3.1.48 \(\int \frac {\text {csch}^4(c+d x)}{(a+b \tanh ^2(c+d x))^3} \, dx\) [48]
Optimal. Leaf size=151 \[ \frac {5 \sqrt {b} (3 a+7 b) \text {ArcTan}\left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a}}\right )}{8 a^{9/2} d}+\frac {(a+3 b) \coth (c+d x)}{a^4 d}-\frac {\coth ^3(c+d x)}{3 a^3 d}+\frac {b (a+b) \tanh (c+d x)}{4 a^3 d \left (a+b \tanh ^2(c+d x)\right )^2}+\frac {b (7 a+11 b) \tanh (c+d x)}{8 a^4 d \left (a+b \tanh ^2(c+d x)\right )} \]
[Out]
(a+3*b)*coth(d*x+c)/a^4/d-1/3*coth(d*x+c)^3/a^3/d+5/8*(3*a+7*b)*arctan(b^(1/2)*tanh(d*x+c)/a^(1/2))*b^(1/2)/a^
(9/2)/d+1/4*b*(a+b)*tanh(d*x+c)/a^3/d/(a+b*tanh(d*x+c)^2)^2+1/8*b*(7*a+11*b)*tanh(d*x+c)/a^4/d/(a+b*tanh(d*x+c
)^2)
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Rubi [A]
time = 0.15, antiderivative size = 151, normalized size of antiderivative = 1.00, number of steps
used = 6, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3744, 467,
1273, 1275, 211} \begin {gather*} \frac {5 \sqrt {b} (3 a+7 b) \text {ArcTan}\left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a}}\right )}{8 a^{9/2} d}+\frac {b (7 a+11 b) \tanh (c+d x)}{8 a^4 d \left (a+b \tanh ^2(c+d x)\right )}+\frac {(a+3 b) \coth (c+d x)}{a^4 d}+\frac {b (a+b) \tanh (c+d x)}{4 a^3 d \left (a+b \tanh ^2(c+d x)\right )^2}-\frac {\coth ^3(c+d x)}{3 a^3 d} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[Csch[c + d*x]^4/(a + b*Tanh[c + d*x]^2)^3,x]
[Out]
(5*Sqrt[b]*(3*a + 7*b)*ArcTan[(Sqrt[b]*Tanh[c + d*x])/Sqrt[a]])/(8*a^(9/2)*d) + ((a + 3*b)*Coth[c + d*x])/(a^4
*d) - Coth[c + d*x]^3/(3*a^3*d) + (b*(a + b)*Tanh[c + d*x])/(4*a^3*d*(a + b*Tanh[c + d*x]^2)^2) + (b*(7*a + 11
*b)*Tanh[c + d*x])/(8*a^4*d*(a + b*Tanh[c + d*x]^2))
Rule 211
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]
Rule 467
Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[(-a)^(m/2 - 1)*(b*c - a*d)*x
*((a + b*x^2)^(p + 1)/(2*b^(m/2 + 1)*(p + 1))), x] + Dist[1/(2*b^(m/2 + 1)*(p + 1)), Int[x^m*(a + b*x^2)^(p +
1)*ExpandToSum[2*b*(p + 1)*Together[(b^(m/2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d)*x^(-m + 2))/(a + b*x^2)]
- ((-a)^(m/2 - 1)*(b*c - a*d))/x^m, x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &
& ILtQ[m/2, 0] && (IntegerQ[p] || EqQ[m + 2*p + 1, 0])
Rule 1273
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/(d + e*x^2))*(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))], 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 1275
Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> In
t[ExpandIntegrand[(f*x)^m*(d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] &&
NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && IGtQ[q, -2]
Rule 3744
Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol] :> With[
{ff = FreeFactors[Tan[e + f*x], x]}, Dist[c*(ff^(m + 1)/f), Subst[Int[x^m*((a + b*(ff*x)^n)^p/(c^2 + ff^2*x^2)
^(m/2 + 1)), x], x, c*(Tan[e + f*x]/ff)], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[m/2]
Rubi steps
\begin {align*} \int \frac {\text {csch}^4(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx &=\frac {\text {Subst}\left (\int \frac {1-x^2}{x^4 \left (a+b x^2\right )^3} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac {b (a+b) \tanh (c+d x)}{4 a^3 d \left (a+b \tanh ^2(c+d x)\right )^2}-\frac {b \text {Subst}\left (\int \frac {-\frac {4}{a b}+\frac {4 (a+b) x^2}{a^2 b}-\frac {3 (a+b) x^4}{a^3}}{x^4 \left (a+b x^2\right )^2} \, dx,x,\tanh (c+d x)\right )}{4 d}\\ &=\frac {b (a+b) \tanh (c+d x)}{4 a^3 d \left (a+b \tanh ^2(c+d x)\right )^2}+\frac {b (7 a+11 b) \tanh (c+d x)}{8 a^4 d \left (a+b \tanh ^2(c+d x)\right )}-\frac {\text {Subst}\left (\int \frac {-8 a b+8 b (a+2 b) x^2-\frac {b^2 (7 a+11 b) x^4}{a}}{x^4 \left (a+b x^2\right )} \, dx,x,\tanh (c+d x)\right )}{8 a^3 b d}\\ &=\frac {b (a+b) \tanh (c+d x)}{4 a^3 d \left (a+b \tanh ^2(c+d x)\right )^2}+\frac {b (7 a+11 b) \tanh (c+d x)}{8 a^4 d \left (a+b \tanh ^2(c+d x)\right )}-\frac {\text {Subst}\left (\int \left (-\frac {8 b}{x^4}+\frac {8 b (a+3 b)}{a x^2}-\frac {5 b^2 (3 a+7 b)}{a \left (a+b x^2\right )}\right ) \, dx,x,\tanh (c+d x)\right )}{8 a^3 b d}\\ &=\frac {(a+3 b) \coth (c+d x)}{a^4 d}-\frac {\coth ^3(c+d x)}{3 a^3 d}+\frac {b (a+b) \tanh (c+d x)}{4 a^3 d \left (a+b \tanh ^2(c+d x)\right )^2}+\frac {b (7 a+11 b) \tanh (c+d x)}{8 a^4 d \left (a+b \tanh ^2(c+d x)\right )}+\frac {(5 b (3 a+7 b)) \text {Subst}\left (\int \frac {1}{a+b x^2} \, dx,x,\tanh (c+d x)\right )}{8 a^4 d}\\ &=\frac {5 \sqrt {b} (3 a+7 b) \tan ^{-1}\left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a}}\right )}{8 a^{9/2} d}+\frac {(a+3 b) \coth (c+d x)}{a^4 d}-\frac {\coth ^3(c+d x)}{3 a^3 d}+\frac {b (a+b) \tanh (c+d x)}{4 a^3 d \left (a+b \tanh ^2(c+d x)\right )^2}+\frac {b (7 a+11 b) \tanh (c+d x)}{8 a^4 d \left (a+b \tanh ^2(c+d x)\right )}\\ \end {align*}
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Mathematica [A]
time = 0.98, size = 149, normalized size = 0.99 \begin {gather*} \frac {15 \sqrt {b} (3 a+7 b) \text {ArcTan}\left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a}}\right )-8 \sqrt {a} \coth (c+d x) \left (-2 a-9 b+a \text {csch}^2(c+d x)\right )+\frac {3 \sqrt {a} b \left (9 a^2+6 a b-11 b^2+\left (9 a^2+20 a b+11 b^2\right ) \cosh (2 (c+d x))\right ) \sinh (2 (c+d x))}{(a-b+(a+b) \cosh (2 (c+d x)))^2}}{24 a^{9/2} d} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[Csch[c + d*x]^4/(a + b*Tanh[c + d*x]^2)^3,x]
[Out]
(15*Sqrt[b]*(3*a + 7*b)*ArcTan[(Sqrt[b]*Tanh[c + d*x])/Sqrt[a]] - 8*Sqrt[a]*Coth[c + d*x]*(-2*a - 9*b + a*Csch
[c + d*x]^2) + (3*Sqrt[a]*b*(9*a^2 + 6*a*b - 11*b^2 + (9*a^2 + 20*a*b + 11*b^2)*Cosh[2*(c + d*x)])*Sinh[2*(c +
d*x)])/(a - b + (a + b)*Cosh[2*(c + d*x)])^2)/(24*a^(9/2)*d)
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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(396\) vs.
\(2(135)=270\).
time = 3.21, size = 397, normalized size = 2.63 Too large to display
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(csch(d*x+c)^4/(a+b*tanh(d*x+c)^2)^3,x,method=_RETURNVERBOSE)
[Out]
1/d*(-1/8/a^4*(1/3*a*tanh(1/2*d*x+1/2*c)^3-3*a*tanh(1/2*d*x+1/2*c)-12*b*tanh(1/2*d*x+1/2*c))-1/24/a^3/tanh(1/2
*d*x+1/2*c)^3-1/8/a^4*(-12*b-3*a)/tanh(1/2*d*x+1/2*c)-2*b/a^4*((-1/8*a*(9*a+13*b)*tanh(1/2*d*x+1/2*c)^7+(-27/8
*a^2-67/8*a*b-11/2*b^2)*tanh(1/2*d*x+1/2*c)^5+(-27/8*a^2-67/8*a*b-11/2*b^2)*tanh(1/2*d*x+1/2*c)^3+(-9/8*a^2-13
/8*a*b)*tanh(1/2*d*x+1/2*c))/(a*tanh(1/2*d*x+1/2*c)^4+2*a*tanh(1/2*d*x+1/2*c)^2+4*b*tanh(1/2*d*x+1/2*c)^2+a)^2
+1/8*(15*a+35*b)*a*(-1/2*(-a+(b*(a+b))^(1/2)-b)/a/(b*(a+b))^(1/2)/((2*(b*(a+b))^(1/2)-a-2*b)*a)^(1/2)*arctanh(
a*tanh(1/2*d*x+1/2*c)/((2*(b*(a+b))^(1/2)-a-2*b)*a)^(1/2))+1/2*(a+(b*(a+b))^(1/2)+b)/a/(b*(a+b))^(1/2)/((2*(b*
(a+b))^(1/2)+a+2*b)*a)^(1/2)*arctan(a*tanh(1/2*d*x+1/2*c)/((2*(b*(a+b))^(1/2)+a+2*b)*a)^(1/2)))))
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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 615 vs.
\(2 (135) = 270\).
time = 0.70, size = 615, normalized size = 4.07 \begin {gather*} \frac {16 \, a^{4} + 147 \, a^{3} b + 351 \, a^{2} b^{2} + 325 \, a b^{3} + 105 \, b^{4} + 2 \, {\left (8 \, a^{4} + 32 \, a^{3} b - 251 \, a^{2} b^{2} - 590 \, a b^{3} - 315 \, b^{4}\right )} e^{\left (-2 \, d x - 2 \, c\right )} - {\left (96 \, a^{4} + 313 \, a^{3} b + 19 \, a^{2} b^{2} - 1725 \, a b^{3} - 1575 \, b^{4}\right )} e^{\left (-4 \, d x - 4 \, c\right )} - 4 \, {\left (56 \, a^{4} + 80 \, a^{3} b - 65 \, a^{2} b^{2} + 400 \, a b^{3} + 525 \, b^{4}\right )} e^{\left (-6 \, d x - 6 \, c\right )} - {\left (176 \, a^{4} + 135 \, a^{3} b + 15 \, a^{2} b^{2} - 1375 \, a b^{3} - 1575 \, b^{4}\right )} e^{\left (-8 \, d x - 8 \, c\right )} - 6 \, {\left (8 \, a^{4} + 45 \, a^{2} b^{2} + 150 \, a b^{3} + 105 \, b^{4}\right )} e^{\left (-10 \, d x - 10 \, c\right )} + 15 \, {\left (3 \, a^{3} b + 13 \, a^{2} b^{2} + 17 \, a b^{3} + 7 \, b^{4}\right )} e^{\left (-12 \, d x - 12 \, c\right )}}{12 \, {\left (a^{7} + 3 \, a^{6} b + 3 \, a^{5} b^{2} + a^{4} b^{3} + {\left (a^{7} - 5 \, a^{6} b - 13 \, a^{5} b^{2} - 7 \, a^{4} b^{3}\right )} e^{\left (-2 \, d x - 2 \, c\right )} - {\left (3 \, a^{7} + a^{6} b - 23 \, a^{5} b^{2} - 21 \, a^{4} b^{3}\right )} e^{\left (-4 \, d x - 4 \, c\right )} - {\left (3 \, a^{7} - 7 \, a^{6} b + 25 \, a^{5} b^{2} + 35 \, a^{4} b^{3}\right )} e^{\left (-6 \, d x - 6 \, c\right )} + {\left (3 \, a^{7} - 7 \, a^{6} b + 25 \, a^{5} b^{2} + 35 \, a^{4} b^{3}\right )} e^{\left (-8 \, d x - 8 \, c\right )} + {\left (3 \, a^{7} + a^{6} b - 23 \, a^{5} b^{2} - 21 \, a^{4} b^{3}\right )} e^{\left (-10 \, d x - 10 \, c\right )} - {\left (a^{7} - 5 \, a^{6} b - 13 \, a^{5} b^{2} - 7 \, a^{4} b^{3}\right )} e^{\left (-12 \, d x - 12 \, c\right )} - {\left (a^{7} + 3 \, a^{6} b + 3 \, a^{5} b^{2} + a^{4} b^{3}\right )} e^{\left (-14 \, d x - 14 \, c\right )}\right )} d} - \frac {5 \, {\left (3 \, a b + 7 \, b^{2}\right )} \arctan \left (\frac {{\left (a + b\right )} e^{\left (-2 \, d x - 2 \, c\right )} + a - b}{2 \, \sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{4} d} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(csch(d*x+c)^4/(a+b*tanh(d*x+c)^2)^3,x, algorithm="maxima")
[Out]
1/12*(16*a^4 + 147*a^3*b + 351*a^2*b^2 + 325*a*b^3 + 105*b^4 + 2*(8*a^4 + 32*a^3*b - 251*a^2*b^2 - 590*a*b^3 -
315*b^4)*e^(-2*d*x - 2*c) - (96*a^4 + 313*a^3*b + 19*a^2*b^2 - 1725*a*b^3 - 1575*b^4)*e^(-4*d*x - 4*c) - 4*(5
6*a^4 + 80*a^3*b - 65*a^2*b^2 + 400*a*b^3 + 525*b^4)*e^(-6*d*x - 6*c) - (176*a^4 + 135*a^3*b + 15*a^2*b^2 - 13
75*a*b^3 - 1575*b^4)*e^(-8*d*x - 8*c) - 6*(8*a^4 + 45*a^2*b^2 + 150*a*b^3 + 105*b^4)*e^(-10*d*x - 10*c) + 15*(
3*a^3*b + 13*a^2*b^2 + 17*a*b^3 + 7*b^4)*e^(-12*d*x - 12*c))/((a^7 + 3*a^6*b + 3*a^5*b^2 + a^4*b^3 + (a^7 - 5*
a^6*b - 13*a^5*b^2 - 7*a^4*b^3)*e^(-2*d*x - 2*c) - (3*a^7 + a^6*b - 23*a^5*b^2 - 21*a^4*b^3)*e^(-4*d*x - 4*c)
- (3*a^7 - 7*a^6*b + 25*a^5*b^2 + 35*a^4*b^3)*e^(-6*d*x - 6*c) + (3*a^7 - 7*a^6*b + 25*a^5*b^2 + 35*a^4*b^3)*e
^(-8*d*x - 8*c) + (3*a^7 + a^6*b - 23*a^5*b^2 - 21*a^4*b^3)*e^(-10*d*x - 10*c) - (a^7 - 5*a^6*b - 13*a^5*b^2 -
7*a^4*b^3)*e^(-12*d*x - 12*c) - (a^7 + 3*a^6*b + 3*a^5*b^2 + a^4*b^3)*e^(-14*d*x - 14*c))*d) - 5/8*(3*a*b + 7
*b^2)*arctan(1/2*((a + b)*e^(-2*d*x - 2*c) + a - b)/sqrt(a*b))/(sqrt(a*b)*a^4*d)
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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 7006 vs.
\(2 (135) = 270\).
time = 0.48, size = 14334, normalized size = 94.93 \begin {gather*} \text {Too large to display} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(csch(d*x+c)^4/(a+b*tanh(d*x+c)^2)^3,x, algorithm="fricas")
[Out]
[1/48*(60*(3*a^3*b + 13*a^2*b^2 + 17*a*b^3 + 7*b^4)*cosh(d*x + c)^12 + 720*(3*a^3*b + 13*a^2*b^2 + 17*a*b^3 +
7*b^4)*cosh(d*x + c)*sinh(d*x + c)^11 + 60*(3*a^3*b + 13*a^2*b^2 + 17*a*b^3 + 7*b^4)*sinh(d*x + c)^12 - 24*(8*
a^4 + 45*a^2*b^2 + 150*a*b^3 + 105*b^4)*cosh(d*x + c)^10 - 24*(8*a^4 + 45*a^2*b^2 + 150*a*b^3 + 105*b^4 - 165*
(3*a^3*b + 13*a^2*b^2 + 17*a*b^3 + 7*b^4)*cosh(d*x + c)^2)*sinh(d*x + c)^10 + 240*(55*(3*a^3*b + 13*a^2*b^2 +
17*a*b^3 + 7*b^4)*cosh(d*x + c)^3 - (8*a^4 + 45*a^2*b^2 + 150*a*b^3 + 105*b^4)*cosh(d*x + c))*sinh(d*x + c)^9
- 4*(176*a^4 + 135*a^3*b + 15*a^2*b^2 - 1375*a*b^3 - 1575*b^4)*cosh(d*x + c)^8 + 4*(7425*(3*a^3*b + 13*a^2*b^2
+ 17*a*b^3 + 7*b^4)*cosh(d*x + c)^4 - 176*a^4 - 135*a^3*b - 15*a^2*b^2 + 1375*a*b^3 + 1575*b^4 - 270*(8*a^4 +
45*a^2*b^2 + 150*a*b^3 + 105*b^4)*cosh(d*x + c)^2)*sinh(d*x + c)^8 + 32*(1485*(3*a^3*b + 13*a^2*b^2 + 17*a*b^
3 + 7*b^4)*cosh(d*x + c)^5 - 90*(8*a^4 + 45*a^2*b^2 + 150*a*b^3 + 105*b^4)*cosh(d*x + c)^3 - (176*a^4 + 135*a^
3*b + 15*a^2*b^2 - 1375*a*b^3 - 1575*b^4)*cosh(d*x + c))*sinh(d*x + c)^7 - 16*(56*a^4 + 80*a^3*b - 65*a^2*b^2
+ 400*a*b^3 + 525*b^4)*cosh(d*x + c)^6 + 16*(3465*(3*a^3*b + 13*a^2*b^2 + 17*a*b^3 + 7*b^4)*cosh(d*x + c)^6 -
315*(8*a^4 + 45*a^2*b^2 + 150*a*b^3 + 105*b^4)*cosh(d*x + c)^4 - 56*a^4 - 80*a^3*b + 65*a^2*b^2 - 400*a*b^3 -
525*b^4 - 7*(176*a^4 + 135*a^3*b + 15*a^2*b^2 - 1375*a*b^3 - 1575*b^4)*cosh(d*x + c)^2)*sinh(d*x + c)^6 + 32*(
1485*(3*a^3*b + 13*a^2*b^2 + 17*a*b^3 + 7*b^4)*cosh(d*x + c)^7 - 189*(8*a^4 + 45*a^2*b^2 + 150*a*b^3 + 105*b^4
)*cosh(d*x + c)^5 - 7*(176*a^4 + 135*a^3*b + 15*a^2*b^2 - 1375*a*b^3 - 1575*b^4)*cosh(d*x + c)^3 - 3*(56*a^4 +
80*a^3*b - 65*a^2*b^2 + 400*a*b^3 + 525*b^4)*cosh(d*x + c))*sinh(d*x + c)^5 - 4*(96*a^4 + 313*a^3*b + 19*a^2*
b^2 - 1725*a*b^3 - 1575*b^4)*cosh(d*x + c)^4 + 4*(7425*(3*a^3*b + 13*a^2*b^2 + 17*a*b^3 + 7*b^4)*cosh(d*x + c)
^8 - 1260*(8*a^4 + 45*a^2*b^2 + 150*a*b^3 + 105*b^4)*cosh(d*x + c)^6 - 70*(176*a^4 + 135*a^3*b + 15*a^2*b^2 -
1375*a*b^3 - 1575*b^4)*cosh(d*x + c)^4 - 96*a^4 - 313*a^3*b - 19*a^2*b^2 + 1725*a*b^3 + 1575*b^4 - 60*(56*a^4
+ 80*a^3*b - 65*a^2*b^2 + 400*a*b^3 + 525*b^4)*cosh(d*x + c)^2)*sinh(d*x + c)^4 + 64*a^4 + 588*a^3*b + 1404*a^
2*b^2 + 1300*a*b^3 + 420*b^4 + 16*(825*(3*a^3*b + 13*a^2*b^2 + 17*a*b^3 + 7*b^4)*cosh(d*x + c)^9 - 180*(8*a^4
+ 45*a^2*b^2 + 150*a*b^3 + 105*b^4)*cosh(d*x + c)^7 - 14*(176*a^4 + 135*a^3*b + 15*a^2*b^2 - 1375*a*b^3 - 1575
*b^4)*cosh(d*x + c)^5 - 20*(56*a^4 + 80*a^3*b - 65*a^2*b^2 + 400*a*b^3 + 525*b^4)*cosh(d*x + c)^3 - (96*a^4 +
313*a^3*b + 19*a^2*b^2 - 1725*a*b^3 - 1575*b^4)*cosh(d*x + c))*sinh(d*x + c)^3 + 8*(8*a^4 + 32*a^3*b - 251*a^2
*b^2 - 590*a*b^3 - 315*b^4)*cosh(d*x + c)^2 + 8*(495*(3*a^3*b + 13*a^2*b^2 + 17*a*b^3 + 7*b^4)*cosh(d*x + c)^1
0 - 135*(8*a^4 + 45*a^2*b^2 + 150*a*b^3 + 105*b^4)*cosh(d*x + c)^8 - 14*(176*a^4 + 135*a^3*b + 15*a^2*b^2 - 13
75*a*b^3 - 1575*b^4)*cosh(d*x + c)^6 - 30*(56*a^4 + 80*a^3*b - 65*a^2*b^2 + 400*a*b^3 + 525*b^4)*cosh(d*x + c)
^4 + 8*a^4 + 32*a^3*b - 251*a^2*b^2 - 590*a*b^3 - 315*b^4 - 3*(96*a^4 + 313*a^3*b + 19*a^2*b^2 - 1725*a*b^3 -
1575*b^4)*cosh(d*x + c)^2)*sinh(d*x + c)^2 + 15*((3*a^4 + 16*a^3*b + 30*a^2*b^2 + 24*a*b^3 + 7*b^4)*cosh(d*x +
c)^14 + 14*(3*a^4 + 16*a^3*b + 30*a^2*b^2 + 24*a*b^3 + 7*b^4)*cosh(d*x + c)*sinh(d*x + c)^13 + (3*a^4 + 16*a^
3*b + 30*a^2*b^2 + 24*a*b^3 + 7*b^4)*sinh(d*x + c)^14 + (3*a^4 - 8*a^3*b - 74*a^2*b^2 - 112*a*b^3 - 49*b^4)*co
sh(d*x + c)^12 + (3*a^4 - 8*a^3*b - 74*a^2*b^2 - 112*a*b^3 - 49*b^4 + 91*(3*a^4 + 16*a^3*b + 30*a^2*b^2 + 24*a
*b^3 + 7*b^4)*cosh(d*x + c)^2)*sinh(d*x + c)^12 + 4*(91*(3*a^4 + 16*a^3*b + 30*a^2*b^2 + 24*a*b^3 + 7*b^4)*cos
h(d*x + c)^3 + 3*(3*a^4 - 8*a^3*b - 74*a^2*b^2 - 112*a*b^3 - 49*b^4)*cosh(d*x + c))*sinh(d*x + c)^11 - (9*a^4
+ 24*a^3*b - 62*a^2*b^2 - 224*a*b^3 - 147*b^4)*cosh(d*x + c)^10 + (1001*(3*a^4 + 16*a^3*b + 30*a^2*b^2 + 24*a*
b^3 + 7*b^4)*cosh(d*x + c)^4 - 9*a^4 - 24*a^3*b + 62*a^2*b^2 + 224*a*b^3 + 147*b^4 + 66*(3*a^4 - 8*a^3*b - 74*
a^2*b^2 - 112*a*b^3 - 49*b^4)*cosh(d*x + c)^2)*sinh(d*x + c)^10 + 2*(1001*(3*a^4 + 16*a^3*b + 30*a^2*b^2 + 24*
a*b^3 + 7*b^4)*cosh(d*x + c)^5 + 110*(3*a^4 - 8*a^3*b - 74*a^2*b^2 - 112*a*b^3 - 49*b^4)*cosh(d*x + c)^3 - 5*(
9*a^4 + 24*a^3*b - 62*a^2*b^2 - 224*a*b^3 - 147*b^4)*cosh(d*x + c))*sinh(d*x + c)^9 - (9*a^4 + 26*a^2*b^2 + 28
0*a*b^3 + 245*b^4)*cosh(d*x + c)^8 + (3003*(3*a^4 + 16*a^3*b + 30*a^2*b^2 + 24*a*b^3 + 7*b^4)*cosh(d*x + c)^6
+ 495*(3*a^4 - 8*a^3*b - 74*a^2*b^2 - 112*a*b^3 - 49*b^4)*cosh(d*x + c)^4 - 9*a^4 - 26*a^2*b^2 - 280*a*b^3 - 2
45*b^4 - 45*(9*a^4 + 24*a^3*b - 62*a^2*b^2 - 224*a*b^3 - 147*b^4)*cosh(d*x + c)^2)*sinh(d*x + c)^8 + 8*(429*(3
*a^4 + 16*a^3*b + 30*a^2*b^2 + 24*a*b^3 + 7*b^4)*cosh(d*x + c)^7 + 99*(3*a^4 - 8*a^3*b - 74*a^2*b^2 - 112*a*b^
3 - 49*b^4)*cosh(d*x + c)^5 - 15*(9*a^4 + 24*a^3*b - 62*a^2*b^2 - 224*a*b^3 - 147*b^4)*cosh(d*x + c)^3 - (9*a^
4 + 26*a^2*b^2 + 280*a*b^3 + 245*b^4)*cosh(d*x + c))*sinh(d*x + c)^7 + (9*a^4 + 26*a^2*b^2 + 280*a*b^3 + 245*b
^4)*cosh(d*x + c)^6 + (3003*(3*a^4 + 16*a^3*b +...
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\operatorname {csch}^{4}{\left (c + d x \right )}}{\left (a + b \tanh ^{2}{\left (c + d x \right )}\right )^{3}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(csch(d*x+c)**4/(a+b*tanh(d*x+c)**2)**3,x)
[Out]
Integral(csch(c + d*x)**4/(a + b*tanh(c + d*x)**2)**3, x)
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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 395 vs.
\(2 (135) = 270\).
time = 1.02, size = 395, normalized size = 2.62 \begin {gather*} -\frac {\frac {6 \, {\left (9 \, a^{3} b e^{\left (6 \, d x + 6 \, c\right )} + 7 \, a^{2} b^{2} e^{\left (6 \, d x + 6 \, c\right )} - 13 \, a b^{3} e^{\left (6 \, d x + 6 \, c\right )} - 11 \, b^{4} e^{\left (6 \, d x + 6 \, c\right )} + 27 \, a^{3} b e^{\left (4 \, d x + 4 \, c\right )} + 15 \, a^{2} b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 5 \, a b^{3} e^{\left (4 \, d x + 4 \, c\right )} + 33 \, b^{4} e^{\left (4 \, d x + 4 \, c\right )} + 27 \, a^{3} b e^{\left (2 \, d x + 2 \, c\right )} + 37 \, a^{2} b^{2} e^{\left (2 \, d x + 2 \, c\right )} - 23 \, a b^{3} e^{\left (2 \, d x + 2 \, c\right )} - 33 \, b^{4} e^{\left (2 \, d x + 2 \, c\right )} + 9 \, a^{3} b + 29 \, a^{2} b^{2} + 31 \, a b^{3} + 11 \, b^{4}\right )}}{{\left (a^{5} + a^{4} b\right )} {\left (a e^{\left (4 \, d x + 4 \, c\right )} + b e^{\left (4 \, d x + 4 \, c\right )} + 2 \, a e^{\left (2 \, d x + 2 \, c\right )} - 2 \, b e^{\left (2 \, d x + 2 \, c\right )} + a + b\right )}^{2}} - \frac {15 \, {\left (3 \, a b + 7 \, b^{2}\right )} \arctan \left (\frac {a e^{\left (2 \, d x + 2 \, c\right )} + b e^{\left (2 \, d x + 2 \, c\right )} + a - b}{2 \, \sqrt {a b}}\right )}{\sqrt {a b} a^{4}} - \frac {16 \, {\left (9 \, b e^{\left (4 \, d x + 4 \, c\right )} - 6 \, a e^{\left (2 \, d x + 2 \, c\right )} - 18 \, b e^{\left (2 \, d x + 2 \, c\right )} + 2 \, a + 9 \, b\right )}}{a^{4} {\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}^{3}}}{24 \, d} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(csch(d*x+c)^4/(a+b*tanh(d*x+c)^2)^3,x, algorithm="giac")
[Out]
-1/24*(6*(9*a^3*b*e^(6*d*x + 6*c) + 7*a^2*b^2*e^(6*d*x + 6*c) - 13*a*b^3*e^(6*d*x + 6*c) - 11*b^4*e^(6*d*x + 6
*c) + 27*a^3*b*e^(4*d*x + 4*c) + 15*a^2*b^2*e^(4*d*x + 4*c) + 5*a*b^3*e^(4*d*x + 4*c) + 33*b^4*e^(4*d*x + 4*c)
+ 27*a^3*b*e^(2*d*x + 2*c) + 37*a^2*b^2*e^(2*d*x + 2*c) - 23*a*b^3*e^(2*d*x + 2*c) - 33*b^4*e^(2*d*x + 2*c) +
9*a^3*b + 29*a^2*b^2 + 31*a*b^3 + 11*b^4)/((a^5 + a^4*b)*(a*e^(4*d*x + 4*c) + b*e^(4*d*x + 4*c) + 2*a*e^(2*d*
x + 2*c) - 2*b*e^(2*d*x + 2*c) + a + b)^2) - 15*(3*a*b + 7*b^2)*arctan(1/2*(a*e^(2*d*x + 2*c) + b*e^(2*d*x + 2
*c) + a - b)/sqrt(a*b))/(sqrt(a*b)*a^4) - 16*(9*b*e^(4*d*x + 4*c) - 6*a*e^(2*d*x + 2*c) - 18*b*e^(2*d*x + 2*c)
+ 2*a + 9*b)/(a^4*(e^(2*d*x + 2*c) - 1)^3))/d
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{{\mathrm {sinh}\left (c+d\,x\right )}^4\,{\left (b\,{\mathrm {tanh}\left (c+d\,x\right )}^2+a\right )}^3} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(1/(sinh(c + d*x)^4*(a + b*tanh(c + d*x)^2)^3),x)
[Out]
int(1/(sinh(c + d*x)^4*(a + b*tanh(c + d*x)^2)^3), x)
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