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

Optimal. Leaf size=252 \[ \frac {\sqrt {\frac {\pi }{2}} \left (\sinh \left (\frac {a}{2 b}\right )+\cosh \left (\frac {a}{2 b}\right )\right ) \sinh \left (\frac {1}{2} \cosh ^{-1}\left (d x^2+1\right )\right ) \text {erf}\left (\frac {\sqrt {a+b \cosh ^{-1}\left (d x^2+1\right )}}{\sqrt {2} \sqrt {b}}\right )}{3 b^{5/2} d x}+\frac {\sqrt {\frac {\pi }{2}} \left (\cosh \left (\frac {a}{2 b}\right )-\sinh \left (\frac {a}{2 b}\right )\right ) \sinh \left (\frac {1}{2} \cosh ^{-1}\left (d x^2+1\right )\right ) \text {erfi}\left (\frac {\sqrt {a+b \cosh ^{-1}\left (d x^2+1\right )}}{\sqrt {2} \sqrt {b}}\right )}{3 b^{5/2} d x}-\frac {x}{3 b^2 \sqrt {a+b \cosh ^{-1}\left (d x^2+1\right )}}-\frac {d x^4+2 x^2}{3 b x \sqrt {d x^2} \sqrt {d x^2+2} \left (a+b \cosh ^{-1}\left (d x^2+1\right )\right )^{3/2}} \]

[Out]

1/6*erfi(1/2*(a+b*arccosh(d*x^2+1))^(1/2)*2^(1/2)/b^(1/2))*(cosh(1/2*a/b)-sinh(1/2*a/b))*sinh(1/2*arccosh(d*x^
2+1))*2^(1/2)*Pi^(1/2)/b^(5/2)/d/x+1/6*erf(1/2*(a+b*arccosh(d*x^2+1))^(1/2)*2^(1/2)/b^(1/2))*(cosh(1/2*a/b)+si
nh(1/2*a/b))*sinh(1/2*arccosh(d*x^2+1))*2^(1/2)*Pi^(1/2)/b^(5/2)/d/x+1/3*(-d*x^4-2*x^2)/b/x/(a+b*arccosh(d*x^2
+1))^(3/2)/(d*x^2)^(1/2)/(d*x^2+2)^(1/2)-1/3*x/b^2/(a+b*arccosh(d*x^2+1))^(1/2)

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Rubi [A]  time = 0.07, antiderivative size = 252, 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 = {5889, 5883} \[ \frac {\sqrt {\frac {\pi }{2}} \left (\sinh \left (\frac {a}{2 b}\right )+\cosh \left (\frac {a}{2 b}\right )\right ) \sinh \left (\frac {1}{2} \cosh ^{-1}\left (d x^2+1\right )\right ) \text {Erf}\left (\frac {\sqrt {a+b \cosh ^{-1}\left (d x^2+1\right )}}{\sqrt {2} \sqrt {b}}\right )}{3 b^{5/2} d x}+\frac {\sqrt {\frac {\pi }{2}} \left (\cosh \left (\frac {a}{2 b}\right )-\sinh \left (\frac {a}{2 b}\right )\right ) \sinh \left (\frac {1}{2} \cosh ^{-1}\left (d x^2+1\right )\right ) \text {Erfi}\left (\frac {\sqrt {a+b \cosh ^{-1}\left (d x^2+1\right )}}{\sqrt {2} \sqrt {b}}\right )}{3 b^{5/2} d x}-\frac {x}{3 b^2 \sqrt {a+b \cosh ^{-1}\left (d x^2+1\right )}}-\frac {d x^4+2 x^2}{3 b x \sqrt {d x^2} \sqrt {d x^2+2} \left (a+b \cosh ^{-1}\left (d x^2+1\right )\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(2*x^2 + d*x^4)/(3*b*x*Sqrt[d*x^2]*Sqrt[2 + d*x^2]*(a + b*ArcCosh[1 + d*x^2])^(3/2)) - x/(3*b^2*Sqrt[a + b*Ar
cCosh[1 + d*x^2]]) + (Sqrt[Pi/2]*Erfi[Sqrt[a + b*ArcCosh[1 + d*x^2]]/(Sqrt[2]*Sqrt[b])]*(Cosh[a/(2*b)] - Sinh[
a/(2*b)])*Sinh[ArcCosh[1 + d*x^2]/2])/(3*b^(5/2)*d*x) + (Sqrt[Pi/2]*Erf[Sqrt[a + b*ArcCosh[1 + d*x^2]]/(Sqrt[2
]*Sqrt[b])]*(Cosh[a/(2*b)] + Sinh[a/(2*b)])*Sinh[ArcCosh[1 + d*x^2]/2])/(3*b^(5/2)*d*x)

Rule 5883

Int[1/Sqrt[(a_.) + ArcCosh[1 + (d_.)*(x_)^2]*(b_.)], x_Symbol] :> Simp[(Sqrt[Pi/2]*(Cosh[a/(2*b)] - Sinh[a/(2*
b)])*Sinh[ArcCosh[1 + d*x^2]/2]*Erfi[Sqrt[a + b*ArcCosh[1 + d*x^2]]/Sqrt[2*b]])/(Sqrt[b]*d*x), x] + Simp[(Sqrt
[Pi/2]*(Cosh[a/(2*b)] + Sinh[a/(2*b)])*Sinh[ArcCosh[1 + d*x^2]/2]*Erf[Sqrt[a + b*ArcCosh[1 + d*x^2]]/Sqrt[2*b]
])/(Sqrt[b]*d*x), x] /; FreeQ[{a, b, d}, x]

Rule 5889

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

Rubi steps

\begin {align*} \int \frac {1}{\left (a+b \cosh ^{-1}\left (1+d x^2\right )\right )^{5/2}} \, dx &=-\frac {2 x^2+d x^4}{3 b x \sqrt {d x^2} \sqrt {2+d x^2} \left (a+b \cosh ^{-1}\left (1+d x^2\right )\right )^{3/2}}-\frac {x}{3 b^2 \sqrt {a+b \cosh ^{-1}\left (1+d x^2\right )}}+\frac {\int \frac {1}{\sqrt {a+b \cosh ^{-1}\left (1+d x^2\right )}} \, dx}{3 b^2}\\ &=-\frac {2 x^2+d x^4}{3 b x \sqrt {d x^2} \sqrt {2+d x^2} \left (a+b \cosh ^{-1}\left (1+d x^2\right )\right )^{3/2}}-\frac {x}{3 b^2 \sqrt {a+b \cosh ^{-1}\left (1+d x^2\right )}}+\frac {\sqrt {\frac {\pi }{2}} \text {erfi}\left (\frac {\sqrt {a+b \cosh ^{-1}\left (1+d x^2\right )}}{\sqrt {2} \sqrt {b}}\right ) \left (\cosh \left (\frac {a}{2 b}\right )-\sinh \left (\frac {a}{2 b}\right )\right ) \sinh \left (\frac {1}{2} \cosh ^{-1}\left (1+d x^2\right )\right )}{3 b^{5/2} d x}+\frac {\sqrt {\frac {\pi }{2}} \text {erf}\left (\frac {\sqrt {a+b \cosh ^{-1}\left (1+d x^2\right )}}{\sqrt {2} \sqrt {b}}\right ) \left (\cosh \left (\frac {a}{2 b}\right )+\sinh \left (\frac {a}{2 b}\right )\right ) \sinh \left (\frac {1}{2} \cosh ^{-1}\left (1+d x^2\right )\right )}{3 b^{5/2} d x}\\ \end {align*}

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Mathematica [A]  time = 1.04, size = 273, normalized size = 1.08 \[ \frac {x \sinh \left (\frac {1}{2} \cosh ^{-1}\left (d x^2+1\right )\right ) \left (\sqrt {2 \pi } \left (\sinh \left (\frac {a}{2 b}\right )+\cosh \left (\frac {a}{2 b}\right )\right ) \left (a+b \cosh ^{-1}\left (d x^2+1\right )\right )^{3/2} \text {erf}\left (\frac {\sqrt {a+b \cosh ^{-1}\left (d x^2+1\right )}}{\sqrt {2} \sqrt {b}}\right )+\sqrt {2 \pi } \left (\cosh \left (\frac {a}{2 b}\right )-\sinh \left (\frac {a}{2 b}\right )\right ) \left (a+b \cosh ^{-1}\left (d x^2+1\right )\right )^{3/2} \text {erfi}\left (\frac {\sqrt {a+b \cosh ^{-1}\left (d x^2+1\right )}}{\sqrt {2} \sqrt {b}}\right )+4 \sqrt {b} \left (-\sinh \left (\frac {1}{2} \cosh ^{-1}\left (d x^2+1\right )\right ) \left (a+b \cosh ^{-1}\left (d x^2+1\right )\right )-b \cosh \left (\frac {1}{2} \cosh ^{-1}\left (d x^2+1\right )\right )\right )\right )}{6 b^{5/2} \sqrt {d x^2} \sqrt {\frac {d x^2}{d x^2+2}} \sqrt {d x^2+2} \left (a+b \cosh ^{-1}\left (d x^2+1\right )\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*Sinh[ArcCosh[1 + d*x^2]/2]*(Sqrt[2*Pi]*(a + b*ArcCosh[1 + d*x^2])^(3/2)*Erfi[Sqrt[a + b*ArcCosh[1 + d*x^2]]
/(Sqrt[2]*Sqrt[b])]*(Cosh[a/(2*b)] - Sinh[a/(2*b)]) + Sqrt[2*Pi]*(a + b*ArcCosh[1 + d*x^2])^(3/2)*Erf[Sqrt[a +
b*ArcCosh[1 + d*x^2]]/(Sqrt[2]*Sqrt[b])]*(Cosh[a/(2*b)] + Sinh[a/(2*b)]) + 4*Sqrt[b]*(-(b*Cosh[ArcCosh[1 + d*
x^2]/2]) - (a + b*ArcCosh[1 + d*x^2])*Sinh[ArcCosh[1 + d*x^2]/2])))/(6*b^(5/2)*Sqrt[d*x^2]*Sqrt[(d*x^2)/(2 + d
*x^2)]*Sqrt[2 + d*x^2]*(a + b*ArcCosh[1 + d*x^2])^(3/2))

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [ab
s(t_nostep)]index.cc index_m i_lex_is_greater Error: Bad Argument ValueEvaluation time: 0.78index.cc index_m o
perator + Error: Bad Argument Value

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maple [F]  time = 0.07, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a +b \,\mathrm {arccosh}\left (d \,x^{2}+1\right )\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*arccosh(d*x^2+1))^(5/2),x)

[Out]

int(1/(a+b*arccosh(d*x^2+1))^(5/2),x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {1}{{\left (a+b\,\mathrm {acosh}\left (d\,x^2+1\right )\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*acosh(d*x^2 + 1))^(5/2),x)

[Out]

int(1/(a + b*acosh(d*x^2 + 1))^(5/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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