3.135 \(\int \frac{(a+b \sinh ^{-1}(c+d x))^2}{(c e+d e x)^4} \, dx\)

Optimal. Leaf size=169 \[ \frac{b^2 \text{PolyLog}\left (2,-e^{\sinh ^{-1}(c+d x)}\right )}{3 d e^4}-\frac{b^2 \text{PolyLog}\left (2,e^{\sinh ^{-1}(c+d x)}\right )}{3 d e^4}-\frac{b \sqrt{(c+d x)^2+1} \left (a+b \sinh ^{-1}(c+d x)\right )}{3 d e^4 (c+d x)^2}-\frac{\left (a+b \sinh ^{-1}(c+d x)\right )^2}{3 d e^4 (c+d x)^3}+\frac{2 b \tanh ^{-1}\left (e^{\sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )}{3 d e^4}-\frac{b^2}{3 d e^4 (c+d x)} \]

[Out]

-b^2/(3*d*e^4*(c + d*x)) - (b*Sqrt[1 + (c + d*x)^2]*(a + b*ArcSinh[c + d*x]))/(3*d*e^4*(c + d*x)^2) - (a + b*A
rcSinh[c + d*x])^2/(3*d*e^4*(c + d*x)^3) + (2*b*(a + b*ArcSinh[c + d*x])*ArcTanh[E^ArcSinh[c + d*x]])/(3*d*e^4
) + (b^2*PolyLog[2, -E^ArcSinh[c + d*x]])/(3*d*e^4) - (b^2*PolyLog[2, E^ArcSinh[c + d*x]])/(3*d*e^4)

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Rubi [A]  time = 0.246695, antiderivative size = 169, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 9, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.391, Rules used = {5865, 12, 5661, 5747, 5760, 4182, 2279, 2391, 30} \[ \frac{b^2 \text{PolyLog}\left (2,-e^{\sinh ^{-1}(c+d x)}\right )}{3 d e^4}-\frac{b^2 \text{PolyLog}\left (2,e^{\sinh ^{-1}(c+d x)}\right )}{3 d e^4}-\frac{b \sqrt{(c+d x)^2+1} \left (a+b \sinh ^{-1}(c+d x)\right )}{3 d e^4 (c+d x)^2}-\frac{\left (a+b \sinh ^{-1}(c+d x)\right )^2}{3 d e^4 (c+d x)^3}+\frac{2 b \tanh ^{-1}\left (e^{\sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )}{3 d e^4}-\frac{b^2}{3 d e^4 (c+d x)} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSinh[c + d*x])^2/(c*e + d*e*x)^4,x]

[Out]

-b^2/(3*d*e^4*(c + d*x)) - (b*Sqrt[1 + (c + d*x)^2]*(a + b*ArcSinh[c + d*x]))/(3*d*e^4*(c + d*x)^2) - (a + b*A
rcSinh[c + d*x])^2/(3*d*e^4*(c + d*x)^3) + (2*b*(a + b*ArcSinh[c + d*x])*ArcTanh[E^ArcSinh[c + d*x]])/(3*d*e^4
) + (b^2*PolyLog[2, -E^ArcSinh[c + d*x]])/(3*d*e^4) - (b^2*PolyLog[2, E^ArcSinh[c + d*x]])/(3*d*e^4)

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
]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 5747

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Simp[
((f*x)^(m + 1)*(d + e*x^2)^(p + 1)*(a + b*ArcSinh[c*x])^n)/(d*f*(m + 1)), x] + (-Dist[(c^2*(m + 2*p + 3))/(f^2
*(m + 1)), Int[(f*x)^(m + 2)*(d + e*x^2)^p*(a + b*ArcSinh[c*x])^n, x], x] - Dist[(b*c*n*d^IntPart[p]*(d + e*x^
2)^FracPart[p])/(f*(m + 1)*(1 + c^2*x^2)^FracPart[p]), Int[(f*x)^(m + 1)*(1 + c^2*x^2)^(p + 1/2)*(a + b*ArcSin
h[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e, f, p}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && LtQ[m, -1] && Int
egerQ[m]

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 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 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 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

\begin{align*} \int \frac{\left (a+b \sinh ^{-1}(c+d x)\right )^2}{(c e+d e x)^4} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (a+b \sinh ^{-1}(x)\right )^2}{e^4 x^4} \, dx,x,c+d x\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \frac{\left (a+b \sinh ^{-1}(x)\right )^2}{x^4} \, dx,x,c+d x\right )}{d e^4}\\ &=-\frac{\left (a+b \sinh ^{-1}(c+d x)\right )^2}{3 d e^4 (c+d x)^3}+\frac{(2 b) \operatorname{Subst}\left (\int \frac{a+b \sinh ^{-1}(x)}{x^3 \sqrt{1+x^2}} \, dx,x,c+d x\right )}{3 d e^4}\\ &=-\frac{b \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )}{3 d e^4 (c+d x)^2}-\frac{\left (a+b \sinh ^{-1}(c+d x)\right )^2}{3 d e^4 (c+d x)^3}-\frac{b \operatorname{Subst}\left (\int \frac{a+b \sinh ^{-1}(x)}{x \sqrt{1+x^2}} \, dx,x,c+d x\right )}{3 d e^4}+\frac{b^2 \operatorname{Subst}\left (\int \frac{1}{x^2} \, dx,x,c+d x\right )}{3 d e^4}\\ &=-\frac{b^2}{3 d e^4 (c+d x)}-\frac{b \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )}{3 d e^4 (c+d x)^2}-\frac{\left (a+b \sinh ^{-1}(c+d x)\right )^2}{3 d e^4 (c+d x)^3}-\frac{b \operatorname{Subst}\left (\int (a+b x) \text{csch}(x) \, dx,x,\sinh ^{-1}(c+d x)\right )}{3 d e^4}\\ &=-\frac{b^2}{3 d e^4 (c+d x)}-\frac{b \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )}{3 d e^4 (c+d x)^2}-\frac{\left (a+b \sinh ^{-1}(c+d x)\right )^2}{3 d e^4 (c+d x)^3}+\frac{2 b \left (a+b \sinh ^{-1}(c+d x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c+d x)}\right )}{3 d e^4}+\frac{b^2 \operatorname{Subst}\left (\int \log \left (1-e^x\right ) \, dx,x,\sinh ^{-1}(c+d x)\right )}{3 d e^4}-\frac{b^2 \operatorname{Subst}\left (\int \log \left (1+e^x\right ) \, dx,x,\sinh ^{-1}(c+d x)\right )}{3 d e^4}\\ &=-\frac{b^2}{3 d e^4 (c+d x)}-\frac{b \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )}{3 d e^4 (c+d x)^2}-\frac{\left (a+b \sinh ^{-1}(c+d x)\right )^2}{3 d e^4 (c+d x)^3}+\frac{2 b \left (a+b \sinh ^{-1}(c+d x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c+d x)}\right )}{3 d e^4}+\frac{b^2 \operatorname{Subst}\left (\int \frac{\log (1-x)}{x} \, dx,x,e^{\sinh ^{-1}(c+d x)}\right )}{3 d e^4}-\frac{b^2 \operatorname{Subst}\left (\int \frac{\log (1+x)}{x} \, dx,x,e^{\sinh ^{-1}(c+d x)}\right )}{3 d e^4}\\ &=-\frac{b^2}{3 d e^4 (c+d x)}-\frac{b \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )}{3 d e^4 (c+d x)^2}-\frac{\left (a+b \sinh ^{-1}(c+d x)\right )^2}{3 d e^4 (c+d x)^3}+\frac{2 b \left (a+b \sinh ^{-1}(c+d x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c+d x)}\right )}{3 d e^4}+\frac{b^2 \text{Li}_2\left (-e^{\sinh ^{-1}(c+d x)}\right )}{3 d e^4}-\frac{b^2 \text{Li}_2\left (e^{\sinh ^{-1}(c+d x)}\right )}{3 d e^4}\\ \end{align*}

Mathematica [A]  time = 1.67858, size = 212, normalized size = 1.25 \[ -\frac{b^2 \left (4 (c+d x)^3 \text{PolyLog}\left (2,-e^{-\sinh ^{-1}(c+d x)}\right )-4 (c+d x)^3 \text{PolyLog}\left (2,e^{-\sinh ^{-1}(c+d x)}\right )+4 (c+d x)^2+4 \sinh ^{-1}(c+d x)^2+\sinh ^{-1}(c+d x) \left (2 \sinh \left (2 \sinh ^{-1}(c+d x)\right )+\left (\sinh \left (3 \sinh ^{-1}(c+d x)\right )-3 (c+d x)\right ) \left (\log \left (1-e^{-\sinh ^{-1}(c+d x)}\right )-\log \left (e^{-\sinh ^{-1}(c+d x)}+1\right )\right )\right )\right )+4 a^2+a b \left (8 \sinh ^{-1}(c+d x)+2 \sinh \left (2 \sinh ^{-1}(c+d x)\right )+\left (\sinh \left (3 \sinh ^{-1}(c+d x)\right )-3 (c+d x)\right ) \log \left (\tanh \left (\frac{1}{2} \sinh ^{-1}(c+d x)\right )\right )\right )}{12 d e^4 (c+d x)^3} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*ArcSinh[c + d*x])^2/(c*e + d*e*x)^4,x]

[Out]

-(4*a^2 + a*b*(8*ArcSinh[c + d*x] + 2*Sinh[2*ArcSinh[c + d*x]] + Log[Tanh[ArcSinh[c + d*x]/2]]*(-3*(c + d*x) +
Sinh[3*ArcSinh[c + d*x]])) + b^2*(4*(c + d*x)^2 + 4*ArcSinh[c + d*x]^2 + 4*(c + d*x)^3*PolyLog[2, -E^(-ArcSin
h[c + d*x])] - 4*(c + d*x)^3*PolyLog[2, E^(-ArcSinh[c + d*x])] + ArcSinh[c + d*x]*(2*Sinh[2*ArcSinh[c + d*x]]
+ (Log[1 - E^(-ArcSinh[c + d*x])] - Log[1 + E^(-ArcSinh[c + d*x])])*(-3*(c + d*x) + Sinh[3*ArcSinh[c + d*x]]))
))/(12*d*e^4*(c + d*x)^3)

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Maple [A]  time = 0.08, size = 310, normalized size = 1.8 \begin{align*} -{\frac{{a}^{2}}{3\,d{e}^{4} \left ( dx+c \right ) ^{3}}}-{\frac{{b}^{2}{\it Arcsinh} \left ( dx+c \right ) }{3\,d{e}^{4} \left ( dx+c \right ) ^{2}}\sqrt{1+ \left ( dx+c \right ) ^{2}}}-{\frac{{b}^{2} \left ({\it Arcsinh} \left ( dx+c \right ) \right ) ^{2}}{3\,d{e}^{4} \left ( dx+c \right ) ^{3}}}-{\frac{{b}^{2}}{3\,d{e}^{4} \left ( dx+c \right ) }}+{\frac{{b}^{2}{\it Arcsinh} \left ( dx+c \right ) }{3\,d{e}^{4}}\ln \left ( 1+dx+c+\sqrt{1+ \left ( dx+c \right ) ^{2}} \right ) }+{\frac{{b}^{2}}{3\,d{e}^{4}}{\it polylog} \left ( 2,-dx-c-\sqrt{1+ \left ( dx+c \right ) ^{2}} \right ) }-{\frac{{b}^{2}{\it Arcsinh} \left ( dx+c \right ) }{3\,d{e}^{4}}\ln \left ( 1-dx-c-\sqrt{1+ \left ( dx+c \right ) ^{2}} \right ) }-{\frac{{b}^{2}}{3\,d{e}^{4}}{\it polylog} \left ( 2,dx+c+\sqrt{1+ \left ( dx+c \right ) ^{2}} \right ) }-{\frac{2\,ab{\it Arcsinh} \left ( dx+c \right ) }{3\,d{e}^{4} \left ( dx+c \right ) ^{3}}}-{\frac{ab}{3\,d{e}^{4} \left ( dx+c \right ) ^{2}}\sqrt{1+ \left ( dx+c \right ) ^{2}}}+{\frac{ab}{3\,d{e}^{4}}{\it Artanh} \left ({\frac{1}{\sqrt{1+ \left ( dx+c \right ) ^{2}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsinh(d*x+c))^2/(d*e*x+c*e)^4,x)

[Out]

-1/3/d*a^2/e^4/(d*x+c)^3-1/3/d*b^2/e^4/(d*x+c)^2*(1+(d*x+c)^2)^(1/2)*arcsinh(d*x+c)-1/3/d*b^2/e^4/(d*x+c)^3*ar
csinh(d*x+c)^2-1/3*b^2/d/e^4/(d*x+c)+1/3/d*b^2/e^4*arcsinh(d*x+c)*ln(1+d*x+c+(1+(d*x+c)^2)^(1/2))+1/3*b^2*poly
log(2,-d*x-c-(1+(d*x+c)^2)^(1/2))/d/e^4-1/3/d*b^2/e^4*arcsinh(d*x+c)*ln(1-d*x-c-(1+(d*x+c)^2)^(1/2))-1/3*b^2*p
olylog(2,d*x+c+(1+(d*x+c)^2)^(1/2))/d/e^4-2/3/d*a*b/e^4/(d*x+c)^3*arcsinh(d*x+c)-1/3/d*a*b/e^4/(d*x+c)^2*(1+(d
*x+c)^2)^(1/2)+1/3/d*a*b/e^4*arctanh(1/(1+(d*x+c)^2)^(1/2))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{b^{2} \log \left (d x + c + \sqrt{d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )^{2}}{3 \,{\left (d^{4} e^{4} x^{3} + 3 \, c d^{3} e^{4} x^{2} + 3 \, c^{2} d^{2} e^{4} x + c^{3} d e^{4}\right )}} - \frac{a^{2}}{3 \,{\left (d^{4} e^{4} x^{3} + 3 \, c d^{3} e^{4} x^{2} + 3 \, c^{2} d^{2} e^{4} x + c^{3} d e^{4}\right )}} + \int \frac{2 \,{\left ({\left (3 \, a b d^{3} + b^{2} d^{3}\right )} x^{3} + 3 \,{\left (c^{3} + c\right )} a b +{\left (c^{3} + c\right )} b^{2} + 3 \,{\left (3 \, a b c d^{2} + b^{2} c d^{2}\right )} x^{2} +{\left (3 \,{\left (3 \, c^{2} d + d\right )} a b +{\left (3 \, c^{2} d + d\right )} b^{2}\right )} x +{\left (b^{2} c^{2} + 3 \,{\left (c^{2} + 1\right )} a b +{\left (3 \, a b d^{2} + b^{2} d^{2}\right )} x^{2} + 2 \,{\left (3 \, a b c d + b^{2} c d\right )} x\right )} \sqrt{d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )} \log \left (d x + c + \sqrt{d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )}{3 \,{\left (d^{7} e^{4} x^{7} + 7 \, c d^{6} e^{4} x^{6} + c^{7} e^{4} + c^{5} e^{4} +{\left (21 \, c^{2} d^{5} e^{4} + d^{5} e^{4}\right )} x^{5} + 5 \,{\left (7 \, c^{3} d^{4} e^{4} + c d^{4} e^{4}\right )} x^{4} + 5 \,{\left (7 \, c^{4} d^{3} e^{4} + 2 \, c^{2} d^{3} e^{4}\right )} x^{3} +{\left (21 \, c^{5} d^{2} e^{4} + 10 \, c^{3} d^{2} e^{4}\right )} x^{2} +{\left (7 \, c^{6} d e^{4} + 5 \, c^{4} d e^{4}\right )} x +{\left (d^{6} e^{4} x^{6} + 6 \, c d^{5} e^{4} x^{5} + c^{6} e^{4} + c^{4} e^{4} +{\left (15 \, c^{2} d^{4} e^{4} + d^{4} e^{4}\right )} x^{4} + 4 \,{\left (5 \, c^{3} d^{3} e^{4} + c d^{3} e^{4}\right )} x^{3} + 3 \,{\left (5 \, c^{4} d^{2} e^{4} + 2 \, c^{2} d^{2} e^{4}\right )} x^{2} + 2 \,{\left (3 \, c^{5} d e^{4} + 2 \, c^{3} d e^{4}\right )} x\right )} \sqrt{d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(d*x+c))^2/(d*e*x+c*e)^4,x, algorithm="maxima")

[Out]

-1/3*b^2*log(d*x + c + sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1))^2/(d^4*e^4*x^3 + 3*c*d^3*e^4*x^2 + 3*c^2*d^2*e^4*x +
c^3*d*e^4) - 1/3*a^2/(d^4*e^4*x^3 + 3*c*d^3*e^4*x^2 + 3*c^2*d^2*e^4*x + c^3*d*e^4) + integrate(2/3*((3*a*b*d^
3 + b^2*d^3)*x^3 + 3*(c^3 + c)*a*b + (c^3 + c)*b^2 + 3*(3*a*b*c*d^2 + b^2*c*d^2)*x^2 + (3*(3*c^2*d + d)*a*b +
(3*c^2*d + d)*b^2)*x + (b^2*c^2 + 3*(c^2 + 1)*a*b + (3*a*b*d^2 + b^2*d^2)*x^2 + 2*(3*a*b*c*d + b^2*c*d)*x)*sqr
t(d^2*x^2 + 2*c*d*x + c^2 + 1))*log(d*x + c + sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1))/(d^7*e^4*x^7 + 7*c*d^6*e^4*x^
6 + c^7*e^4 + c^5*e^4 + (21*c^2*d^5*e^4 + d^5*e^4)*x^5 + 5*(7*c^3*d^4*e^4 + c*d^4*e^4)*x^4 + 5*(7*c^4*d^3*e^4
+ 2*c^2*d^3*e^4)*x^3 + (21*c^5*d^2*e^4 + 10*c^3*d^2*e^4)*x^2 + (7*c^6*d*e^4 + 5*c^4*d*e^4)*x + (d^6*e^4*x^6 +
6*c*d^5*e^4*x^5 + c^6*e^4 + c^4*e^4 + (15*c^2*d^4*e^4 + d^4*e^4)*x^4 + 4*(5*c^3*d^3*e^4 + c*d^3*e^4)*x^3 + 3*(
5*c^4*d^2*e^4 + 2*c^2*d^2*e^4)*x^2 + 2*(3*c^5*d*e^4 + 2*c^3*d*e^4)*x)*sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1)), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\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^{4} e^{4} x^{4} + 4 \, c d^{3} e^{4} x^{3} + 6 \, c^{2} d^{2} e^{4} x^{2} + 4 \, c^{3} d e^{4} x + c^{4} e^{4}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(d*x+c))^2/(d*e*x+c*e)^4,x, algorithm="fricas")

[Out]

integral((b^2*arcsinh(d*x + c)^2 + 2*a*b*arcsinh(d*x + c) + a^2)/(d^4*e^4*x^4 + 4*c*d^3*e^4*x^3 + 6*c^2*d^2*e^
4*x^2 + 4*c^3*d*e^4*x + c^4*e^4), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{a^{2}}{c^{4} + 4 c^{3} d x + 6 c^{2} d^{2} x^{2} + 4 c d^{3} x^{3} + d^{4} x^{4}}\, dx + \int \frac{b^{2} \operatorname{asinh}^{2}{\left (c + d x \right )}}{c^{4} + 4 c^{3} d x + 6 c^{2} d^{2} x^{2} + 4 c d^{3} x^{3} + d^{4} x^{4}}\, dx + \int \frac{2 a b \operatorname{asinh}{\left (c + d x \right )}}{c^{4} + 4 c^{3} d x + 6 c^{2} d^{2} x^{2} + 4 c d^{3} x^{3} + d^{4} x^{4}}\, dx}{e^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asinh(d*x+c))**2/(d*e*x+c*e)**4,x)

[Out]

(Integral(a**2/(c**4 + 4*c**3*d*x + 6*c**2*d**2*x**2 + 4*c*d**3*x**3 + d**4*x**4), x) + Integral(b**2*asinh(c
+ d*x)**2/(c**4 + 4*c**3*d*x + 6*c**2*d**2*x**2 + 4*c*d**3*x**3 + d**4*x**4), x) + Integral(2*a*b*asinh(c + d*
x)/(c**4 + 4*c**3*d*x + 6*c**2*d**2*x**2 + 4*c*d**3*x**3 + d**4*x**4), x))/e**4

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \operatorname{arsinh}\left (d x + c\right ) + a\right )}^{2}}{{\left (d e x + c e\right )}^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(d*x+c))^2/(d*e*x+c*e)^4,x, algorithm="giac")

[Out]

integrate((b*arcsinh(d*x + c) + a)^2/(d*e*x + c*e)^4, x)