3.2.65 \(\int \text {csch}^2(c+d x) (a+b \sinh ^3(c+d x))^3 \, dx\) [165]

Optimal. Leaf size=152 \[ \frac {9}{8} a b^2 x+\frac {3 a^2 b \cosh (c+d x)}{d}-\frac {b^3 \cosh (c+d x)}{d}+\frac {b^3 \cosh ^3(c+d x)}{d}-\frac {3 b^3 \cosh ^5(c+d x)}{5 d}+\frac {b^3 \cosh ^7(c+d x)}{7 d}-\frac {a^3 \coth (c+d x)}{d}-\frac {9 a b^2 \cosh (c+d x) \sinh (c+d x)}{8 d}+\frac {3 a b^2 \cosh (c+d x) \sinh ^3(c+d x)}{4 d} \]

[Out]

9/8*a*b^2*x+3*a^2*b*cosh(d*x+c)/d-b^3*cosh(d*x+c)/d+b^3*cosh(d*x+c)^3/d-3/5*b^3*cosh(d*x+c)^5/d+1/7*b^3*cosh(d
*x+c)^7/d-a^3*coth(d*x+c)/d-9/8*a*b^2*cosh(d*x+c)*sinh(d*x+c)/d+3/4*a*b^2*cosh(d*x+c)*sinh(d*x+c)^3/d

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Rubi [A]
time = 0.10, antiderivative size = 152, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3299, 3852, 8, 2718, 2715, 2713} \begin {gather*} -\frac {a^3 \coth (c+d x)}{d}+\frac {3 a^2 b \cosh (c+d x)}{d}+\frac {3 a b^2 \sinh ^3(c+d x) \cosh (c+d x)}{4 d}-\frac {9 a b^2 \sinh (c+d x) \cosh (c+d x)}{8 d}+\frac {9}{8} a b^2 x+\frac {b^3 \cosh ^7(c+d x)}{7 d}-\frac {3 b^3 \cosh ^5(c+d x)}{5 d}+\frac {b^3 \cosh ^3(c+d x)}{d}-\frac {b^3 \cosh (c+d x)}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(9*a*b^2*x)/8 + (3*a^2*b*Cosh[c + d*x])/d - (b^3*Cosh[c + d*x])/d + (b^3*Cosh[c + d*x]^3)/d - (3*b^3*Cosh[c +
d*x]^5)/(5*d) + (b^3*Cosh[c + d*x]^7)/(7*d) - (a^3*Coth[c + d*x])/d - (9*a*b^2*Cosh[c + d*x]*Sinh[c + d*x])/(8
*d) + (3*a*b^2*Cosh[c + d*x]*Sinh[c + d*x]^3)/(4*d)

Rule 8

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

Rule 2713

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 2715

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] && Integ
erQ[2*n]

Rule 2718

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

Rule 3299

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 3852

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

Rubi steps

\begin {align*} \int \text {csch}^2(c+d x) \left (a+b \sinh ^3(c+d x)\right )^3 \, dx &=-\int \left (-a^3 \text {csch}^2(c+d x)-3 a^2 b \sinh (c+d x)-3 a b^2 \sinh ^4(c+d x)-b^3 \sinh ^7(c+d x)\right ) \, dx\\ &=a^3 \int \text {csch}^2(c+d x) \, dx+\left (3 a^2 b\right ) \int \sinh (c+d x) \, dx+\left (3 a b^2\right ) \int \sinh ^4(c+d x) \, dx+b^3 \int \sinh ^7(c+d x) \, dx\\ &=\frac {3 a^2 b \cosh (c+d x)}{d}+\frac {3 a b^2 \cosh (c+d x) \sinh ^3(c+d x)}{4 d}-\frac {1}{4} \left (9 a b^2\right ) \int \sinh ^2(c+d x) \, dx-\frac {\left (i a^3\right ) \text {Subst}(\int 1 \, dx,x,-i \coth (c+d x))}{d}-\frac {b^3 \text {Subst}\left (\int \left (1-3 x^2+3 x^4-x^6\right ) \, dx,x,\cosh (c+d x)\right )}{d}\\ &=\frac {3 a^2 b \cosh (c+d x)}{d}-\frac {b^3 \cosh (c+d x)}{d}+\frac {b^3 \cosh ^3(c+d x)}{d}-\frac {3 b^3 \cosh ^5(c+d x)}{5 d}+\frac {b^3 \cosh ^7(c+d x)}{7 d}-\frac {a^3 \coth (c+d x)}{d}-\frac {9 a b^2 \cosh (c+d x) \sinh (c+d x)}{8 d}+\frac {3 a b^2 \cosh (c+d x) \sinh ^3(c+d x)}{4 d}+\frac {1}{8} \left (9 a b^2\right ) \int 1 \, dx\\ &=\frac {9}{8} a b^2 x+\frac {3 a^2 b \cosh (c+d x)}{d}-\frac {b^3 \cosh (c+d x)}{d}+\frac {b^3 \cosh ^3(c+d x)}{d}-\frac {3 b^3 \cosh ^5(c+d x)}{5 d}+\frac {b^3 \cosh ^7(c+d x)}{7 d}-\frac {a^3 \coth (c+d x)}{d}-\frac {9 a b^2 \cosh (c+d x) \sinh (c+d x)}{8 d}+\frac {3 a b^2 \cosh (c+d x) \sinh ^3(c+d x)}{4 d}\\ \end {align*}

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Mathematica [A]
time = 0.77, size = 140, normalized size = 0.92 \begin {gather*} \frac {2520 a b^2 c+2520 a b^2 d x+35 b \left (192 a^2-35 b^2\right ) \cosh (c+d x)+245 b^3 \cosh (3 (c+d x))-49 b^3 \cosh (5 (c+d x))+5 b^3 \cosh (7 (c+d x))-1120 a^3 \coth \left (\frac {1}{2} (c+d x)\right )-1680 a b^2 \sinh (2 (c+d x))+210 a b^2 \sinh (4 (c+d x))-1120 a^3 \tanh \left (\frac {1}{2} (c+d x)\right )}{2240 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2520*a*b^2*c + 2520*a*b^2*d*x + 35*b*(192*a^2 - 35*b^2)*Cosh[c + d*x] + 245*b^3*Cosh[3*(c + d*x)] - 49*b^3*Co
sh[5*(c + d*x)] + 5*b^3*Cosh[7*(c + d*x)] - 1120*a^3*Coth[(c + d*x)/2] - 1680*a*b^2*Sinh[2*(c + d*x)] + 210*a*
b^2*Sinh[4*(c + d*x)] - 1120*a^3*Tanh[(c + d*x)/2])/(2240*d)

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Maple [A]
time = 1.90, size = 268, normalized size = 1.76

method	result	size
risch	\(\frac {9 a \,b^{2} x}{8}+\frac {b^{3} {\mathrm e}^{7 d x +7 c}}{896 d}-\frac {7 b^{3} {\mathrm e}^{5 d x +5 c}}{640 d}+\frac {3 a \,b^{2} {\mathrm e}^{4 d x +4 c}}{64 d}+\frac {7 b^{3} {\mathrm e}^{3 d x +3 c}}{128 d}-\frac {3 a \,{\mathrm e}^{2 d x +2 c} b^{2}}{8 d}+\frac {3 b \,{\mathrm e}^{d x +c} a^{2}}{2 d}-\frac {35 b^{3} {\mathrm e}^{d x +c}}{128 d}+\frac {3 b \,{\mathrm e}^{-d x -c} a^{2}}{2 d}-\frac {35 b^{3} {\mathrm e}^{-d x -c}}{128 d}+\frac {3 a \,{\mathrm e}^{-2 d x -2 c} b^{2}}{8 d}+\frac {7 b^{3} {\mathrm e}^{-3 d x -3 c}}{128 d}-\frac {3 a \,b^{2} {\mathrm e}^{-4 d x -4 c}}{64 d}-\frac {7 b^{3} {\mathrm e}^{-5 d x -5 c}}{640 d}+\frac {b^{3} {\mathrm e}^{-7 d x -7 c}}{896 d}-\frac {2 a^{3}}{d \left ({\mathrm e}^{2 d x +2 c}-1\right )}\)	\(268\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

9/8*a*b^2*x+1/896*b^3/d*exp(7*d*x+7*c)-7/640*b^3/d*exp(5*d*x+5*c)+3/64*a*b^2/d*exp(4*d*x+4*c)+7/128*b^3/d*exp(
3*d*x+3*c)-3/8*a/d*exp(2*d*x+2*c)*b^2+3/2*b/d*exp(d*x+c)*a^2-35/128*b^3/d*exp(d*x+c)+3/2*b/d*exp(-d*x-c)*a^2-3
5/128*b^3/d*exp(-d*x-c)+3/8*a/d*exp(-2*d*x-2*c)*b^2+7/128*b^3/d*exp(-3*d*x-3*c)-3/64*a*b^2/d*exp(-4*d*x-4*c)-7
/640*b^3/d*exp(-5*d*x-5*c)+1/896*b^3/d*exp(-7*d*x-7*c)-2*a^3/d/(exp(2*d*x+2*c)-1)

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Maxima [A]
time = 0.29, size = 220, normalized size = 1.45 \begin {gather*} \frac {3}{64} \, a b^{2} {\left (24 \, x + \frac {e^{\left (4 \, d x + 4 \, c\right )}}{d} - \frac {8 \, e^{\left (2 \, d x + 2 \, c\right )}}{d} + \frac {8 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d} - \frac {e^{\left (-4 \, d x - 4 \, c\right )}}{d}\right )} - \frac {1}{4480} \, b^{3} {\left (\frac {{\left (49 \, e^{\left (-2 \, d x - 2 \, c\right )} - 245 \, e^{\left (-4 \, d x - 4 \, c\right )} + 1225 \, e^{\left (-6 \, d x - 6 \, c\right )} - 5\right )} e^{\left (7 \, d x + 7 \, c\right )}}{d} + \frac {1225 \, e^{\left (-d x - c\right )} - 245 \, e^{\left (-3 \, d x - 3 \, c\right )} + 49 \, e^{\left (-5 \, d x - 5 \, c\right )} - 5 \, e^{\left (-7 \, d x - 7 \, 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 {2 \, a^{3}}{d {\left (e^{\left (-2 \, d x - 2 \, c\right )} - 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/64*a*b^2*(24*x + e^(4*d*x + 4*c)/d - 8*e^(2*d*x + 2*c)/d + 8*e^(-2*d*x - 2*c)/d - e^(-4*d*x - 4*c)/d) - 1/44
80*b^3*((49*e^(-2*d*x - 2*c) - 245*e^(-4*d*x - 4*c) + 1225*e^(-6*d*x - 6*c) - 5)*e^(7*d*x + 7*c)/d + (1225*e^(
-d*x - c) - 245*e^(-3*d*x - 3*c) + 49*e^(-5*d*x - 5*c) - 5*e^(-7*d*x - 7*c))/d) + 3/2*a^2*b*(e^(d*x + c)/d + e
^(-d*x - c)/d) + 2*a^3/(d*(e^(-2*d*x - 2*c) - 1))

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 302 vs. \(2 (142) = 284\).
time = 0.49, size = 302, normalized size = 1.99 \begin {gather*} \frac {20 \, b^{3} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{7} + 105 \, a b^{2} \cosh \left (d x + c\right )^{5} + 525 \, a b^{2} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{4} - 945 \, a b^{2} \cosh \left (d x + c\right )^{3} + 2 \, {\left (70 \, b^{3} \cosh \left (d x + c\right )^{3} - 81 \, b^{3} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{5} + 4 \, {\left (35 \, b^{3} \cosh \left (d x + c\right )^{5} - 135 \, b^{3} \cosh \left (d x + c\right )^{3} + 147 \, b^{3} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{3} + 105 \, {\left (10 \, a b^{2} \cosh \left (d x + c\right )^{3} - 27 \, a b^{2} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} - 280 \, {\left (8 \, a^{3} - 3 \, a b^{2}\right )} \cosh \left (d x + c\right ) + 2 \, {\left (10 \, b^{3} \cosh \left (d x + c\right )^{7} - 81 \, b^{3} \cosh \left (d x + c\right )^{5} + 294 \, b^{3} \cosh \left (d x + c\right )^{3} + 1260 \, a b^{2} d x + 1120 \, a^{3} + 105 \, {\left (32 \, a^{2} b - 7 \, b^{3}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )}{2240 \, d \sinh \left (d x + c\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2240*(20*b^3*cosh(d*x + c)*sinh(d*x + c)^7 + 105*a*b^2*cosh(d*x + c)^5 + 525*a*b^2*cosh(d*x + c)*sinh(d*x +
c)^4 - 945*a*b^2*cosh(d*x + c)^3 + 2*(70*b^3*cosh(d*x + c)^3 - 81*b^3*cosh(d*x + c))*sinh(d*x + c)^5 + 4*(35*b
^3*cosh(d*x + c)^5 - 135*b^3*cosh(d*x + c)^3 + 147*b^3*cosh(d*x + c))*sinh(d*x + c)^3 + 105*(10*a*b^2*cosh(d*x
+ c)^3 - 27*a*b^2*cosh(d*x + c))*sinh(d*x + c)^2 - 280*(8*a^3 - 3*a*b^2)*cosh(d*x + c) + 2*(10*b^3*cosh(d*x +
c)^7 - 81*b^3*cosh(d*x + c)^5 + 294*b^3*cosh(d*x + c)^3 + 1260*a*b^2*d*x + 1120*a^3 + 105*(32*a^2*b - 7*b^3)*
cosh(d*x + c))*sinh(d*x + c))/(d*sinh(d*x + c))

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]
time = 0.50, size = 276, normalized size = 1.82 \begin {gather*} \frac {5040 \, {\left (d x + c\right )} a b^{2} + 5 \, b^{3} e^{\left (7 \, d x + 7 \, c\right )} - 49 \, b^{3} e^{\left (5 \, d x + 5 \, c\right )} + 210 \, a b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 245 \, b^{3} e^{\left (3 \, d x + 3 \, c\right )} - 1680 \, a b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 6720 \, a^{2} b e^{\left (d x + c\right )} - 1225 \, b^{3} e^{\left (d x + c\right )} - \frac {{\left (1890 \, a b^{2} e^{\left (5 \, d x + 5 \, c\right )} + 294 \, b^{3} e^{\left (4 \, d x + 4 \, c\right )} - 210 \, a b^{2} e^{\left (3 \, d x + 3 \, c\right )} - 54 \, b^{3} e^{\left (2 \, d x + 2 \, c\right )} + 5 \, b^{3} - 35 \, {\left (192 \, a^{2} b - 35 \, b^{3}\right )} e^{\left (8 \, d x + 8 \, c\right )} + 560 \, {\left (16 \, a^{3} - 3 \, a b^{2}\right )} e^{\left (7 \, d x + 7 \, c\right )} + 210 \, {\left (32 \, a^{2} b - 7 \, b^{3}\right )} e^{\left (6 \, d x + 6 \, c\right )}\right )} e^{\left (-7 \, d x - 7 \, c\right )}}{{\left (e^{\left (d x + c\right )} + 1\right )} {\left (e^{\left (d x + c\right )} - 1\right )}}}{4480 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4480*(5040*(d*x + c)*a*b^2 + 5*b^3*e^(7*d*x + 7*c) - 49*b^3*e^(5*d*x + 5*c) + 210*a*b^2*e^(4*d*x + 4*c) + 24
5*b^3*e^(3*d*x + 3*c) - 1680*a*b^2*e^(2*d*x + 2*c) + 6720*a^2*b*e^(d*x + c) - 1225*b^3*e^(d*x + c) - (1890*a*b
^2*e^(5*d*x + 5*c) + 294*b^3*e^(4*d*x + 4*c) - 210*a*b^2*e^(3*d*x + 3*c) - 54*b^3*e^(2*d*x + 2*c) + 5*b^3 - 35
*(192*a^2*b - 35*b^3)*e^(8*d*x + 8*c) + 560*(16*a^3 - 3*a*b^2)*e^(7*d*x + 7*c) + 210*(32*a^2*b - 7*b^3)*e^(6*d
*x + 6*c))*e^(-7*d*x - 7*c)/((e^(d*x + c) + 1)*(e^(d*x + c) - 1)))/d

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Mupad [B]
time = 0.40, size = 252, normalized size = 1.66 \begin {gather*} \frac {{\mathrm {e}}^{c+d\,x}\,\left (192\,a^2\,b-35\,b^3\right )}{128\,d}-\frac {2\,a^3}{d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}-1\right )}+\frac {7\,b^3\,{\mathrm {e}}^{-3\,c-3\,d\,x}}{128\,d}+\frac {7\,b^3\,{\mathrm {e}}^{3\,c+3\,d\,x}}{128\,d}-\frac {7\,b^3\,{\mathrm {e}}^{-5\,c-5\,d\,x}}{640\,d}-\frac {7\,b^3\,{\mathrm {e}}^{5\,c+5\,d\,x}}{640\,d}+\frac {b^3\,{\mathrm {e}}^{-7\,c-7\,d\,x}}{896\,d}+\frac {b^3\,{\mathrm {e}}^{7\,c+7\,d\,x}}{896\,d}+\frac {{\mathrm {e}}^{-c-d\,x}\,\left (192\,a^2\,b-35\,b^3\right )}{128\,d}+\frac {9\,a\,b^2\,x}{8}+\frac {3\,a\,b^2\,{\mathrm {e}}^{-2\,c-2\,d\,x}}{8\,d}-\frac {3\,a\,b^2\,{\mathrm {e}}^{2\,c+2\,d\,x}}{8\,d}-\frac {3\,a\,b^2\,{\mathrm {e}}^{-4\,c-4\,d\,x}}{64\,d}+\frac {3\,a\,b^2\,{\mathrm {e}}^{4\,c+4\,d\,x}}{64\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(exp(c + d*x)*(192*a^2*b - 35*b^3))/(128*d) - (2*a^3)/(d*(exp(2*c + 2*d*x) - 1)) + (7*b^3*exp(- 3*c - 3*d*x))/
(128*d) + (7*b^3*exp(3*c + 3*d*x))/(128*d) - (7*b^3*exp(- 5*c - 5*d*x))/(640*d) - (7*b^3*exp(5*c + 5*d*x))/(64
0*d) + (b^3*exp(- 7*c - 7*d*x))/(896*d) + (b^3*exp(7*c + 7*d*x))/(896*d) + (exp(- c - d*x)*(192*a^2*b - 35*b^3
))/(128*d) + (9*a*b^2*x)/8 + (3*a*b^2*exp(- 2*c - 2*d*x))/(8*d) - (3*a*b^2*exp(2*c + 2*d*x))/(8*d) - (3*a*b^2*
exp(- 4*c - 4*d*x))/(64*d) + (3*a*b^2*exp(4*c + 4*d*x))/(64*d)

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