3.2.19 \(\int \frac {(a+b \cosh ^{-1}(c+d x))^3}{(c e+d e x)^2} \, dx\) [119]

Optimal. Leaf size=186 \[ -\frac {\left (a+b \cosh ^{-1}(c+d x)\right )^3}{d e^2 (c+d x)}+\frac {6 b \left (a+b \cosh ^{-1}(c+d x)\right )^2 \text {ArcTan}\left (e^{\cosh ^{-1}(c+d x)}\right )}{d e^2}-\frac {6 i b^2 \left (a+b \cosh ^{-1}(c+d x)\right ) \text {PolyLog}\left (2,-i e^{\cosh ^{-1}(c+d x)}\right )}{d e^2}+\frac {6 i b^2 \left (a+b \cosh ^{-1}(c+d x)\right ) \text {PolyLog}\left (2,i e^{\cosh ^{-1}(c+d x)}\right )}{d e^2}+\frac {6 i b^3 \text {PolyLog}\left (3,-i e^{\cosh ^{-1}(c+d x)}\right )}{d e^2}-\frac {6 i b^3 \text {PolyLog}\left (3,i e^{\cosh ^{-1}(c+d x)}\right )}{d e^2} \]

[Out]

-(a+b*arccosh(d*x+c))^3/d/e^2/(d*x+c)+6*b*(a+b*arccosh(d*x+c))^2*arctan(d*x+c+(d*x+c-1)^(1/2)*(d*x+c+1)^(1/2))
/d/e^2-6*I*b^2*(a+b*arccosh(d*x+c))*polylog(2,-I*(d*x+c+(d*x+c-1)^(1/2)*(d*x+c+1)^(1/2)))/d/e^2+6*I*b^2*(a+b*a
rccosh(d*x+c))*polylog(2,I*(d*x+c+(d*x+c-1)^(1/2)*(d*x+c+1)^(1/2)))/d/e^2+6*I*b^3*polylog(3,-I*(d*x+c+(d*x+c-1
)^(1/2)*(d*x+c+1)^(1/2)))/d/e^2-6*I*b^3*polylog(3,I*(d*x+c+(d*x+c-1)^(1/2)*(d*x+c+1)^(1/2)))/d/e^2

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Rubi [A]
time = 0.23, antiderivative size = 186, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {5996, 12, 5883, 5947, 4265, 2611, 2320, 6724} \begin {gather*} \frac {6 b \text {ArcTan}\left (e^{\cosh ^{-1}(c+d x)}\right ) \left (a+b \cosh ^{-1}(c+d x)\right )^2}{d e^2}-\frac {6 i b^2 \text {Li}_2\left (-i e^{\cosh ^{-1}(c+d x)}\right ) \left (a+b \cosh ^{-1}(c+d x)\right )}{d e^2}+\frac {6 i b^2 \text {Li}_2\left (i e^{\cosh ^{-1}(c+d x)}\right ) \left (a+b \cosh ^{-1}(c+d x)\right )}{d e^2}-\frac {\left (a+b \cosh ^{-1}(c+d x)\right )^3}{d e^2 (c+d x)}+\frac {6 i b^3 \text {Li}_3\left (-i e^{\cosh ^{-1}(c+d x)}\right )}{d e^2}-\frac {6 i b^3 \text {Li}_3\left (i e^{\cosh ^{-1}(c+d x)}\right )}{d e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((a + b*ArcCosh[c + d*x])^3/(d*e^2*(c + d*x))) + (6*b*(a + b*ArcCosh[c + d*x])^2*ArcTan[E^ArcCosh[c + d*x]])/
(d*e^2) - ((6*I)*b^2*(a + b*ArcCosh[c + d*x])*PolyLog[2, (-I)*E^ArcCosh[c + d*x]])/(d*e^2) + ((6*I)*b^2*(a + b
*ArcCosh[c + d*x])*PolyLog[2, I*E^ArcCosh[c + d*x]])/(d*e^2) + ((6*I)*b^3*PolyLog[3, (-I)*E^ArcCosh[c + d*x]])
/(d*e^2) - ((6*I)*b^3*PolyLog[3, I*E^ArcCosh[c + d*x]])/(d*e^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 2320

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 2611

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> Simp[(-(
f + g*x)^m)*(PolyLog[2, (-e)*(F^(c*(a + b*x)))^n]/(b*c*n*Log[F])), x] + Dist[g*(m/(b*c*n*Log[F])), Int[(f + g*
x)^(m - 1)*PolyLog[2, (-e)*(F^(c*(a + b*x)))^n], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

Rule 4265

Int[csc[(e_.) + Pi*(k_.) + (Complex[0, fz_])*(f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[-2*(c +
d*x)^m*(ArcTanh[E^((-I)*e + f*fz*x)/E^(I*k*Pi)]/(f*fz*I)), x] + (-Dist[d*(m/(f*fz*I)), Int[(c + d*x)^(m - 1)*
Log[1 - E^((-I)*e + f*fz*x)/E^(I*k*Pi)], x], x] + Dist[d*(m/(f*fz*I)), Int[(c + d*x)^(m - 1)*Log[1 + E^((-I)*e
+ f*fz*x)/E^(I*k*Pi)], x], x]) /; FreeQ[{c, d, e, f, fz}, x] && IntegerQ[2*k] && IGtQ[m, 0]

Rule 5883

Int[((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))^(n_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(d*x)^(m + 1)*((a + b*ArcC
osh[c*x])^n/(d*(m + 1))), x] - Dist[b*c*(n/(d*(m + 1))), Int[(d*x)^(m + 1)*((a + b*ArcCosh[c*x])^(n - 1)/(Sqrt
[1 + c*x]*Sqrt[-1 + c*x])), x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && NeQ[m, -1]

Rule 5947

Int[(((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_))/(Sqrt[(d1_) + (e1_.)*(x_)]*Sqrt[(d2_) + (e2_.)*(x_)]
), x_Symbol] :> Dist[(1/c^(m + 1))*Simp[Sqrt[1 + c*x]/Sqrt[d1 + e1*x]]*Simp[Sqrt[-1 + c*x]/Sqrt[d2 + e2*x]], S
ubst[Int[(a + b*x)^n*Cosh[x]^m, x], x, ArcCosh[c*x]], x] /; FreeQ[{a, b, c, d1, e1, d2, e2}, x] && EqQ[e1, c*d
1] && EqQ[e2, (-c)*d2] && IGtQ[n, 0] && IntegerQ[m]

Rule 5996

Int[((a_.) + ArcCosh[(c_) + (d_.)*(x_)]*(b_.))^(n_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[1/d, Subst[
Int[((d*e - c*f)/d + f*(x/d))^m*(a + b*ArcCosh[x])^n, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x
]

Rule 6724

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

Rubi steps

\begin {align*} \int \frac {\left (a+b \cosh ^{-1}(c+d x)\right )^3}{(c e+d e x)^2} \, dx &=\frac {\text {Subst}\left (\int \frac {\left (a+b \cosh ^{-1}(x)\right )^3}{e^2 x^2} \, dx,x,c+d x\right )}{d}\\ &=\frac {\text {Subst}\left (\int \frac {\left (a+b \cosh ^{-1}(x)\right )^3}{x^2} \, dx,x,c+d x\right )}{d e^2}\\ &=-\frac {\left (a+b \cosh ^{-1}(c+d x)\right )^3}{d e^2 (c+d x)}+\frac {(3 b) \text {Subst}\left (\int \frac {\left (a+b \cosh ^{-1}(x)\right )^2}{\sqrt {-1+x} x \sqrt {1+x}} \, dx,x,c+d x\right )}{d e^2}\\ &=-\frac {\left (a+b \cosh ^{-1}(c+d x)\right )^3}{d e^2 (c+d x)}+\frac {(3 b) \text {Subst}\left (\int (a+b x)^2 \text {sech}(x) \, dx,x,\cosh ^{-1}(c+d x)\right )}{d e^2}\\ &=-\frac {\left (a+b \cosh ^{-1}(c+d x)\right )^3}{d e^2 (c+d x)}+\frac {6 b \left (a+b \cosh ^{-1}(c+d x)\right )^2 \tan ^{-1}\left (e^{\cosh ^{-1}(c+d x)}\right )}{d e^2}-\frac {\left (6 i b^2\right ) \text {Subst}\left (\int (a+b x) \log \left (1-i e^x\right ) \, dx,x,\cosh ^{-1}(c+d x)\right )}{d e^2}+\frac {\left (6 i b^2\right ) \text {Subst}\left (\int (a+b x) \log \left (1+i e^x\right ) \, dx,x,\cosh ^{-1}(c+d x)\right )}{d e^2}\\ &=-\frac {\left (a+b \cosh ^{-1}(c+d x)\right )^3}{d e^2 (c+d x)}+\frac {6 b \left (a+b \cosh ^{-1}(c+d x)\right )^2 \tan ^{-1}\left (e^{\cosh ^{-1}(c+d x)}\right )}{d e^2}-\frac {6 i b^2 \left (a+b \cosh ^{-1}(c+d x)\right ) \text {Li}_2\left (-i e^{\cosh ^{-1}(c+d x)}\right )}{d e^2}+\frac {6 i b^2 \left (a+b \cosh ^{-1}(c+d x)\right ) \text {Li}_2\left (i e^{\cosh ^{-1}(c+d x)}\right )}{d e^2}+\frac {\left (6 i b^3\right ) \text {Subst}\left (\int \text {Li}_2\left (-i e^x\right ) \, dx,x,\cosh ^{-1}(c+d x)\right )}{d e^2}-\frac {\left (6 i b^3\right ) \text {Subst}\left (\int \text {Li}_2\left (i e^x\right ) \, dx,x,\cosh ^{-1}(c+d x)\right )}{d e^2}\\ &=-\frac {\left (a+b \cosh ^{-1}(c+d x)\right )^3}{d e^2 (c+d x)}+\frac {6 b \left (a+b \cosh ^{-1}(c+d x)\right )^2 \tan ^{-1}\left (e^{\cosh ^{-1}(c+d x)}\right )}{d e^2}-\frac {6 i b^2 \left (a+b \cosh ^{-1}(c+d x)\right ) \text {Li}_2\left (-i e^{\cosh ^{-1}(c+d x)}\right )}{d e^2}+\frac {6 i b^2 \left (a+b \cosh ^{-1}(c+d x)\right ) \text {Li}_2\left (i e^{\cosh ^{-1}(c+d x)}\right )}{d e^2}+\frac {\left (6 i b^3\right ) \text {Subst}\left (\int \frac {\text {Li}_2(-i x)}{x} \, dx,x,e^{\cosh ^{-1}(c+d x)}\right )}{d e^2}-\frac {\left (6 i b^3\right ) \text {Subst}\left (\int \frac {\text {Li}_2(i x)}{x} \, dx,x,e^{\cosh ^{-1}(c+d x)}\right )}{d e^2}\\ &=-\frac {\left (a+b \cosh ^{-1}(c+d x)\right )^3}{d e^2 (c+d x)}+\frac {6 b \left (a+b \cosh ^{-1}(c+d x)\right )^2 \tan ^{-1}\left (e^{\cosh ^{-1}(c+d x)}\right )}{d e^2}-\frac {6 i b^2 \left (a+b \cosh ^{-1}(c+d x)\right ) \text {Li}_2\left (-i e^{\cosh ^{-1}(c+d x)}\right )}{d e^2}+\frac {6 i b^2 \left (a+b \cosh ^{-1}(c+d x)\right ) \text {Li}_2\left (i e^{\cosh ^{-1}(c+d x)}\right )}{d e^2}+\frac {6 i b^3 \text {Li}_3\left (-i e^{\cosh ^{-1}(c+d x)}\right )}{d e^2}-\frac {6 i b^3 \text {Li}_3\left (i e^{\cosh ^{-1}(c+d x)}\right )}{d e^2}\\ \end {align*}

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Mathematica [A]
time = 0.85, size = 327, normalized size = 1.76 \begin {gather*} -\frac {\frac {a^3}{c+d x}+\frac {3 a^2 b \cosh ^{-1}(c+d x)}{c+d x}+3 a^2 b \text {ArcTan}\left (\frac {1}{\sqrt {-1+c+d x} \sqrt {1+c+d x}}\right )+3 i a b^2 \left (\cosh ^{-1}(c+d x) \left (-\frac {i \cosh ^{-1}(c+d x)}{c+d x}+2 \log \left (1-i e^{-\cosh ^{-1}(c+d x)}\right )-2 \log \left (1+i e^{-\cosh ^{-1}(c+d x)}\right )\right )+2 \text {PolyLog}\left (2,-i e^{-\cosh ^{-1}(c+d x)}\right )-2 \text {PolyLog}\left (2,i e^{-\cosh ^{-1}(c+d x)}\right )\right )+b^3 \left (\frac {\cosh ^{-1}(c+d x)^3}{c+d x}-3 i \left (-\cosh ^{-1}(c+d x)^2 \left (\log \left (1-i e^{-\cosh ^{-1}(c+d x)}\right )-\log \left (1+i e^{-\cosh ^{-1}(c+d x)}\right )\right )-2 \cosh ^{-1}(c+d x) \left (\text {PolyLog}\left (2,-i e^{-\cosh ^{-1}(c+d x)}\right )-\text {PolyLog}\left (2,i e^{-\cosh ^{-1}(c+d x)}\right )\right )-2 \text {PolyLog}\left (3,-i e^{-\cosh ^{-1}(c+d x)}\right )+2 \text {PolyLog}\left (3,i e^{-\cosh ^{-1}(c+d x)}\right )\right )\right )}{d e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((a^3/(c + d*x) + (3*a^2*b*ArcCosh[c + d*x])/(c + d*x) + 3*a^2*b*ArcTan[1/(Sqrt[-1 + c + d*x]*Sqrt[1 + c + d*
x])] + (3*I)*a*b^2*(ArcCosh[c + d*x]*(((-I)*ArcCosh[c + d*x])/(c + d*x) + 2*Log[1 - I/E^ArcCosh[c + d*x]] - 2*
Log[1 + I/E^ArcCosh[c + d*x]]) + 2*PolyLog[2, (-I)/E^ArcCosh[c + d*x]] - 2*PolyLog[2, I/E^ArcCosh[c + d*x]]) +
b^3*(ArcCosh[c + d*x]^3/(c + d*x) - (3*I)*(-(ArcCosh[c + d*x]^2*(Log[1 - I/E^ArcCosh[c + d*x]] - Log[1 + I/E^
ArcCosh[c + d*x]])) - 2*ArcCosh[c + d*x]*(PolyLog[2, (-I)/E^ArcCosh[c + d*x]] - PolyLog[2, I/E^ArcCosh[c + d*x
]]) - 2*PolyLog[3, (-I)/E^ArcCosh[c + d*x]] + 2*PolyLog[3, I/E^ArcCosh[c + d*x]])))/(d*e^2))

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Maple [F]
time = 0.18, size = 0, normalized size = 0.00 \[\int \frac {\left (a +b \,\mathrm {arccosh}\left (d x +c \right )\right )^{3}}{\left (d e x +c e \right )^{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

-b^3*log(d*x + sqrt(d*x + c + 1)*sqrt(d*x + c - 1) + c)^3/(d^2*x*e^2 + c*d*e^2) - 3*a^2*b*(arcsin(d*e^2/abs(d^
2*x*e^2 + c*d*e^2))*e^(-2)/d + arccosh(d*x + c)/(d^2*x*e^2 + c*d*e^2)) - a^3/(d^2*x*e^2 + c*d*e^2) + integrate
(3*((c^3 - c)*a*b^2 + (c^3 - c)*b^3 + (a*b^2*d^3 + b^3*d^3)*x^3 + 3*(a*b^2*c*d^2 + b^3*c*d^2)*x^2 + (b^3*c^2 +
(c^2 - 1)*a*b^2 + (a*b^2*d^2 + b^3*d^2)*x^2 + 2*(a*b^2*c*d + b^3*c*d)*x)*sqrt(d*x + c + 1)*sqrt(d*x + c - 1)
+ ((3*c^2*d - d)*a*b^2 + (3*c^2*d - d)*b^3)*x)*log(d*x + sqrt(d*x + c + 1)*sqrt(d*x + c - 1) + c)^2/(d^5*x^5*e
^2 + 5*c*d^4*x^4*e^2 + (10*c^2*d^3 - d^3)*x^3*e^2 + (10*c^3*d^2 - 3*c*d^2)*x^2*e^2 + (5*c^4*d - 3*c^2*d)*x*e^2
+ (d^4*x^4*e^2 + 4*c*d^3*x^3*e^2 + (6*c^2*d^2 - d^2)*x^2*e^2 + 2*(2*c^3*d - c*d)*x*e^2 + (c^4 - c^2)*e^2)*sqr
t(d*x + c + 1)*sqrt(d*x + c - 1) + (c^5 - c^3)*e^2), x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Warning, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Ch
eck [abs(t_

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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