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

Optimal. Leaf size=204 \[ \frac{2 e \cosh \left (\frac{2 a}{b}\right ) \text{Chi}\left (\frac{2 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )}{3 b^4 d}-\frac{2 e \sinh \left (\frac{2 a}{b}\right ) \text{Shi}\left (\frac{2 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )}{3 b^4 d}-\frac{e (c+d x)^2}{3 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^2}-\frac{2 e \sqrt{(c+d x)^2+1} (c+d x)}{3 b^3 d \left (a+b \sinh ^{-1}(c+d x)\right )}-\frac{e}{6 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^2}-\frac{e \sqrt{(c+d x)^2+1} (c+d x)}{3 b d \left (a+b \sinh ^{-1}(c+d x)\right )^3} \]

[Out]

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

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Rubi [A]  time = 0.341499, antiderivative size = 204, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 9, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.429, Rules used = {5865, 12, 5667, 5774, 5665, 3303, 3298, 3301, 5675} \[ \frac{2 e \cosh \left (\frac{2 a}{b}\right ) \text{Chi}\left (\frac{2 a}{b}+2 \sinh ^{-1}(c+d x)\right )}{3 b^4 d}-\frac{2 e \sinh \left (\frac{2 a}{b}\right ) \text{Shi}\left (\frac{2 a}{b}+2 \sinh ^{-1}(c+d x)\right )}{3 b^4 d}-\frac{e (c+d x)^2}{3 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^2}-\frac{2 e \sqrt{(c+d x)^2+1} (c+d x)}{3 b^3 d \left (a+b \sinh ^{-1}(c+d x)\right )}-\frac{e}{6 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^2}-\frac{e \sqrt{(c+d x)^2+1} (c+d x)}{3 b d \left (a+b \sinh ^{-1}(c+d x)\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 5667

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 5774

Int[(((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_)*((f_.)*(x_))^(m_.))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp
[((f*x)^m*(a + b*ArcSinh[c*x])^(n + 1))/(b*c*Sqrt[d]*(n + 1)), x] - Dist[(f*m)/(b*c*Sqrt[d]*(n + 1)), 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] && GtQ[d, 0]

Rule 5665

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[1/(b*c^(m + 1)*(n + 1)), Subst[Int[ExpandTrigReduce[(a + b*x)^(n +
1), Sinh[x]^(m - 1)*(m + (m + 1)*Sinh[x]^2), x], x], x, ArcSinh[c*x]], x] /; FreeQ[{a, b, c}, x] && IGtQ[m, 0]
&& GeQ[n, -2] && LtQ[n, -1]

Rule 3303

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 3298

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 3301

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 5675

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

Rubi steps

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

Mathematica [A]  time = 0.790984, size = 181, normalized size = 0.89 \[ \frac{e \left (-\frac{2 b^3 (c+d x) \sqrt{(c+d x)^2+1}}{\left (a+b \sinh ^{-1}(c+d x)\right )^3}+\frac{b^2 \left (-2 (c+d x)^2-1\right )}{\left (a+b \sinh ^{-1}(c+d x)\right )^2}+4 \cosh \left (\frac{2 a}{b}\right ) \text{Chi}\left (2 \left (\frac{a}{b}+\sinh ^{-1}(c+d x)\right )\right )-4 \left (\sinh \left (\frac{2 a}{b}\right ) \text{Shi}\left (2 \left (\frac{a}{b}+\sinh ^{-1}(c+d x)\right )\right )+\log \left (a+b \sinh ^{-1}(c+d x)\right )\right )-\frac{4 b (c+d x) \sqrt{(c+d x)^2+1}}{a+b \sinh ^{-1}(c+d x)}+4 \log \left (a+b \sinh ^{-1}(c+d x)\right )\right )}{6 b^4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(e*((-2*b^3*(c + d*x)*Sqrt[1 + (c + d*x)^2])/(a + b*ArcSinh[c + d*x])^3 + (b^2*(-1 - 2*(c + d*x)^2))/(a + b*Ar
cSinh[c + d*x])^2 - (4*b*(c + d*x)*Sqrt[1 + (c + d*x)^2])/(a + b*ArcSinh[c + d*x]) + 4*Cosh[(2*a)/b]*CoshInteg
ral[2*(a/b + ArcSinh[c + d*x])] + 4*Log[a + b*ArcSinh[c + d*x]] - 4*(Log[a + b*ArcSinh[c + d*x]] + Sinh[(2*a)/
b]*SinhIntegral[2*(a/b + ArcSinh[c + d*x])])))/(6*b^4*d)

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Maple [A]  time = 0.072, size = 333, normalized size = 1.6 \begin{align*}{\frac{1}{d} \left ({\frac{e \left ( 2\,{b}^{2} \left ({\it Arcsinh} \left ( dx+c \right ) \right ) ^{2}+4\,ab{\it Arcsinh} \left ( dx+c \right ) -{\it Arcsinh} \left ( dx+c \right ){b}^{2}+2\,{a}^{2}-ab+{b}^{2} \right ) }{12\,{b}^{3} \left ({b}^{3} \left ({\it Arcsinh} \left ( dx+c \right ) \right ) ^{3}+3\, \left ({\it Arcsinh} \left ( dx+c \right ) \right ) ^{2}a{b}^{2}+3\,{a}^{2}b{\it Arcsinh} \left ( dx+c \right ) +{a}^{3} \right ) } \left ( 2\, \left ( dx+c \right ) ^{2}-2\, \left ( dx+c \right ) \sqrt{1+ \left ( dx+c \right ) ^{2}}+1 \right ) }-{\frac{e}{3\,{b}^{4}}{{\rm e}^{2\,{\frac{a}{b}}}}{\it Ei} \left ( 1,2\,{\it Arcsinh} \left ( dx+c \right ) +2\,{\frac{a}{b}} \right ) }-{\frac{e}{12\,b \left ( a+b{\it Arcsinh} \left ( dx+c \right ) \right ) ^{3}} \left ( 2\, \left ( dx+c \right ) ^{2}+1+2\, \left ( dx+c \right ) \sqrt{1+ \left ( dx+c \right ) ^{2}} \right ) }-{\frac{e}{12\,{b}^{2} \left ( a+b{\it Arcsinh} \left ( dx+c \right ) \right ) ^{2}} \left ( 2\, \left ( dx+c \right ) ^{2}+1+2\, \left ( dx+c \right ) \sqrt{1+ \left ( dx+c \right ) ^{2}} \right ) }-{\frac{e}{6\,{b}^{3} \left ( a+b{\it Arcsinh} \left ( dx+c \right ) \right ) } \left ( 2\, \left ( dx+c \right ) ^{2}+1+2\, \left ( dx+c \right ) \sqrt{1+ \left ( dx+c \right ) ^{2}} \right ) }-{\frac{e}{3\,{b}^{4}}{{\rm e}^{-2\,{\frac{a}{b}}}}{\it Ei} \left ( 1,-2\,{\it Arcsinh} \left ( dx+c \right ) -2\,{\frac{a}{b}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(1/12*(2*(d*x+c)^2-2*(d*x+c)*(1+(d*x+c)^2)^(1/2)+1)*e*(2*b^2*arcsinh(d*x+c)^2+4*a*b*arcsinh(d*x+c)-arcsinh
(d*x+c)*b^2+2*a^2-a*b+b^2)/b^3/(b^3*arcsinh(d*x+c)^3+3*arcsinh(d*x+c)^2*a*b^2+3*a^2*b*arcsinh(d*x+c)+a^3)-1/3*
e/b^4*exp(2*a/b)*Ei(1,2*arcsinh(d*x+c)+2*a/b)-1/12*e/b*(2*(d*x+c)^2+1+2*(d*x+c)*(1+(d*x+c)^2)^(1/2))/(a+b*arcs
inh(d*x+c))^3-1/12*e/b^2*(2*(d*x+c)^2+1+2*(d*x+c)*(1+(d*x+c)^2)^(1/2))/(a+b*arcsinh(d*x+c))^2-1/6*e/b^3*(2*(d*
x+c)^2+1+2*(d*x+c)*(1+(d*x+c)^2)^(1/2))/(a+b*arcsinh(d*x+c))-1/3*e/b^4*exp(-2*a/b)*Ei(1,-2*arcsinh(d*x+c)-2*a/
b))

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Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{d e x + c e}{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}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((d*e*x + c*e)/(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), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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