3.4.27 \(\int \frac {1}{(a-i b \text {ArcSin}(1+i d x^2))^3} \, dx\) [327]
Optimal. Leaf size=272 \[ -\frac {\sqrt {-2 i d x^2+d^2 x^4}}{4 b d x \left (a-i b \text {ArcSin}\left (1+i d x^2\right )\right )^2}-\frac {x}{8 b^2 \left (a-i b \text {ArcSin}\left (1+i d x^2\right )\right )}-\frac {x \text {CosIntegral}\left (\frac {i \left (a-i b \text {ArcSin}\left (1+i d x^2\right )\right )}{2 b}\right ) \left (i \cosh \left (\frac {a}{2 b}\right )+\sinh \left (\frac {a}{2 b}\right )\right )}{16 b^3 \left (\cos \left (\frac {1}{2} \text {ArcSin}\left (1+i d x^2\right )\right )-\sin \left (\frac {1}{2} \text {ArcSin}\left (1+i d x^2\right )\right )\right )}+\frac {x \left (\cosh \left (\frac {a}{2 b}\right )+i \sinh \left (\frac {a}{2 b}\right )\right ) \text {Shi}\left (\frac {a-i b \text {ArcSin}\left (1+i d x^2\right )}{2 b}\right )}{16 b^3 \left (\cos \left (\frac {1}{2} \text {ArcSin}\left (1+i d x^2\right )\right )-\sin \left (\frac {1}{2} \text {ArcSin}\left (1+i d x^2\right )\right )\right )} \]
[Out]
-1/8*x/b^2/(a-I*b*arcsin(1+I*d*x^2))+1/16*x*Shi(1/2*(a-I*b*arcsin(1+I*d*x^2))/b)*(cosh(1/2*a/b)+I*sinh(1/2*a/b
))/b^3/(cos(1/2*arcsin(1+I*d*x^2))-sin(1/2*arcsin(1+I*d*x^2)))-1/16*x*Ci(1/2*I*(a-I*b*arcsin(1+I*d*x^2))/b)*(I
*cosh(1/2*a/b)+sinh(1/2*a/b))/b^3/(cos(1/2*arcsin(1+I*d*x^2))-sin(1/2*arcsin(1+I*d*x^2)))-1/4*(-2*I*d*x^2+d^2*
x^4)^(1/2)/b/d/x/(a-I*b*arcsin(1+I*d*x^2))^2
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Rubi [A]
time = 0.04, antiderivative size = 272, normalized size of antiderivative = 1.00, number of steps
used = 2, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {4912, 4900}
\begin {gather*} -\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 \text {ArcSin}\left (1+i d x^2\right )\right )}{2 b}\right )}{16 b^3 \left (\cos \left (\frac {1}{2} \text {ArcSin}\left (1+i d x^2\right )\right )-\sin \left (\frac {1}{2} \text {ArcSin}\left (1+i d x^2\right )\right )\right )}+\frac {x \left (\cosh \left (\frac {a}{2 b}\right )+i \sinh \left (\frac {a}{2 b}\right )\right ) \text {Shi}\left (\frac {a-i b \text {ArcSin}\left (i d x^2+1\right )}{2 b}\right )}{16 b^3 \left (\cos \left (\frac {1}{2} \text {ArcSin}\left (1+i d x^2\right )\right )-\sin \left (\frac {1}{2} \text {ArcSin}\left (1+i d x^2\right )\right )\right )}-\frac {x}{8 b^2 \left (a-i b \text {ArcSin}\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 \text {ArcSin}\left (1+i d x^2\right )\right )^2} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[(a - I*b*ArcSin[1 + I*d*x^2])^(-3),x]
[Out]
-1/4*Sqrt[(-2*I)*d*x^2 + d^2*x^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*(Co
s[ArcSin[1 + I*d*x^2]/2] - Sin[ArcSin[1 + I*d*x^2]/2])) + (x*(Cosh[a/(2*b)] + I*Sinh[a/(2*b)])*SinhIntegral[(a
- I*b*ArcSin[1 + I*d*x^2])/(2*b)])/(16*b^3*(Cos[ArcSin[1 + I*d*x^2]/2] - Sin[ArcSin[1 + I*d*x^2]/2]))
Rule 4900
Int[((a_.) + ArcSin[(c_) + (d_.)*(x_)^2]*(b_.))^(-1), x_Symbol] :> Simp[(-x)*(c*Cos[a/(2*b)] - Sin[a/(2*b)])*(
CosIntegral[(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[Arc
Sin[c + d*x^2]/2] - c*Sin[ArcSin[c + d*x^2]/2]))), x] /; FreeQ[{a, b, c, d}, x] && EqQ[c^2, 1]
Rule 4912
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]
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 (\cosh \left (\frac {a}{2 b}\right )+i \sinh \left (\frac {a}{2 b}\right )\right ) \text {Shi}\left (\frac {a-i b \sin ^{-1}\left (1+i d x^2\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 )}\\ \end {align*}
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Mathematica [A]
time = 0.56, size = 227, normalized size = 0.83 \begin {gather*} \frac {-\frac {4 b x^2}{a-i b \text {ArcSin}\left (1+i d x^2\right )}+\frac {8 b^2 \sqrt {d x^2 \left (-2 i+d x^2\right )}}{d \left (i a+b \text {ArcSin}\left (1+i d x^2\right )\right )^2}-\frac {2 i x^2 \left (\text {CosIntegral}\left (\frac {1}{2} \left (\frac {i a}{b}+\text {ArcSin}\left (1+i d x^2\right )\right )\right ) \left (\cosh \left (\frac {a}{2 b}\right )-i \sinh \left (\frac {a}{2 b}\right )\right )+\left (\cosh \left (\frac {a}{2 b}\right )+i \sinh \left (\frac {a}{2 b}\right )\right ) \text {Si}\left (\frac {1}{2} \left (\frac {i a}{b}+\text {ArcSin}\left (1+i d x^2\right )\right )\right )\right )}{\cos \left (\frac {1}{2} \text {ArcSin}\left (1+i d x^2\right )\right )-\sin \left (\frac {1}{2} \text {ArcSin}\left (1+i d x^2\right )\right )}}{32 b^3 x} \end {gather*}
Warning: Unable to verify antiderivative.
[In]
Integrate[(a - I*b*ArcSin[1 + I*d*x^2])^(-3),x]
[Out]
((-4*b*x^2)/(a - I*b*ArcSin[1 + I*d*x^2]) + (8*b^2*Sqrt[d*x^2*(-2*I + d*x^2)])/(d*(I*a + b*ArcSin[1 + I*d*x^2]
)^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*a)/b + ArcSin[1 + I*d*x^2])/2]))/(Cos[ArcSin[1 + I*d*x^2]/2] - Sin[A
rcSin[1 + I*d*x^2]/2]))/(32*b^3*x)
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Maple [F]
time = 0.93, size = 0, normalized size = 0.00 \[\int \frac {1}{\left (a +b \arcsinh \left (d \,x^{2}-i\right )\right )^{3}}\, dx\]
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)
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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}
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]
-1/8*((a*d^(11/2) + 2*b*d^(11/2))*x^10 - 2*(3*I*a*d^(9/2) + 7*I*b*d^(9/2))*x^8 - (11*a*d^(7/2) + 36*b*d^(7/2))
*x^6 - 2*(-I*a*d^(5/2) - 20*I*b*d^(5/2))*x^4 - 4*(3*a*d^(3/2) - 4*b*d^(3/2))*x^2 + ((a*d^4 + 2*b*d^4)*x^7 - (3
*I*a*d^3 + 8*I*b*d^3)*x^5 - 2*(2*a*d^2 + 5*b*d^2)*x^3 - 4*(-I*a*d - I*b*d)*x)*(d*x^2 - 2*I)^(3/2) + (3*(a*d^(9
/2) + 2*b*d^(9/2))*x^8 - 6*(2*I*a*d^(7/2) + 5*I*b*d^(7/2))*x^6 - 2*(8*a*d^(5/2) + 25*b*d^(5/2))*x^4 - 10*(-I*a
*d^(3/2) - 3*I*b*d^(3/2))*x^2 + 4*a*sqrt(d) + 4*b*sqrt(d))*(d*x^2 - 2*I) + (b*d^(11/2)*x^10 - 6*I*b*d^(9/2)*x^
8 - 11*b*d^(7/2)*x^6 + 2*I*b*d^(5/2)*x^4 - 12*b*d^(3/2)*x^2 + (b*d^4*x^7 - 3*I*b*d^3*x^5 - 4*b*d^2*x^3 + 4*I*b
*d*x)*(d*x^2 - 2*I)^(3/2) + (3*b*d^(9/2)*x^8 - 12*I*b*d^(7/2)*x^6 - 16*b*d^(5/2)*x^4 + 10*I*b*d^(3/2)*x^2 + 4*
b*sqrt(d))*(d*x^2 - 2*I) + (3*b*d^5*x^9 - 15*I*b*d^4*x^7 - 23*b*d^3*x^5 + 7*I*b*d^2*x^3 - 6*b*d*x)*sqrt(d*x^2
- 2*I) + 8*I*b*sqrt(d))*log(d*x^2 + sqrt(d*x^2 - 2*I)*sqrt(d)*x - I) + (3*(a*d^5 + 2*b*d^5)*x^9 - 3*(5*I*a*d^4
+ 12*I*b*d^4)*x^7 - (23*a*d^3 + 76*b*d^3)*x^5 - (-7*I*a*d^2 - 64*I*b*d^2)*x^3 - 2*(3*a*d - 8*b*d)*x)*sqrt(d*x
^2 - 2*I) + 8*I*a*sqrt(d))/(a^2*b^2*d^(11/2)*x^9 - 6*I*a^2*b^2*d^(9/2)*x^7 - 12*a^2*b^2*d^(7/2)*x^5 + 8*I*a^2*
b^2*d^(5/2)*x^3 + (b^4*d^(11/2)*x^9 - 6*I*b^4*d^(9/2)*x^7 - 12*b^4*d^(7/2)*x^5 + 8*I*b^4*d^(5/2)*x^3 + (b^4*d^
4*x^6 - 3*I*b^4*d^3*x^4 - 3*b^4*d^2*x^2 + I*b^4*d)*(d*x^2 - 2*I)^(3/2) + 3*(b^4*d^(9/2)*x^7 - 4*I*b^4*d^(7/2)*
x^5 - 5*b^4*d^(5/2)*x^3 + 2*I*b^4*d^(3/2)*x)*(d*x^2 - 2*I) + 3*(b^4*d^5*x^8 - 5*I*b^4*d^4*x^6 - 8*b^4*d^3*x^4
+ 4*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 + (a^2*b^2*d^4*x^6 - 3*I*
a^2*b^2*d^3*x^4 - 3*a^2*b^2*d^2*x^2 + I*a^2*b^2*d)*(d*x^2 - 2*I)^(3/2) + 3*(a^2*b^2*d^(9/2)*x^7 - 4*I*a^2*b^2*
d^(7/2)*x^5 - 5*a^2*b^2*d^(5/2)*x^3 + 2*I*a^2*b^2*d^(3/2)*x)*(d*x^2 - 2*I) + 2*(a*b^3*d^(11/2)*x^9 - 6*I*a*b^3
*d^(9/2)*x^7 - 12*a*b^3*d^(7/2)*x^5 + 8*I*a*b^3*d^(5/2)*x^3 + (a*b^3*d^4*x^6 - 3*I*a*b^3*d^3*x^4 - 3*a*b^3*d^2
*x^2 + I*a*b^3*d)*(d*x^2 - 2*I)^(3/2) + 3*(a*b^3*d^(9/2)*x^7 - 4*I*a*b^3*d^(7/2)*x^5 - 5*a*b^3*d^(5/2)*x^3 + 2
*I*a*b^3*d^(3/2)*x)*(d*x^2 - 2*I) + 3*(a*b^3*d^5*x^8 - 5*I*a*b^3*d^4*x^6 - 8*a*b^3*d^3*x^4 + 4*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) + 3*(a^2*b^2*d^5*x^8 - 5*I*a^2*b^2*d^4*x^6 -
8*a^2*b^2*d^3*x^4 + 4*I*a^2*b^2*d^2*x^2)*sqrt(d*x^2 - 2*I)) + integrate(1/8*(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 + 2*(2*d^(9/2)*x^9 - 10*I*d^(7/2)*x^7 - 15*d^(5/2)*x^5 - I*d^(3/2)*x^3 - 11*sqrt(d)*x)*(d*x^2 - 2*I)^(3/
2) + 3*(2*d^5*x^10 - 12*I*d^4*x^8 - 26*d^3*x^6 + 24*I*d^2*x^4 + 3*d*x^2 + 10*I)*(d*x^2 - 2*I) + 2*(2*d^(11/2)*
x^11 - 14*I*d^(9/2)*x^9 - 39*d^(7/2)*x^7 + 61*I*d^(5/2)*x^5 + 61*d^(3/2)*x^3 - 30*I*sqrt(d)*x)*sqrt(d*x^2 - 2*
I) - 48)/(a*b^2*d^6*x^12 - 8*I*a*b^2*d^5*x^10 - 24*a*b^2*d^4*x^8 + 32*I*a*b^2*d^3*x^6 + 16*a*b^2*d^2*x^4 + (a*
b^2*d^4*x^8 - 4*I*a*b^2*d^3*x^6 - 6*a*b^2*d^2*x^4 + 4*I*a*b^2*d*x^2 + a*b^2)*(d*x^2 - 2*I)^2 + 4*(a*b^2*d^(9/2
)*x^9 - 5*I*a*b^2*d^(7/2)*x^7 - 9*a*b^2*d^(5/2)*x^5 + 7*I*a*b^2*d^(3/2)*x^3 + 2*a*b^2*sqrt(d)*x)*(d*x^2 - 2*I)
^(3/2) + 6*(a*b^2*d^5*x^10 - 6*I*a*b^2*d^4*x^8 - 13*a*b^2*d^3*x^6 + 12*I*a*b^2*d^2*x^4 + 4*a*b^2*d*x^2)*(d*x^2
- 2*I) + (b^3*d^6*x^12 - 8*I*b^3*d^5*x^10 - 24*b^3*d^4*x^8 + 32*I*b^3*d^3*x^6 + 16*b^3*d^2*x^4 + (b^3*d^4*x^8
- 4*I*b^3*d^3*x^6 - 6*b^3*d^2*x^4 + 4*I*b^3*d*x^2 + b^3)*(d*x^2 - 2*I)^2 + 4*(b^3*d^(9/2)*x^9 - 5*I*b^3*d^(7/
2)*x^7 - 9*b^3*d^(5/2)*x^5 + 7*I*b^3*d^(3/2)*x^3 + 2*b^3*sqrt(d)*x)*(d*x^2 - 2*I)^(3/2) + 6*(b^3*d^5*x^10 - 6*
I*b^3*d^4*x^8 - 13*b^3*d^3*x^6 + 12*I*b^3*d^2*x^4 + 4*b^3*d*x^2)*(d*x^2 - 2*I) + 4*(b^3*d^(11/2)*x^11 - 7*I*b^
3*d^(9/2)*x^9 - 18*b^3*d^(7/2)*x^7 + 20*I*b^3*d^(5/2)*x^5 + 8*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) + 4*(a*b^2*d^(11/2)*x^11 - 7*I*a*b^2*d^(9/2)*x^9 - 18*a*b^2*d^(7/2)*x^7 + 20*
I*a*b^2*d^(5/2)*x^5 + 8*a*b^2*d^(3/2)*x^3)*sqrt(d*x^2 - 2*I)), x)
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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
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*log(d*x^2 + sqrt(d^2*x^2 - 2*I*d)*x - I) + a*d*x - 8*(b^4*d*log(d*x^2 + sqrt(d^2*x^2 - 2*I*d)*x -
I)^2 + 2*a*b^3*d*log(d*x^2 + sqrt(d^2*x^2 - 2*I*d)*x - I) + a^2*b^2*d)*integral(1/8/(b^3*log(d*x^2 + sqrt(d^2*
x^2 - 2*I*d)*x - I) + a*b^2), x) + 2*sqrt(d^2*x^2 - 2*I*d)*b)/(b^4*d*log(d*x^2 + sqrt(d^2*x^2 - 2*I*d)*x - I)^
2 + 2*a*b^3*d*log(d*x^2 + sqrt(d^2*x^2 - 2*I*d)*x - I) + a^2*b^2*d)
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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}
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 >> Invalid comparison of non-real -I
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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}
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]
Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Warning, need to choose a branch for the root of a polynomial with parameters. This might be wrong.The choi
ce was done
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {1}{{\left (a+b\,\mathrm {asinh}\left (d\,x^2-\mathrm {i}\right )\right )}^3} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(1/(a + b*asinh(d*x^2 - 1i))^3,x)
[Out]
int(1/(a + b*asinh(d*x^2 - 1i))^3, x)
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