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

Optimal. Leaf size=110 \[ -\frac {\left (a+b \cosh ^{-1}(c+d x)\right )^2}{d e^2 (c+d x)}+\frac {4 b \tan ^{-1}\left (e^{\cosh ^{-1}(c+d x)}\right ) \left (a+b \cosh ^{-1}(c+d x)\right )}{d e^2}-\frac {2 i b^2 \text {Li}_2\left (-i e^{\cosh ^{-1}(c+d x)}\right )}{d e^2}+\frac {2 i b^2 \text {Li}_2\left (i e^{\cosh ^{-1}(c+d x)}\right )}{d e^2} \]

[Out]

-(a+b*arccosh(d*x+c))^2/d/e^2/(d*x+c)+4*b*(a+b*arccosh(d*x+c))*arctan(d*x+c+(d*x+c-1)^(1/2)*(d*x+c+1)^(1/2))/d
/e^2-2*I*b^2*polylog(2,-I*(d*x+c+(d*x+c-1)^(1/2)*(d*x+c+1)^(1/2)))/d/e^2+2*I*b^2*polylog(2,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 = 0.24, antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {5866, 12, 5662, 5761, 4180, 2279, 2391} \[ -\frac {2 i b^2 \text {PolyLog}\left (2,-i e^{\cosh ^{-1}(c+d x)}\right )}{d e^2}+\frac {2 i b^2 \text {PolyLog}\left (2,i e^{\cosh ^{-1}(c+d x)}\right )}{d e^2}-\frac {\left (a+b \cosh ^{-1}(c+d x)\right )^2}{d e^2 (c+d x)}+\frac {4 b \tan ^{-1}\left (e^{\cosh ^{-1}(c+d x)}\right ) \left (a+b \cosh ^{-1}(c+d x)\right )}{d e^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a + b*ArcCosh[c + d*x])^2/(d*e^2*(c + d*x))) + (4*b*(a + b*ArcCosh[c + d*x])*ArcTan[E^ArcCosh[c + d*x]])/(d
*e^2) - ((2*I)*b^2*PolyLog[2, (-I)*E^ArcCosh[c + d*x]])/(d*e^2) + ((2*I)*b^2*PolyLog[2, 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 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 4180

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 5662

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))/(Sqr
t[-1 + c*x]*Sqrt[1 + c*x]), x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && NeQ[m, -1]

Rule 5761

Int[(((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_))/(Sqrt[(d1_) + (e1_.)*(x_)]*Sqrt[(d2_) + (e2_.)*(x_)]
), x_Symbol] :> Dist[1/(c^(m + 1)*Sqrt[-(d1*d2)]), Subst[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*d1, 0] && EqQ[e2 + c*d2, 0] && IGtQ[n, 0] && GtQ[d1, 0] &&
LtQ[d2, 0] && IntegerQ[m]

Rule 5866

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
]

Rubi steps

\begin {align*} \int \frac {\left (a+b \cosh ^{-1}(c+d x)\right )^2}{(c e+d e x)^2} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (a+b \cosh ^{-1}(x)\right )^2}{e^2 x^2} \, dx,x,c+d x\right )}{d}\\ &=\frac {\operatorname {Subst}\left (\int \frac {\left (a+b \cosh ^{-1}(x)\right )^2}{x^2} \, dx,x,c+d x\right )}{d e^2}\\ &=-\frac {\left (a+b \cosh ^{-1}(c+d x)\right )^2}{d e^2 (c+d x)}+\frac {(2 b) \operatorname {Subst}\left (\int \frac {a+b \cosh ^{-1}(x)}{\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 )^2}{d e^2 (c+d x)}+\frac {(2 b) \operatorname {Subst}\left (\int (a+b x) \text {sech}(x) \, dx,x,\cosh ^{-1}(c+d x)\right )}{d e^2}\\ &=-\frac {\left (a+b \cosh ^{-1}(c+d x)\right )^2}{d e^2 (c+d x)}+\frac {4 b \left (a+b \cosh ^{-1}(c+d x)\right ) \tan ^{-1}\left (e^{\cosh ^{-1}(c+d x)}\right )}{d e^2}-\frac {\left (2 i b^2\right ) \operatorname {Subst}\left (\int \log \left (1-i e^x\right ) \, dx,x,\cosh ^{-1}(c+d x)\right )}{d e^2}+\frac {\left (2 i b^2\right ) \operatorname {Subst}\left (\int \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 )^2}{d e^2 (c+d x)}+\frac {4 b \left (a+b \cosh ^{-1}(c+d x)\right ) \tan ^{-1}\left (e^{\cosh ^{-1}(c+d x)}\right )}{d e^2}-\frac {\left (2 i b^2\right ) \operatorname {Subst}\left (\int \frac {\log (1-i x)}{x} \, dx,x,e^{\cosh ^{-1}(c+d x)}\right )}{d e^2}+\frac {\left (2 i b^2\right ) \operatorname {Subst}\left (\int \frac {\log (1+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 )^2}{d e^2 (c+d x)}+\frac {4 b \left (a+b \cosh ^{-1}(c+d x)\right ) \tan ^{-1}\left (e^{\cosh ^{-1}(c+d x)}\right )}{d e^2}-\frac {2 i b^2 \text {Li}_2\left (-i e^{\cosh ^{-1}(c+d x)}\right )}{d e^2}+\frac {2 i b^2 \text {Li}_2\left (i e^{\cosh ^{-1}(c+d x)}\right )}{d e^2}\\ \end {align*}

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Mathematica [A]  time = 0.77, size = 161, normalized size = 1.46 \[ \frac {-\frac {a^2}{c+d x}+2 a b \left (2 \tan ^{-1}\left (\tanh \left (\frac {1}{2} \cosh ^{-1}(c+d x)\right )\right )-\frac {\cosh ^{-1}(c+d x)}{c+d x}\right )-i b^2 \left (2 \text {Li}_2\left (-i e^{-\cosh ^{-1}(c+d x)}\right )-2 \text {Li}_2\left (i e^{-\cosh ^{-1}(c+d x)}\right )+\cosh ^{-1}(c+d x) \left (-\frac {i \cosh ^{-1}(c+d x)}{c+d x}+2 \log \left (1-i e^{-\cosh ^{-1}(c+d x)}\right )-2 \log \left (1+i e^{-\cosh ^{-1}(c+d x)}\right )\right )\right )}{d e^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-(a^2/(c + d*x)) + 2*a*b*(-(ArcCosh[c + d*x]/(c + d*x)) + 2*ArcTan[Tanh[ArcCosh[c + d*x]/2]]) - I*b^2*(ArcCos
h[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]]))/(d*e^2)

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fricas [F]  time = 0.62, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{2} \operatorname {arcosh}\left (d x + c\right )^{2} + 2 \, a b \operatorname {arcosh}\left (d x + c\right ) + a^{2}}{d^{2} e^{2} x^{2} + 2 \, c d e^{2} x + c^{2} e^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [ab
s(t_nostep)]sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [A]  time = 0.17, size = 290, normalized size = 2.64 \[ -\frac {a^{2}}{d \,e^{2} \left (d x +c \right )}-\frac {b^{2} \mathrm {arccosh}\left (d x +c \right )^{2}}{d \,e^{2} \left (d x +c \right )}-\frac {2 i b^{2} \mathrm {arccosh}\left (d x +c \right ) \ln \left (1+i \left (d x +c +\sqrt {d x +c -1}\, \sqrt {d x +c +1}\right )\right )}{d \,e^{2}}+\frac {2 i b^{2} \mathrm {arccosh}\left (d x +c \right ) \ln \left (1-i \left (d x +c +\sqrt {d x +c -1}\, \sqrt {d x +c +1}\right )\right )}{d \,e^{2}}-\frac {2 i b^{2} \dilog \left (1+i \left (d x +c +\sqrt {d x +c -1}\, \sqrt {d x +c +1}\right )\right )}{d \,e^{2}}+\frac {2 i b^{2} \dilog \left (1-i \left (d x +c +\sqrt {d x +c -1}\, \sqrt {d x +c +1}\right )\right )}{d \,e^{2}}-\frac {2 a b \,\mathrm {arccosh}\left (d x +c \right )}{d \,e^{2} \left (d x +c \right )}-\frac {2 a b \sqrt {d x +c -1}\, \sqrt {d x +c +1}\, \arctan \left (\frac {1}{\sqrt {\left (d x +c \right )^{2}-1}}\right )}{d \,e^{2} \sqrt {\left (d x +c \right )^{2}-1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/d*a^2/e^2/(d*x+c)-1/d*b^2/e^2*arccosh(d*x+c)^2/(d*x+c)-2*I/d*b^2/e^2*arccosh(d*x+c)*ln(1+I*(d*x+c+(d*x+c-1)
^(1/2)*(d*x+c+1)^(1/2)))+2*I/d*b^2/e^2*arccosh(d*x+c)*ln(1-I*(d*x+c+(d*x+c-1)^(1/2)*(d*x+c+1)^(1/2)))-2*I/d*b^
2/e^2*dilog(1+I*(d*x+c+(d*x+c-1)^(1/2)*(d*x+c+1)^(1/2)))+2*I/d*b^2/e^2*dilog(1-I*(d*x+c+(d*x+c-1)^(1/2)*(d*x+c
+1)^(1/2)))-2/d*a*b/e^2/(d*x+c)*arccosh(d*x+c)-2/d*a*b/e^2*(d*x+c-1)^(1/2)*(d*x+c+1)^(1/2)/((d*x+c)^2-1)^(1/2)
*arctan(1/((d*x+c)^2-1)^(1/2))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -b^{2} {\left (\frac {\log \left (d x + \sqrt {d x + c + 1} \sqrt {d x + c - 1} + c\right )^{2}}{d^{2} e^{2} x + c d e^{2}} - \int \frac {2 \, {\left (d^{2} x^{2} + 2 \, c d x + \sqrt {d x + c + 1} {\left (d x + c\right )} \sqrt {d x + c - 1} + c^{2} - 1\right )} \log \left (d x + \sqrt {d x + c + 1} \sqrt {d x + c - 1} + c\right )}{d^{4} e^{2} x^{4} + 4 \, c d^{3} e^{2} x^{3} + c^{4} e^{2} - c^{2} e^{2} + {\left (6 \, c^{2} d^{2} e^{2} - d^{2} e^{2}\right )} x^{2} + {\left (d^{3} e^{2} x^{3} + 3 \, c d^{2} e^{2} x^{2} + c^{3} e^{2} - c e^{2} + {\left (3 \, c^{2} d e^{2} - d e^{2}\right )} x\right )} \sqrt {d x + c + 1} \sqrt {d x + c - 1} + 2 \, {\left (2 \, c^{3} d e^{2} - c d e^{2}\right )} x}\,{d x}\right )} - 2 \, a b {\left (\frac {\operatorname {arcosh}\left (d x + c\right )}{d^{2} e^{2} x + c d e^{2}} + \frac {\arcsin \left (\frac {d e^{2}}{{\left | d^{2} e^{2} x + c d e^{2} \right |}}\right )}{d e^{2}}\right )} - \frac {a^{2}}{d^{2} e^{2} x + c d e^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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