3.50 \(\int \frac {\cosh ^{-1}(a x)}{(c+d x^2)^{5/2}} \, dx\)

Optimal. Leaf size=180 \[ -\frac {2 \sqrt {a^2 x^2-1} \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a^2 x^2-1}}{a \sqrt {c+d x^2}}\right )}{3 c^2 \sqrt {d} \sqrt {a x-1} \sqrt {a x+1}}+\frac {a \left (1-a^2 x^2\right )}{3 c \sqrt {a x-1} \sqrt {a x+1} \left (a^2 c+d\right ) \sqrt {c+d x^2}}+\frac {2 x \cosh ^{-1}(a x)}{3 c^2 \sqrt {c+d x^2}}+\frac {x \cosh ^{-1}(a x)}{3 c \left (c+d x^2\right )^{3/2}} \]

[Out]

1/3*x*arccosh(a*x)/c/(d*x^2+c)^(3/2)-2/3*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^2/d^(1/2)/(a*x-1)^(1/2)/(a*x+1)^(1/2)+2/3*x*arccosh(a*x)/c^2/(d*x^2+c)^(1/2)+1/3*a*(-a^2*x^2+1)/c/(a^2*c+d
)/(a*x-1)^(1/2)/(a*x+1)^(1/2)/(d*x^2+c)^(1/2)

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Rubi [A]  time = 0.18, antiderivative size = 180, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 10, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.625, Rules used = {192, 191, 5705, 12, 519, 571, 78, 63, 217, 206} \[ -\frac {2 \sqrt {a^2 x^2-1} \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a^2 x^2-1}}{a \sqrt {c+d x^2}}\right )}{3 c^2 \sqrt {d} \sqrt {a x-1} \sqrt {a x+1}}+\frac {a \left (1-a^2 x^2\right )}{3 c \sqrt {a x-1} \sqrt {a x+1} \left (a^2 c+d\right ) \sqrt {c+d x^2}}+\frac {2 x \cosh ^{-1}(a x)}{3 c^2 \sqrt {c+d x^2}}+\frac {x \cosh ^{-1}(a x)}{3 c \left (c+d x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[ArcCosh[a*x]/(c + d*x^2)^(5/2),x]

[Out]

(a*(1 - a^2*x^2))/(3*c*(a^2*c + d)*Sqrt[-1 + a*x]*Sqrt[1 + a*x]*Sqrt[c + d*x^2]) + (x*ArcCosh[a*x])/(3*c*(c +
d*x^2)^(3/2)) + (2*x*ArcCosh[a*x])/(3*c^2*Sqrt[c + d*x^2]) - (2*Sqrt[-1 + a^2*x^2]*ArcTanh[(Sqrt[d]*Sqrt[-1 +
a^2*x^2])/(a*Sqrt[c + d*x^2])])/(3*c^2*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 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

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 192

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

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 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 571

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_))^(r_.), x
_Symbol] :> Dist[1/n, Subst[Int[(a + b*x)^p*(c + d*x)^q*(e + f*x)^r, x], x, x^n], x] /; FreeQ[{a, b, c, d, e,
f, m, n, p, q, r}, x] && EqQ[m - n + 1, 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 )^{5/2}} \, dx &=\frac {x \cosh ^{-1}(a x)}{3 c \left (c+d x^2\right )^{3/2}}+\frac {2 x \cosh ^{-1}(a x)}{3 c^2 \sqrt {c+d x^2}}-a \int \frac {x \left (3 c+2 d x^2\right )}{3 c^2 \sqrt {-1+a x} \sqrt {1+a x} \left (c+d x^2\right )^{3/2}} \, dx\\ &=\frac {x \cosh ^{-1}(a x)}{3 c \left (c+d x^2\right )^{3/2}}+\frac {2 x \cosh ^{-1}(a x)}{3 c^2 \sqrt {c+d x^2}}-\frac {a \int \frac {x \left (3 c+2 d x^2\right )}{\sqrt {-1+a x} \sqrt {1+a x} \left (c+d x^2\right )^{3/2}} \, dx}{3 c^2}\\ &=\frac {x \cosh ^{-1}(a x)}{3 c \left (c+d x^2\right )^{3/2}}+\frac {2 x \cosh ^{-1}(a x)}{3 c^2 \sqrt {c+d x^2}}-\frac {\left (a \sqrt {-1+a^2 x^2}\right ) \int \frac {x \left (3 c+2 d x^2\right )}{\sqrt {-1+a^2 x^2} \left (c+d x^2\right )^{3/2}} \, dx}{3 c^2 \sqrt {-1+a x} \sqrt {1+a x}}\\ &=\frac {x \cosh ^{-1}(a x)}{3 c \left (c+d x^2\right )^{3/2}}+\frac {2 x \cosh ^{-1}(a x)}{3 c^2 \sqrt {c+d x^2}}-\frac {\left (a \sqrt {-1+a^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {3 c+2 d x}{\sqrt {-1+a^2 x} (c+d x)^{3/2}} \, dx,x,x^2\right )}{6 c^2 \sqrt {-1+a x} \sqrt {1+a x}}\\ &=\frac {a \left (1-a^2 x^2\right )}{3 c \left (a^2 c+d\right ) \sqrt {-1+a x} \sqrt {1+a x} \sqrt {c+d x^2}}+\frac {x \cosh ^{-1}(a x)}{3 c \left (c+d x^2\right )^{3/2}}+\frac {2 x \cosh ^{-1}(a x)}{3 c^2 \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 )}{3 c^2 \sqrt {-1+a x} \sqrt {1+a x}}\\ &=\frac {a \left (1-a^2 x^2\right )}{3 c \left (a^2 c+d\right ) \sqrt {-1+a x} \sqrt {1+a x} \sqrt {c+d x^2}}+\frac {x \cosh ^{-1}(a x)}{3 c \left (c+d x^2\right )^{3/2}}+\frac {2 x \cosh ^{-1}(a x)}{3 c^2 \sqrt {c+d x^2}}-\frac {\left (2 \sqrt {-1+a^2 x^2}\right ) \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 )}{3 a c^2 \sqrt {-1+a x} \sqrt {1+a x}}\\ &=\frac {a \left (1-a^2 x^2\right )}{3 c \left (a^2 c+d\right ) \sqrt {-1+a x} \sqrt {1+a x} \sqrt {c+d x^2}}+\frac {x \cosh ^{-1}(a x)}{3 c \left (c+d x^2\right )^{3/2}}+\frac {2 x \cosh ^{-1}(a x)}{3 c^2 \sqrt {c+d x^2}}-\frac {\left (2 \sqrt {-1+a^2 x^2}\right ) \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 )}{3 a c^2 \sqrt {-1+a x} \sqrt {1+a x}}\\ &=\frac {a \left (1-a^2 x^2\right )}{3 c \left (a^2 c+d\right ) \sqrt {-1+a x} \sqrt {1+a x} \sqrt {c+d x^2}}+\frac {x \cosh ^{-1}(a x)}{3 c \left (c+d x^2\right )^{3/2}}+\frac {2 x \cosh ^{-1}(a x)}{3 c^2 \sqrt {c+d x^2}}-\frac {2 \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 )}{3 c^2 \sqrt {d} \sqrt {-1+a x} \sqrt {1+a x}}\\ \end {align*}

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Mathematica [C]  time = 2.21, size = 609, normalized size = 3.38 \[ \frac {-\frac {a c \sqrt {a x+1} \sqrt {a x-1} \left (c+d x^2\right )}{a^2 c+d}+\frac {4 (a x-1)^{3/2} \left (c+d x^2\right ) \sqrt {\frac {(a x+1) \left (a \sqrt {c}-i \sqrt {d}\right )}{(a x-1) \left (a \sqrt {c}+i \sqrt {d}\right )}} \left (a \sqrt {c} \left (-a \sqrt {c}+i \sqrt {d}\right ) \sqrt {\frac {\left (a^2 c+d\right ) \left (c+d x^2\right )}{c d (a x-1)^2}} \sqrt {-\frac {a \left (x+\frac {i \sqrt {c}}{\sqrt {d}}\right )+\frac {i \sqrt {d} x}{\sqrt {c}}-1}{1-a x}} \Pi \left (\frac {2 a \sqrt {c}}{\sqrt {c} a+i \sqrt {d}};\sin ^{-1}\left (\sqrt {-\frac {\frac {i \sqrt {d} x}{\sqrt {c}}+a \left (x+\frac {i \sqrt {c}}{\sqrt {d}}\right )-1}{2-2 a x}}\right )|\frac {4 i a \sqrt {c} \sqrt {d}}{\left (\sqrt {c} a+i \sqrt {d}\right )^2}\right )+\frac {a \left (\sqrt {d}-i a \sqrt {c}\right ) \left (\sqrt {d} x+i \sqrt {c}\right ) \sqrt {\frac {\frac {i a \sqrt {c}}{\sqrt {d}}+a (-x)+\frac {i \sqrt {d} x}{\sqrt {c}}+1}{1-a x}} F\left (\sin ^{-1}\left (\sqrt {-\frac {\frac {i \sqrt {d} x}{\sqrt {c}}+a \left (x+\frac {i \sqrt {c}}{\sqrt {d}}\right )-1}{2-2 a x}}\right )|\frac {4 i a \sqrt {c} \sqrt {d}}{\left (\sqrt {c} a+i \sqrt {d}\right )^2}\right )}{a x-1}\right )}{a \sqrt {a x+1} \left (a^2 c+d\right ) \sqrt {-\frac {a \left (x+\frac {i \sqrt {c}}{\sqrt {d}}\right )+\frac {i \sqrt {d} x}{\sqrt {c}}-1}{1-a x}}}+x \cosh ^{-1}(a x) \left (3 c+2 d x^2\right )}{3 c^2 \left (c+d x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-((a*c*Sqrt[-1 + a*x]*Sqrt[1 + a*x]*(c + d*x^2))/(a^2*c + d)) + x*(3*c + 2*d*x^2)*ArcCosh[a*x] + (4*(-1 + a*x
)^(3/2)*Sqrt[((a*Sqrt[c] - I*Sqrt[d])*(1 + a*x))/((a*Sqrt[c] + I*Sqrt[d])*(-1 + a*x))]*(c + d*x^2)*((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)*S
qrt[1 + a*x]*Sqrt[-((-1 + (I*Sqrt[d]*x)/Sqrt[c] + a*((I*Sqrt[c])/Sqrt[d] + x))/(1 - a*x))]))/(3*c^2*(c + d*x^2
)^(3/2))

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fricas [B]  time = 0.72, size = 613, normalized size = 3.41 \[ \left [\frac {{\left (a^{2} c^{3} + {\left (a^{2} c d^{2} + d^{3}\right )} x^{4} + c^{2} d + 2 \, {\left (a^{2} c^{2} d + c d^{2}\right )} x^{2}\right )} \sqrt {d} \log \left (8 \, a^{4} d^{2} x^{4} + a^{4} c^{2} - 6 \, a^{2} c d + 8 \, {\left (a^{4} c d - a^{2} d^{2}\right )} x^{2} - 4 \, {\left (2 \, a^{3} d x^{2} + a^{3} c - a d\right )} \sqrt {a^{2} x^{2} - 1} \sqrt {d x^{2} + c} \sqrt {d} + d^{2}\right ) + 2 \, {\left (2 \, {\left (a^{2} c d^{2} + d^{3}\right )} x^{3} + 3 \, {\left (a^{2} c^{2} d + c d^{2}\right )} x\right )} \sqrt {d x^{2} + c} \log \left (a x + \sqrt {a^{2} x^{2} - 1}\right ) - 2 \, {\left (a c d^{2} x^{2} + a c^{2} d\right )} \sqrt {a^{2} x^{2} - 1} \sqrt {d x^{2} + c}}{6 \, {\left (a^{2} c^{5} d + c^{4} d^{2} + {\left (a^{2} c^{3} d^{3} + c^{2} d^{4}\right )} x^{4} + 2 \, {\left (a^{2} c^{4} d^{2} + c^{3} d^{3}\right )} x^{2}\right )}}, \frac {{\left (a^{2} c^{3} + {\left (a^{2} c d^{2} + d^{3}\right )} x^{4} + c^{2} d + 2 \, {\left (a^{2} c^{2} d + c d^{2}\right )} x^{2}\right )} \sqrt {-d} \arctan \left (\frac {{\left (2 \, a^{2} d x^{2} + a^{2} c - d\right )} \sqrt {a^{2} x^{2} - 1} \sqrt {d x^{2} + c} \sqrt {-d}}{2 \, {\left (a^{3} d^{2} x^{4} - a c d + {\left (a^{3} c d - a d^{2}\right )} x^{2}\right )}}\right ) + {\left (2 \, {\left (a^{2} c d^{2} + d^{3}\right )} x^{3} + 3 \, {\left (a^{2} c^{2} d + c d^{2}\right )} x\right )} \sqrt {d x^{2} + c} \log \left (a x + \sqrt {a^{2} x^{2} - 1}\right ) - {\left (a c d^{2} x^{2} + a c^{2} d\right )} \sqrt {a^{2} x^{2} - 1} \sqrt {d x^{2} + c}}{3 \, {\left (a^{2} c^{5} d + c^{4} d^{2} + {\left (a^{2} c^{3} d^{3} + c^{2} d^{4}\right )} x^{4} + 2 \, {\left (a^{2} c^{4} d^{2} + c^{3} d^{3}\right )} x^{2}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/6*((a^2*c^3 + (a^2*c*d^2 + d^3)*x^4 + c^2*d + 2*(a^2*c^2*d + c*d^2)*x^2)*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)*sq
rt(d) + d^2) + 2*(2*(a^2*c*d^2 + d^3)*x^3 + 3*(a^2*c^2*d + c*d^2)*x)*sqrt(d*x^2 + c)*log(a*x + sqrt(a^2*x^2 -
1)) - 2*(a*c*d^2*x^2 + a*c^2*d)*sqrt(a^2*x^2 - 1)*sqrt(d*x^2 + c))/(a^2*c^5*d + c^4*d^2 + (a^2*c^3*d^3 + c^2*d
^4)*x^4 + 2*(a^2*c^4*d^2 + c^3*d^3)*x^2), 1/3*((a^2*c^3 + (a^2*c*d^2 + d^3)*x^4 + c^2*d + 2*(a^2*c^2*d + c*d^2
)*x^2)*sqrt(-d)*arctan(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)) + (2*(a^2*c*d^2 + d^3)*x^3 + 3*(a^2*c^2*d + c*d^2)*x)*sqrt(d*x^2 + c)*log(a*x
+ sqrt(a^2*x^2 - 1)) - (a*c*d^2*x^2 + a*c^2*d)*sqrt(a^2*x^2 - 1)*sqrt(d*x^2 + c))/(a^2*c^5*d + c^4*d^2 + (a^2
*c^3*d^3 + c^2*d^4)*x^4 + 2*(a^2*c^4*d^2 + c^3*d^3)*x^2)]

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giac [A]  time = 0.60, size = 190, normalized size = 1.06 \[ -\frac {1}{3} \, a {\left (\frac {2 \, a^{2} {\left | d \right |}}{{\left (a^{2} c d + {\left (\sqrt {a^{2} d} \sqrt {d x^{2} + c} - \sqrt {{\left (d x^{2} + c\right )} a^{2} d - a^{2} c d - d^{2}}\right )}^{2} + d^{2}\right )} c \sqrt {d} {\left | a \right |}} - \frac {{\left | d \right |} \log \left ({\left (\sqrt {a^{2} d} \sqrt {d x^{2} + c} - \sqrt {{\left (d x^{2} + c\right )} a^{2} d - a^{2} c d - d^{2}}\right )}^{2}\right )}{c^{2} d^{\frac {3}{2}} {\left | a \right |}}\right )} + \frac {x {\left (\frac {2 \, d x^{2}}{c^{2}} + \frac {3}{c}\right )} \log \left (a x + \sqrt {a^{2} x^{2} - 1}\right )}{3 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*a*(2*a^2*abs(d)/((a^2*c*d + (sqrt(a^2*d)*sqrt(d*x^2 + c) - sqrt((d*x^2 + c)*a^2*d - a^2*c*d - d^2))^2 + d
^2)*c*sqrt(d)*abs(a)) - abs(d)*log((sqrt(a^2*d)*sqrt(d*x^2 + c) - sqrt((d*x^2 + c)*a^2*d - a^2*c*d - d^2))^2)/
(c^2*d^(3/2)*abs(a))) + 1/3*x*(2*d*x^2/c^2 + 3/c)*log(a*x + sqrt(a^2*x^2 - 1))/(d*x^2 + c)^(3/2)

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maple [F]  time = 0.32, size = 0, normalized size = 0.00 \[ \int \frac {\mathrm {arccosh}\left (a x \right )}{\left (d \,x^{2}+c \right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(arccosh(a*x)/(d*x^2+c)^(5/2),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)^(5/2),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 \[ \int \frac {\mathrm {acosh}\left (a\,x\right )}{{\left (d\,x^2+c\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acosh(a*x)/(c + d*x^2)^(5/2),x)

[Out]

int(acosh(a*x)/(c + d*x^2)^(5/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {acosh}{\left (a x \right )}}{\left (c + d x^{2}\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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