3.380 \(\int \frac{\sinh (c+d x) \tanh (c+d x)}{a+b \sinh (c+d x)} \, dx\)

Optimal. Leaf size=74 \[ \frac{a^2 \log (a+b \sinh (c+d x))}{b d \left (a^2+b^2\right )}-\frac{a \tan ^{-1}(\sinh (c+d x))}{d \left (a^2+b^2\right )}+\frac{b \log (\cosh (c+d x))}{d \left (a^2+b^2\right )} \]

[Out]

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

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Rubi [A]  time = 0.159761, antiderivative size = 74, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {2837, 12, 1629, 635, 203, 260} \[ \frac{a^2 \log (a+b \sinh (c+d x))}{b d \left (a^2+b^2\right )}-\frac{a \tan ^{-1}(\sinh (c+d x))}{d \left (a^2+b^2\right )}+\frac{b \log (\cosh (c+d x))}{d \left (a^2+b^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[(Sinh[c + d*x]*Tanh[c + d*x])/(a + b*Sinh[c + d*x]),x]

[Out]

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

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]

Rule 12

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

Rule 1629

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*
Pq*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rule 635

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[d, Int[1/(a + c*x^2), x], x] + Dist[e, Int[x/
(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] &&  !NiceSqrtQ[-(a*c)]

Rule 203

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

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rubi steps

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

Mathematica [C]  time = 0.085892, size = 78, normalized size = 1.05 \[ \frac{2 a^2 \log (a+b \sinh (c+d x))+b (b+i a) \log (-\sinh (c+d x)+i)+b (b-i a) \log (\sinh (c+d x)+i)}{2 b d \left (a^2+b^2\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sinh[c + d*x]*Tanh[c + d*x])/(a + b*Sinh[c + d*x]),x]

[Out]

(b*(I*a + b)*Log[I - Sinh[c + d*x]] + b*((-I)*a + b)*Log[I + Sinh[c + d*x]] + 2*a^2*Log[a + b*Sinh[c + d*x]])/
(2*b*(a^2 + b^2)*d)

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Maple [B]  time = 0.001, size = 153, normalized size = 2.1 \begin{align*} -{\frac{1}{bd}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }-{\frac{1}{bd}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }+{\frac{{a}^{2}}{bd \left ({a}^{2}+{b}^{2} \right ) }\ln \left ( \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}a-2\,\tanh \left ( 1/2\,dx+c/2 \right ) b-a \right ) }+4\,{\frac{b\ln \left ( \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1 \right ) }{d \left ( 4\,{a}^{2}+4\,{b}^{2} \right ) }}-8\,{\frac{a\arctan \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) }{d \left ( 4\,{a}^{2}+4\,{b}^{2} \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(d*x+c)*tanh(d*x+c)/(a+b*sinh(d*x+c)),x)

[Out]

-1/d/b*ln(tanh(1/2*d*x+1/2*c)+1)-1/d/b*ln(tanh(1/2*d*x+1/2*c)-1)+1/d*a^2/b/(a^2+b^2)*ln(tanh(1/2*d*x+1/2*c)^2*
a-2*tanh(1/2*d*x+1/2*c)*b-a)+4/d/(4*a^2+4*b^2)*b*ln(tanh(1/2*d*x+1/2*c)^2+1)-8/d/(4*a^2+4*b^2)*a*arctan(tanh(1
/2*d*x+1/2*c))

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Maxima [A]  time = 1.62767, size = 149, normalized size = 2.01 \begin{align*} \frac{a^{2} \log \left (-2 \, a e^{\left (-d x - c\right )} + b e^{\left (-2 \, d x - 2 \, c\right )} - b\right )}{{\left (a^{2} b + b^{3}\right )} d} + \frac{2 \, a \arctan \left (e^{\left (-d x - c\right )}\right )}{{\left (a^{2} + b^{2}\right )} d} + \frac{b \log \left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}{{\left (a^{2} + b^{2}\right )} d} + \frac{d x + c}{b d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^2*log(-2*a*e^(-d*x - c) + b*e^(-2*d*x - 2*c) - b)/((a^2*b + b^3)*d) + 2*a*arctan(e^(-d*x - c))/((a^2 + b^2)*
d) + b*log(e^(-2*d*x - 2*c) + 1)/((a^2 + b^2)*d) + (d*x + c)/(b*d)

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Fricas [A]  time = 2.30851, size = 284, normalized size = 3.84 \begin{align*} -\frac{{\left (a^{2} + b^{2}\right )} d x + 2 \, a b \arctan \left (\cosh \left (d x + c\right ) + \sinh \left (d x + c\right )\right ) - a^{2} \log \left (\frac{2 \,{\left (b \sinh \left (d x + c\right ) + a\right )}}{\cosh \left (d x + c\right ) - \sinh \left (d x + c\right )}\right ) - b^{2} \log \left (\frac{2 \, \cosh \left (d x + c\right )}{\cosh \left (d x + c\right ) - \sinh \left (d x + c\right )}\right )}{{\left (a^{2} b + b^{3}\right )} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((a^2 + b^2)*d*x + 2*a*b*arctan(cosh(d*x + c) + sinh(d*x + c)) - a^2*log(2*(b*sinh(d*x + c) + a)/(cosh(d*x +
c) - sinh(d*x + c))) - b^2*log(2*cosh(d*x + c)/(cosh(d*x + c) - sinh(d*x + c))))/((a^2*b + b^3)*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.52435, size = 128, normalized size = 1.73 \begin{align*} \frac{\frac{a^{2} \log \left ({\left | b e^{\left (2 \, d x + 2 \, c\right )} + 2 \, a e^{\left (d x + c\right )} - b \right |}\right )}{a^{2} b + b^{3}} - \frac{d x}{b} - \frac{2 \, a \arctan \left (e^{\left (d x + c\right )}\right )}{a^{2} + b^{2}} + \frac{b \log \left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}{a^{2} + b^{2}}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a^2*log(abs(b*e^(2*d*x + 2*c) + 2*a*e^(d*x + c) - b))/(a^2*b + b^3) - d*x/b - 2*a*arctan(e^(d*x + c))/(a^2 +
b^2) + b*log(e^(2*d*x + 2*c) + 1)/(a^2 + b^2))/d