3.2.90 \(\int \frac {1}{(a+b \cosh ^{-1}(c+d x))^{5/2}} \, dx\) [190]

Optimal. Leaf size=165 \[ -\frac {2 \sqrt {-1+c+d x} \sqrt {1+c+d x}}{3 b d \left (a+b \cosh ^{-1}(c+d x)\right )^{3/2}}-\frac {4 (c+d x)}{3 b^2 d \sqrt {a+b \cosh ^{-1}(c+d x)}}-\frac {2 e^{a/b} \sqrt {\pi } \text {Erf}\left (\frac {\sqrt {a+b \cosh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{3 b^{5/2} d}+\frac {2 e^{-\frac {a}{b}} \sqrt {\pi } \text {Erfi}\left (\frac {\sqrt {a+b \cosh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{3 b^{5/2} d} \]

[Out]

-2/3*exp(a/b)*erf((a+b*arccosh(d*x+c))^(1/2)/b^(1/2))*Pi^(1/2)/b^(5/2)/d+2/3*erfi((a+b*arccosh(d*x+c))^(1/2)/b
^(1/2))*Pi^(1/2)/b^(5/2)/d/exp(a/b)-2/3*(d*x+c-1)^(1/2)*(d*x+c+1)^(1/2)/b/d/(a+b*arccosh(d*x+c))^(3/2)-4/3*(d*
x+c)/b^2/d/(a+b*arccosh(d*x+c))^(1/2)

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Rubi [A]
time = 0.26, antiderivative size = 165, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {5995, 5880, 5951, 5881, 3389, 2211, 2236, 2235} \begin {gather*} -\frac {2 \sqrt {\pi } e^{a/b} \text {Erf}\left (\frac {\sqrt {a+b \cosh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{3 b^{5/2} d}+\frac {2 \sqrt {\pi } e^{-\frac {a}{b}} \text {Erfi}\left (\frac {\sqrt {a+b \cosh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{3 b^{5/2} d}-\frac {4 (c+d x)}{3 b^2 d \sqrt {a+b \cosh ^{-1}(c+d x)}}-\frac {2 \sqrt {c+d x-1} \sqrt {c+d x+1}}{3 b d \left (a+b \cosh ^{-1}(c+d x)\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2211

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Dist[2/d, Subst[Int[F^(g*(e - c*(
f/d)) + f*g*(x^2/d)), x], x, Sqrt[c + d*x]], x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !TrueQ[$UseGamma]

Rule 2235

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[F^a*Sqrt[Pi]*(Erfi[(c + d*x)*Rt[b*Log[F], 2
]]/(2*d*Rt[b*Log[F], 2])), x] /; FreeQ[{F, a, b, c, d}, x] && PosQ[b]

Rule 2236

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[F^a*Sqrt[Pi]*(Erf[(c + d*x)*Rt[(-b)*Log[F],
2]]/(2*d*Rt[(-b)*Log[F], 2])), x] /; FreeQ[{F, a, b, c, d}, x] && NegQ[b]

Rule 3389

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Dist[I/2, Int[(c + d*x)^m/E^(I*(e + f*x))
, x], x] - Dist[I/2, Int[(c + d*x)^m*E^(I*(e + f*x)), x], x] /; FreeQ[{c, d, e, f, m}, x]

Rule 5880

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

Rule 5881

Int[((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[1/(b*c), Subst[Int[x^n*Sinh[-a/b + x/b], x], x
, a + b*ArcCosh[c*x]], x] /; FreeQ[{a, b, c, n}, x]

Rule 5951

Int[(((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))^(n_)*((f_.)*(x_))^(m_.))/(Sqrt[(d1_) + (e1_.)*(x_)]*Sqrt[(d2_) + (e2_
.)*(x_)]), x_Symbol] :> Simp[(f*x)^m*((a + b*ArcCosh[c*x])^(n + 1)/(b*c*(n + 1)))*Simp[Sqrt[1 + c*x]/Sqrt[d1 +
e1*x]]*Simp[Sqrt[-1 + c*x]/Sqrt[d2 + e2*x]], x] - Dist[f*(m/(b*c*(n + 1)))*Simp[Sqrt[1 + c*x]/Sqrt[d1 + e1*x]
]*Simp[Sqrt[-1 + c*x]/Sqrt[d2 + e2*x]], Int[(f*x)^(m - 1)*(a + b*ArcCosh[c*x])^(n + 1), x], x] /; FreeQ[{a, b,
c, d1, e1, d2, e2, f, m}, x] && EqQ[e1, c*d1] && EqQ[e2, (-c)*d2] && LtQ[n, -1]

Rule 5995

Int[((a_.) + ArcCosh[(c_) + (d_.)*(x_)]*(b_.))^(n_.), x_Symbol] :> Dist[1/d, Subst[Int[(a + b*ArcCosh[x])^n, x
], x, c + d*x], x] /; FreeQ[{a, b, c, d, n}, x]

Rubi steps

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

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Mathematica [A]
time = 1.13, size = 219, normalized size = 1.33 \begin {gather*} \frac {e^{-\frac {a+b \cosh ^{-1}(c+d x)}{b}} \left (2 e^{\frac {2 a}{b}+\cosh ^{-1}(c+d x)} \sqrt {\frac {a}{b}+\cosh ^{-1}(c+d x)} \left (a+b \cosh ^{-1}(c+d x)\right ) \Gamma \left (\frac {1}{2},\frac {a}{b}+\cosh ^{-1}(c+d x)\right )-2 \left (e^{a/b} \left (b e^{\cosh ^{-1}(c+d x)} \sqrt {\frac {-1+c+d x}{1+c+d x}} (1+c+d x)+\left (1+e^{2 \cosh ^{-1}(c+d x)}\right ) \left (a+b \cosh ^{-1}(c+d x)\right )\right )+b e^{\cosh ^{-1}(c+d x)} \left (-\frac {a+b \cosh ^{-1}(c+d x)}{b}\right )^{3/2} \Gamma \left (\frac {1}{2},-\frac {a+b \cosh ^{-1}(c+d x)}{b}\right )\right )\right )}{3 b^2 d \left (a+b \cosh ^{-1}(c+d x)\right )^{3/2}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2*E^((2*a)/b + ArcCosh[c + d*x])*Sqrt[a/b + ArcCosh[c + d*x]]*(a + b*ArcCosh[c + d*x])*Gamma[1/2, a/b + ArcCo
sh[c + d*x]] - 2*(E^(a/b)*(b*E^ArcCosh[c + d*x]*Sqrt[(-1 + c + d*x)/(1 + c + d*x)]*(1 + c + d*x) + (1 + E^(2*A
rcCosh[c + d*x]))*(a + b*ArcCosh[c + d*x])) + b*E^ArcCosh[c + d*x]*(-((a + b*ArcCosh[c + d*x])/b))^(3/2)*Gamma
[1/2, -((a + b*ArcCosh[c + d*x])/b)]))/(3*b^2*d*E^((a + b*ArcCosh[c + d*x])/b)*(a + b*ArcCosh[c + d*x])^(3/2))

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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {1}{\left (a +b \,\mathrm {arccosh}\left (d x +c \right )\right )^{\frac {5}{2}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (co
nstant residues)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{{\left (a+b\,\mathrm {acosh}\left (c+d\,x\right )\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*acosh(c + d*x))^(5/2),x)

[Out]

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

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