3.3.43 \(\int (a+b \cosh ^{-1}(1+d x^2)) \, dx\) [243]

Optimal. Leaf size=49 \[ a x-\frac {2 b \sqrt {\frac {d x^2}{2+d x^2}} \left (2+d x^2\right )}{d x}+b x \cosh ^{-1}\left (1+d x^2\right ) \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {6016, 12, 1986, 15, 267} \begin {gather*} a x-\frac {2 b \sqrt {\frac {d x^2}{d x^2+2}} \left (d x^2+2\right )}{d x}+b x \cosh ^{-1}\left (d x^2+1\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[a + b*ArcCosh[1 + d*x^2],x]

[Out]

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

Rule 12

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

Rule 15

Int[(u_.)*((a_.)*(x_)^(n_))^(m_), x_Symbol] :> Dist[a^IntPart[m]*((a*x^n)^FracPart[m]/x^(n*FracPart[m])), Int[
u*x^(m*n), x], x] /; FreeQ[{a, m, n}, x] &&  !IntegerQ[m]

Rule 267

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

Rule 1986

Int[(u_.)*((e_.)*((a_.) + (b_.)*(x_)^(n_.))^(q_.)*((c_) + (d_.)*(x_)^(n_))^(r_.))^(p_), x_Symbol] :> Dist[Simp
[(e*(a + b*x^n)^q*(c + d*x^n)^r)^p/((a + b*x^n)^(p*q)*(c + d*x^n)^(p*r))], Int[u*(a + b*x^n)^(p*q)*(c + d*x^n)
^(p*r), x], x] /; FreeQ[{a, b, c, d, e, n, p, q, r}, x]

Rule 6016

Int[ArcCosh[u_], x_Symbol] :> Simp[x*ArcCosh[u], x] - Int[SimplifyIntegrand[x*(D[u, x]/(Sqrt[-1 + u]*Sqrt[1 +
u])), x], x] /; InverseFunctionFreeQ[u, x] &&  !FunctionOfExponentialQ[u, x]

Rubi steps

\begin {align*} \int \left (a+b \cosh ^{-1}\left (1+d x^2\right )\right ) \, dx &=a x+b \int \cosh ^{-1}\left (1+d x^2\right ) \, dx\\ &=a x+b x \cosh ^{-1}\left (1+d x^2\right )-b \int 2 \sqrt {\frac {d x^2}{2+d x^2}} \, dx\\ &=a x+b x \cosh ^{-1}\left (1+d x^2\right )-(2 b) \int \sqrt {\frac {d x^2}{2+d x^2}} \, dx\\ &=a x+b x \cosh ^{-1}\left (1+d x^2\right )-\frac {\left (2 b \sqrt {\frac {d x^2}{2+d x^2}} \sqrt {2+d x^2}\right ) \int \frac {x}{\sqrt {2+d x^2}} \, dx}{x}\\ &=a x-\frac {2 b \sqrt {\frac {d x^2}{2+d x^2}} \left (2+d x^2\right )}{d x}+b x \cosh ^{-1}\left (1+d x^2\right )\\ \end {align*}

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Mathematica [A]
time = 0.04, size = 37, normalized size = 0.76 \begin {gather*} a x-\frac {2 b x}{\sqrt {\frac {d x^2}{2+d x^2}}}+b x \cosh ^{-1}\left (1+d x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[a + b*ArcCosh[1 + d*x^2],x]

[Out]

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

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Maple [A]
time = 0.01, size = 37, normalized size = 0.76

method	result	size
default	\(a x +b \left (x \,\mathrm {arccosh}\left (d \,x^{2}+1\right )-\frac {2 x \sqrt {d \,x^{2}+2}}{\sqrt {d \,x^{2}}}\right )\)	\(37\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(a+b*arccosh(d*x^2+1),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.31, size = 44, normalized size = 0.90 \begin {gather*} {\left (x \operatorname {arcosh}\left (d x^{2} + 1\right ) - \frac {2 \, {\left (d^{\frac {3}{2}} x^{2} + 2 \, \sqrt {d}\right )}}{\sqrt {d x^{2} + 2} d}\right )} b + a x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.33, size = 63, normalized size = 1.29 \begin {gather*} \frac {b d x^{2} \log \left (d x^{2} + \sqrt {d^{2} x^{4} + 2 \, d x^{2}} + 1\right ) + a d x^{2} - 2 \, \sqrt {d^{2} x^{4} + 2 \, d x^{2}} b}{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b*d*x^2*log(d*x^2 + sqrt(d^2*x^4 + 2*d*x^2) + 1) + a*d*x^2 - 2*sqrt(d^2*x^4 + 2*d*x^2)*b)/(d*x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(a + b*acosh(d*x**2 + 1), x)

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Giac [A]
time = 0.41, size = 62, normalized size = 1.27 \begin {gather*} {\left (x \log \left (d x^{2} + \sqrt {{\left (d x^{2} + 1\right )}^{2} - 1} + 1\right ) + \frac {2 \, \sqrt {2} \mathrm {sgn}\left (x\right )}{\sqrt {d}} - \frac {2 \, \sqrt {d^{2} x^{2} + 2 \, d}}{d \mathrm {sgn}\left (x\right )}\right )} b + a x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x*log(d*x^2 + sqrt((d*x^2 + 1)^2 - 1) + 1) + 2*sqrt(2)*sgn(x)/sqrt(d) - 2*sqrt(d^2*x^2 + 2*d)/(d*sgn(x)))*b +
a*x

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Mupad [B]
time = 0.59, size = 32, normalized size = 0.65 \begin {gather*} a\,x+b\,x\,\mathrm {acosh}\left (d\,x^2+1\right )-\frac {2\,b\,\mathrm {sign}\left (x\right )\,\sqrt {d\,x^2+2}}{\sqrt {d}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(a + b*acosh(d*x^2 + 1),x)

[Out]

a*x + b*x*acosh(d*x^2 + 1) - (2*b*sign(x)*(d*x^2 + 2)^(1/2))/d^(1/2)

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