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

Optimal. Leaf size=281 \[ -\frac {15 \sqrt {\frac {\pi }{2}} b^{5/2} \cosh \left (\frac {1}{2} \cosh ^{-1}\left (d x^2-1\right )\right ) \left (\sinh \left (\frac {a}{2 b}\right )+\cosh \left (\frac {a}{2 b}\right )\right ) \text {erf}\left (\frac {\sqrt {a+b \cosh ^{-1}\left (d x^2-1\right )}}{\sqrt {2} \sqrt {b}}\right )}{d x}-\frac {15 \sqrt {\frac {\pi }{2}} b^{5/2} \cosh \left (\frac {1}{2} \cosh ^{-1}\left (d x^2-1\right )\right ) \left (\cosh \left (\frac {a}{2 b}\right )-\sinh \left (\frac {a}{2 b}\right )\right ) \text {erfi}\left (\frac {\sqrt {a+b \cosh ^{-1}\left (d x^2-1\right )}}{\sqrt {2} \sqrt {b}}\right )}{d x}+\frac {30 b^2 \cosh ^2\left (\frac {1}{2} \cosh ^{-1}\left (d x^2-1\right )\right ) \sqrt {a+b \cosh ^{-1}\left (d x^2-1\right )}}{d x}+x \left (a+b \cosh ^{-1}\left (d x^2-1\right )\right )^{5/2}+\frac {5 b \left (2 x^2-d x^4\right ) \left (a+b \cosh ^{-1}\left (d x^2-1\right )\right )^{3/2}}{x \sqrt {d x^2} \sqrt {d x^2-2}} \]

[Out]

x*(a+b*arccosh(d*x^2-1))^(5/2)-15/2*b^(5/2)*cosh(1/2*arccosh(d*x^2-1))*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))*2^(1/2)*Pi^(1/2)/d/x-15/2*b^(5/2)*cosh(1/2*arccosh(d*x^2-1))*erf
(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))*2^(1/2)*Pi^(1/2)/d/x+5*b*(-d*
x^4+2*x^2)*(a+b*arccosh(d*x^2-1))^(3/2)/x/(d*x^2)^(1/2)/(d*x^2-2)^(1/2)+30*b^2*cosh(1/2*arccosh(d*x^2-1))^2*(a
+b*arccosh(d*x^2-1))^(1/2)/d/x

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Rubi [A]  time = 0.06, antiderivative size = 281, 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 = {5880, 5879} \[ -\frac {15 \sqrt {\frac {\pi }{2}} b^{5/2} \cosh \left (\frac {1}{2} \cosh ^{-1}\left (d x^2-1\right )\right ) \left (\sinh \left (\frac {a}{2 b}\right )+\cosh \left (\frac {a}{2 b}\right )\right ) \text {Erf}\left (\frac {\sqrt {a+b \cosh ^{-1}\left (d x^2-1\right )}}{\sqrt {2} \sqrt {b}}\right )}{d x}-\frac {15 \sqrt {\frac {\pi }{2}} b^{5/2} \cosh \left (\frac {1}{2} \cosh ^{-1}\left (d x^2-1\right )\right ) \left (\cosh \left (\frac {a}{2 b}\right )-\sinh \left (\frac {a}{2 b}\right )\right ) \text {Erfi}\left (\frac {\sqrt {a+b \cosh ^{-1}\left (d x^2-1\right )}}{\sqrt {2} \sqrt {b}}\right )}{d x}+\frac {30 b^2 \cosh ^2\left (\frac {1}{2} \cosh ^{-1}\left (d x^2-1\right )\right ) \sqrt {a+b \cosh ^{-1}\left (d x^2-1\right )}}{d x}+\frac {5 b \left (2 x^2-d x^4\right ) \left (a+b \cosh ^{-1}\left (d x^2-1\right )\right )^{3/2}}{x \sqrt {d x^2} \sqrt {d x^2-2}}+x \left (a+b \cosh ^{-1}\left (d x^2-1\right )\right )^{5/2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*b*(2*x^2 - d*x^4)*(a + b*ArcCosh[-1 + d*x^2])^(3/2))/(x*Sqrt[d*x^2]*Sqrt[-2 + d*x^2]) + x*(a + b*ArcCosh[-1
+ d*x^2])^(5/2) + (30*b^2*Sqrt[a + b*ArcCosh[-1 + d*x^2]]*Cosh[ArcCosh[-1 + d*x^2]/2]^2)/(d*x) - (15*b^(5/2)*
Sqrt[Pi/2]*Cosh[ArcCosh[-1 + d*x^2]/2]*Erfi[Sqrt[a + b*ArcCosh[-1 + d*x^2]]/(Sqrt[2]*Sqrt[b])]*(Cosh[a/(2*b)]
- Sinh[a/(2*b)]))/(d*x) - (15*b^(5/2)*Sqrt[Pi/2]*Cosh[ArcCosh[-1 + d*x^2]/2]*Erf[Sqrt[a + b*ArcCosh[-1 + d*x^2
]]/(Sqrt[2]*Sqrt[b])]*(Cosh[a/(2*b)] + Sinh[a/(2*b)]))/(d*x)

Rule 5879

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

Rule 5880

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

Rubi steps

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

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Mathematica [A]  time = 1.68, size = 277, normalized size = 0.99 \[ \frac {\cosh \left (\frac {1}{2} \cosh ^{-1}\left (d x^2-1\right )\right ) \left (4 \sqrt {a+b \cosh ^{-1}\left (d x^2-1\right )} \left (\left (a^2+15 b^2\right ) \cosh \left (\frac {1}{2} \cosh ^{-1}\left (d x^2-1\right )\right )+b \cosh ^{-1}\left (d x^2-1\right ) \left (2 a \cosh \left (\frac {1}{2} \cosh ^{-1}\left (d x^2-1\right )\right )-5 b \sinh \left (\frac {1}{2} \cosh ^{-1}\left (d x^2-1\right )\right )\right )-5 a b \sinh \left (\frac {1}{2} \cosh ^{-1}\left (d x^2-1\right )\right )+b^2 \cosh \left (\frac {1}{2} \cosh ^{-1}\left (d x^2-1\right )\right ) \cosh ^{-1}\left (d x^2-1\right )^2\right )-15 \sqrt {2 \pi } b^{5/2} \left (\sinh \left (\frac {a}{2 b}\right )+\cosh \left (\frac {a}{2 b}\right )\right ) \text {erf}\left (\frac {\sqrt {a+b \cosh ^{-1}\left (d x^2-1\right )}}{\sqrt {2} \sqrt {b}}\right )-15 \sqrt {2 \pi } b^{5/2} \left (\cosh \left (\frac {a}{2 b}\right )-\sinh \left (\frac {a}{2 b}\right )\right ) \text {erfi}\left (\frac {\sqrt {a+b \cosh ^{-1}\left (d x^2-1\right )}}{\sqrt {2} \sqrt {b}}\right )\right )}{2 d x} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Cosh[ArcCosh[-1 + d*x^2]/2]*(-15*b^(5/2)*Sqrt[2*Pi]*Erfi[Sqrt[a + b*ArcCosh[-1 + d*x^2]]/(Sqrt[2]*Sqrt[b])]*(
Cosh[a/(2*b)] - Sinh[a/(2*b)]) - 15*b^(5/2)*Sqrt[2*Pi]*Erf[Sqrt[a + b*ArcCosh[-1 + d*x^2]]/(Sqrt[2]*Sqrt[b])]*
(Cosh[a/(2*b)] + Sinh[a/(2*b)]) + 4*Sqrt[a + b*ArcCosh[-1 + d*x^2]]*((a^2 + 15*b^2)*Cosh[ArcCosh[-1 + d*x^2]/2
] + b^2*ArcCosh[-1 + d*x^2]^2*Cosh[ArcCosh[-1 + d*x^2]/2] - 5*a*b*Sinh[ArcCosh[-1 + d*x^2]/2] + b*ArcCosh[-1 +
d*x^2]*(2*a*Cosh[ArcCosh[-1 + d*x^2]/2] - 5*b*Sinh[ArcCosh[-1 + d*x^2]/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*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((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(x)]Warning, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):
Check [abs(t_nostep)]index.cc index_m i_lex_is_greater Error: Bad Argument ValuePolynomial exponent overflow.
Error: Bad Argument Value

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((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 \[ \int {\left (b \operatorname {arcosh}\left (d x^{2} - 1\right ) + a\right )}^{\frac {5}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*acosh(d*x^2 - 1))^(5/2),x)

[Out]

int((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 \[ \int \left (a + b \operatorname {acosh}{\left (d x^{2} - 1 \right )}\right )^{\frac {5}{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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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