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

Optimal. Leaf size=75 \[ \frac {b^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a+b}}\right )}{a^2 d \sqrt {a+b}}+\frac {x (a-2 b)}{2 a^2}+\frac {\sinh (c+d x) \cosh (c+d x)}{2 a d} \]

[Out]

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

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Rubi [A]  time = 0.12, antiderivative size = 75, normalized size of antiderivative = 1.00, 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 = {4146, 414, 522, 206, 208} \[ \frac {b^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a+b}}\right )}{a^2 d \sqrt {a+b}}+\frac {x (a-2 b)}{2 a^2}+\frac {\sinh (c+d x) \cosh (c+d x)}{2 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

Rule 414

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(b*x*(a + b*x^n)^(p + 1)*(
c + d*x^n)^(q + 1))/(a*n*(p + 1)*(b*c - a*d)), x] + Dist[1/(a*n*(p + 1)*(b*c - a*d)), Int[(a + b*x^n)^(p + 1)*
(c + d*x^n)^q*Simp[b*c + n*(p + 1)*(b*c - a*d) + d*b*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d,
n, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &&  !( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomial
Q[a, b, c, d, 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 4146

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

Rubi steps

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

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Mathematica [A]  time = 0.24, size = 67, normalized size = 0.89 \[ \frac {\frac {4 b^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a+b}}\right )}{\sqrt {a+b}}+2 (a-2 b) (c+d x)+a \sinh (2 (c+d x))}{4 a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.46, size = 829, normalized size = 11.05 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(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*(b*cosh(d*x + c)^2 + 2*b*cosh(d*x + c)*sin
h(d*x + c) + b*sinh(d*x + c)^2)*sqrt(b/(a + b))*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^2 + a
*b)*cosh(d*x + c)^2 + 2*(a^2 + a*b)*cosh(d*x + c)*sinh(d*x + c) + (a^2 + a*b)*sinh(d*x + c)^2 + a^2 + 3*a*b +
2*b^2)*sqrt(b/(a + b)))/(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*(b*cosh(d*x + c)^2 + 2*b*cosh(d*x + c)*sinh(d*x + c) + b*sinh(d*x + c)
^2)*sqrt(-b/(a + b))*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(-b/(a + b))/b) + 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*co
sh(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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giac [A]  time = 1.57, size = 125, normalized size = 1.67 \[ \frac {\frac {8 \, b^{2} \arctan \left (\frac {a e^{\left (2 \, d x + 2 \, c\right )} + a + 2 \, b}{2 \, \sqrt {-a b - b^{2}}}\right )}{\sqrt {-a b - b^{2}} a^{2}} + \frac {4 \, {\left (d x + c\right )} {\left (a - 2 \, b\right )}}{a^{2}} + \frac {e^{\left (2 \, d x + 2 \, c\right )}}{a} - \frac {{\left (2 \, a e^{\left (2 \, d x + 2 \, c\right )} - 4 \, b e^{\left (2 \, d x + 2 \, c\right )} + a\right )} e^{\left (-2 \, d x - 2 \, c\right )}}{a^{2}}}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.46, size = 278, normalized size = 3.71 \[ \frac {1}{2 d a \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {1}{2 d a \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d a}+\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b}{d \,a^{2}}-\frac {1}{2 d a \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {1}{2 d a \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d a}-\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b}{d \,a^{2}}-\frac {b^{\frac {3}{2}} \ln \left (-\sqrt {a +b}\, \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 \sqrt {b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {a +b}\right )}{2 d \,a^{2} \sqrt {a +b}}+\frac {b^{\frac {3}{2}} \ln \left (\sqrt {a +b}\, \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 \sqrt {b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {a +b}\right )}{2 d \,a^{2} \sqrt {a +b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(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*ln
(tanh(1/2*d*x+1/2*c)-1)*b-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-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*b^(1/2)*tanh(1/2*d*x+1/2*c)-(a+b)^(1/2))+1/2/d/a^2*b^(3/2)/(a+b)^(1/2)*ln((a+b)^(1/2)*tanh(1/2*d*x
+1/2*c)^2+2*b^(1/2)*tanh(1/2*d*x+1/2*c)+(a+b)^(1/2))

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maxima [B]  time = 0.42, size = 352, normalized size = 4.69 \[ \frac {b \log \left (\frac {a e^{\left (-2 \, d x - 2 \, c\right )} + a + 2 \, b - 2 \, \sqrt {{\left (a + b\right )} b}}{a e^{\left (-2 \, d x - 2 \, c\right )} + a + 2 \, b + 2 \, \sqrt {{\left (a + b\right )} b}}\right )}{4 \, \sqrt {{\left (a + b\right )} b} a d} + \frac {d x + c}{2 \, a d} + \frac {e^{\left (2 \, d x + 2 \, c\right )}}{8 \, a d} - \frac {e^{\left (-2 \, d x - 2 \, c\right )}}{8 \, a d} - \frac {b \log \left (a e^{\left (4 \, d x + 4 \, c\right )} + 2 \, {\left (a + 2 \, b\right )} e^{\left (2 \, d x + 2 \, c\right )} + a\right )}{4 \, a^{2} d} + \frac {b \log \left (2 \, {\left (a + 2 \, b\right )} e^{\left (-2 \, d x - 2 \, c\right )} + a e^{\left (-4 \, d x - 4 \, c\right )} + a\right )}{4 \, a^{2} d} + \frac {{\left (a b + 2 \, b^{2}\right )} \log \left (\frac {a e^{\left (2 \, d x + 2 \, c\right )} + a + 2 \, b - 2 \, \sqrt {{\left (a + b\right )} b}}{a e^{\left (2 \, d x + 2 \, c\right )} + a + 2 \, b + 2 \, \sqrt {{\left (a + b\right )} b}}\right )}{8 \, \sqrt {{\left (a + b\right )} b} a^{2} d} - \frac {{\left (a b + 2 \, b^{2}\right )} \log \left (\frac {a e^{\left (-2 \, d x - 2 \, c\right )} + a + 2 \, b - 2 \, \sqrt {{\left (a + b\right )} b}}{a e^{\left (-2 \, d x - 2 \, c\right )} + a + 2 \, b + 2 \, \sqrt {{\left (a + b\right )} b}}\right )}{8 \, \sqrt {{\left (a + b\right )} b} a^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.97, size = 206, normalized size = 2.75 \[ \frac {x\,\left (a-2\,b\right )}{2\,a^2}-\frac {{\mathrm {e}}^{-2\,c-2\,d\,x}}{8\,a\,d}+\frac {{\mathrm {e}}^{2\,c+2\,d\,x}}{8\,a\,d}+\frac {b^{3/2}\,\ln \left (-\frac {4\,b^2\,{\mathrm {e}}^{2\,c+2\,d\,x}}{a^3}-\frac {2\,b^{3/2}\,\left (a\,d+a\,d\,{\mathrm {e}}^{2\,c+2\,d\,x}+2\,b\,d\,{\mathrm {e}}^{2\,c+2\,d\,x}\right )}{a^3\,d\,\sqrt {a+b}}\right )}{2\,a^2\,d\,\sqrt {a+b}}-\frac {b^{3/2}\,\ln \left (\frac {2\,b^{3/2}\,\left (a\,d+a\,d\,{\mathrm {e}}^{2\,c+2\,d\,x}+2\,b\,d\,{\mathrm {e}}^{2\,c+2\,d\,x}\right )}{a^3\,d\,\sqrt {a+b}}-\frac {4\,b^2\,{\mathrm {e}}^{2\,c+2\,d\,x}}{a^3}\right )}{2\,a^2\,d\,\sqrt {a+b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x*(a - 2*b))/(2*a^2) - exp(- 2*c - 2*d*x)/(8*a*d) + exp(2*c + 2*d*x)/(8*a*d) + (b^(3/2)*log(- (4*b^2*exp(2*c
+ 2*d*x))/a^3 - (2*b^(3/2)*(a*d + a*d*exp(2*c + 2*d*x) + 2*b*d*exp(2*c + 2*d*x)))/(a^3*d*(a + b)^(1/2))))/(2*a
^2*d*(a + b)^(1/2)) - (b^(3/2)*log((2*b^(3/2)*(a*d + a*d*exp(2*c + 2*d*x) + 2*b*d*exp(2*c + 2*d*x)))/(a^3*d*(a
+ b)^(1/2)) - (4*b^2*exp(2*c + 2*d*x))/a^3))/(2*a^2*d*(a + b)^(1/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cosh ^{2}{\left (c + d x \right )}}{a + b \operatorname {sech}^{2}{\left (c + d x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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