3.64 \(\int \text{sech}^4(c+d x) (a+b \text{sech}^2(c+d x))^2 \, dx\)
Optimal. Leaf size=80 \[ \frac{b (2 a+3 b) \tanh ^5(c+d x)}{5 d}-\frac{(a+b) (a+3 b) \tanh ^3(c+d x)}{3 d}+\frac{(a+b)^2 \tanh (c+d x)}{d}-\frac{b^2 \tanh ^7(c+d x)}{7 d} \]
[Out]
((a + b)^2*Tanh[c + d*x])/d - ((a + b)*(a + 3*b)*Tanh[c + d*x]^3)/(3*d) + (b*(2*a + 3*b)*Tanh[c + d*x]^5)/(5*d
) - (b^2*Tanh[c + d*x]^7)/(7*d)
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Rubi [A] time = 0.076328, antiderivative size = 80, normalized size of antiderivative = 1.,
number of steps used = 3, number of rules used = 2, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.087, Rules used =
{4146, 373} \[ \frac{b (2 a+3 b) \tanh ^5(c+d x)}{5 d}-\frac{(a+b) (a+3 b) \tanh ^3(c+d x)}{3 d}+\frac{(a+b)^2 \tanh (c+d x)}{d}-\frac{b^2 \tanh ^7(c+d x)}{7 d} \]
Antiderivative was successfully verified.
[In]
Int[Sech[c + d*x]^4*(a + b*Sech[c + d*x]^2)^2,x]
[Out]
((a + b)^2*Tanh[c + d*x])/d - ((a + b)*(a + 3*b)*Tanh[c + d*x]^3)/(3*d) + (b*(2*a + 3*b)*Tanh[c + d*x]^5)/(5*d
) - (b^2*Tanh[c + d*x]^7)/(7*d)
Rule 4146
Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = Fre
eFactors[Tan[e + f*x], x]}, Dist[ff/f, Subst[Int[(1 + ff^2*x^2)^(m/2 - 1)*ExpandToSum[a + b*(1 + ff^2*x^2)^(n/
2), x]^p, x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[m/2] && IntegerQ[n/2]
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]
Rubi steps
\begin{align*} \int \text{sech}^4(c+d x) \left (a+b \text{sech}^2(c+d x)\right )^2 \, dx &=\frac{\operatorname{Subst}\left (\int \left (1-x^2\right ) \left (a+b-b x^2\right )^2 \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \left ((a+b)^2+(-a-3 b) (a+b) x^2+b (2 a+3 b) x^4-b^2 x^6\right ) \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac{(a+b)^2 \tanh (c+d x)}{d}-\frac{(a+b) (a+3 b) \tanh ^3(c+d x)}{3 d}+\frac{b (2 a+3 b) \tanh ^5(c+d x)}{5 d}-\frac{b^2 \tanh ^7(c+d x)}{7 d}\\ \end{align*}
Mathematica [A] time = 0.0297176, size = 144, normalized size = 1.8 \[ -\frac{a^2 \tanh ^3(c+d x)}{3 d}+\frac{a^2 \tanh (c+d x)}{d}+\frac{2 a b \tanh ^5(c+d x)}{5 d}-\frac{4 a b \tanh ^3(c+d x)}{3 d}+\frac{2 a b \tanh (c+d x)}{d}-\frac{b^2 \tanh ^7(c+d x)}{7 d}+\frac{3 b^2 \tanh ^5(c+d x)}{5 d}-\frac{b^2 \tanh ^3(c+d x)}{d}+\frac{b^2 \tanh (c+d x)}{d} \]
Antiderivative was successfully verified.
[In]
Integrate[Sech[c + d*x]^4*(a + b*Sech[c + d*x]^2)^2,x]
[Out]
(a^2*Tanh[c + d*x])/d + (2*a*b*Tanh[c + d*x])/d + (b^2*Tanh[c + d*x])/d - (a^2*Tanh[c + d*x]^3)/(3*d) - (4*a*b
*Tanh[c + d*x]^3)/(3*d) - (b^2*Tanh[c + d*x]^3)/d + (2*a*b*Tanh[c + d*x]^5)/(5*d) + (3*b^2*Tanh[c + d*x]^5)/(5
*d) - (b^2*Tanh[c + d*x]^7)/(7*d)
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Maple [A] time = 0.026, size = 102, normalized size = 1.3 \begin{align*}{\frac{1}{d} \left ({a}^{2} \left ({\frac{2}{3}}+{\frac{ \left ({\rm sech} \left (dx+c\right ) \right ) ^{2}}{3}} \right ) \tanh \left ( dx+c \right ) +2\,ab \left ({\frac{8}{15}}+1/5\, \left ({\rm sech} \left (dx+c\right ) \right ) ^{4}+{\frac{4\, \left ({\rm sech} \left (dx+c\right ) \right ) ^{2}}{15}} \right ) \tanh \left ( dx+c \right ) +{b}^{2} \left ({\frac{16}{35}}+{\frac{ \left ({\rm sech} \left (dx+c\right ) \right ) ^{6}}{7}}+{\frac{6\, \left ({\rm sech} \left (dx+c\right ) \right ) ^{4}}{35}}+{\frac{8\, \left ({\rm sech} \left (dx+c\right ) \right ) ^{2}}{35}} \right ) \tanh \left ( dx+c \right ) \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(sech(d*x+c)^4*(a+b*sech(d*x+c)^2)^2,x)
[Out]
1/d*(a^2*(2/3+1/3*sech(d*x+c)^2)*tanh(d*x+c)+2*a*b*(8/15+1/5*sech(d*x+c)^4+4/15*sech(d*x+c)^2)*tanh(d*x+c)+b^2
*(16/35+1/7*sech(d*x+c)^6+6/35*sech(d*x+c)^4+8/35*sech(d*x+c)^2)*tanh(d*x+c))
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Maxima [B] time = 1.15371, size = 906, normalized size = 11.32 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sech(d*x+c)^4*(a+b*sech(d*x+c)^2)^2,x, algorithm="maxima")
[Out]
32/35*b^2*(7*e^(-2*d*x - 2*c)/(d*(7*e^(-2*d*x - 2*c) + 21*e^(-4*d*x - 4*c) + 35*e^(-6*d*x - 6*c) + 35*e^(-8*d*
x - 8*c) + 21*e^(-10*d*x - 10*c) + 7*e^(-12*d*x - 12*c) + e^(-14*d*x - 14*c) + 1)) + 21*e^(-4*d*x - 4*c)/(d*(7
*e^(-2*d*x - 2*c) + 21*e^(-4*d*x - 4*c) + 35*e^(-6*d*x - 6*c) + 35*e^(-8*d*x - 8*c) + 21*e^(-10*d*x - 10*c) +
7*e^(-12*d*x - 12*c) + e^(-14*d*x - 14*c) + 1)) + 35*e^(-6*d*x - 6*c)/(d*(7*e^(-2*d*x - 2*c) + 21*e^(-4*d*x -
4*c) + 35*e^(-6*d*x - 6*c) + 35*e^(-8*d*x - 8*c) + 21*e^(-10*d*x - 10*c) + 7*e^(-12*d*x - 12*c) + e^(-14*d*x -
14*c) + 1)) + 1/(d*(7*e^(-2*d*x - 2*c) + 21*e^(-4*d*x - 4*c) + 35*e^(-6*d*x - 6*c) + 35*e^(-8*d*x - 8*c) + 21
*e^(-10*d*x - 10*c) + 7*e^(-12*d*x - 12*c) + e^(-14*d*x - 14*c) + 1))) + 32/15*a*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)) +
10*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)) + 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^2*(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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Fricas [B] time = 2.05501, size = 1828, normalized size = 22.85 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sech(d*x+c)^4*(a+b*sech(d*x+c)^2)^2,x, algorithm="fricas")
[Out]
-8/105*(2*(35*a^2 + 14*a*b + 6*b^2)*cosh(d*x + c)^5 + 10*(35*a^2 + 14*a*b + 6*b^2)*cosh(d*x + c)*sinh(d*x + c)
^4 + (35*a^2 - 28*a*b - 12*b^2)*sinh(d*x + c)^5 + 14*(25*a^2 + 34*a*b + 6*b^2)*cosh(d*x + c)^3 + (10*(35*a^2 -
28*a*b - 12*b^2)*cosh(d*x + c)^2 + 105*a^2 + 84*a*b - 84*b^2)*sinh(d*x + c)^3 + 2*(10*(35*a^2 + 14*a*b + 6*b^
2)*cosh(d*x + c)^3 + 21*(25*a^2 + 34*a*b + 6*b^2)*cosh(d*x + c))*sinh(d*x + c)^2 + 28*(25*a^2 + 46*a*b + 24*b^
2)*cosh(d*x + c) + (5*(35*a^2 - 28*a*b - 12*b^2)*cosh(d*x + c)^4 + 63*(5*a^2 + 4*a*b - 4*b^2)*cosh(d*x + c)^2
+ 70*a^2 + 112*a*b + 168*b^2)*sinh(d*x + c))/(d*cosh(d*x + c)^9 + 9*d*cosh(d*x + c)*sinh(d*x + c)^8 + d*sinh(d
*x + c)^9 + 7*d*cosh(d*x + c)^7 + (36*d*cosh(d*x + c)^2 + 7*d)*sinh(d*x + c)^7 + 7*(12*d*cosh(d*x + c)^3 + 7*d
*cosh(d*x + c))*sinh(d*x + c)^6 + 22*d*cosh(d*x + c)^5 + (126*d*cosh(d*x + c)^4 + 147*d*cosh(d*x + c)^2 + 20*d
)*sinh(d*x + c)^5 + (126*d*cosh(d*x + c)^5 + 245*d*cosh(d*x + c)^3 + 110*d*cosh(d*x + c))*sinh(d*x + c)^4 + 42
*d*cosh(d*x + c)^3 + (84*d*cosh(d*x + c)^6 + 245*d*cosh(d*x + c)^4 + 200*d*cosh(d*x + c)^2 + 28*d)*sinh(d*x +
c)^3 + (36*d*cosh(d*x + c)^7 + 147*d*cosh(d*x + c)^5 + 220*d*cosh(d*x + c)^3 + 126*d*cosh(d*x + c))*sinh(d*x +
c)^2 + 56*d*cosh(d*x + c) + (9*d*cosh(d*x + c)^8 + 49*d*cosh(d*x + c)^6 + 100*d*cosh(d*x + c)^4 + 84*d*cosh(d
*x + c)^2 + 14*d)*sinh(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 )^{2} \operatorname{sech}^{4}{\left (c + d x \right )}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sech(d*x+c)**4*(a+b*sech(d*x+c)**2)**2,x)
[Out]
Integral((a + b*sech(c + d*x)**2)**2*sech(c + d*x)**4, x)
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Giac [B] time = 1.15939, size = 266, normalized size = 3.32 \begin{align*} -\frac{4 \,{\left (105 \, a^{2} e^{\left (10 \, d x + 10 \, c\right )} + 455 \, a^{2} e^{\left (8 \, d x + 8 \, c\right )} + 560 \, a b e^{\left (8 \, d x + 8 \, c\right )} + 770 \, a^{2} e^{\left (6 \, d x + 6 \, c\right )} + 1400 \, a b e^{\left (6 \, d x + 6 \, c\right )} + 840 \, b^{2} e^{\left (6 \, d x + 6 \, c\right )} + 630 \, a^{2} e^{\left (4 \, d x + 4 \, c\right )} + 1176 \, a b e^{\left (4 \, d x + 4 \, c\right )} + 504 \, b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 245 \, a^{2} e^{\left (2 \, d x + 2 \, c\right )} + 392 \, a b e^{\left (2 \, d x + 2 \, c\right )} + 168 \, b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 35 \, a^{2} + 56 \, a b + 24 \, b^{2}\right )}}{105 \, d{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{7}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sech(d*x+c)^4*(a+b*sech(d*x+c)^2)^2,x, algorithm="giac")
[Out]
-4/105*(105*a^2*e^(10*d*x + 10*c) + 455*a^2*e^(8*d*x + 8*c) + 560*a*b*e^(8*d*x + 8*c) + 770*a^2*e^(6*d*x + 6*c
) + 1400*a*b*e^(6*d*x + 6*c) + 840*b^2*e^(6*d*x + 6*c) + 630*a^2*e^(4*d*x + 4*c) + 1176*a*b*e^(4*d*x + 4*c) +
504*b^2*e^(4*d*x + 4*c) + 245*a^2*e^(2*d*x + 2*c) + 392*a*b*e^(2*d*x + 2*c) + 168*b^2*e^(2*d*x + 2*c) + 35*a^2
+ 56*a*b + 24*b^2)/(d*(e^(2*d*x + 2*c) + 1)^7)