3.80 \(\int \frac {\sinh ^4(x)}{a+b \tanh (x)} \, dx\)

Optimal. Leaf size=147 \[ -\frac {\cosh ^4(x) (b-a \tanh (x))}{4 \left (a^2-b^2\right )}+\frac {\cosh ^2(x) \left (4 b \left (2 a^2-b^2\right )-a \left (5 a^2-b^2\right ) \tanh (x)\right )}{8 \left (a^2-b^2\right )^2}-\frac {a^4 b \log (a+b \tanh (x))}{\left (a^2-b^2\right )^3}-\frac {a (3 a+b) \log (1-\tanh (x))}{16 (a+b)^3}+\frac {a (3 a-b) \log (\tanh (x)+1)}{16 (a-b)^3} \]

[Out]

-1/16*a*(3*a+b)*ln(1-tanh(x))/(a+b)^3+1/16*a*(3*a-b)*ln(1+tanh(x))/(a-b)^3-a^4*b*ln(a+b*tanh(x))/(a^2-b^2)^3-1
/4*cosh(x)^4*(b-a*tanh(x))/(a^2-b^2)+1/8*cosh(x)^2*(4*b*(2*a^2-b^2)-a*(5*a^2-b^2)*tanh(x))/(a^2-b^2)^2

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Rubi [A]  time = 0.35, antiderivative size = 147, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {3516, 1647, 801} \[ -\frac {a^4 b \log (a+b \tanh (x))}{\left (a^2-b^2\right )^3}-\frac {\cosh ^4(x) (b-a \tanh (x))}{4 \left (a^2-b^2\right )}+\frac {\cosh ^2(x) \left (4 b \left (2 a^2-b^2\right )-a \left (5 a^2-b^2\right ) \tanh (x)\right )}{8 \left (a^2-b^2\right )^2}-\frac {a (3 a+b) \log (1-\tanh (x))}{16 (a+b)^3}+\frac {a (3 a-b) \log (\tanh (x)+1)}{16 (a-b)^3} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[x]^4/(a + b*Tanh[x]),x]

[Out]

-(a*(3*a + b)*Log[1 - Tanh[x]])/(16*(a + b)^3) + (a*(3*a - b)*Log[1 + Tanh[x]])/(16*(a - b)^3) - (a^4*b*Log[a
+ b*Tanh[x]])/(a^2 - b^2)^3 - (Cosh[x]^4*(b - a*Tanh[x]))/(4*(a^2 - b^2)) + (Cosh[x]^2*(4*b*(2*a^2 - b^2) - a*
(5*a^2 - b^2)*Tanh[x]))/(8*(a^2 - b^2)^2)

Rule 801

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
(d + e*x)^m*(f + g*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

Rule 1647

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

Rule 3516

Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[b/f, Subst[Int
[(x^m*(a + x)^n)/(b^2 + x^2)^(m/2 + 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x] && IntegerQ[m/
2]

Rubi steps

\begin {align*} \int \frac {\sinh ^4(x)}{a+b \tanh (x)} \, dx &=-\left (b \operatorname {Subst}\left (\int \frac {x^4}{(a+x) \left (-b^2+x^2\right )^3} \, dx,x,b \tanh (x)\right )\right )\\ &=-\frac {\cosh ^4(x) (b-a \tanh (x))}{4 \left (a^2-b^2\right )}-\frac {\operatorname {Subst}\left (\int \frac {\frac {a^2 b^4}{a^2-b^2}-\frac {3 a b^4 x}{a^2-b^2}+4 b^2 x^2}{(a+x) \left (-b^2+x^2\right )^2} \, dx,x,b \tanh (x)\right )}{4 b}\\ &=-\frac {\cosh ^4(x) (b-a \tanh (x))}{4 \left (a^2-b^2\right )}+\frac {\cosh ^2(x) \left (4 b \left (2 a^2-b^2\right )-a \left (5 a^2-b^2\right ) \tanh (x)\right )}{8 \left (a^2-b^2\right )^2}-\frac {\operatorname {Subst}\left (\int \frac {\frac {a^2 b^4 \left (3 a^2+b^2\right )}{\left (a^2-b^2\right )^2}-\frac {a b^4 \left (5 a^2-b^2\right ) x}{\left (a^2-b^2\right )^2}}{(a+x) \left (-b^2+x^2\right )} \, dx,x,b \tanh (x)\right )}{8 b^3}\\ &=-\frac {\cosh ^4(x) (b-a \tanh (x))}{4 \left (a^2-b^2\right )}+\frac {\cosh ^2(x) \left (4 b \left (2 a^2-b^2\right )-a \left (5 a^2-b^2\right ) \tanh (x)\right )}{8 \left (a^2-b^2\right )^2}-\frac {\operatorname {Subst}\left (\int \left (-\frac {a b^3 (3 a+b)}{2 (a+b)^3 (b-x)}+\frac {8 a^4 b^4}{(a-b)^3 (a+b)^3 (a+x)}-\frac {a (3 a-b) b^3}{2 (a-b)^3 (b+x)}\right ) \, dx,x,b \tanh (x)\right )}{8 b^3}\\ &=-\frac {a (3 a+b) \log (1-\tanh (x))}{16 (a+b)^3}+\frac {a (3 a-b) \log (1+\tanh (x))}{16 (a-b)^3}-\frac {a^4 b \log (a+b \tanh (x))}{\left (a^2-b^2\right )^3}-\frac {\cosh ^4(x) (b-a \tanh (x))}{4 \left (a^2-b^2\right )}+\frac {\cosh ^2(x) \left (4 b \left (2 a^2-b^2\right )-a \left (5 a^2-b^2\right ) \tanh (x)\right )}{8 \left (a^2-b^2\right )^2}\\ \end {align*}

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Mathematica [A]  time = 0.60, size = 144, normalized size = 0.98 \[ \frac {12 a^5 x+a^5 \sinh (4 x)-32 a^4 b \log (a \cosh (x)+b \sinh (x))+24 a^3 b^2 x-2 a^3 b^2 \sinh (4 x)-b \left (a^2-b^2\right )^2 \cosh (4 x)+4 b \left (3 a^4-4 a^2 b^2+b^4\right ) \cosh (2 x)-8 a^3 \left (a^2-b^2\right ) \sinh (2 x)-4 a b^4 x+a b^4 \sinh (4 x)}{32 (a-b)^3 (a+b)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[x]^4/(a + b*Tanh[x]),x]

[Out]

(12*a^5*x + 24*a^3*b^2*x - 4*a*b^4*x + 4*b*(3*a^4 - 4*a^2*b^2 + b^4)*Cosh[2*x] - b*(a^2 - b^2)^2*Cosh[4*x] - 3
2*a^4*b*Log[a*Cosh[x] + b*Sinh[x]] - 8*a^3*(a^2 - b^2)*Sinh[2*x] + a^5*Sinh[4*x] - 2*a^3*b^2*Sinh[4*x] + a*b^4
*Sinh[4*x])/(32*(a - b)^3*(a + b)^3)

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fricas [B]  time = 0.55, size = 1226, normalized size = 8.34 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)^4/(a+b*tanh(x)),x, algorithm="fricas")

[Out]

1/64*((a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5)*cosh(x)^8 + 8*(a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3 +
a*b^4 - b^5)*cosh(x)*sinh(x)^7 + (a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5)*sinh(x)^8 - 4*(2*a^5 - 3
*a^4*b - 2*a^3*b^2 + 4*a^2*b^3 - b^5)*cosh(x)^6 - 4*(2*a^5 - 3*a^4*b - 2*a^3*b^2 + 4*a^2*b^3 - b^5 - 7*(a^5 -
a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5)*cosh(x)^2)*sinh(x)^6 + 8*(3*a^5 + 8*a^4*b + 6*a^3*b^2 - a*b^4)*x*
cosh(x)^4 + 8*(7*(a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5)*cosh(x)^3 - 3*(2*a^5 - 3*a^4*b - 2*a^3*b^
2 + 4*a^2*b^3 - b^5)*cosh(x))*sinh(x)^5 - a^5 - a^4*b + 2*a^3*b^2 + 2*a^2*b^3 - a*b^4 - b^5 + 2*(35*(a^5 - a^4
*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5)*cosh(x)^4 - 30*(2*a^5 - 3*a^4*b - 2*a^3*b^2 + 4*a^2*b^3 - b^5)*cosh(
x)^2 + 4*(3*a^5 + 8*a^4*b + 6*a^3*b^2 - a*b^4)*x)*sinh(x)^4 + 8*(7*(a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^
4 - b^5)*cosh(x)^5 - 10*(2*a^5 - 3*a^4*b - 2*a^3*b^2 + 4*a^2*b^3 - b^5)*cosh(x)^3 + 4*(3*a^5 + 8*a^4*b + 6*a^3
*b^2 - a*b^4)*x*cosh(x))*sinh(x)^3 + 4*(2*a^5 + 3*a^4*b - 2*a^3*b^2 - 4*a^2*b^3 + b^5)*cosh(x)^2 + 4*(7*(a^5 -
a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5)*cosh(x)^6 + 2*a^5 + 3*a^4*b - 2*a^3*b^2 - 4*a^2*b^3 + b^5 - 15*(
2*a^5 - 3*a^4*b - 2*a^3*b^2 + 4*a^2*b^3 - b^5)*cosh(x)^4 + 12*(3*a^5 + 8*a^4*b + 6*a^3*b^2 - a*b^4)*x*cosh(x)^
2)*sinh(x)^2 - 64*(a^4*b*cosh(x)^4 + 4*a^4*b*cosh(x)^3*sinh(x) + 6*a^4*b*cosh(x)^2*sinh(x)^2 + 4*a^4*b*cosh(x)
*sinh(x)^3 + a^4*b*sinh(x)^4)*log(2*(a*cosh(x) + b*sinh(x))/(cosh(x) - sinh(x))) + 8*((a^5 - a^4*b - 2*a^3*b^2
+ 2*a^2*b^3 + a*b^4 - b^5)*cosh(x)^7 - 3*(2*a^5 - 3*a^4*b - 2*a^3*b^2 + 4*a^2*b^3 - b^5)*cosh(x)^5 + 4*(3*a^5
+ 8*a^4*b + 6*a^3*b^2 - a*b^4)*x*cosh(x)^3 + (2*a^5 + 3*a^4*b - 2*a^3*b^2 - 4*a^2*b^3 + b^5)*cosh(x))*sinh(x)
)/((a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*cosh(x)^4 + 4*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*cosh(x)^3*sinh(x) + 6
*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*cosh(x)^2*sinh(x)^2 + 4*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*cosh(x)*sinh(
x)^3 + (a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*sinh(x)^4)

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giac [A]  time = 0.14, size = 214, normalized size = 1.46 \[ -\frac {a^{4} b \log \left ({\left | a e^{\left (2 \, x\right )} + b e^{\left (2 \, x\right )} + a - b \right |}\right )}{a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}} + \frac {{\left (3 \, a^{2} - a b\right )} x}{8 \, {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )}} - \frac {{\left (18 \, a^{2} e^{\left (4 \, x\right )} - 6 \, a b e^{\left (4 \, x\right )} - 8 \, a^{2} e^{\left (2 \, x\right )} + 12 \, a b e^{\left (2 \, x\right )} - 4 \, b^{2} e^{\left (2 \, x\right )} + a^{2} - 2 \, a b + b^{2}\right )} e^{\left (-4 \, x\right )}}{64 \, {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )}} + \frac {a e^{\left (4 \, x\right )} + b e^{\left (4 \, x\right )} - 8 \, a e^{\left (2 \, x\right )} - 4 \, b e^{\left (2 \, x\right )}}{64 \, {\left (a^{2} + 2 \, a b + b^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)^4/(a+b*tanh(x)),x, algorithm="giac")

[Out]

-a^4*b*log(abs(a*e^(2*x) + b*e^(2*x) + a - b))/(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6) + 1/8*(3*a^2 - a*b)*x/(a^3
- 3*a^2*b + 3*a*b^2 - b^3) - 1/64*(18*a^2*e^(4*x) - 6*a*b*e^(4*x) - 8*a^2*e^(2*x) + 12*a*b*e^(2*x) - 4*b^2*e^(
2*x) + a^2 - 2*a*b + b^2)*e^(-4*x)/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) + 1/64*(a*e^(4*x) + b*e^(4*x) - 8*a*e^(2*x)
- 4*b*e^(2*x))/(a^2 + 2*a*b + b^2)

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maple [B]  time = 0.12, size = 320, normalized size = 2.18 \[ -\frac {a^{4} b \ln \left (a \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )+2 \tanh \left (\frac {x}{2}\right ) b +a \right )}{\left (a -b \right )^{3} \left (a +b \right )^{3}}+\frac {8}{\left (32 a +32 b \right ) \left (\tanh \left (\frac {x}{2}\right )-1\right )^{4}}+\frac {32}{\left (64 a +64 b \right ) \left (\tanh \left (\frac {x}{2}\right )-1\right )^{3}}-\frac {a}{8 \left (a +b \right )^{2} \left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}+\frac {b}{8 \left (a +b \right )^{2} \left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}-\frac {3 a}{8 \left (a +b \right )^{2} \left (\tanh \left (\frac {x}{2}\right )-1\right )}-\frac {b}{8 \left (a +b \right )^{2} \left (\tanh \left (\frac {x}{2}\right )-1\right )}-\frac {3 a^{2} \ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{8 \left (a +b \right )^{3}}-\frac {a \ln \left (\tanh \left (\frac {x}{2}\right )-1\right ) b}{8 \left (a +b \right )^{3}}-\frac {8}{\left (32 a -32 b \right ) \left (\tanh \left (\frac {x}{2}\right )+1\right )^{4}}+\frac {32}{\left (64 a -64 b \right ) \left (\tanh \left (\frac {x}{2}\right )+1\right )^{3}}+\frac {a}{8 \left (a -b \right )^{2} \left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}+\frac {b}{8 \left (a -b \right )^{2} \left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}-\frac {3 a}{8 \left (a -b \right )^{2} \left (\tanh \left (\frac {x}{2}\right )+1\right )}+\frac {b}{8 \left (a -b \right )^{2} \left (\tanh \left (\frac {x}{2}\right )+1\right )}+\frac {3 a^{2} \ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{8 \left (a -b \right )^{3}}-\frac {a \ln \left (\tanh \left (\frac {x}{2}\right )+1\right ) b}{8 \left (a -b \right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(x)^4/(a+b*tanh(x)),x)

[Out]

-a^4*b/(a-b)^3/(a+b)^3*ln(a*tanh(1/2*x)^2+2*tanh(1/2*x)*b+a)+8/(32*a+32*b)/(tanh(1/2*x)-1)^4+32/(64*a+64*b)/(t
anh(1/2*x)-1)^3-1/8/(a+b)^2/(tanh(1/2*x)-1)^2*a+1/8/(a+b)^2/(tanh(1/2*x)-1)^2*b-3/8/(a+b)^2/(tanh(1/2*x)-1)*a-
1/8/(a+b)^2/(tanh(1/2*x)-1)*b-3/8*a^2/(a+b)^3*ln(tanh(1/2*x)-1)-1/8*a/(a+b)^3*ln(tanh(1/2*x)-1)*b-8/(32*a-32*b
)/(tanh(1/2*x)+1)^4+32/(64*a-64*b)/(tanh(1/2*x)+1)^3+1/8/(a-b)^2/(tanh(1/2*x)+1)^2*a+1/8/(a-b)^2/(tanh(1/2*x)+
1)^2*b-3/8/(a-b)^2/(tanh(1/2*x)+1)*a+1/8/(a-b)^2/(tanh(1/2*x)+1)*b+3/8*a^2/(a-b)^3*ln(tanh(1/2*x)+1)-1/8*a/(a-
b)^3*ln(tanh(1/2*x)+1)*b

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maxima [A]  time = 0.33, size = 163, normalized size = 1.11 \[ -\frac {a^{4} b \log \left (-{\left (a - b\right )} e^{\left (-2 \, x\right )} - a - b\right )}{a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}} + \frac {{\left (3 \, a^{2} + a b\right )} x}{8 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )}} - \frac {{\left (4 \, {\left (2 \, a + b\right )} e^{\left (-2 \, x\right )} - a - b\right )} e^{\left (4 \, x\right )}}{64 \, {\left (a^{2} + 2 \, a b + b^{2}\right )}} + \frac {4 \, {\left (2 \, a - b\right )} e^{\left (-2 \, x\right )} - {\left (a - b\right )} e^{\left (-4 \, x\right )}}{64 \, {\left (a^{2} - 2 \, a b + b^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)^4/(a+b*tanh(x)),x, algorithm="maxima")

[Out]

-a^4*b*log(-(a - b)*e^(-2*x) - a - b)/(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6) + 1/8*(3*a^2 + a*b)*x/(a^3 + 3*a^2*b
+ 3*a*b^2 + b^3) - 1/64*(4*(2*a + b)*e^(-2*x) - a - b)*e^(4*x)/(a^2 + 2*a*b + b^2) + 1/64*(4*(2*a - b)*e^(-2*
x) - (a - b)*e^(-4*x))/(a^2 - 2*a*b + b^2)

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mupad [B]  time = 1.67, size = 135, normalized size = 0.92 \[ \frac {{\mathrm {e}}^{4\,x}}{64\,a+64\,b}-\frac {{\mathrm {e}}^{-4\,x}}{64\,a-64\,b}+\frac {{\mathrm {e}}^{-2\,x}\,\left (2\,a-b\right )}{16\,{\left (a-b\right )}^2}-\frac {{\mathrm {e}}^{2\,x}\,\left (2\,a+b\right )}{16\,{\left (a+b\right )}^2}-\frac {a^4\,b\,\ln \left (a-b+a\,{\mathrm {e}}^{2\,x}+b\,{\mathrm {e}}^{2\,x}\right )}{a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6}+\frac {a\,x\,\left (3\,a-b\right )}{8\,{\left (a-b\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(x)^4/(a + b*tanh(x)),x)

[Out]

exp(4*x)/(64*a + 64*b) - exp(-4*x)/(64*a - 64*b) + (exp(-2*x)*(2*a - b))/(16*(a - b)^2) - (exp(2*x)*(2*a + b))
/(16*(a + b)^2) - (a^4*b*log(a - b + a*exp(2*x) + b*exp(2*x)))/(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2) + (a*x*(3*a
- b))/(8*(a - b)^3)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sinh ^{4}{\relax (x )}}{a + b \tanh {\relax (x )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)**4/(a+b*tanh(x)),x)

[Out]

Integral(sinh(x)**4/(a + b*tanh(x)), x)

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