3.251 \(\int \frac{\text{csch}^3(c+d x)}{a+b \sinh (c+d x)} \, dx\)
Optimal. Leaf size=113 \[ \frac{2 b^3 \tanh ^{-1}\left (\frac{b-a \tanh \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2+b^2}}\right )}{a^3 d \sqrt{a^2+b^2}}+\frac{\left (a^2-2 b^2\right ) \tanh ^{-1}(\cosh (c+d x))}{2 a^3 d}+\frac{b \coth (c+d x)}{a^2 d}-\frac{\coth (c+d x) \text{csch}(c+d x)}{2 a d} \]
[Out]
((a^2 - 2*b^2)*ArcTanh[Cosh[c + d*x]])/(2*a^3*d) + (2*b^3*ArcTanh[(b - a*Tanh[(c + d*x)/2])/Sqrt[a^2 + b^2]])/
(a^3*Sqrt[a^2 + b^2]*d) + (b*Coth[c + d*x])/(a^2*d) - (Coth[c + d*x]*Csch[c + d*x])/(2*a*d)
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Rubi [A] time = 0.404259, antiderivative size = 113, normalized size of antiderivative = 1.,
number of steps used = 7, number of rules used = 7, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used =
{2802, 3055, 3001, 3770, 2660, 618, 204} \[ \frac{2 b^3 \tanh ^{-1}\left (\frac{b-a \tanh \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2+b^2}}\right )}{a^3 d \sqrt{a^2+b^2}}+\frac{\left (a^2-2 b^2\right ) \tanh ^{-1}(\cosh (c+d x))}{2 a^3 d}+\frac{b \coth (c+d x)}{a^2 d}-\frac{\coth (c+d x) \text{csch}(c+d x)}{2 a d} \]
Antiderivative was successfully verified.
[In]
Int[Csch[c + d*x]^3/(a + b*Sinh[c + d*x]),x]
[Out]
((a^2 - 2*b^2)*ArcTanh[Cosh[c + d*x]])/(2*a^3*d) + (2*b^3*ArcTanh[(b - a*Tanh[(c + d*x)/2])/Sqrt[a^2 + b^2]])/
(a^3*Sqrt[a^2 + b^2]*d) + (b*Coth[c + d*x])/(a^2*d) - (Coth[c + d*x]*Csch[c + d*x])/(2*a*d)
Rule 2802
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -S
imp[(b^2*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 -
b^2)), x] + Dist[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n
*Simp[a*(b*c - a*d)*(m + 1) + b^2*d*(m + n + 2) - (b^2*c + b*(b*c - a*d)*(m + 1))*Sin[e + f*x] - b^2*d*(m + n
+ 3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] &
& NeQ[c^2 - d^2, 0] && LtQ[m, -1] && IntegersQ[2*m, 2*n] && ((EqQ[a, 0] && IntegerQ[m] && !IntegerQ[n]) || !
(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] && !IntegerQ[m]) || EqQ[a, 0])))
Rule 3055
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*s
in[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 - a*b*B + a^2*C)*Cos[e +
f*x]*(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2)), x] + Dis
t[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(m + 1)*(b
*c - a*d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - a*b*B + a^2*C) + (m + 1)*(b*
c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A*b^2 - a*b*B + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x] /;
FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && Lt
Q[m, -1] && ((EqQ[a, 0] && IntegerQ[m] && !IntegerQ[n]) || !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&
!IntegerQ[m]) || EqQ[a, 0])))
Rule 3001
Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)])), x_Symbol] :> Dist[(A*b - a*B)/(b*c - a*d), Int[1/(a + b*Sin[e + f*x]), x], x] + Dist[(B*c - A
*d)/(b*c - a*d), Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0]
&& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
Rule 3770
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Rule 2660
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 618
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 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])
Rubi steps
\begin{align*} \int \frac{\text{csch}^3(c+d x)}{a+b \sinh (c+d x)} \, dx &=-\frac{\coth (c+d x) \text{csch}(c+d x)}{2 a d}+\frac{i \int \frac{\text{csch}^2(c+d x) \left (2 i b+i a \sinh (c+d x)+i b \sinh ^2(c+d x)\right )}{a+b \sinh (c+d x)} \, dx}{2 a}\\ &=\frac{b \coth (c+d x)}{a^2 d}-\frac{\coth (c+d x) \text{csch}(c+d x)}{2 a d}-\frac{\int \frac{\text{csch}(c+d x) \left (a^2-2 b^2+a b \sinh (c+d x)\right )}{a+b \sinh (c+d x)} \, dx}{2 a^2}\\ &=\frac{b \coth (c+d x)}{a^2 d}-\frac{\coth (c+d x) \text{csch}(c+d x)}{2 a d}-\frac{b^3 \int \frac{1}{a+b \sinh (c+d x)} \, dx}{a^3}-\frac{\left (a^2-2 b^2\right ) \int \text{csch}(c+d x) \, dx}{2 a^3}\\ &=\frac{\left (a^2-2 b^2\right ) \tanh ^{-1}(\cosh (c+d x))}{2 a^3 d}+\frac{b \coth (c+d x)}{a^2 d}-\frac{\coth (c+d x) \text{csch}(c+d x)}{2 a d}+\frac{\left (2 i b^3\right ) \operatorname{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 )}{a^3 d}\\ &=\frac{\left (a^2-2 b^2\right ) \tanh ^{-1}(\cosh (c+d x))}{2 a^3 d}+\frac{b \coth (c+d x)}{a^2 d}-\frac{\coth (c+d x) \text{csch}(c+d x)}{2 a d}-\frac{\left (4 i b^3\right ) \operatorname{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 )}{a^3 d}\\ &=\frac{\left (a^2-2 b^2\right ) \tanh ^{-1}(\cosh (c+d x))}{2 a^3 d}+\frac{2 b^3 \tanh ^{-1}\left (\frac{b-a \tanh \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2+b^2}}\right )}{a^3 \sqrt{a^2+b^2} d}+\frac{b \coth (c+d x)}{a^2 d}-\frac{\coth (c+d x) \text{csch}(c+d x)}{2 a d}\\ \end{align*}
Mathematica [A] time = 1.94352, size = 145, normalized size = 1.28 \[ -\frac{4 \left (a^2-2 b^2\right ) \log \left (\tanh \left (\frac{1}{2} (c+d x)\right )\right )+\frac{16 b^3 \tan ^{-1}\left (\frac{b-a \tanh \left (\frac{1}{2} (c+d x)\right )}{\sqrt{-a^2-b^2}}\right )}{\sqrt{-a^2-b^2}}+a^2 \text{csch}^2\left (\frac{1}{2} (c+d x)\right )+a^2 \text{sech}^2\left (\frac{1}{2} (c+d x)\right )-4 a b \tanh \left (\frac{1}{2} (c+d x)\right )-4 a b \coth \left (\frac{1}{2} (c+d x)\right )}{8 a^3 d} \]
Antiderivative was successfully verified.
[In]
Integrate[Csch[c + d*x]^3/(a + b*Sinh[c + d*x]),x]
[Out]
-((16*b^3*ArcTan[(b - a*Tanh[(c + d*x)/2])/Sqrt[-a^2 - b^2]])/Sqrt[-a^2 - b^2] - 4*a*b*Coth[(c + d*x)/2] + a^2
*Csch[(c + d*x)/2]^2 + 4*(a^2 - 2*b^2)*Log[Tanh[(c + d*x)/2]] + a^2*Sech[(c + d*x)/2]^2 - 4*a*b*Tanh[(c + d*x)
/2])/(8*a^3*d)
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Maple [A] time = 0.003, size = 164, normalized size = 1.5 \begin{align*}{\frac{1}{8\,da} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}}+{\frac{b}{2\,d{a}^{2}}\tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-2\,{\frac{{b}^{3}}{d{a}^{3}\sqrt{{a}^{2}+{b}^{2}}}{\it Artanh} \left ( 1/2\,{\frac{2\,a\tanh \left ( 1/2\,dx+c/2 \right ) -2\,b}{\sqrt{{a}^{2}+{b}^{2}}}} \right ) }-{\frac{1}{8\,da} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-2}}-{\frac{1}{2\,da}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) }+{\frac{{b}^{2}}{d{a}^{3}}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) }+{\frac{b}{2\,d{a}^{2}} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(csch(d*x+c)^3/(a+b*sinh(d*x+c)),x)
[Out]
1/8/d/a*tanh(1/2*d*x+1/2*c)^2+1/2/d/a^2*tanh(1/2*d*x+1/2*c)*b-2/d/a^3*b^3/(a^2+b^2)^(1/2)*arctanh(1/2*(2*a*tan
h(1/2*d*x+1/2*c)-2*b)/(a^2+b^2)^(1/2))-1/8/d/a/tanh(1/2*d*x+1/2*c)^2-1/2/d/a*ln(tanh(1/2*d*x+1/2*c))+1/d/a^3*l
n(tanh(1/2*d*x+1/2*c))*b^2+1/2/d*b/a^2/tanh(1/2*d*x+1/2*c)
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Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(csch(d*x+c)^3/(a+b*sinh(d*x+c)),x, algorithm="maxima")
[Out]
Exception raised: ValueError
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Fricas [B] time = 3.02522, size = 2844, normalized size = 25.17 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(csch(d*x+c)^3/(a+b*sinh(d*x+c)),x, algorithm="fricas")
[Out]
-1/2*(4*a^3*b + 4*a*b^3 + 2*(a^4 + a^2*b^2)*cosh(d*x + c)^3 + 2*(a^4 + a^2*b^2)*sinh(d*x + c)^3 - 4*(a^3*b + a
*b^3)*cosh(d*x + c)^2 - 2*(2*a^3*b + 2*a*b^3 - 3*(a^4 + a^2*b^2)*cosh(d*x + c))*sinh(d*x + c)^2 - 2*(b^3*cosh(
d*x + c)^4 + 4*b^3*cosh(d*x + c)*sinh(d*x + c)^3 + b^3*sinh(d*x + c)^4 - 2*b^3*cosh(d*x + c)^2 + b^3 + 2*(3*b^
3*cosh(d*x + c)^2 - b^3)*sinh(d*x + c)^2 + 4*(b^3*cosh(d*x + c)^3 - b^3*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*cosh(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*s
inh(d*x + c)^2 + 2*a*cosh(d*x + c) + 2*(b*cosh(d*x + c) + a)*sinh(d*x + c) - b)) + 2*(a^4 + a^2*b^2)*cosh(d*x
+ c) - ((a^4 - a^2*b^2 - 2*b^4)*cosh(d*x + c)^4 + 4*(a^4 - a^2*b^2 - 2*b^4)*cosh(d*x + c)*sinh(d*x + c)^3 + (a
^4 - a^2*b^2 - 2*b^4)*sinh(d*x + c)^4 + a^4 - a^2*b^2 - 2*b^4 - 2*(a^4 - a^2*b^2 - 2*b^4)*cosh(d*x + c)^2 - 2*
(a^4 - a^2*b^2 - 2*b^4 - 3*(a^4 - a^2*b^2 - 2*b^4)*cosh(d*x + c)^2)*sinh(d*x + c)^2 + 4*((a^4 - a^2*b^2 - 2*b^
4)*cosh(d*x + c)^3 - (a^4 - a^2*b^2 - 2*b^4)*cosh(d*x + c))*sinh(d*x + c))*log(cosh(d*x + c) + sinh(d*x + c) +
1) + ((a^4 - a^2*b^2 - 2*b^4)*cosh(d*x + c)^4 + 4*(a^4 - a^2*b^2 - 2*b^4)*cosh(d*x + c)*sinh(d*x + c)^3 + (a^
4 - a^2*b^2 - 2*b^4)*sinh(d*x + c)^4 + a^4 - a^2*b^2 - 2*b^4 - 2*(a^4 - a^2*b^2 - 2*b^4)*cosh(d*x + c)^2 - 2*(
a^4 - a^2*b^2 - 2*b^4 - 3*(a^4 - a^2*b^2 - 2*b^4)*cosh(d*x + c)^2)*sinh(d*x + c)^2 + 4*((a^4 - a^2*b^2 - 2*b^4
)*cosh(d*x + c)^3 - (a^4 - a^2*b^2 - 2*b^4)*cosh(d*x + c))*sinh(d*x + c))*log(cosh(d*x + c) + sinh(d*x + c) -
1) + 2*(a^4 + a^2*b^2 + 3*(a^4 + a^2*b^2)*cosh(d*x + c)^2 - 4*(a^3*b + a*b^3)*cosh(d*x + c))*sinh(d*x + c))/((
a^5 + a^3*b^2)*d*cosh(d*x + c)^4 + 4*(a^5 + a^3*b^2)*d*cosh(d*x + c)*sinh(d*x + c)^3 + (a^5 + a^3*b^2)*d*sinh(
d*x + c)^4 - 2*(a^5 + a^3*b^2)*d*cosh(d*x + c)^2 + 2*(3*(a^5 + a^3*b^2)*d*cosh(d*x + c)^2 - (a^5 + a^3*b^2)*d)
*sinh(d*x + c)^2 + (a^5 + a^3*b^2)*d + 4*((a^5 + a^3*b^2)*d*cosh(d*x + c)^3 - (a^5 + a^3*b^2)*d*cosh(d*x + c))
*sinh(d*x + c))
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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(csch(d*x+c)**3/(a+b*sinh(d*x+c)),x)
[Out]
Timed out
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Giac [A] time = 1.2314, size = 248, normalized size = 2.19 \begin{align*} -\frac{b^{3} \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 )}{\sqrt{a^{2} + b^{2}} a^{3} d} + \frac{{\left (a^{2} - 2 \, b^{2}\right )} \log \left (e^{\left (d x + c\right )} + 1\right )}{2 \, a^{3} d} - \frac{{\left (a^{2} - 2 \, b^{2}\right )} \log \left ({\left | e^{\left (d x + c\right )} - 1 \right |}\right )}{2 \, a^{3} d} - \frac{a e^{\left (3 \, d x + 3 \, c\right )} - 2 \, b e^{\left (2 \, d x + 2 \, c\right )} + a e^{\left (d x + c\right )} + 2 \, b}{a^{2} d{\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}^{2}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(csch(d*x+c)^3/(a+b*sinh(d*x+c)),x, algorithm="giac")
[Out]
-b^3*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)))/(sqrt(
a^2 + b^2)*a^3*d) + 1/2*(a^2 - 2*b^2)*log(e^(d*x + c) + 1)/(a^3*d) - 1/2*(a^2 - 2*b^2)*log(abs(e^(d*x + c) - 1
))/(a^3*d) - (a*e^(3*d*x + 3*c) - 2*b*e^(2*d*x + 2*c) + a*e^(d*x + c) + 2*b)/(a^2*d*(e^(2*d*x + 2*c) - 1)^2)