3.422 \(\int \frac {1}{(a+b \sin ^{-1}(1+d x^2))^{5/2}} \, dx\)

Optimal. Leaf size=261 \[ \frac {\sqrt {\pi } x \left (\cos \left (\frac {a}{2 b}\right )-\sin \left (\frac {a}{2 b}\right )\right ) C\left (\frac {\sqrt {a+b \sin ^{-1}\left (d x^2+1\right )}}{\sqrt {b} \sqrt {\pi }}\right )}{3 b^{5/2} \left (\cos \left (\frac {1}{2} \sin ^{-1}\left (d x^2+1\right )\right )-\sin \left (\frac {1}{2} \sin ^{-1}\left (d x^2+1\right )\right )\right )}+\frac {\sqrt {\pi } x \left (\sin \left (\frac {a}{2 b}\right )+\cos \left (\frac {a}{2 b}\right )\right ) S\left (\frac {\sqrt {a+b \sin ^{-1}\left (d x^2+1\right )}}{\sqrt {b} \sqrt {\pi }}\right )}{3 b^{5/2} \left (\cos \left (\frac {1}{2} \sin ^{-1}\left (d x^2+1\right )\right )-\sin \left (\frac {1}{2} \sin ^{-1}\left (d x^2+1\right )\right )\right )}+\frac {x}{3 b^2 \sqrt {a+b \sin ^{-1}\left (d x^2+1\right )}}-\frac {\sqrt {-d^2 x^4-2 d x^2}}{3 b d x \left (a+b \sin ^{-1}\left (d x^2+1\right )\right )^{3/2}} \]

[Out]

1/3*x*FresnelC((a+b*arcsin(d*x^2+1))^(1/2)/b^(1/2)/Pi^(1/2))*(cos(1/2*a/b)-sin(1/2*a/b))*Pi^(1/2)/b^(5/2)/(cos
(1/2*arcsin(d*x^2+1))-sin(1/2*arcsin(d*x^2+1)))+1/3*x*FresnelS((a+b*arcsin(d*x^2+1))^(1/2)/b^(1/2)/Pi^(1/2))*(
cos(1/2*a/b)+sin(1/2*a/b))*Pi^(1/2)/b^(5/2)/(cos(1/2*arcsin(d*x^2+1))-sin(1/2*arcsin(d*x^2+1)))-1/3*(-d^2*x^4-
2*d*x^2)^(1/2)/b/d/x/(a+b*arcsin(d*x^2+1))^(3/2)+1/3*x/b^2/(a+b*arcsin(d*x^2+1))^(1/2)

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Rubi [A]  time = 0.07, antiderivative size = 261, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {4828, 4819} \[ \frac {\sqrt {\pi } x \left (\cos \left (\frac {a}{2 b}\right )-\sin \left (\frac {a}{2 b}\right )\right ) \text {FresnelC}\left (\frac {\sqrt {a+b \sin ^{-1}\left (d x^2+1\right )}}{\sqrt {\pi } \sqrt {b}}\right )}{3 b^{5/2} \left (\cos \left (\frac {1}{2} \sin ^{-1}\left (d x^2+1\right )\right )-\sin \left (\frac {1}{2} \sin ^{-1}\left (d x^2+1\right )\right )\right )}+\frac {\sqrt {\pi } x \left (\sin \left (\frac {a}{2 b}\right )+\cos \left (\frac {a}{2 b}\right )\right ) S\left (\frac {\sqrt {a+b \sin ^{-1}\left (d x^2+1\right )}}{\sqrt {b} \sqrt {\pi }}\right )}{3 b^{5/2} \left (\cos \left (\frac {1}{2} \sin ^{-1}\left (d x^2+1\right )\right )-\sin \left (\frac {1}{2} \sin ^{-1}\left (d x^2+1\right )\right )\right )}+\frac {x}{3 b^2 \sqrt {a+b \sin ^{-1}\left (d x^2+1\right )}}-\frac {\sqrt {-d^2 x^4-2 d x^2}}{3 b d x \left (a+b \sin ^{-1}\left (d x^2+1\right )\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSin[1 + d*x^2])^(-5/2),x]

[Out]

-Sqrt[-2*d*x^2 - d^2*x^4]/(3*b*d*x*(a + b*ArcSin[1 + d*x^2])^(3/2)) + x/(3*b^2*Sqrt[a + b*ArcSin[1 + d*x^2]])
+ (Sqrt[Pi]*x*FresnelC[Sqrt[a + b*ArcSin[1 + d*x^2]]/(Sqrt[b]*Sqrt[Pi])]*(Cos[a/(2*b)] - Sin[a/(2*b)]))/(3*b^(
5/2)*(Cos[ArcSin[1 + d*x^2]/2] - Sin[ArcSin[1 + d*x^2]/2])) + (Sqrt[Pi]*x*FresnelS[Sqrt[a + b*ArcSin[1 + d*x^2
]]/(Sqrt[b]*Sqrt[Pi])]*(Cos[a/(2*b)] + Sin[a/(2*b)]))/(3*b^(5/2)*(Cos[ArcSin[1 + d*x^2]/2] - Sin[ArcSin[1 + d*
x^2]/2]))

Rule 4819

Int[1/Sqrt[(a_.) + ArcSin[(c_) + (d_.)*(x_)^2]*(b_.)], x_Symbol] :> -Simp[(Sqrt[Pi]*x*(Cos[a/(2*b)] - c*Sin[a/
(2*b)])*FresnelC[(1*Sqrt[a + b*ArcSin[c + d*x^2]])/(Sqrt[b*c]*Sqrt[Pi])])/(Sqrt[b*c]*(Cos[ArcSin[c + d*x^2]/2]
- c*Sin[ArcSin[c + d*x^2]/2])), x] - Simp[(Sqrt[Pi]*x*(Cos[a/(2*b)] + c*Sin[a/(2*b)])*FresnelS[(1/(Sqrt[b*c]*
Sqrt[Pi]))*Sqrt[a + b*ArcSin[c + d*x^2]]])/(Sqrt[b*c]*(Cos[ArcSin[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 )^{5/2}} \, dx &=-\frac {\sqrt {-2 d x^2-d^2 x^4}}{3 b d x \left (a+b \sin ^{-1}\left (1+d x^2\right )\right )^{3/2}}+\frac {x}{3 b^2 \sqrt {a+b \sin ^{-1}\left (1+d x^2\right )}}-\frac {\int \frac {1}{\sqrt {a+b \sin ^{-1}\left (1+d x^2\right )}} \, dx}{3 b^2}\\ &=-\frac {\sqrt {-2 d x^2-d^2 x^4}}{3 b d x \left (a+b \sin ^{-1}\left (1+d x^2\right )\right )^{3/2}}+\frac {x}{3 b^2 \sqrt {a+b \sin ^{-1}\left (1+d x^2\right )}}+\frac {\sqrt {\pi } x C\left (\frac {\sqrt {a+b \sin ^{-1}\left (1+d x^2\right )}}{\sqrt {b} \sqrt {\pi }}\right ) \left (\cos \left (\frac {a}{2 b}\right )-\sin \left (\frac {a}{2 b}\right )\right )}{3 b^{5/2} \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 {\sqrt {\pi } x S\left (\frac {\sqrt {a+b \sin ^{-1}\left (1+d x^2\right )}}{\sqrt {b} \sqrt {\pi }}\right ) \left (\cos \left (\frac {a}{2 b}\right )+\sin \left (\frac {a}{2 b}\right )\right )}{3 b^{5/2} \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 = 0.52, size = 247, normalized size = 0.95 \[ \frac {x \left (\frac {\sqrt {\pi } \left (\cos \left (\frac {a}{2 b}\right )-\sin \left (\frac {a}{2 b}\right )\right ) C\left (\frac {\sqrt {a+b \sin ^{-1}\left (d x^2+1\right )}}{\sqrt {b} \sqrt {\pi }}\right )}{\sqrt {b} \left (\cos \left (\frac {1}{2} \sin ^{-1}\left (d x^2+1\right )\right )-\sin \left (\frac {1}{2} \sin ^{-1}\left (d x^2+1\right )\right )\right )}+\frac {\sqrt {\pi } \left (\sin \left (\frac {a}{2 b}\right )+\cos \left (\frac {a}{2 b}\right )\right ) S\left (\frac {\sqrt {a+b \sin ^{-1}\left (d x^2+1\right )}}{\sqrt {b} \sqrt {\pi }}\right )}{\sqrt {b} \left (\cos \left (\frac {1}{2} \sin ^{-1}\left (d x^2+1\right )\right )-\sin \left (\frac {1}{2} \sin ^{-1}\left (d x^2+1\right )\right )\right )}+\frac {b \left (d x^2+2\right )}{\sqrt {-d x^2 \left (d x^2+2\right )} \left (a+b \sin ^{-1}\left (d x^2+1\right )\right )^{3/2}}+\frac {1}{\sqrt {a+b \sin ^{-1}\left (d x^2+1\right )}}\right )}{3 b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcSin[1 + d*x^2])^(-5/2),x]

[Out]

(x*((b*(2 + d*x^2))/(Sqrt[-(d*x^2*(2 + d*x^2))]*(a + b*ArcSin[1 + d*x^2])^(3/2)) + 1/Sqrt[a + b*ArcSin[1 + d*x
^2]] + (Sqrt[Pi]*FresnelC[Sqrt[a + b*ArcSin[1 + d*x^2]]/(Sqrt[b]*Sqrt[Pi])]*(Cos[a/(2*b)] - Sin[a/(2*b)]))/(Sq
rt[b]*(Cos[ArcSin[1 + d*x^2]/2] - Sin[ArcSin[1 + d*x^2]/2])) + (Sqrt[Pi]*FresnelS[Sqrt[a + b*ArcSin[1 + d*x^2]
]/(Sqrt[b]*Sqrt[Pi])]*(Cos[a/(2*b)] + Sin[a/(2*b)]))/(Sqrt[b]*(Cos[ArcSin[1 + d*x^2]/2] - Sin[ArcSin[1 + d*x^2
]/2]))))/(3*b^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))^(5/2),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 \[ \int \frac {1}{{\left (b \arcsin \left (d x^{2} + 1\right ) + a\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*arcsin(d*x^2+1))^(5/2),x, algorithm="giac")

[Out]

integrate((b*arcsin(d*x^2 + 1) + a)^(-5/2), x)

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maple [F]  time = 0.07, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a +b \arcsin \left (d \,x^{2}+1\right )\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*arcsin(d*x^2+1))^(5/2),x)

[Out]

int(1/(a+b*arcsin(d*x^2+1))^(5/2),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))^(5/2),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 \[ \int \frac {1}{{\left (a+b\,\mathrm {asin}\left (d\,x^2+1\right )\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*asin(d*x^2 + 1))^(5/2),x)

[Out]

int(1/(a + b*asin(d*x^2 + 1))^(5/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a + b \operatorname {asin}{\left (d x^{2} + 1 \right )}\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*asin(d*x**2+1))**(5/2),x)

[Out]

Integral((a + b*asin(d*x**2 + 1))**(-5/2), x)

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