3.2.78 \(\int \frac {1}{(a+b \sinh ^{-1}(c+d x))^4} \, dx\) [178]

Optimal. Leaf size=160 \[ -\frac {\sqrt {1+(c+d x)^2}}{3 b d \left (a+b \sinh ^{-1}(c+d x)\right )^3}-\frac {c+d x}{6 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^2}-\frac {\sqrt {1+(c+d x)^2}}{6 b^3 d \left (a+b \sinh ^{-1}(c+d x)\right )}-\frac {\text {Chi}\left (\frac {a+b \sinh ^{-1}(c+d x)}{b}\right ) \sinh \left (\frac {a}{b}\right )}{6 b^4 d}+\frac {\cosh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a+b \sinh ^{-1}(c+d x)}{b}\right )}{6 b^4 d} \]

[Out]

1/6*(-d*x-c)/b^2/d/(a+b*arcsinh(d*x+c))^2+1/6*cosh(a/b)*Shi((a+b*arcsinh(d*x+c))/b)/b^4/d-1/6*Chi((a+b*arcsinh
(d*x+c))/b)*sinh(a/b)/b^4/d-1/3*(1+(d*x+c)^2)^(1/2)/b/d/(a+b*arcsinh(d*x+c))^3-1/6*(1+(d*x+c)^2)^(1/2)/b^3/d/(
a+b*arcsinh(d*x+c))

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Rubi [A]
time = 0.19, antiderivative size = 160, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.583, Rules used = {5858, 5773, 5818, 5819, 3384, 3379, 3382} \begin {gather*} -\frac {\sinh \left (\frac {a}{b}\right ) \text {Chi}\left (\frac {a+b \sinh ^{-1}(c+d x)}{b}\right )}{6 b^4 d}+\frac {\cosh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a+b \sinh ^{-1}(c+d x)}{b}\right )}{6 b^4 d}-\frac {\sqrt {(c+d x)^2+1}}{6 b^3 d \left (a+b \sinh ^{-1}(c+d x)\right )}-\frac {c+d x}{6 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^2}-\frac {\sqrt {(c+d x)^2+1}}{3 b d \left (a+b \sinh ^{-1}(c+d x)\right )^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSinh[c + d*x])^(-4),x]

[Out]

-1/3*Sqrt[1 + (c + d*x)^2]/(b*d*(a + b*ArcSinh[c + d*x])^3) - (c + d*x)/(6*b^2*d*(a + b*ArcSinh[c + d*x])^2) -
Sqrt[1 + (c + d*x)^2]/(6*b^3*d*(a + b*ArcSinh[c + d*x])) - (CoshIntegral[(a + b*ArcSinh[c + d*x])/b]*Sinh[a/b
])/(6*b^4*d) + (Cosh[a/b]*SinhIntegral[(a + b*ArcSinh[c + d*x])/b])/(6*b^4*d)

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 5773

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[Sqrt[1 + c^2*x^2]*((a + b*ArcSinh[c*x])^(n + 1
)/(b*c*(n + 1))), x] - Dist[c/(b*(n + 1)), Int[x*((a + b*ArcSinh[c*x])^(n + 1)/Sqrt[1 + c^2*x^2]), x], x] /; F
reeQ[{a, b, c}, x] && LtQ[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 5819

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

Rule 5858

Int[((a_.) + ArcSinh[(c_) + (d_.)*(x_)]*(b_.))^(n_.), x_Symbol] :> Dist[1/d, Subst[Int[(a + b*ArcSinh[x])^n, x
], x, c + d*x], x] /; FreeQ[{a, b, c, d, n}, x]

Rubi steps

\begin {align*} \int \frac {1}{\left (a+b \sinh ^{-1}(c+d x)\right )^4} \, dx &=\frac {\text {Subst}\left (\int \frac {1}{\left (a+b \sinh ^{-1}(x)\right )^4} \, dx,x,c+d x\right )}{d}\\ &=-\frac {\sqrt {1+(c+d x)^2}}{3 b d \left (a+b \sinh ^{-1}(c+d x)\right )^3}+\frac {\text {Subst}\left (\int \frac {x}{\sqrt {1+x^2} \left (a+b \sinh ^{-1}(x)\right )^3} \, dx,x,c+d x\right )}{3 b d}\\ &=-\frac {\sqrt {1+(c+d x)^2}}{3 b d \left (a+b \sinh ^{-1}(c+d x)\right )^3}-\frac {c+d x}{6 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^2}+\frac {\text {Subst}\left (\int \frac {1}{\left (a+b \sinh ^{-1}(x)\right )^2} \, dx,x,c+d x\right )}{6 b^2 d}\\ &=-\frac {\sqrt {1+(c+d x)^2}}{3 b d \left (a+b \sinh ^{-1}(c+d x)\right )^3}-\frac {c+d x}{6 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^2}-\frac {\sqrt {1+(c+d x)^2}}{6 b^3 d \left (a+b \sinh ^{-1}(c+d x)\right )}+\frac {\text {Subst}\left (\int \frac {x}{\sqrt {1+x^2} \left (a+b \sinh ^{-1}(x)\right )} \, dx,x,c+d x\right )}{6 b^3 d}\\ &=-\frac {\sqrt {1+(c+d x)^2}}{3 b d \left (a+b \sinh ^{-1}(c+d x)\right )^3}-\frac {c+d x}{6 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^2}-\frac {\sqrt {1+(c+d x)^2}}{6 b^3 d \left (a+b \sinh ^{-1}(c+d x)\right )}+\frac {\text {Subst}\left (\int \frac {\sinh (x)}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{6 b^3 d}\\ &=-\frac {\sqrt {1+(c+d x)^2}}{3 b d \left (a+b \sinh ^{-1}(c+d x)\right )^3}-\frac {c+d x}{6 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^2}-\frac {\sqrt {1+(c+d x)^2}}{6 b^3 d \left (a+b \sinh ^{-1}(c+d x)\right )}+\frac {\cosh \left (\frac {a}{b}\right ) \text {Subst}\left (\int \frac {\sinh \left (\frac {a}{b}+x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{6 b^3 d}-\frac {\sinh \left (\frac {a}{b}\right ) \text {Subst}\left (\int \frac {\cosh \left (\frac {a}{b}+x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{6 b^3 d}\\ &=-\frac {\sqrt {1+(c+d x)^2}}{3 b d \left (a+b \sinh ^{-1}(c+d x)\right )^3}-\frac {c+d x}{6 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^2}-\frac {\sqrt {1+(c+d x)^2}}{6 b^3 d \left (a+b \sinh ^{-1}(c+d x)\right )}-\frac {\text {Chi}\left (\frac {a}{b}+\sinh ^{-1}(c+d x)\right ) \sinh \left (\frac {a}{b}\right )}{6 b^4 d}+\frac {\cosh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a}{b}+\sinh ^{-1}(c+d x)\right )}{6 b^4 d}\\ \end {align*}

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Mathematica [A]
time = 0.31, size = 130, normalized size = 0.81 \begin {gather*} -\frac {\frac {2 b^3 \sqrt {1+(c+d x)^2}}{\left (a+b \sinh ^{-1}(c+d x)\right )^3}+\frac {b^2 (c+d x)}{\left (a+b \sinh ^{-1}(c+d x)\right )^2}+\frac {b \sqrt {1+(c+d x)^2}}{a+b \sinh ^{-1}(c+d x)}+\text {Chi}\left (\frac {a}{b}+\sinh ^{-1}(c+d x)\right ) \sinh \left (\frac {a}{b}\right )-\cosh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a}{b}+\sinh ^{-1}(c+d x)\right )}{6 b^4 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/6*((2*b^3*Sqrt[1 + (c + d*x)^2])/(a + b*ArcSinh[c + d*x])^3 + (b^2*(c + d*x))/(a + b*ArcSinh[c + d*x])^2 +
(b*Sqrt[1 + (c + d*x)^2])/(a + b*ArcSinh[c + d*x]) + CoshIntegral[a/b + ArcSinh[c + d*x]]*Sinh[a/b] - Cosh[a/b
]*SinhIntegral[a/b + ArcSinh[c + d*x]])/(b^4*d)

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Maple [A]
time = 3.34, size = 272, normalized size = 1.70

method	result	size
derivativedivides	\(\frac {\frac {\left (-\sqrt {1+\left (d x +c \right )^{2}}+d x +c \right ) \left (b^{2} \arcsinh \left (d x +c \right )^{2}+2 a b \arcsinh \left (d x +c \right )-b^{2} \arcsinh \left (d x +c \right )+a^{2}-a b +2 b^{2}\right )}{12 b^{3} \left (b^{3} \arcsinh \left (d x +c \right )^{3}+3 a \,b^{2} \arcsinh \left (d x +c \right )^{2}+3 a^{2} b \arcsinh \left (d x +c \right )+a^{3}\right )}+\frac {{\mathrm e}^{\frac {a}{b}} \expIntegral \left (1, \arcsinh \left (d x +c \right )+\frac {a}{b}\right )}{12 b^{4}}-\frac {d x +c +\sqrt {1+\left (d x +c \right )^{2}}}{6 b \left (a +b \arcsinh \left (d x +c \right )\right )^{3}}-\frac {d x +c +\sqrt {1+\left (d x +c \right )^{2}}}{12 b^{2} \left (a +b \arcsinh \left (d x +c \right )\right )^{2}}-\frac {d x +c +\sqrt {1+\left (d x +c \right )^{2}}}{12 b^{3} \left (a +b \arcsinh \left (d x +c \right )\right )}-\frac {{\mathrm e}^{-\frac {a}{b}} \expIntegral \left (1, -\arcsinh \left (d x +c \right )-\frac {a}{b}\right )}{12 b^{4}}}{d}\)	\(272\)
default	\(\frac {\frac {\left (-\sqrt {1+\left (d x +c \right )^{2}}+d x +c \right ) \left (b^{2} \arcsinh \left (d x +c \right )^{2}+2 a b \arcsinh \left (d x +c \right )-b^{2} \arcsinh \left (d x +c \right )+a^{2}-a b +2 b^{2}\right )}{12 b^{3} \left (b^{3} \arcsinh \left (d x +c \right )^{3}+3 a \,b^{2} \arcsinh \left (d x +c \right )^{2}+3 a^{2} b \arcsinh \left (d x +c \right )+a^{3}\right )}+\frac {{\mathrm e}^{\frac {a}{b}} \expIntegral \left (1, \arcsinh \left (d x +c \right )+\frac {a}{b}\right )}{12 b^{4}}-\frac {d x +c +\sqrt {1+\left (d x +c \right )^{2}}}{6 b \left (a +b \arcsinh \left (d x +c \right )\right )^{3}}-\frac {d x +c +\sqrt {1+\left (d x +c \right )^{2}}}{12 b^{2} \left (a +b \arcsinh \left (d x +c \right )\right )^{2}}-\frac {d x +c +\sqrt {1+\left (d x +c \right )^{2}}}{12 b^{3} \left (a +b \arcsinh \left (d x +c \right )\right )}-\frac {{\mathrm e}^{-\frac {a}{b}} \expIntegral \left (1, -\arcsinh \left (d x +c \right )-\frac {a}{b}\right )}{12 b^{4}}}{d}\)	\(272\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*arcsinh(d*x+c))^4,x,method=_RETURNVERBOSE)

[Out]

1/d*(1/12*(-(1+(d*x+c)^2)^(1/2)+d*x+c)*(b^2*arcsinh(d*x+c)^2+2*a*b*arcsinh(d*x+c)-b^2*arcsinh(d*x+c)+a^2-a*b+2
*b^2)/b^3/(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)+1/12/b^4*exp(a/b)*Ei(1,ar
csinh(d*x+c)+a/b)-1/6/b*(d*x+c+(1+(d*x+c)^2)^(1/2))/(a+b*arcsinh(d*x+c))^3-1/12/b^2*(d*x+c+(1+(d*x+c)^2)^(1/2)
)/(a+b*arcsinh(d*x+c))^2-1/12/b^3*(d*x+c+(1+(d*x+c)^2)^(1/2))/(a+b*arcsinh(d*x+c))-1/12/b^4*exp(-a/b)*Ei(1,-ar
csinh(d*x+c)-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(1/(a+b*arcsinh(d*x+c))^4,x, algorithm="maxima")

[Out]

-1/6*((a^2*d^11 + a*b*d^11 + 2*b^2*d^11)*x^11 + 11*(a^2*c*d^10 + a*b*c*d^10 + 2*b^2*c*d^10)*x^10 + 5*((11*c^2*
d^9 + d^9)*a^2 + (11*c^2*d^9 + d^9)*a*b + 2*(11*c^2*d^9 + d^9)*b^2)*x^9 + 15*((11*c^3*d^8 + 3*c*d^8)*a^2 + (11
*c^3*d^8 + 3*c*d^8)*a*b + 2*(11*c^3*d^8 + 3*c*d^8)*b^2)*x^8 + 10*((33*c^4*d^7 + 18*c^2*d^7 + d^7)*a^2 + (33*c^
4*d^7 + 18*c^2*d^7 + d^7)*a*b + 2*(33*c^4*d^7 + 18*c^2*d^7 + d^7)*b^2)*x^7 + 14*((33*c^5*d^6 + 30*c^3*d^6 + 5*
c*d^6)*a^2 + (33*c^5*d^6 + 30*c^3*d^6 + 5*c*d^6)*a*b + 2*(33*c^5*d^6 + 30*c^3*d^6 + 5*c*d^6)*b^2)*x^6 + 2*((23
1*c^6*d^5 + 315*c^4*d^5 + 105*c^2*d^5 + 5*d^5)*a^2 + (231*c^6*d^5 + 315*c^4*d^5 + 105*c^2*d^5 + 5*d^5)*a*b + 2
*(231*c^6*d^5 + 315*c^4*d^5 + 105*c^2*d^5 + 5*d^5)*b^2)*x^5 + 10*((33*c^7*d^4 + 63*c^5*d^4 + 35*c^3*d^4 + 5*c*
d^4)*a^2 + (33*c^7*d^4 + 63*c^5*d^4 + 35*c^3*d^4 + 5*c*d^4)*a*b + 2*(33*c^7*d^4 + 63*c^5*d^4 + 35*c^3*d^4 + 5*
c*d^4)*b^2)*x^4 + 5*((33*c^8*d^3 + 84*c^6*d^3 + 70*c^4*d^3 + 20*c^2*d^3 + d^3)*a^2 + (33*c^8*d^3 + 84*c^6*d^3
+ 70*c^4*d^3 + 20*c^2*d^3 + d^3)*a*b + 2*(33*c^8*d^3 + 84*c^6*d^3 + 70*c^4*d^3 + 20*c^2*d^3 + d^3)*b^2)*x^3 +
((a^2*d^6 + a*b*d^6 + 2*b^2*d^6)*x^6 + 6*(a^2*c*d^5 + a*b*c*d^5 + 2*b^2*c*d^5)*x^5 + (15*a*b*c^2*d^4 + (15*c^2
*d^4 + d^4)*a^2 + 2*(15*c^2*d^4 + d^4)*b^2)*x^4 + 4*(5*a*b*c^3*d^3 + (5*c^3*d^3 + c*d^3)*a^2 + 2*(5*c^3*d^3 +
c*d^3)*b^2)*x^3 + (c^6 + c^4 + 3*c^2 + 3)*a^2 + (c^6 - c^2)*a*b + 2*(c^6 + c^4)*b^2 + (3*(5*c^4*d^2 + 2*c^2*d^
2 + d^2)*a^2 + (15*c^4*d^2 - d^2)*a*b + 6*(5*c^4*d^2 + 2*c^2*d^2)*b^2)*x^2 + 2*((3*c^5*d + 2*c^3*d + 3*c*d)*a^
2 + (3*c^5*d - c*d)*a*b + 2*(3*c^5*d + 2*c^3*d)*b^2)*x)*(d^2*x^2 + 2*c*d*x + c^2 + 1)^(5/2) + (5*(a^2*d^7 + a*
b*d^7 + 2*b^2*d^7)*x^7 + 35*(a^2*c*d^6 + a*b*c*d^6 + 2*b^2*c*d^6)*x^6 + (3*(35*c^2*d^5 + 3*d^5)*a^2 + 5*(21*c^
2*d^5 + d^5)*a*b + 6*(35*c^2*d^5 + 3*d^5)*b^2)*x^5 + 5*((35*c^3*d^4 + 9*c*d^4)*a^2 + 5*(7*c^3*d^4 + c*d^4)*a*b
+ 2*(35*c^3*d^4 + 9*c*d^4)*b^2)*x^4 + (5*(35*c^4*d^3 + 18*c^2*d^3 + 2*d^3)*a^2 + (175*c^4*d^3 + 50*c^2*d^3 -
2*d^3)*a*b + 2*(175*c^4*d^3 + 90*c^2*d^3 + 4*d^3)*b^2)*x^3 + (5*c^7 + 9*c^5 + 10*c^3 + 6*c)*a^2 + (5*c^7 + 5*c
^5 - 2*c^3 - 2*c)*a*b + 2*(5*c^7 + 9*c^5 + 4*c^3)*b^2 + (15*(7*c^5*d^2 + 6*c^3*d^2 + 2*c*d^2)*a^2 + (105*c^5*d
^2 + 50*c^3*d^2 - 6*c*d^2)*a*b + 6*(35*c^5*d^2 + 30*c^3*d^2 + 4*c*d^2)*b^2)*x^2 + ((35*c^6*d + 45*c^4*d + 30*c
^2*d + 6*d)*a^2 + (35*c^6*d + 25*c^4*d - 6*c^2*d - 2*d)*a*b + 2*(35*c^6*d + 45*c^4*d + 12*c^2*d)*b^2)*x)*(d^2*
x^2 + 2*c*d*x + c^2 + 1)^2 + (c^11 + 5*c^9 + 10*c^7 + 10*c^5 + 5*c^3 + c)*a^2 + (c^11 + 5*c^9 + 10*c^7 + 10*c^
5 + 5*c^3 + c)*a*b + 2*(c^11 + 5*c^9 + 10*c^7 + 10*c^5 + 5*c^3 + c)*b^2 + 5*((11*c^9*d^2 + 36*c^7*d^2 + 42*c^5
*d^2 + 20*c^3*d^2 + 3*c*d^2)*a^2 + (11*c^9*d^2 + 36*c^7*d^2 + 42*c^5*d^2 + 20*c^3*d^2 + 3*c*d^2)*a*b + 2*(11*c
^9*d^2 + 36*c^7*d^2 + 42*c^5*d^2 + 20*c^3*d^2 + 3*c*d^2)*b^2)*x^2 + (b^2*d^11*x^11 + 11*b^2*c*d^10*x^10 + 5*(1
1*c^2*d^9 + d^9)*b^2*x^9 + 15*(11*c^3*d^8 + 3*c*d^8)*b^2*x^8 + 10*(33*c^4*d^7 + 18*c^2*d^7 + d^7)*b^2*x^7 + 14
*(33*c^5*d^6 + 30*c^3*d^6 + 5*c*d^6)*b^2*x^6 + 2*(231*c^6*d^5 + 315*c^4*d^5 + 105*c^2*d^5 + 5*d^5)*b^2*x^5 + 1
0*(33*c^7*d^4 + 63*c^5*d^4 + 35*c^3*d^4 + 5*c*d^4)*b^2*x^4 + 5*(33*c^8*d^3 + 84*c^6*d^3 + 70*c^4*d^3 + 20*c^2*
d^3 + d^3)*b^2*x^3 + 5*(11*c^9*d^2 + 36*c^7*d^2 + 42*c^5*d^2 + 20*c^3*d^2 + 3*c*d^2)*b^2*x^2 + (11*c^10*d + 45
*c^8*d + 70*c^6*d + 50*c^4*d + 15*c^2*d + d)*b^2*x + (b^2*d^6*x^6 + 6*b^2*c*d^5*x^5 + (15*c^2*d^4 + d^4)*b^2*x
^4 + 4*(5*c^3*d^3 + c*d^3)*b^2*x^3 + 3*(5*c^4*d^2 + 2*c^2*d^2 + d^2)*b^2*x^2 + 2*(3*c^5*d + 2*c^3*d + 3*c*d)*b
^2*x + (c^6 + c^4 + 3*c^2 + 3)*b^2)*(d^2*x^2 + 2*c*d*x + c^2 + 1)^(5/2) + (5*b^2*d^7*x^7 + 35*b^2*c*d^6*x^6 +
3*(35*c^2*d^5 + 3*d^5)*b^2*x^5 + 5*(35*c^3*d^4 + 9*c*d^4)*b^2*x^4 + 5*(35*c^4*d^3 + 18*c^2*d^3 + 2*d^3)*b^2*x^
3 + 15*(7*c^5*d^2 + 6*c^3*d^2 + 2*c*d^2)*b^2*x^2 + (35*c^6*d + 45*c^4*d + 30*c^2*d + 6*d)*b^2*x + (5*c^7 + 9*c
^5 + 10*c^3 + 6*c)*b^2)*(d^2*x^2 + 2*c*d*x + c^2 + 1)^2 + (c^11 + 5*c^9 + 10*c^7 + 10*c^5 + 5*c^3 + c)*b^2 + (
10*b^2*d^8*x^8 + 80*b^2*c*d^7*x^7 + 2*(140*c^2*d^6 + 13*d^6)*b^2*x^6 + 4*(140*c^3*d^5 + 39*c*d^5)*b^2*x^5 + 2*
(350*c^4*d^4 + 195*c^2*d^4 + 11*d^4)*b^2*x^4 + 8*(70*c^5*d^3 + 65*c^3*d^3 + 11*c*d^3)*b^2*x^3 + (280*c^6*d^2 +
390*c^4*d^2 + 132*c^2*d^2 + 3*d^2)*b^2*x^2 + 2*(40*c^7*d + 78*c^5*d + 44*c^3*d + 3*c*d)*b^2*x + (10*c^8 + 26*
c^6 + 22*c^4 + 3*c^2 - 3)*b^2)*(d^2*x^2 + 2*c*d*x + c^2 + 1)^(3/2) + 2*(5*b^2*d^9*x^9 + 45*b^2*c*d^8*x^8 + (18
0*c^2*d^7 + 17*d^7)*b^2*x^7 + 7*(60*c^3*d^6 + 17*c*d^6)*b^2*x^6 + 3*(210*c^4*d^5 + 119*c^2*d^5 + 6*d^5)*b^2*x^
5 + 5*(126*c^5*d^4 + 119*c^3*d^4 + 18*c*d^4)*b^2*x^4 + 5*(84*c^6*d^3 + 119*c^4*d^3 + 36*c^2*d^3 + d^3)*b^2*x^3
+ 3*(60*c^7*d^2 + 119*c^5*d^2 + 60*c^3*d^2 + 5*c*d^2)*b^2*x^2 + (45*c^8*d + 119*c^6*d + 90*c^4*d + 15*c^2*d -
d)*b^2*x + (5*c^9 + 17*c^7 + 18*c^5 + 5*c^3 - c)*b^2)*(d^2*x^2 + 2*c*d*x + c^2 + 1) + (5*b^2*d^10*x^10 + 50*b
^2*c*d^9*x^9 + 3*(75*c^2*d^8 + 7*d^8)*b^2*x^8 + 24*(25*c^3*d^7 + 7*c*d^7)*b^2*x^7 + (1050*c^4*d^6 + 588*c^2*d^
6 + 31*d^6)*b^2*x^6 + 6*(210*c^5*d^5 + 196*c^3*...

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

[Out]

integral(1/(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*arcsi
nh(d*x + c) + a^4), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (a + b \operatorname {asinh}{\left (c + d x \right )}\right )^{4}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*asinh(d*x+c))**4,x)

[Out]

Integral((a + b*asinh(c + d*x))**(-4), 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(1/(a+b*arcsinh(d*x+c))^4,x, algorithm="giac")

[Out]

integrate((b*arcsinh(d*x + c) + a)^(-4), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{{\left (a+b\,\mathrm {asinh}\left (c+d\,x\right )\right )}^4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*asinh(c + d*x))^4,x)

[Out]

int(1/(a + b*asinh(c + d*x))^4, x)

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