3.32 \(\int \frac{(a+b \text{csch}^{-1}(c x))^3}{x^5} \, dx\)

Optimal. Leaf size=204 \[ \frac{9 b^2 c^2 \left (a+b \text{csch}^{-1}(c x)\right )}{32 x^2}-\frac{3 b^2 \left (a+b \text{csch}^{-1}(c x)\right )}{32 x^4}-\frac{9 b c^3 \sqrt{\frac{1}{c^2 x^2}+1} \left (a+b \text{csch}^{-1}(c x)\right )^2}{32 x}+\frac{3 b c \sqrt{\frac{1}{c^2 x^2}+1} \left (a+b \text{csch}^{-1}(c x)\right )^2}{16 x^3}+\frac{3}{32} c^4 \left (a+b \text{csch}^{-1}(c x)\right )^3-\frac{\left (a+b \text{csch}^{-1}(c x)\right )^3}{4 x^4}-\frac{45 b^3 c^3 \sqrt{\frac{1}{c^2 x^2}+1}}{256 x}+\frac{3 b^3 c \sqrt{\frac{1}{c^2 x^2}+1}}{128 x^3}+\frac{45}{256} b^3 c^4 \text{csch}^{-1}(c x) \]

[Out]

(3*b^3*c*Sqrt[1 + 1/(c^2*x^2)])/(128*x^3) - (45*b^3*c^3*Sqrt[1 + 1/(c^2*x^2)])/(256*x) + (45*b^3*c^4*ArcCsch[c
*x])/256 - (3*b^2*(a + b*ArcCsch[c*x]))/(32*x^4) + (9*b^2*c^2*(a + b*ArcCsch[c*x]))/(32*x^2) + (3*b*c*Sqrt[1 +
1/(c^2*x^2)]*(a + b*ArcCsch[c*x])^2)/(16*x^3) - (9*b*c^3*Sqrt[1 + 1/(c^2*x^2)]*(a + b*ArcCsch[c*x])^2)/(32*x)
+ (3*c^4*(a + b*ArcCsch[c*x])^3)/32 - (a + b*ArcCsch[c*x])^3/(4*x^4)

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Rubi [A]  time = 0.181361, antiderivative size = 204, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 6, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.429, Rules used = {6286, 5446, 3311, 32, 2635, 8} \[ \frac{9 b^2 c^2 \left (a+b \text{csch}^{-1}(c x)\right )}{32 x^2}-\frac{3 b^2 \left (a+b \text{csch}^{-1}(c x)\right )}{32 x^4}-\frac{9 b c^3 \sqrt{\frac{1}{c^2 x^2}+1} \left (a+b \text{csch}^{-1}(c x)\right )^2}{32 x}+\frac{3 b c \sqrt{\frac{1}{c^2 x^2}+1} \left (a+b \text{csch}^{-1}(c x)\right )^2}{16 x^3}+\frac{3}{32} c^4 \left (a+b \text{csch}^{-1}(c x)\right )^3-\frac{\left (a+b \text{csch}^{-1}(c x)\right )^3}{4 x^4}-\frac{45 b^3 c^3 \sqrt{\frac{1}{c^2 x^2}+1}}{256 x}+\frac{3 b^3 c \sqrt{\frac{1}{c^2 x^2}+1}}{128 x^3}+\frac{45}{256} b^3 c^4 \text{csch}^{-1}(c x) \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcCsch[c*x])^3/x^5,x]

[Out]

(3*b^3*c*Sqrt[1 + 1/(c^2*x^2)])/(128*x^3) - (45*b^3*c^3*Sqrt[1 + 1/(c^2*x^2)])/(256*x) + (45*b^3*c^4*ArcCsch[c
*x])/256 - (3*b^2*(a + b*ArcCsch[c*x]))/(32*x^4) + (9*b^2*c^2*(a + b*ArcCsch[c*x]))/(32*x^2) + (3*b*c*Sqrt[1 +
1/(c^2*x^2)]*(a + b*ArcCsch[c*x])^2)/(16*x^3) - (9*b*c^3*Sqrt[1 + 1/(c^2*x^2)]*(a + b*ArcCsch[c*x])^2)/(32*x)
+ (3*c^4*(a + b*ArcCsch[c*x])^3)/32 - (a + b*ArcCsch[c*x])^3/(4*x^4)

Rule 6286

Int[((a_.) + ArcCsch[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> -Dist[(c^(m + 1))^(-1), Subst[Int[(a + b
*x)^n*Csch[x]^(m + 1)*Coth[x], x], x, ArcCsch[c*x]], x] /; FreeQ[{a, b, c}, x] && IntegerQ[n] && IntegerQ[m] &
& (GtQ[n, 0] || LtQ[m, -1])

Rule 5446

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

Rule 3311

Int[((c_.) + (d_.)*(x_))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(d*m*(c + d*x)^(m - 1)*(
b*Sin[e + f*x])^n)/(f^2*n^2), x] + (Dist[(b^2*(n - 1))/n, Int[(c + d*x)^m*(b*Sin[e + f*x])^(n - 2), x], x] - D
ist[(d^2*m*(m - 1))/(f^2*n^2), Int[(c + d*x)^(m - 2)*(b*Sin[e + f*x])^n, x], x] - Simp[(b*(c + d*x)^m*Cos[e +
f*x]*(b*Sin[e + f*x])^(n - 1))/(f*n), x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n, 1] && GtQ[m, 1]

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin{align*} \int \frac{\left (a+b \text{csch}^{-1}(c x)\right )^3}{x^5} \, dx &=-\left (c^4 \operatorname{Subst}\left (\int (a+b x)^3 \cosh (x) \sinh ^3(x) \, dx,x,\text{csch}^{-1}(c x)\right )\right )\\ &=-\frac{\left (a+b \text{csch}^{-1}(c x)\right )^3}{4 x^4}+\frac{1}{4} \left (3 b c^4\right ) \operatorname{Subst}\left (\int (a+b x)^2 \sinh ^4(x) \, dx,x,\text{csch}^{-1}(c x)\right )\\ &=-\frac{3 b^2 \left (a+b \text{csch}^{-1}(c x)\right )}{32 x^4}+\frac{3 b c \sqrt{1+\frac{1}{c^2 x^2}} \left (a+b \text{csch}^{-1}(c x)\right )^2}{16 x^3}-\frac{\left (a+b \text{csch}^{-1}(c x)\right )^3}{4 x^4}-\frac{1}{16} \left (9 b c^4\right ) \operatorname{Subst}\left (\int (a+b x)^2 \sinh ^2(x) \, dx,x,\text{csch}^{-1}(c x)\right )+\frac{1}{32} \left (3 b^3 c^4\right ) \operatorname{Subst}\left (\int \sinh ^4(x) \, dx,x,\text{csch}^{-1}(c x)\right )\\ &=\frac{3 b^3 c \sqrt{1+\frac{1}{c^2 x^2}}}{128 x^3}-\frac{3 b^2 \left (a+b \text{csch}^{-1}(c x)\right )}{32 x^4}+\frac{9 b^2 c^2 \left (a+b \text{csch}^{-1}(c x)\right )}{32 x^2}+\frac{3 b c \sqrt{1+\frac{1}{c^2 x^2}} \left (a+b \text{csch}^{-1}(c x)\right )^2}{16 x^3}-\frac{9 b c^3 \sqrt{1+\frac{1}{c^2 x^2}} \left (a+b \text{csch}^{-1}(c x)\right )^2}{32 x}-\frac{\left (a+b \text{csch}^{-1}(c x)\right )^3}{4 x^4}+\frac{1}{32} \left (9 b c^4\right ) \operatorname{Subst}\left (\int (a+b x)^2 \, dx,x,\text{csch}^{-1}(c x)\right )-\frac{1}{128} \left (9 b^3 c^4\right ) \operatorname{Subst}\left (\int \sinh ^2(x) \, dx,x,\text{csch}^{-1}(c x)\right )-\frac{1}{32} \left (9 b^3 c^4\right ) \operatorname{Subst}\left (\int \sinh ^2(x) \, dx,x,\text{csch}^{-1}(c x)\right )\\ &=\frac{3 b^3 c \sqrt{1+\frac{1}{c^2 x^2}}}{128 x^3}-\frac{45 b^3 c^3 \sqrt{1+\frac{1}{c^2 x^2}}}{256 x}-\frac{3 b^2 \left (a+b \text{csch}^{-1}(c x)\right )}{32 x^4}+\frac{9 b^2 c^2 \left (a+b \text{csch}^{-1}(c x)\right )}{32 x^2}+\frac{3 b c \sqrt{1+\frac{1}{c^2 x^2}} \left (a+b \text{csch}^{-1}(c x)\right )^2}{16 x^3}-\frac{9 b c^3 \sqrt{1+\frac{1}{c^2 x^2}} \left (a+b \text{csch}^{-1}(c x)\right )^2}{32 x}+\frac{3}{32} c^4 \left (a+b \text{csch}^{-1}(c x)\right )^3-\frac{\left (a+b \text{csch}^{-1}(c x)\right )^3}{4 x^4}+\frac{1}{256} \left (9 b^3 c^4\right ) \operatorname{Subst}\left (\int 1 \, dx,x,\text{csch}^{-1}(c x)\right )+\frac{1}{64} \left (9 b^3 c^4\right ) \operatorname{Subst}\left (\int 1 \, dx,x,\text{csch}^{-1}(c x)\right )\\ &=\frac{3 b^3 c \sqrt{1+\frac{1}{c^2 x^2}}}{128 x^3}-\frac{45 b^3 c^3 \sqrt{1+\frac{1}{c^2 x^2}}}{256 x}+\frac{45}{256} b^3 c^4 \text{csch}^{-1}(c x)-\frac{3 b^2 \left (a+b \text{csch}^{-1}(c x)\right )}{32 x^4}+\frac{9 b^2 c^2 \left (a+b \text{csch}^{-1}(c x)\right )}{32 x^2}+\frac{3 b c \sqrt{1+\frac{1}{c^2 x^2}} \left (a+b \text{csch}^{-1}(c x)\right )^2}{16 x^3}-\frac{9 b c^3 \sqrt{1+\frac{1}{c^2 x^2}} \left (a+b \text{csch}^{-1}(c x)\right )^2}{32 x}+\frac{3}{32} c^4 \left (a+b \text{csch}^{-1}(c x)\right )^3-\frac{\left (a+b \text{csch}^{-1}(c x)\right )^3}{4 x^4}\\ \end{align*}

Mathematica [A]  time = 0.336004, size = 277, normalized size = 1.36 \[ \frac{9 b c^4 x^4 \left (8 a^2+5 b^2\right ) \sinh ^{-1}\left (\frac{1}{c x}\right )-24 b \text{csch}^{-1}(c x) \left (8 a^2+2 a b c x \sqrt{\frac{1}{c^2 x^2}+1} \left (3 c^2 x^2-2\right )+b^2 \left (1-3 c^2 x^2\right )\right )-72 a^2 b c^3 x^3 \sqrt{\frac{1}{c^2 x^2}+1}+48 a^2 b c x \sqrt{\frac{1}{c^2 x^2}+1}-64 a^3+72 a b^2 c^2 x^2+24 b^2 \text{csch}^{-1}(c x)^2 \left (a \left (3 c^4 x^4-8\right )+b c x \sqrt{\frac{1}{c^2 x^2}+1} \left (2-3 c^2 x^2\right )\right )-24 a b^2-45 b^3 c^3 x^3 \sqrt{\frac{1}{c^2 x^2}+1}+6 b^3 c x \sqrt{\frac{1}{c^2 x^2}+1}+8 b^3 \left (3 c^4 x^4-8\right ) \text{csch}^{-1}(c x)^3}{256 x^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcCsch[c*x])^3/x^5,x]

[Out]

(-64*a^3 - 24*a*b^2 + 48*a^2*b*c*Sqrt[1 + 1/(c^2*x^2)]*x + 6*b^3*c*Sqrt[1 + 1/(c^2*x^2)]*x + 72*a*b^2*c^2*x^2
- 72*a^2*b*c^3*Sqrt[1 + 1/(c^2*x^2)]*x^3 - 45*b^3*c^3*Sqrt[1 + 1/(c^2*x^2)]*x^3 - 24*b*(8*a^2 + b^2*(1 - 3*c^2
*x^2) + 2*a*b*c*Sqrt[1 + 1/(c^2*x^2)]*x*(-2 + 3*c^2*x^2))*ArcCsch[c*x] + 24*b^2*(b*c*Sqrt[1 + 1/(c^2*x^2)]*x*(
2 - 3*c^2*x^2) + a*(-8 + 3*c^4*x^4))*ArcCsch[c*x]^2 + 8*b^3*(-8 + 3*c^4*x^4)*ArcCsch[c*x]^3 + 9*b*(8*a^2 + 5*b
^2)*c^4*x^4*ArcSinh[1/(c*x)])/(256*x^4)

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Maple [F]  time = 0.183, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( a+b{\rm arccsch} \left (cx\right ) \right ) ^{3}}{{x}^{5}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arccsch(c*x))^3/x^5,x)

[Out]

int((a+b*arccsch(c*x))^3/x^5,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arccsch(c*x))^3/x^5,x, algorithm="maxima")

[Out]

3/64*a^2*b*((3*c^5*log(c*x*sqrt(1/(c^2*x^2) + 1) + 1) - 3*c^5*log(c*x*sqrt(1/(c^2*x^2) + 1) - 1) - 2*(3*c^8*x^
3*(1/(c^2*x^2) + 1)^(3/2) - 5*c^6*x*sqrt(1/(c^2*x^2) + 1))/(c^4*x^4*(1/(c^2*x^2) + 1)^2 - 2*c^2*x^2*(1/(c^2*x^
2) + 1) + 1))/c - 16*arccsch(c*x)/x^4) - 1/4*b^3*log(sqrt(c^2*x^2 + 1) + 1)^3/x^4 - 1/4*a^3/x^4 - integrate(1/
4*(4*b^3*log(c)^3 - 12*a*b^2*log(c)^2 + 4*(b^3*c^2*x^2 + b^3)*log(x)^3 + 4*(b^3*c^2*log(c)^3 - 3*a*b^2*c^2*log
(c)^2)*x^2 + 12*(b^3*log(c) - a*b^2 + (b^3*c^2*log(c) - a*b^2*c^2)*x^2)*log(x)^2 + 3*(4*b^3*log(c) - 4*a*b^2 +
4*(b^3*c^2*log(c) - a*b^2*c^2)*x^2 + 4*(b^3*c^2*x^2 + b^3)*log(x) + sqrt(c^2*x^2 + 1)*(4*b^3*log(c) - 4*a*b^2
+ (b^3*c^2*(4*log(c) - 1) - 4*a*b^2*c^2)*x^2 + 4*(b^3*c^2*x^2 + b^3)*log(x)))*log(sqrt(c^2*x^2 + 1) + 1)^2 +
12*(b^3*log(c)^2 - 2*a*b^2*log(c) + (b^3*c^2*log(c)^2 - 2*a*b^2*c^2*log(c))*x^2)*log(x) - 12*(b^3*log(c)^2 - 2
*a*b^2*log(c) + (b^3*c^2*log(c)^2 - 2*a*b^2*c^2*log(c))*x^2 + (b^3*c^2*x^2 + b^3)*log(x)^2 + 2*(b^3*log(c) - a
*b^2 + (b^3*c^2*log(c) - a*b^2*c^2)*x^2)*log(x) + (b^3*log(c)^2 - 2*a*b^2*log(c) + (b^3*c^2*log(c)^2 - 2*a*b^2
*c^2*log(c))*x^2 + (b^3*c^2*x^2 + b^3)*log(x)^2 + 2*(b^3*log(c) - a*b^2 + (b^3*c^2*log(c) - a*b^2*c^2)*x^2)*lo
g(x))*sqrt(c^2*x^2 + 1))*log(sqrt(c^2*x^2 + 1) + 1) + 4*(b^3*log(c)^3 - 3*a*b^2*log(c)^2 + (b^3*c^2*x^2 + b^3)
*log(x)^3 + (b^3*c^2*log(c)^3 - 3*a*b^2*c^2*log(c)^2)*x^2 + 3*(b^3*log(c) - a*b^2 + (b^3*c^2*log(c) - a*b^2*c^
2)*x^2)*log(x)^2 + 3*(b^3*log(c)^2 - 2*a*b^2*log(c) + (b^3*c^2*log(c)^2 - 2*a*b^2*c^2*log(c))*x^2)*log(x))*sqr
t(c^2*x^2 + 1))/(c^2*x^7 + x^5 + (c^2*x^7 + x^5)*sqrt(c^2*x^2 + 1)), x)

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Fricas [A]  time = 2.47163, size = 747, normalized size = 3.66 \begin{align*} \frac{72 \, a b^{2} c^{2} x^{2} + 8 \,{\left (3 \, b^{3} c^{4} x^{4} - 8 \, b^{3}\right )} \log \left (\frac{c x \sqrt{\frac{c^{2} x^{2} + 1}{c^{2} x^{2}}} + 1}{c x}\right )^{3} - 64 \, a^{3} - 24 \, a b^{2} + 24 \,{\left (3 \, a b^{2} c^{4} x^{4} - 8 \, a b^{2} -{\left (3 \, b^{3} c^{3} x^{3} - 2 \, b^{3} c x\right )} \sqrt{\frac{c^{2} x^{2} + 1}{c^{2} x^{2}}}\right )} \log \left (\frac{c x \sqrt{\frac{c^{2} x^{2} + 1}{c^{2} x^{2}}} + 1}{c x}\right )^{2} + 3 \,{\left (3 \,{\left (8 \, a^{2} b + 5 \, b^{3}\right )} c^{4} x^{4} + 24 \, b^{3} c^{2} x^{2} - 64 \, a^{2} b - 8 \, b^{3} - 16 \,{\left (3 \, a b^{2} c^{3} x^{3} - 2 \, a b^{2} c x\right )} \sqrt{\frac{c^{2} x^{2} + 1}{c^{2} x^{2}}}\right )} \log \left (\frac{c x \sqrt{\frac{c^{2} x^{2} + 1}{c^{2} x^{2}}} + 1}{c x}\right ) - 3 \,{\left (3 \,{\left (8 \, a^{2} b + 5 \, b^{3}\right )} c^{3} x^{3} - 2 \,{\left (8 \, a^{2} b + b^{3}\right )} c x\right )} \sqrt{\frac{c^{2} x^{2} + 1}{c^{2} x^{2}}}}{256 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arccsch(c*x))^3/x^5,x, algorithm="fricas")

[Out]

1/256*(72*a*b^2*c^2*x^2 + 8*(3*b^3*c^4*x^4 - 8*b^3)*log((c*x*sqrt((c^2*x^2 + 1)/(c^2*x^2)) + 1)/(c*x))^3 - 64*
a^3 - 24*a*b^2 + 24*(3*a*b^2*c^4*x^4 - 8*a*b^2 - (3*b^3*c^3*x^3 - 2*b^3*c*x)*sqrt((c^2*x^2 + 1)/(c^2*x^2)))*lo
g((c*x*sqrt((c^2*x^2 + 1)/(c^2*x^2)) + 1)/(c*x))^2 + 3*(3*(8*a^2*b + 5*b^3)*c^4*x^4 + 24*b^3*c^2*x^2 - 64*a^2*
b - 8*b^3 - 16*(3*a*b^2*c^3*x^3 - 2*a*b^2*c*x)*sqrt((c^2*x^2 + 1)/(c^2*x^2)))*log((c*x*sqrt((c^2*x^2 + 1)/(c^2
*x^2)) + 1)/(c*x)) - 3*(3*(8*a^2*b + 5*b^3)*c^3*x^3 - 2*(8*a^2*b + b^3)*c*x)*sqrt((c^2*x^2 + 1)/(c^2*x^2)))/x^
4

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b \operatorname{acsch}{\left (c x \right )}\right )^{3}}{x^{5}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*acsch(c*x))**3/x**5,x)

[Out]

Integral((a + b*acsch(c*x))**3/x**5, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \operatorname{arcsch}\left (c x\right ) + a\right )}^{3}}{x^{5}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arccsch(c*x))^3/x^5,x, algorithm="giac")

[Out]

integrate((b*arccsch(c*x) + a)^3/x^5, x)