### 3.27 $$\int \frac{\sinh ^2(c+d x)}{a+b \tanh ^2(c+d x)} \, dx$$

Optimal. Leaf size=78 $-\frac{\sqrt{a} \sqrt{b} \tan ^{-1}\left (\frac{\sqrt{b} \tanh (c+d x)}{\sqrt{a}}\right )}{d (a+b)^2}+\frac{\sinh (c+d x) \cosh (c+d x)}{2 d (a+b)}-\frac{x (a-b)}{2 (a+b)^2}$

[Out]

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

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Rubi [A]  time = 0.104404, antiderivative size = 78, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 23, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.217, Rules used = {3663, 471, 522, 206, 205} $-\frac{\sqrt{a} \sqrt{b} \tan ^{-1}\left (\frac{\sqrt{b} \tanh (c+d x)}{\sqrt{a}}\right )}{d (a+b)^2}+\frac{\sinh (c+d x) \cosh (c+d x)}{2 d (a+b)}-\frac{x (a-b)}{2 (a+b)^2}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

Rule 3663

Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol] :> With[
{ff = FreeFactors[Tan[e + f*x], x]}, Dist[(c*ff^(m + 1))/f, Subst[Int[(x^m*(a + b*(ff*x)^n)^p)/(c^2 + ff^2*x^2
)^(m/2 + 1), x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[m/2]

Rule 471

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(e^(n -
1)*(e*x)^(m - n + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(n*(b*c - a*d)*(p + 1)), x] - Dist[e^n/(n*(b*c -
a*d)*(p + 1)), Int[(e*x)^(m - n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(m - n + 1) + d*(m + n*(p + q + 1)
+ 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[p, -1] && GeQ[n
, m - n + 1] && GtQ[m - n + 1, 0] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 522

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[(b*e - a*f
)/(b*c - a*d), Int[1/(a + b*x^n), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[1/(c + d*x^n), x], x] /; FreeQ[{a
, b, c, d, e, f, n}, x]

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])

Rule 205

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

Rubi steps

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

Mathematica [A]  time = 0.142357, size = 67, normalized size = 0.86 $\frac{-2 (a-b) (c+d x)+(a+b) \sinh (2 (c+d x))-4 \sqrt{a} \sqrt{b} \tan ^{-1}\left (\frac{\sqrt{b} \tanh (c+d x)}{\sqrt{a}}\right )}{4 d (a+b)^2}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [B]  time = 0.076, size = 605, normalized size = 7.8 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/d/(8*a+8*b)/(tanh(1/2*d*x+1/2*c)+1)^2+8/d/(16*a+16*b)/(tanh(1/2*d*x+1/2*c)+1)-1/2/d/(a+b)^2*ln(tanh(1/2*d*x
+1/2*c)+1)*a+1/2/d/(a+b)^2*ln(tanh(1/2*d*x+1/2*c)+1)*b+1/d*a^2*b/(a+b)^2/(b*(a+b))^(1/2)/((2*(b*(a+b))^(1/2)-a
-2*b)*a)^(1/2)*arctanh(a*tanh(1/2*d*x+1/2*c)/((2*(b*(a+b))^(1/2)-a-2*b)*a)^(1/2))-1/d*a*b/(a+b)^2/((2*(b*(a+b)
)^(1/2)-a-2*b)*a)^(1/2)*arctanh(a*tanh(1/2*d*x+1/2*c)/((2*(b*(a+b))^(1/2)-a-2*b)*a)^(1/2))+1/d*a*b^2/(a+b)^2/(
b*(a+b))^(1/2)/((2*(b*(a+b))^(1/2)-a-2*b)*a)^(1/2)*arctanh(a*tanh(1/2*d*x+1/2*c)/((2*(b*(a+b))^(1/2)-a-2*b)*a)
^(1/2))+1/d*a^2*b/(a+b)^2/(b*(a+b))^(1/2)/((2*(b*(a+b))^(1/2)+a+2*b)*a)^(1/2)*arctan(a*tanh(1/2*d*x+1/2*c)/((2
*(b*(a+b))^(1/2)+a+2*b)*a)^(1/2))+1/d*a*b/(a+b)^2/((2*(b*(a+b))^(1/2)+a+2*b)*a)^(1/2)*arctan(a*tanh(1/2*d*x+1/
2*c)/((2*(b*(a+b))^(1/2)+a+2*b)*a)^(1/2))+1/d*a*b^2/(a+b)^2/(b*(a+b))^(1/2)/((2*(b*(a+b))^(1/2)+a+2*b)*a)^(1/2
)*arctan(a*tanh(1/2*d*x+1/2*c)/((2*(b*(a+b))^(1/2)+a+2*b)*a)^(1/2))+4/d/(8*a+8*b)/(tanh(1/2*d*x+1/2*c)-1)^2+8/
d/(16*a+16*b)/(tanh(1/2*d*x+1/2*c)-1)+1/2/d/(a+b)^2*ln(tanh(1/2*d*x+1/2*c)-1)*a-1/2/d/(a+b)^2*ln(tanh(1/2*d*x+
1/2*c)-1)*b

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.44108, size = 2475, normalized size = 31.73 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.65799, size = 238, normalized size = 3.05 \begin{align*} -\frac{\frac{4 \,{\left (a - b\right )} d x}{a^{2} + 2 \, a b + b^{2}} + \frac{8 \, a b \arctan \left (\frac{a e^{\left (2 \, d x + 2 \, c\right )} + b e^{\left (2 \, d x + 2 \, c\right )} + a - b}{2 \, \sqrt{a b}}\right )}{{\left (a^{2} + 2 \, a b + b^{2}\right )} \sqrt{a b}} - \frac{{\left (2 \, a e^{\left (2 \, d x + 2 \, c\right )} - 2 \, b e^{\left (2 \, d x + 2 \, c\right )} - a - b\right )} e^{\left (-2 \, d x\right )}}{a^{2} e^{\left (2 \, c\right )} + 2 \, a b e^{\left (2 \, c\right )} + b^{2} e^{\left (2 \, c\right )}} - \frac{e^{\left (2 \, d x + 8 \, c\right )}}{a e^{\left (6 \, c\right )} + b e^{\left (6 \, c\right )}}}{8 \, d} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(4*(a - b)*d*x/(a^2 + 2*a*b + b^2) + 8*a*b*arctan(1/2*(a*e^(2*d*x + 2*c) + b*e^(2*d*x + 2*c) + a - b)/sqr
t(a*b))/((a^2 + 2*a*b + b^2)*sqrt(a*b)) - (2*a*e^(2*d*x + 2*c) - 2*b*e^(2*d*x + 2*c) - a - b)*e^(-2*d*x)/(a^2*
e^(2*c) + 2*a*b*e^(2*c) + b^2*e^(2*c)) - e^(2*d*x + 8*c)/(a*e^(6*c) + b*e^(6*c)))/d