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

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Rubi [A]  time = 0.06, antiderivative size = 48, 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} \[ -\frac {a \tanh ^3(c+d x)}{3 d}-\frac {a \tanh (c+d x)}{d}+a x+\frac {b \tanh ^5(c+d x)}{5 d} \]

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 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 ^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*}

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Mathematica [A]  time = 0.03, size = 57, normalized size = 1.19 \[ \frac {a \tanh ^{-1}(\tanh (c+d x))}{d}-\frac {a \tanh ^3(c+d x)}{3 d}-\frac {a \tanh (c+d x)}{d}+\frac {b \tanh ^5(c+d x)}{5 d} \]

Antiderivative was successfully verified.

[In]

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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fricas [B]  time = 0.42, size = 327, normalized size = 6.81 \[ \frac {{\left (15 \, a d x + 20 \, a - 3 \, b\right )} \cosh \left (d x + c\right )^{5} + 5 \, {\left (15 \, a d x + 20 \, a - 3 \, b\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{4} - {\left (20 \, a - 3 \, b\right )} \sinh \left (d x + c\right )^{5} + 5 \, {\left (15 \, a d x + 20 \, a - 3 \, b\right )} \cosh \left (d x + c\right )^{3} - 5 \, {\left (2 \, {\left (20 \, a - 3 \, b\right )} \cosh \left (d x + c\right )^{2} + 8 \, a + 3 \, b\right )} \sinh \left (d x + c\right )^{3} + 5 \, {\left (2 \, {\left (15 \, a d x + 20 \, a - 3 \, b\right )} \cosh \left (d x + c\right )^{3} + 3 \, {\left (15 \, a d x + 20 \, a - 3 \, b\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} + 10 \, {\left (15 \, a d x + 20 \, a - 3 \, b\right )} \cosh \left (d x + c\right ) - 5 \, {\left ({\left (20 \, a - 3 \, b\right )} \cosh \left (d x + c\right )^{4} + 3 \, {\left (8 \, a + 3 \, b\right )} \cosh \left (d x + c\right )^{2} + 4 \, a - 6 \, b\right )} \sinh \left (d x + c\right )}{15 \, {\left (d \cosh \left (d x + c\right )^{5} + 5 \, d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{4} + 5 \, d \cosh \left (d x + c\right )^{3} + 5 \, {\left (2 \, d \cosh \left (d x + c\right )^{3} + 3 \, d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} + 10 \, d \cosh \left (d x + c\right )\right )}} \]

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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giac [B]  time = 0.20, size = 105, normalized size = 2.19 \[ \frac {15 \, a d x + \frac {2 \, {\left (30 \, a e^{\left (8 \, d x + 8 \, c\right )} - 15 \, b e^{\left (8 \, d x + 8 \, c\right )} + 90 \, a e^{\left (6 \, d x + 6 \, c\right )} + 110 \, a e^{\left (4 \, d x + 4 \, c\right )} - 30 \, b e^{\left (4 \, d x + 4 \, c\right )} + 70 \, a e^{\left (2 \, d x + 2 \, c\right )} + 20 \, a - 3 \, b\right )}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{5}}}{15 \, d} \]

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

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maple [B]  time = 0.33, size = 98, normalized size = 2.04 \[ \frac {a \left (d x +c -\tanh \left (d x +c \right )-\frac {\left (\tanh ^{3}\left (d x +c \right )\right )}{3}\right )+b \left (-\frac {\sinh ^{3}\left (d x +c \right )}{2 \cosh \left (d x +c \right )^{5}}-\frac {3 \sinh \left (d x +c \right )}{8 \cosh \left (d x +c \right )^{5}}+\frac {3 \left (\frac {8}{15}+\frac {\mathrm {sech}\left (d x +c \right )^{4}}{5}+\frac {4 \mathrm {sech}\left (d x +c \right )^{2}}{15}\right ) \tanh \left (d x +c \right )}{8}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sech(d*x+c)^2)*tanh(d*x+c)^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 = 0.61, size = 92, normalized size = 1.92 \[ \frac {b \tanh \left (d x + c\right )^{5}}{5 \, d} + \frac {1}{3} \, a {\left (3 \, x + \frac {3 \, c}{d} - \frac {4 \, {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + 2\right )}}{d {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} + 1\right )}}\right )} \]

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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mupad [B]  time = 1.54, size = 433, normalized size = 9.02 \[ a\,x+\frac {\frac {2\,\left (2\,a-3\,b\right )}{15\,d}+\frac {4\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (a+b\right )}{5\,d}+\frac {2\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (2\,a-b\right )}{5\,d}}{3\,{\mathrm {e}}^{2\,c+2\,d\,x}+3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}+1}+\frac {\frac {2\,\left (2\,a-b\right )}{5\,d}+\frac {8\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (a+b\right )}{5\,d}+\frac {8\,{\mathrm {e}}^{6\,c+6\,d\,x}\,\left (a+b\right )}{5\,d}+\frac {4\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (2\,a-3\,b\right )}{5\,d}+\frac {2\,{\mathrm {e}}^{8\,c+8\,d\,x}\,\left (2\,a-b\right )}{5\,d}}{5\,{\mathrm {e}}^{2\,c+2\,d\,x}+10\,{\mathrm {e}}^{4\,c+4\,d\,x}+10\,{\mathrm {e}}^{6\,c+6\,d\,x}+5\,{\mathrm {e}}^{8\,c+8\,d\,x}+{\mathrm {e}}^{10\,c+10\,d\,x}+1}+\frac {\frac {2\,\left (a+b\right )}{5\,d}+\frac {2\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (2\,a-b\right )}{5\,d}}{2\,{\mathrm {e}}^{2\,c+2\,d\,x}+{\mathrm {e}}^{4\,c+4\,d\,x}+1}+\frac {\frac {2\,\left (a+b\right )}{5\,d}+\frac {6\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (a+b\right )}{5\,d}+\frac {2\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (2\,a-3\,b\right )}{5\,d}+\frac {2\,{\mathrm {e}}^{6\,c+6\,d\,x}\,\left (2\,a-b\right )}{5\,d}}{4\,{\mathrm {e}}^{2\,c+2\,d\,x}+6\,{\mathrm {e}}^{4\,c+4\,d\,x}+4\,{\mathrm {e}}^{6\,c+6\,d\,x}+{\mathrm {e}}^{8\,c+8\,d\,x}+1}+\frac {2\,\left (2\,a-b\right )}{5\,d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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