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

Optimal. Leaf size=249 \[ -\frac{30 b^2 \cos ^2\left (\frac{1}{2} \cos ^{-1}\left (d x^2-1\right )\right ) \sqrt{a+b \cos ^{-1}\left (d x^2-1\right )}}{d x}-\frac{5 b \sqrt{2 d x^2-d^2 x^4} \left (a+b \cos ^{-1}\left (d x^2-1\right )\right )^{3/2}}{d x}+\frac{30 \sqrt{\pi } \cos \left (\frac{a}{2 b}\right ) \cos \left (\frac{1}{2} \cos ^{-1}\left (d x^2-1\right )\right ) \text{FresnelC}\left (\frac{\sqrt{\frac{1}{b}} \sqrt{a+b \cos ^{-1}\left (d x^2-1\right )}}{\sqrt{\pi }}\right )}{\left (\frac{1}{b}\right )^{5/2} d x}+\frac{30 \sqrt{\pi } \sin \left (\frac{a}{2 b}\right ) \cos \left (\frac{1}{2} \cos ^{-1}\left (d x^2-1\right )\right ) S\left (\frac{\sqrt{\frac{1}{b}} \sqrt{a+b \cos ^{-1}\left (d x^2-1\right )}}{\sqrt{\pi }}\right )}{\left (\frac{1}{b}\right )^{5/2} d x}+x \left (a+b \cos ^{-1}\left (d x^2-1\right )\right )^{5/2} \]

[Out]

(-5*b*Sqrt[2*d*x^2 - d^2*x^4]*(a + b*ArcCos[-1 + d*x^2])^(3/2))/(d*x) + x*(a + b*ArcCos[-1 + d*x^2])^(5/2) - (
30*b^2*Sqrt[a + b*ArcCos[-1 + d*x^2]]*Cos[ArcCos[-1 + d*x^2]/2]^2)/(d*x) + (30*Sqrt[Pi]*Cos[a/(2*b)]*Cos[ArcCo
s[-1 + d*x^2]/2]*FresnelC[(Sqrt[b^(-1)]*Sqrt[a + b*ArcCos[-1 + d*x^2]])/Sqrt[Pi]])/((b^(-1))^(5/2)*d*x) + (30*
Sqrt[Pi]*Cos[ArcCos[-1 + d*x^2]/2]*FresnelS[(Sqrt[b^(-1)]*Sqrt[a + b*ArcCos[-1 + d*x^2]])/Sqrt[Pi]]*Sin[a/(2*b
)])/((b^(-1))^(5/2)*d*x)

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Rubi [A]  time = 0.0552374, antiderivative size = 249, 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 = {4815, 4813} \[ -\frac{30 b^2 \cos ^2\left (\frac{1}{2} \cos ^{-1}\left (d x^2-1\right )\right ) \sqrt{a+b \cos ^{-1}\left (d x^2-1\right )}}{d x}-\frac{5 b \sqrt{2 d x^2-d^2 x^4} \left (a+b \cos ^{-1}\left (d x^2-1\right )\right )^{3/2}}{d x}+\frac{30 \sqrt{\pi } \cos \left (\frac{a}{2 b}\right ) \cos \left (\frac{1}{2} \cos ^{-1}\left (d x^2-1\right )\right ) \text{FresnelC}\left (\frac{\sqrt{\frac{1}{b}} \sqrt{a+b \cos ^{-1}\left (d x^2-1\right )}}{\sqrt{\pi }}\right )}{\left (\frac{1}{b}\right )^{5/2} d x}+\frac{30 \sqrt{\pi } \sin \left (\frac{a}{2 b}\right ) \cos \left (\frac{1}{2} \cos ^{-1}\left (d x^2-1\right )\right ) S\left (\frac{\sqrt{\frac{1}{b}} \sqrt{a+b \cos ^{-1}\left (d x^2-1\right )}}{\sqrt{\pi }}\right )}{\left (\frac{1}{b}\right )^{5/2} d x}+x \left (a+b \cos ^{-1}\left (d x^2-1\right )\right )^{5/2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*b*Sqrt[2*d*x^2 - d^2*x^4]*(a + b*ArcCos[-1 + d*x^2])^(3/2))/(d*x) + x*(a + b*ArcCos[-1 + d*x^2])^(5/2) - (
30*b^2*Sqrt[a + b*ArcCos[-1 + d*x^2]]*Cos[ArcCos[-1 + d*x^2]/2]^2)/(d*x) + (30*Sqrt[Pi]*Cos[a/(2*b)]*Cos[ArcCo
s[-1 + d*x^2]/2]*FresnelC[(Sqrt[b^(-1)]*Sqrt[a + b*ArcCos[-1 + d*x^2]])/Sqrt[Pi]])/((b^(-1))^(5/2)*d*x) + (30*
Sqrt[Pi]*Cos[ArcCos[-1 + d*x^2]/2]*FresnelS[(Sqrt[b^(-1)]*Sqrt[a + b*ArcCos[-1 + d*x^2]])/Sqrt[Pi]]*Sin[a/(2*b
)])/((b^(-1))^(5/2)*d*x)

Rule 4815

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

Rule 4813

Int[Sqrt[(a_.) + ArcCos[-1 + (d_.)*(x_)^2]*(b_.)], x_Symbol] :> Simp[(2*Sqrt[a + b*ArcCos[-1 + d*x^2]]*Cos[(1/
2)*ArcCos[-1 + d*x^2]]^2)/(d*x), x] + (-Simp[(2*Sqrt[Pi]*Cos[a/(2*b)]*Cos[ArcCos[-1 + d*x^2]/2]*FresnelC[Sqrt[
1/(Pi*b)]*Sqrt[a + b*ArcCos[-1 + d*x^2]]])/(Sqrt[1/b]*d*x), x] - Simp[(2*Sqrt[Pi]*Sin[a/(2*b)]*Cos[ArcCos[-1 +
d*x^2]/2]*FresnelS[Sqrt[1/(Pi*b)]*Sqrt[a + b*ArcCos[-1 + d*x^2]]])/(Sqrt[1/b]*d*x), x]) /; FreeQ[{a, b, d}, x
]

Rubi steps

\begin{align*} \int \left (a+b \cos ^{-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 \cos ^{-1}\left (-1+d x^2\right )\right )^{3/2}}{d x}+x \left (a+b \cos ^{-1}\left (-1+d x^2\right )\right )^{5/2}-\left (15 b^2\right ) \int \sqrt{a+b \cos ^{-1}\left (-1+d x^2\right )} \, dx\\ &=-\frac{5 b \sqrt{2 d x^2-d^2 x^4} \left (a+b \cos ^{-1}\left (-1+d x^2\right )\right )^{3/2}}{d x}+x \left (a+b \cos ^{-1}\left (-1+d x^2\right )\right )^{5/2}-\frac{30 b^2 \sqrt{a+b \cos ^{-1}\left (-1+d x^2\right )} \cos ^2\left (\frac{1}{2} \cos ^{-1}\left (-1+d x^2\right )\right )}{d x}+\frac{30 \sqrt{\pi } \cos \left (\frac{a}{2 b}\right ) \cos \left (\frac{1}{2} \cos ^{-1}\left (-1+d x^2\right )\right ) C\left (\frac{\sqrt{\frac{1}{b}} \sqrt{a+b \cos ^{-1}\left (-1+d x^2\right )}}{\sqrt{\pi }}\right )}{\left (\frac{1}{b}\right )^{5/2} d x}+\frac{30 \sqrt{\pi } \cos \left (\frac{1}{2} \cos ^{-1}\left (-1+d x^2\right )\right ) S\left (\frac{\sqrt{\frac{1}{b}} \sqrt{a+b \cos ^{-1}\left (-1+d x^2\right )}}{\sqrt{\pi }}\right ) \sin \left (\frac{a}{2 b}\right )}{\left (\frac{1}{b}\right )^{5/2} d x}\\ \end{align*}

Mathematica [A]  time = 2.14601, size = 256, normalized size = 1.03 \[ \frac{2 \cos \left (\frac{1}{2} \cos ^{-1}\left (d x^2-1\right )\right ) \left (\sqrt{a+b \cos ^{-1}\left (d x^2-1\right )} \left (\left (a^2-15 b^2\right ) \cos \left (\frac{1}{2} \cos ^{-1}\left (d x^2-1\right )\right )+b \cos ^{-1}\left (d x^2-1\right ) \left (2 a \cos \left (\frac{1}{2} \cos ^{-1}\left (d x^2-1\right )\right )-5 b \sin \left (\frac{1}{2} \cos ^{-1}\left (d x^2-1\right )\right )\right )-5 a b \sin \left (\frac{1}{2} \cos ^{-1}\left (d x^2-1\right )\right )+b^2 \cos \left (\frac{1}{2} \cos ^{-1}\left (d x^2-1\right )\right ) \cos ^{-1}\left (d x^2-1\right )^2\right )+\frac{15 \sqrt{\pi } \cos \left (\frac{a}{2 b}\right ) \text{FresnelC}\left (\frac{\sqrt{\frac{1}{b}} \sqrt{a+b \cos ^{-1}\left (d x^2-1\right )}}{\sqrt{\pi }}\right )}{\left (\frac{1}{b}\right )^{5/2}}+\frac{15 \sqrt{\pi } \sin \left (\frac{a}{2 b}\right ) S\left (\frac{\sqrt{\frac{1}{b}} \sqrt{a+b \cos ^{-1}\left (d x^2-1\right )}}{\sqrt{\pi }}\right )}{\left (\frac{1}{b}\right )^{5/2}}\right )}{d x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.066, size = 0, normalized size = 0. \begin{align*} \int \left ( a+b\arccos \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*arccos(d*x^2-1))^(5/2),x)

[Out]

int((a+b*arccos(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 \arccos \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*arccos(d*x^2-1))^(5/2),x, algorithm="maxima")

[Out]

integrate((b*arccos(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*arccos(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*acos(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 \arccos \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*arccos(d*x^2-1))^(5/2),x, algorithm="giac")

[Out]

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