3.404 \(\int \frac {\cosh ^3(c+d x) \sinh ^3(c+d x)}{a+b \sinh (c+d x)} \, dx\)

Optimal. Leaf size=141 \[ \frac {a^2 \left (a^2+b^2\right ) \sinh (c+d x)}{b^5 d}-\frac {a \left (a^2+b^2\right ) \sinh ^2(c+d x)}{2 b^4 d}+\frac {\left (a^2+b^2\right ) \sinh ^3(c+d x)}{3 b^3 d}-\frac {a^3 \left (a^2+b^2\right ) \log (a+b \sinh (c+d x))}{b^6 d}-\frac {a \sinh ^4(c+d x)}{4 b^2 d}+\frac {\sinh ^5(c+d x)}{5 b d} \]

[Out]

-a^3*(a^2+b^2)*ln(a+b*sinh(d*x+c))/b^6/d+a^2*(a^2+b^2)*sinh(d*x+c)/b^5/d-1/2*a*(a^2+b^2)*sinh(d*x+c)^2/b^4/d+1
/3*(a^2+b^2)*sinh(d*x+c)^3/b^3/d-1/4*a*sinh(d*x+c)^4/b^2/d+1/5*sinh(d*x+c)^5/b/d

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Rubi [A]  time = 0.22, antiderivative size = 141, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {2837, 12, 894} \[ \frac {\left (a^2+b^2\right ) \sinh ^3(c+d x)}{3 b^3 d}-\frac {a \left (a^2+b^2\right ) \sinh ^2(c+d x)}{2 b^4 d}+\frac {a^2 \left (a^2+b^2\right ) \sinh (c+d x)}{b^5 d}-\frac {a^3 \left (a^2+b^2\right ) \log (a+b \sinh (c+d x))}{b^6 d}-\frac {a \sinh ^4(c+d x)}{4 b^2 d}+\frac {\sinh ^5(c+d x)}{5 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a^3*(a^2 + b^2)*Log[a + b*Sinh[c + d*x]])/(b^6*d)) + (a^2*(a^2 + b^2)*Sinh[c + d*x])/(b^5*d) - (a*(a^2 + b^
2)*Sinh[c + d*x]^2)/(2*b^4*d) + ((a^2 + b^2)*Sinh[c + d*x]^3)/(3*b^3*d) - (a*Sinh[c + d*x]^4)/(4*b^2*d) + Sinh
[c + d*x]^5/(5*b*d)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 894

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIn
tegrand[(d + e*x)^m*(f + g*x)^n*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] &&
NeQ[c*d^2 + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && IntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

Rule 2837

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)
*(x_)])^(n_.), x_Symbol] :> Dist[1/(b^p*f), Subst[Int[(a + x)^m*(c + (d*x)/b)^n*(b^2 - x^2)^((p - 1)/2), x], x
, b*Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IntegerQ[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rubi steps

\begin {align*} \int \frac {\cosh ^3(c+d x) \sinh ^3(c+d x)}{a+b \sinh (c+d x)} \, dx &=-\frac {\operatorname {Subst}\left (\int \frac {x^3 \left (-b^2-x^2\right )}{b^3 (a+x)} \, dx,x,b \sinh (c+d x)\right )}{b^3 d}\\ &=-\frac {\operatorname {Subst}\left (\int \frac {x^3 \left (-b^2-x^2\right )}{a+x} \, dx,x,b \sinh (c+d x)\right )}{b^6 d}\\ &=-\frac {\operatorname {Subst}\left (\int \left (-a^2 \left (a^2+b^2\right )+a \left (a^2+b^2\right ) x-\left (a^2+b^2\right ) x^2+a x^3-x^4+\frac {a^3 \left (a^2+b^2\right )}{a+x}\right ) \, dx,x,b \sinh (c+d x)\right )}{b^6 d}\\ &=-\frac {a^3 \left (a^2+b^2\right ) \log (a+b \sinh (c+d x))}{b^6 d}+\frac {a^2 \left (a^2+b^2\right ) \sinh (c+d x)}{b^5 d}-\frac {a \left (a^2+b^2\right ) \sinh ^2(c+d x)}{2 b^4 d}+\frac {\left (a^2+b^2\right ) \sinh ^3(c+d x)}{3 b^3 d}-\frac {a \sinh ^4(c+d x)}{4 b^2 d}+\frac {\sinh ^5(c+d x)}{5 b d}\\ \end {align*}

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Mathematica [A]  time = 0.36, size = 123, normalized size = 0.87 \[ -\frac {-\frac {60 a^2 \left (a^2+b^2\right ) \sinh (c+d x)}{b^5}+\frac {30 a \left (a^2+b^2\right ) \sinh ^2(c+d x)}{b^4}-\frac {20 \left (a^2+b^2\right ) \sinh ^3(c+d x)}{b^3}+\frac {60 a^3 \left (a^2+b^2\right ) \log (a+b \sinh (c+d x))}{b^6}+\frac {15 a \sinh ^4(c+d x)}{b^2}-\frac {12 \sinh ^5(c+d x)}{b}}{60 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/60*((60*a^3*(a^2 + b^2)*Log[a + b*Sinh[c + d*x]])/b^6 - (60*a^2*(a^2 + b^2)*Sinh[c + d*x])/b^5 + (30*a*(a^2
+ b^2)*Sinh[c + d*x]^2)/b^4 - (20*(a^2 + b^2)*Sinh[c + d*x]^3)/b^3 + (15*a*Sinh[c + d*x]^4)/b^2 - (12*Sinh[c
+ d*x]^5)/b)/d

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fricas [B]  time = 0.48, size = 1660, normalized size = 11.77 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(d*x+c)^3*sinh(d*x+c)^3/(a+b*sinh(d*x+c)),x, algorithm="fricas")

[Out]

1/960*(6*b^5*cosh(d*x + c)^10 + 6*b^5*sinh(d*x + c)^10 - 15*a*b^4*cosh(d*x + c)^9 + 15*(4*b^5*cosh(d*x + c) -
a*b^4)*sinh(d*x + c)^9 + 10*(4*a^2*b^3 + b^5)*cosh(d*x + c)^8 + 5*(54*b^5*cosh(d*x + c)^2 - 27*a*b^4*cosh(d*x
+ c) + 8*a^2*b^3 + 2*b^5)*sinh(d*x + c)^8 + 960*(a^5 + a^3*b^2)*d*x*cosh(d*x + c)^5 - 60*(2*a^3*b^2 + a*b^4)*c
osh(d*x + c)^7 + 20*(36*b^5*cosh(d*x + c)^3 - 27*a*b^4*cosh(d*x + c)^2 - 6*a^3*b^2 - 3*a*b^4 + 4*(4*a^2*b^3 +
b^5)*cosh(d*x + c))*sinh(d*x + c)^7 + 60*(8*a^4*b + 6*a^2*b^3 - b^5)*cosh(d*x + c)^6 + 20*(63*b^5*cosh(d*x + c
)^4 - 63*a*b^4*cosh(d*x + c)^3 + 24*a^4*b + 18*a^2*b^3 - 3*b^5 + 14*(4*a^2*b^3 + b^5)*cosh(d*x + c)^2 - 21*(2*
a^3*b^2 + a*b^4)*cosh(d*x + c))*sinh(d*x + c)^6 - 15*a*b^4*cosh(d*x + c) + 2*(756*b^5*cosh(d*x + c)^5 - 945*a*
b^4*cosh(d*x + c)^4 + 280*(4*a^2*b^3 + b^5)*cosh(d*x + c)^3 + 480*(a^5 + a^3*b^2)*d*x - 630*(2*a^3*b^2 + a*b^4
)*cosh(d*x + c)^2 + 180*(8*a^4*b + 6*a^2*b^3 - b^5)*cosh(d*x + c))*sinh(d*x + c)^5 - 6*b^5 - 60*(8*a^4*b + 6*a
^2*b^3 - b^5)*cosh(d*x + c)^4 + 10*(126*b^5*cosh(d*x + c)^6 - 189*a*b^4*cosh(d*x + c)^5 - 48*a^4*b - 36*a^2*b^
3 + 6*b^5 + 70*(4*a^2*b^3 + b^5)*cosh(d*x + c)^4 + 480*(a^5 + a^3*b^2)*d*x*cosh(d*x + c) - 210*(2*a^3*b^2 + a*
b^4)*cosh(d*x + c)^3 + 90*(8*a^4*b + 6*a^2*b^3 - b^5)*cosh(d*x + c)^2)*sinh(d*x + c)^4 - 60*(2*a^3*b^2 + a*b^4
)*cosh(d*x + c)^3 + 20*(36*b^5*cosh(d*x + c)^7 - 63*a*b^4*cosh(d*x + c)^6 + 28*(4*a^2*b^3 + b^5)*cosh(d*x + c)
^5 - 6*a^3*b^2 - 3*a*b^4 + 480*(a^5 + a^3*b^2)*d*x*cosh(d*x + c)^2 - 105*(2*a^3*b^2 + a*b^4)*cosh(d*x + c)^4 +
60*(8*a^4*b + 6*a^2*b^3 - b^5)*cosh(d*x + c)^3 - 12*(8*a^4*b + 6*a^2*b^3 - b^5)*cosh(d*x + c))*sinh(d*x + c)^
3 - 10*(4*a^2*b^3 + b^5)*cosh(d*x + c)^2 + 10*(27*b^5*cosh(d*x + c)^8 - 54*a*b^4*cosh(d*x + c)^7 + 28*(4*a^2*b
^3 + b^5)*cosh(d*x + c)^6 + 960*(a^5 + a^3*b^2)*d*x*cosh(d*x + c)^3 - 126*(2*a^3*b^2 + a*b^4)*cosh(d*x + c)^5
- 4*a^2*b^3 - b^5 + 90*(8*a^4*b + 6*a^2*b^3 - b^5)*cosh(d*x + c)^4 - 36*(8*a^4*b + 6*a^2*b^3 - b^5)*cosh(d*x +
c)^2 - 18*(2*a^3*b^2 + a*b^4)*cosh(d*x + c))*sinh(d*x + c)^2 - 960*((a^5 + a^3*b^2)*cosh(d*x + c)^5 + 5*(a^5
+ a^3*b^2)*cosh(d*x + c)^4*sinh(d*x + c) + 10*(a^5 + a^3*b^2)*cosh(d*x + c)^3*sinh(d*x + c)^2 + 10*(a^5 + a^3*
b^2)*cosh(d*x + c)^2*sinh(d*x + c)^3 + 5*(a^5 + a^3*b^2)*cosh(d*x + c)*sinh(d*x + c)^4 + (a^5 + a^3*b^2)*sinh(
d*x + c)^5)*log(2*(b*sinh(d*x + c) + a)/(cosh(d*x + c) - sinh(d*x + c))) + 5*(12*b^5*cosh(d*x + c)^9 - 27*a*b^
4*cosh(d*x + c)^8 + 16*(4*a^2*b^3 + b^5)*cosh(d*x + c)^7 + 960*(a^5 + a^3*b^2)*d*x*cosh(d*x + c)^4 - 84*(2*a^3
*b^2 + a*b^4)*cosh(d*x + c)^6 + 72*(8*a^4*b + 6*a^2*b^3 - b^5)*cosh(d*x + c)^5 - 3*a*b^4 - 48*(8*a^4*b + 6*a^2
*b^3 - b^5)*cosh(d*x + c)^3 - 36*(2*a^3*b^2 + a*b^4)*cosh(d*x + c)^2 - 4*(4*a^2*b^3 + b^5)*cosh(d*x + c))*sinh
(d*x + c))/(b^6*d*cosh(d*x + c)^5 + 5*b^6*d*cosh(d*x + c)^4*sinh(d*x + c) + 10*b^6*d*cosh(d*x + c)^3*sinh(d*x
+ c)^2 + 10*b^6*d*cosh(d*x + c)^2*sinh(d*x + c)^3 + 5*b^6*d*cosh(d*x + c)*sinh(d*x + c)^4 + b^6*d*sinh(d*x + c
)^5)

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giac [A]  time = 0.36, size = 258, normalized size = 1.83 \[ \frac {\frac {6 \, b^{4} {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{5} - 15 \, a b^{3} {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{4} + 40 \, a^{2} b^{2} {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{3} + 40 \, b^{4} {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{3} - 120 \, a^{3} b {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{2} - 120 \, a b^{3} {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{2} + 480 \, a^{4} {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )} + 480 \, a^{2} b^{2} {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}}{b^{5}} - \frac {960 \, {\left (a^{5} + a^{3} b^{2}\right )} \log \left ({\left | b {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )} + 2 \, a \right |}\right )}{b^{6}}}{960 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/960*((6*b^4*(e^(d*x + c) - e^(-d*x - c))^5 - 15*a*b^3*(e^(d*x + c) - e^(-d*x - c))^4 + 40*a^2*b^2*(e^(d*x +
c) - e^(-d*x - c))^3 + 40*b^4*(e^(d*x + c) - e^(-d*x - c))^3 - 120*a^3*b*(e^(d*x + c) - e^(-d*x - c))^2 - 120*
a*b^3*(e^(d*x + c) - e^(-d*x - c))^2 + 480*a^4*(e^(d*x + c) - e^(-d*x - c)) + 480*a^2*b^2*(e^(d*x + c) - e^(-d
*x - c)))/b^5 - 960*(a^5 + a^3*b^2)*log(abs(b*(e^(d*x + c) - e^(-d*x - c)) + 2*a))/b^6)/d

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maple [B]  time = 0.14, size = 804, normalized size = 5.70 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(d*x+c)^3*sinh(d*x+c)^3/(a+b*sinh(d*x+c)),x)

[Out]

-1/2/d/b^4/(tanh(1/2*d*x+1/2*c)+1)^2*a^3-1/d*a^4/b^5/(tanh(1/2*d*x+1/2*c)+1)-1/4/d/b^2/(tanh(1/2*d*x+1/2*c)-1)
^4*a-1/3/d/b^3/(tanh(1/2*d*x+1/2*c)-1)^3*a^2-1/2/d/b^4/(tanh(1/2*d*x+1/2*c)-1)^2*a^3-1/d*a^4/b^5/(tanh(1/2*d*x
+1/2*c)-1)-3/8/d/b^2/(tanh(1/2*d*x+1/2*c)-1)*a+3/8/d/b^2/(tanh(1/2*d*x+1/2*c)+1)*a+1/d*a^3/b^4*ln(tanh(1/2*d*x
+1/2*c)+1)-1/d*a^3/b^4*ln(tanh(1/2*d*x+1/2*c)^2*a-2*tanh(1/2*d*x+1/2*c)*b-a)-5/8/d/b^2/(tanh(1/2*d*x+1/2*c)-1)
^2*a-1/d/b^3/(tanh(1/2*d*x+1/2*c)-1)*a^2+1/d*a^3/b^4*ln(tanh(1/2*d*x+1/2*c)-1)-5/8/d/b^2/(tanh(1/2*d*x+1/2*c)+
1)^2*a-1/d/b^3/(tanh(1/2*d*x+1/2*c)+1)*a^2+1/d*a^5/b^6*ln(tanh(1/2*d*x+1/2*c)+1)-1/d*a^5/b^6*ln(tanh(1/2*d*x+1
/2*c)^2*a-2*tanh(1/2*d*x+1/2*c)*b-a)+1/d*a^5/b^6*ln(tanh(1/2*d*x+1/2*c)-1)-1/4/d/b^2/(tanh(1/2*d*x+1/2*c)+1)^4
*a-1/3/d/b^3/(tanh(1/2*d*x+1/2*c)+1)^3*a^2-1/2/d/b^2/(tanh(1/2*d*x+1/2*c)-1)^3*a-1/2/d/b^3/(tanh(1/2*d*x+1/2*c
)-1)^2*a^2-1/2/d/b^4/(tanh(1/2*d*x+1/2*c)-1)*a^3+1/2/d/b^2/(tanh(1/2*d*x+1/2*c)+1)^3*a+1/2/d/b^3/(tanh(1/2*d*x
+1/2*c)+1)^2*a^2+1/2/d/b^4/(tanh(1/2*d*x+1/2*c)+1)*a^3-1/5/d/b/(tanh(1/2*d*x+1/2*c)-1)^5-1/5/d/b/(tanh(1/2*d*x
+1/2*c)+1)^5-1/2/d/b/(tanh(1/2*d*x+1/2*c)-1)^4+1/2/d/b/(tanh(1/2*d*x+1/2*c)+1)^4-3/8/d/b/(tanh(1/2*d*x+1/2*c)-
1)^2+3/8/d/b/(tanh(1/2*d*x+1/2*c)+1)^2-7/12/d/b/(tanh(1/2*d*x+1/2*c)-1)^3-7/12/d/b/(tanh(1/2*d*x+1/2*c)+1)^3

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maxima [B]  time = 0.36, size = 300, normalized size = 2.13 \[ -\frac {{\left (15 \, a b^{3} e^{\left (-d x - c\right )} - 6 \, b^{4} - 10 \, {\left (4 \, a^{2} b^{2} + b^{4}\right )} e^{\left (-2 \, d x - 2 \, c\right )} + 60 \, {\left (2 \, a^{3} b + a b^{3}\right )} e^{\left (-3 \, d x - 3 \, c\right )} - 60 \, {\left (8 \, a^{4} + 6 \, a^{2} b^{2} - b^{4}\right )} e^{\left (-4 \, d x - 4 \, c\right )}\right )} e^{\left (5 \, d x + 5 \, c\right )}}{960 \, b^{5} d} - \frac {{\left (a^{5} + a^{3} b^{2}\right )} {\left (d x + c\right )}}{b^{6} d} - \frac {15 \, a b^{3} e^{\left (-4 \, d x - 4 \, c\right )} + 6 \, b^{4} e^{\left (-5 \, d x - 5 \, c\right )} + 60 \, {\left (8 \, a^{4} + 6 \, a^{2} b^{2} - b^{4}\right )} e^{\left (-d x - c\right )} + 60 \, {\left (2 \, a^{3} b + a b^{3}\right )} e^{\left (-2 \, d x - 2 \, c\right )} + 10 \, {\left (4 \, a^{2} b^{2} + b^{4}\right )} e^{\left (-3 \, d x - 3 \, c\right )}}{960 \, b^{5} d} - \frac {{\left (a^{5} + a^{3} b^{2}\right )} \log \left (-2 \, a e^{\left (-d x - c\right )} + b e^{\left (-2 \, d x - 2 \, c\right )} - b\right )}{b^{6} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/960*(15*a*b^3*e^(-d*x - c) - 6*b^4 - 10*(4*a^2*b^2 + b^4)*e^(-2*d*x - 2*c) + 60*(2*a^3*b + a*b^3)*e^(-3*d*x
- 3*c) - 60*(8*a^4 + 6*a^2*b^2 - b^4)*e^(-4*d*x - 4*c))*e^(5*d*x + 5*c)/(b^5*d) - (a^5 + a^3*b^2)*(d*x + c)/(
b^6*d) - 1/960*(15*a*b^3*e^(-4*d*x - 4*c) + 6*b^4*e^(-5*d*x - 5*c) + 60*(8*a^4 + 6*a^2*b^2 - b^4)*e^(-d*x - c)
+ 60*(2*a^3*b + a*b^3)*e^(-2*d*x - 2*c) + 10*(4*a^2*b^2 + b^4)*e^(-3*d*x - 3*c))/(b^5*d) - (a^5 + a^3*b^2)*lo
g(-2*a*e^(-d*x - c) + b*e^(-2*d*x - 2*c) - b)/(b^6*d)

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mupad [B]  time = 0.75, size = 307, normalized size = 2.18 \[ \frac {{\mathrm {e}}^{5\,c+5\,d\,x}}{160\,b\,d}-\frac {{\mathrm {e}}^{-5\,c-5\,d\,x}}{160\,b\,d}-\frac {a\,{\mathrm {e}}^{-4\,c-4\,d\,x}}{64\,b^2\,d}-\frac {a\,{\mathrm {e}}^{4\,c+4\,d\,x}}{64\,b^2\,d}-\frac {\ln \left (2\,a\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c-b+b\,{\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}\right )\,\left (a^5+a^3\,b^2\right )}{b^6\,d}-\frac {{\mathrm {e}}^{-c-d\,x}\,\left (8\,a^4+6\,a^2\,b^2-b^4\right )}{16\,b^5\,d}+\frac {a^3\,x\,\left (a^2+b^2\right )}{b^6}-\frac {{\mathrm {e}}^{-2\,c-2\,d\,x}\,\left (2\,a^3+a\,b^2\right )}{16\,b^4\,d}-\frac {{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (2\,a^3+a\,b^2\right )}{16\,b^4\,d}+\frac {{\mathrm {e}}^{c+d\,x}\,\left (8\,a^4+6\,a^2\,b^2-b^4\right )}{16\,b^5\,d}-\frac {{\mathrm {e}}^{-3\,c-3\,d\,x}\,\left (4\,a^2+b^2\right )}{96\,b^3\,d}+\frac {{\mathrm {e}}^{3\,c+3\,d\,x}\,\left (4\,a^2+b^2\right )}{96\,b^3\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cosh(c + d*x)^3*sinh(c + d*x)^3)/(a + b*sinh(c + d*x)),x)

[Out]

exp(5*c + 5*d*x)/(160*b*d) - exp(- 5*c - 5*d*x)/(160*b*d) - (a*exp(- 4*c - 4*d*x))/(64*b^2*d) - (a*exp(4*c + 4
*d*x))/(64*b^2*d) - (log(2*a*exp(d*x)*exp(c) - b + b*exp(2*c)*exp(2*d*x))*(a^5 + a^3*b^2))/(b^6*d) - (exp(- c
- d*x)*(8*a^4 - b^4 + 6*a^2*b^2))/(16*b^5*d) + (a^3*x*(a^2 + b^2))/b^6 - (exp(- 2*c - 2*d*x)*(a*b^2 + 2*a^3))/
(16*b^4*d) - (exp(2*c + 2*d*x)*(a*b^2 + 2*a^3))/(16*b^4*d) + (exp(c + d*x)*(8*a^4 - b^4 + 6*a^2*b^2))/(16*b^5*
d) - (exp(- 3*c - 3*d*x)*(4*a^2 + b^2))/(96*b^3*d) + (exp(3*c + 3*d*x)*(4*a^2 + b^2))/(96*b^3*d)

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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(cosh(d*x+c)**3*sinh(d*x+c)**3/(a+b*sinh(d*x+c)),x)

[Out]

Timed out

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