3.259 \(\int \frac{\sinh ^8(c+d x)}{(a-b \sinh ^4(c+d x))^3} \, dx\)
Optimal. Leaf size=319 \[ -\frac{\left (2 \sqrt{a}-5 \sqrt{b}\right ) \tanh ^{-1}\left (\frac{\sqrt{\sqrt{a}-\sqrt{b}} \tanh (c+d x)}{\sqrt [4]{a}}\right )}{64 a^{3/4} b^{3/2} d \left (\sqrt{a}-\sqrt{b}\right )^{5/2}}+\frac{\left (2 \sqrt{a}+5 \sqrt{b}\right ) \tanh ^{-1}\left (\frac{\sqrt{\sqrt{a}+\sqrt{b}} \tanh (c+d x)}{\sqrt [4]{a}}\right )}{64 a^{3/4} b^{3/2} d \left (\sqrt{a}+\sqrt{b}\right )^{5/2}}+\frac{\tanh ^9(c+d x)}{8 a d \left ((a-b) \tanh ^4(c+d x)-2 a \tanh ^2(c+d x)+a\right )^2}-\frac{\tanh ^3(c+d x)}{32 a b d (a-b)}-\frac{(a+5 b) \tanh (c+d x)}{32 a b d (a-b)^2}-\frac{\tanh ^5(c+d x) \text{sech}^2(c+d x)}{32 a b d \left ((a-b) \tanh ^4(c+d x)-2 a \tanh ^2(c+d x)+a\right )} \]
[Out]
-((2*Sqrt[a] - 5*Sqrt[b])*ArcTanh[(Sqrt[Sqrt[a] - Sqrt[b]]*Tanh[c + d*x])/a^(1/4)])/(64*a^(3/4)*(Sqrt[a] - Sqr
t[b])^(5/2)*b^(3/2)*d) + ((2*Sqrt[a] + 5*Sqrt[b])*ArcTanh[(Sqrt[Sqrt[a] + Sqrt[b]]*Tanh[c + d*x])/a^(1/4)])/(6
4*a^(3/4)*(Sqrt[a] + Sqrt[b])^(5/2)*b^(3/2)*d) - ((a + 5*b)*Tanh[c + d*x])/(32*a*(a - b)^2*b*d) - Tanh[c + d*x
]^3/(32*a*(a - b)*b*d) + Tanh[c + d*x]^9/(8*a*d*(a - 2*a*Tanh[c + d*x]^2 + (a - b)*Tanh[c + d*x]^4)^2) - (Sech
[c + d*x]^2*Tanh[c + d*x]^5)/(32*a*b*d*(a - 2*a*Tanh[c + d*x]^2 + (a - b)*Tanh[c + d*x]^4))
________________________________________________________________________________________
Rubi [A] time = 0.48526, antiderivative size = 319, normalized size of antiderivative = 1.,
number of steps used = 9, number of rules used = 7, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.292, Rules used =
{3217, 1275, 12, 1120, 1279, 1166, 208} \[ -\frac{\left (2 \sqrt{a}-5 \sqrt{b}\right ) \tanh ^{-1}\left (\frac{\sqrt{\sqrt{a}-\sqrt{b}} \tanh (c+d x)}{\sqrt [4]{a}}\right )}{64 a^{3/4} b^{3/2} d \left (\sqrt{a}-\sqrt{b}\right )^{5/2}}+\frac{\left (2 \sqrt{a}+5 \sqrt{b}\right ) \tanh ^{-1}\left (\frac{\sqrt{\sqrt{a}+\sqrt{b}} \tanh (c+d x)}{\sqrt [4]{a}}\right )}{64 a^{3/4} b^{3/2} d \left (\sqrt{a}+\sqrt{b}\right )^{5/2}}+\frac{\tanh ^9(c+d x)}{8 a d \left ((a-b) \tanh ^4(c+d x)-2 a \tanh ^2(c+d x)+a\right )^2}-\frac{\tanh ^3(c+d x)}{32 a b d (a-b)}-\frac{(a+5 b) \tanh (c+d x)}{32 a b d (a-b)^2}-\frac{\tanh ^5(c+d x) \text{sech}^2(c+d x)}{32 a b d \left ((a-b) \tanh ^4(c+d x)-2 a \tanh ^2(c+d x)+a\right )} \]
Antiderivative was successfully verified.
[In]
Int[Sinh[c + d*x]^8/(a - b*Sinh[c + d*x]^4)^3,x]
[Out]
-((2*Sqrt[a] - 5*Sqrt[b])*ArcTanh[(Sqrt[Sqrt[a] - Sqrt[b]]*Tanh[c + d*x])/a^(1/4)])/(64*a^(3/4)*(Sqrt[a] - Sqr
t[b])^(5/2)*b^(3/2)*d) + ((2*Sqrt[a] + 5*Sqrt[b])*ArcTanh[(Sqrt[Sqrt[a] + Sqrt[b]]*Tanh[c + d*x])/a^(1/4)])/(6
4*a^(3/4)*(Sqrt[a] + Sqrt[b])^(5/2)*b^(3/2)*d) - ((a + 5*b)*Tanh[c + d*x])/(32*a*(a - b)^2*b*d) - Tanh[c + d*x
]^3/(32*a*(a - b)*b*d) + Tanh[c + d*x]^9/(8*a*d*(a - 2*a*Tanh[c + d*x]^2 + (a - b)*Tanh[c + d*x]^4)^2) - (Sech
[c + d*x]^2*Tanh[c + d*x]^5)/(32*a*b*d*(a - 2*a*Tanh[c + d*x]^2 + (a - b)*Tanh[c + d*x]^4))
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 1275
Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Simp[(f*
(f*x)^(m - 1)*(a + b*x^2 + c*x^4)^(p + 1)*(b*d - 2*a*e - (b*e - 2*c*d)*x^2))/(2*(p + 1)*(b^2 - 4*a*c)), x] - D
ist[f^2/(2*(p + 1)*(b^2 - 4*a*c)), Int[(f*x)^(m - 2)*(a + b*x^2 + c*x^4)^(p + 1)*Simp[(m - 1)*(b*d - 2*a*e) -
(4*p + 4 + m + 1)*(b*e - 2*c*d)*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[
p, -1] && GtQ[m, 1] && IntegerQ[2*p] && (IntegerQ[p] || IntegerQ[m])
Rule 12
Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] && !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]
Rule 1120
Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> -Simp[(d^3*(d*x)^(m - 3)*(2*a +
b*x^2)*(a + b*x^2 + c*x^4)^(p + 1))/(2*(p + 1)*(b^2 - 4*a*c)), x] + Dist[d^4/(2*(p + 1)*(b^2 - 4*a*c)), Int[(
d*x)^(m - 4)*(2*a*(m - 3) + b*(m + 4*p + 3)*x^2)*(a + b*x^2 + c*x^4)^(p + 1), x], x] /; FreeQ[{a, b, c, d}, x]
&& NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1] && GtQ[m, 3] && IntegerQ[2*p] && (IntegerQ[p] || IntegerQ[m])
Rule 1279
Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[(e*f
*(f*x)^(m - 1)*(a + b*x^2 + c*x^4)^(p + 1))/(c*(m + 4*p + 3)), x] - Dist[f^2/(c*(m + 4*p + 3)), Int[(f*x)^(m -
2)*(a + b*x^2 + c*x^4)^p*Simp[a*e*(m - 1) + (b*e*(m + 2*p + 1) - c*d*(m + 4*p + 3))*x^2, x], x], x] /; FreeQ[
{a, b, c, d, e, f, p}, x] && NeQ[b^2 - 4*a*c, 0] && GtQ[m, 1] && NeQ[m + 4*p + 3, 0] && IntegerQ[2*p] && (Inte
gerQ[p] || IntegerQ[m])
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 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]
Rubi steps
\begin{align*} \int \frac{\sinh ^8(c+d x)}{\left (a-b \sinh ^4(c+d x)\right )^3} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^8 \left (1-x^2\right )}{\left (a-2 a x^2+(a-b) x^4\right )^3} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac{\tanh ^9(c+d x)}{8 a 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{2 b x^8}{\left (a-2 a x^2+(a-b) x^4\right )^2} \, dx,x,\tanh (c+d x)\right )}{16 a b d}\\ &=\frac{\tanh ^9(c+d x)}{8 a 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{x^8}{\left (a-2 a x^2+(a-b) x^4\right )^2} \, dx,x,\tanh (c+d x)\right )}{8 a d}\\ &=\frac{\tanh ^9(c+d x)}{8 a d \left (a-2 a \tanh ^2(c+d x)+(a-b) \tanh ^4(c+d x)\right )^2}-\frac{\text{sech}^2(c+d x) \tanh ^5(c+d x)}{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{x^4 \left (10 a-6 a x^2\right )}{a-2 a x^2+(a-b) x^4} \, dx,x,\tanh (c+d x)\right )}{64 a^2 b d}\\ &=-\frac{\tanh ^3(c+d x)}{32 a (a-b) b d}+\frac{\tanh ^9(c+d x)}{8 a d \left (a-2 a \tanh ^2(c+d x)+(a-b) \tanh ^4(c+d x)\right )^2}-\frac{\text{sech}^2(c+d x) \tanh ^5(c+d x)}{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{x^2 \left (-18 a^2+6 a (a+5 b) x^2\right )}{a-2 a x^2+(a-b) x^4} \, dx,x,\tanh (c+d x)\right )}{192 a^2 (a-b) b d}\\ &=-\frac{(a+5 b) \tanh (c+d x)}{32 a (a-b)^2 b d}-\frac{\tanh ^3(c+d x)}{32 a (a-b) b d}+\frac{\tanh ^9(c+d x)}{8 a d \left (a-2 a \tanh ^2(c+d x)+(a-b) \tanh ^4(c+d x)\right )^2}-\frac{\text{sech}^2(c+d x) \tanh ^5(c+d x)}{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{6 a^2 (a+5 b)+6 a^2 (a-13 b) x^2}{a-2 a x^2+(a-b) x^4} \, dx,x,\tanh (c+d x)\right )}{192 a^2 (a-b)^2 b d}\\ &=-\frac{(a+5 b) \tanh (c+d x)}{32 a (a-b)^2 b d}-\frac{\tanh ^3(c+d x)}{32 a (a-b) b d}+\frac{\tanh ^9(c+d x)}{8 a d \left (a-2 a \tanh ^2(c+d x)+(a-b) \tanh ^4(c+d x)\right )^2}-\frac{\text{sech}^2(c+d x) \tanh ^5(c+d x)}{32 a b d \left (a-2 a \tanh ^2(c+d x)+(a-b) \tanh ^4(c+d x)\right )}-\frac{\left (2 a+3 \sqrt{a} \sqrt{b}-5 b\right ) \operatorname{Subst}\left (\int \frac{1}{-a+\sqrt{a} \sqrt{b}+(a-b) x^2} \, dx,x,\tanh (c+d x)\right )}{64 \sqrt{a} \left (\sqrt{a}+\sqrt{b}\right )^2 b^{3/2} d}+\frac{\left (\left (2 \sqrt{a}-5 \sqrt{b}\right ) \left (\sqrt{a}+\sqrt{b}\right )^3\right ) \operatorname{Subst}\left (\int \frac{1}{-a-\sqrt{a} \sqrt{b}+(a-b) x^2} \, dx,x,\tanh (c+d x)\right )}{64 \sqrt{a} (a-b)^2 b^{3/2} d}\\ &=-\frac{\left (2 \sqrt{a}-5 \sqrt{b}\right ) \tanh ^{-1}\left (\frac{\sqrt{\sqrt{a}-\sqrt{b}} \tanh (c+d x)}{\sqrt [4]{a}}\right )}{64 a^{3/4} \left (\sqrt{a}-\sqrt{b}\right )^{5/2} b^{3/2} d}+\frac{\left (2 \sqrt{a}+5 \sqrt{b}\right ) \tanh ^{-1}\left (\frac{\sqrt{\sqrt{a}+\sqrt{b}} \tanh (c+d x)}{\sqrt [4]{a}}\right )}{64 a^{3/4} \left (\sqrt{a}+\sqrt{b}\right )^{5/2} b^{3/2} d}-\frac{(a+5 b) \tanh (c+d x)}{32 a (a-b)^2 b d}-\frac{\tanh ^3(c+d x)}{32 a (a-b) b d}+\frac{\tanh ^9(c+d x)}{8 a d \left (a-2 a \tanh ^2(c+d x)+(a-b) \tanh ^4(c+d x)\right )^2}-\frac{\text{sech}^2(c+d x) \tanh ^5(c+d x)}{32 a b d \left (a-2 a \tanh ^2(c+d x)+(a-b) \tanh ^4(c+d x)\right )}\\ \end{align*}
Mathematica [A] time = 3.95342, size = 331, normalized size = 1.04 \[ \frac{\frac{\left (2 a^{3/2} \sqrt{b}-8 \sqrt{a} b^{3/2}+a b+5 b^2\right ) \tanh ^{-1}\left (\frac{\left (\sqrt{a}+\sqrt{b}\right ) \tanh (c+d x)}{\sqrt{\sqrt{a} \sqrt{b}+a}}\right )}{\sqrt{a} \sqrt{\sqrt{a} \sqrt{b}+a}}+\frac{\sqrt{b} \left (2 \sqrt{a}-5 \sqrt{b}\right ) \left (\sqrt{a}+\sqrt{b}\right )^2 \tan ^{-1}\left (\frac{\left (\sqrt{a}-\sqrt{b}\right ) \tanh (c+d x)}{\sqrt{\sqrt{a} \sqrt{b}-a}}\right )}{\sqrt{a} \sqrt{\sqrt{a} \sqrt{b}-a}}+\frac{8 b \sinh (2 (c+d x)) ((5 b-2 a) \cosh (2 (c+d x))+5 a-14 b)}{8 a+4 b \cosh (2 (c+d x))-b \cosh (4 (c+d x))-3 b}+\frac{64 a b (a-b) (\sinh (4 (c+d x))-6 \sinh (2 (c+d x)))}{(-8 a-4 b \cosh (2 (c+d x))+b \cosh (4 (c+d x))+3 b)^2}}{64 b^2 d (a-b)^2} \]
Antiderivative was successfully verified.
[In]
Integrate[Sinh[c + d*x]^8/(a - b*Sinh[c + d*x]^4)^3,x]
[Out]
(((2*Sqrt[a] - 5*Sqrt[b])*(Sqrt[a] + Sqrt[b])^2*Sqrt[b]*ArcTan[((Sqrt[a] - Sqrt[b])*Tanh[c + d*x])/Sqrt[-a + S
qrt[a]*Sqrt[b]]])/(Sqrt[a]*Sqrt[-a + Sqrt[a]*Sqrt[b]]) + ((2*a^(3/2)*Sqrt[b] + a*b - 8*Sqrt[a]*b^(3/2) + 5*b^2
)*ArcTanh[((Sqrt[a] + Sqrt[b])*Tanh[c + d*x])/Sqrt[a + Sqrt[a]*Sqrt[b]]])/(Sqrt[a]*Sqrt[a + Sqrt[a]*Sqrt[b]])
+ (8*b*(5*a - 14*b + (-2*a + 5*b)*Cosh[2*(c + d*x)])*Sinh[2*(c + d*x)])/(8*a - 3*b + 4*b*Cosh[2*(c + d*x)] - b
*Cosh[4*(c + d*x)]) + (64*a*(a - b)*b*(-6*Sinh[2*(c + d*x)] + Sinh[4*(c + d*x)]))/(-8*a + 3*b - 4*b*Cosh[2*(c
+ d*x)] + b*Cosh[4*(c + d*x)])^2)/(64*(a - b)^2*b^2*d)
________________________________________________________________________________________
Maple [C] time = 0.092, size = 2236, normalized size = 7. \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(sinh(d*x+c)^8/(a-b*sinh(d*x+c)^4)^3,x)
[Out]
-1/16/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)-5/16/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*a/(
a^2-2*a*b+b^2)*tanh(1/2*d*x+1/2*c)+5/16/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
+49/16/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)^3-9/16/d/(tanh(1/2*d*x+1/2*c)^8*a-4*ta
nh(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^2-165/16/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+9/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+5/16/d/(tanh(1/2*d*x+1/2*c)^8*a-4*tan
h(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+377/16/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-49/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+5/16/d/(tanh(1/2*d*x+1/2*c)^8*a-4*tan
h(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+377/16/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-49/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)^9-9/16/d/(tanh(1/2*d*x+1/2*c)^8*a-4*tan
h(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^2-165/16/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)^
11*a+9/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+5/16/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*a^2
/b/(a^2-2*a*b+b^2)*tanh(1/2*d*x+1/2*c)^13+49/16/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)^13-1/16/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)^15-5/16/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)^15-1/128/d/b/(a^2-2*a*b+b^2)*sum(((a+5*b)*_R^6+(5*a-47*b)*_R^4+(-5*
a+47*b)*_R^2-a-5*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))
________________________________________________________________________________________
Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sinh(d*x+c)^8/(a-b*sinh(d*x+c)^4)^3,x, algorithm="maxima")
[Out]
-1/8*(2*a*b^2 - 5*b^3 + (a*b^2*e^(14*c) - 4*b^3*e^(14*c))*e^(14*d*x) - (32*a^2*b*e^(12*c) - 58*a*b^2*e^(12*c)
- b^3*e^(12*c))*e^(12*d*x) + 3*(48*a^2*b*e^(10*c) - 73*a*b^2*e^(10*c) + 20*b^3*e^(10*c))*e^(10*d*x) + (256*a^3
*e^(8*c) - 832*a^2*b*e^(8*c) + 550*a*b^2*e^(8*c) - 175*b^3*e^(8*c))*e^(8*d*x) + (112*a^2*b*e^(6*c) - 533*a*b^2
*e^(6*c) + 220*b^3*e^(6*c))*e^(6*d*x) - (32*a^2*b*e^(4*c) - 158*a*b^2*e^(4*c) + 141*b^3*e^(4*c))*e^(4*d*x) - (
17*a*b^2*e^(2*c) - 44*b^3*e^(2*c))*e^(2*d*x))/(a^2*b^4*d - 2*a*b^5*d + b^6*d + (a^2*b^4*d*e^(16*c) - 2*a*b^5*d
*e^(16*c) + b^6*d*e^(16*c))*e^(16*d*x) - 8*(a^2*b^4*d*e^(14*c) - 2*a*b^5*d*e^(14*c) + b^6*d*e^(14*c))*e^(14*d*
x) - 4*(8*a^3*b^3*d*e^(12*c) - 23*a^2*b^4*d*e^(12*c) + 22*a*b^5*d*e^(12*c) - 7*b^6*d*e^(12*c))*e^(12*d*x) + 8*
(16*a^3*b^3*d*e^(10*c) - 39*a^2*b^4*d*e^(10*c) + 30*a*b^5*d*e^(10*c) - 7*b^6*d*e^(10*c))*e^(10*d*x) + 2*(128*a
^4*b^2*d*e^(8*c) - 352*a^3*b^3*d*e^(8*c) + 355*a^2*b^4*d*e^(8*c) - 166*a*b^5*d*e^(8*c) + 35*b^6*d*e^(8*c))*e^(
8*d*x) + 8*(16*a^3*b^3*d*e^(6*c) - 39*a^2*b^4*d*e^(6*c) + 30*a*b^5*d*e^(6*c) - 7*b^6*d*e^(6*c))*e^(6*d*x) - 4*
(8*a^3*b^3*d*e^(4*c) - 23*a^2*b^4*d*e^(4*c) + 22*a*b^5*d*e^(4*c) - 7*b^6*d*e^(4*c))*e^(4*d*x) - 8*(a^2*b^4*d*e
^(2*c) - 2*a*b^5*d*e^(2*c) + b^6*d*e^(2*c))*e^(2*d*x)) - 1/256*integrate(64*((a*e^(6*c) - 4*b*e^(6*c))*e^(6*d*
x) + (a*e^(2*c) - 4*b*e^(2*c))*e^(2*d*x) + 18*b*e^(4*d*x + 4*c))/(a^2*b^2 - 2*a*b^3 + b^4 + (a^2*b^2*e^(8*c) -
2*a*b^3*e^(8*c) + b^4*e^(8*c))*e^(8*d*x) - 4*(a^2*b^2*e^(6*c) - 2*a*b^3*e^(6*c) + b^4*e^(6*c))*e^(6*d*x) - 2*
(8*a^3*b*e^(4*c) - 19*a^2*b^2*e^(4*c) + 14*a*b^3*e^(4*c) - 3*b^4*e^(4*c))*e^(4*d*x) - 4*(a^2*b^2*e^(2*c) - 2*a
*b^3*e^(2*c) + b^4*e^(2*c))*e^(2*d*x)), x)
________________________________________________________________________________________
Fricas [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(sinh(d*x+c)^8/(a-b*sinh(d*x+c)^4)^3,x, algorithm="fricas")
[Out]
Timed out
________________________________________________________________________________________
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(sinh(d*x+c)**8/(a-b*sinh(d*x+c)**4)**3,x)
[Out]
Timed out
________________________________________________________________________________________
Giac [A] time = 17.4165, size = 527, normalized size = 1.65 \begin{align*} -\frac{a b^{2} e^{\left (14 \, d x + 14 \, c\right )} - 4 \, b^{3} e^{\left (14 \, d x + 14 \, c\right )} - 32 \, a^{2} b e^{\left (12 \, d x + 12 \, c\right )} + 58 \, a b^{2} e^{\left (12 \, d x + 12 \, c\right )} + b^{3} e^{\left (12 \, d x + 12 \, c\right )} + 144 \, a^{2} b e^{\left (10 \, d x + 10 \, c\right )} - 219 \, a b^{2} e^{\left (10 \, d x + 10 \, c\right )} + 60 \, b^{3} e^{\left (10 \, d x + 10 \, c\right )} + 256 \, a^{3} e^{\left (8 \, d x + 8 \, c\right )} - 832 \, a^{2} b e^{\left (8 \, d x + 8 \, c\right )} + 550 \, a b^{2} e^{\left (8 \, d x + 8 \, c\right )} - 175 \, b^{3} e^{\left (8 \, d x + 8 \, c\right )} + 112 \, a^{2} b e^{\left (6 \, d x + 6 \, c\right )} - 533 \, a b^{2} e^{\left (6 \, d x + 6 \, c\right )} + 220 \, b^{3} e^{\left (6 \, d x + 6 \, c\right )} - 32 \, a^{2} b e^{\left (4 \, d x + 4 \, c\right )} + 158 \, a b^{2} e^{\left (4 \, d x + 4 \, c\right )} - 141 \, b^{3} e^{\left (4 \, d x + 4 \, c\right )} - 17 \, a b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 44 \, b^{3} e^{\left (2 \, d x + 2 \, c\right )} + 2 \, a b^{2} - 5 \, b^{3}}{8 \,{\left (a^{2} b^{2} d - 2 \, a b^{3} d + b^{4} d\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}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sinh(d*x+c)^8/(a-b*sinh(d*x+c)^4)^3,x, algorithm="giac")
[Out]
-1/8*(a*b^2*e^(14*d*x + 14*c) - 4*b^3*e^(14*d*x + 14*c) - 32*a^2*b*e^(12*d*x + 12*c) + 58*a*b^2*e^(12*d*x + 12
*c) + b^3*e^(12*d*x + 12*c) + 144*a^2*b*e^(10*d*x + 10*c) - 219*a*b^2*e^(10*d*x + 10*c) + 60*b^3*e^(10*d*x + 1
0*c) + 256*a^3*e^(8*d*x + 8*c) - 832*a^2*b*e^(8*d*x + 8*c) + 550*a*b^2*e^(8*d*x + 8*c) - 175*b^3*e^(8*d*x + 8*
c) + 112*a^2*b*e^(6*d*x + 6*c) - 533*a*b^2*e^(6*d*x + 6*c) + 220*b^3*e^(6*d*x + 6*c) - 32*a^2*b*e^(4*d*x + 4*c
) + 158*a*b^2*e^(4*d*x + 4*c) - 141*b^3*e^(4*d*x + 4*c) - 17*a*b^2*e^(2*d*x + 2*c) + 44*b^3*e^(2*d*x + 2*c) +
2*a*b^2 - 5*b^3)/((a^2*b^2*d - 2*a*b^3*d + b^4*d)*(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)