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

Optimal. Leaf size=28 \[ \frac {a \tanh (c+d x)}{d}+\frac {b \tanh ^3(c+d x)}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {3675} \[ \frac {a \tanh (c+d x)}{d}+\frac {b \tanh ^3(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.01, size = 28, normalized size = 1.00 \[ \frac {a \tanh (c+d x)}{d}+\frac {b \tanh ^3(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.38, size = 159, normalized size = 5.68 \[ -\frac {4 \, {\left ({\left (3 \, a + 2 \, b\right )} \cosh \left (d x + c\right )^{2} + 2 \, b \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + {\left (3 \, a + 2 \, b\right )} \sinh \left (d x + c\right )^{2} + 3 \, a\right )}}{3 \, {\left (d \cosh \left (d x + c\right )^{4} + 4 \, d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + d \sinh \left (d x + c\right )^{4} + 4 \, d \cosh \left (d x + c\right )^{2} + 2 \, {\left (3 \, d \cosh \left (d x + c\right )^{2} + 2 \, d\right )} \sinh \left (d x + c\right )^{2} + 4 \, {\left (d \cosh \left (d x + c\right )^{3} + d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right ) + 3 \, d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/3*((3*a + 2*b)*cosh(d*x + c)^2 + 2*b*cosh(d*x + c)*sinh(d*x + c) + (3*a + 2*b)*sinh(d*x + c)^2 + 3*a)/(d*co
sh(d*x + c)^4 + 4*d*cosh(d*x + c)*sinh(d*x + c)^3 + d*sinh(d*x + c)^4 + 4*d*cosh(d*x + c)^2 + 2*(3*d*cosh(d*x
+ c)^2 + 2*d)*sinh(d*x + c)^2 + 4*(d*cosh(d*x + c)^3 + d*cosh(d*x + c))*sinh(d*x + c) + 3*d)

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giac [B]  time = 0.16, size = 59, normalized size = 2.11 \[ -\frac {2 \, {\left (3 \, a e^{\left (4 \, d x + 4 \, c\right )} + 3 \, b e^{\left (4 \, d x + 4 \, c\right )} + 6 \, a e^{\left (2 \, d x + 2 \, c\right )} + 3 \, a + b\right )}}{3 \, d {\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*(3*a*e^(4*d*x + 4*c) + 3*b*e^(4*d*x + 4*c) + 6*a*e^(2*d*x + 2*c) + 3*a + b)/(d*(e^(2*d*x + 2*c) + 1)^3)

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maple [A]  time = 0.37, size = 53, normalized size = 1.89 \[ \frac {a \tanh \left (d x +c \right )+b \left (-\frac {\sinh \left (d x +c \right )}{2 \cosh \left (d x +c \right )^{3}}+\frac {\left (\frac {2}{3}+\frac {\mathrm {sech}\left (d x +c \right )^{2}}{3}\right ) \tanh \left (d x +c \right )}{2}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.34, size = 34, normalized size = 1.21 \[ \frac {b \tanh \left (d x + c\right )^{3}}{3 \, d} + \frac {2 \, a}{d {\left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.22, size = 59, normalized size = 2.11 \[ -\frac {2\,\left (3\,a+b+6\,a\,{\mathrm {e}}^{2\,c+2\,d\,x}+3\,a\,{\mathrm {e}}^{4\,c+4\,d\,x}+3\,b\,{\mathrm {e}}^{4\,c+4\,d\,x}\right )}{3\,d\,{\left ({\mathrm {e}}^{2\,c+2\,d\,x}+1\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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