a*b^2*d*x^2 + 8*a*b^2)*(d*x^2
+ 2*I)^2 + (32*a*b^2*d^(9/2)*x^9 + 160*I*a*b^2*d^(7/2)*x^7 - 288*a*b^2*d^(5/2)*x^5 - 224*I*a*b^2*d^(3/2)*x^3 +
64*a*b^2*sqrt(d)*x)*(d*x^2 + 2*I)^(3/2) + (48*a*b^2*d^5*x^10 + 288*I*a*b^2*d^4*x^8 - 624*a*b^2*d^3*x^6 - 576*
I*a*b^2*d^2*x^4 + 192*a*b^2*d*x^2)*(d*x^2 + 2*I) + (8*b^3*d^6*x^12 + 64*I*b^3*d^5*x^10 - 192*b^3*d^4*x^8 - 256
*I*b^3*d^3*x^6 + 128*b^3*d^2*x^4 + (8*b^3*d^4*x^8 + 32*I*b^3*d^3*x^6 - 48*b^3*d^2*x^4 - 32*I*b^3*d*x^2 + 8*b^3
)*(d*x^2 + 2*I)^2 + (32*b^3*d^(9/2)*x^9 + 160*I*b^3*d^(7/2)*x^7 - 288*b^3*d^(5/2)*x^5 - 224*I*b^3*d^(3/2)*x^3
+ 64*b^3*sqrt(d)*x)*(d*x^2 + 2*I)^(3/2) + (48*b^3*d^5*x^10 + 288*I*b^3*d^4*x^8 - 624*b^3*d^3*x^6 - 576*I*b^3*d
^2*x^4 + 192*b^3*d*x^2)*(d*x^2 + 2*I) + (32*b^3*d^(11/2)*x^11 + 224*I*b^3*d^(9/2)*x^9 - 576*b^3*d^(7/2)*x^7 -
640*I*b^3*d^(5/2)*x^5 + 256*b^3*d^(3/2)*x^3)*sqrt(d*x^2 + 2*I))*log(d*x^2 + sqrt(d*x^2 + 2*I)*sqrt(d)*x + I) +
(32*a*b^2*d^(11/2)*x^11 + 224*I*a*b^2*d^(9/2)*x^9 - 576*a*b^2*d^(7/2)*x^7 - 640*I*a*b^2*d^(5/2)*x^5 + 256*a*b
^2*d^(3/2)*x^3)*sqrt(d*x^2 + 2*I)), x)
________________________________________________________________________________________
Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{b d x^{2} \log \left (d x^{2} + \sqrt{d^{2} x^{4} + 2 i \, d x^{2}} + i\right ) + a d x^{2} - 8 \,{\left (b^{4} d x \log \left (d x^{2} + \sqrt{d^{2} x^{4} + 2 i \, d x^{2}} + i\right )^{2} + 2 \, a b^{3} d x \log \left (d x^{2} + \sqrt{d^{2} x^{4} + 2 i \, d x^{2}} + i\right ) + a^{2} b^{2} d x\right )}{\rm integral}\left (\frac{1}{8 \,{\left (b^{3} \log \left (d x^{2} + \sqrt{d^{2} x^{4} + 2 i \, d x^{2}} + i\right ) + a b^{2}\right )}}, x\right ) + 2 \, \sqrt{d^{2} x^{4} + 2 i \, d x^{2}} b}{8 \,{\left (b^{4} d x \log \left (d x^{2} + \sqrt{d^{2} x^{4} + 2 i \, d x^{2}} + i\right )^{2} + 2 \, a b^{3} d x \log \left (d x^{2} + \sqrt{d^{2} x^{4} + 2 i \, d x^{2}} + i\right ) + a^{2} b^{2} d x\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(a+b*arcsinh(I+d*x^2))^3,x, algorithm="fricas")
[Out]
-1/8*(b*d*x^2*log(d*x^2 + sqrt(d^2*x^4 + 2*I*d*x^2) + I) + a*d*x^2 - 8*(b^4*d*x*log(d*x^2 + sqrt(d^2*x^4 + 2*I
*d*x^2) + I)^2 + 2*a*b^3*d*x*log(d*x^2 + sqrt(d^2*x^4 + 2*I*d*x^2) + I) + a^2*b^2*d*x)*integral(1/8/(b^3*log(d
*x^2 + sqrt(d^2*x^4 + 2*I*d*x^2) + I) + a*b^2), x) + 2*sqrt(d^2*x^4 + 2*I*d*x^2)*b)/(b^4*d*x*log(d*x^2 + sqrt(
d^2*x^4 + 2*I*d*x^2) + I)^2 + 2*a*b^3*d*x*log(d*x^2 + sqrt(d^2*x^4 + 2*I*d*x^2) + I) + a^2*b^2*d*x)
________________________________________________________________________________________
Sympy [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(a+b*asinh(I+d*x**2))**3,x)
[Out]
Exception raised: TypeError
________________________________________________________________________________________
Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b \operatorname{arsinh}\left (d x^{2} + i\right ) + a\right )}^{3}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(a+b*arcsinh(I+d*x^2))^3,x, algorithm="giac")
[Out]
integrate((b*arcsinh(d*x^2 + I) + a)^(-3), x)