### 3.579 $$\int \frac{a+b \text{csch}^2(x)}{c+d \sinh (x)} \, dx$$

Optimal. Leaf size=69 $-\frac{2 \left (a c^2+b d^2\right ) \tanh ^{-1}\left (\frac{d-c \tanh \left (\frac{x}{2}\right )}{\sqrt{c^2+d^2}}\right )}{c^2 \sqrt{c^2+d^2}}+\frac{b d \tanh ^{-1}(\cosh (x))}{c^2}-\frac{b \coth (x)}{c}$

[Out]

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

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Rubi [A]  time = 0.26344, antiderivative size = 69, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 17, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.412, Rules used = {4233, 3056, 3001, 3770, 2660, 618, 206} $-\frac{2 \left (a c^2+b d^2\right ) \tanh ^{-1}\left (\frac{d-c \tanh \left (\frac{x}{2}\right )}{\sqrt{c^2+d^2}}\right )}{c^2 \sqrt{c^2+d^2}}+\frac{b d \tanh ^{-1}(\cosh (x))}{c^2}-\frac{b \coth (x)}{c}$

Antiderivative was successfully veriﬁed.

[In]

Int[(a + b*Csch[x]^2)/(c + d*Sinh[x]),x]

[Out]

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

Rule 4233

Int[(csc[(a_.) + (b_.)*(x_)]^2*(C_.) + (A_))*(u_), x_Symbol] :> Int[(ActivateTrig[u]*(C + A*Sin[a + b*x]^2))/S
in[a + b*x]^2, x] /; FreeQ[{a, b, A, C}, x] && KnownSineIntegrandQ[u, x]

Rule 3056

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*s
in[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 + 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] + Dist[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), I
nt[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[a*(m + 1)*(b*c - a*d)*(A + C) + d*(A*b^2 + a^2*C)*
(m + n + 2) - (c*(A*b^2 + a^2*C) + b*(m + 1)*(b*c - a*d)*(A + C))*Sin[e + f*x] - d*(A*b^2 + a^2*C)*(m + n + 3)
*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]
&& NeQ[c^2 - d^2, 0] && LtQ[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 206

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

Rubi steps

\begin{align*} \int \frac{a+b \text{csch}^2(x)}{c+d \sinh (x)} \, dx &=-\int \frac{\text{csch}^2(x) \left (-b-a \sinh ^2(x)\right )}{c+d \sinh (x)} \, dx\\ &=-\frac{b \coth (x)}{c}-\frac{i \int \frac{\text{csch}(x) (-i b d+i a c \sinh (x))}{c+d \sinh (x)} \, dx}{c}\\ &=-\frac{b \coth (x)}{c}-\frac{(b d) \int \text{csch}(x) \, dx}{c^2}+\left (a+\frac{b d^2}{c^2}\right ) \int \frac{1}{c+d \sinh (x)} \, dx\\ &=\frac{b d \tanh ^{-1}(\cosh (x))}{c^2}-\frac{b \coth (x)}{c}+\left (2 \left (a+\frac{b d^2}{c^2}\right )\right ) \operatorname{Subst}\left (\int \frac{1}{c+2 d x-c x^2} \, dx,x,\tanh \left (\frac{x}{2}\right )\right )\\ &=\frac{b d \tanh ^{-1}(\cosh (x))}{c^2}-\frac{b \coth (x)}{c}-\left (4 \left (a+\frac{b d^2}{c^2}\right )\right ) \operatorname{Subst}\left (\int \frac{1}{4 \left (c^2+d^2\right )-x^2} \, dx,x,2 d-2 c \tanh \left (\frac{x}{2}\right )\right )\\ &=\frac{b d \tanh ^{-1}(\cosh (x))}{c^2}-\frac{2 \left (a+\frac{b d^2}{c^2}\right ) \tanh ^{-1}\left (\frac{d-c \tanh \left (\frac{x}{2}\right )}{\sqrt{c^2+d^2}}\right )}{\sqrt{c^2+d^2}}-\frac{b \coth (x)}{c}\\ \end{align*}

Mathematica [A]  time = 0.410293, size = 125, normalized size = 1.81 $-\frac{\text{csch}\left (\frac{x}{2}\right ) \text{sech}\left (\frac{x}{2}\right ) \left (\sinh (x) \left (b d \sqrt{-c^2-d^2} \log \left (\tanh \left (\frac{x}{2}\right )\right )-2 \left (a c^2+b d^2\right ) \tan ^{-1}\left (\frac{d-c \tanh \left (\frac{x}{2}\right )}{\sqrt{-c^2-d^2}}\right )\right )+b c \sqrt{-c^2-d^2} \cosh (x)\right )}{2 c^2 \sqrt{-c^2-d^2}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(a + b*Csch[x]^2)/(c + d*Sinh[x]),x]

[Out]

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

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Maple [A]  time = 0.036, size = 112, normalized size = 1.6 \begin{align*} -{\frac{b}{2\,c}\tanh \left ({\frac{x}{2}} \right ) }-{\frac{b}{2\,c} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{-1}}-{\frac{bd}{{c}^{2}}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) }+2\,{\frac{a}{\sqrt{{c}^{2}+{d}^{2}}}{\it Artanh} \left ( 1/2\,{\frac{2\,c\tanh \left ( x/2 \right ) -2\,d}{\sqrt{{c}^{2}+{d}^{2}}}} \right ) }+2\,{\frac{b{d}^{2}}{{c}^{2}\sqrt{{c}^{2}+{d}^{2}}}{\it Artanh} \left ( 1/2\,{\frac{2\,c\tanh \left ( x/2 \right ) -2\,d}{\sqrt{{c}^{2}+{d}^{2}}}} \right ) } \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*csch(x)^2)/(c+d*sinh(x)),x)

[Out]

-1/2*b/c*tanh(1/2*x)-1/2*b/c/tanh(1/2*x)-1/c^2*b*d*ln(tanh(1/2*x))+2/(c^2+d^2)^(1/2)*arctanh(1/2*(2*c*tanh(1/2
*x)-2*d)/(c^2+d^2)^(1/2))*a+2/c^2/(c^2+d^2)^(1/2)*arctanh(1/2*(2*c*tanh(1/2*x)-2*d)/(c^2+d^2)^(1/2))*b*d^2

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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((a+b*csch(x)^2)/(c+d*sinh(x)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 6.54043, size = 1037, normalized size = 15.03 \begin{align*} \frac{2 \, b c^{3} + 2 \, b c d^{2} +{\left (a c^{2} + b d^{2} -{\left (a c^{2} + b d^{2}\right )} \cosh \left (x\right )^{2} - 2 \,{\left (a c^{2} + b d^{2}\right )} \cosh \left (x\right ) \sinh \left (x\right ) -{\left (a c^{2} + b d^{2}\right )} \sinh \left (x\right )^{2}\right )} \sqrt{c^{2} + d^{2}} \log \left (\frac{d^{2} \cosh \left (x\right )^{2} + d^{2} \sinh \left (x\right )^{2} + 2 \, c d \cosh \left (x\right ) + 2 \, c^{2} + d^{2} + 2 \,{\left (d^{2} \cosh \left (x\right ) + c d\right )} \sinh \left (x\right ) - 2 \, \sqrt{c^{2} + d^{2}}{\left (d \cosh \left (x\right ) + d \sinh \left (x\right ) + c\right )}}{d \cosh \left (x\right )^{2} + d \sinh \left (x\right )^{2} + 2 \, c \cosh \left (x\right ) + 2 \,{\left (d \cosh \left (x\right ) + c\right )} \sinh \left (x\right ) - d}\right ) +{\left (b c^{2} d + b d^{3} -{\left (b c^{2} d + b d^{3}\right )} \cosh \left (x\right )^{2} - 2 \,{\left (b c^{2} d + b d^{3}\right )} \cosh \left (x\right ) \sinh \left (x\right ) -{\left (b c^{2} d + b d^{3}\right )} \sinh \left (x\right )^{2}\right )} \log \left (\cosh \left (x\right ) + \sinh \left (x\right ) + 1\right ) -{\left (b c^{2} d + b d^{3} -{\left (b c^{2} d + b d^{3}\right )} \cosh \left (x\right )^{2} - 2 \,{\left (b c^{2} d + b d^{3}\right )} \cosh \left (x\right ) \sinh \left (x\right ) -{\left (b c^{2} d + b d^{3}\right )} \sinh \left (x\right )^{2}\right )} \log \left (\cosh \left (x\right ) + \sinh \left (x\right ) - 1\right )}{c^{4} + c^{2} d^{2} -{\left (c^{4} + c^{2} d^{2}\right )} \cosh \left (x\right )^{2} - 2 \,{\left (c^{4} + c^{2} d^{2}\right )} \cosh \left (x\right ) \sinh \left (x\right ) -{\left (c^{4} + c^{2} d^{2}\right )} \sinh \left (x\right )^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*csch(x)^2)/(c+d*sinh(x)),x, algorithm="fricas")

[Out]

(2*b*c^3 + 2*b*c*d^2 + (a*c^2 + b*d^2 - (a*c^2 + b*d^2)*cosh(x)^2 - 2*(a*c^2 + b*d^2)*cosh(x)*sinh(x) - (a*c^2
+ b*d^2)*sinh(x)^2)*sqrt(c^2 + d^2)*log((d^2*cosh(x)^2 + d^2*sinh(x)^2 + 2*c*d*cosh(x) + 2*c^2 + d^2 + 2*(d^2
*cosh(x) + c*d)*sinh(x) - 2*sqrt(c^2 + d^2)*(d*cosh(x) + d*sinh(x) + c))/(d*cosh(x)^2 + d*sinh(x)^2 + 2*c*cosh
(x) + 2*(d*cosh(x) + c)*sinh(x) - d)) + (b*c^2*d + b*d^3 - (b*c^2*d + b*d^3)*cosh(x)^2 - 2*(b*c^2*d + b*d^3)*c
osh(x)*sinh(x) - (b*c^2*d + b*d^3)*sinh(x)^2)*log(cosh(x) + sinh(x) + 1) - (b*c^2*d + b*d^3 - (b*c^2*d + b*d^3
)*cosh(x)^2 - 2*(b*c^2*d + b*d^3)*cosh(x)*sinh(x) - (b*c^2*d + b*d^3)*sinh(x)^2)*log(cosh(x) + sinh(x) - 1))/(
c^4 + c^2*d^2 - (c^4 + c^2*d^2)*cosh(x)^2 - 2*(c^4 + c^2*d^2)*cosh(x)*sinh(x) - (c^4 + c^2*d^2)*sinh(x)^2)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{a + b \operatorname{csch}^{2}{\left (x \right )}}{c + d \sinh{\left (x \right )}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*csch(x)**2)/(c+d*sinh(x)),x)

[Out]

Integral((a + b*csch(x)**2)/(c + d*sinh(x)), x)

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Giac [A]  time = 1.17998, size = 147, normalized size = 2.13 \begin{align*} \frac{b d \log \left (e^{x} + 1\right )}{c^{2}} - \frac{b d \log \left ({\left | e^{x} - 1 \right |}\right )}{c^{2}} + \frac{{\left (a c^{2} + b d^{2}\right )} \log \left (\frac{{\left | 2 \, d e^{x} + 2 \, c - 2 \, \sqrt{c^{2} + d^{2}} \right |}}{{\left | 2 \, d e^{x} + 2 \, c + 2 \, \sqrt{c^{2} + d^{2}} \right |}}\right )}{\sqrt{c^{2} + d^{2}} c^{2}} - \frac{2 \, b}{c{\left (e^{\left (2 \, x\right )} - 1\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*csch(x)^2)/(c+d*sinh(x)),x, algorithm="giac")

[Out]

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