3.49 \(\int \frac {\cosh ^{-1}(a x)}{(c+d x^2)^{3/2}} \, dx\)

Optimal. Leaf size=96 \[ \frac {x \cosh ^{-1}(a x)}{c \sqrt {c+d x^2}}-\frac {\sqrt {a^2 x^2-1} \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a^2 x^2-1}}{a \sqrt {c+d x^2}}\right )}{c \sqrt {d} \sqrt {a x-1} \sqrt {a x+1}} \]

[Out]

-arctanh(d^(1/2)*(a^2*x^2-1)^(1/2)/a/(d*x^2+c)^(1/2))*(a^2*x^2-1)^(1/2)/c/d^(1/2)/(a*x-1)^(1/2)/(a*x+1)^(1/2)+
x*arccosh(a*x)/c/(d*x^2+c)^(1/2)

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Rubi [A]  time = 0.19, antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 8, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {191, 5705, 12, 519, 444, 63, 217, 206} \[ \frac {x \cosh ^{-1}(a x)}{c \sqrt {c+d x^2}}-\frac {\sqrt {a^2 x^2-1} \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a^2 x^2-1}}{a \sqrt {c+d x^2}}\right )}{c \sqrt {d} \sqrt {a x-1} \sqrt {a x+1}} \]

Antiderivative was successfully verified.

[In]

Int[ArcCosh[a*x]/(c + d*x^2)^(3/2),x]

[Out]

(x*ArcCosh[a*x])/(c*Sqrt[c + d*x^2]) - (Sqrt[-1 + a^2*x^2]*ArcTanh[(Sqrt[d]*Sqrt[-1 + a^2*x^2])/(a*Sqrt[c + d*
x^2])])/(c*Sqrt[d]*Sqrt[-1 + a*x]*Sqrt[1 + a*x])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 191

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^(p + 1))/a, x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 444

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && EqQ[m
- n + 1, 0]

Rule 519

Int[(u_.)*((c_) + (d_.)*(x_)^(n_.))^(q_.)*((a1_) + (b1_.)*(x_)^(non2_.))^(p_)*((a2_) + (b2_.)*(x_)^(non2_.))^(
p_), x_Symbol] :> Dist[((a1 + b1*x^(n/2))^FracPart[p]*(a2 + b2*x^(n/2))^FracPart[p])/(a1*a2 + b1*b2*x^n)^FracP
art[p], Int[u*(a1*a2 + b1*b2*x^n)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a1, b1, a2, b2, c, d, n, p, q}, x] && EqQ[
non2, n/2] && EqQ[a2*b1 + a1*b2, 0] &&  !(EqQ[n, 2] && IGtQ[q, 0])

Rule 5705

Int[((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> With[{u = IntHide[(d + e*x^2
)^p, x]}, Dist[a + b*ArcCosh[c*x], u, x] - Dist[b*c, Int[SimplifyIntegrand[u/(Sqrt[1 + c*x]*Sqrt[-1 + c*x]), x
], x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[c^2*d + e, 0] && (IGtQ[p, 0] || ILtQ[p + 1/2, 0])

Rubi steps

\begin {align*} \int \frac {\cosh ^{-1}(a x)}{\left (c+d x^2\right )^{3/2}} \, dx &=\frac {x \cosh ^{-1}(a x)}{c \sqrt {c+d x^2}}-a \int \frac {x}{c \sqrt {-1+a x} \sqrt {1+a x} \sqrt {c+d x^2}} \, dx\\ &=\frac {x \cosh ^{-1}(a x)}{c \sqrt {c+d x^2}}-\frac {a \int \frac {x}{\sqrt {-1+a x} \sqrt {1+a x} \sqrt {c+d x^2}} \, dx}{c}\\ &=\frac {x \cosh ^{-1}(a x)}{c \sqrt {c+d x^2}}-\frac {\left (a \sqrt {-1+a^2 x^2}\right ) \int \frac {x}{\sqrt {-1+a^2 x^2} \sqrt {c+d x^2}} \, dx}{c \sqrt {-1+a x} \sqrt {1+a x}}\\ &=\frac {x \cosh ^{-1}(a x)}{c \sqrt {c+d x^2}}-\frac {\left (a \sqrt {-1+a^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {-1+a^2 x} \sqrt {c+d x}} \, dx,x,x^2\right )}{2 c \sqrt {-1+a x} \sqrt {1+a x}}\\ &=\frac {x \cosh ^{-1}(a x)}{c \sqrt {c+d x^2}}-\frac {\sqrt {-1+a^2 x^2} \operatorname {Subst}\left (\int \frac {1}{\sqrt {c+\frac {d}{a^2}+\frac {d x^2}{a^2}}} \, dx,x,\sqrt {-1+a^2 x^2}\right )}{a c \sqrt {-1+a x} \sqrt {1+a x}}\\ &=\frac {x \cosh ^{-1}(a x)}{c \sqrt {c+d x^2}}-\frac {\sqrt {-1+a^2 x^2} \operatorname {Subst}\left (\int \frac {1}{1-\frac {d x^2}{a^2}} \, dx,x,\frac {\sqrt {-1+a^2 x^2}}{\sqrt {c+d x^2}}\right )}{a c \sqrt {-1+a x} \sqrt {1+a x}}\\ &=\frac {x \cosh ^{-1}(a x)}{c \sqrt {c+d x^2}}-\frac {\sqrt {-1+a^2 x^2} \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {-1+a^2 x^2}}{a \sqrt {c+d x^2}}\right )}{c \sqrt {d} \sqrt {-1+a x} \sqrt {1+a x}}\\ \end {align*}

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Mathematica [C]  time = 3.23, size = 551, normalized size = 5.74 \[ \frac {x \cosh ^{-1}(a x)+\frac {2 (a x-1)^{3/2} \sqrt {\frac {(a x+1) \left (a \sqrt {c}-i \sqrt {d}\right )}{(a x-1) \left (a \sqrt {c}+i \sqrt {d}\right )}} \left (a \sqrt {c} \left (-a \sqrt {c}+i \sqrt {d}\right ) \sqrt {\frac {\left (a^2 c+d\right ) \left (c+d x^2\right )}{c d (a x-1)^2}} \sqrt {-\frac {a \left (x+\frac {i \sqrt {c}}{\sqrt {d}}\right )+\frac {i \sqrt {d} x}{\sqrt {c}}-1}{1-a x}} \Pi \left (\frac {2 a \sqrt {c}}{\sqrt {c} a+i \sqrt {d}};\sin ^{-1}\left (\sqrt {-\frac {\frac {i \sqrt {d} x}{\sqrt {c}}+a \left (x+\frac {i \sqrt {c}}{\sqrt {d}}\right )-1}{2-2 a x}}\right )|\frac {4 i a \sqrt {c} \sqrt {d}}{\left (\sqrt {c} a+i \sqrt {d}\right )^2}\right )+\frac {a \left (\sqrt {d}-i a \sqrt {c}\right ) \left (\sqrt {d} x+i \sqrt {c}\right ) \sqrt {\frac {\frac {i a \sqrt {c}}{\sqrt {d}}+a (-x)+\frac {i \sqrt {d} x}{\sqrt {c}}+1}{1-a x}} F\left (\sin ^{-1}\left (\sqrt {-\frac {\frac {i \sqrt {d} x}{\sqrt {c}}+a \left (x+\frac {i \sqrt {c}}{\sqrt {d}}\right )-1}{2-2 a x}}\right )|\frac {4 i a \sqrt {c} \sqrt {d}}{\left (\sqrt {c} a+i \sqrt {d}\right )^2}\right )}{a x-1}\right )}{a \sqrt {a x+1} \left (a^2 c+d\right ) \sqrt {-\frac {a \left (x+\frac {i \sqrt {c}}{\sqrt {d}}\right )+\frac {i \sqrt {d} x}{\sqrt {c}}-1}{1-a x}}}}{c \sqrt {c+d x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcCosh[a*x]/(c + d*x^2)^(3/2),x]

[Out]

(x*ArcCosh[a*x] + (2*(-1 + a*x)^(3/2)*Sqrt[((a*Sqrt[c] - I*Sqrt[d])*(1 + a*x))/((a*Sqrt[c] + I*Sqrt[d])*(-1 +
a*x))]*((a*((-I)*a*Sqrt[c] + Sqrt[d])*(I*Sqrt[c] + Sqrt[d]*x)*Sqrt[(1 + (I*a*Sqrt[c])/Sqrt[d] - a*x + (I*Sqrt[
d]*x)/Sqrt[c])/(1 - a*x)]*EllipticF[ArcSin[Sqrt[-((-1 + (I*Sqrt[d]*x)/Sqrt[c] + a*((I*Sqrt[c])/Sqrt[d] + x))/(
2 - 2*a*x))]], ((4*I)*a*Sqrt[c]*Sqrt[d])/(a*Sqrt[c] + I*Sqrt[d])^2])/(-1 + a*x) + a*Sqrt[c]*(-(a*Sqrt[c]) + I*
Sqrt[d])*Sqrt[((a^2*c + d)*(c + d*x^2))/(c*d*(-1 + a*x)^2)]*Sqrt[-((-1 + (I*Sqrt[d]*x)/Sqrt[c] + a*((I*Sqrt[c]
)/Sqrt[d] + x))/(1 - a*x))]*EllipticPi[(2*a*Sqrt[c])/(a*Sqrt[c] + I*Sqrt[d]), ArcSin[Sqrt[-((-1 + (I*Sqrt[d]*x
)/Sqrt[c] + a*((I*Sqrt[c])/Sqrt[d] + x))/(2 - 2*a*x))]], ((4*I)*a*Sqrt[c]*Sqrt[d])/(a*Sqrt[c] + I*Sqrt[d])^2])
)/(a*(a^2*c + d)*Sqrt[1 + a*x]*Sqrt[-((-1 + (I*Sqrt[d]*x)/Sqrt[c] + a*((I*Sqrt[c])/Sqrt[d] + x))/(1 - a*x))]))
/(c*Sqrt[c + d*x^2])

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fricas [A]  time = 0.59, size = 296, normalized size = 3.08 \[ \left [\frac {4 \, \sqrt {d x^{2} + c} d x \log \left (a x + \sqrt {a^{2} x^{2} - 1}\right ) + {\left (d x^{2} + c\right )} \sqrt {d} \log \left (8 \, a^{4} d^{2} x^{4} + a^{4} c^{2} - 6 \, a^{2} c d + 8 \, {\left (a^{4} c d - a^{2} d^{2}\right )} x^{2} - 4 \, {\left (2 \, a^{3} d x^{2} + a^{3} c - a d\right )} \sqrt {a^{2} x^{2} - 1} \sqrt {d x^{2} + c} \sqrt {d} + d^{2}\right )}{4 \, {\left (c d^{2} x^{2} + c^{2} d\right )}}, \frac {2 \, \sqrt {d x^{2} + c} d x \log \left (a x + \sqrt {a^{2} x^{2} - 1}\right ) + {\left (d x^{2} + c\right )} \sqrt {-d} \arctan \left (\frac {{\left (2 \, a^{2} d x^{2} + a^{2} c - d\right )} \sqrt {a^{2} x^{2} - 1} \sqrt {d x^{2} + c} \sqrt {-d}}{2 \, {\left (a^{3} d^{2} x^{4} - a c d + {\left (a^{3} c d - a d^{2}\right )} x^{2}\right )}}\right )}{2 \, {\left (c d^{2} x^{2} + c^{2} d\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccosh(a*x)/(d*x^2+c)^(3/2),x, algorithm="fricas")

[Out]

[1/4*(4*sqrt(d*x^2 + c)*d*x*log(a*x + sqrt(a^2*x^2 - 1)) + (d*x^2 + c)*sqrt(d)*log(8*a^4*d^2*x^4 + a^4*c^2 - 6
*a^2*c*d + 8*(a^4*c*d - a^2*d^2)*x^2 - 4*(2*a^3*d*x^2 + a^3*c - a*d)*sqrt(a^2*x^2 - 1)*sqrt(d*x^2 + c)*sqrt(d)
+ d^2))/(c*d^2*x^2 + c^2*d), 1/2*(2*sqrt(d*x^2 + c)*d*x*log(a*x + sqrt(a^2*x^2 - 1)) + (d*x^2 + c)*sqrt(-d)*a
rctan(1/2*(2*a^2*d*x^2 + a^2*c - d)*sqrt(a^2*x^2 - 1)*sqrt(d*x^2 + c)*sqrt(-d)/(a^3*d^2*x^4 - a*c*d + (a^3*c*d
- a*d^2)*x^2)))/(c*d^2*x^2 + c^2*d)]

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giac [A]  time = 0.62, size = 82, normalized size = 0.85 \[ \frac {x \log \left (a x + \sqrt {a^{2} x^{2} - 1}\right )}{\sqrt {d x^{2} + c} c} + \frac {a \log \left ({\left | -\sqrt {a^{2} x^{2} - 1} \sqrt {d} + \sqrt {a^{2} c + {\left (a^{2} x^{2} - 1\right )} d + d} \right |}\right )}{c \sqrt {d} {\left | a \right |}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccosh(a*x)/(d*x^2+c)^(3/2),x, algorithm="giac")

[Out]

x*log(a*x + sqrt(a^2*x^2 - 1))/(sqrt(d*x^2 + c)*c) + a*log(abs(-sqrt(a^2*x^2 - 1)*sqrt(d) + sqrt(a^2*c + (a^2*
x^2 - 1)*d + d)))/(c*sqrt(d)*abs(a))

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maple [F]  time = 0.33, size = 0, normalized size = 0.00 \[ \int \frac {\mathrm {arccosh}\left (a x \right )}{\left (d \,x^{2}+c \right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arccosh(a*x)/(d*x^2+c)^(3/2),x)

[Out]

int(arccosh(a*x)/(d*x^2+c)^(3/2),x)

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccosh(a*x)/(d*x^2+c)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(d-a^2*c>0)', see `assume?` for
more details)Is d-a^2*c zero or nonzero?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\mathrm {acosh}\left (a\,x\right )}{{\left (d\,x^2+c\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acosh(a*x)/(c + d*x^2)^(3/2),x)

[Out]

int(acosh(a*x)/(c + d*x^2)^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {acosh}{\left (a x \right )}}{\left (c + d x^{2}\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(acosh(a*x)/(d*x**2+c)**(3/2),x)

[Out]

Integral(acosh(a*x)/(c + d*x**2)**(3/2), x)

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