3.9 \(\int (a+b \text {csch}^2(c+d x))^{5/2} \, dx\)

Optimal. Leaf size=174 \[ \frac {a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a} \coth (c+d x)}{\sqrt {a+b \coth ^2(c+d x)-b}}\right )}{d}-\frac {\sqrt {b} \left (15 a^2-10 a b+3 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \coth (c+d x)}{\sqrt {a+b \coth ^2(c+d x)-b}}\right )}{8 d}-\frac {b \coth (c+d x) \left (a+b \coth ^2(c+d x)-b\right )^{3/2}}{4 d}-\frac {b (7 a-3 b) \coth (c+d x) \sqrt {a+b \coth ^2(c+d x)-b}}{8 d} \]

[Out]

a^(5/2)*arctanh(coth(d*x+c)*a^(1/2)/(a-b+b*coth(d*x+c)^2)^(1/2))/d-1/4*b*coth(d*x+c)*(a-b+b*coth(d*x+c)^2)^(3/
2)/d-1/8*(15*a^2-10*a*b+3*b^2)*arctanh(coth(d*x+c)*b^(1/2)/(a-b+b*coth(d*x+c)^2)^(1/2))*b^(1/2)/d-1/8*(7*a-3*b
)*b*coth(d*x+c)*(a-b+b*coth(d*x+c)^2)^(1/2)/d

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Rubi [A]  time = 0.19, antiderivative size = 174, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {4128, 416, 528, 523, 217, 206, 377} \[ -\frac {\sqrt {b} \left (15 a^2-10 a b+3 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \coth (c+d x)}{\sqrt {a+b \coth ^2(c+d x)-b}}\right )}{8 d}+\frac {a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a} \coth (c+d x)}{\sqrt {a+b \coth ^2(c+d x)-b}}\right )}{d}-\frac {b \coth (c+d x) \left (a+b \coth ^2(c+d x)-b\right )^{3/2}}{4 d}-\frac {b (7 a-3 b) \coth (c+d x) \sqrt {a+b \coth ^2(c+d x)-b}}{8 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Csch[c + d*x]^2)^(5/2),x]

[Out]

(a^(5/2)*ArcTanh[(Sqrt[a]*Coth[c + d*x])/Sqrt[a - b + b*Coth[c + d*x]^2]])/d - (Sqrt[b]*(15*a^2 - 10*a*b + 3*b
^2)*ArcTanh[(Sqrt[b]*Coth[c + d*x])/Sqrt[a - b + b*Coth[c + d*x]^2]])/(8*d) - ((7*a - 3*b)*b*Coth[c + d*x]*Sqr
t[a - b + b*Coth[c + d*x]^2])/(8*d) - (b*Coth[c + d*x]*(a - b + b*Coth[c + d*x]^2)^(3/2))/(4*d)

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 377

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 416

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(d*x*(a + b*x^n)^(p + 1)*(c
+ d*x^n)^(q - 1))/(b*(n*(p + q) + 1)), x] + Dist[1/(b*(n*(p + q) + 1)), Int[(a + b*x^n)^p*(c + d*x^n)^(q - 2)
*Simp[c*(b*c*(n*(p + q) + 1) - a*d) + d*(b*c*(n*(p + 2*q - 1) + 1) - a*d*(n*(q - 1) + 1))*x^n, x], x], x] /; F
reeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && GtQ[q, 1] && NeQ[n*(p + q) + 1, 0] &&  !IGtQ[p, 1] && IntB
inomialQ[a, b, c, d, n, p, q, x]

Rule 523

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*Sqrt[(c_) + (d_.)*(x_)^(n_)]), x_Symbol] :> Dist[f/b, I
nt[1/Sqrt[c + d*x^n], x], x] + Dist[(b*e - a*f)/b, Int[1/((a + b*x^n)*Sqrt[c + d*x^n]), x], x] /; FreeQ[{a, b,
c, d, e, f, n}, x]

Rule 528

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

Rule 4128

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist
[ff/f, Subst[Int[(a + b + b*ff^2*x^2)^p/(1 + ff^2*x^2), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p},
x] && NeQ[a + b, 0] && NeQ[p, -1]

Rubi steps

\begin {align*} \int \left (a+b \text {csch}^2(c+d x)\right )^{5/2} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (a-b+b x^2\right )^{5/2}}{1-x^2} \, dx,x,\coth (c+d x)\right )}{d}\\ &=-\frac {b \coth (c+d x) \left (a-b+b \coth ^2(c+d x)\right )^{3/2}}{4 d}-\frac {\operatorname {Subst}\left (\int \frac {\sqrt {a-b+b x^2} \left (-(4 a-3 b) (a-b)-(7 a-3 b) b x^2\right )}{1-x^2} \, dx,x,\coth (c+d x)\right )}{4 d}\\ &=-\frac {(7 a-3 b) b \coth (c+d x) \sqrt {a-b+b \coth ^2(c+d x)}}{8 d}-\frac {b \coth (c+d x) \left (a-b+b \coth ^2(c+d x)\right )^{3/2}}{4 d}+\frac {\operatorname {Subst}\left (\int \frac {(a-b) \left (8 a^2-7 a b+3 b^2\right )+b \left (15 a^2-10 a b+3 b^2\right ) x^2}{\left (1-x^2\right ) \sqrt {a-b+b x^2}} \, dx,x,\coth (c+d x)\right )}{8 d}\\ &=-\frac {(7 a-3 b) b \coth (c+d x) \sqrt {a-b+b \coth ^2(c+d x)}}{8 d}-\frac {b \coth (c+d x) \left (a-b+b \coth ^2(c+d x)\right )^{3/2}}{4 d}+\frac {a^3 \operatorname {Subst}\left (\int \frac {1}{\left (1-x^2\right ) \sqrt {a-b+b x^2}} \, dx,x,\coth (c+d x)\right )}{d}-\frac {\left (b \left (15 a^2-10 a b+3 b^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a-b+b x^2}} \, dx,x,\coth (c+d x)\right )}{8 d}\\ &=-\frac {(7 a-3 b) b \coth (c+d x) \sqrt {a-b+b \coth ^2(c+d x)}}{8 d}-\frac {b \coth (c+d x) \left (a-b+b \coth ^2(c+d x)\right )^{3/2}}{4 d}+\frac {a^3 \operatorname {Subst}\left (\int \frac {1}{1-a x^2} \, dx,x,\frac {\coth (c+d x)}{\sqrt {a-b+b \coth ^2(c+d x)}}\right )}{d}-\frac {\left (b \left (15 a^2-10 a b+3 b^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {\coth (c+d x)}{\sqrt {a-b+b \coth ^2(c+d x)}}\right )}{8 d}\\ &=\frac {a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a} \coth (c+d x)}{\sqrt {a-b+b \coth ^2(c+d x)}}\right )}{d}-\frac {\sqrt {b} \left (15 a^2-10 a b+3 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \coth (c+d x)}{\sqrt {a-b+b \coth ^2(c+d x)}}\right )}{8 d}-\frac {(7 a-3 b) b \coth (c+d x) \sqrt {a-b+b \coth ^2(c+d x)}}{8 d}-\frac {b \coth (c+d x) \left (a-b+b \coth ^2(c+d x)\right )^{3/2}}{4 d}\\ \end {align*}

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Mathematica [A]  time = 4.90, size = 231, normalized size = 1.33 \[ \frac {\sinh ^5(c+d x) \left (a+b \text {csch}^2(c+d x)\right )^{5/2} \left (16 \sqrt {2} a^{5/2} \log \left (\sqrt {a \cosh (2 (c+d x))-a+2 b}+\sqrt {2} \sqrt {a} \cosh (c+d x)\right )-2 \sqrt {2} \sqrt {b} \left (15 a^2-10 a b+3 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {b} \cosh (c+d x)}{\sqrt {a \cosh (2 (c+d x))-a+2 b}}\right )+b \coth (c+d x) \text {csch}^3(c+d x) \sqrt {a \cosh (2 (c+d x))-a+2 b} ((3 b-9 a) \cosh (2 (c+d x))+9 a-7 b)\right )}{4 d (a \cosh (2 (c+d x))-a+2 b)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Csch[c + d*x]^2)^(5/2),x]

[Out]

((a + b*Csch[c + d*x]^2)^(5/2)*(-2*Sqrt[2]*Sqrt[b]*(15*a^2 - 10*a*b + 3*b^2)*ArcTanh[(Sqrt[2]*Sqrt[b]*Cosh[c +
d*x])/Sqrt[-a + 2*b + a*Cosh[2*(c + d*x)]]] + b*Sqrt[-a + 2*b + a*Cosh[2*(c + d*x)]]*(9*a - 7*b + (-9*a + 3*b
)*Cosh[2*(c + d*x)])*Coth[c + d*x]*Csch[c + d*x]^3 + 16*Sqrt[2]*a^(5/2)*Log[Sqrt[2]*Sqrt[a]*Cosh[c + d*x] + Sq
rt[-a + 2*b + a*Cosh[2*(c + d*x)]]])*Sinh[c + d*x]^5)/(4*d*(-a + 2*b + a*Cosh[2*(c + d*x)])^(5/2))

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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((a+b*csch(d*x+c)^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((a+b*csch(d*x+c)^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(exp(d*x+c)^2-1)]Error: Bad Argument Type

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maple [F]  time = 0.62, size = 0, normalized size = 0.00 \[ \int \left (a +b \mathrm {csch}\left (d x +c \right )^{2}\right )^{\frac {5}{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*csch(d*x+c)^2)^(5/2),x)

[Out]

int((a+b*csch(d*x+c)^2)^(5/2),x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*csch(d*x+c)^2)^(5/2),x, algorithm="maxima")

[Out]

integrate((b*csch(d*x + c)^2 + a)^(5/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (a+\frac {b}{{\mathrm {sinh}\left (c+d\,x\right )}^2}\right )}^{5/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b/sinh(c + d*x)^2)^(5/2),x)

[Out]

int((a + b/sinh(c + d*x)^2)^(5/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \operatorname {csch}^{2}{\left (c + d x \right )}\right )^{\frac {5}{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*csch(d*x+c)**2)**(5/2),x)

[Out]

Integral((a + b*csch(c + d*x)**2)**(5/2), x)

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