3.506 \(\int \frac {\coth ^5(e+f x)}{(a+b \sinh ^2(e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=208 \[ -\frac {(8 a-7 b) \text {csch}^2(e+f x)}{8 a^2 f \left (a+b \sinh ^2(e+f x)\right )^{3/2}}-\frac {\left (8 a^2-40 a b+35 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b \sinh ^2(e+f x)}}{\sqrt {a}}\right )}{8 a^{9/2} f}+\frac {8 a^2-40 a b+35 b^2}{8 a^4 f \sqrt {a+b \sinh ^2(e+f x)}}+\frac {8 a^2-40 a b+35 b^2}{24 a^3 f \left (a+b \sinh ^2(e+f x)\right )^{3/2}}-\frac {\text {csch}^4(e+f x)}{4 a f \left (a+b \sinh ^2(e+f x)\right )^{3/2}} \]

[Out]

-1/8*(8*a^2-40*a*b+35*b^2)*arctanh((a+b*sinh(f*x+e)^2)^(1/2)/a^(1/2))/a^(9/2)/f+1/24*(8*a^2-40*a*b+35*b^2)/a^3
/f/(a+b*sinh(f*x+e)^2)^(3/2)-1/8*(8*a-7*b)*csch(f*x+e)^2/a^2/f/(a+b*sinh(f*x+e)^2)^(3/2)-1/4*csch(f*x+e)^4/a/f
/(a+b*sinh(f*x+e)^2)^(3/2)+1/8*(8*a^2-40*a*b+35*b^2)/a^4/f/(a+b*sinh(f*x+e)^2)^(1/2)

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Rubi [A]  time = 0.22, antiderivative size = 208, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3194, 89, 78, 51, 63, 208} \[ \frac {8 a^2-40 a b+35 b^2}{8 a^4 f \sqrt {a+b \sinh ^2(e+f x)}}+\frac {8 a^2-40 a b+35 b^2}{24 a^3 f \left (a+b \sinh ^2(e+f x)\right )^{3/2}}-\frac {\left (8 a^2-40 a b+35 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b \sinh ^2(e+f x)}}{\sqrt {a}}\right )}{8 a^{9/2} f}-\frac {(8 a-7 b) \text {csch}^2(e+f x)}{8 a^2 f \left (a+b \sinh ^2(e+f x)\right )^{3/2}}-\frac {\text {csch}^4(e+f x)}{4 a f \left (a+b \sinh ^2(e+f x)\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Coth[e + f*x]^5/(a + b*Sinh[e + f*x]^2)^(5/2),x]

[Out]

-((8*a^2 - 40*a*b + 35*b^2)*ArcTanh[Sqrt[a + b*Sinh[e + f*x]^2]/Sqrt[a]])/(8*a^(9/2)*f) + (8*a^2 - 40*a*b + 35
*b^2)/(24*a^3*f*(a + b*Sinh[e + f*x]^2)^(3/2)) - ((8*a - 7*b)*Csch[e + f*x]^2)/(8*a^2*f*(a + b*Sinh[e + f*x]^2
)^(3/2)) - Csch[e + f*x]^4/(4*a*f*(a + b*Sinh[e + f*x]^2)^(3/2)) + (8*a^2 - 40*a*b + 35*b^2)/(8*a^4*f*Sqrt[a +
b*Sinh[e + f*x]^2])

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

Rule 89

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((b*c - a*
d)^2*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d^2*(d*e - c*f)*(n + 1)), x] - Dist[1/(d^2*(d*e - c*f)*(n + 1)), In
t[(c + d*x)^(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*(p + 1)) - 2*a*b*d*(d*e*
(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ
[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rule 3194

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.)*tan[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> With[{ff = Free
Factors[Sin[e + f*x]^2, x]}, Dist[ff^((m + 1)/2)/(2*f), Subst[Int[(x^((m - 1)/2)*(a + b*ff*x)^p)/(1 - ff*x)^((
m + 1)/2), x], x, Sin[e + f*x]^2/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]

Rubi steps

\begin {align*} \int \frac {\coth ^5(e+f x)}{\left (a+b \sinh ^2(e+f x)\right )^{5/2}} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {(1+x)^2}{x^3 (a+b x)^{5/2}} \, dx,x,\sinh ^2(e+f x)\right )}{2 f}\\ &=-\frac {\text {csch}^4(e+f x)}{4 a f \left (a+b \sinh ^2(e+f x)\right )^{3/2}}+\frac {\operatorname {Subst}\left (\int \frac {\frac {1}{2} (8 a-7 b)+2 a x}{x^2 (a+b x)^{5/2}} \, dx,x,\sinh ^2(e+f x)\right )}{4 a f}\\ &=-\frac {(8 a-7 b) \text {csch}^2(e+f x)}{8 a^2 f \left (a+b \sinh ^2(e+f x)\right )^{3/2}}-\frac {\text {csch}^4(e+f x)}{4 a f \left (a+b \sinh ^2(e+f x)\right )^{3/2}}+\frac {\left (8 a^2-40 a b+35 b^2\right ) \operatorname {Subst}\left (\int \frac {1}{x (a+b x)^{5/2}} \, dx,x,\sinh ^2(e+f x)\right )}{16 a^2 f}\\ &=\frac {8 a^2-40 a b+35 b^2}{24 a^3 f \left (a+b \sinh ^2(e+f x)\right )^{3/2}}-\frac {(8 a-7 b) \text {csch}^2(e+f x)}{8 a^2 f \left (a+b \sinh ^2(e+f x)\right )^{3/2}}-\frac {\text {csch}^4(e+f x)}{4 a f \left (a+b \sinh ^2(e+f x)\right )^{3/2}}+\frac {\left (8 a^2-40 a b+35 b^2\right ) \operatorname {Subst}\left (\int \frac {1}{x (a+b x)^{3/2}} \, dx,x,\sinh ^2(e+f x)\right )}{16 a^3 f}\\ &=\frac {8 a^2-40 a b+35 b^2}{24 a^3 f \left (a+b \sinh ^2(e+f x)\right )^{3/2}}-\frac {(8 a-7 b) \text {csch}^2(e+f x)}{8 a^2 f \left (a+b \sinh ^2(e+f x)\right )^{3/2}}-\frac {\text {csch}^4(e+f x)}{4 a f \left (a+b \sinh ^2(e+f x)\right )^{3/2}}+\frac {8 a^2-40 a b+35 b^2}{8 a^4 f \sqrt {a+b \sinh ^2(e+f x)}}+\frac {\left (8 a^2-40 a b+35 b^2\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,\sinh ^2(e+f x)\right )}{16 a^4 f}\\ &=\frac {8 a^2-40 a b+35 b^2}{24 a^3 f \left (a+b \sinh ^2(e+f x)\right )^{3/2}}-\frac {(8 a-7 b) \text {csch}^2(e+f x)}{8 a^2 f \left (a+b \sinh ^2(e+f x)\right )^{3/2}}-\frac {\text {csch}^4(e+f x)}{4 a f \left (a+b \sinh ^2(e+f x)\right )^{3/2}}+\frac {8 a^2-40 a b+35 b^2}{8 a^4 f \sqrt {a+b \sinh ^2(e+f x)}}+\frac {\left (8 a^2-40 a b+35 b^2\right ) \operatorname {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b \sinh ^2(e+f x)}\right )}{8 a^4 b f}\\ &=-\frac {\left (8 a^2-40 a b+35 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b \sinh ^2(e+f x)}}{\sqrt {a}}\right )}{8 a^{9/2} f}+\frac {8 a^2-40 a b+35 b^2}{24 a^3 f \left (a+b \sinh ^2(e+f x)\right )^{3/2}}-\frac {(8 a-7 b) \text {csch}^2(e+f x)}{8 a^2 f \left (a+b \sinh ^2(e+f x)\right )^{3/2}}-\frac {\text {csch}^4(e+f x)}{4 a f \left (a+b \sinh ^2(e+f x)\right )^{3/2}}+\frac {8 a^2-40 a b+35 b^2}{8 a^4 f \sqrt {a+b \sinh ^2(e+f x)}}\\ \end {align*}

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Mathematica [C]  time = 0.43, size = 117, normalized size = 0.56 \[ -\frac {\text {csch}^2(e+f x) \left (\left (-8 a^2+40 a b-35 b^2\right ) \, _2F_1\left (-\frac {3}{2},1;-\frac {1}{2};\frac {b \sinh ^2(e+f x)}{a}+1\right )+3 a \text {csch}^2(e+f x) \left (2 a \text {csch}^2(e+f x)+8 a-7 b\right )\right )}{24 a^3 f \sqrt {a+b \sinh ^2(e+f x)} \left (a \text {csch}^2(e+f x)+b\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[Coth[e + f*x]^5/(a + b*Sinh[e + f*x]^2)^(5/2),x]

[Out]

-1/24*(Csch[e + f*x]^2*(3*a*Csch[e + f*x]^2*(8*a - 7*b + 2*a*Csch[e + f*x]^2) + (-8*a^2 + 40*a*b - 35*b^2)*Hyp
ergeometric2F1[-3/2, 1, -1/2, 1 + (b*Sinh[e + f*x]^2)/a]))/(a^3*f*(b + a*Csch[e + f*x]^2)*Sqrt[a + b*Sinh[e +
f*x]^2])

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(f*x+e)^5/(a+b*sinh(f*x+e)^2)^(5/2),x, algorithm="fricas")

[Out]

Timed out

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(f*x+e)^5/(a+b*sinh(f*x+e)^2)^(5/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [ab
s(t_nostep)]Evaluation time: 9.86Error: Bad Argument Type

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maple [C]  time = 0.26, size = 73, normalized size = 0.35 \[ \frac {\mathit {`\,int/indef0`\,}\left (\frac {\cosh ^{4}\left (f x +e \right )}{\left (b^{2} \left (\sinh ^{4}\left (f x +e \right )\right )+2 a b \left (\sinh ^{2}\left (f x +e \right )\right )+a^{2}\right ) \sinh \left (f x +e \right )^{5} \sqrt {a +b \left (\sinh ^{2}\left (f x +e \right )\right )}}, \sinh \left (f x +e \right )\right )}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(coth(f*x+e)^5/(a+b*sinh(f*x+e)^2)^(5/2),x)

[Out]

`int/indef0`(cosh(f*x+e)^4/(b^2*sinh(f*x+e)^4+2*a*b*sinh(f*x+e)^2+a^2)/sinh(f*x+e)^5/(a+b*sinh(f*x+e)^2)^(1/2)
,sinh(f*x+e))/f

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\coth \left (f x + e\right )^{5}}{{\left (b \sinh \left (f x + e\right )^{2} + a\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(f*x+e)^5/(a+b*sinh(f*x+e)^2)^(5/2),x, algorithm="maxima")

[Out]

integrate(coth(f*x + e)^5/(b*sinh(f*x + e)^2 + a)^(5/2), x)

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mupad [F(-1)]  time = 0.00, size = -1, normalized size = -0.00 \[ \text {Hanged} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(coth(e + f*x)^5/(a + b*sinh(e + f*x)^2)^(5/2),x)

[Out]

\text{Hanged}

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(f*x+e)**5/(a+b*sinh(f*x+e)**2)**(5/2),x)

[Out]

Timed out

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