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

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

[Out]

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

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Rubi [A]  time = 0.0638269, antiderivative size = 48, normalized size of antiderivative = 1., 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} \[ -\frac{a \tanh ^3(c+d x)}{3 d}-\frac{a \tanh (c+d x)}{d}+a x+\frac{b \tanh ^5(c+d x)}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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])

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 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])

Rubi steps

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

Mathematica [A]  time = 0.0243839, size = 57, normalized size = 1.19 \[ -\frac{a \tanh ^3(c+d x)}{3 d}+\frac{a \tanh ^{-1}(\tanh (c+d x))}{d}-\frac{a \tanh (c+d x)}{d}+\frac{b \tanh ^5(c+d x)}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.034, size = 98, normalized size = 2. \begin{align*}{\frac{1}{d} \left ( a \left ( dx+c-\tanh \left ( dx+c \right ) -{\frac{ \left ( \tanh \left ( dx+c \right ) \right ) ^{3}}{3}} \right ) +b \left ( -{\frac{ \left ( \sinh \left ( dx+c \right ) \right ) ^{3}}{2\, \left ( \cosh \left ( dx+c \right ) \right ) ^{5}}}-{\frac{3\,\sinh \left ( dx+c \right ) }{8\, \left ( \cosh \left ( dx+c \right ) \right ) ^{5}}}+{\frac{3\,\tanh \left ( dx+c \right ) }{8} \left ({\frac{8}{15}}+{\frac{ \left ({\rm sech} \left (dx+c\right ) \right ) ^{4}}{5}}+{\frac{4\, \left ({\rm sech} \left (dx+c\right ) \right ) ^{2}}{15}} \right ) } \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(a*(d*x+c-tanh(d*x+c)-1/3*tanh(d*x+c)^3)+b*(-1/2*sinh(d*x+c)^3/cosh(d*x+c)^5-3/8*sinh(d*x+c)/cosh(d*x+c)^5
+3/8*(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 = 1.19444, size = 124, normalized size = 2.58 \begin{align*} \frac{b \tanh \left (d x + c\right )^{5}}{5 \, d} + \frac{1}{3} \, a{\left (3 \, x + \frac{3 \, c}{d} - \frac{4 \,{\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + 2\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 )}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*b*tanh(d*x + c)^5/d + 1/3*a*(3*x + 3*c/d - 4*(3*e^(-2*d*x - 2*c) + 3*e^(-4*d*x - 4*c) + 2)/(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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Fricas [B]  time = 2.07739, size = 876, normalized size = 18.25 \begin{align*} \frac{{\left (15 \, a d x + 20 \, a - 3 \, b\right )} \cosh \left (d x + c\right )^{5} + 5 \,{\left (15 \, a d x + 20 \, a - 3 \, b\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{4} -{\left (20 \, a - 3 \, b\right )} \sinh \left (d x + c\right )^{5} + 5 \,{\left (15 \, a d x + 20 \, a - 3 \, b\right )} \cosh \left (d x + c\right )^{3} - 5 \,{\left (2 \,{\left (20 \, a - 3 \, b\right )} \cosh \left (d x + c\right )^{2} + 8 \, a + 3 \, b\right )} \sinh \left (d x + c\right )^{3} + 5 \,{\left (2 \,{\left (15 \, a d x + 20 \, a - 3 \, b\right )} \cosh \left (d x + c\right )^{3} + 3 \,{\left (15 \, a d x + 20 \, a - 3 \, b\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} + 10 \,{\left (15 \, a d x + 20 \, a - 3 \, b\right )} \cosh \left (d x + c\right ) - 5 \,{\left ({\left (20 \, a - 3 \, b\right )} \cosh \left (d x + c\right )^{4} + 3 \,{\left (8 \, a + 3 \, b\right )} \cosh \left (d x + c\right )^{2} + 4 \, a - 6 \, b\right )} \sinh \left (d x + c\right )}{15 \,{\left (d \cosh \left (d x + c\right )^{5} + 5 \, d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{4} + 5 \, d \cosh \left (d x + c\right )^{3} + 5 \,{\left (2 \, d \cosh \left (d x + c\right )^{3} + 3 \, d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} + 10 \, d \cosh \left (d x + c\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*((15*a*d*x + 20*a - 3*b)*cosh(d*x + c)^5 + 5*(15*a*d*x + 20*a - 3*b)*cosh(d*x + c)*sinh(d*x + c)^4 - (20*
a - 3*b)*sinh(d*x + c)^5 + 5*(15*a*d*x + 20*a - 3*b)*cosh(d*x + c)^3 - 5*(2*(20*a - 3*b)*cosh(d*x + c)^2 + 8*a
+ 3*b)*sinh(d*x + c)^3 + 5*(2*(15*a*d*x + 20*a - 3*b)*cosh(d*x + c)^3 + 3*(15*a*d*x + 20*a - 3*b)*cosh(d*x +
c))*sinh(d*x + c)^2 + 10*(15*a*d*x + 20*a - 3*b)*cosh(d*x + c) - 5*((20*a - 3*b)*cosh(d*x + c)^4 + 3*(8*a + 3*
b)*cosh(d*x + c)^2 + 4*a - 6*b)*sinh(d*x + c))/(d*cosh(d*x + c)^5 + 5*d*cosh(d*x + c)*sinh(d*x + c)^4 + 5*d*co
sh(d*x + c)^3 + 5*(2*d*cosh(d*x + c)^3 + 3*d*cosh(d*x + c))*sinh(d*x + c)^2 + 10*d*cosh(d*x + c))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a + b \operatorname{sech}^{2}{\left (c + d x \right )}\right ) \tanh ^{4}{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.317, size = 142, normalized size = 2.96 \begin{align*} \frac{15 \, a d x + \frac{2 \,{\left (30 \, a e^{\left (8 \, d x + 8 \, c\right )} - 15 \, b e^{\left (8 \, d x + 8 \, c\right )} + 90 \, a e^{\left (6 \, d x + 6 \, c\right )} + 110 \, a e^{\left (4 \, d x + 4 \, c\right )} - 30 \, b e^{\left (4 \, d x + 4 \, c\right )} + 70 \, a e^{\left (2 \, d x + 2 \, c\right )} + 20 \, a - 3 \, b\right )}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{5}}}{15 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(15*a*d*x + 2*(30*a*e^(8*d*x + 8*c) - 15*b*e^(8*d*x + 8*c) + 90*a*e^(6*d*x + 6*c) + 110*a*e^(4*d*x + 4*c)
- 30*b*e^(4*d*x + 4*c) + 70*a*e^(2*d*x + 2*c) + 20*a - 3*b)/(e^(2*d*x + 2*c) + 1)^5)/d