3.414 \(\int \frac{\sinh (c+d x) \tanh ^2(c+d x)}{a+b \sinh (c+d x)} \, dx\)

Optimal. Leaf size=121 \[ \frac{2 a^3 \tanh ^{-1}\left (\frac{b-a \tanh \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2+b^2}}\right )}{b d \left (a^2+b^2\right )^{3/2}}-\frac{b \tanh (c+d x)}{d \left (a^2+b^2\right )}+\frac{a \text{sech}(c+d x)}{d \left (a^2+b^2\right )}+\frac{a^2 x}{b \left (a^2+b^2\right )}+\frac{b x}{a^2+b^2} \]

[Out]

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

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Rubi [A]  time = 0.20893, antiderivative size = 121, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.296, Rules used = {2902, 2606, 8, 3473, 2735, 2660, 618, 204} \[ \frac{2 a^3 \tanh ^{-1}\left (\frac{b-a \tanh \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2+b^2}}\right )}{b d \left (a^2+b^2\right )^{3/2}}-\frac{b \tanh (c+d x)}{d \left (a^2+b^2\right )}+\frac{a \text{sech}(c+d x)}{d \left (a^2+b^2\right )}+\frac{a^2 x}{b \left (a^2+b^2\right )}+\frac{b x}{a^2+b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2902

Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_))/((a_) + (b_.)*sin[(e_.) + (f_.
)*(x_)]), x_Symbol] :> Dist[(a*d^2)/(a^2 - b^2), Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^(n - 2), x], x] + (-D
ist[(b*d)/(a^2 - b^2), Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^(n - 1), x], x] - Dist[(a^2*d^2)/(g^2*(a^2 - b^
2)), Int[((g*Cos[e + f*x])^(p + 2)*(d*Sin[e + f*x])^(n - 2))/(a + b*Sin[e + f*x]), x], x]) /; FreeQ[{a, b, d,
e, f, g}, x] && NeQ[a^2 - b^2, 0] && IntegersQ[2*n, 2*p] && LtQ[p, -1] && GtQ[n, 1]

Rule 2606

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 3473

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(b*Tan[c + d*x])^(n - 1))/(d*(n - 1)), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

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

Mathematica [A]  time = 0.502191, size = 96, normalized size = 0.79 \[ \frac{\frac{2 a^3 \tan ^{-1}\left (\frac{b-a \tanh \left (\frac{1}{2} (c+d x)\right )}{\sqrt{-a^2-b^2}}\right )}{b \left (-a^2-b^2\right )^{3/2}}+\frac{\text{sech}(c+d x) (a-b \sinh (c+d x))}{a^2+b^2}+\frac{c+d x}{b}}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.003, size = 158, normalized size = 1.3 \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 ) }-2\,{\frac{{a}^{3}}{bd \left ({a}^{2}+{b}^{2} \right ) ^{3/2}}{\it Artanh} \left ( 1/2\,{\frac{2\,a\tanh \left ( 1/2\,dx+c/2 \right ) -2\,b}{\sqrt{{a}^{2}+{b}^{2}}}} \right ) }-2\,{\frac{\tanh \left ( 1/2\,dx+c/2 \right ) b}{d \left ({a}^{2}+{b}^{2} \right ) \left ( \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1 \right ) }}+2\,{\frac{a}{d \left ({a}^{2}+{b}^{2} \right ) \left ( \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1 \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(d*x+c)*tanh(d*x+c)^2/(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)-2/d/b*a^3/(a^2+b^2)^(3/2)*arctanh(1/2*(2*a*tan
h(1/2*d*x+1/2*c)-2*b)/(a^2+b^2)^(1/2))-2/d/(a^2+b^2)/(tanh(1/2*d*x+1/2*c)^2+1)*tanh(1/2*d*x+1/2*c)*b+2/d/(a^2+
b^2)/(tanh(1/2*d*x+1/2*c)^2+1)*a

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.53876, size = 1126, normalized size = 9.31 \begin{align*} \frac{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} d x \cosh \left (d x + c\right )^{2} +{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} d x \sinh \left (d x + c\right )^{2} + 2 \, a^{2} b^{2} + 2 \, b^{4} +{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} d x +{\left (a^{3} \cosh \left (d x + c\right )^{2} + 2 \, a^{3} \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + a^{3} \sinh \left (d x + c\right )^{2} + a^{3}\right )} \sqrt{a^{2} + b^{2}} \log \left (\frac{b^{2} \cosh \left (d x + c\right )^{2} + b^{2} \sinh \left (d x + c\right )^{2} + 2 \, a b \cosh \left (d x + c\right ) + 2 \, a^{2} + b^{2} + 2 \,{\left (b^{2} \cosh \left (d x + c\right ) + a b\right )} \sinh \left (d x + c\right ) + 2 \, \sqrt{a^{2} + b^{2}}{\left (b \cosh \left (d x + c\right ) + b \sinh \left (d x + c\right ) + a\right )}}{b \cosh \left (d x + c\right )^{2} + b \sinh \left (d x + c\right )^{2} + 2 \, a \cosh \left (d x + c\right ) + 2 \,{\left (b \cosh \left (d x + c\right ) + a\right )} \sinh \left (d x + c\right ) - b}\right ) + 2 \,{\left (a^{3} b + a b^{3}\right )} \cosh \left (d x + c\right ) + 2 \,{\left (a^{3} b + a b^{3} +{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} d x \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )}{{\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} d \cosh \left (d x + c\right )^{2} + 2 \,{\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} d \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) +{\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} d \sinh \left (d x + c\right )^{2} +{\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((a^4 + 2*a^2*b^2 + b^4)*d*x*cosh(d*x + c)^2 + (a^4 + 2*a^2*b^2 + b^4)*d*x*sinh(d*x + c)^2 + 2*a^2*b^2 + 2*b^4
+ (a^4 + 2*a^2*b^2 + b^4)*d*x + (a^3*cosh(d*x + c)^2 + 2*a^3*cosh(d*x + c)*sinh(d*x + c) + a^3*sinh(d*x + c)^
2 + a^3)*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*sinh(d*x + c)^2 + 2*a*cosh(d*x + c) + 2*(b*cosh(d*x + c) + a)*sinh(d*x + c) - b)) + 2*(a^3*b +
a*b^3)*cosh(d*x + c) + 2*(a^3*b + a*b^3 + (a^4 + 2*a^2*b^2 + b^4)*d*x*cosh(d*x + c))*sinh(d*x + c))/((a^4*b +
2*a^2*b^3 + b^5)*d*cosh(d*x + c)^2 + 2*(a^4*b + 2*a^2*b^3 + b^5)*d*cosh(d*x + c)*sinh(d*x + c) + (a^4*b + 2*a
^2*b^3 + b^5)*d*sinh(d*x + c)^2 + (a^4*b + 2*a^2*b^3 + b^5)*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 ^{2}{\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)**2/(a+b*sinh(d*x+c)),x)

[Out]

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

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Giac [A]  time = 1.43235, size = 182, normalized size = 1.5 \begin{align*} \frac{\frac{a^{3} \log \left (\frac{{\left | -2 \, b e^{\left (d x + 2 \, c\right )} - 2 \, a e^{c} - 2 \, \sqrt{a^{2} + b^{2}} e^{c} \right |}}{{\left | -2 \, b e^{\left (d x + 2 \, c\right )} - 2 \, a e^{c} + 2 \, \sqrt{a^{2} + b^{2}} e^{c} \right |}}\right )}{{\left (a^{2} b + b^{3}\right )} \sqrt{a^{2} + b^{2}}} + \frac{d x}{b} + \frac{2 \,{\left (a e^{\left (d x + c\right )} + b\right )}}{{\left (a^{2} + b^{2}\right )}{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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