3.260 \(\int \frac {\sinh ^6(c+d x)}{(a-b \sinh ^4(c+d x))^3} \, dx\)

Optimal. Leaf size=345 \[ \frac {\left (-10 \sqrt {a} \sqrt {b}+4 a+3 b\right ) \tanh ^{-1}\left (\frac {\sqrt {\sqrt {a}-\sqrt {b}} \tanh (c+d x)}{\sqrt [4]{a}}\right )}{64 a^{5/4} b^{3/2} d \left (\sqrt {a}-\sqrt {b}\right )^{5/2}}-\frac {\left (10 \sqrt {a} \sqrt {b}+4 a+3 b\right ) \tanh ^{-1}\left (\frac {\sqrt {\sqrt {a}+\sqrt {b}} \tanh (c+d x)}{\sqrt [4]{a}}\right )}{64 a^{5/4} b^{3/2} d \left (\sqrt {a}+\sqrt {b}\right )^{5/2}}+\frac {\tanh (c+d x) \left (\frac {2 a \left (a^2-a b-8 b^2\right )}{(a-b)^3}-\frac {\left (2 a^2+15 a b+3 b^2\right ) \tanh ^2(c+d x)}{(a-b)^2}\right )}{32 a b d \left ((a-b) \tanh ^4(c+d x)-2 a \tanh ^2(c+d x)+a\right )}+\frac {\tanh (c+d x) \left (a (a+3 b)-\left (a^2+6 a b+b^2\right ) \tanh ^2(c+d x)\right )}{8 d (a-b)^3 \left ((a-b) \tanh ^4(c+d x)-2 a \tanh ^2(c+d x)+a\right )^2} \]

[Out]

1/64*arctanh((a^(1/2)-b^(1/2))^(1/2)*tanh(d*x+c)/a^(1/4))*(4*a+3*b-10*a^(1/2)*b^(1/2))/a^(5/4)/b^(3/2)/d/(a^(1
/2)-b^(1/2))^(5/2)-1/64*arctanh((a^(1/2)+b^(1/2))^(1/2)*tanh(d*x+c)/a^(1/4))*(4*a+3*b+10*a^(1/2)*b^(1/2))/a^(5
/4)/b^(3/2)/d/(a^(1/2)+b^(1/2))^(5/2)+1/8*tanh(d*x+c)*(a*(a+3*b)-(a^2+6*a*b+b^2)*tanh(d*x+c)^2)/(a-b)^3/d/(a-2
*a*tanh(d*x+c)^2+(a-b)*tanh(d*x+c)^4)^2+1/32*tanh(d*x+c)*(2*a*(a^2-a*b-8*b^2)/(a-b)^3-(2*a^2+15*a*b+3*b^2)*tan
h(d*x+c)^2/(a-b)^2)/a/b/d/(a-2*a*tanh(d*x+c)^2+(a-b)*tanh(d*x+c)^4)

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Rubi [A]  time = 0.73, antiderivative size = 345, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {3217, 1333, 1678, 1166, 208} \[ \frac {\left (-10 \sqrt {a} \sqrt {b}+4 a+3 b\right ) \tanh ^{-1}\left (\frac {\sqrt {\sqrt {a}-\sqrt {b}} \tanh (c+d x)}{\sqrt [4]{a}}\right )}{64 a^{5/4} b^{3/2} d \left (\sqrt {a}-\sqrt {b}\right )^{5/2}}-\frac {\left (10 \sqrt {a} \sqrt {b}+4 a+3 b\right ) \tanh ^{-1}\left (\frac {\sqrt {\sqrt {a}+\sqrt {b}} \tanh (c+d x)}{\sqrt [4]{a}}\right )}{64 a^{5/4} b^{3/2} d \left (\sqrt {a}+\sqrt {b}\right )^{5/2}}+\frac {\tanh (c+d x) \left (\frac {2 a \left (a^2-a b-8 b^2\right )}{(a-b)^3}-\frac {\left (2 a^2+15 a b+3 b^2\right ) \tanh ^2(c+d x)}{(a-b)^2}\right )}{32 a b d \left ((a-b) \tanh ^4(c+d x)-2 a \tanh ^2(c+d x)+a\right )}+\frac {\tanh (c+d x) \left (a (a+3 b)-\left (a^2+6 a b+b^2\right ) \tanh ^2(c+d x)\right )}{8 d (a-b)^3 \left ((a-b) \tanh ^4(c+d x)-2 a \tanh ^2(c+d x)+a\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((4*a - 10*Sqrt[a]*Sqrt[b] + 3*b)*ArcTanh[(Sqrt[Sqrt[a] - Sqrt[b]]*Tanh[c + d*x])/a^(1/4)])/(64*a^(5/4)*(Sqrt[
a] - Sqrt[b])^(5/2)*b^(3/2)*d) - ((4*a + 10*Sqrt[a]*Sqrt[b] + 3*b)*ArcTanh[(Sqrt[Sqrt[a] + Sqrt[b]]*Tanh[c + d
*x])/a^(1/4)])/(64*a^(5/4)*(Sqrt[a] + Sqrt[b])^(5/2)*b^(3/2)*d) + (Tanh[c + d*x]*(a*(a + 3*b) - (a^2 + 6*a*b +
b^2)*Tanh[c + d*x]^2))/(8*(a - b)^3*d*(a - 2*a*Tanh[c + d*x]^2 + (a - b)*Tanh[c + d*x]^4)^2) + (Tanh[c + d*x]
*((2*a*(a^2 - a*b - 8*b^2))/(a - b)^3 - ((2*a^2 + 15*a*b + 3*b^2)*Tanh[c + d*x]^2)/(a - b)^2))/(32*a*b*d*(a -
2*a*Tanh[c + d*x]^2 + (a - b)*Tanh[c + d*x]^4))

Rule 208

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

Rule 1166

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Di
st[e/2 + (2*c*d - b*e)/(2*q), Int[1/(b/2 - q/2 + c*x^2), x], x] + Dist[e/2 - (2*c*d - b*e)/(2*q), Int[1/(b/2 +
q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - a*e^2, 0] && PosQ[b^
2 - 4*a*c]

Rule 1333

Int[(x_)^(m_)*((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> With[{f = Coe
ff[PolynomialRemainder[x^m*(d + e*x^2)^q, a + b*x^2 + c*x^4, x], x, 0], g = Coeff[PolynomialRemainder[x^m*(d +
e*x^2)^q, a + b*x^2 + c*x^4, x], x, 2]}, Simp[(x*(a + b*x^2 + c*x^4)^(p + 1)*(a*b*g - f*(b^2 - 2*a*c) - c*(b*
f - 2*a*g)*x^2))/(2*a*(p + 1)*(b^2 - 4*a*c)), x] + Dist[1/(2*a*(p + 1)*(b^2 - 4*a*c)), Int[(a + b*x^2 + c*x^4)
^(p + 1)*Simp[ExpandToSum[2*a*(p + 1)*(b^2 - 4*a*c)*PolynomialQuotient[x^m*(d + e*x^2)^q, a + b*x^2 + c*x^4, x
] + b^2*f*(2*p + 3) - 2*a*c*f*(4*p + 5) - a*b*g + c*(4*p + 7)*(b*f - 2*a*g)*x^2, x], x], x], x]] /; FreeQ[{a,
b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1] && IGtQ[q, 1] && IGtQ[m/2, 0]

Rule 1678

Int[(Pq_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> With[{d = Coeff[PolynomialRemainder[Pq, a +
b*x^2 + c*x^4, x], x, 0], e = Coeff[PolynomialRemainder[Pq, a + b*x^2 + c*x^4, x], x, 2]}, Simp[(x*(a + b*x^2
+ c*x^4)^(p + 1)*(a*b*e - d*(b^2 - 2*a*c) - c*(b*d - 2*a*e)*x^2))/(2*a*(p + 1)*(b^2 - 4*a*c)), x] + Dist[1/(2*
a*(p + 1)*(b^2 - 4*a*c)), Int[(a + b*x^2 + c*x^4)^(p + 1)*ExpandToSum[2*a*(p + 1)*(b^2 - 4*a*c)*PolynomialQuot
ient[Pq, a + b*x^2 + c*x^4, x] + b^2*d*(2*p + 3) - 2*a*c*d*(4*p + 5) - a*b*e + c*(4*p + 7)*(b*d - 2*a*e)*x^2,
x], x], x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x^2] && Expon[Pq, x^2] > 1 && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1
]

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]

Rubi steps

\begin {align*} \int \frac {\sinh ^6(c+d x)}{\left (a-b \sinh ^4(c+d x)\right )^3} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {x^6 \left (1-x^2\right )^2}{\left (a-2 a x^2+(a-b) x^4\right )^3} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac {\tanh (c+d x) \left (a (a+3 b)-\left (a^2+6 a b+b^2\right ) \tanh ^2(c+d x)\right )}{8 (a-b)^3 d \left (a-2 a \tanh ^2(c+d x)+(a-b) \tanh ^4(c+d x)\right )^2}-\frac {\operatorname {Subst}\left (\int \frac {\frac {2 a^3 b (a+3 b)}{(a-b)^3}+\frac {2 a^2 b \left (5 a^2+6 a b-3 b^2\right ) x^2}{(a-b)^3}-\frac {32 a^2 b^2 x^4}{(a-b)^2}-\frac {16 a^2 b x^6}{a-b}}{\left (a-2 a x^2+(a-b) x^4\right )^2} \, dx,x,\tanh (c+d x)\right )}{16 a^2 b d}\\ &=\frac {\tanh (c+d x) \left (a (a+3 b)-\left (a^2+6 a b+b^2\right ) \tanh ^2(c+d x)\right )}{8 (a-b)^3 d \left (a-2 a \tanh ^2(c+d x)+(a-b) \tanh ^4(c+d x)\right )^2}+\frac {\tanh (c+d x) \left (\frac {2 a \left (a^2-a b-8 b^2\right )}{(a-b)^3}-\frac {\left (2 a^2+15 a b+3 b^2\right ) \tanh ^2(c+d x)}{(a-b)^2}\right )}{32 a b d \left (a-2 a \tanh ^2(c+d x)+(a-b) \tanh ^4(c+d x)\right )}+\frac {\operatorname {Subst}\left (\int \frac {-\frac {8 a^4 b (a+2 b)}{(a-b)^2}-\frac {4 a^3 b \left (2 a^2-17 a b+3 b^2\right ) x^2}{(a-b)^2}}{a-2 a x^2+(a-b) x^4} \, dx,x,\tanh (c+d x)\right )}{128 a^4 b^2 d}\\ &=\frac {\tanh (c+d x) \left (a (a+3 b)-\left (a^2+6 a b+b^2\right ) \tanh ^2(c+d x)\right )}{8 (a-b)^3 d \left (a-2 a \tanh ^2(c+d x)+(a-b) \tanh ^4(c+d x)\right )^2}+\frac {\tanh (c+d x) \left (\frac {2 a \left (a^2-a b-8 b^2\right )}{(a-b)^3}-\frac {\left (2 a^2+15 a b+3 b^2\right ) \tanh ^2(c+d x)}{(a-b)^2}\right )}{32 a b d \left (a-2 a \tanh ^2(c+d x)+(a-b) \tanh ^4(c+d x)\right )}-\frac {\left (\left (\sqrt {a}+\sqrt {b}\right ) \left (4 a-10 \sqrt {a} \sqrt {b}+3 b\right )\right ) \operatorname {Subst}\left (\int \frac {1}{-a-\sqrt {a} \sqrt {b}+(a-b) x^2} \, dx,x,\tanh (c+d x)\right )}{64 a \left (\sqrt {a}-\sqrt {b}\right )^2 b^{3/2} d}+\frac {\left (-\frac {2 a^3 b \left (2 a^2-17 a b+3 b^2\right )}{(a-b)^2}-\frac {-\frac {16 a^4 b (a+2 b)}{a-b}-\frac {8 a^4 b \left (2 a^2-17 a b+3 b^2\right )}{(a-b)^2}}{4 \sqrt {a} \sqrt {b}}\right ) \operatorname {Subst}\left (\int \frac {1}{-a+\sqrt {a} \sqrt {b}+(a-b) x^2} \, dx,x,\tanh (c+d x)\right )}{128 a^4 b^2 d}\\ &=\frac {\left (4 a-10 \sqrt {a} \sqrt {b}+3 b\right ) \tanh ^{-1}\left (\frac {\sqrt {\sqrt {a}-\sqrt {b}} \tanh (c+d x)}{\sqrt [4]{a}}\right )}{64 a^{5/4} \left (\sqrt {a}-\sqrt {b}\right )^{5/2} b^{3/2} d}-\frac {\left (4 a+10 \sqrt {a} \sqrt {b}+3 b\right ) \tanh ^{-1}\left (\frac {\sqrt {\sqrt {a}+\sqrt {b}} \tanh (c+d x)}{\sqrt [4]{a}}\right )}{64 a^{5/4} \left (\sqrt {a}+\sqrt {b}\right )^{5/2} b^{3/2} d}+\frac {\tanh (c+d x) \left (a (a+3 b)-\left (a^2+6 a b+b^2\right ) \tanh ^2(c+d x)\right )}{8 (a-b)^3 d \left (a-2 a \tanh ^2(c+d x)+(a-b) \tanh ^4(c+d x)\right )^2}+\frac {\tanh (c+d x) \left (\frac {2 a \left (a^2-a b-8 b^2\right )}{(a-b)^3}-\frac {\left (2 a^2+15 a b+3 b^2\right ) \tanh ^2(c+d x)}{(a-b)^2}\right )}{32 a b d \left (a-2 a \tanh ^2(c+d x)+(a-b) \tanh ^4(c+d x)\right )}\\ \end {align*}

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Mathematica [A]  time = 3.37, size = 351, normalized size = 1.02 \[ -\frac {\frac {4 \sinh (2 (c+d x)) \left (4 a^2+3 b (a+b) \cosh (2 (c+d x))-19 a b-3 b^2\right )}{a b (8 a+4 b \cosh (2 (c+d x))-b \cosh (4 (c+d x))-3 b)}+\frac {\left (10 \sqrt {a} \sqrt {b}+4 a+3 b\right ) \left (\sqrt {a}-\sqrt {b}\right )^2 \tanh ^{-1}\left (\frac {\left (\sqrt {a}+\sqrt {b}\right ) \tanh (c+d x)}{\sqrt {\sqrt {a} \sqrt {b}+a}}\right )}{a b^{3/2} \sqrt {\sqrt {a} \sqrt {b}+a}}+\frac {\left (\sqrt {a}+\sqrt {b}\right )^2 \left (-10 \sqrt {a} \sqrt {b}+4 a+3 b\right ) \tan ^{-1}\left (\frac {\left (\sqrt {a}-\sqrt {b}\right ) \tanh (c+d x)}{\sqrt {\sqrt {a} \sqrt {b}-a}}\right )}{a b^{3/2} \sqrt {\sqrt {a} \sqrt {b}-a}}-\frac {128 (a-b) \sinh (2 (c+d x)) (2 a-b \cosh (2 (c+d x))+b)}{b (-8 a-4 b \cosh (2 (c+d x))+b \cosh (4 (c+d x))+3 b)^2}}{64 d (a-b)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/64*(((Sqrt[a] + Sqrt[b])^2*(4*a - 10*Sqrt[a]*Sqrt[b] + 3*b)*ArcTan[((Sqrt[a] - Sqrt[b])*Tanh[c + d*x])/Sqrt
[-a + Sqrt[a]*Sqrt[b]]])/(a*Sqrt[-a + Sqrt[a]*Sqrt[b]]*b^(3/2)) + ((Sqrt[a] - Sqrt[b])^2*(4*a + 10*Sqrt[a]*Sqr
t[b] + 3*b)*ArcTanh[((Sqrt[a] + Sqrt[b])*Tanh[c + d*x])/Sqrt[a + Sqrt[a]*Sqrt[b]]])/(a*Sqrt[a + Sqrt[a]*Sqrt[b
]]*b^(3/2)) + (4*(4*a^2 - 19*a*b - 3*b^2 + 3*b*(a + b)*Cosh[2*(c + d*x)])*Sinh[2*(c + d*x)])/(a*b*(8*a - 3*b +
4*b*Cosh[2*(c + d*x)] - b*Cosh[4*(c + d*x)])) - (128*(a - b)*(2*a + b - b*Cosh[2*(c + d*x)])*Sinh[2*(c + d*x)
])/(b*(-8*a + 3*b - 4*b*Cosh[2*(c + d*x)] + b*Cosh[4*(c + d*x)])^2))/((a - b)^2*d)

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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(sinh(d*x+c)^6/(a-b*sinh(d*x+c)^4)^3,x, algorithm="fricas")

[Out]

Timed out

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giac [A]  time = 2.54, size = 451, normalized size = 1.31 \[ \frac {4 \, a^{2} b e^{\left (14 \, d x + 14 \, c\right )} - 13 \, a b^{2} e^{\left (14 \, d x + 14 \, c\right )} + 3 \, b^{3} e^{\left (14 \, d x + 14 \, c\right )} - 24 \, a^{2} b e^{\left (12 \, d x + 12 \, c\right )} + 99 \, a b^{2} e^{\left (12 \, d x + 12 \, c\right )} - 21 \, b^{3} e^{\left (12 \, d x + 12 \, c\right )} + 64 \, a^{3} e^{\left (10 \, d x + 10 \, c\right )} + 68 \, a^{2} b e^{\left (10 \, d x + 10 \, c\right )} - 225 \, a b^{2} e^{\left (10 \, d x + 10 \, c\right )} + 63 \, b^{3} e^{\left (10 \, d x + 10 \, c\right )} - 384 \, a^{3} e^{\left (8 \, d x + 8 \, c\right )} - 96 \, a^{2} b e^{\left (8 \, d x + 8 \, c\right )} + 183 \, a b^{2} e^{\left (8 \, d x + 8 \, c\right )} - 105 \, b^{3} e^{\left (8 \, d x + 8 \, c\right )} - 64 \, a^{3} e^{\left (6 \, d x + 6 \, c\right )} - 452 \, a^{2} b e^{\left (6 \, d x + 6 \, c\right )} + 9 \, a b^{2} e^{\left (6 \, d x + 6 \, c\right )} + 105 \, b^{3} e^{\left (6 \, d x + 6 \, c\right )} + 120 \, a^{2} b e^{\left (4 \, d x + 4 \, c\right )} - 87 \, a b^{2} e^{\left (4 \, d x + 4 \, c\right )} - 63 \, b^{3} e^{\left (4 \, d x + 4 \, c\right )} - 4 \, a^{2} b e^{\left (2 \, d x + 2 \, c\right )} + 37 \, a b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 21 \, b^{3} e^{\left (2 \, d x + 2 \, c\right )} - 3 \, a b^{2} - 3 \, b^{3}}{16 \, {\left (a^{3} b - 2 \, a^{2} b^{2} + a b^{3}\right )} {\left (b e^{\left (8 \, d x + 8 \, c\right )} - 4 \, b e^{\left (6 \, d x + 6 \, c\right )} - 16 \, a e^{\left (4 \, d x + 4 \, c\right )} + 6 \, b e^{\left (4 \, d x + 4 \, c\right )} - 4 \, b e^{\left (2 \, d x + 2 \, c\right )} + b\right )}^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(4*a^2*b*e^(14*d*x + 14*c) - 13*a*b^2*e^(14*d*x + 14*c) + 3*b^3*e^(14*d*x + 14*c) - 24*a^2*b*e^(12*d*x +
12*c) + 99*a*b^2*e^(12*d*x + 12*c) - 21*b^3*e^(12*d*x + 12*c) + 64*a^3*e^(10*d*x + 10*c) + 68*a^2*b*e^(10*d*x
+ 10*c) - 225*a*b^2*e^(10*d*x + 10*c) + 63*b^3*e^(10*d*x + 10*c) - 384*a^3*e^(8*d*x + 8*c) - 96*a^2*b*e^(8*d*x
+ 8*c) + 183*a*b^2*e^(8*d*x + 8*c) - 105*b^3*e^(8*d*x + 8*c) - 64*a^3*e^(6*d*x + 6*c) - 452*a^2*b*e^(6*d*x +
6*c) + 9*a*b^2*e^(6*d*x + 6*c) + 105*b^3*e^(6*d*x + 6*c) + 120*a^2*b*e^(4*d*x + 4*c) - 87*a*b^2*e^(4*d*x + 4*c
) - 63*b^3*e^(4*d*x + 4*c) - 4*a^2*b*e^(2*d*x + 2*c) + 37*a*b^2*e^(2*d*x + 2*c) + 21*b^3*e^(2*d*x + 2*c) - 3*a
*b^2 - 3*b^3)/((a^3*b - 2*a^2*b^2 + a*b^3)*(b*e^(8*d*x + 8*c) - 4*b*e^(6*d*x + 6*c) - 16*a*e^(4*d*x + 4*c) + 6
*b*e^(4*d*x + 4*c) - 4*b*e^(2*d*x + 2*c) + b)^2*d)

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maple [C]  time = 0.16, size = 2681, normalized size = 7.77 \[ \text {Expression too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5/8/d/(tanh(1/2*d*x+1/2*c)^8*a-4*tanh(1/2*d*x+1/2*c)^6*a+6*tanh(1/2*d*x+1/2*c)^4*a-16*b*tanh(1/2*d*x+1/2*c)^4
-4*tanh(1/2*d*x+1/2*c)^2*a+a)^2/b/(a^2-2*a*b+b^2)*tanh(1/2*d*x+1/2*c)^7*a^2-5/8/d/(tanh(1/2*d*x+1/2*c)^8*a-4*t
anh(1/2*d*x+1/2*c)^6*a+6*tanh(1/2*d*x+1/2*c)^4*a-16*b*tanh(1/2*d*x+1/2*c)^4-4*tanh(1/2*d*x+1/2*c)^2*a+a)^2/b/(
a^2-2*a*b+b^2)*tanh(1/2*d*x+1/2*c)^9*a^2+9/8/d/(tanh(1/2*d*x+1/2*c)^8*a-4*tanh(1/2*d*x+1/2*c)^6*a+6*tanh(1/2*d
*x+1/2*c)^4*a-16*b*tanh(1/2*d*x+1/2*c)^4-4*tanh(1/2*d*x+1/2*c)^2*a+a)^2/b/(a^2-2*a*b+b^2)*tanh(1/2*d*x+1/2*c)^
11*a^2-5/8/d/(tanh(1/2*d*x+1/2*c)^8*a-4*tanh(1/2*d*x+1/2*c)^6*a+6*tanh(1/2*d*x+1/2*c)^4*a-16*b*tanh(1/2*d*x+1/
2*c)^4-4*tanh(1/2*d*x+1/2*c)^2*a+a)^2*a^2/b/(a^2-2*a*b+b^2)*tanh(1/2*d*x+1/2*c)^13+1/8/d/(tanh(1/2*d*x+1/2*c)^
8*a-4*tanh(1/2*d*x+1/2*c)^6*a+6*tanh(1/2*d*x+1/2*c)^4*a-16*b*tanh(1/2*d*x+1/2*c)^4-4*tanh(1/2*d*x+1/2*c)^2*a+a
)^2*a^2/b/(a^2-2*a*b+b^2)*tanh(1/2*d*x+1/2*c)^15+1/8/d/(tanh(1/2*d*x+1/2*c)^8*a-4*tanh(1/2*d*x+1/2*c)^6*a+6*ta
nh(1/2*d*x+1/2*c)^4*a-16*b*tanh(1/2*d*x+1/2*c)^4-4*tanh(1/2*d*x+1/2*c)^2*a+a)^2*a^2/b/(a^2-2*a*b+b^2)*tanh(1/2
*d*x+1/2*c)-5/8/d/(tanh(1/2*d*x+1/2*c)^8*a-4*tanh(1/2*d*x+1/2*c)^6*a+6*tanh(1/2*d*x+1/2*c)^4*a-16*b*tanh(1/2*d
*x+1/2*c)^4-4*tanh(1/2*d*x+1/2*c)^2*a+a)^2*a^2/b/(a^2-2*a*b+b^2)*tanh(1/2*d*x+1/2*c)^3+9/8/d/(tanh(1/2*d*x+1/2
*c)^8*a-4*tanh(1/2*d*x+1/2*c)^6*a+6*tanh(1/2*d*x+1/2*c)^4*a-16*b*tanh(1/2*d*x+1/2*c)^4-4*tanh(1/2*d*x+1/2*c)^2
*a+a)^2/b/(a^2-2*a*b+b^2)*tanh(1/2*d*x+1/2*c)^5*a^2+12/d/(tanh(1/2*d*x+1/2*c)^8*a-4*tanh(1/2*d*x+1/2*c)^6*a+6*
tanh(1/2*d*x+1/2*c)^4*a-16*b*tanh(1/2*d*x+1/2*c)^4-4*tanh(1/2*d*x+1/2*c)^2*a+a)^2/a*b^2/(a^2-2*a*b+b^2)*tanh(1
/2*d*x+1/2*c)^7+12/d/(tanh(1/2*d*x+1/2*c)^8*a-4*tanh(1/2*d*x+1/2*c)^6*a+6*tanh(1/2*d*x+1/2*c)^4*a-16*b*tanh(1/
2*d*x+1/2*c)^4-4*tanh(1/2*d*x+1/2*c)^2*a+a)^2/a*b^2/(a^2-2*a*b+b^2)*tanh(1/2*d*x+1/2*c)^9-1/64/d/a/b/(a^2-2*a*
b+b^2)*sum((a*(-a-2*b)*_R^6+(-5*a^2+32*a*b-6*b^2)*_R^4+(5*a^2-32*a*b+6*b^2)*_R^2+a^2+2*a*b)/(_R^7*a-3*_R^5*a+3
*_R^3*a-8*_R^3*b-_R*a)*ln(tanh(1/2*d*x+1/2*c)-_R),_R=RootOf(a*_Z^8-4*a*_Z^6+(6*a-16*b)*_Z^4-4*a*_Z^2+a))+1/4/d
/(tanh(1/2*d*x+1/2*c)^8*a-4*tanh(1/2*d*x+1/2*c)^6*a+6*tanh(1/2*d*x+1/2*c)^4*a-16*b*tanh(1/2*d*x+1/2*c)^4-4*tan
h(1/2*d*x+1/2*c)^2*a+a)^2*b/(a^2-2*a*b+b^2)*tanh(1/2*d*x+1/2*c)^13+1/4/d/(tanh(1/2*d*x+1/2*c)^8*a-4*tanh(1/2*d
*x+1/2*c)^6*a+6*tanh(1/2*d*x+1/2*c)^4*a-16*b*tanh(1/2*d*x+1/2*c)^4-4*tanh(1/2*d*x+1/2*c)^2*a+a)^2*b/(a^2-2*a*b
+b^2)*tanh(1/2*d*x+1/2*c)^3+1/4/d/(tanh(1/2*d*x+1/2*c)^8*a-4*tanh(1/2*d*x+1/2*c)^6*a+6*tanh(1/2*d*x+1/2*c)^4*a
-16*b*tanh(1/2*d*x+1/2*c)^4-4*tanh(1/2*d*x+1/2*c)^2*a+a)^2*a/(a^2-2*a*b+b^2)*tanh(1/2*d*x+1/2*c)-3/d/(tanh(1/2
*d*x+1/2*c)^8*a-4*tanh(1/2*d*x+1/2*c)^6*a+6*tanh(1/2*d*x+1/2*c)^4*a-16*b*tanh(1/2*d*x+1/2*c)^4-4*tanh(1/2*d*x+
1/2*c)^2*a+a)^2*a/(a^2-2*a*b+b^2)*tanh(1/2*d*x+1/2*c)^3+19/2/d/(tanh(1/2*d*x+1/2*c)^8*a-4*tanh(1/2*d*x+1/2*c)^
6*a+6*tanh(1/2*d*x+1/2*c)^4*a-16*b*tanh(1/2*d*x+1/2*c)^4-4*tanh(1/2*d*x+1/2*c)^2*a+a)^2/(a^2-2*a*b+b^2)*tanh(1
/2*d*x+1/2*c)^5*a-35/4/d/(tanh(1/2*d*x+1/2*c)^8*a-4*tanh(1/2*d*x+1/2*c)^6*a+6*tanh(1/2*d*x+1/2*c)^4*a-16*b*tan
h(1/2*d*x+1/2*c)^4-4*tanh(1/2*d*x+1/2*c)^2*a+a)^2*b/(a^2-2*a*b+b^2)*tanh(1/2*d*x+1/2*c)^5-27/4/d/(tanh(1/2*d*x
+1/2*c)^8*a-4*tanh(1/2*d*x+1/2*c)^6*a+6*tanh(1/2*d*x+1/2*c)^4*a-16*b*tanh(1/2*d*x+1/2*c)^4-4*tanh(1/2*d*x+1/2*
c)^2*a+a)^2/(a^2-2*a*b+b^2)*tanh(1/2*d*x+1/2*c)^7*a+41/2/d/(tanh(1/2*d*x+1/2*c)^8*a-4*tanh(1/2*d*x+1/2*c)^6*a+
6*tanh(1/2*d*x+1/2*c)^4*a-16*b*tanh(1/2*d*x+1/2*c)^4-4*tanh(1/2*d*x+1/2*c)^2*a+a)^2*b/(a^2-2*a*b+b^2)*tanh(1/2
*d*x+1/2*c)^7-27/4/d/(tanh(1/2*d*x+1/2*c)^8*a-4*tanh(1/2*d*x+1/2*c)^6*a+6*tanh(1/2*d*x+1/2*c)^4*a-16*b*tanh(1/
2*d*x+1/2*c)^4-4*tanh(1/2*d*x+1/2*c)^2*a+a)^2/(a^2-2*a*b+b^2)*tanh(1/2*d*x+1/2*c)^9*a+41/2/d/(tanh(1/2*d*x+1/2
*c)^8*a-4*tanh(1/2*d*x+1/2*c)^6*a+6*tanh(1/2*d*x+1/2*c)^4*a-16*b*tanh(1/2*d*x+1/2*c)^4-4*tanh(1/2*d*x+1/2*c)^2
*a+a)^2*b/(a^2-2*a*b+b^2)*tanh(1/2*d*x+1/2*c)^9+19/2/d/(tanh(1/2*d*x+1/2*c)^8*a-4*tanh(1/2*d*x+1/2*c)^6*a+6*ta
nh(1/2*d*x+1/2*c)^4*a-16*b*tanh(1/2*d*x+1/2*c)^4-4*tanh(1/2*d*x+1/2*c)^2*a+a)^2/(a^2-2*a*b+b^2)*tanh(1/2*d*x+1
/2*c)^11*a-35/4/d/(tanh(1/2*d*x+1/2*c)^8*a-4*tanh(1/2*d*x+1/2*c)^6*a+6*tanh(1/2*d*x+1/2*c)^4*a-16*b*tanh(1/2*d
*x+1/2*c)^4-4*tanh(1/2*d*x+1/2*c)^2*a+a)^2*b/(a^2-2*a*b+b^2)*tanh(1/2*d*x+1/2*c)^11-3/d/(tanh(1/2*d*x+1/2*c)^8
*a-4*tanh(1/2*d*x+1/2*c)^6*a+6*tanh(1/2*d*x+1/2*c)^4*a-16*b*tanh(1/2*d*x+1/2*c)^4-4*tanh(1/2*d*x+1/2*c)^2*a+a)
^2*a/(a^2-2*a*b+b^2)*tanh(1/2*d*x+1/2*c)^13+1/4/d/(tanh(1/2*d*x+1/2*c)^8*a-4*tanh(1/2*d*x+1/2*c)^6*a+6*tanh(1/
2*d*x+1/2*c)^4*a-16*b*tanh(1/2*d*x+1/2*c)^4-4*tanh(1/2*d*x+1/2*c)^2*a+a)^2*a/(a^2-2*a*b+b^2)*tanh(1/2*d*x+1/2*
c)^15

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(3*a*b^2 + 3*b^3 - (4*a^2*b*e^(14*c) - 13*a*b^2*e^(14*c) + 3*b^3*e^(14*c))*e^(14*d*x) + 3*(8*a^2*b*e^(12
*c) - 33*a*b^2*e^(12*c) + 7*b^3*e^(12*c))*e^(12*d*x) - (64*a^3*e^(10*c) + 68*a^2*b*e^(10*c) - 225*a*b^2*e^(10*
c) + 63*b^3*e^(10*c))*e^(10*d*x) + 3*(128*a^3*e^(8*c) + 32*a^2*b*e^(8*c) - 61*a*b^2*e^(8*c) + 35*b^3*e^(8*c))*
e^(8*d*x) + (64*a^3*e^(6*c) + 452*a^2*b*e^(6*c) - 9*a*b^2*e^(6*c) - 105*b^3*e^(6*c))*e^(6*d*x) - 3*(40*a^2*b*e
^(4*c) - 29*a*b^2*e^(4*c) - 21*b^3*e^(4*c))*e^(4*d*x) + (4*a^2*b*e^(2*c) - 37*a*b^2*e^(2*c) - 21*b^3*e^(2*c))*
e^(2*d*x))/(a^3*b^3*d - 2*a^2*b^4*d + a*b^5*d + (a^3*b^3*d*e^(16*c) - 2*a^2*b^4*d*e^(16*c) + a*b^5*d*e^(16*c))
*e^(16*d*x) - 8*(a^3*b^3*d*e^(14*c) - 2*a^2*b^4*d*e^(14*c) + a*b^5*d*e^(14*c))*e^(14*d*x) - 4*(8*a^4*b^2*d*e^(
12*c) - 23*a^3*b^3*d*e^(12*c) + 22*a^2*b^4*d*e^(12*c) - 7*a*b^5*d*e^(12*c))*e^(12*d*x) + 8*(16*a^4*b^2*d*e^(10
*c) - 39*a^3*b^3*d*e^(10*c) + 30*a^2*b^4*d*e^(10*c) - 7*a*b^5*d*e^(10*c))*e^(10*d*x) + 2*(128*a^5*b*d*e^(8*c)
- 352*a^4*b^2*d*e^(8*c) + 355*a^3*b^3*d*e^(8*c) - 166*a^2*b^4*d*e^(8*c) + 35*a*b^5*d*e^(8*c))*e^(8*d*x) + 8*(1
6*a^4*b^2*d*e^(6*c) - 39*a^3*b^3*d*e^(6*c) + 30*a^2*b^4*d*e^(6*c) - 7*a*b^5*d*e^(6*c))*e^(6*d*x) - 4*(8*a^4*b^
2*d*e^(4*c) - 23*a^3*b^3*d*e^(4*c) + 22*a^2*b^4*d*e^(4*c) - 7*a*b^5*d*e^(4*c))*e^(4*d*x) - 8*(a^3*b^3*d*e^(2*c
) - 2*a^2*b^4*d*e^(2*c) + a*b^5*d*e^(2*c))*e^(2*d*x)) + 1/64*integrate(8*((4*a^2*e^(6*c) - 13*a*b*e^(6*c) + 3*
b^2*e^(6*c))*e^(6*d*x) + 6*(7*a*b*e^(4*c) - b^2*e^(4*c))*e^(4*d*x) + (4*a^2*e^(2*c) - 13*a*b*e^(2*c) + 3*b^2*e
^(2*c))*e^(2*d*x))/(a^3*b^2 - 2*a^2*b^3 + a*b^4 + (a^3*b^2*e^(8*c) - 2*a^2*b^3*e^(8*c) + a*b^4*e^(8*c))*e^(8*d
*x) - 4*(a^3*b^2*e^(6*c) - 2*a^2*b^3*e^(6*c) + a*b^4*e^(6*c))*e^(6*d*x) - 2*(8*a^4*b*e^(4*c) - 19*a^3*b^2*e^(4
*c) + 14*a^2*b^3*e^(4*c) - 3*a*b^4*e^(4*c))*e^(4*d*x) - 4*(a^3*b^2*e^(2*c) - 2*a^2*b^3*e^(2*c) + a*b^4*e^(2*c)
)*e^(2*d*x)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\mathrm {sinh}\left (c+d\,x\right )}^6}{{\left (a-b\,{\mathrm {sinh}\left (c+d\,x\right )}^4\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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