3.221 \(\int \text{csch}^6(c+d x) (a+b \sinh ^4(c+d x))^3 \, dx\)

Optimal. Leaf size=148 \[ -\frac{a^2 (a+3 b) \coth (c+d x)}{d}-\frac{a^3 \coth ^5(c+d x)}{5 d}+\frac{2 a^3 \coth ^3(c+d x)}{3 d}+\frac{b^2 (24 a+11 b) \sinh (c+d x) \cosh (c+d x)}{16 d}-\frac{1}{16} b^2 x (24 a+5 b)+\frac{b^3 \sinh (c+d x) \cosh ^5(c+d x)}{6 d}-\frac{13 b^3 \sinh (c+d x) \cosh ^3(c+d x)}{24 d} \]

[Out]

-(b^2*(24*a + 5*b)*x)/16 - (a^2*(a + 3*b)*Coth[c + d*x])/d + (2*a^3*Coth[c + d*x]^3)/(3*d) - (a^3*Coth[c + d*x
]^5)/(5*d) + (b^2*(24*a + 11*b)*Cosh[c + d*x]*Sinh[c + d*x])/(16*d) - (13*b^3*Cosh[c + d*x]^3*Sinh[c + d*x])/(
24*d) + (b^3*Cosh[c + d*x]^5*Sinh[c + d*x])/(6*d)

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Rubi [A]  time = 0.355256, antiderivative size = 148, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {3217, 1259, 1805, 1802, 207} \[ -\frac{a^2 (a+3 b) \coth (c+d x)}{d}-\frac{a^3 \coth ^5(c+d x)}{5 d}+\frac{2 a^3 \coth ^3(c+d x)}{3 d}+\frac{b^2 (24 a+11 b) \sinh (c+d x) \cosh (c+d x)}{16 d}-\frac{1}{16} b^2 x (24 a+5 b)+\frac{b^3 \sinh (c+d x) \cosh ^5(c+d x)}{6 d}-\frac{13 b^3 \sinh (c+d x) \cosh ^3(c+d x)}{24 d} \]

Antiderivative was successfully verified.

[In]

Int[Csch[c + d*x]^6*(a + b*Sinh[c + d*x]^4)^3,x]

[Out]

-(b^2*(24*a + 5*b)*x)/16 - (a^2*(a + 3*b)*Coth[c + d*x])/d + (2*a^3*Coth[c + d*x]^3)/(3*d) - (a^3*Coth[c + d*x
]^5)/(5*d) + (b^2*(24*a + 11*b)*Cosh[c + d*x]*Sinh[c + d*x])/(16*d) - (13*b^3*Cosh[c + d*x]^3*Sinh[c + d*x])/(
24*d) + (b^3*Cosh[c + d*x]^5*Sinh[c + d*x])/(6*d)

Rule 3217

Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^4)^(p_.), x_Symbol] :> With[{ff = FreeF
actors[Tan[e + f*x], x]}, Dist[ff^(m + 1)/f, Subst[Int[(x^m*(a + 2*a*ff^2*x^2 + (a + b)*ff^4*x^4)^p)/(1 + ff^2
*x^2)^(m/2 + 2*p + 1), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[m/2] && IntegerQ[p]

Rule 1259

Int[(x_)^(m_)*((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Simp[((-d)^(
m/2 - 1)*(c*d^2 - b*d*e + a*e^2)^p*x*(d + e*x^2)^(q + 1))/(2*e^(2*p + m/2)*(q + 1)), x] + Dist[(-d)^(m/2 - 1)/
(2*e^(2*p)*(q + 1)), Int[x^m*(d + e*x^2)^(q + 1)*ExpandToSum[Together[(1*(2*(-d)^(-(m/2) + 1)*e^(2*p)*(q + 1)*
(a + b*x^2 + c*x^4)^p - ((c*d^2 - b*d*e + a*e^2)^p/(e^(m/2)*x^m))*(d + e*(2*q + 3)*x^2)))/(d + e*x^2)], x], x]
, x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && ILtQ[q, -1] && ILtQ[m/2, 0]

Rule 1805

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[(c*x)^m*Pq,
a + b*x^2, x], f = Coeff[PolynomialRemainder[(c*x)^m*Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[
(c*x)^m*Pq, a + b*x^2, x], x, 1]}, Simp[((a*g - b*f*x)*(a + b*x^2)^(p + 1))/(2*a*b*(p + 1)), x] + Dist[1/(2*a*
(p + 1)), Int[(c*x)^m*(a + b*x^2)^(p + 1)*ExpandToSum[(2*a*(p + 1)*Q)/(c*x)^m + (f*(2*p + 3))/(c*x)^m, x], x],
x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && LtQ[p, -1] && ILtQ[m, 0]

Rule 1802

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*Pq*(a + b*x
^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rule 207

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

Rubi steps

\begin{align*} \int \text{csch}^6(c+d x) \left (a+b \sinh ^4(c+d x)\right )^3 \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (a-2 a x^2+(a+b) x^4\right )^3}{x^6 \left (1-x^2\right )^4} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac{b^3 \cosh ^5(c+d x) \sinh (c+d x)}{6 d}-\frac{\operatorname{Subst}\left (\int \frac{-6 a^3+30 a^3 x^2-6 a^2 (10 a+3 b) x^4+\left (60 a^3+54 a^2 b+b^3\right ) x^6-6 (5 a-b) (a+b)^2 x^8+6 (a+b)^3 x^{10}}{x^6 \left (1-x^2\right )^3} \, dx,x,\tanh (c+d x)\right )}{6 d}\\ &=-\frac{13 b^3 \cosh ^3(c+d x) \sinh (c+d x)}{24 d}+\frac{b^3 \cosh ^5(c+d x) \sinh (c+d x)}{6 d}+\frac{\operatorname{Subst}\left (\int \frac{24 a^3-96 a^3 x^2+72 a^2 (2 a+b) x^4-3 \left (32 a^3+48 a^2 b-3 b^3\right ) x^6+24 (a+b)^3 x^8}{x^6 \left (1-x^2\right )^2} \, dx,x,\tanh (c+d x)\right )}{24 d}\\ &=\frac{b^2 (24 a+11 b) \cosh (c+d x) \sinh (c+d x)}{16 d}-\frac{13 b^3 \cosh ^3(c+d x) \sinh (c+d x)}{24 d}+\frac{b^3 \cosh ^5(c+d x) \sinh (c+d x)}{6 d}-\frac{\operatorname{Subst}\left (\int \frac{-48 a^3+144 a^3 x^2-144 a^2 (a+b) x^4+3 \left (16 a^3+48 a^2 b+24 a b^2+5 b^3\right ) x^6}{x^6 \left (1-x^2\right )} \, dx,x,\tanh (c+d x)\right )}{48 d}\\ &=\frac{b^2 (24 a+11 b) \cosh (c+d x) \sinh (c+d x)}{16 d}-\frac{13 b^3 \cosh ^3(c+d x) \sinh (c+d x)}{24 d}+\frac{b^3 \cosh ^5(c+d x) \sinh (c+d x)}{6 d}-\frac{\operatorname{Subst}\left (\int \left (-\frac{48 a^3}{x^6}+\frac{96 a^3}{x^4}-\frac{48 a^2 (a+3 b)}{x^2}-\frac{3 b^2 (24 a+5 b)}{-1+x^2}\right ) \, dx,x,\tanh (c+d x)\right )}{48 d}\\ &=-\frac{a^2 (a+3 b) \coth (c+d x)}{d}+\frac{2 a^3 \coth ^3(c+d x)}{3 d}-\frac{a^3 \coth ^5(c+d x)}{5 d}+\frac{b^2 (24 a+11 b) \cosh (c+d x) \sinh (c+d x)}{16 d}-\frac{13 b^3 \cosh ^3(c+d x) \sinh (c+d x)}{24 d}+\frac{b^3 \cosh ^5(c+d x) \sinh (c+d x)}{6 d}+\frac{\left (b^2 (24 a+5 b)\right ) \operatorname{Subst}\left (\int \frac{1}{-1+x^2} \, dx,x,\tanh (c+d x)\right )}{16 d}\\ &=-\frac{1}{16} b^2 (24 a+5 b) x-\frac{a^2 (a+3 b) \coth (c+d x)}{d}+\frac{2 a^3 \coth ^3(c+d x)}{3 d}-\frac{a^3 \coth ^5(c+d x)}{5 d}+\frac{b^2 (24 a+11 b) \cosh (c+d x) \sinh (c+d x)}{16 d}-\frac{13 b^3 \cosh ^3(c+d x) \sinh (c+d x)}{24 d}+\frac{b^3 \cosh ^5(c+d x) \sinh (c+d x)}{6 d}\\ \end{align*}

Mathematica [A]  time = 1.15998, size = 110, normalized size = 0.74 \[ \frac{5 b^2 (9 (16 a+5 b) \sinh (2 (c+d x))-288 a c-288 a d x-9 b \sinh (4 (c+d x))+b \sinh (6 (c+d x))-60 b c-60 b d x)-64 a^2 \coth (c+d x) \left (3 a \text{csch}^4(c+d x)-4 a \text{csch}^2(c+d x)+8 a+45 b\right )}{960 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Csch[c + d*x]^6*(a + b*Sinh[c + d*x]^4)^3,x]

[Out]

(-64*a^2*Coth[c + d*x]*(8*a + 45*b - 4*a*Csch[c + d*x]^2 + 3*a*Csch[c + d*x]^4) + 5*b^2*(-288*a*c - 60*b*c - 2
88*a*d*x - 60*b*d*x + 9*(16*a + 5*b)*Sinh[2*(c + d*x)] - 9*b*Sinh[4*(c + d*x)] + b*Sinh[6*(c + d*x)]))/(960*d)

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Maple [A]  time = 0.046, size = 126, normalized size = 0.9 \begin{align*}{\frac{1}{d} \left ({a}^{3} \left ( -{\frac{8}{15}}-{\frac{ \left ({\rm csch} \left (dx+c\right ) \right ) ^{4}}{5}}+{\frac{4\, \left ({\rm csch} \left (dx+c\right ) \right ) ^{2}}{15}} \right ){\rm coth} \left (dx+c\right )-3\,{a}^{2}b{\rm coth} \left (dx+c\right )+3\,a{b}^{2} \left ( 1/2\,\cosh \left ( dx+c \right ) \sinh \left ( dx+c \right ) -1/2\,dx-c/2 \right ) +{b}^{3} \left ( \left ({\frac{ \left ( \sinh \left ( dx+c \right ) \right ) ^{5}}{6}}-{\frac{5\, \left ( \sinh \left ( dx+c \right ) \right ) ^{3}}{24}}+{\frac{5\,\sinh \left ( dx+c \right ) }{16}} \right ) \cosh \left ( dx+c \right ) -{\frac{5\,dx}{16}}-{\frac{5\,c}{16}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csch(d*x+c)^6*(a+b*sinh(d*x+c)^4)^3,x)

[Out]

1/d*(a^3*(-8/15-1/5*csch(d*x+c)^4+4/15*csch(d*x+c)^2)*coth(d*x+c)-3*a^2*b*coth(d*x+c)+3*a*b^2*(1/2*cosh(d*x+c)
*sinh(d*x+c)-1/2*d*x-1/2*c)+b^3*((1/6*sinh(d*x+c)^5-5/24*sinh(d*x+c)^3+5/16*sinh(d*x+c))*cosh(d*x+c)-5/16*d*x-
5/16*c))

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Maxima [B]  time = 1.06571, size = 485, normalized size = 3.28 \begin{align*} -\frac{3}{8} \, a b^{2}{\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}{384} \, b^{3}{\left (\frac{{\left (9 \, e^{\left (-2 \, d x - 2 \, c\right )} - 45 \, e^{\left (-4 \, d x - 4 \, c\right )} - 1\right )} e^{\left (6 \, d x + 6 \, c\right )}}{d} + \frac{120 \,{\left (d x + c\right )}}{d} + \frac{45 \, e^{\left (-2 \, d x - 2 \, c\right )} - 9 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )}}{d}\right )} - \frac{16}{15} \, a^{3}{\left (\frac{5 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d{\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} - 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} - 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} - 1\right )}} - \frac{10 \, e^{\left (-4 \, d x - 4 \, c\right )}}{d{\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} - 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} - 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} - 1\right )}} - \frac{1}{d{\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} - 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} - 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} - 1\right )}}\right )} + \frac{6 \, a^{2} b}{d{\left (e^{\left (-2 \, d x - 2 \, c\right )} - 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(d*x+c)^6*(a+b*sinh(d*x+c)^4)^3,x, algorithm="maxima")

[Out]

-3/8*a*b^2*(4*x - e^(2*d*x + 2*c)/d + e^(-2*d*x - 2*c)/d) - 1/384*b^3*((9*e^(-2*d*x - 2*c) - 45*e^(-4*d*x - 4*
c) - 1)*e^(6*d*x + 6*c)/d + 120*(d*x + c)/d + (45*e^(-2*d*x - 2*c) - 9*e^(-4*d*x - 4*c) + e^(-6*d*x - 6*c))/d)
- 16/15*a^3*(5*e^(-2*d*x - 2*c)/(d*(5*e^(-2*d*x - 2*c) - 10*e^(-4*d*x - 4*c) + 10*e^(-6*d*x - 6*c) - 5*e^(-8*
d*x - 8*c) + e^(-10*d*x - 10*c) - 1)) - 10*e^(-4*d*x - 4*c)/(d*(5*e^(-2*d*x - 2*c) - 10*e^(-4*d*x - 4*c) + 10*
e^(-6*d*x - 6*c) - 5*e^(-8*d*x - 8*c) + e^(-10*d*x - 10*c) - 1)) - 1/(d*(5*e^(-2*d*x - 2*c) - 10*e^(-4*d*x - 4
*c) + 10*e^(-6*d*x - 6*c) - 5*e^(-8*d*x - 8*c) + e^(-10*d*x - 10*c) - 1))) + 6*a^2*b/(d*(e^(-2*d*x - 2*c) - 1)
)

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Fricas [B]  time = 1.5227, size = 2006, normalized size = 13.55 \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)^6*(a+b*sinh(d*x+c)^4)^3,x, algorithm="fricas")

[Out]

1/1920*(5*b^3*cosh(d*x + c)^11 + 55*b^3*cosh(d*x + c)*sinh(d*x + c)^10 - 70*b^3*cosh(d*x + c)^9 + 15*(55*b^3*c
osh(d*x + c)^3 - 42*b^3*cosh(d*x + c))*sinh(d*x + c)^8 + 20*(36*a*b^2 + 25*b^3)*cosh(d*x + c)^7 + 70*(33*b^3*c
osh(d*x + c)^5 - 84*b^3*cosh(d*x + c)^3 + 2*(36*a*b^2 + 25*b^3)*cosh(d*x + c))*sinh(d*x + c)^6 - (1024*a^3 + 5
760*a^2*b + 3600*a*b^2 + 1625*b^3)*cosh(d*x + c)^5 + 8*(128*a^3 + 720*a^2*b - 15*(24*a*b^2 + 5*b^3)*d*x)*sinh(
d*x + c)^5 + 5*(330*b^3*cosh(d*x + c)^7 - 1764*b^3*cosh(d*x + c)^5 + 140*(36*a*b^2 + 25*b^3)*cosh(d*x + c)^3 -
(1024*a^3 + 5760*a^2*b + 3600*a*b^2 + 1625*b^3)*cosh(d*x + c))*sinh(d*x + c)^4 + 20*(256*a^3 + 864*a^2*b + 32
4*a*b^2 + 125*b^3)*cosh(d*x + c)^3 - 40*(128*a^3 + 720*a^2*b - 15*(24*a*b^2 + 5*b^3)*d*x - 2*(128*a^3 + 720*a^
2*b - 15*(24*a*b^2 + 5*b^3)*d*x)*cosh(d*x + c)^2)*sinh(d*x + c)^3 + 5*(55*b^3*cosh(d*x + c)^9 - 504*b^3*cosh(d
*x + c)^7 + 84*(36*a*b^2 + 25*b^3)*cosh(d*x + c)^5 - 2*(1024*a^3 + 5760*a^2*b + 3600*a*b^2 + 1625*b^3)*cosh(d*
x + c)^3 + 12*(256*a^3 + 864*a^2*b + 324*a*b^2 + 125*b^3)*cosh(d*x + c))*sinh(d*x + c)^2 - 10*(1024*a^3 + 1152
*a^2*b + 360*a*b^2 + 131*b^3)*cosh(d*x + c) + 40*((128*a^3 + 720*a^2*b - 15*(24*a*b^2 + 5*b^3)*d*x)*cosh(d*x +
c)^4 + 256*a^3 + 1440*a^2*b - 30*(24*a*b^2 + 5*b^3)*d*x - 3*(128*a^3 + 720*a^2*b - 15*(24*a*b^2 + 5*b^3)*d*x)
*cosh(d*x + c)^2)*sinh(d*x + c))/(d*sinh(d*x + c)^5 + 5*(2*d*cosh(d*x + c)^2 - d)*sinh(d*x + c)^3 + 5*(d*cosh(
d*x + c)^4 - 3*d*cosh(d*x + c)^2 + 2*d)*sinh(d*x + c))

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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)**6*(a+b*sinh(d*x+c)**4)**3,x)

[Out]

Timed out

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Giac [B]  time = 1.70671, size = 414, normalized size = 2.8 \begin{align*} -\frac{{\left (24 \, a b^{2} + 5 \, b^{3}\right )}{\left (d x + c\right )}}{16 \, d} + \frac{{\left (528 \, a b^{2} e^{\left (6 \, d x + 6 \, c\right )} + 110 \, b^{3} e^{\left (6 \, d x + 6 \, c\right )} - 144 \, a b^{2} e^{\left (4 \, d x + 4 \, c\right )} - 45 \, b^{3} e^{\left (4 \, d x + 4 \, c\right )} + 9 \, b^{3} e^{\left (2 \, d x + 2 \, c\right )} - b^{3}\right )} e^{\left (-6 \, d x - 6 \, c\right )}}{384 \, d} + \frac{b^{3} d^{2} e^{\left (6 \, d x + 6 \, c\right )} - 9 \, b^{3} d^{2} e^{\left (4 \, d x + 4 \, c\right )} + 144 \, a b^{2} d^{2} e^{\left (2 \, d x + 2 \, c\right )} + 45 \, b^{3} d^{2} e^{\left (2 \, d x + 2 \, c\right )}}{384 \, d^{3}} - \frac{2 \,{\left (45 \, a^{2} b e^{\left (8 \, d x + 8 \, c\right )} - 180 \, a^{2} b e^{\left (6 \, d x + 6 \, c\right )} + 80 \, a^{3} e^{\left (4 \, d x + 4 \, c\right )} + 270 \, a^{2} b e^{\left (4 \, d x + 4 \, c\right )} - 40 \, a^{3} e^{\left (2 \, d x + 2 \, c\right )} - 180 \, a^{2} b e^{\left (2 \, d x + 2 \, c\right )} + 8 \, a^{3} + 45 \, a^{2} b\right )}}{15 \, d{\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(d*x+c)^6*(a+b*sinh(d*x+c)^4)^3,x, algorithm="giac")

[Out]

-1/16*(24*a*b^2 + 5*b^3)*(d*x + c)/d + 1/384*(528*a*b^2*e^(6*d*x + 6*c) + 110*b^3*e^(6*d*x + 6*c) - 144*a*b^2*
e^(4*d*x + 4*c) - 45*b^3*e^(4*d*x + 4*c) + 9*b^3*e^(2*d*x + 2*c) - b^3)*e^(-6*d*x - 6*c)/d + 1/384*(b^3*d^2*e^
(6*d*x + 6*c) - 9*b^3*d^2*e^(4*d*x + 4*c) + 144*a*b^2*d^2*e^(2*d*x + 2*c) + 45*b^3*d^2*e^(2*d*x + 2*c))/d^3 -
2/15*(45*a^2*b*e^(8*d*x + 8*c) - 180*a^2*b*e^(6*d*x + 6*c) + 80*a^3*e^(4*d*x + 4*c) + 270*a^2*b*e^(4*d*x + 4*c
) - 40*a^3*e^(2*d*x + 2*c) - 180*a^2*b*e^(2*d*x + 2*c) + 8*a^3 + 45*a^2*b)/(d*(e^(2*d*x + 2*c) - 1)^5)