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

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

[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]))

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Rubi [A]  time = 0.0657306, antiderivative size = 261, normalized size of antiderivative = 1., 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 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 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]

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*}

Mathematica [A]  time = 0.495018, size = 247, normalized size = 0.95 \[ \frac{x \left (\frac{\sqrt{\pi } \left (\cos \left (\frac{a}{2 b}\right )-\sin \left (\frac{a}{2 b}\right )\right ) \text{FresnelC}\left (\frac{\sqrt{a+b \sin ^{-1}\left (d x^2+1\right )}}{\sqrt{\pi } \sqrt{b}}\right )}{\sqrt{b} \left (\cos \left (\frac{1}{2} \sin ^{-1}\left (d x^2+1\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (d x^2+1\right )\right )\right )}+\frac{\sqrt{\pi } \left (\sin \left (\frac{a}{2 b}\right )+\cos \left (\frac{a}{2 b}\right )\right ) S\left (\frac{\sqrt{a+b \sin ^{-1}\left (d x^2+1\right )}}{\sqrt{b} \sqrt{\pi }}\right )}{\sqrt{b} \left (\cos \left (\frac{1}{2} \sin ^{-1}\left (d x^2+1\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (d x^2+1\right )\right )\right )}+\frac{b \left (d x^2+2\right )}{\sqrt{-d x^2 \left (d x^2+2\right )} \left (a+b \sin ^{-1}\left (d x^2+1\right )\right )^{3/2}}+\frac{1}{\sqrt{a+b \sin ^{-1}\left (d x^2+1\right )}}\right )}{3 b^2} \]

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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Maple [F]  time = 0.059, size = 0, normalized size = 0. \begin{align*} \int \left ( a+b\arcsin \left ( d{x}^{2}+1 \right ) \right ) ^{-{\frac{5}{2}}}\, dx \end{align*}

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]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b \arcsin \left (d x^{2} + 1\right ) + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

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]

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

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Fricas [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: UnboundLocalError} \end{align*}

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: UnboundLocalError

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (a + b \operatorname{asin}{\left (d x^{2} + 1 \right )}\right )^{\frac{5}{2}}}\, dx \end{align*}

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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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b \arcsin \left (d x^{2} + 1\right ) + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

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)