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

Optimal. Leaf size=234 \[ \frac {24 b^3 \text {Li}_3\left (-e^{\sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )}{d e^2}-\frac {24 b^3 \text {Li}_3\left (e^{\sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )}{d e^2}-\frac {12 b^2 \text {Li}_2\left (-e^{\sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )^2}{d e^2}+\frac {12 b^2 \text {Li}_2\left (e^{\sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )^2}{d e^2}-\frac {\left (a+b \sinh ^{-1}(c+d x)\right )^4}{d e^2 (c+d x)}-\frac {8 b \tanh ^{-1}\left (e^{\sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )^3}{d e^2}-\frac {24 b^4 \text {Li}_4\left (-e^{\sinh ^{-1}(c+d x)}\right )}{d e^2}+\frac {24 b^4 \text {Li}_4\left (e^{\sinh ^{-1}(c+d x)}\right )}{d e^2} \]

[Out]

-(a+b*arcsinh(d*x+c))^4/d/e^2/(d*x+c)-8*b*(a+b*arcsinh(d*x+c))^3*arctanh(d*x+c+(1+(d*x+c)^2)^(1/2))/d/e^2-12*b
^2*(a+b*arcsinh(d*x+c))^2*polylog(2,-d*x-c-(1+(d*x+c)^2)^(1/2))/d/e^2+12*b^2*(a+b*arcsinh(d*x+c))^2*polylog(2,
d*x+c+(1+(d*x+c)^2)^(1/2))/d/e^2+24*b^3*(a+b*arcsinh(d*x+c))*polylog(3,-d*x-c-(1+(d*x+c)^2)^(1/2))/d/e^2-24*b^
3*(a+b*arcsinh(d*x+c))*polylog(3,d*x+c+(1+(d*x+c)^2)^(1/2))/d/e^2-24*b^4*polylog(4,-d*x-c-(1+(d*x+c)^2)^(1/2))
/d/e^2+24*b^4*polylog(4,d*x+c+(1+(d*x+c)^2)^(1/2))/d/e^2

________________________________________________________________________________________

Rubi [A]  time = 0.32, antiderivative size = 234, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 9, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {5865, 12, 5661, 5760, 4182, 2531, 6609, 2282, 6589} \[ \frac {24 b^3 \text {PolyLog}\left (3,-e^{\sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )}{d e^2}-\frac {24 b^3 \text {PolyLog}\left (3,e^{\sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )}{d e^2}-\frac {12 b^2 \text {PolyLog}\left (2,-e^{\sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )^2}{d e^2}+\frac {12 b^2 \text {PolyLog}\left (2,e^{\sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )^2}{d e^2}-\frac {24 b^4 \text {PolyLog}\left (4,-e^{\sinh ^{-1}(c+d x)}\right )}{d e^2}+\frac {24 b^4 \text {PolyLog}\left (4,e^{\sinh ^{-1}(c+d x)}\right )}{d e^2}-\frac {\left (a+b \sinh ^{-1}(c+d x)\right )^4}{d e^2 (c+d x)}-\frac {8 b \tanh ^{-1}\left (e^{\sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )^3}{d e^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a + b*ArcSinh[c + d*x])^4/(d*e^2*(c + d*x))) - (8*b*(a + b*ArcSinh[c + d*x])^3*ArcTanh[E^ArcSinh[c + d*x]])
/(d*e^2) - (12*b^2*(a + b*ArcSinh[c + d*x])^2*PolyLog[2, -E^ArcSinh[c + d*x]])/(d*e^2) + (12*b^2*(a + b*ArcSin
h[c + d*x])^2*PolyLog[2, E^ArcSinh[c + d*x]])/(d*e^2) + (24*b^3*(a + b*ArcSinh[c + d*x])*PolyLog[3, -E^ArcSinh
[c + d*x]])/(d*e^2) - (24*b^3*(a + b*ArcSinh[c + d*x])*PolyLog[3, E^ArcSinh[c + d*x]])/(d*e^2) - (24*b^4*PolyL
og[4, -E^ArcSinh[c + d*x]])/(d*e^2) + (24*b^4*PolyLog[4, 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 2282

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 2531

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 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
]

Rule 6589

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]

Rule 6609

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

Rubi steps

\begin {align*} \int \frac {\left (a+b \sinh ^{-1}(c+d x)\right )^4}{(c e+d e x)^2} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (a+b \sinh ^{-1}(x)\right )^4}{e^2 x^2} \, dx,x,c+d x\right )}{d}\\ &=\frac {\operatorname {Subst}\left (\int \frac {\left (a+b \sinh ^{-1}(x)\right )^4}{x^2} \, dx,x,c+d x\right )}{d e^2}\\ &=-\frac {\left (a+b \sinh ^{-1}(c+d x)\right )^4}{d e^2 (c+d x)}+\frac {(4 b) \operatorname {Subst}\left (\int \frac {\left (a+b \sinh ^{-1}(x)\right )^3}{x \sqrt {1+x^2}} \, dx,x,c+d x\right )}{d e^2}\\ &=-\frac {\left (a+b \sinh ^{-1}(c+d x)\right )^4}{d e^2 (c+d x)}+\frac {(4 b) \operatorname {Subst}\left (\int (a+b x)^3 \text {csch}(x) \, dx,x,\sinh ^{-1}(c+d x)\right )}{d e^2}\\ &=-\frac {\left (a+b \sinh ^{-1}(c+d x)\right )^4}{d e^2 (c+d x)}-\frac {8 b \left (a+b \sinh ^{-1}(c+d x)\right )^3 \tanh ^{-1}\left (e^{\sinh ^{-1}(c+d x)}\right )}{d e^2}-\frac {\left (12 b^2\right ) \operatorname {Subst}\left (\int (a+b x)^2 \log \left (1-e^x\right ) \, dx,x,\sinh ^{-1}(c+d x)\right )}{d e^2}+\frac {\left (12 b^2\right ) \operatorname {Subst}\left (\int (a+b x)^2 \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 )^4}{d e^2 (c+d x)}-\frac {8 b \left (a+b \sinh ^{-1}(c+d x)\right )^3 \tanh ^{-1}\left (e^{\sinh ^{-1}(c+d x)}\right )}{d e^2}-\frac {12 b^2 \left (a+b \sinh ^{-1}(c+d x)\right )^2 \text {Li}_2\left (-e^{\sinh ^{-1}(c+d x)}\right )}{d e^2}+\frac {12 b^2 \left (a+b \sinh ^{-1}(c+d x)\right )^2 \text {Li}_2\left (e^{\sinh ^{-1}(c+d x)}\right )}{d e^2}+\frac {\left (24 b^3\right ) \operatorname {Subst}\left (\int (a+b x) \text {Li}_2\left (-e^x\right ) \, dx,x,\sinh ^{-1}(c+d x)\right )}{d e^2}-\frac {\left (24 b^3\right ) \operatorname {Subst}\left (\int (a+b x) \text {Li}_2\left (e^x\right ) \, dx,x,\sinh ^{-1}(c+d x)\right )}{d e^2}\\ &=-\frac {\left (a+b \sinh ^{-1}(c+d x)\right )^4}{d e^2 (c+d x)}-\frac {8 b \left (a+b \sinh ^{-1}(c+d x)\right )^3 \tanh ^{-1}\left (e^{\sinh ^{-1}(c+d x)}\right )}{d e^2}-\frac {12 b^2 \left (a+b \sinh ^{-1}(c+d x)\right )^2 \text {Li}_2\left (-e^{\sinh ^{-1}(c+d x)}\right )}{d e^2}+\frac {12 b^2 \left (a+b \sinh ^{-1}(c+d x)\right )^2 \text {Li}_2\left (e^{\sinh ^{-1}(c+d x)}\right )}{d e^2}+\frac {24 b^3 \left (a+b \sinh ^{-1}(c+d x)\right ) \text {Li}_3\left (-e^{\sinh ^{-1}(c+d x)}\right )}{d e^2}-\frac {24 b^3 \left (a+b \sinh ^{-1}(c+d x)\right ) \text {Li}_3\left (e^{\sinh ^{-1}(c+d x)}\right )}{d e^2}-\frac {\left (24 b^4\right ) \operatorname {Subst}\left (\int \text {Li}_3\left (-e^x\right ) \, dx,x,\sinh ^{-1}(c+d x)\right )}{d e^2}+\frac {\left (24 b^4\right ) \operatorname {Subst}\left (\int \text {Li}_3\left (e^x\right ) \, dx,x,\sinh ^{-1}(c+d x)\right )}{d e^2}\\ &=-\frac {\left (a+b \sinh ^{-1}(c+d x)\right )^4}{d e^2 (c+d x)}-\frac {8 b \left (a+b \sinh ^{-1}(c+d x)\right )^3 \tanh ^{-1}\left (e^{\sinh ^{-1}(c+d x)}\right )}{d e^2}-\frac {12 b^2 \left (a+b \sinh ^{-1}(c+d x)\right )^2 \text {Li}_2\left (-e^{\sinh ^{-1}(c+d x)}\right )}{d e^2}+\frac {12 b^2 \left (a+b \sinh ^{-1}(c+d x)\right )^2 \text {Li}_2\left (e^{\sinh ^{-1}(c+d x)}\right )}{d e^2}+\frac {24 b^3 \left (a+b \sinh ^{-1}(c+d x)\right ) \text {Li}_3\left (-e^{\sinh ^{-1}(c+d x)}\right )}{d e^2}-\frac {24 b^3 \left (a+b \sinh ^{-1}(c+d x)\right ) \text {Li}_3\left (e^{\sinh ^{-1}(c+d x)}\right )}{d e^2}-\frac {\left (24 b^4\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_3(-x)}{x} \, dx,x,e^{\sinh ^{-1}(c+d x)}\right )}{d e^2}+\frac {\left (24 b^4\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_3(x)}{x} \, dx,x,e^{\sinh ^{-1}(c+d x)}\right )}{d e^2}\\ &=-\frac {\left (a+b \sinh ^{-1}(c+d x)\right )^4}{d e^2 (c+d x)}-\frac {8 b \left (a+b \sinh ^{-1}(c+d x)\right )^3 \tanh ^{-1}\left (e^{\sinh ^{-1}(c+d x)}\right )}{d e^2}-\frac {12 b^2 \left (a+b \sinh ^{-1}(c+d x)\right )^2 \text {Li}_2\left (-e^{\sinh ^{-1}(c+d x)}\right )}{d e^2}+\frac {12 b^2 \left (a+b \sinh ^{-1}(c+d x)\right )^2 \text {Li}_2\left (e^{\sinh ^{-1}(c+d x)}\right )}{d e^2}+\frac {24 b^3 \left (a+b \sinh ^{-1}(c+d x)\right ) \text {Li}_3\left (-e^{\sinh ^{-1}(c+d x)}\right )}{d e^2}-\frac {24 b^3 \left (a+b \sinh ^{-1}(c+d x)\right ) \text {Li}_3\left (e^{\sinh ^{-1}(c+d x)}\right )}{d e^2}-\frac {24 b^4 \text {Li}_4\left (-e^{\sinh ^{-1}(c+d x)}\right )}{d e^2}+\frac {24 b^4 \text {Li}_4\left (e^{\sinh ^{-1}(c+d x)}\right )}{d e^2}\\ \end {align*}

________________________________________________________________________________________

Mathematica [B]  time = 1.60, size = 501, normalized size = 2.14 \[ \frac {-\frac {2 a^4}{c+d x}+4 a^3 b \left (2 \log \left (\frac {2 \sinh ^2\left (\frac {1}{2} \sinh ^{-1}(c+d x)\right )}{c+d x}\right )-\frac {2 \sinh ^{-1}(c+d x)}{c+d x}\right )+12 a^2 b^2 \left (2 \text {Li}_2\left (-e^{-\sinh ^{-1}(c+d x)}\right )-2 \text {Li}_2\left (e^{-\sinh ^{-1}(c+d x)}\right )+\sinh ^{-1}(c+d x) \left (-\frac {\sinh ^{-1}(c+d x)}{c+d x}+2 \log \left (1-e^{-\sinh ^{-1}(c+d x)}\right )-2 \log \left (e^{-\sinh ^{-1}(c+d x)}+1\right )\right )\right )+8 a b^3 \left (6 \sinh ^{-1}(c+d x) \text {Li}_2\left (-e^{-\sinh ^{-1}(c+d x)}\right )-6 \sinh ^{-1}(c+d x) \text {Li}_2\left (e^{-\sinh ^{-1}(c+d x)}\right )+6 \text {Li}_3\left (-e^{-\sinh ^{-1}(c+d x)}\right )-6 \text {Li}_3\left (e^{-\sinh ^{-1}(c+d x)}\right )-\frac {\sinh ^{-1}(c+d x)^3}{c+d x}+3 \sinh ^{-1}(c+d x)^2 \log \left (1-e^{-\sinh ^{-1}(c+d x)}\right )-3 \sinh ^{-1}(c+d x)^2 \log \left (e^{-\sinh ^{-1}(c+d x)}+1\right )\right )+b^4 \left (24 \sinh ^{-1}(c+d x)^2 \text {Li}_2\left (-e^{-\sinh ^{-1}(c+d x)}\right )+24 \sinh ^{-1}(c+d x)^2 \text {Li}_2\left (e^{\sinh ^{-1}(c+d x)}\right )+48 \sinh ^{-1}(c+d x) \text {Li}_3\left (-e^{-\sinh ^{-1}(c+d x)}\right )-48 \sinh ^{-1}(c+d x) \text {Li}_3\left (e^{\sinh ^{-1}(c+d x)}\right )+48 \text {Li}_4\left (-e^{-\sinh ^{-1}(c+d x)}\right )+48 \text {Li}_4\left (e^{\sinh ^{-1}(c+d x)}\right )-\frac {2 \sinh ^{-1}(c+d x)^4}{c+d x}-2 \sinh ^{-1}(c+d x)^4-8 \sinh ^{-1}(c+d x)^3 \log \left (e^{-\sinh ^{-1}(c+d x)}+1\right )+8 \sinh ^{-1}(c+d x)^3 \log \left (1-e^{\sinh ^{-1}(c+d x)}\right )+\pi ^4\right )}{2 d e^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((-2*a^4)/(c + d*x) + 4*a^3*b*((-2*ArcSinh[c + d*x])/(c + d*x) + 2*Log[(2*Sinh[ArcSinh[c + d*x]/2]^2)/(c + d*x
)]) + 12*a^2*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^(-ArcSinh[c + d*x])]) + 2*PolyLog[2, -E^(-ArcSinh[c + d*x])] - 2*PolyLog[2, E^(-ArcSinh[c + d*x])]) + 8*a
*b^3*(-(ArcSinh[c + d*x]^3/(c + d*x)) + 3*ArcSinh[c + d*x]^2*Log[1 - E^(-ArcSinh[c + d*x])] - 3*ArcSinh[c + d*
x]^2*Log[1 + E^(-ArcSinh[c + d*x])] + 6*ArcSinh[c + d*x]*PolyLog[2, -E^(-ArcSinh[c + d*x])] - 6*ArcSinh[c + d*
x]*PolyLog[2, E^(-ArcSinh[c + d*x])] + 6*PolyLog[3, -E^(-ArcSinh[c + d*x])] - 6*PolyLog[3, E^(-ArcSinh[c + d*x
])]) + b^4*(Pi^4 - 2*ArcSinh[c + d*x]^4 - (2*ArcSinh[c + d*x]^4)/(c + d*x) - 8*ArcSinh[c + d*x]^3*Log[1 + E^(-
ArcSinh[c + d*x])] + 8*ArcSinh[c + d*x]^3*Log[1 - E^ArcSinh[c + d*x]] + 24*ArcSinh[c + d*x]^2*PolyLog[2, -E^(-
ArcSinh[c + d*x])] + 24*ArcSinh[c + d*x]^2*PolyLog[2, E^ArcSinh[c + d*x]] + 48*ArcSinh[c + d*x]*PolyLog[3, -E^
(-ArcSinh[c + d*x])] - 48*ArcSinh[c + d*x]*PolyLog[3, E^ArcSinh[c + d*x]] + 48*PolyLog[4, -E^(-ArcSinh[c + d*x
])] + 48*PolyLog[4, E^ArcSinh[c + d*x]]))/(2*d*e^2)

________________________________________________________________________________________

fricas [F]  time = 0.65, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{4} \operatorname {arsinh}\left (d x + c\right )^{4} + 4 \, a b^{3} \operatorname {arsinh}\left (d x + c\right )^{3} + 6 \, a^{2} b^{2} \operatorname {arsinh}\left (d x + c\right )^{2} + 4 \, a^{3} b \operatorname {arsinh}\left (d x + c\right ) + a^{4}}{d^{2} e^{2} x^{2} + 2 \, c d e^{2} x + c^{2} e^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b^4*arcsinh(d*x + c)^4 + 4*a*b^3*arcsinh(d*x + c)^3 + 6*a^2*b^2*arcsinh(d*x + c)^2 + 4*a^3*b*arcsinh
(d*x + c) + a^4)/(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 \[ \int \frac {{\left (b \operatorname {arsinh}\left (d x + c\right ) + a\right )}^{4}}{{\left (d e x + c e\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [B]  time = 0.20, size = 820, normalized size = 3.50 \[ -\frac {a^{4}}{d \,e^{2} \left (d x +c \right )}-\frac {b^{4} \arcsinh \left (d x +c \right )^{4}}{d \,e^{2} \left (d x +c \right )}-\frac {4 b^{4} \arcsinh \left (d x +c \right )^{3} \ln \left (1+d x +c +\sqrt {1+\left (d x +c \right )^{2}}\right )}{d \,e^{2}}-\frac {12 b^{4} \arcsinh \left (d x +c \right )^{2} \polylog \left (2, -d x -c -\sqrt {1+\left (d x +c \right )^{2}}\right )}{d \,e^{2}}+\frac {24 b^{4} \arcsinh \left (d x +c \right ) \polylog \left (3, -d x -c -\sqrt {1+\left (d x +c \right )^{2}}\right )}{d \,e^{2}}-\frac {24 b^{4} \polylog \left (4, -d x -c -\sqrt {1+\left (d x +c \right )^{2}}\right )}{d \,e^{2}}+\frac {4 b^{4} \arcsinh \left (d x +c \right )^{3} \ln \left (1-d x -c -\sqrt {1+\left (d x +c \right )^{2}}\right )}{d \,e^{2}}+\frac {12 b^{4} \arcsinh \left (d x +c \right )^{2} \polylog \left (2, d x +c +\sqrt {1+\left (d x +c \right )^{2}}\right )}{d \,e^{2}}-\frac {24 b^{4} \arcsinh \left (d x +c \right ) \polylog \left (3, d x +c +\sqrt {1+\left (d x +c \right )^{2}}\right )}{d \,e^{2}}+\frac {24 b^{4} \polylog \left (4, d x +c +\sqrt {1+\left (d x +c \right )^{2}}\right )}{d \,e^{2}}-\frac {4 a \,b^{3} \arcsinh \left (d x +c \right )^{3}}{d \,e^{2} \left (d x +c \right )}-\frac {12 a \,b^{3} \arcsinh \left (d x +c \right )^{2} \ln \left (1+d x +c +\sqrt {1+\left (d x +c \right )^{2}}\right )}{d \,e^{2}}-\frac {24 a \,b^{3} \arcsinh \left (d x +c \right ) \polylog \left (2, -d x -c -\sqrt {1+\left (d x +c \right )^{2}}\right )}{d \,e^{2}}+\frac {24 a \,b^{3} \polylog \left (3, -d x -c -\sqrt {1+\left (d x +c \right )^{2}}\right )}{d \,e^{2}}+\frac {12 a \,b^{3} \arcsinh \left (d x +c \right )^{2} \ln \left (1-d x -c -\sqrt {1+\left (d x +c \right )^{2}}\right )}{d \,e^{2}}+\frac {24 a \,b^{3} \arcsinh \left (d x +c \right ) \polylog \left (2, d x +c +\sqrt {1+\left (d x +c \right )^{2}}\right )}{d \,e^{2}}-\frac {24 a \,b^{3} \polylog \left (3, d x +c +\sqrt {1+\left (d x +c \right )^{2}}\right )}{d \,e^{2}}-\frac {6 a^{2} b^{2} \arcsinh \left (d x +c \right )^{2}}{d \,e^{2} \left (d x +c \right )}-\frac {12 a^{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 {12 a^{2} b^{2} \polylog \left (2, -d x -c -\sqrt {1+\left (d x +c \right )^{2}}\right )}{d \,e^{2}}+\frac {12 a^{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 {12 a^{2} b^{2} \polylog \left (2, d x +c +\sqrt {1+\left (d x +c \right )^{2}}\right )}{d \,e^{2}}-\frac {4 a^{3} b \arcsinh \left (d x +c \right )}{d \,e^{2} \left (d x +c \right )}-\frac {4 a^{3} 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))^4/(d*e*x+c*e)^2,x)

[Out]

-1/d*a^4/e^2/(d*x+c)-1/d*b^4/e^2*arcsinh(d*x+c)^4/(d*x+c)-4/d*b^4/e^2*arcsinh(d*x+c)^3*ln(1+d*x+c+(1+(d*x+c)^2
)^(1/2))-12/d*b^4/e^2*arcsinh(d*x+c)^2*polylog(2,-d*x-c-(1+(d*x+c)^2)^(1/2))+24/d*b^4/e^2*arcsinh(d*x+c)*polyl
og(3,-d*x-c-(1+(d*x+c)^2)^(1/2))-24*b^4*polylog(4,-d*x-c-(1+(d*x+c)^2)^(1/2))/d/e^2+4/d*b^4/e^2*arcsinh(d*x+c)
^3*ln(1-d*x-c-(1+(d*x+c)^2)^(1/2))+12/d*b^4/e^2*arcsinh(d*x+c)^2*polylog(2,d*x+c+(1+(d*x+c)^2)^(1/2))-24/d*b^4
/e^2*arcsinh(d*x+c)*polylog(3,d*x+c+(1+(d*x+c)^2)^(1/2))+24*b^4*polylog(4,d*x+c+(1+(d*x+c)^2)^(1/2))/d/e^2-4/d
*a*b^3/e^2*arcsinh(d*x+c)^3/(d*x+c)-12/d*a*b^3/e^2*arcsinh(d*x+c)^2*ln(1+d*x+c+(1+(d*x+c)^2)^(1/2))-24/d*a*b^3
/e^2*arcsinh(d*x+c)*polylog(2,-d*x-c-(1+(d*x+c)^2)^(1/2))+24/d*a*b^3/e^2*polylog(3,-d*x-c-(1+(d*x+c)^2)^(1/2))
+12/d*a*b^3/e^2*arcsinh(d*x+c)^2*ln(1-d*x-c-(1+(d*x+c)^2)^(1/2))+24/d*a*b^3/e^2*arcsinh(d*x+c)*polylog(2,d*x+c
+(1+(d*x+c)^2)^(1/2))-24/d*a*b^3/e^2*polylog(3,d*x+c+(1+(d*x+c)^2)^(1/2))-6/d*a^2*b^2/e^2*arcsinh(d*x+c)^2/(d*
x+c)-12/d*a^2*b^2/e^2*arcsinh(d*x+c)*ln(1+d*x+c+(1+(d*x+c)^2)^(1/2))-12/d*a^2*b^2/e^2*polylog(2,-d*x-c-(1+(d*x
+c)^2)^(1/2))+12/d*a^2*b^2/e^2*arcsinh(d*x+c)*ln(1-d*x-c-(1+(d*x+c)^2)^(1/2))+12/d*a^2*b^2/e^2*polylog(2,d*x+c
+(1+(d*x+c)^2)^(1/2))-4/d*a^3*b/e^2/(d*x+c)*arcsinh(d*x+c)-4/d*a^3*b/e^2*arctanh(1/(1+(d*x+c)^2)^(1/2))

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {b^{4} \log \left (d x + c + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )^{4}}{d^{2} e^{2} x + c d e^{2}} - 4 \, a^{3} b {\left (\frac {\operatorname {arsinh}\left (d x + c\right )}{d^{2} e^{2} x + c d e^{2}} + \frac {\operatorname {arsinh}\left (\frac {d e^{2}}{{\left | d^{2} e^{2} x + c d e^{2} \right |}}\right )}{d e^{2}}\right )} - \frac {a^{4}}{d^{2} e^{2} x + c d e^{2}} + \int \frac {2 \, {\left (2 \, {\left ({\left (c^{3} + c\right )} a b^{3} + {\left (c^{3} + c\right )} b^{4} + {\left (a b^{3} d^{3} + b^{4} d^{3}\right )} x^{3} + 3 \, {\left (a b^{3} c d^{2} + b^{4} c d^{2}\right )} x^{2} + {\left ({\left (3 \, c^{2} d + d\right )} a b^{3} + {\left (3 \, c^{2} d + d\right )} b^{4}\right )} x + {\left (b^{4} c^{2} + {\left (c^{2} + 1\right )} a b^{3} + {\left (a b^{3} d^{2} + b^{4} d^{2}\right )} x^{2} + 2 \, {\left (a b^{3} c d + b^{4} c d\right )} x\right )} \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )} \log \left (d x + c + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )^{3} + 3 \, {\left (a^{2} b^{2} d^{3} x^{3} + 3 \, a^{2} b^{2} c d^{2} x^{2} + {\left (3 \, c^{2} d + d\right )} a^{2} b^{2} x + {\left (c^{3} + c\right )} a^{2} b^{2} + {\left (a^{2} b^{2} d^{2} x^{2} + 2 \, a^{2} b^{2} c d x + {\left (c^{2} + 1\right )} a^{2} b^{2}\right )} \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )} \log \left (d x + c + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )^{2}\right )}}{d^{5} e^{2} x^{5} + 5 \, c d^{4} e^{2} x^{4} + c^{5} e^{2} + c^{3} e^{2} + {\left (10 \, c^{2} d^{3} e^{2} + d^{3} e^{2}\right )} x^{3} + {\left (10 \, c^{3} d^{2} e^{2} + 3 \, c d^{2} e^{2}\right )} x^{2} + {\left (5 \, c^{4} d e^{2} + 3 \, c^{2} d e^{2}\right )} x + {\left (d^{4} e^{2} x^{4} + 4 \, c d^{3} e^{2} x^{3} + c^{4} e^{2} + c^{2} e^{2} + {\left (6 \, c^{2} d^{2} e^{2} + d^{2} e^{2}\right )} x^{2} + 2 \, {\left (2 \, c^{3} d e^{2} + c d e^{2}\right )} x\right )} \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-b^4*log(d*x + c + sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1))^4/(d^2*e^2*x + c*d*e^2) - 4*a^3*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^4/(d^2*e^2*x + c*d*e^2) + integrate(2
*(2*((c^3 + c)*a*b^3 + (c^3 + c)*b^4 + (a*b^3*d^3 + b^4*d^3)*x^3 + 3*(a*b^3*c*d^2 + b^4*c*d^2)*x^2 + ((3*c^2*d
+ d)*a*b^3 + (3*c^2*d + d)*b^4)*x + (b^4*c^2 + (c^2 + 1)*a*b^3 + (a*b^3*d^2 + b^4*d^2)*x^2 + 2*(a*b^3*c*d + b
^4*c*d)*x)*sqrt(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))^3 + 3*(a^2*b^2*
d^3*x^3 + 3*a^2*b^2*c*d^2*x^2 + (3*c^2*d + d)*a^2*b^2*x + (c^3 + c)*a^2*b^2 + (a^2*b^2*d^2*x^2 + 2*a^2*b^2*c*d
*x + (c^2 + 1)*a^2*b^2)*sqrt(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))^2)
/(d^5*e^2*x^5 + 5*c*d^4*e^2*x^4 + c^5*e^2 + c^3*e^2 + (10*c^2*d^3*e^2 + d^3*e^2)*x^3 + (10*c^3*d^2*e^2 + 3*c*d
^2*e^2)*x^2 + (5*c^4*d*e^2 + 3*c^2*d*e^2)*x + (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)*sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1)), x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a+b\,\mathrm {asinh}\left (c+d\,x\right )\right )}^4}{{\left (c\,e+d\,e\,x\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {a^{4}}{c^{2} + 2 c d x + d^{2} x^{2}}\, dx + \int \frac {b^{4} \operatorname {asinh}^{4}{\left (c + d x \right )}}{c^{2} + 2 c d x + d^{2} x^{2}}\, dx + \int \frac {4 a b^{3} \operatorname {asinh}^{3}{\left (c + d x \right )}}{c^{2} + 2 c d x + d^{2} x^{2}}\, dx + \int \frac {6 a^{2} b^{2} \operatorname {asinh}^{2}{\left (c + d x \right )}}{c^{2} + 2 c d x + d^{2} x^{2}}\, dx + \int \frac {4 a^{3} b \operatorname {asinh}{\left (c + d x \right )}}{c^{2} + 2 c d x + d^{2} x^{2}}\, dx}{e^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________