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

Optimal. Leaf size=238 \[ -\frac {\sqrt {-d^2 x^4-2 d x^2}}{b d x \sqrt {a+b \sin ^{-1}\left (d x^2+1\right )}}-\frac {\sqrt {\pi } \left (\frac {1}{b}\right )^{3/2} x \left (\sin \left (\frac {a}{2 b}\right )+\cos \left (\frac {a}{2 b}\right )\right ) C\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \sin ^{-1}\left (d x^2+1\right )}}{\sqrt {\pi }}\right )}{\cos \left (\frac {1}{2} \sin ^{-1}\left (d x^2+1\right )\right )-\sin \left (\frac {1}{2} \sin ^{-1}\left (d x^2+1\right )\right )}+\frac {\sqrt {\pi } \left (\frac {1}{b}\right )^{3/2} x \left (\cos \left (\frac {a}{2 b}\right )-\sin \left (\frac {a}{2 b}\right )\right ) S\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \sin ^{-1}\left (d x^2+1\right )}}{\sqrt {\pi }}\right )}{\cos \left (\frac {1}{2} \sin ^{-1}\left (d x^2+1\right )\right )-\sin \left (\frac {1}{2} \sin ^{-1}\left (d x^2+1\right )\right )} \]

[Out]

(1/b)^(3/2)*x*FresnelS((1/b)^(1/2)*(a+b*arcsin(d*x^2+1))^(1/2)/Pi^(1/2))*(cos(1/2*a/b)-sin(1/2*a/b))*Pi^(1/2)/
(cos(1/2*arcsin(d*x^2+1))-sin(1/2*arcsin(d*x^2+1)))-(1/b)^(3/2)*x*FresnelC((1/b)^(1/2)*(a+b*arcsin(d*x^2+1))^(
1/2)/Pi^(1/2))*(cos(1/2*a/b)+sin(1/2*a/b))*Pi^(1/2)/(cos(1/2*arcsin(d*x^2+1))-sin(1/2*arcsin(d*x^2+1)))-(-d^2*
x^4-2*d*x^2)^(1/2)/b/d/x/(a+b*arcsin(d*x^2+1))^(1/2)

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Rubi [A]  time = 0.04, antiderivative size = 238, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {4822} \[ -\frac {\sqrt {-d^2 x^4-2 d x^2}}{b d x \sqrt {a+b \sin ^{-1}\left (d x^2+1\right )}}-\frac {\sqrt {\pi } \left (\frac {1}{b}\right )^{3/2} x \left (\sin \left (\frac {a}{2 b}\right )+\cos \left (\frac {a}{2 b}\right )\right ) \text {FresnelC}\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \sin ^{-1}\left (d x^2+1\right )}}{\sqrt {\pi }}\right )}{\cos \left (\frac {1}{2} \sin ^{-1}\left (d x^2+1\right )\right )-\sin \left (\frac {1}{2} \sin ^{-1}\left (d x^2+1\right )\right )}+\frac {\sqrt {\pi } \left (\frac {1}{b}\right )^{3/2} x \left (\cos \left (\frac {a}{2 b}\right )-\sin \left (\frac {a}{2 b}\right )\right ) S\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \sin ^{-1}\left (d x^2+1\right )}}{\sqrt {\pi }}\right )}{\cos \left (\frac {1}{2} \sin ^{-1}\left (d x^2+1\right )\right )-\sin \left (\frac {1}{2} \sin ^{-1}\left (d x^2+1\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[-2*d*x^2 - d^2*x^4]/(b*d*x*Sqrt[a + b*ArcSin[1 + d*x^2]])) + ((b^(-1))^(3/2)*Sqrt[Pi]*x*FresnelS[(Sqrt[
b^(-1)]*Sqrt[a + b*ArcSin[1 + d*x^2]])/Sqrt[Pi]]*(Cos[a/(2*b)] - Sin[a/(2*b)]))/(Cos[ArcSin[1 + d*x^2]/2] - Si
n[ArcSin[1 + d*x^2]/2]) - ((b^(-1))^(3/2)*Sqrt[Pi]*x*FresnelC[(Sqrt[b^(-1)]*Sqrt[a + b*ArcSin[1 + d*x^2]])/Sqr
t[Pi]]*(Cos[a/(2*b)] + Sin[a/(2*b)]))/(Cos[ArcSin[1 + d*x^2]/2] - Sin[ArcSin[1 + d*x^2]/2])

Rule 4822

Int[((a_.) + ArcSin[(c_) + (d_.)*(x_)^2]*(b_.))^(-3/2), x_Symbol] :> -Simp[Sqrt[-2*c*d*x^2 - d^2*x^4]/(b*d*x*S
qrt[a + b*ArcSin[c + d*x^2]]), x] + (-Simp[((c/b)^(3/2)*Sqrt[Pi]*x*(Cos[a/(2*b)] + c*Sin[a/(2*b)])*FresnelC[Sq
rt[c/(Pi*b)]*Sqrt[a + b*ArcSin[c + d*x^2]]])/(Cos[(1/2)*ArcSin[c + d*x^2]] - c*Sin[ArcSin[c + d*x^2]/2]), x] +
Simp[((c/b)^(3/2)*Sqrt[Pi]*x*(Cos[a/(2*b)] - c*Sin[a/(2*b)])*FresnelS[Sqrt[c/(Pi*b)]*Sqrt[a + b*ArcSin[c + d*
x^2]]])/(Cos[(1/2)*ArcSin[c + d*x^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+b \sin ^{-1}\left (1+d x^2\right )\right )^{3/2}} \, dx &=-\frac {\sqrt {-2 d x^2-d^2 x^4}}{b d x \sqrt {a+b \sin ^{-1}\left (1+d x^2\right )}}+\frac {\left (\frac {1}{b}\right )^{3/2} \sqrt {\pi } x S\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \sin ^{-1}\left (1+d x^2\right )}}{\sqrt {\pi }}\right ) \left (\cos \left (\frac {a}{2 b}\right )-\sin \left (\frac {a}{2 b}\right )\right )}{\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 )}-\frac {\left (\frac {1}{b}\right )^{3/2} \sqrt {\pi } x C\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \sin ^{-1}\left (1+d x^2\right )}}{\sqrt {\pi }}\right ) \left (\cos \left (\frac {a}{2 b}\right )+\sin \left (\frac {a}{2 b}\right )\right )}{\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 )}\\ \end {align*}

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Mathematica [A]  time = 0.69, size = 238, normalized size = 1.00 \[ -\frac {\sqrt {-d^2 x^4-2 d x^2}}{b d x \sqrt {a+b \sin ^{-1}\left (d x^2+1\right )}}-\frac {\sqrt {\pi } \left (\frac {1}{b}\right )^{3/2} x \left (\sin \left (\frac {a}{2 b}\right )+\cos \left (\frac {a}{2 b}\right )\right ) C\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \sin ^{-1}\left (d x^2+1\right )}}{\sqrt {\pi }}\right )}{\cos \left (\frac {1}{2} \sin ^{-1}\left (d x^2+1\right )\right )-\sin \left (\frac {1}{2} \sin ^{-1}\left (d x^2+1\right )\right )}+\frac {\sqrt {\pi } \left (\frac {1}{b}\right )^{3/2} x \left (\cos \left (\frac {a}{2 b}\right )-\sin \left (\frac {a}{2 b}\right )\right ) S\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \sin ^{-1}\left (d x^2+1\right )}}{\sqrt {\pi }}\right )}{\cos \left (\frac {1}{2} \sin ^{-1}\left (d x^2+1\right )\right )-\sin \left (\frac {1}{2} \sin ^{-1}\left (d x^2+1\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[-2*d*x^2 - d^2*x^4]/(b*d*x*Sqrt[a + b*ArcSin[1 + d*x^2]])) + ((b^(-1))^(3/2)*Sqrt[Pi]*x*FresnelS[(Sqrt[
b^(-1)]*Sqrt[a + b*ArcSin[1 + d*x^2]])/Sqrt[Pi]]*(Cos[a/(2*b)] - Sin[a/(2*b)]))/(Cos[ArcSin[1 + d*x^2]/2] - Si
n[ArcSin[1 + d*x^2]/2]) - ((b^(-1))^(3/2)*Sqrt[Pi]*x*FresnelC[(Sqrt[b^(-1)]*Sqrt[a + b*ArcSin[1 + d*x^2]])/Sqr
t[Pi]]*(Cos[a/(2*b)] + Sin[a/(2*b)]))/(Cos[ArcSin[1 + d*x^2]/2] - Sin[ArcSin[1 + d*x^2]/2])

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (co
nstant residues)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(1/(a+b*arcsin(d*x^2+1))^(3/2),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))^(3/2),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 \[ \int \frac {1}{{\left (a+b\,\mathrm {asin}\left (d\,x^2+1\right )\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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