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

Optimal. Leaf size=100 \[ -\frac {\left (a+b \sinh ^{-1}(c+d x)\right )^2}{d e^2 (c+d x)}-\frac {4 b \tanh ^{-1}\left (e^{\sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )}{d e^2}-\frac {2 b^2 \text {Li}_2\left (-e^{\sinh ^{-1}(c+d x)}\right )}{d e^2}+\frac {2 b^2 \text {Li}_2\left (e^{\sinh ^{-1}(c+d x)}\right )}{d e^2} \]

[Out]

-(a+b*arcsinh(d*x+c))^2/d/e^2/(d*x+c)-4*b*(a+b*arcsinh(d*x+c))*arctanh(d*x+c+(1+(d*x+c)^2)^(1/2))/d/e^2-2*b^2*
polylog(2,-d*x-c-(1+(d*x+c)^2)^(1/2))/d/e^2+2*b^2*polylog(2,d*x+c+(1+(d*x+c)^2)^(1/2))/d/e^2

________________________________________________________________________________________

Rubi [A]  time = 0.18, antiderivative size = 100, 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 = {5865, 12, 5661, 5760, 4182, 2279, 2391} \[ -\frac {2 b^2 \text {PolyLog}\left (2,-e^{\sinh ^{-1}(c+d x)}\right )}{d e^2}+\frac {2 b^2 \text {PolyLog}\left (2,e^{\sinh ^{-1}(c+d x)}\right )}{d e^2}-\frac {\left (a+b \sinh ^{-1}(c+d x)\right )^2}{d e^2 (c+d x)}-\frac {4 b \tanh ^{-1}\left (e^{\sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )}{d e^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a + b*ArcSinh[c + d*x])^2/(d*e^2*(c + d*x))) - (4*b*(a + b*ArcSinh[c + d*x])*ArcTanh[E^ArcSinh[c + d*x]])/(
d*e^2) - (2*b^2*PolyLog[2, -E^ArcSinh[c + d*x]])/(d*e^2) + (2*b^2*PolyLog[2, E^ArcSinh[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 4182

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

Rule 5661

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcS
inh[c*x])^n)/(d*(m + 1)), x] - Dist[(b*c*n)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcSinh[c*x])^(n - 1))/Sqrt
[1 + c^2*x^2], x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && NeQ[m, -1]

Rule 5760

Int[(((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Dist[1/(c^(m
+ 1)*Sqrt[d]), Subst[Int[(a + b*x)^n*Sinh[x]^m, x], x, ArcSinh[c*x]], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[
e, c^2*d] && GtQ[d, 0] && IGtQ[n, 0] && IntegerQ[m]

Rule 5865

Int[((a_.) + ArcSinh[(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*ArcSinh[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 \sinh ^{-1}(c+d x)\right )^2}{(c e+d e x)^2} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (a+b \sinh ^{-1}(x)\right )^2}{e^2 x^2} \, dx,x,c+d x\right )}{d}\\ &=\frac {\operatorname {Subst}\left (\int \frac {\left (a+b \sinh ^{-1}(x)\right )^2}{x^2} \, dx,x,c+d x\right )}{d e^2}\\ &=-\frac {\left (a+b \sinh ^{-1}(c+d x)\right )^2}{d e^2 (c+d x)}+\frac {(2 b) \operatorname {Subst}\left (\int \frac {a+b \sinh ^{-1}(x)}{x \sqrt {1+x^2}} \, dx,x,c+d x\right )}{d e^2}\\ &=-\frac {\left (a+b \sinh ^{-1}(c+d x)\right )^2}{d e^2 (c+d x)}+\frac {(2 b) \operatorname {Subst}\left (\int (a+b x) \text {csch}(x) \, dx,x,\sinh ^{-1}(c+d x)\right )}{d e^2}\\ &=-\frac {\left (a+b \sinh ^{-1}(c+d x)\right )^2}{d e^2 (c+d x)}-\frac {4 b \left (a+b \sinh ^{-1}(c+d x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c+d x)}\right )}{d e^2}-\frac {\left (2 b^2\right ) \operatorname {Subst}\left (\int \log \left (1-e^x\right ) \, dx,x,\sinh ^{-1}(c+d x)\right )}{d e^2}+\frac {\left (2 b^2\right ) \operatorname {Subst}\left (\int \log \left (1+e^x\right ) \, dx,x,\sinh ^{-1}(c+d x)\right )}{d e^2}\\ &=-\frac {\left (a+b \sinh ^{-1}(c+d x)\right )^2}{d e^2 (c+d x)}-\frac {4 b \left (a+b \sinh ^{-1}(c+d x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c+d x)}\right )}{d e^2}-\frac {\left (2 b^2\right ) \operatorname {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{\sinh ^{-1}(c+d x)}\right )}{d e^2}+\frac {\left (2 b^2\right ) \operatorname {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{\sinh ^{-1}(c+d x)}\right )}{d e^2}\\ &=-\frac {\left (a+b \sinh ^{-1}(c+d x)\right )^2}{d e^2 (c+d x)}-\frac {4 b \left (a+b \sinh ^{-1}(c+d x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c+d x)}\right )}{d e^2}-\frac {2 b^2 \text {Li}_2\left (-e^{\sinh ^{-1}(c+d x)}\right )}{d e^2}+\frac {2 b^2 \text {Li}_2\left (e^{\sinh ^{-1}(c+d x)}\right )}{d e^2}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.60, size = 154, normalized size = 1.54 \[ \frac {-\frac {a^2}{c+d x}+a b \left (2 \log \left (\frac {2 \sinh ^2\left (\frac {1}{2} \sinh ^{-1}(c+d x)\right )}{c+d x}\right )-\frac {2 \sinh ^{-1}(c+d x)}{c+d x}\right )+b^2 \left (2 \text {Li}_2\left (-e^{-\sinh ^{-1}(c+d x)}\right )-2 \text {Li}_2\left (e^{-\sinh ^{-1}(c+d x)}\right )+\sinh ^{-1}(c+d x) \left (-\frac {\sinh ^{-1}(c+d x)}{c+d x}+2 \log \left (1-e^{-\sinh ^{-1}(c+d x)}\right )-2 \log \left (e^{-\sinh ^{-1}(c+d x)}+1\right )\right )\right )}{d e^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-(a^2/(c + d*x)) + a*b*((-2*ArcSinh[c + d*x])/(c + d*x) + 2*Log[(2*Sinh[ArcSinh[c + d*x]/2]^2)/(c + d*x)]) +
b^2*(ArcSinh[c + d*x]*(-(ArcSinh[c + d*x]/(c + d*x)) + 2*Log[1 - E^(-ArcSinh[c + d*x])] - 2*Log[1 + E^(-ArcSin
h[c + d*x])]) + 2*PolyLog[2, -E^(-ArcSinh[c + d*x])] - 2*PolyLog[2, E^(-ArcSinh[c + d*x])]))/(d*e^2)

________________________________________________________________________________________

fricas [F]  time = 0.55, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{2} \operatorname {arsinh}\left (d x + c\right )^{2} + 2 \, a b \operatorname {arsinh}\left (d x + c\right ) + a^{2}}{d^{2} e^{2} x^{2} + 2 \, c d e^{2} x + c^{2} e^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.14, size = 229, normalized size = 2.29 \[ -\frac {a^{2}}{d \,e^{2} \left (d x +c \right )}-\frac {b^{2} \arcsinh \left (d x +c \right )^{2}}{d \,e^{2} \left (d x +c \right )}-\frac {2 b^{2} \arcsinh \left (d x +c \right ) \ln \left (1+d x +c +\sqrt {1+\left (d x +c \right )^{2}}\right )}{d \,e^{2}}-\frac {2 b^{2} \polylog \left (2, -d x -c -\sqrt {1+\left (d x +c \right )^{2}}\right )}{d \,e^{2}}+\frac {2 b^{2} \arcsinh \left (d x +c \right ) \ln \left (1-d x -c -\sqrt {1+\left (d x +c \right )^{2}}\right )}{d \,e^{2}}+\frac {2 b^{2} \polylog \left (2, d x +c +\sqrt {1+\left (d x +c \right )^{2}}\right )}{d \,e^{2}}-\frac {2 a b \arcsinh \left (d x +c \right )}{d \,e^{2} \left (d x +c \right )}-\frac {2 a b \arctanh \left (\frac {1}{\sqrt {1+\left (d x +c \right )^{2}}}\right )}{d \,e^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsinh(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*arcsinh(d*x+c)^2/(d*x+c)-2/d*b^2/e^2*arcsinh(d*x+c)*ln(1+d*x+c+(1+(d*x+c)^2)^
(1/2))-2*b^2*polylog(2,-d*x-c-(1+(d*x+c)^2)^(1/2))/d/e^2+2/d*b^2/e^2*arcsinh(d*x+c)*ln(1-d*x-c-(1+(d*x+c)^2)^(
1/2))+2*b^2*polylog(2,d*x+c+(1+(d*x+c)^2)^(1/2))/d/e^2-2/d*a*b/e^2/(d*x+c)*arcsinh(d*x+c)-2/d*a*b/e^2*arctanh(
1/(1+(d*x+c)^2)^(1/2))

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

________________________________________________________________________________________