3.2.71 \(\int \frac {c e+d e x}{(a+b \sinh ^{-1}(c+d x))^3} \, dx\) [171]

Optimal. Leaf size=156 \[ -\frac {e (c+d x) \sqrt {1+(c+d x)^2}}{2 b d \left (a+b \sinh ^{-1}(c+d x)\right )^2}-\frac {e}{2 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )}-\frac {e (c+d x)^2}{b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )}-\frac {e \text {Chi}\left (\frac {2 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right ) \sinh \left (\frac {2 a}{b}\right )}{b^3 d}+\frac {e \cosh \left (\frac {2 a}{b}\right ) \text {Shi}\left (\frac {2 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )}{b^3 d} \]

[Out]

-1/2*e/b^2/d/(a+b*arcsinh(d*x+c))-e*(d*x+c)^2/b^2/d/(a+b*arcsinh(d*x+c))+e*cosh(2*a/b)*Shi(2*(a+b*arcsinh(d*x+
c))/b)/b^3/d-e*Chi(2*(a+b*arcsinh(d*x+c))/b)*sinh(2*a/b)/b^3/d-1/2*e*(d*x+c)*(1+(d*x+c)^2)^(1/2)/b/d/(a+b*arcs
inh(d*x+c))^2

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Rubi [A]
time = 0.22, antiderivative size = 156, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 10, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.476, Rules used = {5859, 12, 5779, 5818, 5780, 5556, 3384, 3379, 3382, 5783} \begin {gather*} -\frac {e \sinh \left (\frac {2 a}{b}\right ) \text {Chi}\left (\frac {2 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )}{b^3 d}+\frac {e \cosh \left (\frac {2 a}{b}\right ) \text {Shi}\left (\frac {2 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )}{b^3 d}-\frac {e (c+d x)^2}{b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )}-\frac {e}{2 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )}-\frac {e \sqrt {(c+d x)^2+1} (c+d x)}{2 b d \left (a+b \sinh ^{-1}(c+d x)\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c*e + d*e*x)/(a + b*ArcSinh[c + d*x])^3,x]

[Out]

-1/2*(e*(c + d*x)*Sqrt[1 + (c + d*x)^2])/(b*d*(a + b*ArcSinh[c + d*x])^2) - e/(2*b^2*d*(a + b*ArcSinh[c + d*x]
)) - (e*(c + d*x)^2)/(b^2*d*(a + b*ArcSinh[c + d*x])) - (e*CoshIntegral[(2*(a + b*ArcSinh[c + d*x]))/b]*Sinh[(
2*a)/b])/(b^3*d) + (e*Cosh[(2*a)/b]*SinhIntegral[(2*(a + b*ArcSinh[c + d*x]))/b])/(b^3*d)

Rule 12

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

Rule 3379

Int[sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[I*(SinhIntegral[c*f*(fz/
d) + f*fz*x]/d), x] /; FreeQ[{c, d, e, f, fz}, x] && EqQ[d*e - c*f*fz*I, 0]

Rule 3382

Int[sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CoshIntegral[c*f*(fz/d)
+ f*fz*x]/d, x] /; FreeQ[{c, d, e, f, fz}, x] && EqQ[d*(e - Pi/2) - c*f*fz*I, 0]

Rule 3384

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Dist[Cos[(d*e - c*f)/d], Int[Sin[c*(f/d) + f*x]
/(c + d*x), x], x] + Dist[Sin[(d*e - c*f)/d], Int[Cos[c*(f/d) + f*x]/(c + d*x), x], x] /; FreeQ[{c, d, e, f},
x] && NeQ[d*e - c*f, 0]

Rule 5556

Int[Cosh[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sinh[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Int
[ExpandTrigReduce[(c + d*x)^m, Sinh[a + b*x]^n*Cosh[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n,
0] && IGtQ[p, 0]

Rule 5779

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

Rule 5780

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Dist[1/(b*c^(m + 1)), Subst[Int[x^n*Sinh
[-a/b + x/b]^m*Cosh[-a/b + x/b], x], x, a + b*ArcSinh[c*x]], x] /; FreeQ[{a, b, c, n}, x] && IGtQ[m, 0]

Rule 5783

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[(1/(b*c*(n + 1)))*S
imp[Sqrt[1 + c^2*x^2]/Sqrt[d + e*x^2]]*(a + b*ArcSinh[c*x])^(n + 1), x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ
[e, c^2*d] && NeQ[n, -1]

Rule 5818

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

Rule 5859

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 {c e+d e x}{\left (a+b \sinh ^{-1}(c+d x)\right )^3} \, dx &=\frac {\text {Subst}\left (\int \frac {e x}{\left (a+b \sinh ^{-1}(x)\right )^3} \, dx,x,c+d x\right )}{d}\\ &=\frac {e \text {Subst}\left (\int \frac {x}{\left (a+b \sinh ^{-1}(x)\right )^3} \, dx,x,c+d x\right )}{d}\\ &=-\frac {e (c+d x) \sqrt {1+(c+d x)^2}}{2 b d \left (a+b \sinh ^{-1}(c+d x)\right )^2}+\frac {e \text {Subst}\left (\int \frac {1}{\sqrt {1+x^2} \left (a+b \sinh ^{-1}(x)\right )^2} \, dx,x,c+d x\right )}{2 b d}+\frac {e \text {Subst}\left (\int \frac {x^2}{\sqrt {1+x^2} \left (a+b \sinh ^{-1}(x)\right )^2} \, dx,x,c+d x\right )}{b d}\\ &=-\frac {e (c+d x) \sqrt {1+(c+d x)^2}}{2 b d \left (a+b \sinh ^{-1}(c+d x)\right )^2}-\frac {e}{2 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )}-\frac {e (c+d x)^2}{b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )}+\frac {(2 e) \text {Subst}\left (\int \frac {x}{a+b \sinh ^{-1}(x)} \, dx,x,c+d x\right )}{b^2 d}\\ &=-\frac {e (c+d x) \sqrt {1+(c+d x)^2}}{2 b d \left (a+b \sinh ^{-1}(c+d x)\right )^2}-\frac {e}{2 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )}-\frac {e (c+d x)^2}{b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )}+\frac {(2 e) \text {Subst}\left (\int \frac {\cosh (x) \sinh (x)}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{b^2 d}\\ &=-\frac {e (c+d x) \sqrt {1+(c+d x)^2}}{2 b d \left (a+b \sinh ^{-1}(c+d x)\right )^2}-\frac {e}{2 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )}-\frac {e (c+d x)^2}{b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )}+\frac {(2 e) \text {Subst}\left (\int \frac {\sinh (2 x)}{2 (a+b x)} \, dx,x,\sinh ^{-1}(c+d x)\right )}{b^2 d}\\ &=-\frac {e (c+d x) \sqrt {1+(c+d x)^2}}{2 b d \left (a+b \sinh ^{-1}(c+d x)\right )^2}-\frac {e}{2 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )}-\frac {e (c+d x)^2}{b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )}+\frac {e \text {Subst}\left (\int \frac {\sinh (2 x)}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{b^2 d}\\ &=-\frac {e (c+d x) \sqrt {1+(c+d x)^2}}{2 b d \left (a+b \sinh ^{-1}(c+d x)\right )^2}-\frac {e}{2 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )}-\frac {e (c+d x)^2}{b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )}+\frac {\left (e \cosh \left (\frac {2 a}{b}\right )\right ) \text {Subst}\left (\int \frac {\sinh \left (\frac {2 a}{b}+2 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{b^2 d}-\frac {\left (e \sinh \left (\frac {2 a}{b}\right )\right ) \text {Subst}\left (\int \frac {\cosh \left (\frac {2 a}{b}+2 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{b^2 d}\\ &=-\frac {e (c+d x) \sqrt {1+(c+d x)^2}}{2 b d \left (a+b \sinh ^{-1}(c+d x)\right )^2}-\frac {e}{2 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )}-\frac {e (c+d x)^2}{b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )}-\frac {e \text {Chi}\left (\frac {2 a}{b}+2 \sinh ^{-1}(c+d x)\right ) \sinh \left (\frac {2 a}{b}\right )}{b^3 d}+\frac {e \cosh \left (\frac {2 a}{b}\right ) \text {Shi}\left (\frac {2 a}{b}+2 \sinh ^{-1}(c+d x)\right )}{b^3 d}\\ \end {align*}

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Mathematica [A]
time = 0.24, size = 120, normalized size = 0.77 \begin {gather*} \frac {e \left (-\frac {b^2 (c+d x) \sqrt {1+(c+d x)^2}}{\left (a+b \sinh ^{-1}(c+d x)\right )^2}+\frac {b \left (-1-2 (c+d x)^2\right )}{a+b \sinh ^{-1}(c+d x)}-2 \text {Chi}\left (2 \left (\frac {a}{b}+\sinh ^{-1}(c+d x)\right )\right ) \sinh \left (\frac {2 a}{b}\right )+2 \cosh \left (\frac {2 a}{b}\right ) \text {Shi}\left (2 \left (\frac {a}{b}+\sinh ^{-1}(c+d x)\right )\right )\right )}{2 b^3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(c*e + d*e*x)/(a + b*ArcSinh[c + d*x])^3,x]

[Out]

(e*(-((b^2*(c + d*x)*Sqrt[1 + (c + d*x)^2])/(a + b*ArcSinh[c + d*x])^2) + (b*(-1 - 2*(c + d*x)^2))/(a + b*ArcS
inh[c + d*x]) - 2*CoshIntegral[2*(a/b + ArcSinh[c + d*x])]*Sinh[(2*a)/b] + 2*Cosh[(2*a)/b]*SinhIntegral[2*(a/b
+ ArcSinh[c + d*x])]))/(2*b^3*d)

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Maple [A]
time = 5.26, size = 239, normalized size = 1.53

method	result	size
derivativedivides	\(\frac {-\frac {\left (-2 \left (d x +c \right ) \sqrt {1+\left (d x +c \right )^{2}}+2 \left (d x +c \right )^{2}+1\right ) e \left (2 b \arcsinh \left (d x +c \right )+2 a -b \right )}{8 b^{2} \left (b^{2} \arcsinh \left (d x +c \right )^{2}+2 a b \arcsinh \left (d x +c \right )+a^{2}\right )}+\frac {e \,{\mathrm e}^{\frac {2 a}{b}} \expIntegral \left (1, 2 \arcsinh \left (d x +c \right )+\frac {2 a}{b}\right )}{2 b^{3}}-\frac {e \left (2 \left (d x +c \right )^{2}+1+2 \left (d x +c \right ) \sqrt {1+\left (d x +c \right )^{2}}\right )}{8 b \left (a +b \arcsinh \left (d x +c \right )\right )^{2}}-\frac {e \left (2 \left (d x +c \right )^{2}+1+2 \left (d x +c \right ) \sqrt {1+\left (d x +c \right )^{2}}\right )}{4 b^{2} \left (a +b \arcsinh \left (d x +c \right )\right )}-\frac {e \,{\mathrm e}^{-\frac {2 a}{b}} \expIntegral \left (1, -2 \arcsinh \left (d x +c \right )-\frac {2 a}{b}\right )}{2 b^{3}}}{d}\)	\(239\)
default	\(\frac {-\frac {\left (-2 \left (d x +c \right ) \sqrt {1+\left (d x +c \right )^{2}}+2 \left (d x +c \right )^{2}+1\right ) e \left (2 b \arcsinh \left (d x +c \right )+2 a -b \right )}{8 b^{2} \left (b^{2} \arcsinh \left (d x +c \right )^{2}+2 a b \arcsinh \left (d x +c \right )+a^{2}\right )}+\frac {e \,{\mathrm e}^{\frac {2 a}{b}} \expIntegral \left (1, 2 \arcsinh \left (d x +c \right )+\frac {2 a}{b}\right )}{2 b^{3}}-\frac {e \left (2 \left (d x +c \right )^{2}+1+2 \left (d x +c \right ) \sqrt {1+\left (d x +c \right )^{2}}\right )}{8 b \left (a +b \arcsinh \left (d x +c \right )\right )^{2}}-\frac {e \left (2 \left (d x +c \right )^{2}+1+2 \left (d x +c \right ) \sqrt {1+\left (d x +c \right )^{2}}\right )}{4 b^{2} \left (a +b \arcsinh \left (d x +c \right )\right )}-\frac {e \,{\mathrm e}^{-\frac {2 a}{b}} \expIntegral \left (1, -2 \arcsinh \left (d x +c \right )-\frac {2 a}{b}\right )}{2 b^{3}}}{d}\)	\(239\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*e*x+c*e)/(a+b*arcsinh(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

1/d*(-1/8*(-2*(d*x+c)*(1+(d*x+c)^2)^(1/2)+2*(d*x+c)^2+1)*e*(2*b*arcsinh(d*x+c)+2*a-b)/b^2/(b^2*arcsinh(d*x+c)^
2+2*a*b*arcsinh(d*x+c)+a^2)+1/2*e/b^3*exp(2*a/b)*Ei(1,2*arcsinh(d*x+c)+2*a/b)-1/8/b*e*(2*(d*x+c)^2+1+2*(d*x+c)
*(1+(d*x+c)^2)^(1/2))/(a+b*arcsinh(d*x+c))^2-1/4/b^2*e*(2*(d*x+c)^2+1+2*(d*x+c)*(1+(d*x+c)^2)^(1/2))/(a+b*arcs
inh(d*x+c))-1/2/b^3*e*exp(-2*a/b)*Ei(1,-2*arcsinh(d*x+c)-2*a/b))

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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((d*e*x+c*e)/(a+b*arcsinh(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/2*((2*a*d^8 + b*d^8)*x^8*e + 8*(2*a*c*d^7 + b*c*d^7)*x^7*e + (2*(28*c^2*d^6 + 3*d^6)*a + (28*c^2*d^6 + 3*d^
6)*b)*x^6*e + 2*(2*(28*c^3*d^5 + 9*c*d^5)*a + (28*c^3*d^5 + 9*c*d^5)*b)*x^5*e + (2*(70*c^4*d^4 + 45*c^2*d^4 +
3*d^4)*a + (70*c^4*d^4 + 45*c^2*d^4 + 3*d^4)*b)*x^4*e + 4*(2*(14*c^5*d^3 + 15*c^3*d^3 + 3*c*d^3)*a + (14*c^5*d
^3 + 15*c^3*d^3 + 3*c*d^3)*b)*x^3*e + (2*(28*c^6*d^2 + 45*c^4*d^2 + 18*c^2*d^2 + d^2)*a + (28*c^6*d^2 + 45*c^4
*d^2 + 18*c^2*d^2 + d^2)*b)*x^2*e + 2*(2*(4*c^7*d + 9*c^5*d + 6*c^3*d + c*d)*a + (4*c^7*d + 9*c^5*d + 6*c^3*d
+ c*d)*b)*x*e + ((2*a*d^5 + b*d^5)*x^5*e + 5*(2*a*c*d^4 + b*c*d^4)*x^4*e + (2*(10*c^2*d^3 + d^3)*a + (10*c^2*d
^3 + d^3)*b)*x^3*e + (2*(10*c^3*d^2 + 3*c*d^2)*a + (10*c^3*d^2 + 3*c*d^2)*b)*x^2*e + (2*(5*c^4*d + 3*c^2*d)*a
+ (5*c^4*d + 3*c^2*d)*b)*x*e + (2*(c^5 + c^3)*a + (c^5 + c^3)*b)*e)*(d^2*x^2 + 2*c*d*x + c^2 + 1)^(3/2) + (3*(
2*a*d^6 + b*d^6)*x^6*e + 18*(2*a*c*d^5 + b*c*d^5)*x^5*e + 5*(2*(9*c^2*d^4 + d^4)*a + (9*c^2*d^4 + d^4)*b)*x^4*
e + 20*(2*(3*c^3*d^3 + c*d^3)*a + (3*c^3*d^3 + c*d^3)*b)*x^3*e + (5*(18*c^4*d^2 + 12*c^2*d^2 + d^2)*a + (45*c^
4*d^2 + 30*c^2*d^2 + 2*d^2)*b)*x^2*e + 2*((18*c^5*d + 20*c^3*d + 5*c*d)*a + (9*c^5*d + 10*c^3*d + 2*c*d)*b)*x*
e + ((6*c^6 + 10*c^4 + 5*c^2 + 1)*a + (3*c^6 + 5*c^4 + 2*c^2)*b)*e)*(d^2*x^2 + 2*c*d*x + c^2 + 1) + (2*(c^8 +
3*c^6 + 3*c^4 + c^2)*a + (c^8 + 3*c^6 + 3*c^4 + c^2)*b)*e + (2*b*d^8*x^8*e + 16*b*c*d^7*x^7*e + 2*(28*c^2*d^6
+ 3*d^6)*b*x^6*e + 4*(28*c^3*d^5 + 9*c*d^5)*b*x^5*e + 2*(70*c^4*d^4 + 45*c^2*d^4 + 3*d^4)*b*x^4*e + 8*(14*c^5*
d^3 + 15*c^3*d^3 + 3*c*d^3)*b*x^3*e + 2*(28*c^6*d^2 + 45*c^4*d^2 + 18*c^2*d^2 + d^2)*b*x^2*e + 4*(4*c^7*d + 9*
c^5*d + 6*c^3*d + c*d)*b*x*e + 2*(c^8 + 3*c^6 + 3*c^4 + c^2)*b*e + 2*(b*d^5*x^5*e + 5*b*c*d^4*x^4*e + (10*c^2*
d^3 + d^3)*b*x^3*e + (10*c^3*d^2 + 3*c*d^2)*b*x^2*e + (5*c^4*d + 3*c^2*d)*b*x*e + (c^5 + c^3)*b*e)*(d^2*x^2 +
2*c*d*x + c^2 + 1)^(3/2) + (6*b*d^6*x^6*e + 36*b*c*d^5*x^5*e + 10*(9*c^2*d^4 + d^4)*b*x^4*e + 40*(3*c^3*d^3 +
c*d^3)*b*x^3*e + 5*(18*c^4*d^2 + 12*c^2*d^2 + d^2)*b*x^2*e + 2*(18*c^5*d + 20*c^3*d + 5*c*d)*b*x*e + (6*c^6 +
10*c^4 + 5*c^2 + 1)*b*e)*(d^2*x^2 + 2*c*d*x + c^2 + 1) + (6*b*d^7*x^7*e + 42*b*c*d^6*x^6*e + 14*(9*c^2*d^5 + d
^5)*b*x^5*e + 70*(3*c^3*d^4 + c*d^4)*b*x^4*e + (210*c^4*d^3 + 140*c^2*d^3 + 11*d^3)*b*x^3*e + (126*c^5*d^2 + 1
40*c^3*d^2 + 33*c*d^2)*b*x^2*e + (42*c^6*d + 70*c^4*d + 33*c^2*d + 3*d)*b*x*e + (6*c^7 + 14*c^5 + 11*c^3 + 3*c
)*b*e)*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*(2*a*d^7 + b*d
^7)*x^7*e + 21*(2*a*c*d^6 + b*c*d^6)*x^6*e + 7*(2*(9*c^2*d^5 + d^5)*a + (9*c^2*d^5 + d^5)*b)*x^5*e + 35*(2*(3*
c^3*d^4 + c*d^4)*a + (3*c^3*d^4 + c*d^4)*b)*x^4*e + ((210*c^4*d^3 + 140*c^2*d^3 + 11*d^3)*a + 5*(21*c^4*d^3 +
14*c^2*d^3 + d^3)*b)*x^3*e + ((126*c^5*d^2 + 140*c^3*d^2 + 33*c*d^2)*a + (63*c^5*d^2 + 70*c^3*d^2 + 15*c*d^2)*
b)*x^2*e + ((42*c^6*d + 70*c^4*d + 33*c^2*d + 3*d)*a + (21*c^6*d + 35*c^4*d + 15*c^2*d + d)*b)*x*e + ((6*c^7 +
14*c^5 + 11*c^3 + 3*c)*a + (3*c^7 + 7*c^5 + 5*c^3 + c)*b)*e)*sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1))/(a^2*b^2*d^7*
x^6 + 6*a^2*b^2*c*d^6*x^5 + 3*(5*c^2*d^5 + d^5)*a^2*b^2*x^4 + 4*(5*c^3*d^4 + 3*c*d^4)*a^2*b^2*x^3 + 3*(5*c^4*d
^3 + 6*c^2*d^3 + d^3)*a^2*b^2*x^2 + 6*(c^5*d^2 + 2*c^3*d^2 + c*d^2)*a^2*b^2*x + (c^6*d + 3*c^4*d + 3*c^2*d + d
)*a^2*b^2 + (b^4*d^7*x^6 + 6*b^4*c*d^6*x^5 + 3*(5*c^2*d^5 + d^5)*b^4*x^4 + 4*(5*c^3*d^4 + 3*c*d^4)*b^4*x^3 + 3
*(5*c^4*d^3 + 6*c^2*d^3 + d^3)*b^4*x^2 + 6*(c^5*d^2 + 2*c^3*d^2 + c*d^2)*b^4*x + (c^6*d + 3*c^4*d + 3*c^2*d +
d)*b^4 + (b^4*d^4*x^3 + 3*b^4*c*d^3*x^2 + 3*b^4*c^2*d^2*x + b^4*c^3*d)*(d^2*x^2 + 2*c*d*x + c^2 + 1)^(3/2) + 3
*(b^4*d^5*x^4 + 4*b^4*c*d^4*x^3 + (6*c^2*d^3 + d^3)*b^4*x^2 + 2*(2*c^3*d^2 + c*d^2)*b^4*x + (c^4*d + c^2*d)*b^
4)*(d^2*x^2 + 2*c*d*x + c^2 + 1) + 3*(b^4*d^6*x^5 + 5*b^4*c*d^5*x^4 + 2*(5*c^2*d^4 + d^4)*b^4*x^3 + 2*(5*c^3*d
^3 + 3*c*d^3)*b^4*x^2 + (5*c^4*d^2 + 6*c^2*d^2 + d^2)*b^4*x + (c^5*d + 2*c^3*d + c*d)*b^4)*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 + (a^2*b^2*d^4*x^3 + 3*a^2*b^2*c*d^3*x^2 +
3*a^2*b^2*c^2*d^2*x + a^2*b^2*c^3*d)*(d^2*x^2 + 2*c*d*x + c^2 + 1)^(3/2) + 3*(a^2*b^2*d^5*x^4 + 4*a^2*b^2*c*d^
4*x^3 + (6*c^2*d^3 + d^3)*a^2*b^2*x^2 + 2*(2*c^3*d^2 + c*d^2)*a^2*b^2*x + (c^4*d + c^2*d)*a^2*b^2)*(d^2*x^2 +
2*c*d*x + c^2 + 1) + 2*(a*b^3*d^7*x^6 + 6*a*b^3*c*d^6*x^5 + 3*(5*c^2*d^5 + d^5)*a*b^3*x^4 + 4*(5*c^3*d^4 + 3*c
*d^4)*a*b^3*x^3 + 3*(5*c^4*d^3 + 6*c^2*d^3 + d^3)*a*b^3*x^2 + 6*(c^5*d^2 + 2*c^3*d^2 + c*d^2)*a*b^3*x + (c^6*d
+ 3*c^4*d + 3*c^2*d + d)*a*b^3 + (a*b^3*d^4*x^3 + 3*a*b^3*c*d^3*x^2 + 3*a*b^3*c^2*d^2*x + a*b^3*c^3*d)*(d^2*x
^2 + 2*c*d*x + c^2 + 1)^(3/2) + 3*(a*b^3*d^5*x^4 + 4*a*b^3*c*d^4*x^3 + (6*c^2*d^3 + d^3)*a*b^3*x^2 + 2*(2*c^3*
d^2 + c*d^2)*a*b^3*x + (c^4*d + c^2*d)*a*b^3)*(d^2*x^2 + 2*c*d*x + c^2 + 1) + 3*(a*b^3*d^6*x^5 + 5*a*b^3*c*d^5
*x^4 + 2*(5*c^2*d^4 + d^4)*a*b^3*x^3 + 2*(5*c^3*d^3 + 3*c*d^3)*a*b^3*x^2 + (5*c^4*d^2 + 6*c^2*d^2 + d^2)*a*b^3
*x + (c^5*d + 2*c^3*d + c*d)*a*b^3)*sqrt(d^2*x^...

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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((d*e*x+c*e)/(a+b*arcsinh(d*x+c))^3,x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)/(a+b*asinh(d*x+c))**3,x)

[Out]

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

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Giac [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((d*e*x+c*e)/(a+b*arcsinh(d*x+c))^3,x, algorithm="giac")

[Out]

integrate((d*e*x + c*e)/(b*arcsinh(d*x + c) + a)^3, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {c\,e+d\,e\,x}{{\left (a+b\,\mathrm {asinh}\left (c+d\,x\right )\right )}^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*e + d*e*x)/(a + b*asinh(c + d*x))^3,x)

[Out]

int((c*e + d*e*x)/(a + b*asinh(c + d*x))^3, x)

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