### 3.73 $$\int \frac{\sinh ^4(c+d x)}{a+b \tanh ^3(c+d x)} \, dx$$

Optimal. Leaf size=491 $-\frac{a^2 b \left (a^2+2 b^2\right ) \log \left (a+b \tanh ^3(c+d x)\right )}{d \left (a^2-b^2\right )^3}+\frac{a^{2/3} \sqrt [3]{b} \left (7 a^2 b^2+3 a^{2/3} b^{4/3} \left (2 a^2+b^2\right )+a^4+b^4\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} \tanh (c+d x)+b^{2/3} \tanh ^2(c+d x)\right )}{6 d \left (a^2-b^2\right )^3}-\frac{a^{2/3} \sqrt [3]{b} \left (7 a^2 b^2+3 a^{2/3} b^{4/3} \left (2 a^2+b^2\right )+a^4+b^4\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} \tanh (c+d x)\right )}{3 d \left (a^2-b^2\right )^3}-\frac{a^{2/3} \sqrt [3]{b} \left (3 a^{4/3} b^{2/3}+a^2-b^2\right ) \tan ^{-1}\left (\frac{\sqrt [3]{a}-2 \sqrt [3]{b} \tanh (c+d x)}{\sqrt{3} \sqrt [3]{a}}\right )}{\sqrt{3} d \left (a^{2/3} b^{2/3}+a^{4/3}+b^{4/3}\right )^3}-\frac{5 a-b}{16 d (a+b)^2 (1-\tanh (c+d x))}+\frac{5 a+b}{16 d (a-b)^2 (\tanh (c+d x)+1)}+\frac{1}{16 d (a+b) (1-\tanh (c+d x))^2}-\frac{1}{16 d (a-b) (\tanh (c+d x)+1)^2}-\frac{3 a (a-5 b) \log (1-\tanh (c+d x))}{16 d (a+b)^3}+\frac{3 a (a+5 b) \log (\tanh (c+d x)+1)}{16 d (a-b)^3}$

[Out]

-((a^(2/3)*b^(1/3)*(a^2 + 3*a^(4/3)*b^(2/3) - b^2)*ArcTan[(a^(1/3) - 2*b^(1/3)*Tanh[c + d*x])/(Sqrt[3]*a^(1/3)
)])/(Sqrt[3]*(a^(4/3) + a^(2/3)*b^(2/3) + b^(4/3))^3*d)) - (3*a*(a - 5*b)*Log[1 - Tanh[c + d*x]])/(16*(a + b)^
3*d) + (3*a*(a + 5*b)*Log[1 + Tanh[c + d*x]])/(16*(a - b)^3*d) - (a^(2/3)*b^(1/3)*(a^4 + 7*a^2*b^2 + b^4 + 3*a
^(2/3)*b^(4/3)*(2*a^2 + b^2))*Log[a^(1/3) + b^(1/3)*Tanh[c + d*x]])/(3*(a^2 - b^2)^3*d) + (a^(2/3)*b^(1/3)*(a^
4 + 7*a^2*b^2 + b^4 + 3*a^(2/3)*b^(4/3)*(2*a^2 + b^2))*Log[a^(2/3) - a^(1/3)*b^(1/3)*Tanh[c + d*x] + b^(2/3)*T
anh[c + d*x]^2])/(6*(a^2 - b^2)^3*d) - (a^2*b*(a^2 + 2*b^2)*Log[a + b*Tanh[c + d*x]^3])/((a^2 - b^2)^3*d) + 1/
(16*(a + b)*d*(1 - Tanh[c + d*x])^2) - (5*a - b)/(16*(a + b)^2*d*(1 - Tanh[c + d*x])) - 1/(16*(a - b)*d*(1 + T
anh[c + d*x])^2) + (5*a + b)/(16*(a - b)^2*d*(1 + Tanh[c + d*x]))

________________________________________________________________________________________

Rubi [A]  time = 0.891549, antiderivative size = 491, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 10, integrand size = 23, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.435, Rules used = {3663, 6725, 1871, 1860, 31, 634, 617, 204, 628, 260} $-\frac{a^2 b \left (a^2+2 b^2\right ) \log \left (a+b \tanh ^3(c+d x)\right )}{d \left (a^2-b^2\right )^3}+\frac{a^{2/3} \sqrt [3]{b} \left (7 a^2 b^2+3 a^{2/3} b^{4/3} \left (2 a^2+b^2\right )+a^4+b^4\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} \tanh (c+d x)+b^{2/3} \tanh ^2(c+d x)\right )}{6 d \left (a^2-b^2\right )^3}-\frac{a^{2/3} \sqrt [3]{b} \left (7 a^2 b^2+3 a^{2/3} b^{4/3} \left (2 a^2+b^2\right )+a^4+b^4\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} \tanh (c+d x)\right )}{3 d \left (a^2-b^2\right )^3}-\frac{a^{2/3} \sqrt [3]{b} \left (3 a^{4/3} b^{2/3}+a^2-b^2\right ) \tan ^{-1}\left (\frac{\sqrt [3]{a}-2 \sqrt [3]{b} \tanh (c+d x)}{\sqrt{3} \sqrt [3]{a}}\right )}{\sqrt{3} d \left (a^{2/3} b^{2/3}+a^{4/3}+b^{4/3}\right )^3}-\frac{5 a-b}{16 d (a+b)^2 (1-\tanh (c+d x))}+\frac{5 a+b}{16 d (a-b)^2 (\tanh (c+d x)+1)}+\frac{1}{16 d (a+b) (1-\tanh (c+d x))^2}-\frac{1}{16 d (a-b) (\tanh (c+d x)+1)^2}-\frac{3 a (a-5 b) \log (1-\tanh (c+d x))}{16 d (a+b)^3}+\frac{3 a (a+5 b) \log (\tanh (c+d x)+1)}{16 d (a-b)^3}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-((a^(2/3)*b^(1/3)*(a^2 + 3*a^(4/3)*b^(2/3) - b^2)*ArcTan[(a^(1/3) - 2*b^(1/3)*Tanh[c + d*x])/(Sqrt[3]*a^(1/3)
)])/(Sqrt[3]*(a^(4/3) + a^(2/3)*b^(2/3) + b^(4/3))^3*d)) - (3*a*(a - 5*b)*Log[1 - Tanh[c + d*x]])/(16*(a + b)^
3*d) + (3*a*(a + 5*b)*Log[1 + Tanh[c + d*x]])/(16*(a - b)^3*d) - (a^(2/3)*b^(1/3)*(a^4 + 7*a^2*b^2 + b^4 + 3*a
^(2/3)*b^(4/3)*(2*a^2 + b^2))*Log[a^(1/3) + b^(1/3)*Tanh[c + d*x]])/(3*(a^2 - b^2)^3*d) + (a^(2/3)*b^(1/3)*(a^
4 + 7*a^2*b^2 + b^4 + 3*a^(2/3)*b^(4/3)*(2*a^2 + b^2))*Log[a^(2/3) - a^(1/3)*b^(1/3)*Tanh[c + d*x] + b^(2/3)*T
anh[c + d*x]^2])/(6*(a^2 - b^2)^3*d) - (a^2*b*(a^2 + 2*b^2)*Log[a + b*Tanh[c + d*x]^3])/((a^2 - b^2)^3*d) + 1/
(16*(a + b)*d*(1 - Tanh[c + d*x])^2) - (5*a - b)/(16*(a + b)^2*d*(1 - Tanh[c + d*x])) - 1/(16*(a - b)*d*(1 + T
anh[c + d*x])^2) + (5*a + b)/(16*(a - b)^2*d*(1 + Tanh[c + d*x]))

Rule 3663

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]

Rule 6725

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rule 1871

Int[(P2_)/((a_) + (b_.)*(x_)^3), x_Symbol] :> With[{A = Coeff[P2, x, 0], B = Coeff[P2, x, 1], C = Coeff[P2, x,
2]}, Int[(A + B*x)/(a + b*x^3), x] + Dist[C, Int[x^2/(a + b*x^3), x], x] /; EqQ[a*B^3 - b*A^3, 0] ||  !Ration
alQ[a/b]] /; FreeQ[{a, b}, x] && PolyQ[P2, x, 2]

Rule 1860

Int[((A_) + (B_.)*(x_))/((a_) + (b_.)*(x_)^3), x_Symbol] :> With[{r = Numerator[Rt[a/b, 3]], s = Denominator[R
t[a/b, 3]]}, -Dist[(r*(B*r - A*s))/(3*a*s), Int[1/(r + s*x), x], x] + Dist[r/(3*a*s), Int[(r*(B*r + 2*A*s) + s
*(B*r - A*s)*x)/(r^2 - r*s*x + s^2*x^2), x], x]] /; FreeQ[{a, b, A, B}, x] && NeQ[a*B^3 - b*A^3, 0] && PosQ[a/
b]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 634

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

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

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

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rubi steps

\begin{align*} \int \frac{\sinh ^4(c+d x)}{a+b \tanh ^3(c+d x)} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^4}{\left (1-x^2\right )^3 \left (a+b x^3\right )} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \left (-\frac{1}{8 (a+b) (-1+x)^3}+\frac{-5 a+b}{16 (a+b)^2 (-1+x)^2}-\frac{3 a (a-5 b)}{16 (a+b)^3 (-1+x)}+\frac{1}{8 (a-b) (1+x)^3}+\frac{-5 a-b}{16 (a-b)^2 (1+x)^2}+\frac{3 a (a+5 b)}{16 (a-b)^3 (1+x)}+\frac{a b \left (-3 a b \left (2 a^2+b^2\right )+\left (a^4+7 a^2 b^2+b^4\right ) x-3 a b \left (a^2+2 b^2\right ) x^2\right )}{\left (a^2-b^2\right )^3 \left (a+b x^3\right )}\right ) \, dx,x,\tanh (c+d x)\right )}{d}\\ &=-\frac{3 a (a-5 b) \log (1-\tanh (c+d x))}{16 (a+b)^3 d}+\frac{3 a (a+5 b) \log (1+\tanh (c+d x))}{16 (a-b)^3 d}+\frac{1}{16 (a+b) d (1-\tanh (c+d x))^2}-\frac{5 a-b}{16 (a+b)^2 d (1-\tanh (c+d x))}-\frac{1}{16 (a-b) d (1+\tanh (c+d x))^2}+\frac{5 a+b}{16 (a-b)^2 d (1+\tanh (c+d x))}+\frac{(a b) \operatorname{Subst}\left (\int \frac{-3 a b \left (2 a^2+b^2\right )+\left (a^4+7 a^2 b^2+b^4\right ) x-3 a b \left (a^2+2 b^2\right ) x^2}{a+b x^3} \, dx,x,\tanh (c+d x)\right )}{\left (a^2-b^2\right )^3 d}\\ &=-\frac{3 a (a-5 b) \log (1-\tanh (c+d x))}{16 (a+b)^3 d}+\frac{3 a (a+5 b) \log (1+\tanh (c+d x))}{16 (a-b)^3 d}+\frac{1}{16 (a+b) d (1-\tanh (c+d x))^2}-\frac{5 a-b}{16 (a+b)^2 d (1-\tanh (c+d x))}-\frac{1}{16 (a-b) d (1+\tanh (c+d x))^2}+\frac{5 a+b}{16 (a-b)^2 d (1+\tanh (c+d x))}+\frac{(a b) \operatorname{Subst}\left (\int \frac{-3 a b \left (2 a^2+b^2\right )+\left (a^4+7 a^2 b^2+b^4\right ) x}{a+b x^3} \, dx,x,\tanh (c+d x)\right )}{\left (a^2-b^2\right )^3 d}-\frac{\left (3 a^2 b^2 \left (a^2+2 b^2\right )\right ) \operatorname{Subst}\left (\int \frac{x^2}{a+b x^3} \, dx,x,\tanh (c+d x)\right )}{\left (a^2-b^2\right )^3 d}\\ &=-\frac{3 a (a-5 b) \log (1-\tanh (c+d x))}{16 (a+b)^3 d}+\frac{3 a (a+5 b) \log (1+\tanh (c+d x))}{16 (a-b)^3 d}-\frac{a^2 b \left (a^2+2 b^2\right ) \log \left (a+b \tanh ^3(c+d x)\right )}{\left (a^2-b^2\right )^3 d}+\frac{1}{16 (a+b) d (1-\tanh (c+d x))^2}-\frac{5 a-b}{16 (a+b)^2 d (1-\tanh (c+d x))}-\frac{1}{16 (a-b) d (1+\tanh (c+d x))^2}+\frac{5 a+b}{16 (a-b)^2 d (1+\tanh (c+d x))}+\frac{\left (\sqrt [3]{a} b^{2/3}\right ) \operatorname{Subst}\left (\int \frac{\sqrt [3]{a} \left (-6 a b^{4/3} \left (2 a^2+b^2\right )+\sqrt [3]{a} \left (a^4+7 a^2 b^2+b^4\right )\right )+\sqrt [3]{b} \left (3 a b^{4/3} \left (2 a^2+b^2\right )+\sqrt [3]{a} \left (a^4+7 a^2 b^2+b^4\right )\right ) x}{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2} \, dx,x,\tanh (c+d x)\right )}{3 \left (a^2-b^2\right )^3 d}-\frac{\left (a^{2/3} b^{2/3} \left (a^4+7 a^2 b^2+b^4+3 a^{2/3} b^{4/3} \left (2 a^2+b^2\right )\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt [3]{a}+\sqrt [3]{b} x} \, dx,x,\tanh (c+d x)\right )}{3 \left (a^2-b^2\right )^3 d}\\ &=-\frac{3 a (a-5 b) \log (1-\tanh (c+d x))}{16 (a+b)^3 d}+\frac{3 a (a+5 b) \log (1+\tanh (c+d x))}{16 (a-b)^3 d}-\frac{a^{2/3} \sqrt [3]{b} \left (a^4+7 a^2 b^2+b^4+3 a^{2/3} b^{4/3} \left (2 a^2+b^2\right )\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} \tanh (c+d x)\right )}{3 \left (a^2-b^2\right )^3 d}-\frac{a^2 b \left (a^2+2 b^2\right ) \log \left (a+b \tanh ^3(c+d x)\right )}{\left (a^2-b^2\right )^3 d}+\frac{1}{16 (a+b) d (1-\tanh (c+d x))^2}-\frac{5 a-b}{16 (a+b)^2 d (1-\tanh (c+d x))}-\frac{1}{16 (a-b) d (1+\tanh (c+d x))^2}+\frac{5 a+b}{16 (a-b)^2 d (1+\tanh (c+d x))}+\frac{\left (a b^{2/3} \left (a^2+3 a^{4/3} b^{2/3}-b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2} \, dx,x,\tanh (c+d x)\right )}{2 \left (a^{4/3}+a^{2/3} b^{2/3}+b^{4/3}\right )^3 d}+\frac{\left (a^{2/3} \sqrt [3]{b} \left (a^4+7 a^2 b^2+b^4+3 a^{2/3} b^{4/3} \left (2 a^2+b^2\right )\right )\right ) \operatorname{Subst}\left (\int \frac{-\sqrt [3]{a} \sqrt [3]{b}+2 b^{2/3} x}{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2} \, dx,x,\tanh (c+d x)\right )}{6 \left (a^2-b^2\right )^3 d}\\ &=-\frac{3 a (a-5 b) \log (1-\tanh (c+d x))}{16 (a+b)^3 d}+\frac{3 a (a+5 b) \log (1+\tanh (c+d x))}{16 (a-b)^3 d}-\frac{a^{2/3} \sqrt [3]{b} \left (a^4+7 a^2 b^2+b^4+3 a^{2/3} b^{4/3} \left (2 a^2+b^2\right )\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} \tanh (c+d x)\right )}{3 \left (a^2-b^2\right )^3 d}+\frac{a^{2/3} \sqrt [3]{b} \left (a^4+7 a^2 b^2+b^4+3 a^{2/3} b^{4/3} \left (2 a^2+b^2\right )\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} \tanh (c+d x)+b^{2/3} \tanh ^2(c+d x)\right )}{6 \left (a^2-b^2\right )^3 d}-\frac{a^2 b \left (a^2+2 b^2\right ) \log \left (a+b \tanh ^3(c+d x)\right )}{\left (a^2-b^2\right )^3 d}+\frac{1}{16 (a+b) d (1-\tanh (c+d x))^2}-\frac{5 a-b}{16 (a+b)^2 d (1-\tanh (c+d x))}-\frac{1}{16 (a-b) d (1+\tanh (c+d x))^2}+\frac{5 a+b}{16 (a-b)^2 d (1+\tanh (c+d x))}+\frac{\left (a^{2/3} \sqrt [3]{b} \left (a^2+3 a^{4/3} b^{2/3}-b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,1-\frac{2 \sqrt [3]{b} \tanh (c+d x)}{\sqrt [3]{a}}\right )}{\left (a^{4/3}+a^{2/3} b^{2/3}+b^{4/3}\right )^3 d}\\ &=-\frac{a^{2/3} \sqrt [3]{b} \left (a^2+3 a^{4/3} b^{2/3}-b^2\right ) \tan ^{-1}\left (\frac{1-\frac{2 \sqrt [3]{b} \tanh (c+d x)}{\sqrt [3]{a}}}{\sqrt{3}}\right )}{\sqrt{3} \left (a^{4/3}+a^{2/3} b^{2/3}+b^{4/3}\right )^3 d}-\frac{3 a (a-5 b) \log (1-\tanh (c+d x))}{16 (a+b)^3 d}+\frac{3 a (a+5 b) \log (1+\tanh (c+d x))}{16 (a-b)^3 d}-\frac{a^{2/3} \sqrt [3]{b} \left (a^4+7 a^2 b^2+b^4+3 a^{2/3} b^{4/3} \left (2 a^2+b^2\right )\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} \tanh (c+d x)\right )}{3 \left (a^2-b^2\right )^3 d}+\frac{a^{2/3} \sqrt [3]{b} \left (a^4+7 a^2 b^2+b^4+3 a^{2/3} b^{4/3} \left (2 a^2+b^2\right )\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} \tanh (c+d x)+b^{2/3} \tanh ^2(c+d x)\right )}{6 \left (a^2-b^2\right )^3 d}-\frac{a^2 b \left (a^2+2 b^2\right ) \log \left (a+b \tanh ^3(c+d x)\right )}{\left (a^2-b^2\right )^3 d}+\frac{1}{16 (a+b) d (1-\tanh (c+d x))^2}-\frac{5 a-b}{16 (a+b)^2 d (1-\tanh (c+d x))}-\frac{1}{16 (a-b) d (1+\tanh (c+d x))^2}+\frac{5 a+b}{16 (a-b)^2 d (1+\tanh (c+d x))}\\ \end{align*}

Mathematica [C]  time = 4.37695, size = 645, normalized size = 1.31 $\frac{3 \left (-8 a \left (a^2 b+a^3+2 a b^2+2 b^3\right ) \sinh (2 (c+d x))+a (a-b) \left (12 \left (a^2-6 a b+5 b^2\right ) (c+d x)+(a+b)^2 \sinh (4 (c+d x))\right )+4 b \left (5 a^2 b+5 a^3+a b^2+b^3\right ) \cosh (2 (c+d x))+b (-(a-b)) (a+b)^2 \cosh (4 (c+d x))\right )-32 a b \text{RootSum}\left [\text{\#1}^3 a+3 \text{\#1}^2 a+\text{\#1}^3 b-3 \text{\#1}^2 b+3 \text{\#1} a+3 \text{\#1} b+a-b\& ,\frac{-10 \text{\#1}^2 a^2 b \log \left (e^{2 (c+d x)}-\text{\#1}\right )+20 \text{\#1}^2 a^2 b c+20 \text{\#1}^2 a^2 b d x+5 \text{\#1}^2 a^3 \log \left (e^{2 (c+d x)}-\text{\#1}\right )-10 \text{\#1}^2 a^3 c-10 \text{\#1}^2 a^3 d x+10 \text{\#1}^2 a b^2 \log \left (e^{2 (c+d x)}-\text{\#1}\right )-20 \text{\#1}^2 a b^2 c-20 \text{\#1}^2 a b^2 d x-2 \text{\#1}^2 b^3 \log \left (e^{2 (c+d x)}-\text{\#1}\right )+4 \text{\#1}^2 b^3 c+4 \text{\#1}^2 b^3 d x-2 \text{\#1} a^2 b \log \left (e^{2 (c+d x)}-\text{\#1}\right )+4 \text{\#1} a^2 b c+4 \text{\#1} a^2 b d x+3 a^3 \log \left (e^{2 (c+d x)}-\text{\#1}\right )+4 \text{\#1} a^3 \log \left (e^{2 (c+d x)}-\text{\#1}\right )-8 \text{\#1} a^3 c-8 \text{\#1} a^3 d x+6 a b^2 \log \left (e^{2 (c+d x)}-\text{\#1}\right )-4 \text{\#1} a b^2 \log \left (e^{2 (c+d x)}-\text{\#1}\right )+8 \text{\#1} a b^2 c+8 \text{\#1} a b^2 d x+2 \text{\#1} b^3 \log \left (e^{2 (c+d x)}-\text{\#1}\right )-4 \text{\#1} b^3 c-4 \text{\#1} b^3 d x-6 a^3 c-6 a^3 d x-12 a b^2 c-12 a b^2 d x}{\text{\#1}^2 a-\text{\#1}^2 b+2 \text{\#1} a+2 \text{\#1} b+a-b}\& \right ]}{96 d (a-b)^2 (a+b)^3}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(-32*a*b*RootSum[a - b + 3*a*#1 + 3*b*#1 + 3*a*#1^2 - 3*b*#1^2 + a*#1^3 + b*#1^3 & , (-6*a^3*c - 12*a*b^2*c -
6*a^3*d*x - 12*a*b^2*d*x + 3*a^3*Log[E^(2*(c + d*x)) - #1] + 6*a*b^2*Log[E^(2*(c + d*x)) - #1] - 8*a^3*c*#1 +
4*a^2*b*c*#1 + 8*a*b^2*c*#1 - 4*b^3*c*#1 - 8*a^3*d*x*#1 + 4*a^2*b*d*x*#1 + 8*a*b^2*d*x*#1 - 4*b^3*d*x*#1 + 4*a
^3*Log[E^(2*(c + d*x)) - #1]*#1 - 2*a^2*b*Log[E^(2*(c + d*x)) - #1]*#1 - 4*a*b^2*Log[E^(2*(c + d*x)) - #1]*#1
+ 2*b^3*Log[E^(2*(c + d*x)) - #1]*#1 - 10*a^3*c*#1^2 + 20*a^2*b*c*#1^2 - 20*a*b^2*c*#1^2 + 4*b^3*c*#1^2 - 10*a
^3*d*x*#1^2 + 20*a^2*b*d*x*#1^2 - 20*a*b^2*d*x*#1^2 + 4*b^3*d*x*#1^2 + 5*a^3*Log[E^(2*(c + d*x)) - #1]*#1^2 -
10*a^2*b*Log[E^(2*(c + d*x)) - #1]*#1^2 + 10*a*b^2*Log[E^(2*(c + d*x)) - #1]*#1^2 - 2*b^3*Log[E^(2*(c + d*x))
- #1]*#1^2)/(a - b + 2*a*#1 + 2*b*#1 + a*#1^2 - b*#1^2) & ] + 3*(4*b*(5*a^3 + 5*a^2*b + a*b^2 + b^3)*Cosh[2*(c
+ d*x)] - (a - b)*b*(a + b)^2*Cosh[4*(c + d*x)] - 8*a*(a^3 + a^2*b + 2*a*b^2 + 2*b^3)*Sinh[2*(c + d*x)] + a*(
a - b)*(12*(a^2 - 6*a*b + 5*b^2)*(c + d*x) + (a + b)^2*Sinh[4*(c + d*x)])))/(96*(a - b)^2*(a + b)^3*d)

________________________________________________________________________________________

Maple [C]  time = 0.128, size = 603, normalized size = 1.2 \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)^4/(a+b*tanh(d*x+c)^3),x)

[Out]

-8/d/(32*a-32*b)/(tanh(1/2*d*x+1/2*c)+1)^4+32/d/(64*a-64*b)/(tanh(1/2*d*x+1/2*c)+1)^3+1/8/d/(a-b)^2/(tanh(1/2*
d*x+1/2*c)+1)^2*a+5/8/d/(a-b)^2/(tanh(1/2*d*x+1/2*c)+1)^2*b-3/8/d/(a-b)^2/(tanh(1/2*d*x+1/2*c)+1)*a-3/8/d/(a-b
)^2/(tanh(1/2*d*x+1/2*c)+1)*b+3/8/d*a^2/(a-b)^3*ln(tanh(1/2*d*x+1/2*c)+1)+15/8/d*a/(a-b)^3*ln(tanh(1/2*d*x+1/2
*c)+1)*b+8/d/(32*a+32*b)/(tanh(1/2*d*x+1/2*c)-1)^4+32/d/(64*a+64*b)/(tanh(1/2*d*x+1/2*c)-1)^3-1/8/d/(a+b)^2/(t
anh(1/2*d*x+1/2*c)-1)^2*a+5/8/d/(a+b)^2/(tanh(1/2*d*x+1/2*c)-1)^2*b-3/8/d/(a+b)^2/(tanh(1/2*d*x+1/2*c)-1)*a+3/
8/d/(a+b)^2/(tanh(1/2*d*x+1/2*c)-1)*b-3/8/d/(a+b)^3*ln(tanh(1/2*d*x+1/2*c)-1)*a^2+15/8/d/(a+b)^3*ln(tanh(1/2*d
*x+1/2*c)-1)*a*b-1/3/d*a*b/(a-b)^3/(a+b)^3*sum((3*a^2*(a^2+2*b^2)*_R^5+3*a*b*(-2*a^2-b^2)*_R^4+2*(4*a^4+13*a^2
*b^2+b^4)*_R^3+12*a*b*(a^2+2*b^2)*_R^2+(a^4-8*a^2*b^2-2*b^4)*_R+6*a^3*b+3*a*b^3)/(_R^5*a+2*_R^3*a+4*_R^2*b+_R*
a)*ln(tanh(1/2*d*x+1/2*c)-_R),_R=RootOf(_Z^6*a+3*_Z^4*a+8*_Z^3*b+3*_Z^2*a+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)^4/(a+b*tanh(d*x+c)^3),x, algorithm="maxima")

[Out]

-6*a^4*b*(integrate(((a + b)*e^(4*d*x + 4*c) + 3*(a - b)*e^(2*d*x + 2*c) + 3*a + 3*b)*e^(2*d*x + 2*c)/((a + b)
*e^(6*d*x + 6*c) + 3*(a - b)*e^(4*d*x + 4*c) + 3*(a + b)*e^(2*d*x + 2*c) + a - b), x)/(a^6 - 3*a^4*b^2 + 3*a^2
*b^4 - b^6) - (d*x + c)/((a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*d)) - 12*a^2*b^3*(integrate(((a + b)*e^(4*d*x + 4
*c) + 3*(a - b)*e^(2*d*x + 2*c) + 3*a + 3*b)*e^(2*d*x + 2*c)/((a + b)*e^(6*d*x + 6*c) + 3*(a - b)*e^(4*d*x + 4
*c) + 3*(a + b)*e^(2*d*x + 2*c) + a - b), x)/(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6) - (d*x + c)/((a^6 - 3*a^4*b^2
+ 3*a^2*b^4 - b^6)*d)) + 10*a^4*b*integrate(e^(4*d*x + 4*c)/((a + b)*e^(6*d*x + 6*c) + 3*(a - b)*e^(4*d*x + 4
*c) + 3*(a + b)*e^(2*d*x + 2*c) + a - b), x)/(a^5 + a^4*b - 2*a^3*b^2 - 2*a^2*b^3 + a*b^4 + b^5) - 20*a^3*b^2*
integrate(e^(4*d*x + 4*c)/((a + b)*e^(6*d*x + 6*c) + 3*(a - b)*e^(4*d*x + 4*c) + 3*(a + b)*e^(2*d*x + 2*c) + a
- b), x)/(a^5 + a^4*b - 2*a^3*b^2 - 2*a^2*b^3 + a*b^4 + b^5) + 20*a^2*b^3*integrate(e^(4*d*x + 4*c)/((a + b)*
e^(6*d*x + 6*c) + 3*(a - b)*e^(4*d*x + 4*c) + 3*(a + b)*e^(2*d*x + 2*c) + a - b), x)/(a^5 + a^4*b - 2*a^3*b^2
- 2*a^2*b^3 + a*b^4 + b^5) - 4*a*b^4*integrate(e^(4*d*x + 4*c)/((a + b)*e^(6*d*x + 6*c) + 3*(a - b)*e^(4*d*x +
4*c) + 3*(a + b)*e^(2*d*x + 2*c) + a - b), x)/(a^5 + a^4*b - 2*a^3*b^2 - 2*a^2*b^3 + a*b^4 + b^5) + 8*a^4*b*i
ntegrate(e^(2*d*x + 2*c)/((a + b)*e^(6*d*x + 6*c) + 3*(a - b)*e^(4*d*x + 4*c) + 3*(a + b)*e^(2*d*x + 2*c) + a
- b), x)/(a^5 + a^4*b - 2*a^3*b^2 - 2*a^2*b^3 + a*b^4 + b^5) - 4*a^3*b^2*integrate(e^(2*d*x + 2*c)/((a + b)*e^
(6*d*x + 6*c) + 3*(a - b)*e^(4*d*x + 4*c) + 3*(a + b)*e^(2*d*x + 2*c) + a - b), x)/(a^5 + a^4*b - 2*a^3*b^2 -
2*a^2*b^3 + a*b^4 + b^5) - 8*a^2*b^3*integrate(e^(2*d*x + 2*c)/((a + b)*e^(6*d*x + 6*c) + 3*(a - b)*e^(4*d*x +
4*c) + 3*(a + b)*e^(2*d*x + 2*c) + a - b), x)/(a^5 + a^4*b - 2*a^3*b^2 - 2*a^2*b^3 + a*b^4 + b^5) + 4*a*b^4*i
ntegrate(e^(2*d*x + 2*c)/((a + b)*e^(6*d*x + 6*c) + 3*(a - b)*e^(4*d*x + 4*c) + 3*(a + b)*e^(2*d*x + 2*c) + a
- b), x)/(a^5 + a^4*b - 2*a^3*b^2 - 2*a^2*b^3 + a*b^4 + b^5) - 1/64*(a^4 + 2*a^3*b - 2*a*b^3 - b^4 - 24*(a^4*d
*e^(4*c) - 7*a^3*b*d*e^(4*c) + 11*a^2*b^2*d*e^(4*c) - 5*a*b^3*d*e^(4*c))*x*e^(4*d*x) - (a^4*e^(8*c) - 2*a^2*b^
2*e^(8*c) + b^4*e^(8*c))*e^(8*d*x) + 4*(2*a^4*e^(6*c) - 3*a^3*b*e^(6*c) - a^2*b^2*e^(6*c) + 3*a*b^3*e^(6*c) -
b^4*e^(6*c))*e^(6*d*x) - 4*(2*a^4*e^(2*c) + 7*a^3*b*e^(2*c) + 9*a^2*b^2*e^(2*c) + 5*a*b^3*e^(2*c) + b^4*e^(2*c
))*e^(2*d*x))*e^(-4*d*x)/(a^5*d*e^(4*c) + a^4*b*d*e^(4*c) - 2*a^3*b^2*d*e^(4*c) - 2*a^2*b^3*d*e^(4*c) + a*b^4*
d*e^(4*c) + b^5*d*e^(4*c))

________________________________________________________________________________________

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)^4/(a+b*tanh(d*x+c)^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)**4/(a+b*tanh(d*x+c)**3),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 2.33021, size = 489, normalized size = 1. \begin{align*} \frac{\frac{24 \,{\left (a^{2} + 5 \, a b\right )} d x}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} - \frac{{\left (18 \, a^{2} e^{\left (4 \, d x + 4 \, c\right )} + 90 \, a b e^{\left (4 \, d x + 4 \, c\right )} - 8 \, a^{2} e^{\left (2 \, d x + 2 \, c\right )} + 4 \, a b e^{\left (2 \, d x + 2 \, c\right )} + 4 \, b^{2} e^{\left (2 \, d x + 2 \, c\right )} + a^{2} - 2 \, a b + b^{2}\right )} e^{\left (-4 \, d x\right )}}{a^{3} e^{\left (4 \, c\right )} - 3 \, a^{2} b e^{\left (4 \, c\right )} + 3 \, a b^{2} e^{\left (4 \, c\right )} - b^{3} e^{\left (4 \, c\right )}} - \frac{64 \,{\left (a^{4} b + 2 \, a^{2} b^{3}\right )} \log \left ({\left | a e^{\left (6 \, d x + 6 \, c\right )} + b e^{\left (6 \, d x + 6 \, c\right )} + 3 \, a e^{\left (4 \, d x + 4 \, c\right )} - 3 \, b e^{\left (4 \, d x + 4 \, c\right )} + 3 \, a e^{\left (2 \, d x + 2 \, c\right )} + 3 \, b e^{\left (2 \, d x + 2 \, c\right )} + a - b \right |}\right )}{a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}} + \frac{a e^{\left (4 \, d x + 24 \, c\right )} + b e^{\left (4 \, d x + 24 \, c\right )} - 8 \, a e^{\left (2 \, d x + 22 \, c\right )} + 4 \, b e^{\left (2 \, d x + 22 \, c\right )}}{a^{2} e^{\left (20 \, c\right )} + 2 \, a b e^{\left (20 \, c\right )} + b^{2} e^{\left (20 \, c\right )}}}{64 \, d} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/64*(24*(a^2 + 5*a*b)*d*x/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) - (18*a^2*e^(4*d*x + 4*c) + 90*a*b*e^(4*d*x + 4*c)
- 8*a^2*e^(2*d*x + 2*c) + 4*a*b*e^(2*d*x + 2*c) + 4*b^2*e^(2*d*x + 2*c) + a^2 - 2*a*b + b^2)*e^(-4*d*x)/(a^3*e
^(4*c) - 3*a^2*b*e^(4*c) + 3*a*b^2*e^(4*c) - b^3*e^(4*c)) - 64*(a^4*b + 2*a^2*b^3)*log(abs(a*e^(6*d*x + 6*c) +
b*e^(6*d*x + 6*c) + 3*a*e^(4*d*x + 4*c) - 3*b*e^(4*d*x + 4*c) + 3*a*e^(2*d*x + 2*c) + 3*b*e^(2*d*x + 2*c) + a
- b))/(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6) + (a*e^(4*d*x + 24*c) + b*e^(4*d*x + 24*c) - 8*a*e^(2*d*x + 22*c) +
4*b*e^(2*d*x + 22*c))/(a^2*e^(20*c) + 2*a*b*e^(20*c) + b^2*e^(20*c)))/d