3.88 \(\int \text {sech}^4(c+d x) (a+b \tanh ^2(c+d x)) \, dx\)

Optimal. Leaf size=48 \[ -\frac {(a-b) \tanh ^3(c+d x)}{3 d}+\frac {a \tanh (c+d x)}{d}-\frac {b \tanh ^5(c+d x)}{5 d} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {3675, 373} \[ -\frac {(a-b) \tanh ^3(c+d x)}{3 d}+\frac {a \tanh (c+d x)}{d}-\frac {b \tanh ^5(c+d x)}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[Sech[c + d*x]^4*(a + b*Tanh[c + d*x]^2),x]

[Out]

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

Rule 373

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

Rule 3675

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol] :> With[
{ff = FreeFactors[Tan[e + f*x], x]}, Dist[ff/(c^(m - 1)*f), Subst[Int[(c^2 + ff^2*x^2)^(m/2 - 1)*(a + b*(ff*x)
^n)^p, x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[m/2] && (IntegersQ[n, p
] || IGtQ[m, 0] || IGtQ[p, 0] || EqQ[n^2, 4] || EqQ[n^2, 16])

Rubi steps

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

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Mathematica [A]  time = 0.05, size = 86, normalized size = 1.79 \[ -\frac {a \tanh ^3(c+d x)}{3 d}+\frac {a \tanh (c+d x)}{d}+\frac {2 b \tanh (c+d x)}{15 d}-\frac {b \tanh (c+d x) \text {sech}^4(c+d x)}{5 d}+\frac {b \tanh (c+d x) \text {sech}^2(c+d x)}{15 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sech[c + d*x]^4*(a + b*Tanh[c + d*x]^2),x]

[Out]

(a*Tanh[c + d*x])/d + (2*b*Tanh[c + d*x])/(15*d) + (b*Sech[c + d*x]^2*Tanh[c + d*x])/(15*d) - (b*Sech[c + d*x]
^4*Tanh[c + d*x])/(5*d) - (a*Tanh[c + d*x]^3)/(3*d)

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fricas [B]  time = 0.39, size = 345, normalized size = 7.19 \[ -\frac {8 \, {\left (2 \, {\left (5 \, a + 4 \, b\right )} \cosh \left (d x + c\right )^{3} + 6 \, {\left (5 \, a + 4 \, b\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{2} + {\left (5 \, a + 7 \, b\right )} \sinh \left (d x + c\right )^{3} + 30 \, a \cosh \left (d x + c\right ) + {\left (3 \, {\left (5 \, a + 7 \, b\right )} \cosh \left (d x + c\right )^{2} + 5 \, a - 5 \, b\right )} \sinh \left (d x + c\right )\right )}}{15 \, {\left (d \cosh \left (d x + c\right )^{7} + 7 \, d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{6} + d \sinh \left (d x + c\right )^{7} + 5 \, d \cosh \left (d x + c\right )^{5} + {\left (21 \, d \cosh \left (d x + c\right )^{2} + 5 \, d\right )} \sinh \left (d x + c\right )^{5} + 5 \, {\left (7 \, d \cosh \left (d x + c\right )^{3} + 5 \, d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{4} + 11 \, d \cosh \left (d x + c\right )^{3} + {\left (35 \, d \cosh \left (d x + c\right )^{4} + 50 \, d \cosh \left (d x + c\right )^{2} + 9 \, d\right )} \sinh \left (d x + c\right )^{3} + {\left (21 \, d \cosh \left (d x + c\right )^{5} + 50 \, d \cosh \left (d x + c\right )^{3} + 33 \, d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} + 15 \, d \cosh \left (d x + c\right ) + {\left (7 \, d \cosh \left (d x + c\right )^{6} + 25 \, d \cosh \left (d x + c\right )^{4} + 27 \, d \cosh \left (d x + c\right )^{2} + 5 \, d\right )} \sinh \left (d x + c\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(d*x+c)^4*(a+b*tanh(d*x+c)^2),x, algorithm="fricas")

[Out]

-8/15*(2*(5*a + 4*b)*cosh(d*x + c)^3 + 6*(5*a + 4*b)*cosh(d*x + c)*sinh(d*x + c)^2 + (5*a + 7*b)*sinh(d*x + c)
^3 + 30*a*cosh(d*x + c) + (3*(5*a + 7*b)*cosh(d*x + c)^2 + 5*a - 5*b)*sinh(d*x + c))/(d*cosh(d*x + c)^7 + 7*d*
cosh(d*x + c)*sinh(d*x + c)^6 + d*sinh(d*x + c)^7 + 5*d*cosh(d*x + c)^5 + (21*d*cosh(d*x + c)^2 + 5*d)*sinh(d*
x + c)^5 + 5*(7*d*cosh(d*x + c)^3 + 5*d*cosh(d*x + c))*sinh(d*x + c)^4 + 11*d*cosh(d*x + c)^3 + (35*d*cosh(d*x
+ c)^4 + 50*d*cosh(d*x + c)^2 + 9*d)*sinh(d*x + c)^3 + (21*d*cosh(d*x + c)^5 + 50*d*cosh(d*x + c)^3 + 33*d*co
sh(d*x + c))*sinh(d*x + c)^2 + 15*d*cosh(d*x + c) + (7*d*cosh(d*x + c)^6 + 25*d*cosh(d*x + c)^4 + 27*d*cosh(d*
x + c)^2 + 5*d)*sinh(d*x + c))

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giac [B]  time = 0.16, size = 95, normalized size = 1.98 \[ -\frac {4 \, {\left (15 \, a e^{\left (6 \, d x + 6 \, c\right )} + 15 \, b e^{\left (6 \, d x + 6 \, c\right )} + 35 \, a e^{\left (4 \, d x + 4 \, c\right )} - 5 \, b e^{\left (4 \, d x + 4 \, c\right )} + 25 \, a e^{\left (2 \, d x + 2 \, c\right )} + 5 \, b e^{\left (2 \, d x + 2 \, c\right )} + 5 \, a + b\right )}}{15 \, d {\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(d*x+c)^4*(a+b*tanh(d*x+c)^2),x, algorithm="giac")

[Out]

-4/15*(15*a*e^(6*d*x + 6*c) + 15*b*e^(6*d*x + 6*c) + 35*a*e^(4*d*x + 4*c) - 5*b*e^(4*d*x + 4*c) + 25*a*e^(2*d*
x + 2*c) + 5*b*e^(2*d*x + 2*c) + 5*a + b)/(d*(e^(2*d*x + 2*c) + 1)^5)

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maple [A]  time = 0.36, size = 75, normalized size = 1.56 \[ \frac {a \left (\frac {2}{3}+\frac {\mathrm {sech}\left (d x +c \right )^{2}}{3}\right ) \tanh \left (d x +c \right )+b \left (-\frac {\sinh \left (d x +c \right )}{4 \cosh \left (d x +c \right )^{5}}+\frac {\left (\frac {8}{15}+\frac {\mathrm {sech}\left (d x +c \right )^{4}}{5}+\frac {4 \mathrm {sech}\left (d x +c \right )^{2}}{15}\right ) \tanh \left (d x +c \right )}{4}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sech(d*x+c)^4*(a+b*tanh(d*x+c)^2),x)

[Out]

1/d*(a*(2/3+1/3*sech(d*x+c)^2)*tanh(d*x+c)+b*(-1/4*sinh(d*x+c)/cosh(d*x+c)^5+1/4*(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 = 0.35, size = 371, normalized size = 7.73 \[ \frac {4}{15} \, b {\left (\frac {5 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d {\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} + 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} + 1\right )}} - \frac {5 \, e^{\left (-4 \, d x - 4 \, c\right )}}{d {\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} + 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} + 1\right )}} + \frac {15 \, e^{\left (-6 \, d x - 6 \, c\right )}}{d {\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} + 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} + 1\right )}} + \frac {1}{d {\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} + 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} + 1\right )}}\right )} + \frac {4}{3} \, a {\left (\frac {3 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} + 1\right )}} + \frac {1}{d {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} + 1\right )}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(d*x+c)^4*(a+b*tanh(d*x+c)^2),x, algorithm="maxima")

[Out]

4/15*b*(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)) - 5*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)) + 15*e^(-6*d*x - 6*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))) + 4/3*a*(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)))

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mupad [B]  time = 0.17, size = 304, normalized size = 6.33 \[ -\frac {\frac {8\,\left (a-b\right )}{15\,d}+\frac {4\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (a+b\right )}{5\,d}}{3\,{\mathrm {e}}^{2\,c+2\,d\,x}+3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}+1}-\frac {\frac {8\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (a+b\right )}{5\,d}+\frac {8\,{\mathrm {e}}^{6\,c+6\,d\,x}\,\left (a+b\right )}{5\,d}+\frac {16\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (a-b\right )}{5\,d}}{5\,{\mathrm {e}}^{2\,c+2\,d\,x}+10\,{\mathrm {e}}^{4\,c+4\,d\,x}+10\,{\mathrm {e}}^{6\,c+6\,d\,x}+5\,{\mathrm {e}}^{8\,c+8\,d\,x}+{\mathrm {e}}^{10\,c+10\,d\,x}+1}-\frac {\frac {2\,\left (a+b\right )}{5\,d}+\frac {6\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (a+b\right )}{5\,d}+\frac {8\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (a-b\right )}{5\,d}}{4\,{\mathrm {e}}^{2\,c+2\,d\,x}+6\,{\mathrm {e}}^{4\,c+4\,d\,x}+4\,{\mathrm {e}}^{6\,c+6\,d\,x}+{\mathrm {e}}^{8\,c+8\,d\,x}+1}-\frac {2\,\left (a+b\right )}{5\,d\,\left (2\,{\mathrm {e}}^{2\,c+2\,d\,x}+{\mathrm {e}}^{4\,c+4\,d\,x}+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*tanh(c + d*x)^2)/cosh(c + d*x)^4,x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \tanh ^{2}{\left (c + d x \right )}\right ) \operatorname {sech}^{4}{\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(d*x+c)**4*(a+b*tanh(d*x+c)**2),x)

[Out]

Integral((a + b*tanh(c + d*x)**2)*sech(c + d*x)**4, x)

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