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

Optimal. Leaf size=280 \[ \frac {15 \sqrt {\frac {\pi }{2}} b^{5/2} \left (\sinh \left (\frac {a}{2 b}\right )+\cosh \left (\frac {a}{2 b}\right )\right ) \sinh \left (\frac {1}{2} \cosh ^{-1}\left (d x^2+1\right )\right ) \text {erf}\left (\frac {\sqrt {a+b \cosh ^{-1}\left (d x^2+1\right )}}{\sqrt {2} \sqrt {b}}\right )}{d x}-\frac {15 \sqrt {\frac {\pi }{2}} b^{5/2} \left (\cosh \left (\frac {a}{2 b}\right )-\sinh \left (\frac {a}{2 b}\right )\right ) \sinh \left (\frac {1}{2} \cosh ^{-1}\left (d x^2+1\right )\right ) \text {erfi}\left (\frac {\sqrt {a+b \cosh ^{-1}\left (d x^2+1\right )}}{\sqrt {2} \sqrt {b}}\right )}{d x}+\frac {30 b^2 \sinh ^2\left (\frac {1}{2} \cosh ^{-1}\left (d x^2+1\right )\right ) \sqrt {a+b \cosh ^{-1}\left (d x^2+1\right )}}{d x}+x \left (a+b \cosh ^{-1}\left (d x^2+1\right )\right )^{5/2}-\frac {5 b \left (d x^4+2 x^2\right ) \left (a+b \cosh ^{-1}\left (d x^2+1\right )\right )^{3/2}}{x \sqrt {d x^2} \sqrt {d x^2+2}} \]

[Out]

x*(a+b*arccosh(d*x^2+1))^(5/2)-15/2*b^(5/2)*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)/d/x+15/2*b^(5/2)*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))*sinh(1/2*arccosh(d*x^2+1))*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*sinh(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 = 0.11, antiderivative size = 280, 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, 5878} \[ \frac {15 \sqrt {\frac {\pi }{2}} b^{5/2} \left (\sinh \left (\frac {a}{2 b}\right )+\cosh \left (\frac {a}{2 b}\right )\right ) \sinh \left (\frac {1}{2} \cosh ^{-1}\left (d x^2+1\right )\right ) \text {Erf}\left (\frac {\sqrt {a+b \cosh ^{-1}\left (d x^2+1\right )}}{\sqrt {2} \sqrt {b}}\right )}{d x}-\frac {15 \sqrt {\frac {\pi }{2}} b^{5/2} \left (\cosh \left (\frac {a}{2 b}\right )-\sinh \left (\frac {a}{2 b}\right )\right ) \sinh \left (\frac {1}{2} \cosh ^{-1}\left (d x^2+1\right )\right ) \text {Erfi}\left (\frac {\sqrt {a+b \cosh ^{-1}\left (d x^2+1\right )}}{\sqrt {2} \sqrt {b}}\right )}{d x}+\frac {30 b^2 \sinh ^2\left (\frac {1}{2} \cosh ^{-1}\left (d x^2+1\right )\right ) \sqrt {a+b \cosh ^{-1}\left (d x^2+1\right )}}{d x}-\frac {5 b \left (d x^4+2 x^2\right ) \left (a+b \cosh ^{-1}\left (d x^2+1\right )\right )^{3/2}}{x \sqrt {d x^2} \sqrt {d x^2+2}}+x \left (a+b \cosh ^{-1}\left (d x^2+1\right )\right )^{5/2} \]

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) - (15*b^(5/2)*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])/(d*x) + (15*b^(5/2)*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])/(d*x) + (30*b^2*Sqrt[a + b*Arc
Cosh[1 + d*x^2]]*Sinh[ArcCosh[1 + d*x^2]/2]^2)/(d*x)

Rule 5878

Int[Sqrt[(a_.) + ArcCosh[1 + (d_.)*(x_)^2]*(b_.)], x_Symbol] :> Simp[(2*Sqrt[a + b*ArcCosh[1 + d*x^2]]*Sinh[(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)])*Sinh[(1/2)*Ar
cCosh[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]*(Cos
h[a/(2*b)] - Sinh[a/(2*b)])*Sinh[(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 {15 b^{5/2} \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 )}{d x}+\frac {15 b^{5/2} \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 )}{d x}+\frac {30 b^2 \sqrt {a+b \cosh ^{-1}\left (1+d x^2\right )} \sinh ^2\left (\frac {1}{2} \cosh ^{-1}\left (1+d x^2\right )\right )}{d x}\\ \end {align*}

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Mathematica [A]  time = 3.61, size = 311, normalized size = 1.11 \[ \frac {x \sinh \left (\frac {1}{2} \cosh ^{-1}\left (d x^2+1\right )\right ) \left (4 \sqrt {a+b \cosh ^{-1}\left (d x^2+1\right )} \left (\left (a^2+15 b^2\right ) \sinh \left (\frac {1}{2} \cosh ^{-1}\left (d x^2+1\right )\right )-5 a b \cosh \left (\frac {1}{2} \cosh ^{-1}\left (d x^2+1\right )\right )-b \cosh ^{-1}\left (d x^2+1\right ) \left (5 b \cosh \left (\frac {1}{2} \cosh ^{-1}\left (d x^2+1\right )\right )-2 a \sinh \left (\frac {1}{2} \cosh ^{-1}\left (d x^2+1\right )\right )\right )+b^2 \cosh ^{-1}\left (d x^2+1\right )^2 \sinh \left (\frac {1}{2} \cosh ^{-1}\left (d x^2+1\right )\right )\right )+15 \sqrt {2 \pi } b^{5/2} \left (\sinh \left (\frac {a}{2 b}\right )+\cosh \left (\frac {a}{2 b}\right )\right ) \text {erf}\left (\frac {\sqrt {a+b \cosh ^{-1}\left (d x^2+1\right )}}{\sqrt {2} \sqrt {b}}\right )-15 \sqrt {2 \pi } b^{5/2} \left (\cosh \left (\frac {a}{2 b}\right )-\sinh \left (\frac {a}{2 b}\right )\right ) \text {erfi}\left (\frac {\sqrt {a+b \cosh ^{-1}\left (d x^2+1\right )}}{\sqrt {2} \sqrt {b}}\right )\right )}{2 \sqrt {d x^2} \sqrt {\frac {d x^2}{d x^2+2}} \sqrt {d x^2+2}} \]

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]*(-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]]*(-5*a*b*Cosh[ArcCosh[1 + d*x^2]/2] + (a^2 +
15*b^2)*Sinh[ArcCosh[1 + d*x^2]/2] + b^2*ArcCosh[1 + d*x^2]^2*Sinh[ArcCosh[1 + d*x^2]/2] - b*ArcCosh[1 + d*x^2
]*(5*b*Cosh[ArcCosh[1 + d*x^2]/2] - 2*a*Sinh[ArcCosh[1 + d*x^2]/2]))))/(2*Sqrt[d*x^2]*Sqrt[(d*x^2)/(2 + d*x^2)
]*Sqrt[2 + d*x^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((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 Valueindex.cc index_m operator + Er
ror: 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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