3.25 \(\int \frac{\cosh ^2(c+d x)}{a+b \text{csch}(c+d x)} \, dx\)

Optimal. Leaf size=95 \[ \frac{2 b \sqrt{a^2+b^2} \tanh ^{-1}\left (\frac{a-b \tanh \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2+b^2}}\right )}{a^3 d}+\frac{x \left (a^2+2 b^2\right )}{2 a^3}-\frac{\cosh (c+d x) (2 b-a \sinh (c+d x))}{2 a^2 d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.249733, antiderivative size = 95, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {3872, 2865, 2735, 2660, 618, 206} \[ \frac{2 b \sqrt{a^2+b^2} \tanh ^{-1}\left (\frac{a-b \tanh \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2+b^2}}\right )}{a^3 d}+\frac{x \left (a^2+2 b^2\right )}{2 a^3}-\frac{\cosh (c+d x) (2 b-a \sinh (c+d x))}{2 a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3872

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[((g*C
os[e + f*x])^p*(b + a*Sin[e + f*x])^m)/Sin[e + f*x]^m, x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]

Rule 2865

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

Rule 2735

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

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

Mathematica [A]  time = 0.520966, size = 109, normalized size = 1.15 \[ \frac{8 b \sqrt{-a^2-b^2} \tan ^{-1}\left (\frac{a-b \tanh \left (\frac{1}{2} (c+d x)\right )}{\sqrt{-a^2-b^2}}\right )+a^2 \sinh (2 (c+d x))+2 a^2 c+2 a^2 d x-4 a b \cosh (c+d x)+4 b^2 c+4 b^2 d x}{4 a^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^2*c + 4*b^2*c + 2*a^2*d*x + 4*b^2*d*x + 8*b*Sqrt[-a^2 - b^2]*ArcTan[(a - b*Tanh[(c + d*x)/2])/Sqrt[-a^2 -
b^2]] - 4*a*b*Cosh[c + d*x] + a^2*Sinh[2*(c + d*x)])/(4*a^3*d)

________________________________________________________________________________________

Maple [B]  time = 0.049, size = 260, normalized size = 2.7 \begin{align*} -{\frac{1}{2\,da} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-2}}+{\frac{1}{2\,da} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}-{\frac{b}{d{a}^{2}} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}+{\frac{1}{2\,da}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }+{\frac{{b}^{2}}{d{a}^{3}}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }-2\,{\frac{b\sqrt{{a}^{2}+{b}^{2}}}{d{a}^{3}}{\it Artanh} \left ( 1/2\,{\frac{2\,b\tanh \left ( 1/2\,dx+c/2 \right ) -2\,a}{\sqrt{{a}^{2}+{b}^{2}}}} \right ) }+{\frac{1}{2\,da} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-2}}+{\frac{1}{2\,da} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}+{\frac{b}{d{a}^{2}} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}-{\frac{1}{2\,da}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }-{\frac{{b}^{2}}{d{a}^{3}}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/d/a/(tanh(1/2*d*x+1/2*c)+1)^2+1/2/d/a/(tanh(1/2*d*x+1/2*c)+1)-1/d/a^2/(tanh(1/2*d*x+1/2*c)+1)*b+1/2/d/a*l
n(tanh(1/2*d*x+1/2*c)+1)+1/d/a^3*ln(tanh(1/2*d*x+1/2*c)+1)*b^2-2/d*b*(a^2+b^2)^(1/2)/a^3*arctanh(1/2*(2*b*tanh
(1/2*d*x+1/2*c)-2*a)/(a^2+b^2)^(1/2))+1/2/d/a/(tanh(1/2*d*x+1/2*c)-1)^2+1/2/d/a/(tanh(1/2*d*x+1/2*c)-1)+1/d/a^
2/(tanh(1/2*d*x+1/2*c)-1)*b-1/2/d/a*ln(tanh(1/2*d*x+1/2*c)-1)-1/d/a^3*ln(tanh(1/2*d*x+1/2*c)-1)*b^2

________________________________________________________________________________________

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [B]  time = 1.73539, size = 1150, normalized size = 12.11 \begin{align*} \frac{a^{2} \cosh \left (d x + c\right )^{4} + a^{2} \sinh \left (d x + c\right )^{4} + 4 \,{\left (a^{2} + 2 \, b^{2}\right )} d x \cosh \left (d x + c\right )^{2} - 4 \, a b \cosh \left (d x + c\right )^{3} + 4 \,{\left (a^{2} \cosh \left (d x + c\right ) - a b\right )} \sinh \left (d x + c\right )^{3} - 4 \, a b \cosh \left (d x + c\right ) + 2 \,{\left (3 \, a^{2} \cosh \left (d x + c\right )^{2} + 2 \,{\left (a^{2} + 2 \, b^{2}\right )} d x - 6 \, a b \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} + 8 \,{\left (b \cosh \left (d x + c\right )^{2} + 2 \, b \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + b \sinh \left (d x + c\right )^{2}\right )} \sqrt{a^{2} + b^{2}} \log \left (\frac{a^{2} \cosh \left (d x + c\right )^{2} + a^{2} \sinh \left (d x + c\right )^{2} + 2 \, a b \cosh \left (d x + c\right ) + a^{2} + 2 \, b^{2} + 2 \,{\left (a^{2} \cosh \left (d x + c\right ) + a b\right )} \sinh \left (d x + c\right ) + 2 \, \sqrt{a^{2} + b^{2}}{\left (a \cosh \left (d x + c\right ) + a \sinh \left (d x + c\right ) + b\right )}}{a \cosh \left (d x + c\right )^{2} + a \sinh \left (d x + c\right )^{2} + 2 \, b \cosh \left (d x + c\right ) + 2 \,{\left (a \cosh \left (d x + c\right ) + b\right )} \sinh \left (d x + c\right ) - a}\right ) - a^{2} + 4 \,{\left (a^{2} \cosh \left (d x + c\right )^{3} + 2 \,{\left (a^{2} + 2 \, b^{2}\right )} d x \cosh \left (d x + c\right ) - 3 \, a b \cosh \left (d x + c\right )^{2} - a b\right )} \sinh \left (d x + c\right )}{8 \,{\left (a^{3} d \cosh \left (d x + c\right )^{2} + 2 \, a^{3} d \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + a^{3} d \sinh \left (d x + c\right )^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(a^2*cosh(d*x + c)^4 + a^2*sinh(d*x + c)^4 + 4*(a^2 + 2*b^2)*d*x*cosh(d*x + c)^2 - 4*a*b*cosh(d*x + c)^3 +
4*(a^2*cosh(d*x + c) - a*b)*sinh(d*x + c)^3 - 4*a*b*cosh(d*x + c) + 2*(3*a^2*cosh(d*x + c)^2 + 2*(a^2 + 2*b^2
)*d*x - 6*a*b*cosh(d*x + c))*sinh(d*x + c)^2 + 8*(b*cosh(d*x + c)^2 + 2*b*cosh(d*x + c)*sinh(d*x + c) + b*sinh
(d*x + c)^2)*sqrt(a^2 + b^2)*log((a^2*cosh(d*x + c)^2 + a^2*sinh(d*x + c)^2 + 2*a*b*cosh(d*x + c) + a^2 + 2*b^
2 + 2*(a^2*cosh(d*x + c) + a*b)*sinh(d*x + c) + 2*sqrt(a^2 + b^2)*(a*cosh(d*x + c) + a*sinh(d*x + c) + b))/(a*
cosh(d*x + c)^2 + a*sinh(d*x + c)^2 + 2*b*cosh(d*x + c) + 2*(a*cosh(d*x + c) + b)*sinh(d*x + c) - a)) - a^2 +
4*(a^2*cosh(d*x + c)^3 + 2*(a^2 + 2*b^2)*d*x*cosh(d*x + c) - 3*a*b*cosh(d*x + c)^2 - a*b)*sinh(d*x + c))/(a^3*
d*cosh(d*x + c)^2 + 2*a^3*d*cosh(d*x + c)*sinh(d*x + c) + a^3*d*sinh(d*x + c)^2)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 1.17657, size = 223, normalized size = 2.35 \begin{align*} \frac{{\left (a^{2} + 2 \, b^{2}\right )}{\left (d x + c\right )}}{2 \, a^{3} d} - \frac{{\left (4 \, a b e^{\left (d x + c\right )} + a^{2}\right )} e^{\left (-2 \, d x - 2 \, c\right )}}{8 \, a^{3} d} - \frac{{\left (a^{2} b + b^{3}\right )} \log \left (\frac{{\left | 2 \, a e^{\left (d x + c\right )} + 2 \, b - 2 \, \sqrt{a^{2} + b^{2}} \right |}}{{\left | 2 \, a e^{\left (d x + c\right )} + 2 \, b + 2 \, \sqrt{a^{2} + b^{2}} \right |}}\right )}{\sqrt{a^{2} + b^{2}} a^{3} d} + \frac{a d e^{\left (2 \, d x + 2 \, c\right )} - 4 \, b d e^{\left (d x + c\right )}}{8 \, a^{2} d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(a^2 + 2*b^2)*(d*x + c)/(a^3*d) - 1/8*(4*a*b*e^(d*x + c) + a^2)*e^(-2*d*x - 2*c)/(a^3*d) - (a^2*b + b^3)*l
og(abs(2*a*e^(d*x + c) + 2*b - 2*sqrt(a^2 + b^2))/abs(2*a*e^(d*x + c) + 2*b + 2*sqrt(a^2 + b^2)))/(sqrt(a^2 +
b^2)*a^3*d) + 1/8*(a*d*e^(2*d*x + 2*c) - 4*b*d*e^(d*x + c))/(a^2*d^2)