3.3.41 \(\int (a+b \cosh ^{-1}(1+d x^2))^3 \, dx\) [241]

Optimal. Leaf size=125 \[ 24 a b^2 x-\frac {48 b^3 \sqrt {\frac {d x^2}{2+d x^2}} \left (2+d x^2\right )}{d x}+24 b^3 x \cosh ^{-1}\left (1+d x^2\right )-\frac {6 b \left (2 x^2+d x^4\right ) \left (a+b \cosh ^{-1}\left (1+d x^2\right )\right )^2}{x \sqrt {d x^2} \sqrt {2+d x^2}}+x \left (a+b \cosh ^{-1}\left (1+d x^2\right )\right )^3 \]

[Out]

24*a*b^2*x+24*b^3*x*arccosh(d*x^2+1)+x*(a+b*arccosh(d*x^2+1))^3-48*b^3*(d*x^2+2)*(d*x^2/(d*x^2+2))^(1/2)/d/x-6
*b*(d*x^4+2*x^2)*(a+b*arccosh(d*x^2+1))^2/x/(d*x^2)^(1/2)/(d*x^2+2)^(1/2)

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Rubi [A]
time = 0.04, antiderivative size = 125, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {6001, 6016, 12, 1986, 15, 267} \begin {gather*} 24 a b^2 x+x \left (a+b \cosh ^{-1}\left (d x^2+1\right )\right )^3-\frac {6 b \left (d x^4+2 x^2\right ) \left (a+b \cosh ^{-1}\left (d x^2+1\right )\right )^2}{x \sqrt {d x^2} \sqrt {d x^2+2}}-\frac {48 b^3 \sqrt {\frac {d x^2}{d x^2+2}} \left (d x^2+2\right )}{d x}+24 b^3 x \cosh ^{-1}\left (d x^2+1\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcCosh[1 + d*x^2])^3,x]

[Out]

24*a*b^2*x - (48*b^3*Sqrt[(d*x^2)/(2 + d*x^2)]*(2 + d*x^2))/(d*x) + 24*b^3*x*ArcCosh[1 + d*x^2] - (6*b*(2*x^2
+ d*x^4)*(a + b*ArcCosh[1 + d*x^2])^2)/(x*Sqrt[d*x^2]*Sqrt[2 + d*x^2]) + x*(a + b*ArcCosh[1 + d*x^2])^3

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 6001

Int[((a_.) + ArcCosh[(c_) + (d_.)*(x_)^2]*(b_.))^(n_), x_Symbol] :> Simp[x*(a + b*ArcCosh[c + d*x^2])^n, x] +
(Dist[4*b^2*n*(n - 1), Int[(a + b*ArcCosh[c + d*x^2])^(n - 2), x], x] - Simp[2*b*n*(2*c*d*x^2 + d^2*x^4)*((a +
b*ArcCosh[c + d*x^2])^(n - 1)/(d*x*Sqrt[-1 + c + d*x^2]*Sqrt[1 + c + d*x^2])), x]) /; FreeQ[{a, b, c, d}, x]
&& EqQ[c^2, 1] && GtQ[n, 1]

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 )^3 \, dx &=-\frac {6 b \left (2 x^2+d x^4\right ) \left (a+b \cosh ^{-1}\left (1+d x^2\right )\right )^2}{x \sqrt {d x^2} \sqrt {2+d x^2}}+x \left (a+b \cosh ^{-1}\left (1+d x^2\right )\right )^3+\left (24 b^2\right ) \int \left (a+b \cosh ^{-1}\left (1+d x^2\right )\right ) \, dx\\ &=24 a b^2 x-\frac {6 b \left (2 x^2+d x^4\right ) \left (a+b \cosh ^{-1}\left (1+d x^2\right )\right )^2}{x \sqrt {d x^2} \sqrt {2+d x^2}}+x \left (a+b \cosh ^{-1}\left (1+d x^2\right )\right )^3+\left (24 b^3\right ) \int \cosh ^{-1}\left (1+d x^2\right ) \, dx\\ &=24 a b^2 x+24 b^3 x \cosh ^{-1}\left (1+d x^2\right )-\frac {6 b \left (2 x^2+d x^4\right ) \left (a+b \cosh ^{-1}\left (1+d x^2\right )\right )^2}{x \sqrt {d x^2} \sqrt {2+d x^2}}+x \left (a+b \cosh ^{-1}\left (1+d x^2\right )\right )^3-\left (24 b^3\right ) \int 2 \sqrt {\frac {d x^2}{2+d x^2}} \, dx\\ &=24 a b^2 x+24 b^3 x \cosh ^{-1}\left (1+d x^2\right )-\frac {6 b \left (2 x^2+d x^4\right ) \left (a+b \cosh ^{-1}\left (1+d x^2\right )\right )^2}{x \sqrt {d x^2} \sqrt {2+d x^2}}+x \left (a+b \cosh ^{-1}\left (1+d x^2\right )\right )^3-\left (48 b^3\right ) \int \sqrt {\frac {d x^2}{2+d x^2}} \, dx\\ &=24 a b^2 x+24 b^3 x \cosh ^{-1}\left (1+d x^2\right )-\frac {6 b \left (2 x^2+d x^4\right ) \left (a+b \cosh ^{-1}\left (1+d x^2\right )\right )^2}{x \sqrt {d x^2} \sqrt {2+d x^2}}+x \left (a+b \cosh ^{-1}\left (1+d x^2\right )\right )^3-\frac {\left (48 b^3 \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}\\ &=24 a b^2 x-\frac {48 b^3 \sqrt {\frac {d x^2}{2+d x^2}} \left (2+d x^2\right )}{d x}+24 b^3 x \cosh ^{-1}\left (1+d x^2\right )-\frac {6 b \left (2 x^2+d x^4\right ) \left (a+b \cosh ^{-1}\left (1+d x^2\right )\right )^2}{x \sqrt {d x^2} \sqrt {2+d x^2}}+x \left (a+b \cosh ^{-1}\left (1+d x^2\right )\right )^3\\ \end {align*}

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Mathematica [A]
time = 0.08, size = 171, normalized size = 1.37 \begin {gather*} \frac {a \left (a^2+24 b^2\right ) d x^2-6 b \left (a^2+8 b^2\right ) \sqrt {d x^2} \sqrt {2+d x^2}+3 b \left (a^2 d x^2+8 b^2 d x^2-4 a b \sqrt {d x^2} \sqrt {2+d x^2}\right ) \cosh ^{-1}\left (1+d x^2\right )+3 b^2 \left (a d x^2-2 b \sqrt {d x^2} \sqrt {2+d x^2}\right ) \cosh ^{-1}\left (1+d x^2\right )^2+b^3 d x^2 \cosh ^{-1}\left (1+d x^2\right )^3}{d x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcCosh[1 + d*x^2])^3,x]

[Out]

(a*(a^2 + 24*b^2)*d*x^2 - 6*b*(a^2 + 8*b^2)*Sqrt[d*x^2]*Sqrt[2 + d*x^2] + 3*b*(a^2*d*x^2 + 8*b^2*d*x^2 - 4*a*b
*Sqrt[d*x^2]*Sqrt[2 + d*x^2])*ArcCosh[1 + d*x^2] + 3*b^2*(a*d*x^2 - 2*b*Sqrt[d*x^2]*Sqrt[2 + d*x^2])*ArcCosh[1
+ d*x^2]^2 + b^3*d*x^2*ArcCosh[1 + d*x^2]^3)/(d*x)

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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \left (a +b \,\mathrm {arccosh}\left (d \,x^{2}+1\right )\right )^{3}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arccosh(d*x^2+1))^3,x)

[Out]

int((a+b*arccosh(d*x^2+1))^3,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((a+b*arccosh(d*x^2+1))^3,x, algorithm="maxima")

[Out]

3*a*b^2*x*arccosh(d*x^2 + 1)^2 + 12*a*b^2*d*(2*x/d - (d^(3/2)*x^2 + 2*sqrt(d))*log(d*x^2 + sqrt(d*x^2 + 2)*sqr
t(d*x^2) + 1)/(sqrt(d*x^2 + 2)*d^2)) + 3*(x*arccosh(d*x^2 + 1) - 2*(d^(3/2)*x^2 + 2*sqrt(d))/(sqrt(d*x^2 + 2)*
d))*a^2*b + (x*log(d*x^2 + sqrt(d*x^2 + 2)*sqrt(d)*x + 1)^3 - integrate(6*(d^2*x^4 + 2*d*x^2 + (d^(3/2)*x^3 +
sqrt(d)*x)*sqrt(d*x^2 + 2))*log(d*x^2 + sqrt(d*x^2 + 2)*sqrt(d)*x + 1)^2/(d^2*x^4 + 3*d*x^2 + (d^(3/2)*x^3 + 2
*sqrt(d)*x)*sqrt(d*x^2 + 2) + 2), x))*b^3 + a^3*x

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Fricas [A]
time = 0.37, size = 210, normalized size = 1.68 \begin {gather*} \frac {b^{3} d x^{2} \log \left (d x^{2} + \sqrt {d^{2} x^{4} + 2 \, d x^{2}} + 1\right )^{3} + {\left (a^{3} + 24 \, a b^{2}\right )} d x^{2} + 3 \, {\left (a b^{2} d x^{2} - 2 \, \sqrt {d^{2} x^{4} + 2 \, d x^{2}} b^{3}\right )} \log \left (d x^{2} + \sqrt {d^{2} x^{4} + 2 \, d x^{2}} + 1\right )^{2} + 3 \, {\left ({\left (a^{2} b + 8 \, b^{3}\right )} d x^{2} - 4 \, \sqrt {d^{2} x^{4} + 2 \, d x^{2}} a b^{2}\right )} \log \left (d x^{2} + \sqrt {d^{2} x^{4} + 2 \, d x^{2}} + 1\right ) - 6 \, \sqrt {d^{2} x^{4} + 2 \, d x^{2}} {\left (a^{2} b + 8 \, b^{3}\right )}}{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arccosh(d*x^2+1))^3,x, algorithm="fricas")

[Out]

(b^3*d*x^2*log(d*x^2 + sqrt(d^2*x^4 + 2*d*x^2) + 1)^3 + (a^3 + 24*a*b^2)*d*x^2 + 3*(a*b^2*d*x^2 - 2*sqrt(d^2*x
^4 + 2*d*x^2)*b^3)*log(d*x^2 + sqrt(d^2*x^4 + 2*d*x^2) + 1)^2 + 3*((a^2*b + 8*b^3)*d*x^2 - 4*sqrt(d^2*x^4 + 2*
d*x^2)*a*b^2)*log(d*x^2 + sqrt(d^2*x^4 + 2*d*x^2) + 1) - 6*sqrt(d^2*x^4 + 2*d*x^2)*(a^2*b + 8*b^3))/(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 )^{3}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*acosh(d*x**2+1))**3,x)

[Out]

Integral((a + b*acosh(d*x**2 + 1))**3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arccosh(d*x^2+1))^3,x, algorithm="giac")

[Out]

Exception raised: RuntimeError >> An error occurred running a Giac command:INPUT:sage2OUTPUT:Warning, integrat
ion of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [sign(sageVARx)]
index.cc index

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (a+b\,\mathrm {acosh}\left (d\,x^2+1\right )\right )}^3 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*acosh(d*x^2 + 1))^3,x)

[Out]

int((a + b*acosh(d*x^2 + 1))^3, x)

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