3.95 \(\int (a+b \cos ^{-1}(-1+d x^2))^{3/2} \, dx\)

Optimal. Leaf size=207 \[ -\frac {3 b \sqrt {2 d x^2-d^2 x^4} \sqrt {a+b \cos ^{-1}\left (d x^2-1\right )}}{d x}-\frac {6 \sqrt {\pi } \sin \left (\frac {a}{2 b}\right ) \cos \left (\frac {1}{2} \cos ^{-1}\left (d x^2-1\right )\right ) C\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \cos ^{-1}\left (d x^2-1\right )}}{\sqrt {\pi }}\right )}{\left (\frac {1}{b}\right )^{3/2} d x}+\frac {6 \sqrt {\pi } \cos \left (\frac {a}{2 b}\right ) \cos \left (\frac {1}{2} \cos ^{-1}\left (d x^2-1\right )\right ) S\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \cos ^{-1}\left (d x^2-1\right )}}{\sqrt {\pi }}\right )}{\left (\frac {1}{b}\right )^{3/2} d x}+x \left (a+b \cos ^{-1}\left (d x^2-1\right )\right )^{3/2} \]

[Out]

x*(a+b*arccos(d*x^2-1))^(3/2)+6*cos(1/2*a/b)*cos(1/2*arccos(d*x^2-1))*FresnelS((1/b)^(1/2)*(a+b*arccos(d*x^2-1
))^(1/2)/Pi^(1/2))*Pi^(1/2)/(1/b)^(3/2)/d/x-6*cos(1/2*arccos(d*x^2-1))*FresnelC((1/b)^(1/2)*(a+b*arccos(d*x^2-
1))^(1/2)/Pi^(1/2))*sin(1/2*a/b)*Pi^(1/2)/(1/b)^(3/2)/d/x-3*b*(-d^2*x^4+2*d*x^2)^(1/2)*(a+b*arccos(d*x^2-1))^(
1/2)/d/x

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Rubi [A]  time = 0.04, antiderivative size = 207, 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 = {4815, 4821} \[ -\frac {3 b \sqrt {2 d x^2-d^2 x^4} \sqrt {a+b \cos ^{-1}\left (d x^2-1\right )}}{d x}-\frac {6 \sqrt {\pi } \sin \left (\frac {a}{2 b}\right ) \cos \left (\frac {1}{2} \cos ^{-1}\left (d x^2-1\right )\right ) \text {FresnelC}\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \cos ^{-1}\left (d x^2-1\right )}}{\sqrt {\pi }}\right )}{\left (\frac {1}{b}\right )^{3/2} d x}+\frac {6 \sqrt {\pi } \cos \left (\frac {a}{2 b}\right ) \cos \left (\frac {1}{2} \cos ^{-1}\left (d x^2-1\right )\right ) S\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \cos ^{-1}\left (d x^2-1\right )}}{\sqrt {\pi }}\right )}{\left (\frac {1}{b}\right )^{3/2} d x}+x \left (a+b \cos ^{-1}\left (d x^2-1\right )\right )^{3/2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*b*Sqrt[2*d*x^2 - d^2*x^4]*Sqrt[a + b*ArcCos[-1 + d*x^2]])/(d*x) + x*(a + b*ArcCos[-1 + d*x^2])^(3/2) + (6*
Sqrt[Pi]*Cos[a/(2*b)]*Cos[ArcCos[-1 + d*x^2]/2]*FresnelS[(Sqrt[b^(-1)]*Sqrt[a + b*ArcCos[-1 + d*x^2]])/Sqrt[Pi
]])/((b^(-1))^(3/2)*d*x) - (6*Sqrt[Pi]*Cos[ArcCos[-1 + d*x^2]/2]*FresnelC[(Sqrt[b^(-1)]*Sqrt[a + b*ArcCos[-1 +
d*x^2]])/Sqrt[Pi]]*Sin[a/(2*b)])/((b^(-1))^(3/2)*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 4821

Int[1/Sqrt[(a_.) + ArcCos[-1 + (d_.)*(x_)^2]*(b_.)], x_Symbol] :> Simp[(2*Sqrt[Pi/b]*Sin[a/(2*b)]*Cos[ArcCos[-
1 + d*x^2]/2]*FresnelC[Sqrt[1/(Pi*b)]*Sqrt[a + b*ArcCos[-1 + d*x^2]]])/(d*x), x] - Simp[(2*Sqrt[Pi/b]*Cos[a/(2
*b)]*Cos[ArcCos[-1 + d*x^2]/2]*FresnelS[Sqrt[1/(Pi*b)]*Sqrt[a + b*ArcCos[-1 + d*x^2]]])/(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 )^{3/2} \, dx &=-\frac {3 b \sqrt {2 d x^2-d^2 x^4} \sqrt {a+b \cos ^{-1}\left (-1+d x^2\right )}}{d x}+x \left (a+b \cos ^{-1}\left (-1+d x^2\right )\right )^{3/2}-\left (3 b^2\right ) \int \frac {1}{\sqrt {a+b \cos ^{-1}\left (-1+d x^2\right )}} \, dx\\ &=-\frac {3 b \sqrt {2 d x^2-d^2 x^4} \sqrt {a+b \cos ^{-1}\left (-1+d x^2\right )}}{d x}+x \left (a+b \cos ^{-1}\left (-1+d x^2\right )\right )^{3/2}+\frac {6 \sqrt {\pi } \cos \left (\frac {a}{2 b}\right ) \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 )}{\left (\frac {1}{b}\right )^{3/2} d x}-\frac {6 \sqrt {\pi } \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 ) \sin \left (\frac {a}{2 b}\right )}{\left (\frac {1}{b}\right )^{3/2} d x}\\ \end {align*}

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Mathematica [A]  time = 0.56, size = 200, normalized size = 0.97 \[ \frac {2 \cos \left (\frac {1}{2} \cos ^{-1}\left (d x^2-1\right )\right ) \left (-3 \sqrt {\pi } \sin \left (\frac {a}{2 b}\right ) C\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \cos ^{-1}\left (d x^2-1\right )}}{\sqrt {\pi }}\right )+3 \sqrt {\pi } \cos \left (\frac {a}{2 b}\right ) S\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \cos ^{-1}\left (d x^2-1\right )}}{\sqrt {\pi }}\right )+\left (\frac {1}{b}\right )^{3/2} \sqrt {a+b \cos ^{-1}\left (d x^2-1\right )} \left (a \cos \left (\frac {1}{2} \cos ^{-1}\left (d x^2-1\right )\right )+b \cos ^{-1}\left (d x^2-1\right ) \cos \left (\frac {1}{2} \cos ^{-1}\left (d x^2-1\right )\right )-3 b \sin \left (\frac {1}{2} \cos ^{-1}\left (d x^2-1\right )\right )\right )\right )}{\left (\frac {1}{b}\right )^{3/2} d x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Cos[ArcCos[-1 + d*x^2]/2]*(3*Sqrt[Pi]*Cos[a/(2*b)]*FresnelS[(Sqrt[b^(-1)]*Sqrt[a + b*ArcCos[-1 + d*x^2]])/S
qrt[Pi]] - 3*Sqrt[Pi]*FresnelC[(Sqrt[b^(-1)]*Sqrt[a + b*ArcCos[-1 + d*x^2]])/Sqrt[Pi]]*Sin[a/(2*b)] + (b^(-1))
^(3/2)*Sqrt[a + b*ArcCos[-1 + d*x^2]]*(a*Cos[ArcCos[-1 + d*x^2]/2] + b*ArcCos[-1 + d*x^2]*Cos[ArcCos[-1 + d*x^
2]/2] - 3*b*Sin[ArcCos[-1 + d*x^2]/2])))/((b^(-1))^(3/2)*d*x)

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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((a+b*arccos(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 \[ \int {\left (b \arccos \left (d x^{2} - 1\right ) + a\right )}^{\frac {3}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.08, size = 0, normalized size = 0.00 \[ \int \left (a +b \arccos \left (d \,x^{2}-1\right )\right )^{\frac {3}{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arccos(d*x^2-1))^(3/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\left (a+b\,\mathrm {acos}\left (d\,x^2-1\right )\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*acos(d*x^2 - 1))^(3/2),x)

[Out]

int((a + b*acos(d*x^2 - 1))^(3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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