3.154 \(\int \text {csch}(c+d x) (a+b \sinh ^3(c+d x))^2 \, dx\)

Optimal. Leaf size=88 \[ -\frac {a^2 \tanh ^{-1}(\cosh (c+d x))}{d}+\frac {a b \sinh (c+d x) \cosh (c+d x)}{d}-a b x+\frac {b^2 \cosh ^5(c+d x)}{5 d}-\frac {2 b^2 \cosh ^3(c+d x)}{3 d}+\frac {b^2 \cosh (c+d x)}{d} \]

[Out]

-a*b*x-a^2*arctanh(cosh(d*x+c))/d+b^2*cosh(d*x+c)/d-2/3*b^2*cosh(d*x+c)^3/d+1/5*b^2*cosh(d*x+c)^5/d+a*b*cosh(d
*x+c)*sinh(d*x+c)/d

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Rubi [A]  time = 0.10, antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3220, 3770, 2635, 8, 2633} \[ -\frac {a^2 \tanh ^{-1}(\cosh (c+d x))}{d}+\frac {a b \sinh (c+d x) \cosh (c+d x)}{d}-a b x+\frac {b^2 \cosh ^5(c+d x)}{5 d}-\frac {2 b^2 \cosh ^3(c+d x)}{3 d}+\frac {b^2 \cosh (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Int[Csch[c + d*x]*(a + b*Sinh[c + d*x]^3)^2,x]

[Out]

-(a*b*x) - (a^2*ArcTanh[Cosh[c + d*x]])/d + (b^2*Cosh[c + d*x])/d - (2*b^2*Cosh[c + d*x]^3)/(3*d) + (b^2*Cosh[
c + d*x]^5)/(5*d) + (a*b*Cosh[c + d*x]*Sinh[c + d*x])/d

Rule 8

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

Rule 2633

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

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 3220

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^(n_))^(p_.), x_Symbol] :> Int[ExpandTr
ig[sin[e + f*x]^m*(a + b*sin[e + f*x]^n)^p, x], x] /; FreeQ[{a, b, e, f}, x] && IntegersQ[m, p] && (EqQ[n, 4]
|| GtQ[p, 0] || (EqQ[p, -1] && IntegerQ[n]))

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin {align*} \int \text {csch}(c+d x) \left (a+b \sinh ^3(c+d x)\right )^2 \, dx &=i \int \left (-i a^2 \text {csch}(c+d x)-2 i a b \sinh ^2(c+d x)-i b^2 \sinh ^5(c+d x)\right ) \, dx\\ &=a^2 \int \text {csch}(c+d x) \, dx+(2 a b) \int \sinh ^2(c+d x) \, dx+b^2 \int \sinh ^5(c+d x) \, dx\\ &=-\frac {a^2 \tanh ^{-1}(\cosh (c+d x))}{d}+\frac {a b \cosh (c+d x) \sinh (c+d x)}{d}-(a b) \int 1 \, dx+\frac {b^2 \operatorname {Subst}\left (\int \left (1-2 x^2+x^4\right ) \, dx,x,\cosh (c+d x)\right )}{d}\\ &=-a b x-\frac {a^2 \tanh ^{-1}(\cosh (c+d x))}{d}+\frac {b^2 \cosh (c+d x)}{d}-\frac {2 b^2 \cosh ^3(c+d x)}{3 d}+\frac {b^2 \cosh ^5(c+d x)}{5 d}+\frac {a b \cosh (c+d x) \sinh (c+d x)}{d}\\ \end {align*}

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Mathematica [A]  time = 0.31, size = 96, normalized size = 1.09 \[ \frac {120 a \left (b \sinh (2 (c+d x))-2 \left (-a \log \left (\sinh \left (\frac {1}{2} (c+d x)\right )\right )+a \log \left (\cosh \left (\frac {1}{2} (c+d x)\right )\right )+b c+b d x\right )\right )+150 b^2 \cosh (c+d x)-25 b^2 \cosh (3 (c+d x))+3 b^2 \cosh (5 (c+d x))}{240 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Csch[c + d*x]*(a + b*Sinh[c + d*x]^3)^2,x]

[Out]

(150*b^2*Cosh[c + d*x] - 25*b^2*Cosh[3*(c + d*x)] + 3*b^2*Cosh[5*(c + d*x)] + 120*a*(-2*(b*c + b*d*x + a*Log[C
osh[(c + d*x)/2]] - a*Log[Sinh[(c + d*x)/2]]) + b*Sinh[2*(c + d*x)]))/(240*d)

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fricas [B]  time = 0.49, size = 1052, normalized size = 11.95 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(d*x+c)*(a+b*sinh(d*x+c)^3)^2,x, algorithm="fricas")

[Out]

1/480*(3*b^2*cosh(d*x + c)^10 + 30*b^2*cosh(d*x + c)*sinh(d*x + c)^9 + 3*b^2*sinh(d*x + c)^10 - 25*b^2*cosh(d*
x + c)^8 - 480*a*b*d*x*cosh(d*x + c)^5 + 120*a*b*cosh(d*x + c)^7 + 5*(27*b^2*cosh(d*x + c)^2 - 5*b^2)*sinh(d*x
+ c)^8 + 150*b^2*cosh(d*x + c)^6 + 40*(9*b^2*cosh(d*x + c)^3 - 5*b^2*cosh(d*x + c) + 3*a*b)*sinh(d*x + c)^7 +
10*(63*b^2*cosh(d*x + c)^4 - 70*b^2*cosh(d*x + c)^2 + 84*a*b*cosh(d*x + c) + 15*b^2)*sinh(d*x + c)^6 + 150*b^
2*cosh(d*x + c)^4 + 4*(189*b^2*cosh(d*x + c)^5 - 350*b^2*cosh(d*x + c)^3 - 120*a*b*d*x + 630*a*b*cosh(d*x + c)
^2 + 225*b^2*cosh(d*x + c))*sinh(d*x + c)^5 - 120*a*b*cosh(d*x + c)^3 + 10*(63*b^2*cosh(d*x + c)^6 - 175*b^2*c
osh(d*x + c)^4 - 240*a*b*d*x*cosh(d*x + c) + 420*a*b*cosh(d*x + c)^3 + 225*b^2*cosh(d*x + c)^2 + 15*b^2)*sinh(
d*x + c)^4 - 25*b^2*cosh(d*x + c)^2 + 40*(9*b^2*cosh(d*x + c)^7 - 35*b^2*cosh(d*x + c)^5 - 120*a*b*d*x*cosh(d*
x + c)^2 + 105*a*b*cosh(d*x + c)^4 + 75*b^2*cosh(d*x + c)^3 + 15*b^2*cosh(d*x + c) - 3*a*b)*sinh(d*x + c)^3 +
5*(27*b^2*cosh(d*x + c)^8 - 140*b^2*cosh(d*x + c)^6 - 960*a*b*d*x*cosh(d*x + c)^3 + 504*a*b*cosh(d*x + c)^5 +
450*b^2*cosh(d*x + c)^4 + 180*b^2*cosh(d*x + c)^2 - 72*a*b*cosh(d*x + c) - 5*b^2)*sinh(d*x + c)^2 + 3*b^2 - 48
0*(a^2*cosh(d*x + c)^5 + 5*a^2*cosh(d*x + c)^4*sinh(d*x + c) + 10*a^2*cosh(d*x + c)^3*sinh(d*x + c)^2 + 10*a^2
*cosh(d*x + c)^2*sinh(d*x + c)^3 + 5*a^2*cosh(d*x + c)*sinh(d*x + c)^4 + a^2*sinh(d*x + c)^5)*log(cosh(d*x + c
) + sinh(d*x + c) + 1) + 480*(a^2*cosh(d*x + c)^5 + 5*a^2*cosh(d*x + c)^4*sinh(d*x + c) + 10*a^2*cosh(d*x + c)
^3*sinh(d*x + c)^2 + 10*a^2*cosh(d*x + c)^2*sinh(d*x + c)^3 + 5*a^2*cosh(d*x + c)*sinh(d*x + c)^4 + a^2*sinh(d
*x + c)^5)*log(cosh(d*x + c) + sinh(d*x + c) - 1) + 10*(3*b^2*cosh(d*x + c)^9 - 20*b^2*cosh(d*x + c)^7 - 240*a
*b*d*x*cosh(d*x + c)^4 + 84*a*b*cosh(d*x + c)^6 + 90*b^2*cosh(d*x + c)^5 + 60*b^2*cosh(d*x + c)^3 - 36*a*b*cos
h(d*x + c)^2 - 5*b^2*cosh(d*x + c))*sinh(d*x + c))/(d*cosh(d*x + c)^5 + 5*d*cosh(d*x + c)^4*sinh(d*x + c) + 10
*d*cosh(d*x + c)^3*sinh(d*x + c)^2 + 10*d*cosh(d*x + c)^2*sinh(d*x + c)^3 + 5*d*cosh(d*x + c)*sinh(d*x + c)^4
+ d*sinh(d*x + c)^5)

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giac [A]  time = 0.21, size = 154, normalized size = 1.75 \[ -\frac {480 \, {\left (d x + c\right )} a b - 3 \, b^{2} e^{\left (5 \, d x + 5 \, c\right )} + 25 \, b^{2} e^{\left (3 \, d x + 3 \, c\right )} - 120 \, a b e^{\left (2 \, d x + 2 \, c\right )} - 150 \, b^{2} e^{\left (d x + c\right )} + 480 \, a^{2} \log \left (e^{\left (d x + c\right )} + 1\right ) - 480 \, a^{2} \log \left ({\left | e^{\left (d x + c\right )} - 1 \right |}\right ) - {\left (150 \, b^{2} e^{\left (4 \, d x + 4 \, c\right )} - 120 \, a b e^{\left (3 \, d x + 3 \, c\right )} - 25 \, b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 3 \, b^{2}\right )} e^{\left (-5 \, d x - 5 \, c\right )}}{480 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(d*x+c)*(a+b*sinh(d*x+c)^3)^2,x, algorithm="giac")

[Out]

-1/480*(480*(d*x + c)*a*b - 3*b^2*e^(5*d*x + 5*c) + 25*b^2*e^(3*d*x + 3*c) - 120*a*b*e^(2*d*x + 2*c) - 150*b^2
*e^(d*x + c) + 480*a^2*log(e^(d*x + c) + 1) - 480*a^2*log(abs(e^(d*x + c) - 1)) - (150*b^2*e^(4*d*x + 4*c) - 1
20*a*b*e^(3*d*x + 3*c) - 25*b^2*e^(2*d*x + 2*c) + 3*b^2)*e^(-5*d*x - 5*c))/d

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maple [A]  time = 0.12, size = 76, normalized size = 0.86 \[ \frac {-2 a^{2} \arctanh \left ({\mathrm e}^{d x +c}\right )+2 a b \left (\frac {\cosh \left (d x +c \right ) \sinh \left (d x +c \right )}{2}-\frac {d x}{2}-\frac {c}{2}\right )+b^{2} \left (\frac {8}{15}+\frac {\left (\sinh ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\sinh ^{2}\left (d x +c \right )\right )}{15}\right ) \cosh \left (d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csch(d*x+c)*(a+b*sinh(d*x+c)^3)^2,x)

[Out]

1/d*(-2*a^2*arctanh(exp(d*x+c))+2*a*b*(1/2*cosh(d*x+c)*sinh(d*x+c)-1/2*d*x-1/2*c)+b^2*(8/15+1/5*sinh(d*x+c)^4-
4/15*sinh(d*x+c)^2)*cosh(d*x+c))

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maxima [A]  time = 0.31, size = 140, normalized size = 1.59 \[ -\frac {1}{4} \, a b {\left (4 \, x - \frac {e^{\left (2 \, d x + 2 \, c\right )}}{d} + \frac {e^{\left (-2 \, d x - 2 \, c\right )}}{d}\right )} + \frac {1}{480} \, b^{2} {\left (\frac {3 \, e^{\left (5 \, d x + 5 \, c\right )}}{d} - \frac {25 \, e^{\left (3 \, d x + 3 \, c\right )}}{d} + \frac {150 \, e^{\left (d x + c\right )}}{d} + \frac {150 \, e^{\left (-d x - c\right )}}{d} - \frac {25 \, e^{\left (-3 \, d x - 3 \, c\right )}}{d} + \frac {3 \, e^{\left (-5 \, d x - 5 \, c\right )}}{d}\right )} + \frac {a^{2} \log \left (\tanh \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(d*x+c)*(a+b*sinh(d*x+c)^3)^2,x, algorithm="maxima")

[Out]

-1/4*a*b*(4*x - e^(2*d*x + 2*c)/d + e^(-2*d*x - 2*c)/d) + 1/480*b^2*(3*e^(5*d*x + 5*c)/d - 25*e^(3*d*x + 3*c)/
d + 150*e^(d*x + c)/d + 150*e^(-d*x - c)/d - 25*e^(-3*d*x - 3*c)/d + 3*e^(-5*d*x - 5*c)/d) + a^2*log(tanh(1/2*
d*x + 1/2*c))/d

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mupad [B]  time = 0.21, size = 177, normalized size = 2.01 \[ \frac {5\,b^2\,{\mathrm {e}}^{c+d\,x}}{16\,d}-\frac {2\,\mathrm {atan}\left (\frac {a^2\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\,\sqrt {-d^2}}{d\,\sqrt {a^4}}\right )\,\sqrt {a^4}}{\sqrt {-d^2}}-a\,b\,x+\frac {5\,b^2\,{\mathrm {e}}^{-c-d\,x}}{16\,d}-\frac {5\,b^2\,{\mathrm {e}}^{-3\,c-3\,d\,x}}{96\,d}-\frac {5\,b^2\,{\mathrm {e}}^{3\,c+3\,d\,x}}{96\,d}+\frac {b^2\,{\mathrm {e}}^{-5\,c-5\,d\,x}}{160\,d}+\frac {b^2\,{\mathrm {e}}^{5\,c+5\,d\,x}}{160\,d}-\frac {a\,b\,{\mathrm {e}}^{-2\,c-2\,d\,x}}{4\,d}+\frac {a\,b\,{\mathrm {e}}^{2\,c+2\,d\,x}}{4\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*sinh(c + d*x)^3)^2/sinh(c + d*x),x)

[Out]

(5*b^2*exp(c + d*x))/(16*d) - (2*atan((a^2*exp(d*x)*exp(c)*(-d^2)^(1/2))/(d*(a^4)^(1/2)))*(a^4)^(1/2))/(-d^2)^
(1/2) - a*b*x + (5*b^2*exp(- c - d*x))/(16*d) - (5*b^2*exp(- 3*c - 3*d*x))/(96*d) - (5*b^2*exp(3*c + 3*d*x))/(
96*d) + (b^2*exp(- 5*c - 5*d*x))/(160*d) + (b^2*exp(5*c + 5*d*x))/(160*d) - (a*b*exp(- 2*c - 2*d*x))/(4*d) + (
a*b*exp(2*c + 2*d*x))/(4*d)

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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(csch(d*x+c)*(a+b*sinh(d*x+c)**3)**2,x)

[Out]

Timed out

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