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

Optimal. Leaf size=277 \[ -15 b^2 x \sqrt{a+b \sin ^{-1}\left (d x^2+1\right )}+\frac{5 b \sqrt{-d^2 x^4-2 d x^2} \left (a+b \sin ^{-1}\left (d x^2+1\right )\right )^{3/2}}{d x}+\frac{15 \sqrt{\pi } x \left (\sin \left (\frac{a}{2 b}\right )+\cos \left (\frac{a}{2 b}\right )\right ) \text{FresnelC}\left (\frac{\sqrt{\frac{1}{b}} \sqrt{a+b \sin ^{-1}\left (d x^2+1\right )}}{\sqrt{\pi }}\right )}{\left (\frac{1}{b}\right )^{5/2} \left (\cos \left (\frac{1}{2} \sin ^{-1}\left (d x^2+1\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (d x^2+1\right )\right )\right )}-\frac{15 \sqrt{\pi } x \left (\cos \left (\frac{a}{2 b}\right )-\sin \left (\frac{a}{2 b}\right )\right ) S\left (\frac{\sqrt{\frac{1}{b}} \sqrt{a+b \sin ^{-1}\left (d x^2+1\right )}}{\sqrt{\pi }}\right )}{\left (\frac{1}{b}\right )^{5/2} \left (\cos \left (\frac{1}{2} \sin ^{-1}\left (d x^2+1\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (d x^2+1\right )\right )\right )}+x \left (a+b \sin ^{-1}\left (d x^2+1\right )\right )^{5/2} \]

[Out]

-15*b^2*x*Sqrt[a + b*ArcSin[1 + d*x^2]] + (5*b*Sqrt[-2*d*x^2 - d^2*x^4]*(a + b*ArcSin[1 + d*x^2])^(3/2))/(d*x)
+ x*(a + b*ArcSin[1 + d*x^2])^(5/2) - (15*Sqrt[Pi]*x*FresnelS[(Sqrt[b^(-1)]*Sqrt[a + b*ArcSin[1 + d*x^2]])/Sq
rt[Pi]]*(Cos[a/(2*b)] - Sin[a/(2*b)]))/((b^(-1))^(5/2)*(Cos[ArcSin[1 + d*x^2]/2] - Sin[ArcSin[1 + d*x^2]/2]))
+ (15*Sqrt[Pi]*x*FresnelC[(Sqrt[b^(-1)]*Sqrt[a + b*ArcSin[1 + d*x^2]])/Sqrt[Pi]]*(Cos[a/(2*b)] + Sin[a/(2*b)])
)/((b^(-1))^(5/2)*(Cos[ArcSin[1 + d*x^2]/2] - Sin[ArcSin[1 + d*x^2]/2]))

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Rubi [A]  time = 0.104452, antiderivative size = 277, 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 = {4814, 4811} \[ -15 b^2 x \sqrt{a+b \sin ^{-1}\left (d x^2+1\right )}+\frac{5 b \sqrt{-d^2 x^4-2 d x^2} \left (a+b \sin ^{-1}\left (d x^2+1\right )\right )^{3/2}}{d x}+\frac{15 \sqrt{\pi } x \left (\sin \left (\frac{a}{2 b}\right )+\cos \left (\frac{a}{2 b}\right )\right ) \text{FresnelC}\left (\frac{\sqrt{\frac{1}{b}} \sqrt{a+b \sin ^{-1}\left (d x^2+1\right )}}{\sqrt{\pi }}\right )}{\left (\frac{1}{b}\right )^{5/2} \left (\cos \left (\frac{1}{2} \sin ^{-1}\left (d x^2+1\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (d x^2+1\right )\right )\right )}-\frac{15 \sqrt{\pi } x \left (\cos \left (\frac{a}{2 b}\right )-\sin \left (\frac{a}{2 b}\right )\right ) S\left (\frac{\sqrt{\frac{1}{b}} \sqrt{a+b \sin ^{-1}\left (d x^2+1\right )}}{\sqrt{\pi }}\right )}{\left (\frac{1}{b}\right )^{5/2} \left (\cos \left (\frac{1}{2} \sin ^{-1}\left (d x^2+1\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (d x^2+1\right )\right )\right )}+x \left (a+b \sin ^{-1}\left (d x^2+1\right )\right )^{5/2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-15*b^2*x*Sqrt[a + b*ArcSin[1 + d*x^2]] + (5*b*Sqrt[-2*d*x^2 - d^2*x^4]*(a + b*ArcSin[1 + d*x^2])^(3/2))/(d*x)
+ x*(a + b*ArcSin[1 + d*x^2])^(5/2) - (15*Sqrt[Pi]*x*FresnelS[(Sqrt[b^(-1)]*Sqrt[a + b*ArcSin[1 + d*x^2]])/Sq
rt[Pi]]*(Cos[a/(2*b)] - Sin[a/(2*b)]))/((b^(-1))^(5/2)*(Cos[ArcSin[1 + d*x^2]/2] - Sin[ArcSin[1 + d*x^2]/2]))
+ (15*Sqrt[Pi]*x*FresnelC[(Sqrt[b^(-1)]*Sqrt[a + b*ArcSin[1 + d*x^2]])/Sqrt[Pi]]*(Cos[a/(2*b)] + Sin[a/(2*b)])
)/((b^(-1))^(5/2)*(Cos[ArcSin[1 + d*x^2]/2] - Sin[ArcSin[1 + d*x^2]/2]))

Rule 4814

Int[((a_.) + ArcSin[(c_) + (d_.)*(x_)^2]*(b_.))^(n_), x_Symbol] :> Simp[x*(a + b*ArcSin[c + d*x^2])^n, x] + (-
Dist[4*b^2*n*(n - 1), Int[(a + b*ArcSin[c + d*x^2])^(n - 2), x], x] + Simp[(2*b*n*Sqrt[-2*c*d*x^2 - d^2*x^4]*(
a + b*ArcSin[c + d*x^2])^(n - 1))/(d*x), x]) /; FreeQ[{a, b, c, d}, x] && EqQ[c^2, 1] && GtQ[n, 1]

Rule 4811

Int[Sqrt[(a_.) + ArcSin[(c_) + (d_.)*(x_)^2]*(b_.)], x_Symbol] :> Simp[x*Sqrt[a + b*ArcSin[c + d*x^2]], x] + (
-Simp[(Sqrt[Pi]*x*(Cos[a/(2*b)] + c*Sin[a/(2*b)])*FresnelC[Sqrt[c/(Pi*b)]*Sqrt[a + b*ArcSin[c + d*x^2]]])/(Sqr
t[c/b]*(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[Sqrt[c/(Pi*b)]*Sqrt[a + b*ArcSin[c + d*x^2]]])/(Sqrt[c/b]*(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 \left (a+b \sin ^{-1}\left (1+d x^2\right )\right )^{5/2} \, dx &=\frac{5 b \sqrt{-2 d x^2-d^2 x^4} \left (a+b \sin ^{-1}\left (1+d x^2\right )\right )^{3/2}}{d x}+x \left (a+b \sin ^{-1}\left (1+d x^2\right )\right )^{5/2}-\left (15 b^2\right ) \int \sqrt{a+b \sin ^{-1}\left (1+d x^2\right )} \, dx\\ &=-15 b^2 x \sqrt{a+b \sin ^{-1}\left (1+d x^2\right )}+\frac{5 b \sqrt{-2 d x^2-d^2 x^4} \left (a+b \sin ^{-1}\left (1+d x^2\right )\right )^{3/2}}{d x}+x \left (a+b \sin ^{-1}\left (1+d x^2\right )\right )^{5/2}-\frac{15 \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 )}{\left (\frac{1}{b}\right )^{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{15 \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 )}{\left (\frac{1}{b}\right )^{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 = 0.26077, size = 269, normalized size = 0.97 \[ -\frac{15 x \left (-\sqrt{\pi } \left (\sin \left (\frac{a}{2 b}\right )+\cos \left (\frac{a}{2 b}\right )\right ) \text{FresnelC}\left (\frac{\sqrt{\frac{1}{b}} \sqrt{a+b \sin ^{-1}\left (d x^2+1\right )}}{\sqrt{\pi }}\right )+\sqrt{\pi } \left (\cos \left (\frac{a}{2 b}\right )-\sin \left (\frac{a}{2 b}\right )\right ) S\left (\frac{\sqrt{\frac{1}{b}} \sqrt{a+b \sin ^{-1}\left (d x^2+1\right )}}{\sqrt{\pi }}\right )+\sqrt{\frac{1}{b}} \left (\cos \left (\frac{1}{2} \sin ^{-1}\left (d x^2+1\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (d x^2+1\right )\right )\right ) \sqrt{a+b \sin ^{-1}\left (d x^2+1\right )}\right )}{\left (\frac{1}{b}\right )^{5/2} \left (\cos \left (\frac{1}{2} \sin ^{-1}\left (d x^2+1\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (d x^2+1\right )\right )\right )}+x \left (a+b \sin ^{-1}\left (d x^2+1\right )\right )^{5/2}+\frac{5 b \sqrt{-d x^2 \left (d x^2+2\right )} \left (a+b \sin ^{-1}\left (d x^2+1\right )\right )^{3/2}}{d x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.069, 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((a+b*arcsin(d*x^2+1))^(5/2),x)

[Out]

int((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{\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((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((a+b*arcsin(d*x^2+1))^(5/2),x, algorithm="fricas")

[Out]

Exception raised: UnboundLocalError

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\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((a+b*arcsin(d*x^2+1))^(5/2),x, algorithm="giac")

[Out]

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