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

Optimal. Leaf size=83 \[ \frac{b \left (3 a^2-3 a b+b^2\right ) \cosh (c+d x)}{d}-\frac{a^3 \tanh ^{-1}(\cosh (c+d x))}{d}+\frac{b^2 (3 a-2 b) \cosh ^3(c+d x)}{3 d}+\frac{b^3 \cosh ^5(c+d x)}{5 d} \]

[Out]

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

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Rubi [A]  time = 0.0869544, antiderivative size = 83, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {3186, 390, 206} \[ \frac{b \left (3 a^2-3 a b+b^2\right ) \cosh (c+d x)}{d}-\frac{a^3 \tanh ^{-1}(\cosh (c+d x))}{d}+\frac{b^2 (3 a-2 b) \cosh ^3(c+d x)}{3 d}+\frac{b^3 \cosh ^5(c+d x)}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3186

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

Rule 390

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Int[PolynomialDivide[(a + b*x^n)
^p, (c + d*x^n)^(-q), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && IGtQ[p, 0] && ILt
Q[q, 0] && GeQ[p, -q]

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])

Rubi steps

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

Mathematica [A]  time = 0.214731, size = 83, normalized size = 1. \[ \frac{30 b \left (24 a^2-18 a b+5 b^2\right ) \cosh (c+d x)+3 \left (80 a^3 \log \left (\tanh \left (\frac{1}{2} (c+d x)\right )\right )+b^3 \cosh (5 (c+d x))\right )+5 b^2 (12 a-5 b) \cosh (3 (c+d x))}{240 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(30*b*(24*a^2 - 18*a*b + 5*b^2)*Cosh[c + d*x] + 5*(12*a - 5*b)*b^2*Cosh[3*(c + d*x)] + 3*(b^3*Cosh[5*(c + d*x)
] + 80*a^3*Log[Tanh[(c + d*x)/2]]))/(240*d)

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Maple [A]  time = 0.03, size = 86, normalized size = 1. \begin{align*}{\frac{1}{d} \left ( -2\,{a}^{3}{\it Artanh} \left ({{\rm e}^{dx+c}} \right ) +3\,{a}^{2}b\cosh \left ( dx+c \right ) +3\,a{b}^{2} \left ( -2/3+1/3\, \left ( \sinh \left ( dx+c \right ) \right ) ^{2} \right ) \cosh \left ( dx+c \right ) +{b}^{3} \left ({\frac{8}{15}}+{\frac{ \left ( \sinh \left ( dx+c \right ) \right ) ^{4}}{5}}-{\frac{4\, \left ( \sinh \left ( dx+c \right ) \right ) ^{2}}{15}} \right ) \cosh \left ( dx+c \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(-2*a^3*arctanh(exp(d*x+c))+3*a^2*b*cosh(d*x+c)+3*a*b^2*(-2/3+1/3*sinh(d*x+c)^2)*cosh(d*x+c)+b^3*(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 [B]  time = 1.06502, size = 261, normalized size = 3.14 \begin{align*} \frac{1}{480} \, b^{3}{\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{1}{8} \, a b^{2}{\left (\frac{e^{\left (3 \, d x + 3 \, c\right )}}{d} - \frac{9 \, e^{\left (d x + c\right )}}{d} - \frac{9 \, e^{\left (-d x - c\right )}}{d} + \frac{e^{\left (-3 \, d x - 3 \, c\right )}}{d}\right )} + \frac{3}{2} \, a^{2} b{\left (\frac{e^{\left (d x + c\right )}}{d} + \frac{e^{\left (-d x - c\right )}}{d}\right )} + \frac{a^{3} \log \left (\tanh \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/480*b^3*(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) + 1/8*a*b^2*(e^(3*d*x + 3*c)/d - 9*e^(d*x + c)/d - 9*e^(-d*x - c)/d + e^(-3*d
*x - 3*c)/d) + 3/2*a^2*b*(e^(d*x + c)/d + e^(-d*x - c)/d) + a^3*log(tanh(1/2*d*x + 1/2*c))/d

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Fricas [B]  time = 2.06531, size = 2889, normalized size = 34.81 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/480*(3*b^3*cosh(d*x + c)^10 + 30*b^3*cosh(d*x + c)*sinh(d*x + c)^9 + 3*b^3*sinh(d*x + c)^10 + 5*(12*a*b^2 -
5*b^3)*cosh(d*x + c)^8 + 5*(27*b^3*cosh(d*x + c)^2 + 12*a*b^2 - 5*b^3)*sinh(d*x + c)^8 + 40*(9*b^3*cosh(d*x +
c)^3 + (12*a*b^2 - 5*b^3)*cosh(d*x + c))*sinh(d*x + c)^7 + 30*(24*a^2*b - 18*a*b^2 + 5*b^3)*cosh(d*x + c)^6 +
10*(63*b^3*cosh(d*x + c)^4 + 72*a^2*b - 54*a*b^2 + 15*b^3 + 14*(12*a*b^2 - 5*b^3)*cosh(d*x + c)^2)*sinh(d*x +
c)^6 + 4*(189*b^3*cosh(d*x + c)^5 + 70*(12*a*b^2 - 5*b^3)*cosh(d*x + c)^3 + 45*(24*a^2*b - 18*a*b^2 + 5*b^3)*c
osh(d*x + c))*sinh(d*x + c)^5 + 30*(24*a^2*b - 18*a*b^2 + 5*b^3)*cosh(d*x + c)^4 + 10*(63*b^3*cosh(d*x + c)^6
+ 35*(12*a*b^2 - 5*b^3)*cosh(d*x + c)^4 + 72*a^2*b - 54*a*b^2 + 15*b^3 + 45*(24*a^2*b - 18*a*b^2 + 5*b^3)*cosh
(d*x + c)^2)*sinh(d*x + c)^4 + 40*(9*b^3*cosh(d*x + c)^7 + 7*(12*a*b^2 - 5*b^3)*cosh(d*x + c)^5 + 15*(24*a^2*b
- 18*a*b^2 + 5*b^3)*cosh(d*x + c)^3 + 3*(24*a^2*b - 18*a*b^2 + 5*b^3)*cosh(d*x + c))*sinh(d*x + c)^3 + 3*b^3
+ 5*(12*a*b^2 - 5*b^3)*cosh(d*x + c)^2 + 5*(27*b^3*cosh(d*x + c)^8 + 28*(12*a*b^2 - 5*b^3)*cosh(d*x + c)^6 + 9
0*(24*a^2*b - 18*a*b^2 + 5*b^3)*cosh(d*x + c)^4 + 12*a*b^2 - 5*b^3 + 36*(24*a^2*b - 18*a*b^2 + 5*b^3)*cosh(d*x
+ c)^2)*sinh(d*x + c)^2 - 480*(a^3*cosh(d*x + c)^5 + 5*a^3*cosh(d*x + c)^4*sinh(d*x + c) + 10*a^3*cosh(d*x +
c)^3*sinh(d*x + c)^2 + 10*a^3*cosh(d*x + c)^2*sinh(d*x + c)^3 + 5*a^3*cosh(d*x + c)*sinh(d*x + c)^4 + a^3*sinh
(d*x + c)^5)*log(cosh(d*x + c) + sinh(d*x + c) + 1) + 480*(a^3*cosh(d*x + c)^5 + 5*a^3*cosh(d*x + c)^4*sinh(d*
x + c) + 10*a^3*cosh(d*x + c)^3*sinh(d*x + c)^2 + 10*a^3*cosh(d*x + c)^2*sinh(d*x + c)^3 + 5*a^3*cosh(d*x + c)
*sinh(d*x + c)^4 + a^3*sinh(d*x + c)^5)*log(cosh(d*x + c) + sinh(d*x + c) - 1) + 10*(3*b^3*cosh(d*x + c)^9 + 4
*(12*a*b^2 - 5*b^3)*cosh(d*x + c)^7 + 18*(24*a^2*b - 18*a*b^2 + 5*b^3)*cosh(d*x + c)^5 + 12*(24*a^2*b - 18*a*b
^2 + 5*b^3)*cosh(d*x + c)^3 + (12*a*b^2 - 5*b^3)*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*co
sh(d*x + c)*sinh(d*x + c)^4 + d*sinh(d*x + c)^5)

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

[Out]

Timed out

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Giac [B]  time = 1.32194, size = 311, normalized size = 3.75 \begin{align*} -\frac{a^{3} \log \left (e^{\left (d x + c\right )} + 1\right )}{d} + \frac{a^{3} \log \left ({\left | e^{\left (d x + c\right )} - 1 \right |}\right )}{d} + \frac{{\left (720 \, a^{2} b e^{\left (4 \, d x + 4 \, c\right )} - 540 \, a b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 150 \, b^{3} e^{\left (4 \, d x + 4 \, c\right )} + 60 \, a b^{2} e^{\left (2 \, d x + 2 \, c\right )} - 25 \, b^{3} e^{\left (2 \, d x + 2 \, c\right )} + 3 \, b^{3}\right )} e^{\left (-5 \, d x - 5 \, c\right )}}{480 \, d} + \frac{3 \, b^{3} d^{4} e^{\left (5 \, d x + 5 \, c\right )} + 60 \, a b^{2} d^{4} e^{\left (3 \, d x + 3 \, c\right )} - 25 \, b^{3} d^{4} e^{\left (3 \, d x + 3 \, c\right )} + 720 \, a^{2} b d^{4} e^{\left (d x + c\right )} - 540 \, a b^{2} d^{4} e^{\left (d x + c\right )} + 150 \, b^{3} d^{4} e^{\left (d x + c\right )}}{480 \, d^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a^3*log(e^(d*x + c) + 1)/d + a^3*log(abs(e^(d*x + c) - 1))/d + 1/480*(720*a^2*b*e^(4*d*x + 4*c) - 540*a*b^2*e
^(4*d*x + 4*c) + 150*b^3*e^(4*d*x + 4*c) + 60*a*b^2*e^(2*d*x + 2*c) - 25*b^3*e^(2*d*x + 2*c) + 3*b^3)*e^(-5*d*
x - 5*c)/d + 1/480*(3*b^3*d^4*e^(5*d*x + 5*c) + 60*a*b^2*d^4*e^(3*d*x + 3*c) - 25*b^3*d^4*e^(3*d*x + 3*c) + 72
0*a^2*b*d^4*e^(d*x + c) - 540*a*b^2*d^4*e^(d*x + c) + 150*b^3*d^4*e^(d*x + c))/d^5