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

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

[Out]

x*(a+b*arccos(d*x^2+1))^(5/2)-30*cos(1/2*a/b)*FresnelS((1/b)^(1/2)*(a+b*arccos(d*x^2+1))^(1/2)/Pi^(1/2))*sin(1
/2*arccos(d*x^2+1))*Pi^(1/2)/(1/b)^(5/2)/d/x+30*FresnelC((1/b)^(1/2)*(a+b*arccos(d*x^2+1))^(1/2)/Pi^(1/2))*sin
(1/2*a/b)*sin(1/2*arccos(d*x^2+1))*Pi^(1/2)/(1/b)^(5/2)/d/x-5*b*(a+b*arccos(d*x^2+1))^(3/2)*(-d^2*x^4-2*d*x^2)
^(1/2)/d/x+30*b^2*sin(1/2*arccos(d*x^2+1))^2*(a+b*arccos(d*x^2+1))^(1/2)/d/x

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Rubi [A]  time = 0.09, antiderivative size = 249, 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, 4812} \[ \frac {30 b^2 \sin ^2\left (\frac {1}{2} \cos ^{-1}\left (d x^2+1\right )\right ) \sqrt {a+b \cos ^{-1}\left (d x^2+1\right )}}{d x}-\frac {5 b \sqrt {-d^2 x^4-2 d x^2} \left (a+b \cos ^{-1}\left (d x^2+1\right )\right )^{3/2}}{d x}+\frac {30 \sqrt {\pi } \sin \left (\frac {a}{2 b}\right ) \sin \left (\frac {1}{2} \cos ^{-1}\left (d x^2+1\right )\right ) \text {FresnelC}\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \cos ^{-1}\left (d x^2+1\right )}}{\sqrt {\pi }}\right )}{\left (\frac {1}{b}\right )^{5/2} d x}-\frac {30 \sqrt {\pi } \cos \left (\frac {a}{2 b}\right ) \sin \left (\frac {1}{2} \cos ^{-1}\left (d x^2+1\right )\right ) S\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \cos ^{-1}\left (d x^2+1\right )}}{\sqrt {\pi }}\right )}{\left (\frac {1}{b}\right )^{5/2} d x}+x \left (a+b \cos ^{-1}\left (d x^2+1\right )\right )^{5/2} \]

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

Rule 4812

Int[Sqrt[(a_.) + ArcCos[1 + (d_.)*(x_)^2]*(b_.)], x_Symbol] :> Simp[(-2*Sqrt[a + b*ArcCos[1 + d*x^2]]*Sin[ArcC
os[1 + d*x^2]/2]^2)/(d*x), x] + (-Simp[(2*Sqrt[Pi]*Sin[a/(2*b)]*Sin[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]*Cos[a/(2*b)]*Sin[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]

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]

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 \sqrt {\pi } \cos \left (\frac {a}{2 b}\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 {1}{2} \cos ^{-1}\left (1+d x^2\right )\right )}{\left (\frac {1}{b}\right )^{5/2} d x}+\frac {30 \sqrt {\pi } 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 ) \sin \left (\frac {1}{2} \cos ^{-1}\left (1+d x^2\right )\right )}{\left (\frac {1}{b}\right )^{5/2} d x}+\frac {30 b^2 \sqrt {a+b \cos ^{-1}\left (1+d x^2\right )} \sin ^2\left (\frac {1}{2} \cos ^{-1}\left (1+d x^2\right )\right )}{d x}\\ \end {align*}

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Mathematica [A]  time = 2.75, size = 256, normalized size = 1.03 \[ -\frac {2 \sin \left (\frac {1}{2} \cos ^{-1}\left (d x^2+1\right )\right ) \left (\sqrt {a+b \cos ^{-1}\left (d x^2+1\right )} \left (\left (a^2-15 b^2\right ) \sin \left (\frac {1}{2} \cos ^{-1}\left (d x^2+1\right )\right )+5 a b \cos \left (\frac {1}{2} \cos ^{-1}\left (d x^2+1\right )\right )+b \cos ^{-1}\left (d x^2+1\right ) \left (2 a \sin \left (\frac {1}{2} \cos ^{-1}\left (d x^2+1\right )\right )+5 b \cos \left (\frac {1}{2} \cos ^{-1}\left (d x^2+1\right )\right )\right )+b^2 \cos ^{-1}\left (d x^2+1\right )^2 \sin \left (\frac {1}{2} \cos ^{-1}\left (d x^2+1\right )\right )\right )-\frac {15 \sqrt {\pi } \sin \left (\frac {a}{2 b}\right ) C\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \cos ^{-1}\left (d x^2+1\right )}}{\sqrt {\pi }}\right )}{\left (\frac {1}{b}\right )^{5/2}}+\frac {15 \sqrt {\pi } \cos \left (\frac {a}{2 b}\right ) S\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \cos ^{-1}\left (d x^2+1\right )}}{\sqrt {\pi }}\right )}{\left (\frac {1}{b}\right )^{5/2}}\right )}{d x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sin[ArcCos[1 + d*x^2]/2]*((15*Sqrt[Pi]*Cos[a/(2*b)]*FresnelS[(Sqrt[b^(-1)]*Sqrt[a + b*ArcCos[1 + d*x^2]])/
Sqrt[Pi]])/(b^(-1))^(5/2) - (15*Sqrt[Pi]*FresnelC[(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]]*(5*a*b*Cos[ArcCos[1 + d*x^2]/2] + (a^2 - 15*b^2)*Sin[A
rcCos[1 + d*x^2]/2] + b^2*ArcCos[1 + d*x^2]^2*Sin[ArcCos[1 + d*x^2]/2] + b*ArcCos[1 + d*x^2]*(5*b*Cos[ArcCos[1
+ d*x^2]/2] + 2*a*Sin[ArcCos[1 + d*x^2]/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))^(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 \[ \int {\left (b \arccos \left (d x^{2} + 1\right ) + a\right )}^{\frac {5}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.09, size = 0, normalized size = 0.00 \[ \int \left (a +b \arccos \left (d \,x^{2}+1\right )\right )^{\frac {5}{2}}\, dx \]

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(-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((a+b*arccos(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 \[ \int {\left (a+b\,\mathrm {acos}\left (d\,x^2+1\right )\right )}^{5/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*acos(d*x^2 + 1))^(5/2),x)

[Out]

int((a + b*acos(d*x^2 + 1))^(5/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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