3.407 \(\int \frac {1}{(a+b \sin ^{-1}(1+d x^2))^3} \, dx\)

Optimal. Leaf size=227 \[ \frac {x \left (\cos \left (\frac {a}{2 b}\right )-\sin \left (\frac {a}{2 b}\right )\right ) \text {Ci}\left (\frac {a+b \sin ^{-1}\left (d x^2+1\right )}{2 b}\right )}{16 b^3 \left (\cos \left (\frac {1}{2} \sin ^{-1}\left (d x^2+1\right )\right )-\sin \left (\frac {1}{2} \sin ^{-1}\left (d x^2+1\right )\right )\right )}+\frac {x \left (\sin \left (\frac {a}{2 b}\right )+\cos \left (\frac {a}{2 b}\right )\right ) \text {Si}\left (\frac {a+b \sin ^{-1}\left (d x^2+1\right )}{2 b}\right )}{16 b^3 \left (\cos \left (\frac {1}{2} \sin ^{-1}\left (d x^2+1\right )\right )-\sin \left (\frac {1}{2} \sin ^{-1}\left (d x^2+1\right )\right )\right )}+\frac {x}{8 b^2 \left (a+b \sin ^{-1}\left (d x^2+1\right )\right )}-\frac {\sqrt {-d^2 x^4-2 d x^2}}{4 b d x \left (a+b \sin ^{-1}\left (d x^2+1\right )\right )^2} \]

[Out]

1/8*x/b^2/(a+b*arcsin(d*x^2+1))+1/16*x*Ci(1/2*(a+b*arcsin(d*x^2+1))/b)*(cos(1/2*a/b)-sin(1/2*a/b))/b^3/(cos(1/
2*arcsin(d*x^2+1))-sin(1/2*arcsin(d*x^2+1)))+1/16*x*Si(1/2*(a+b*arcsin(d*x^2+1))/b)*(cos(1/2*a/b)+sin(1/2*a/b)
)/b^3/(cos(1/2*arcsin(d*x^2+1))-sin(1/2*arcsin(d*x^2+1)))-1/4*(-d^2*x^4-2*d*x^2)^(1/2)/b/d/x/(a+b*arcsin(d*x^2
+1))^2

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Rubi [A]  time = 0.05, antiderivative size = 227, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {4828, 4816} \[ \frac {x \left (\cos \left (\frac {a}{2 b}\right )-\sin \left (\frac {a}{2 b}\right )\right ) \text {CosIntegral}\left (\frac {a+b \sin ^{-1}\left (d x^2+1\right )}{2 b}\right )}{16 b^3 \left (\cos \left (\frac {1}{2} \sin ^{-1}\left (d x^2+1\right )\right )-\sin \left (\frac {1}{2} \sin ^{-1}\left (d x^2+1\right )\right )\right )}+\frac {x \left (\sin \left (\frac {a}{2 b}\right )+\cos \left (\frac {a}{2 b}\right )\right ) \text {Si}\left (\frac {a+b \sin ^{-1}\left (d x^2+1\right )}{2 b}\right )}{16 b^3 \left (\cos \left (\frac {1}{2} \sin ^{-1}\left (d x^2+1\right )\right )-\sin \left (\frac {1}{2} \sin ^{-1}\left (d x^2+1\right )\right )\right )}+\frac {x}{8 b^2 \left (a+b \sin ^{-1}\left (d x^2+1\right )\right )}-\frac {\sqrt {-d^2 x^4-2 d x^2}}{4 b d x \left (a+b \sin ^{-1}\left (d x^2+1\right )\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSin[1 + d*x^2])^(-3),x]

[Out]

-Sqrt[-2*d*x^2 - d^2*x^4]/(4*b*d*x*(a + b*ArcSin[1 + d*x^2])^2) + x/(8*b^2*(a + b*ArcSin[1 + d*x^2])) + (x*Cos
Integral[(a + b*ArcSin[1 + d*x^2])/(2*b)]*(Cos[a/(2*b)] - Sin[a/(2*b)]))/(16*b^3*(Cos[ArcSin[1 + d*x^2]/2] - S
in[ArcSin[1 + d*x^2]/2])) + (x*(Cos[a/(2*b)] + Sin[a/(2*b)])*SinIntegral[(a + b*ArcSin[1 + d*x^2])/(2*b)])/(16
*b^3*(Cos[ArcSin[1 + d*x^2]/2] - Sin[ArcSin[1 + d*x^2]/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]

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]

Rubi steps

\begin {align*} \int \frac {1}{\left (a+b \sin ^{-1}\left (1+d x^2\right )\right )^3} \, dx &=-\frac {\sqrt {-2 d x^2-d^2 x^4}}{4 b d x \left (a+b \sin ^{-1}\left (1+d x^2\right )\right )^2}+\frac {x}{8 b^2 \left (a+b \sin ^{-1}\left (1+d x^2\right )\right )}-\frac {\int \frac {1}{a+b \sin ^{-1}\left (1+d x^2\right )} \, dx}{8 b^2}\\ &=-\frac {\sqrt {-2 d x^2-d^2 x^4}}{4 b d x \left (a+b \sin ^{-1}\left (1+d x^2\right )\right )^2}+\frac {x}{8 b^2 \left (a+b \sin ^{-1}\left (1+d x^2\right )\right )}+\frac {x \text {Ci}\left (\frac {a+b \sin ^{-1}\left (1+d x^2\right )}{2 b}\right ) \left (\cos \left (\frac {a}{2 b}\right )-\sin \left (\frac {a}{2 b}\right )\right )}{16 b^3 \left (\cos \left (\frac {1}{2} \sin ^{-1}\left (1+d x^2\right )\right )-\sin \left (\frac {1}{2} \sin ^{-1}\left (1+d x^2\right )\right )\right )}+\frac {x \left (\cos \left (\frac {a}{2 b}\right )+\sin \left (\frac {a}{2 b}\right )\right ) \text {Si}\left (\frac {a+b \sin ^{-1}\left (1+d x^2\right )}{2 b}\right )}{16 b^3 \left (\cos \left (\frac {1}{2} \sin ^{-1}\left (1+d x^2\right )\right )-\sin \left (\frac {1}{2} \sin ^{-1}\left (1+d x^2\right )\right )\right )}\\ \end {align*}

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Mathematica [A]  time = 0.56, size = 187, normalized size = 0.82 \[ \frac {x \left (\left (\cos \left (\frac {a}{2 b}\right )-\sin \left (\frac {a}{2 b}\right )\right ) \text {Ci}\left (\frac {1}{2} \left (\frac {a}{b}+\sin ^{-1}\left (d x^2+1\right )\right )\right )+\left (\sin \left (\frac {a}{2 b}\right )+\cos \left (\frac {a}{2 b}\right )\right ) \text {Si}\left (\frac {1}{2} \left (\frac {a}{b}+\sin ^{-1}\left (d x^2+1\right )\right )\right )\right )}{16 b^3 \left (\cos \left (\frac {1}{2} \sin ^{-1}\left (d x^2+1\right )\right )-\sin \left (\frac {1}{2} \sin ^{-1}\left (d x^2+1\right )\right )\right )}+\frac {x}{8 b^2 \left (a+b \sin ^{-1}\left (d x^2+1\right )\right )}-\frac {\sqrt {-d x^2 \left (d x^2+2\right )}}{4 b d x \left (a+b \sin ^{-1}\left (d x^2+1\right )\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*Sqrt[-(d*x^2*(2 + d*x^2))]/(b*d*x*(a + b*ArcSin[1 + d*x^2])^2) + x/(8*b^2*(a + b*ArcSin[1 + d*x^2])) + (x
*(CosIntegral[(a/b + ArcSin[1 + d*x^2])/2]*(Cos[a/(2*b)] - Sin[a/(2*b)]) + (Cos[a/(2*b)] + Sin[a/(2*b)])*SinIn
tegral[(a/b + ArcSin[1 + d*x^2])/2]))/(16*b^3*(Cos[ArcSin[1 + d*x^2]/2] - Sin[ArcSin[1 + d*x^2]/2]))

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fricas [F]  time = 0.50, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {1}{b^{3} \arcsin \left (d x^{2} + 1\right )^{3} + 3 \, a b^{2} \arcsin \left (d x^{2} + 1\right )^{2} + 3 \, a^{2} b \arcsin \left (d x^{2} + 1\right ) + a^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(1/(b^3*arcsin(d*x^2 + 1)^3 + 3*a*b^2*arcsin(d*x^2 + 1)^2 + 3*a^2*b*arcsin(d*x^2 + 1) + a^3), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b \arcsin \left (d x^{2} + 1\right ) + a\right )}^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*arcsin(d*x^2 + 1) + a)^(-3), x)

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maple [F]  time = 0.07, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a +b \arcsin \left (d \,x^{2}+1\right )\right )^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*arcsin(d*x^2+1))^3,x)

[Out]

int(1/(a+b*arcsin(d*x^2+1))^3,x)

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: Error executing code in Maxima: sign: argument cannot be imaginary
; found sqrt((-_SAGE_VAR_d*_SAGE_VAR_x^2)-2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*asin(d*x^2 + 1))^3,x)

[Out]

int(1/(a + b*asin(d*x^2 + 1))^3, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a + b \operatorname {asin}{\left (d x^{2} + 1 \right )}\right )^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*asin(d*x**2+1))**3,x)

[Out]

Integral((a + b*asin(d*x**2 + 1))**(-3), x)

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