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

Optimal. Leaf size=32 \[ -\frac {a \tanh (c+d x)}{d}+a x+\frac {b \tanh ^3(c+d x)}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 32, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {4141, 1802, 206} \[ -\frac {a \tanh (c+d x)}{d}+a x+\frac {b \tanh ^3(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

a*x - (a*Tanh[c + d*x])/d + (b*Tanh[c + d*x]^3)/(3*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 1802

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*Pq*(a + b*x
^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rule 4141

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

Rubi steps

\begin {align*} \int \left (a+b \text {sech}^2(c+d x)\right ) \tanh ^2(c+d x) \, dx &=\frac {\operatorname {Subst}\left (\int \frac {x^2 \left (a+b \left (1-x^2\right )\right )}{1-x^2} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac {\operatorname {Subst}\left (\int \left (-a+b x^2+\frac {a}{1-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}+\frac {a \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=a x-\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 = 0.02, size = 41, normalized size = 1.28 \[ \frac {a \tanh ^{-1}(\tanh (c+d x))}{d}-\frac {a \tanh (c+d x)}{d}+\frac {b \tanh ^3(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.40, size = 155, normalized size = 4.84 \[ \frac {{\left (3 \, a d x + 3 \, a - b\right )} \cosh \left (d x + c\right )^{3} + 3 \, {\left (3 \, a d x + 3 \, a - b\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{2} - {\left (3 \, a - b\right )} \sinh \left (d x + c\right )^{3} + 3 \, {\left (3 \, a d x + 3 \, a - b\right )} \cosh \left (d x + c\right ) - 3 \, {\left ({\left (3 \, a - b\right )} \cosh \left (d x + c\right )^{2} + a + b\right )} \sinh \left (d x + c\right )}{3 \, {\left (d \cosh \left (d x + c\right )^{3} + 3 \, d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{2} + 3 \, d \cosh \left (d x + c\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*((3*a*d*x + 3*a - b)*cosh(d*x + c)^3 + 3*(3*a*d*x + 3*a - b)*cosh(d*x + c)*sinh(d*x + c)^2 - (3*a - b)*sin
h(d*x + c)^3 + 3*(3*a*d*x + 3*a - b)*cosh(d*x + c) - 3*((3*a - b)*cosh(d*x + c)^2 + a + b)*sinh(d*x + c))/(d*c
osh(d*x + c)^3 + 3*d*cosh(d*x + c)*sinh(d*x + c)^2 + 3*d*cosh(d*x + c))

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giac [B]  time = 0.16, size = 69, normalized size = 2.16 \[ \frac {3 \, a d x + \frac {2 \, {\left (3 \, a e^{\left (4 \, d x + 4 \, c\right )} - 3 \, b e^{\left (4 \, d x + 4 \, c\right )} + 6 \, a e^{\left (2 \, d x + 2 \, c\right )} + 3 \, a - b\right )}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{3}}}{3 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.33, size = 60, normalized size = 1.88 \[ \frac {a \left (d x +c -\tanh \left (d x +c \right )\right )+b \left (-\frac {\sinh \left (d x +c \right )}{2 \cosh \left (d x +c \right )^{3}}+\frac {\left (\frac {2}{3}+\frac {\mathrm {sech}\left (d x +c \right )^{2}}{3}\right ) \tanh \left (d x +c \right )}{2}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(a*(d*x+c-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 = 0.40, size = 42, normalized size = 1.31 \[ \frac {b \tanh \left (d x + c\right )^{3}}{3 \, d} + a {\left (x + \frac {c}{d} - \frac {2}{d {\left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.50, size = 163, normalized size = 5.09 \[ \frac {\frac {2\,\left (a+b\right )}{3\,d}+\frac {2\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (a-b\right )}{3\,d}}{2\,{\mathrm {e}}^{2\,c+2\,d\,x}+{\mathrm {e}}^{4\,c+4\,d\,x}+1}+a\,x+\frac {\frac {2\,\left (a-b\right )}{3\,d}+\frac {4\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (a+b\right )}{3\,d}+\frac {2\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (a-b\right )}{3\,d}}{3\,{\mathrm {e}}^{2\,c+2\,d\,x}+3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}+1}+\frac {2\,\left (a-b\right )}{3\,d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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