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

Optimal. Leaf size=101 \[ \frac {x \left (a+b \cosh ^{-1}(c x)\right )}{d \sqrt {d+e x^2}}-\frac {b \sqrt {c^2 x^2-1} \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {c^2 x^2-1}}{c \sqrt {d+e x^2}}\right )}{d \sqrt {e} \sqrt {c x-1} \sqrt {c x+1}} \]

[Out]

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

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Rubi [A]  time = 0.19, antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 8, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {191, 5705, 12, 519, 444, 63, 217, 206} \[ \frac {x \left (a+b \cosh ^{-1}(c x)\right )}{d \sqrt {d+e x^2}}-\frac {b \sqrt {c^2 x^2-1} \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {c^2 x^2-1}}{c \sqrt {d+e x^2}}\right )}{d \sqrt {e} \sqrt {c x-1} \sqrt {c x+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(a + b*ArcCosh[c*x]))/(d*Sqrt[d + e*x^2]) - (b*Sqrt[-1 + c^2*x^2]*ArcTanh[(Sqrt[e]*Sqrt[-1 + c^2*x^2])/(c*S
qrt[d + e*x^2])])/(d*Sqrt[e]*Sqrt[-1 + c*x]*Sqrt[1 + c*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 {a+b \cosh ^{-1}(c x)}{\left (d+e x^2\right )^{3/2}} \, dx &=\frac {x \left (a+b \cosh ^{-1}(c x)\right )}{d \sqrt {d+e x^2}}-(b c) \int \frac {x}{d \sqrt {-1+c x} \sqrt {1+c x} \sqrt {d+e x^2}} \, dx\\ &=\frac {x \left (a+b \cosh ^{-1}(c x)\right )}{d \sqrt {d+e x^2}}-\frac {(b c) \int \frac {x}{\sqrt {-1+c x} \sqrt {1+c x} \sqrt {d+e x^2}} \, dx}{d}\\ &=\frac {x \left (a+b \cosh ^{-1}(c x)\right )}{d \sqrt {d+e x^2}}-\frac {\left (b c \sqrt {-1+c^2 x^2}\right ) \int \frac {x}{\sqrt {-1+c^2 x^2} \sqrt {d+e x^2}} \, dx}{d \sqrt {-1+c x} \sqrt {1+c x}}\\ &=\frac {x \left (a+b \cosh ^{-1}(c x)\right )}{d \sqrt {d+e x^2}}-\frac {\left (b c \sqrt {-1+c^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {-1+c^2 x} \sqrt {d+e x}} \, dx,x,x^2\right )}{2 d \sqrt {-1+c x} \sqrt {1+c x}}\\ &=\frac {x \left (a+b \cosh ^{-1}(c x)\right )}{d \sqrt {d+e x^2}}-\frac {\left (b \sqrt {-1+c^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {d+\frac {e}{c^2}+\frac {e x^2}{c^2}}} \, dx,x,\sqrt {-1+c^2 x^2}\right )}{c d \sqrt {-1+c x} \sqrt {1+c x}}\\ &=\frac {x \left (a+b \cosh ^{-1}(c x)\right )}{d \sqrt {d+e x^2}}-\frac {\left (b \sqrt {-1+c^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{1-\frac {e x^2}{c^2}} \, dx,x,\frac {\sqrt {-1+c^2 x^2}}{\sqrt {d+e x^2}}\right )}{c d \sqrt {-1+c x} \sqrt {1+c x}}\\ &=\frac {x \left (a+b \cosh ^{-1}(c x)\right )}{d \sqrt {d+e x^2}}-\frac {b \sqrt {-1+c^2 x^2} \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {-1+c^2 x^2}}{c \sqrt {d+e x^2}}\right )}{d \sqrt {e} \sqrt {-1+c x} \sqrt {1+c x}}\\ \end {align*}

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Mathematica [C]  time = 3.36, size = 556, normalized size = 5.50 \[ \frac {a x+\frac {2 b (c x-1)^{3/2} \sqrt {\frac {(c x+1) \left (c \sqrt {d}-i \sqrt {e}\right )}{(c x-1) \left (c \sqrt {d}+i \sqrt {e}\right )}} \left (c \sqrt {d} \left (-c \sqrt {d}+i \sqrt {e}\right ) \sqrt {\frac {\left (c^2 d+e\right ) \left (d+e x^2\right )}{d e (c x-1)^2}} \sqrt {-\frac {c \left (x+\frac {i \sqrt {d}}{\sqrt {e}}\right )+\frac {i \sqrt {e} x}{\sqrt {d}}-1}{1-c x}} \Pi \left (\frac {2 c \sqrt {d}}{\sqrt {d} c+i \sqrt {e}};\sin ^{-1}\left (\sqrt {-\frac {\frac {i \sqrt {e} x}{\sqrt {d}}+c \left (x+\frac {i \sqrt {d}}{\sqrt {e}}\right )-1}{2-2 c x}}\right )|\frac {4 i c \sqrt {d} \sqrt {e}}{\left (\sqrt {d} c+i \sqrt {e}\right )^2}\right )+\frac {c \left (\sqrt {e}-i c \sqrt {d}\right ) \left (\sqrt {e} x+i \sqrt {d}\right ) \sqrt {\frac {\frac {i c \sqrt {d}}{\sqrt {e}}+c (-x)+\frac {i \sqrt {e} x}{\sqrt {d}}+1}{1-c x}} F\left (\sin ^{-1}\left (\sqrt {-\frac {\frac {i \sqrt {e} x}{\sqrt {d}}+c \left (x+\frac {i \sqrt {d}}{\sqrt {e}}\right )-1}{2-2 c x}}\right )|\frac {4 i c \sqrt {d} \sqrt {e}}{\left (\sqrt {d} c+i \sqrt {e}\right )^2}\right )}{c x-1}\right )}{c \sqrt {c x+1} \left (c^2 d+e\right ) \sqrt {-\frac {c \left (x+\frac {i \sqrt {d}}{\sqrt {e}}\right )+\frac {i \sqrt {e} x}{\sqrt {d}}-1}{1-c x}}}+b x \cosh ^{-1}(c x)}{d \sqrt {d+e x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.43, size = 332, normalized size = 3.29 \[ \left [\frac {4 \, \sqrt {e x^{2} + d} b e x \log \left (c x + \sqrt {c^{2} x^{2} - 1}\right ) + 4 \, \sqrt {e x^{2} + d} a e x + {\left (b e x^{2} + b d\right )} \sqrt {e} \log \left (8 \, c^{4} e^{2} x^{4} + c^{4} d^{2} - 6 \, c^{2} d e + 8 \, {\left (c^{4} d e - c^{2} e^{2}\right )} x^{2} - 4 \, {\left (2 \, c^{3} e x^{2} + c^{3} d - c e\right )} \sqrt {c^{2} x^{2} - 1} \sqrt {e x^{2} + d} \sqrt {e} + e^{2}\right )}{4 \, {\left (d e^{2} x^{2} + d^{2} e\right )}}, \frac {2 \, \sqrt {e x^{2} + d} b e x \log \left (c x + \sqrt {c^{2} x^{2} - 1}\right ) + 2 \, \sqrt {e x^{2} + d} a e x + {\left (b e x^{2} + b d\right )} \sqrt {-e} \arctan \left (\frac {{\left (2 \, c^{2} e x^{2} + c^{2} d - e\right )} \sqrt {c^{2} x^{2} - 1} \sqrt {e x^{2} + d} \sqrt {-e}}{2 \, {\left (c^{3} e^{2} x^{4} - c d e + {\left (c^{3} d e - c e^{2}\right )} x^{2}\right )}}\right )}{2 \, {\left (d e^{2} x^{2} + d^{2} e\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*arccosh(c*x) + a)/(e*x^2 + d)^(3/2), x)

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maple [F]  time = 0.64, size = 0, normalized size = 0.00 \[ \int \frac {a +b \,\mathrm {arccosh}\left (c x \right )}{\left (e \,x^{2}+d \right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((a+b*arccosh(c*x))/(e*x^2+d)^(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((a+b*arccosh(c*x))/(e*x^2+d)^(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(e-c^2*d>0)', see `assume?` for
more details)Is e-c^2*d zero or nonzero?

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*acosh(c*x))/(d + e*x^2)^(3/2),x)

[Out]

int((a + b*acosh(c*x))/(d + e*x^2)^(3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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