3.27 \(\int \frac{\sinh ^2(c+d x)}{a+b \text{sech}^2(c+d x)} \, dx\)

Optimal. Leaf size=75 \[ \frac{\sqrt{b} \sqrt{a+b} \tanh ^{-1}\left (\frac{\sqrt{b} \tanh (c+d x)}{\sqrt{a+b}}\right )}{a^2 d}-\frac{x (a+2 b)}{2 a^2}+\frac{\sinh (c+d x) \cosh (c+d x)}{2 a d} \]

[Out]

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

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Rubi [A]  time = 0.110428, antiderivative size = 75, 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 = {4132, 471, 522, 206, 208} \[ \frac{\sqrt{b} \sqrt{a+b} \tanh ^{-1}\left (\frac{\sqrt{b} \tanh (c+d x)}{\sqrt{a+b}}\right )}{a^2 d}-\frac{x (a+2 b)}{2 a^2}+\frac{\sinh (c+d x) \cosh (c+d x)}{2 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4132

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*sin[(e_.) + (f_.)*(x_)]^(m_), x_Symbol] :> With[{ff = Fr
eeFactors[Tan[e + f*x], x]}, Dist[ff^(m + 1)/f, Subst[Int[(x^m*ExpandToSum[a + b*(1 + ff^2*x^2)^(n/2), x]^p)/(
1 + ff^2*x^2)^(m/2 + 1), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[m/2] && Integer
Q[n/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 208

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

Rubi steps

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

Mathematica [B]  time = 0.903549, size = 236, normalized size = 3.15 \[ \frac{\text{sech}^2(c+d x) (a \cosh (2 (c+d x))+a+2 b) \left (\frac{\frac{\left (a^2+8 a b+8 b^2\right ) (\cosh (2 c)-\sinh (2 c)) \tanh ^{-1}\left (\frac{(\cosh (2 c)-\sinh (2 c)) \text{sech}(d x) ((a+2 b) \sinh (d x)-a \sinh (2 c+d x))}{2 \sqrt{a+b} \sqrt{b (\cosh (c)-\sinh (c))^4}}\right )}{d \sqrt{a+b} \sqrt{b (\cosh (c)-\sinh (c))^4}}-4 x (a+2 b)+\frac{2 a \sinh (2 c) \cosh (2 d x)}{d}+\frac{2 a \cosh (2 c) \sinh (2 d x)}{d}}{a^2}-\frac{\tanh ^{-1}\left (\frac{\sqrt{b} \tanh (c+d x)}{\sqrt{a+b}}\right )}{\sqrt{b} d \sqrt{a+b}}\right )}{16 \left (a+b \text{sech}^2(c+d x)\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + 2*b + a*Cosh[2*(c + d*x)])*Sech[c + d*x]^2*(-(ArcTanh[(Sqrt[b]*Tanh[c + d*x])/Sqrt[a + b]]/(Sqrt[b]*Sqrt
[a + b]*d)) + (-4*(a + 2*b)*x + ((a^2 + 8*a*b + 8*b^2)*ArcTanh[(Sech[d*x]*(Cosh[2*c] - Sinh[2*c])*((a + 2*b)*S
inh[d*x] - a*Sinh[2*c + d*x]))/(2*Sqrt[a + b]*Sqrt[b*(Cosh[c] - Sinh[c])^4])]*(Cosh[2*c] - Sinh[2*c]))/(Sqrt[a
+ b]*d*Sqrt[b*(Cosh[c] - Sinh[c])^4]) + (2*a*Cosh[2*d*x]*Sinh[2*c])/d + (2*a*Cosh[2*c]*Sinh[2*d*x])/d)/a^2))/
(16*(a + b*Sech[c + d*x]^2))

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Maple [B]  time = 0.068, size = 383, normalized size = 5.1 \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{1}{2\,da}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }-{\frac{b}{d{a}^{2}}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }+{\frac{1}{2\,da}\sqrt{b}\ln \left ( \sqrt{a+b} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}+2\,\tanh \left ( 1/2\,dx+c/2 \right ) \sqrt{b}+\sqrt{a+b} \right ){\frac{1}{\sqrt{a+b}}}}-{\frac{1}{2\,da}\sqrt{b}\ln \left ( \sqrt{a+b} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}-2\,\tanh \left ( 1/2\,dx+c/2 \right ) \sqrt{b}+\sqrt{a+b} \right ){\frac{1}{\sqrt{a+b}}}}+{\frac{1}{2\,d{a}^{2}}{b}^{{\frac{3}{2}}}\ln \left ( \sqrt{a+b} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}+2\,\tanh \left ( 1/2\,dx+c/2 \right ) \sqrt{b}+\sqrt{a+b} \right ){\frac{1}{\sqrt{a+b}}}}-{\frac{1}{2\,d{a}^{2}}{b}^{{\frac{3}{2}}}\ln \left ( \sqrt{a+b} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}-2\,\tanh \left ( 1/2\,dx+c/2 \right ) \sqrt{b}+\sqrt{a+b} \right ){\frac{1}{\sqrt{a+b}}}}+{\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{1}{2\,da}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }+{\frac{b}{d{a}^{2}}\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(sinh(d*x+c)^2/(a+b*sech(d*x+c)^2),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)-1/2/d/a*ln(tanh(1/2*d*x+1/2*c)+1)-1/d/a^2*l
n(tanh(1/2*d*x+1/2*c)+1)*b+1/2/d*b^(1/2)/a/(a+b)^(1/2)*ln((a+b)^(1/2)*tanh(1/2*d*x+1/2*c)^2+2*tanh(1/2*d*x+1/2
*c)*b^(1/2)+(a+b)^(1/2))-1/2/d*b^(1/2)/a/(a+b)^(1/2)*ln((a+b)^(1/2)*tanh(1/2*d*x+1/2*c)^2-2*tanh(1/2*d*x+1/2*c
)*b^(1/2)+(a+b)^(1/2))+1/2/d*b^(3/2)/a^2/(a+b)^(1/2)*ln((a+b)^(1/2)*tanh(1/2*d*x+1/2*c)^2+2*tanh(1/2*d*x+1/2*c
)*b^(1/2)+(a+b)^(1/2))-1/2/d*b^(3/2)/a^2/(a+b)^(1/2)*ln((a+b)^(1/2)*tanh(1/2*d*x+1/2*c)^2-2*tanh(1/2*d*x+1/2*c
)*b^(1/2)+(a+b)^(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)+1/2/d/a*ln(tanh(1/2*d
*x+1/2*c)-1)+1/d/a^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*sech(d*x+c)^2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.85626, size = 2132, normalized size = 28.43 \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*sech(d*x+c)^2),x, algorithm="fricas")

[Out]

[-1/8*(4*(a + 2*b)*d*x*cosh(d*x + c)^2 - a*cosh(d*x + c)^4 - 4*a*cosh(d*x + c)*sinh(d*x + c)^3 - a*sinh(d*x +
c)^4 + 2*(2*(a + 2*b)*d*x - 3*a*cosh(d*x + c)^2)*sinh(d*x + c)^2 - 4*sqrt(a*b + b^2)*(cosh(d*x + c)^2 + 2*cosh
(d*x + c)*sinh(d*x + c) + sinh(d*x + c)^2)*log((a^2*cosh(d*x + c)^4 + 4*a^2*cosh(d*x + c)*sinh(d*x + c)^3 + a^
2*sinh(d*x + c)^4 + 2*(a^2 + 2*a*b)*cosh(d*x + c)^2 + 2*(3*a^2*cosh(d*x + c)^2 + a^2 + 2*a*b)*sinh(d*x + c)^2
+ a^2 + 8*a*b + 8*b^2 + 4*(a^2*cosh(d*x + c)^3 + (a^2 + 2*a*b)*cosh(d*x + c))*sinh(d*x + c) - 4*(a*cosh(d*x +
c)^2 + 2*a*cosh(d*x + c)*sinh(d*x + c) + a*sinh(d*x + c)^2 + a + 2*b)*sqrt(a*b + b^2))/(a*cosh(d*x + c)^4 + 4*
a*cosh(d*x + c)*sinh(d*x + c)^3 + a*sinh(d*x + c)^4 + 2*(a + 2*b)*cosh(d*x + c)^2 + 2*(3*a*cosh(d*x + c)^2 + a
+ 2*b)*sinh(d*x + c)^2 + 4*(a*cosh(d*x + c)^3 + (a + 2*b)*cosh(d*x + c))*sinh(d*x + c) + a)) + 4*(2*(a + 2*b)
*d*x*cosh(d*x + c) - a*cosh(d*x + c)^3)*sinh(d*x + c) + a)/(a^2*d*cosh(d*x + c)^2 + 2*a^2*d*cosh(d*x + c)*sinh
(d*x + c) + a^2*d*sinh(d*x + c)^2), -1/8*(4*(a + 2*b)*d*x*cosh(d*x + c)^2 - a*cosh(d*x + c)^4 - 4*a*cosh(d*x +
c)*sinh(d*x + c)^3 - a*sinh(d*x + c)^4 + 2*(2*(a + 2*b)*d*x - 3*a*cosh(d*x + c)^2)*sinh(d*x + c)^2 - 8*sqrt(-
a*b - b^2)*(cosh(d*x + c)^2 + 2*cosh(d*x + c)*sinh(d*x + c) + sinh(d*x + c)^2)*arctan(1/2*(a*cosh(d*x + c)^2 +
2*a*cosh(d*x + c)*sinh(d*x + c) + a*sinh(d*x + c)^2 + a + 2*b)*sqrt(-a*b - b^2)/(a*b + b^2)) + 4*(2*(a + 2*b)
*d*x*cosh(d*x + c) - a*cosh(d*x + c)^3)*sinh(d*x + c) + a)/(a^2*d*cosh(d*x + c)^2 + 2*a^2*d*cosh(d*x + c)*sinh
(d*x + c) + a^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 \operatorname{sech}^{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*sech(d*x+c)**2),x)

[Out]

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

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Giac [B]  time = 1.21563, size = 186, normalized size = 2.48 \begin{align*} -\frac{{\left (d x + c\right )}{\left (a + 2 \, b\right )}}{2 \, a^{2} d} + \frac{e^{\left (2 \, d x + 2 \, c\right )}}{8 \, a d} + \frac{{\left (2 \, a e^{\left (2 \, d x + 2 \, c\right )} + 4 \, b e^{\left (2 \, d x + 2 \, c\right )} - a\right )} e^{\left (-2 \, d x - 2 \, c\right )}}{8 \, a^{2} d} + \frac{{\left (a b + b^{2}\right )} \arctan \left (\frac{a e^{\left (2 \, d x + 2 \, c\right )} + a + 2 \, b}{2 \, \sqrt{-a b - b^{2}}}\right )}{\sqrt{-a b - b^{2}} a^{2} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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