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

Optimal. Leaf size=53 \[ -\frac {2 b (a+b) \tanh ^3(c+d x)}{3 d}+\frac {(a+b)^2 \tanh (c+d x)}{d}+\frac {b^2 \tanh ^5(c+d x)}{5 d} \]

[Out]

(a+b)^2*tanh(d*x+c)/d-2/3*b*(a+b)*tanh(d*x+c)^3/d+1/5*b^2*tanh(d*x+c)^5/d

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Rubi [A]  time = 0.07, antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {4146, 194} \[ -\frac {2 b (a+b) \tanh ^3(c+d x)}{3 d}+\frac {(a+b)^2 \tanh (c+d x)}{d}+\frac {b^2 \tanh ^5(c+d x)}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b)^2*Tanh[c + d*x])/d - (2*b*(a + b)*Tanh[c + d*x]^3)/(3*d) + (b^2*Tanh[c + d*x]^5)/(5*d)

Rule 194

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^n)^p, x], x] /; FreeQ[{a, b}, x]
&& IGtQ[n, 0] && IGtQ[p, 0]

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 \text {sech}^2(c+d x) \left (a+b \text {sech}^2(c+d x)\right )^2 \, dx &=\frac {\operatorname {Subst}\left (\int \left (a+b-b x^2\right )^2 \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac {\operatorname {Subst}\left (\int \left (a^2 \left (1+\frac {b (2 a+b)}{a^2}\right )-2 a b \left (1+\frac {b}{a}\right ) x^2+b^2 x^4\right ) \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac {(a+b)^2 \tanh (c+d x)}{d}-\frac {2 b (a+b) \tanh ^3(c+d x)}{3 d}+\frac {b^2 \tanh ^5(c+d x)}{5 d}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 93, normalized size = 1.75 \[ \frac {a^2 \tanh (c+d x)}{d}-\frac {2 a b \tanh ^3(c+d x)}{3 d}+\frac {2 a b \tanh (c+d x)}{d}+\frac {b^2 \tanh ^5(c+d x)}{5 d}-\frac {2 b^2 \tanh ^3(c+d x)}{3 d}+\frac {b^2 \tanh (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*Tanh[c + d*x])/d + (2*a*b*Tanh[c + d*x])/d + (b^2*Tanh[c + d*x])/d - (2*a*b*Tanh[c + d*x]^3)/(3*d) - (2*b
^2*Tanh[c + d*x]^3)/(3*d) + (b^2*Tanh[c + d*x]^5)/(5*d)

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fricas [B]  time = 0.41, size = 404, normalized size = 7.62 \[ -\frac {4 \, {\left ({\left (15 \, a^{2} + 10 \, a b + 4 \, b^{2}\right )} \cosh \left (d x + c\right )^{4} - 8 \, {\left (5 \, a b + 2 \, b^{2}\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + {\left (15 \, a^{2} + 10 \, a b + 4 \, b^{2}\right )} \sinh \left (d x + c\right )^{4} + 20 \, {\left (3 \, a^{2} + 4 \, a b + b^{2}\right )} \cosh \left (d x + c\right )^{2} + 2 \, {\left (3 \, {\left (15 \, a^{2} + 10 \, a b + 4 \, b^{2}\right )} \cosh \left (d x + c\right )^{2} + 30 \, a^{2} + 40 \, a b + 10 \, b^{2}\right )} \sinh \left (d x + c\right )^{2} + 45 \, a^{2} + 70 \, a b + 40 \, b^{2} - 8 \, {\left ({\left (5 \, a b + 2 \, b^{2}\right )} \cosh \left (d x + c\right )^{3} + 5 \, {\left (a b + b^{2}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )\right )}}{15 \, {\left (d \cosh \left (d x + c\right )^{6} + 6 \, d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{5} + d \sinh \left (d x + c\right )^{6} + 6 \, d \cosh \left (d x + c\right )^{4} + 3 \, {\left (5 \, d \cosh \left (d x + c\right )^{2} + 2 \, d\right )} \sinh \left (d x + c\right )^{4} + 4 \, {\left (5 \, d \cosh \left (d x + c\right )^{3} + 4 \, d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{3} + 15 \, d \cosh \left (d x + c\right )^{2} + 3 \, {\left (5 \, d \cosh \left (d x + c\right )^{4} + 12 \, d \cosh \left (d x + c\right )^{2} + 5 \, d\right )} \sinh \left (d x + c\right )^{2} + 2 \, {\left (3 \, d \cosh \left (d x + c\right )^{5} + 8 \, d \cosh \left (d x + c\right )^{3} + 5 \, d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right ) + 10 \, d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/15*((15*a^2 + 10*a*b + 4*b^2)*cosh(d*x + c)^4 - 8*(5*a*b + 2*b^2)*cosh(d*x + c)*sinh(d*x + c)^3 + (15*a^2 +
10*a*b + 4*b^2)*sinh(d*x + c)^4 + 20*(3*a^2 + 4*a*b + b^2)*cosh(d*x + c)^2 + 2*(3*(15*a^2 + 10*a*b + 4*b^2)*c
osh(d*x + c)^2 + 30*a^2 + 40*a*b + 10*b^2)*sinh(d*x + c)^2 + 45*a^2 + 70*a*b + 40*b^2 - 8*((5*a*b + 2*b^2)*cos
h(d*x + c)^3 + 5*(a*b + b^2)*cosh(d*x + c))*sinh(d*x + c))/(d*cosh(d*x + c)^6 + 6*d*cosh(d*x + c)*sinh(d*x + c
)^5 + d*sinh(d*x + c)^6 + 6*d*cosh(d*x + c)^4 + 3*(5*d*cosh(d*x + c)^2 + 2*d)*sinh(d*x + c)^4 + 4*(5*d*cosh(d*
x + c)^3 + 4*d*cosh(d*x + c))*sinh(d*x + c)^3 + 15*d*cosh(d*x + c)^2 + 3*(5*d*cosh(d*x + c)^4 + 12*d*cosh(d*x
+ c)^2 + 5*d)*sinh(d*x + c)^2 + 2*(3*d*cosh(d*x + c)^5 + 8*d*cosh(d*x + c)^3 + 5*d*cosh(d*x + c))*sinh(d*x + c
) + 10*d)

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giac [B]  time = 0.17, size = 156, normalized size = 2.94 \[ -\frac {2 \, {\left (15 \, a^{2} e^{\left (8 \, d x + 8 \, c\right )} + 60 \, a^{2} e^{\left (6 \, d x + 6 \, c\right )} + 60 \, a b e^{\left (6 \, d x + 6 \, c\right )} + 90 \, a^{2} e^{\left (4 \, d x + 4 \, c\right )} + 140 \, a b e^{\left (4 \, d x + 4 \, c\right )} + 80 \, b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 60 \, a^{2} e^{\left (2 \, d x + 2 \, c\right )} + 100 \, a b e^{\left (2 \, d x + 2 \, c\right )} + 40 \, b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 15 \, a^{2} + 20 \, a b + 8 \, b^{2}\right )}}{15 \, d {\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/15*(15*a^2*e^(8*d*x + 8*c) + 60*a^2*e^(6*d*x + 6*c) + 60*a*b*e^(6*d*x + 6*c) + 90*a^2*e^(4*d*x + 4*c) + 140
*a*b*e^(4*d*x + 4*c) + 80*b^2*e^(4*d*x + 4*c) + 60*a^2*e^(2*d*x + 2*c) + 100*a*b*e^(2*d*x + 2*c) + 40*b^2*e^(2
*d*x + 2*c) + 15*a^2 + 20*a*b + 8*b^2)/(d*(e^(2*d*x + 2*c) + 1)^5)

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maple [A]  time = 0.42, size = 70, normalized size = 1.32 \[ \frac {a^{2} \tanh \left (d x +c \right )+2 a b \left (\frac {2}{3}+\frac {\mathrm {sech}\left (d x +c \right )^{2}}{3}\right ) \tanh \left (d x +c \right )+b^{2} \left (\frac {8}{15}+\frac {\mathrm {sech}\left (d x +c \right )^{4}}{5}+\frac {4 \mathrm {sech}\left (d x +c \right )^{2}}{15}\right ) \tanh \left (d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(a^2*tanh(d*x+c)+2*a*b*(2/3+1/3*sech(d*x+c)^2)*tanh(d*x+c)+b^2*(8/15+1/5*sech(d*x+c)^4+4/15*sech(d*x+c)^2)
*tanh(d*x+c))

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maxima [B]  time = 0.33, size = 324, normalized size = 6.11 \[ \frac {16}{15} \, b^{2} {\left (\frac {5 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d {\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} + 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} + 1\right )}} + \frac {10 \, e^{\left (-4 \, d x - 4 \, c\right )}}{d {\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} + 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} + 1\right )}} + \frac {1}{d {\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} + 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} + 1\right )}}\right )} + \frac {8}{3} \, a b {\left (\frac {3 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} + 1\right )}} + \frac {1}{d {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} + 1\right )}}\right )} + \frac {2 \, a^{2}}{d {\left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

16/15*b^2*(5*e^(-2*d*x - 2*c)/(d*(5*e^(-2*d*x - 2*c) + 10*e^(-4*d*x - 4*c) + 10*e^(-6*d*x - 6*c) + 5*e^(-8*d*x
- 8*c) + e^(-10*d*x - 10*c) + 1)) + 10*e^(-4*d*x - 4*c)/(d*(5*e^(-2*d*x - 2*c) + 10*e^(-4*d*x - 4*c) + 10*e^(
-6*d*x - 6*c) + 5*e^(-8*d*x - 8*c) + e^(-10*d*x - 10*c) + 1)) + 1/(d*(5*e^(-2*d*x - 2*c) + 10*e^(-4*d*x - 4*c)
+ 10*e^(-6*d*x - 6*c) + 5*e^(-8*d*x - 8*c) + e^(-10*d*x - 10*c) + 1))) + 8/3*a*b*(3*e^(-2*d*x - 2*c)/(d*(3*e^
(-2*d*x - 2*c) + 3*e^(-4*d*x - 4*c) + e^(-6*d*x - 6*c) + 1)) + 1/(d*(3*e^(-2*d*x - 2*c) + 3*e^(-4*d*x - 4*c) +
e^(-6*d*x - 6*c) + 1))) + 2*a^2/(d*(e^(-2*d*x - 2*c) + 1))

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mupad [B]  time = 1.46, size = 452, normalized size = 8.53 \[ -\frac {\frac {2\,a\,\left (a+2\,b\right )}{5\,d}+\frac {2\,a^2\,{\mathrm {e}}^{2\,c+2\,d\,x}}{5\,d}}{2\,{\mathrm {e}}^{2\,c+2\,d\,x}+{\mathrm {e}}^{4\,c+4\,d\,x}+1}-\frac {\frac {2\,a^2}{5\,d}+\frac {2\,a^2\,{\mathrm {e}}^{8\,c+8\,d\,x}}{5\,d}+\frac {4\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (3\,a^2+8\,a\,b+8\,b^2\right )}{5\,d}+\frac {8\,a\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (a+2\,b\right )}{5\,d}+\frac {8\,a\,{\mathrm {e}}^{6\,c+6\,d\,x}\,\left (a+2\,b\right )}{5\,d}}{5\,{\mathrm {e}}^{2\,c+2\,d\,x}+10\,{\mathrm {e}}^{4\,c+4\,d\,x}+10\,{\mathrm {e}}^{6\,c+6\,d\,x}+5\,{\mathrm {e}}^{8\,c+8\,d\,x}+{\mathrm {e}}^{10\,c+10\,d\,x}+1}-\frac {\frac {2\,a\,\left (a+2\,b\right )}{5\,d}+\frac {2\,a^2\,{\mathrm {e}}^{6\,c+6\,d\,x}}{5\,d}+\frac {2\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (3\,a^2+8\,a\,b+8\,b^2\right )}{5\,d}+\frac {6\,a\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (a+2\,b\right )}{5\,d}}{4\,{\mathrm {e}}^{2\,c+2\,d\,x}+6\,{\mathrm {e}}^{4\,c+4\,d\,x}+4\,{\mathrm {e}}^{6\,c+6\,d\,x}+{\mathrm {e}}^{8\,c+8\,d\,x}+1}-\frac {\frac {2\,\left (3\,a^2+8\,a\,b+8\,b^2\right )}{15\,d}+\frac {2\,a^2\,{\mathrm {e}}^{4\,c+4\,d\,x}}{5\,d}+\frac {4\,a\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (a+2\,b\right )}{5\,d}}{3\,{\mathrm {e}}^{2\,c+2\,d\,x}+3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}+1}-\frac {2\,a^2}{5\,d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- ((2*a*(a + 2*b))/(5*d) + (2*a^2*exp(2*c + 2*d*x))/(5*d))/(2*exp(2*c + 2*d*x) + exp(4*c + 4*d*x) + 1) - ((2*a
^2)/(5*d) + (2*a^2*exp(8*c + 8*d*x))/(5*d) + (4*exp(4*c + 4*d*x)*(8*a*b + 3*a^2 + 8*b^2))/(5*d) + (8*a*exp(2*c
+ 2*d*x)*(a + 2*b))/(5*d) + (8*a*exp(6*c + 6*d*x)*(a + 2*b))/(5*d))/(5*exp(2*c + 2*d*x) + 10*exp(4*c + 4*d*x)
+ 10*exp(6*c + 6*d*x) + 5*exp(8*c + 8*d*x) + exp(10*c + 10*d*x) + 1) - ((2*a*(a + 2*b))/(5*d) + (2*a^2*exp(6*
c + 6*d*x))/(5*d) + (2*exp(2*c + 2*d*x)*(8*a*b + 3*a^2 + 8*b^2))/(5*d) + (6*a*exp(4*c + 4*d*x)*(a + 2*b))/(5*d
))/(4*exp(2*c + 2*d*x) + 6*exp(4*c + 4*d*x) + 4*exp(6*c + 6*d*x) + exp(8*c + 8*d*x) + 1) - ((2*(8*a*b + 3*a^2
+ 8*b^2))/(15*d) + (2*a^2*exp(4*c + 4*d*x))/(5*d) + (4*a*exp(2*c + 2*d*x)*(a + 2*b))/(5*d))/(3*exp(2*c + 2*d*x
) + 3*exp(4*c + 4*d*x) + exp(6*c + 6*d*x) + 1) - (2*a^2)/(5*d*(exp(2*c + 2*d*x) + 1))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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