\(\int \frac {1}{(a+b \sinh (c+d x))^3} \, dx\) [103]

Optimal result

Integrand size = 12, antiderivative size = 127 \[ \int \frac {1}{(a+b \sinh (c+d x))^3} \, dx=-\frac {\left (2 a^2-b^2\right ) \text {arctanh}\left (\frac {b-a \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{5/2} d}-\frac {b \cosh (c+d x)}{2 \left (a^2+b^2\right ) d (a+b \sinh (c+d x))^2}-\frac {3 a b \cosh (c+d x)}{2 \left (a^2+b^2\right )^2 d (a+b \sinh (c+d x))} \]

[Out]

-(2*a^2-b^2)*arctanh((b-a*tanh(1/2*d*x+1/2*c))/(a^2+b^2)^(1/2))/(a^2+b^2)^(5/2)/d-1/2*b*cosh(d*x+c)/(a^2+b^2)/
d/(a+b*sinh(d*x+c))^2-3/2*a*b*cosh(d*x+c)/(a^2+b^2)^2/d/(a+b*sinh(d*x+c))

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {2743, 2833, 12, 2739, 632, 210} \[ \int \frac {1}{(a+b \sinh (c+d x))^3} \, dx=-\frac {\left (2 a^2-b^2\right ) \text {arctanh}\left (\frac {b-a \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2+b^2}}\right )}{d \left (a^2+b^2\right )^{5/2}}-\frac {3 a b \cosh (c+d x)}{2 d \left (a^2+b^2\right )^2 (a+b \sinh (c+d x))}-\frac {b \cosh (c+d x)}{2 d \left (a^2+b^2\right ) (a+b \sinh (c+d x))^2} \]

[In]

Int[(a + b*Sinh[c + d*x])^(-3),x]

[Out]

-(((2*a^2 - b^2)*ArcTanh[(b - a*Tanh[(c + d*x)/2])/Sqrt[a^2 + b^2]])/((a^2 + b^2)^(5/2)*d)) - (b*Cosh[c + d*x]
)/(2*(a^2 + b^2)*d*(a + b*Sinh[c + d*x])^2) - (3*a*b*Cosh[c + d*x])/(2*(a^2 + b^2)^2*d*(a + b*Sinh[c + d*x]))

Rule 12

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

Rule 210

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

Rule 632

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

Rule 2739

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[2*(e/d), Subst[Int[1/(a + 2*b*e*x + a*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] &&
NeQ[a^2 - b^2, 0]

Rule 2743

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((a + b*Sin[c + d*x])^(n
+ 1)/(d*(n + 1)*(a^2 - b^2))), x] + Dist[1/((n + 1)*(a^2 - b^2)), Int[(a + b*Sin[c + d*x])^(n + 1)*Simp[a*(n +
1) - b*(n + 2)*Sin[c + d*x], x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && LtQ[n, -1] && Integ
erQ[2*n]

Rule 2833

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

Rubi steps \begin{align*} \text {integral}& = -\frac {b \cosh (c+d x)}{2 \left (a^2+b^2\right ) d (a+b \sinh (c+d x))^2}-\frac {\int \frac {-2 a+b \sinh (c+d x)}{(a+b \sinh (c+d x))^2} \, dx}{2 \left (a^2+b^2\right )} \\ & = -\frac {b \cosh (c+d x)}{2 \left (a^2+b^2\right ) d (a+b \sinh (c+d x))^2}-\frac {3 a b \cosh (c+d x)}{2 \left (a^2+b^2\right )^2 d (a+b \sinh (c+d x))}+\frac {\int \frac {2 a^2-b^2}{a+b \sinh (c+d x)} \, dx}{2 \left (a^2+b^2\right )^2} \\ & = -\frac {b \cosh (c+d x)}{2 \left (a^2+b^2\right ) d (a+b \sinh (c+d x))^2}-\frac {3 a b \cosh (c+d x)}{2 \left (a^2+b^2\right )^2 d (a+b \sinh (c+d x))}+\frac {\left (2 a^2-b^2\right ) \int \frac {1}{a+b \sinh (c+d x)} \, dx}{2 \left (a^2+b^2\right )^2} \\ & = -\frac {b \cosh (c+d x)}{2 \left (a^2+b^2\right ) d (a+b \sinh (c+d x))^2}-\frac {3 a b \cosh (c+d x)}{2 \left (a^2+b^2\right )^2 d (a+b \sinh (c+d x))}-\frac {\left (i \left (2 a^2-b^2\right )\right ) \text {Subst}\left (\int \frac {1}{a-2 i b x+a x^2} \, dx,x,\tan \left (\frac {1}{2} (i c+i d x)\right )\right )}{\left (a^2+b^2\right )^2 d} \\ & = -\frac {b \cosh (c+d x)}{2 \left (a^2+b^2\right ) d (a+b \sinh (c+d x))^2}-\frac {3 a b \cosh (c+d x)}{2 \left (a^2+b^2\right )^2 d (a+b \sinh (c+d x))}+\frac {\left (2 i \left (2 a^2-b^2\right )\right ) \text {Subst}\left (\int \frac {1}{-4 \left (a^2+b^2\right )-x^2} \, dx,x,-2 i b+2 a \tan \left (\frac {1}{2} (i c+i d x)\right )\right )}{\left (a^2+b^2\right )^2 d} \\ & = -\frac {\left (2 a^2-b^2\right ) \text {arctanh}\left (\frac {b-a \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{5/2} d}-\frac {b \cosh (c+d x)}{2 \left (a^2+b^2\right ) d (a+b \sinh (c+d x))^2}-\frac {3 a b \cosh (c+d x)}{2 \left (a^2+b^2\right )^2 d (a+b \sinh (c+d x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.24 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.92 \[ \int \frac {1}{(a+b \sinh (c+d x))^3} \, dx=\frac {\frac {2 \left (2 a^2-b^2\right ) \arctan \left (\frac {b-a \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2-b^2}}\right )}{\sqrt {-a^2-b^2}}-\frac {b \cosh (c+d x) \left (4 a^2+b^2+3 a b \sinh (c+d x)\right )}{(a+b \sinh (c+d x))^2}}{2 \left (a^2+b^2\right )^2 d} \]

[In]

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

[Out]

((2*(2*a^2 - b^2)*ArcTan[(b - a*Tanh[(c + d*x)/2])/Sqrt[-a^2 - b^2]])/Sqrt[-a^2 - b^2] - (b*Cosh[c + d*x]*(4*a
^2 + b^2 + 3*a*b*Sinh[c + d*x]))/(a + b*Sinh[c + d*x])^2)/(2*(a^2 + b^2)^2*d)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(279\) vs. \(2(118)=236\).

Time = 1.04 (sec) , antiderivative size = 280, normalized size of antiderivative = 2.20

method	result	size
derivativedivides	\(\frac {-\frac {2 \left (-\frac {b^{2} \left (5 a^{2}+2 b^{2}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{2 a \left (a^{4}+2 a^{2} b^{2}+b^{4}\right )}-\frac {b \left (4 a^{4}-7 a^{2} b^{2}-2 b^{4}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{2 \left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) a^{2}}+\frac {b^{2} \left (11 a^{2}+2 b^{2}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) a}+\frac {b \left (4 a^{2}+b^{2}\right )}{2 a^{4}+4 a^{2} b^{2}+2 b^{4}}\right )}{\left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -2 b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-a \right )^{2}}+\frac {\left (2 a^{2}-b^{2}\right ) \operatorname {arctanh}\left (\frac {2 a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{\left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) \sqrt {a^{2}+b^{2}}}}{d}\)	\(280\)
default	\(\frac {-\frac {2 \left (-\frac {b^{2} \left (5 a^{2}+2 b^{2}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{2 a \left (a^{4}+2 a^{2} b^{2}+b^{4}\right )}-\frac {b \left (4 a^{4}-7 a^{2} b^{2}-2 b^{4}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{2 \left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) a^{2}}+\frac {b^{2} \left (11 a^{2}+2 b^{2}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) a}+\frac {b \left (4 a^{2}+b^{2}\right )}{2 a^{4}+4 a^{2} b^{2}+2 b^{4}}\right )}{\left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -2 b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-a \right )^{2}}+\frac {\left (2 a^{2}-b^{2}\right ) \operatorname {arctanh}\left (\frac {2 a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{\left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) \sqrt {a^{2}+b^{2}}}}{d}\)	\(280\)
risch	\(\frac {2 a^{2} b \,{\mathrm e}^{3 d x +3 c}-b^{3} {\mathrm e}^{3 d x +3 c}+6 \,{\mathrm e}^{2 d x +2 c} a^{3}-3 \,{\mathrm e}^{2 d x +2 c} a \,b^{2}-10 \,{\mathrm e}^{d x +c} a^{2} b -{\mathrm e}^{d x +c} b^{3}+3 a \,b^{2}}{d \left (a^{2}+b^{2}\right )^{2} \left (b \,{\mathrm e}^{2 d x +2 c}+2 a \,{\mathrm e}^{d x +c}-b \right )^{2}}+\frac {\ln \left ({\mathrm e}^{d x +c}+\frac {\left (a^{2}+b^{2}\right )^{\frac {5}{2}} a -a^{6}-3 a^{4} b^{2}-3 a^{2} b^{4}-b^{6}}{b \left (a^{2}+b^{2}\right )^{\frac {5}{2}}}\right ) a^{2}}{\left (a^{2}+b^{2}\right )^{\frac {5}{2}} d}-\frac {\ln \left ({\mathrm e}^{d x +c}+\frac {\left (a^{2}+b^{2}\right )^{\frac {5}{2}} a -a^{6}-3 a^{4} b^{2}-3 a^{2} b^{4}-b^{6}}{b \left (a^{2}+b^{2}\right )^{\frac {5}{2}}}\right ) b^{2}}{2 \left (a^{2}+b^{2}\right )^{\frac {5}{2}} d}-\frac {\ln \left ({\mathrm e}^{d x +c}+\frac {\left (a^{2}+b^{2}\right )^{\frac {5}{2}} a +a^{6}+3 a^{4} b^{2}+3 a^{2} b^{4}+b^{6}}{b \left (a^{2}+b^{2}\right )^{\frac {5}{2}}}\right ) a^{2}}{\left (a^{2}+b^{2}\right )^{\frac {5}{2}} d}+\frac {\ln \left ({\mathrm e}^{d x +c}+\frac {\left (a^{2}+b^{2}\right )^{\frac {5}{2}} a +a^{6}+3 a^{4} b^{2}+3 a^{2} b^{4}+b^{6}}{b \left (a^{2}+b^{2}\right )^{\frac {5}{2}}}\right ) b^{2}}{2 \left (a^{2}+b^{2}\right )^{\frac {5}{2}} d}\)	\(424\)

[In]

int(1/(a+b*sinh(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

1/d*(-2*(-1/2*b^2*(5*a^2+2*b^2)/a/(a^4+2*a^2*b^2+b^4)*tanh(1/2*d*x+1/2*c)^3-1/2*b*(4*a^4-7*a^2*b^2-2*b^4)/(a^4
+2*a^2*b^2+b^4)/a^2*tanh(1/2*d*x+1/2*c)^2+1/2*b^2*(11*a^2+2*b^2)/(a^4+2*a^2*b^2+b^4)/a*tanh(1/2*d*x+1/2*c)+1/2
*b*(4*a^2+b^2)/(a^4+2*a^2*b^2+b^4))/(tanh(1/2*d*x+1/2*c)^2*a-2*b*tanh(1/2*d*x+1/2*c)-a)^2+(2*a^2-b^2)/(a^4+2*a
^2*b^2+b^4)/(a^2+b^2)^(1/2)*arctanh(1/2*(2*a*tanh(1/2*d*x+1/2*c)-2*b)/(a^2+b^2)^(1/2)))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1347 vs. \(2 (120) = 240\).

Time = 0.30 (sec) , antiderivative size = 1347, normalized size of antiderivative = 10.61 \[ \int \frac {1}{(a+b \sinh (c+d x))^3} \, dx=\text {Too large to display} \]

[In]

integrate(1/(a+b*sinh(d*x+c))^3,x, algorithm="fricas")

[Out]

1/2*(6*a^3*b^2 + 6*a*b^4 + 2*(2*a^4*b + a^2*b^3 - b^5)*cosh(d*x + c)^3 + 2*(2*a^4*b + a^2*b^3 - b^5)*sinh(d*x
+ c)^3 + 6*(2*a^5 + a^3*b^2 - a*b^4)*cosh(d*x + c)^2 + 6*(2*a^5 + a^3*b^2 - a*b^4 + (2*a^4*b + a^2*b^3 - b^5)*
cosh(d*x + c))*sinh(d*x + c)^2 - ((2*a^2*b^2 - b^4)*cosh(d*x + c)^4 + (2*a^2*b^2 - b^4)*sinh(d*x + c)^4 + 2*a^
2*b^2 - b^4 + 4*(2*a^3*b - a*b^3)*cosh(d*x + c)^3 + 4*(2*a^3*b - a*b^3 + (2*a^2*b^2 - b^4)*cosh(d*x + c))*sinh
(d*x + c)^3 + 2*(4*a^4 - 4*a^2*b^2 + b^4)*cosh(d*x + c)^2 + 2*(4*a^4 - 4*a^2*b^2 + b^4 + 3*(2*a^2*b^2 - b^4)*c
osh(d*x + c)^2 + 6*(2*a^3*b - a*b^3)*cosh(d*x + c))*sinh(d*x + c)^2 - 4*(2*a^3*b - a*b^3)*cosh(d*x + c) - 4*(2
*a^3*b - a*b^3 - (2*a^2*b^2 - b^4)*cosh(d*x + c)^3 - 3*(2*a^3*b - a*b^3)*cosh(d*x + c)^2 - (4*a^4 - 4*a^2*b^2
+ b^4)*cosh(d*x + c))*sinh(d*x + c))*sqrt(a^2 + b^2)*log((b^2*cosh(d*x + c)^2 + b^2*sinh(d*x + c)^2 + 2*a*b*co
sh(d*x + c) + 2*a^2 + b^2 + 2*(b^2*cosh(d*x + c) + a*b)*sinh(d*x + c) + 2*sqrt(a^2 + b^2)*(b*cosh(d*x + c) + b
*sinh(d*x + c) + a))/(b*cosh(d*x + c)^2 + b*sinh(d*x + c)^2 + 2*a*cosh(d*x + c) + 2*(b*cosh(d*x + c) + a)*sinh
(d*x + c) - b)) - 2*(10*a^4*b + 11*a^2*b^3 + b^5)*cosh(d*x + c) - 2*(10*a^4*b + 11*a^2*b^3 + b^5 - 3*(2*a^4*b
+ a^2*b^3 - b^5)*cosh(d*x + c)^2 - 6*(2*a^5 + a^3*b^2 - a*b^4)*cosh(d*x + c))*sinh(d*x + c))/((a^6*b^2 + 3*a^4
*b^4 + 3*a^2*b^6 + b^8)*d*cosh(d*x + c)^4 + (a^6*b^2 + 3*a^4*b^4 + 3*a^2*b^6 + b^8)*d*sinh(d*x + c)^4 + 4*(a^7
*b + 3*a^5*b^3 + 3*a^3*b^5 + a*b^7)*d*cosh(d*x + c)^3 + 2*(2*a^8 + 5*a^6*b^2 + 3*a^4*b^4 - a^2*b^6 - b^8)*d*co
sh(d*x + c)^2 + 4*((a^6*b^2 + 3*a^4*b^4 + 3*a^2*b^6 + b^8)*d*cosh(d*x + c) + (a^7*b + 3*a^5*b^3 + 3*a^3*b^5 +
a*b^7)*d)*sinh(d*x + c)^3 - 4*(a^7*b + 3*a^5*b^3 + 3*a^3*b^5 + a*b^7)*d*cosh(d*x + c) + 2*(3*(a^6*b^2 + 3*a^4*
b^4 + 3*a^2*b^6 + b^8)*d*cosh(d*x + c)^2 + 6*(a^7*b + 3*a^5*b^3 + 3*a^3*b^5 + a*b^7)*d*cosh(d*x + c) + (2*a^8
+ 5*a^6*b^2 + 3*a^4*b^4 - a^2*b^6 - b^8)*d)*sinh(d*x + c)^2 + (a^6*b^2 + 3*a^4*b^4 + 3*a^2*b^6 + b^8)*d + 4*((
a^6*b^2 + 3*a^4*b^4 + 3*a^2*b^6 + b^8)*d*cosh(d*x + c)^3 + 3*(a^7*b + 3*a^5*b^3 + 3*a^3*b^5 + a*b^7)*d*cosh(d*
x + c)^2 + (2*a^8 + 5*a^6*b^2 + 3*a^4*b^4 - a^2*b^6 - b^8)*d*cosh(d*x + c) - (a^7*b + 3*a^5*b^3 + 3*a^3*b^5 +
a*b^7)*d)*sinh(d*x + c))

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{(a+b \sinh (c+d x))^3} \, dx=\text {Timed out} \]

[In]

integrate(1/(a+b*sinh(d*x+c))**3,x)

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 315 vs. \(2 (120) = 240\).

Time = 0.28 (sec) , antiderivative size = 315, normalized size of antiderivative = 2.48 \[ \int \frac {1}{(a+b \sinh (c+d x))^3} \, dx=\frac {{\left (2 \, a^{2} - b^{2}\right )} \log \left (\frac {b e^{\left (-d x - c\right )} - a - \sqrt {a^{2} + b^{2}}}{b e^{\left (-d x - c\right )} - a + \sqrt {a^{2} + b^{2}}}\right )}{2 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {a^{2} + b^{2}} d} - \frac {3 \, a b^{2} + {\left (10 \, a^{2} b + b^{3}\right )} e^{\left (-d x - c\right )} + 3 \, {\left (2 \, a^{3} - a b^{2}\right )} e^{\left (-2 \, d x - 2 \, c\right )} - {\left (2 \, a^{2} b - b^{3}\right )} e^{\left (-3 \, d x - 3 \, c\right )}}{{\left (a^{4} b^{2} + 2 \, a^{2} b^{4} + b^{6} + 4 \, {\left (a^{5} b + 2 \, a^{3} b^{3} + a b^{5}\right )} e^{\left (-d x - c\right )} + 2 \, {\left (2 \, a^{6} + 3 \, a^{4} b^{2} - b^{6}\right )} e^{\left (-2 \, d x - 2 \, c\right )} - 4 \, {\left (a^{5} b + 2 \, a^{3} b^{3} + a b^{5}\right )} e^{\left (-3 \, d x - 3 \, c\right )} + {\left (a^{4} b^{2} + 2 \, a^{2} b^{4} + b^{6}\right )} e^{\left (-4 \, d x - 4 \, c\right )}\right )} d} \]

[In]

integrate(1/(a+b*sinh(d*x+c))^3,x, algorithm="maxima")

[Out]

1/2*(2*a^2 - b^2)*log((b*e^(-d*x - c) - a - sqrt(a^2 + b^2))/(b*e^(-d*x - c) - a + sqrt(a^2 + b^2)))/((a^4 + 2
*a^2*b^2 + b^4)*sqrt(a^2 + b^2)*d) - (3*a*b^2 + (10*a^2*b + b^3)*e^(-d*x - c) + 3*(2*a^3 - a*b^2)*e^(-2*d*x -
2*c) - (2*a^2*b - b^3)*e^(-3*d*x - 3*c))/((a^4*b^2 + 2*a^2*b^4 + b^6 + 4*(a^5*b + 2*a^3*b^3 + a*b^5)*e^(-d*x -
c) + 2*(2*a^6 + 3*a^4*b^2 - b^6)*e^(-2*d*x - 2*c) - 4*(a^5*b + 2*a^3*b^3 + a*b^5)*e^(-3*d*x - 3*c) + (a^4*b^2
+ 2*a^2*b^4 + b^6)*e^(-4*d*x - 4*c))*d)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 231, normalized size of antiderivative = 1.82 \[ \int \frac {1}{(a+b \sinh (c+d x))^3} \, dx=\frac {\frac {{\left (2 \, a^{2} - b^{2}\right )} \log \left (\frac {{\left | 2 \, b e^{\left (d x + c\right )} + 2 \, a - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, b e^{\left (d x + c\right )} + 2 \, a + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {a^{2} + b^{2}}} + \frac {2 \, {\left (2 \, a^{2} b e^{\left (3 \, d x + 3 \, c\right )} - b^{3} e^{\left (3 \, d x + 3 \, c\right )} + 6 \, a^{3} e^{\left (2 \, d x + 2 \, c\right )} - 3 \, a b^{2} e^{\left (2 \, d x + 2 \, c\right )} - 10 \, a^{2} b e^{\left (d x + c\right )} - b^{3} e^{\left (d x + c\right )} + 3 \, a b^{2}\right )}}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} {\left (b e^{\left (2 \, d x + 2 \, c\right )} + 2 \, a e^{\left (d x + c\right )} - b\right )}^{2}}}{2 \, d} \]

[In]

integrate(1/(a+b*sinh(d*x+c))^3,x, algorithm="giac")

[Out]

1/2*((2*a^2 - b^2)*log(abs(2*b*e^(d*x + c) + 2*a - 2*sqrt(a^2 + b^2))/abs(2*b*e^(d*x + c) + 2*a + 2*sqrt(a^2 +
b^2)))/((a^4 + 2*a^2*b^2 + b^4)*sqrt(a^2 + b^2)) + 2*(2*a^2*b*e^(3*d*x + 3*c) - b^3*e^(3*d*x + 3*c) + 6*a^3*e
^(2*d*x + 2*c) - 3*a*b^2*e^(2*d*x + 2*c) - 10*a^2*b*e^(d*x + c) - b^3*e^(d*x + c) + 3*a*b^2)/((a^4 + 2*a^2*b^2
+ b^4)*(b*e^(2*d*x + 2*c) + 2*a*e^(d*x + c) - b)^2))/d

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(a+b \sinh (c+d x))^3} \, dx=\int \frac {1}{{\left (a+b\,\mathrm {sinh}\left (c+d\,x\right )\right )}^3} \,d x \]

[In]

int(1/(a + b*sinh(c + d*x))^3,x)

[Out]

int(1/(a + b*sinh(c + d*x))^3, x)