\(\int (a+b \text {arccosh}(1+d x^2))^{5/2} \, dx\) [254]

Optimal result

Integrand size = 16, antiderivative size = 280 \[ \int \left (a+b \text {arccosh}\left (1+d x^2\right )\right )^{5/2} \, dx=-\frac {5 b \left (2 x^2+d x^4\right ) \left (a+b \text {arccosh}\left (1+d x^2\right )\right )^{3/2}}{x \sqrt {d x^2} \sqrt {2+d x^2}}+x \left (a+b \text {arccosh}\left (1+d x^2\right )\right )^{5/2}-\frac {15 b^{5/2} \sqrt {\frac {\pi }{2}} \text {erfi}\left (\frac {\sqrt {a+b \text {arccosh}\left (1+d x^2\right )}}{\sqrt {2} \sqrt {b}}\right ) \left (\cosh \left (\frac {a}{2 b}\right )-\sinh \left (\frac {a}{2 b}\right )\right ) \sinh \left (\frac {1}{2} \text {arccosh}\left (1+d x^2\right )\right )}{d x}+\frac {15 b^{5/2} \sqrt {\frac {\pi }{2}} \text {erf}\left (\frac {\sqrt {a+b \text {arccosh}\left (1+d x^2\right )}}{\sqrt {2} \sqrt {b}}\right ) \left (\cosh \left (\frac {a}{2 b}\right )+\sinh \left (\frac {a}{2 b}\right )\right ) \sinh \left (\frac {1}{2} \text {arccosh}\left (1+d x^2\right )\right )}{d x}+\frac {30 b^2 \sqrt {a+b \text {arccosh}\left (1+d x^2\right )} \sinh ^2\left (\frac {1}{2} \text {arccosh}\left (1+d x^2\right )\right )}{d x} \]

[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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 280, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {6001, 5999} \[ \int \left (a+b \text {arccosh}\left (1+d x^2\right )\right )^{5/2} \, dx=\frac {15 \sqrt {\frac {\pi }{2}} b^{5/2} \left (\sinh \left (\frac {a}{2 b}\right )+\cosh \left (\frac {a}{2 b}\right )\right ) \sinh \left (\frac {1}{2} \text {arccosh}\left (d x^2+1\right )\right ) \text {erf}\left (\frac {\sqrt {a+b \text {arccosh}\left (d x^2+1\right )}}{\sqrt {2} \sqrt {b}}\right )}{d x}-\frac {15 \sqrt {\frac {\pi }{2}} b^{5/2} \left (\cosh \left (\frac {a}{2 b}\right )-\sinh \left (\frac {a}{2 b}\right )\right ) \sinh \left (\frac {1}{2} \text {arccosh}\left (d x^2+1\right )\right ) \text {erfi}\left (\frac {\sqrt {a+b \text {arccosh}\left (d x^2+1\right )}}{\sqrt {2} \sqrt {b}}\right )}{d x}+\frac {30 b^2 \sinh ^2\left (\frac {1}{2} \text {arccosh}\left (d x^2+1\right )\right ) \sqrt {a+b \text {arccosh}\left (d x^2+1\right )}}{d x}+x \left (a+b \text {arccosh}\left (d x^2+1\right )\right )^{5/2}-\frac {5 b \left (d x^4+2 x^2\right ) \left (a+b \text {arccosh}\left (d x^2+1\right )\right )^{3/2}}{x \sqrt {d x^2} \sqrt {d x^2+2}} \]

[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 5999

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)*Arc
Cosh[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)])*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 6001

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

Mathematica [A] (verified)

Time = 2.03 (sec) , antiderivative size = 311, normalized size of antiderivative = 1.11 \[ \int \left (a+b \text {arccosh}\left (1+d x^2\right )\right )^{5/2} \, dx=\frac {x \sinh \left (\frac {1}{2} \text {arccosh}\left (1+d x^2\right )\right ) \left (-15 b^{5/2} \sqrt {2 \pi } \text {erfi}\left (\frac {\sqrt {a+b \text {arccosh}\left (1+d x^2\right )}}{\sqrt {2} \sqrt {b}}\right ) \left (\cosh \left (\frac {a}{2 b}\right )-\sinh \left (\frac {a}{2 b}\right )\right )+15 b^{5/2} \sqrt {2 \pi } \text {erf}\left (\frac {\sqrt {a+b \text {arccosh}\left (1+d x^2\right )}}{\sqrt {2} \sqrt {b}}\right ) \left (\cosh \left (\frac {a}{2 b}\right )+\sinh \left (\frac {a}{2 b}\right )\right )+4 \sqrt {a+b \text {arccosh}\left (1+d x^2\right )} \left (-5 a b \cosh \left (\frac {1}{2} \text {arccosh}\left (1+d x^2\right )\right )+\left (a^2+15 b^2\right ) \sinh \left (\frac {1}{2} \text {arccosh}\left (1+d x^2\right )\right )+b^2 \text {arccosh}\left (1+d x^2\right )^2 \sinh \left (\frac {1}{2} \text {arccosh}\left (1+d x^2\right )\right )-b \text {arccosh}\left (1+d x^2\right ) \left (5 b \cosh \left (\frac {1}{2} \text {arccosh}\left (1+d x^2\right )\right )-2 a \sinh \left (\frac {1}{2} \text {arccosh}\left (1+d x^2\right )\right )\right )\right )\right )}{2 \sqrt {d x^2} \sqrt {\frac {d x^2}{2+d x^2}} \sqrt {2+d x^2}} \]

[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])

Maple [F]

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

[In]

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

[Out]

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

Fricas [F(-2)]

Exception generated. \[ \int \left (a+b \text {arccosh}\left (1+d x^2\right )\right )^{5/2} \, dx=\text {Exception raised: TypeError} \]

[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)

Sympy [F(-1)]

Timed out. \[ \int \left (a+b \text {arccosh}\left (1+d x^2\right )\right )^{5/2} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

\[ \int \left (a+b \text {arccosh}\left (1+d x^2\right )\right )^{5/2} \, dx=\int { {\left (b \operatorname {arcosh}\left (d x^{2} + 1\right ) + a\right )}^{\frac {5}{2}} \,d x } \]

[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)

Giac [F(-2)]

Exception generated. \[ \int \left (a+b \text {arccosh}\left (1+d x^2\right )\right )^{5/2} \, dx=\text {Exception raised: TypeError} \]

[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,sageVARx):;OUTP
UT:index.cc index_m operator + Error: Bad Argument Value

Mupad [F(-1)]

Timed out. \[ \int \left (a+b \text {arccosh}\left (1+d x^2\right )\right )^{5/2} \, dx=\int {\left (a+b\,\mathrm {acosh}\left (d\,x^2+1\right )\right )}^{5/2} \,d x \]

[In]

int((a + b*acosh(d*x^2 + 1))^(5/2),x)

[Out]

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