3.2.10 \(\int \frac {(a+b \cosh ^{-1}(c+d x))^2}{(c e+d e x)^2} \, dx\) [110]

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

[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 = 0.16, 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 = {5996, 12, 5883, 5947, 4265, 2317, 2438} \begin {gather*} \frac {4 b \text {ArcTan}\left (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 )^2}{d e^2 (c+d x)}-\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 {gather*}

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 2317

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 2438

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 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
]

Rubi steps

\begin {align*} \int \frac {\left (a+b \cosh ^{-1}(c+d x)\right )^2}{(c e+d e x)^2} \, dx &=\frac {\text {Subst}\left (\int \frac {\left (a+b \cosh ^{-1}(x)\right )^2}{e^2 x^2} \, dx,x,c+d x\right )}{d}\\ &=\frac {\text {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) \text {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) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 = 0.54, size = 161, normalized size = 1.46 \begin {gather*} \frac {-\frac {a^2}{c+d x}+2 a b \left (-\frac {\cosh ^{-1}(c+d x)}{c+d x}+2 \text {ArcTan}\left (\tanh \left (\frac {1}{2} \cosh ^{-1}(c+d x)\right )\right )\right )-i 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 )}{d e^2} \end {gather*}

Antiderivative was successfully verified.

[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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Maple [A]
time = 19.84, size = 270, normalized size = 2.45

method	result	size
derivativedivides	\(\frac {-\frac {a^{2}}{e^{2} \left (d x +c \right )}-\frac {b^{2} \mathrm {arccosh}\left (d x +c \right )^{2}}{e^{2} \left (d x +c \right )}-\frac {2 i b^{2} \mathrm {arccosh}\left (d x +c \right ) \ln \left (1+i \left (d x +c +\sqrt {d x +c -1}\, \sqrt {d x +c +1}\right )\right )}{e^{2}}+\frac {2 i b^{2} \mathrm {arccosh}\left (d x +c \right ) \ln \left (1-i \left (d x +c +\sqrt {d x +c -1}\, \sqrt {d x +c +1}\right )\right )}{e^{2}}-\frac {2 i b^{2} \dilog \left (1+i \left (d x +c +\sqrt {d x +c -1}\, \sqrt {d x +c +1}\right )\right )}{e^{2}}+\frac {2 i b^{2} \dilog \left (1-i \left (d x +c +\sqrt {d x +c -1}\, \sqrt {d x +c +1}\right )\right )}{e^{2}}-\frac {2 a b \,\mathrm {arccosh}\left (d x +c \right )}{e^{2} \left (d x +c \right )}-\frac {2 a b \sqrt {d x +c -1}\, \sqrt {d x +c +1}\, \arctan \left (\frac {1}{\sqrt {\left (d x +c \right )^{2}-1}}\right )}{e^{2} \sqrt {\left (d x +c \right )^{2}-1}}}{d}\)	\(270\)
default	\(\frac {-\frac {a^{2}}{e^{2} \left (d x +c \right )}-\frac {b^{2} \mathrm {arccosh}\left (d x +c \right )^{2}}{e^{2} \left (d x +c \right )}-\frac {2 i b^{2} \mathrm {arccosh}\left (d x +c \right ) \ln \left (1+i \left (d x +c +\sqrt {d x +c -1}\, \sqrt {d x +c +1}\right )\right )}{e^{2}}+\frac {2 i b^{2} \mathrm {arccosh}\left (d x +c \right ) \ln \left (1-i \left (d x +c +\sqrt {d x +c -1}\, \sqrt {d x +c +1}\right )\right )}{e^{2}}-\frac {2 i b^{2} \dilog \left (1+i \left (d x +c +\sqrt {d x +c -1}\, \sqrt {d x +c +1}\right )\right )}{e^{2}}+\frac {2 i b^{2} \dilog \left (1-i \left (d x +c +\sqrt {d x +c -1}\, \sqrt {d x +c +1}\right )\right )}{e^{2}}-\frac {2 a b \,\mathrm {arccosh}\left (d x +c \right )}{e^{2} \left (d x +c \right )}-\frac {2 a b \sqrt {d x +c -1}\, \sqrt {d x +c +1}\, \arctan \left (\frac {1}{\sqrt {\left (d x +c \right )^{2}-1}}\right )}{e^{2} \sqrt {\left (d x +c \right )^{2}-1}}}{d}\)	\(270\)

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,method=_RETURNVERBOSE)

[Out]

1/d*(-a^2/e^2/(d*x+c)-b^2/e^2*arccosh(d*x+c)^2/(d*x+c)-2*I*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*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*b^2/e^2*dil
og(1+I*(d*x+c+(d*x+c-1)^(1/2)*(d*x+c+1)^(1/2)))+2*I*b^2/e^2*dilog(1-I*(d*x+c+(d*x+c-1)^(1/2)*(d*x+c+1)^(1/2)))
-2*a*b/e^2/(d*x+c)*arccosh(d*x+c)-2*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 \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))^2/(d*e*x+c*e)^2,x, algorithm="maxima")

[Out]

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

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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))^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)*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^{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}} \end {gather*}

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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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))^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,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 )}^2}{{\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))^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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