3.102 \(\int \text {sech}^2(c+d x) (a+b \tanh ^2(c+d x))^3 \, dx\)

Optimal. Leaf size=67 \[ \frac {a^3 \tanh (c+d x)}{d}+\frac {a^2 b \tanh ^3(c+d x)}{d}+\frac {3 a b^2 \tanh ^5(c+d x)}{5 d}+\frac {b^3 \tanh ^7(c+d x)}{7 d} \]

[Out]

a^3*tanh(d*x+c)/d+a^2*b*tanh(d*x+c)^3/d+3/5*a*b^2*tanh(d*x+c)^5/d+1/7*b^3*tanh(d*x+c)^7/d

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Rubi [A]  time = 0.06, antiderivative size = 67, normalized size of antiderivative = 1.00, 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 = {3675, 194} \[ \frac {a^2 b \tanh ^3(c+d x)}{d}+\frac {a^3 \tanh (c+d x)}{d}+\frac {3 a b^2 \tanh ^5(c+d x)}{5 d}+\frac {b^3 \tanh ^7(c+d x)}{7 d} \]

Antiderivative was successfully verified.

[In]

Int[Sech[c + d*x]^2*(a + b*Tanh[c + d*x]^2)^3,x]

[Out]

(a^3*Tanh[c + d*x])/d + (a^2*b*Tanh[c + d*x]^3)/d + (3*a*b^2*Tanh[c + d*x]^5)/(5*d) + (b^3*Tanh[c + d*x]^7)/(7
*d)

Rule 194

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^n)^p, x], x] /; FreeQ[{a, b}, x]
&& IGtQ[n, 0] && IGtQ[p, 0]

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 )^3 \, dx &=\frac {\operatorname {Subst}\left (\int \left (a+b x^2\right )^3 \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac {\operatorname {Subst}\left (\int \left (a^3+3 a^2 b x^2+3 a b^2 x^4+b^3 x^6\right ) \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac {a^3 \tanh (c+d x)}{d}+\frac {a^2 b \tanh ^3(c+d x)}{d}+\frac {3 a b^2 \tanh ^5(c+d x)}{5 d}+\frac {b^3 \tanh ^7(c+d x)}{7 d}\\ \end {align*}

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Mathematica [A]  time = 0.18, size = 67, normalized size = 1.00 \[ \frac {a^3 \tanh (c+d x)}{d}+\frac {a^2 b \tanh ^3(c+d x)}{d}+\frac {3 a b^2 \tanh ^5(c+d x)}{5 d}+\frac {b^3 \tanh ^7(c+d x)}{7 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sech[c + d*x]^2*(a + b*Tanh[c + d*x]^2)^3,x]

[Out]

(a^3*Tanh[c + d*x])/d + (a^2*b*Tanh[c + d*x]^3)/d + (3*a*b^2*Tanh[c + d*x]^5)/(5*d) + (b^3*Tanh[c + d*x]^7)/(7
*d)

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fricas [B]  time = 0.40, size = 786, normalized size = 11.73 \[ -\frac {4 \, {\left ({\left (35 \, a^{3} + 70 \, a^{2} b + 63 \, a b^{2} + 20 \, b^{3}\right )} \cosh \left (d x + c\right )^{6} + 6 \, {\left (35 \, a^{2} b + 42 \, a b^{2} + 15 \, b^{3}\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{5} + {\left (35 \, a^{3} + 70 \, a^{2} b + 63 \, a b^{2} + 20 \, b^{3}\right )} \sinh \left (d x + c\right )^{6} + 14 \, {\left (15 \, a^{3} + 20 \, a^{2} b + 9 \, a b^{2}\right )} \cosh \left (d x + c\right )^{4} + {\left (210 \, a^{3} + 280 \, a^{2} b + 126 \, a b^{2} + 15 \, {\left (35 \, a^{3} + 70 \, a^{2} b + 63 \, a b^{2} + 20 \, b^{3}\right )} \cosh \left (d x + c\right )^{2}\right )} \sinh \left (d x + c\right )^{4} + 4 \, {\left (5 \, {\left (35 \, a^{2} b + 42 \, a b^{2} + 15 \, b^{3}\right )} \cosh \left (d x + c\right )^{3} + 28 \, {\left (5 \, a^{2} b + 3 \, a b^{2}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{3} + 350 \, a^{3} + 280 \, a^{2} b + 210 \, a b^{2} + 7 \, {\left (75 \, a^{3} + 70 \, a^{2} b + 39 \, a b^{2} + 20 \, b^{3}\right )} \cosh \left (d x + c\right )^{2} + {\left (15 \, {\left (35 \, a^{3} + 70 \, a^{2} b + 63 \, a b^{2} + 20 \, b^{3}\right )} \cosh \left (d x + c\right )^{4} + 525 \, a^{3} + 490 \, a^{2} b + 273 \, a b^{2} + 140 \, b^{3} + 84 \, {\left (15 \, a^{3} + 20 \, a^{2} b + 9 \, a b^{2}\right )} \cosh \left (d x + c\right )^{2}\right )} \sinh \left (d x + c\right )^{2} + 2 \, {\left (3 \, {\left (35 \, a^{2} b + 42 \, a b^{2} + 15 \, b^{3}\right )} \cosh \left (d x + c\right )^{5} + 56 \, {\left (5 \, a^{2} b + 3 \, a b^{2}\right )} \cosh \left (d x + c\right )^{3} + 7 \, {\left (25 \, a^{2} b + 6 \, a b^{2} + 5 \, b^{3}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )\right )}}{35 \, {\left (d \cosh \left (d x + c\right )^{8} + 8 \, d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{7} + d \sinh \left (d x + c\right )^{8} + 8 \, d \cosh \left (d x + c\right )^{6} + 4 \, {\left (7 \, d \cosh \left (d x + c\right )^{2} + 2 \, d\right )} \sinh \left (d x + c\right )^{6} + 4 \, {\left (14 \, d \cosh \left (d x + c\right )^{3} + 9 \, d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{5} + 28 \, d \cosh \left (d x + c\right )^{4} + 2 \, {\left (35 \, d \cosh \left (d x + c\right )^{4} + 60 \, d \cosh \left (d x + c\right )^{2} + 14 \, d\right )} \sinh \left (d x + c\right )^{4} + 8 \, {\left (7 \, d \cosh \left (d x + c\right )^{5} + 15 \, d \cosh \left (d x + c\right )^{3} + 7 \, d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{3} + 56 \, d \cosh \left (d x + c\right )^{2} + 4 \, {\left (7 \, d \cosh \left (d x + c\right )^{6} + 30 \, d \cosh \left (d x + c\right )^{4} + 42 \, d \cosh \left (d x + c\right )^{2} + 14 \, d\right )} \sinh \left (d x + c\right )^{2} + 4 \, {\left (2 \, d \cosh \left (d x + c\right )^{7} + 9 \, d \cosh \left (d x + c\right )^{5} + 14 \, d \cosh \left (d x + c\right )^{3} + 7 \, d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right ) + 35 \, 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)^3,x, algorithm="fricas")

[Out]

-4/35*((35*a^3 + 70*a^2*b + 63*a*b^2 + 20*b^3)*cosh(d*x + c)^6 + 6*(35*a^2*b + 42*a*b^2 + 15*b^3)*cosh(d*x + c
)*sinh(d*x + c)^5 + (35*a^3 + 70*a^2*b + 63*a*b^2 + 20*b^3)*sinh(d*x + c)^6 + 14*(15*a^3 + 20*a^2*b + 9*a*b^2)
*cosh(d*x + c)^4 + (210*a^3 + 280*a^2*b + 126*a*b^2 + 15*(35*a^3 + 70*a^2*b + 63*a*b^2 + 20*b^3)*cosh(d*x + c)
^2)*sinh(d*x + c)^4 + 4*(5*(35*a^2*b + 42*a*b^2 + 15*b^3)*cosh(d*x + c)^3 + 28*(5*a^2*b + 3*a*b^2)*cosh(d*x +
c))*sinh(d*x + c)^3 + 350*a^3 + 280*a^2*b + 210*a*b^2 + 7*(75*a^3 + 70*a^2*b + 39*a*b^2 + 20*b^3)*cosh(d*x + c
)^2 + (15*(35*a^3 + 70*a^2*b + 63*a*b^2 + 20*b^3)*cosh(d*x + c)^4 + 525*a^3 + 490*a^2*b + 273*a*b^2 + 140*b^3
+ 84*(15*a^3 + 20*a^2*b + 9*a*b^2)*cosh(d*x + c)^2)*sinh(d*x + c)^2 + 2*(3*(35*a^2*b + 42*a*b^2 + 15*b^3)*cosh
(d*x + c)^5 + 56*(5*a^2*b + 3*a*b^2)*cosh(d*x + c)^3 + 7*(25*a^2*b + 6*a*b^2 + 5*b^3)*cosh(d*x + c))*sinh(d*x
+ c))/(d*cosh(d*x + c)^8 + 8*d*cosh(d*x + c)*sinh(d*x + c)^7 + d*sinh(d*x + c)^8 + 8*d*cosh(d*x + c)^6 + 4*(7*
d*cosh(d*x + c)^2 + 2*d)*sinh(d*x + c)^6 + 4*(14*d*cosh(d*x + c)^3 + 9*d*cosh(d*x + c))*sinh(d*x + c)^5 + 28*d
*cosh(d*x + c)^4 + 2*(35*d*cosh(d*x + c)^4 + 60*d*cosh(d*x + c)^2 + 14*d)*sinh(d*x + c)^4 + 8*(7*d*cosh(d*x +
c)^5 + 15*d*cosh(d*x + c)^3 + 7*d*cosh(d*x + c))*sinh(d*x + c)^3 + 56*d*cosh(d*x + c)^2 + 4*(7*d*cosh(d*x + c)
^6 + 30*d*cosh(d*x + c)^4 + 42*d*cosh(d*x + c)^2 + 14*d)*sinh(d*x + c)^2 + 4*(2*d*cosh(d*x + c)^7 + 9*d*cosh(d
*x + c)^5 + 14*d*cosh(d*x + c)^3 + 7*d*cosh(d*x + c))*sinh(d*x + c) + 35*d)

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giac [B]  time = 0.39, size = 347, normalized size = 5.18 \[ -\frac {2 \, {\left (35 \, a^{3} e^{\left (12 \, d x + 12 \, c\right )} + 105 \, a^{2} b e^{\left (12 \, d x + 12 \, c\right )} + 105 \, a b^{2} e^{\left (12 \, d x + 12 \, c\right )} + 35 \, b^{3} e^{\left (12 \, d x + 12 \, c\right )} + 210 \, a^{3} e^{\left (10 \, d x + 10 \, c\right )} + 420 \, a^{2} b e^{\left (10 \, d x + 10 \, c\right )} + 210 \, a b^{2} e^{\left (10 \, d x + 10 \, c\right )} + 525 \, a^{3} e^{\left (8 \, d x + 8 \, c\right )} + 665 \, a^{2} b e^{\left (8 \, d x + 8 \, c\right )} + 315 \, a b^{2} e^{\left (8 \, d x + 8 \, c\right )} + 175 \, b^{3} e^{\left (8 \, d x + 8 \, c\right )} + 700 \, a^{3} e^{\left (6 \, d x + 6 \, c\right )} + 560 \, a^{2} b e^{\left (6 \, d x + 6 \, c\right )} + 420 \, a b^{2} e^{\left (6 \, d x + 6 \, c\right )} + 525 \, a^{3} e^{\left (4 \, d x + 4 \, c\right )} + 315 \, a^{2} b e^{\left (4 \, d x + 4 \, c\right )} + 231 \, a b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 105 \, b^{3} e^{\left (4 \, d x + 4 \, c\right )} + 210 \, a^{3} e^{\left (2 \, d x + 2 \, c\right )} + 140 \, a^{2} b e^{\left (2 \, d x + 2 \, c\right )} + 42 \, a b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 35 \, a^{3} + 35 \, a^{2} b + 21 \, a b^{2} + 5 \, b^{3}\right )}}{35 \, d {\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(d*x+c)^2*(a+b*tanh(d*x+c)^2)^3,x, algorithm="giac")

[Out]

-2/35*(35*a^3*e^(12*d*x + 12*c) + 105*a^2*b*e^(12*d*x + 12*c) + 105*a*b^2*e^(12*d*x + 12*c) + 35*b^3*e^(12*d*x
+ 12*c) + 210*a^3*e^(10*d*x + 10*c) + 420*a^2*b*e^(10*d*x + 10*c) + 210*a*b^2*e^(10*d*x + 10*c) + 525*a^3*e^(
8*d*x + 8*c) + 665*a^2*b*e^(8*d*x + 8*c) + 315*a*b^2*e^(8*d*x + 8*c) + 175*b^3*e^(8*d*x + 8*c) + 700*a^3*e^(6*
d*x + 6*c) + 560*a^2*b*e^(6*d*x + 6*c) + 420*a*b^2*e^(6*d*x + 6*c) + 525*a^3*e^(4*d*x + 4*c) + 315*a^2*b*e^(4*
d*x + 4*c) + 231*a*b^2*e^(4*d*x + 4*c) + 105*b^3*e^(4*d*x + 4*c) + 210*a^3*e^(2*d*x + 2*c) + 140*a^2*b*e^(2*d*
x + 2*c) + 42*a*b^2*e^(2*d*x + 2*c) + 35*a^3 + 35*a^2*b + 21*a*b^2 + 5*b^3)/(d*(e^(2*d*x + 2*c) + 1)^7)

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maple [B]  time = 0.55, size = 227, normalized size = 3.39 \[ \frac {a^{3} \tanh \left (d x +c \right )+3 a^{2} b \left (-\frac {\sinh \left (d x +c \right )}{2 \cosh \left (d x +c \right )^{3}}+\frac {\left (\frac {2}{3}+\frac {\mathrm {sech}\left (d x +c \right )^{2}}{3}\right ) \tanh \left (d x +c \right )}{2}\right )+3 a \,b^{2} \left (-\frac {\sinh ^{3}\left (d x +c \right )}{2 \cosh \left (d x +c \right )^{5}}-\frac {3 \sinh \left (d x +c \right )}{8 \cosh \left (d x +c \right )^{5}}+\frac {3 \left (\frac {8}{15}+\frac {\mathrm {sech}\left (d x +c \right )^{4}}{5}+\frac {4 \mathrm {sech}\left (d x +c \right )^{2}}{15}\right ) \tanh \left (d x +c \right )}{8}\right )+b^{3} \left (-\frac {\sinh ^{5}\left (d x +c \right )}{2 \cosh \left (d x +c \right )^{7}}-\frac {5 \left (\sinh ^{3}\left (d x +c \right )\right )}{8 \cosh \left (d x +c \right )^{7}}-\frac {5 \sinh \left (d x +c \right )}{16 \cosh \left (d x +c \right )^{7}}+\frac {5 \left (\frac {16}{35}+\frac {\mathrm {sech}\left (d x +c \right )^{6}}{7}+\frac {6 \mathrm {sech}\left (d x +c \right )^{4}}{35}+\frac {8 \mathrm {sech}\left (d x +c \right )^{2}}{35}\right ) \tanh \left (d x +c \right )}{16}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sech(d*x+c)^2*(a+b*tanh(d*x+c)^2)^3,x)

[Out]

1/d*(a^3*tanh(d*x+c)+3*a^2*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))+3*a*b^2*
(-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))+b^3*(-1/2*sinh(d*x+c)^5/cosh(d*x+c)^7-5/8*sinh(d*x+c)^3/cosh(d*x+c)^7-5/16*sinh(d*x+c)/cosh(d*x
+c)^7+5/16*(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 [A]  time = 0.33, size = 71, normalized size = 1.06 \[ \frac {b^{3} \tanh \left (d x + c\right )^{7}}{7 \, d} + \frac {3 \, a b^{2} \tanh \left (d x + c\right )^{5}}{5 \, d} + \frac {a^{2} b \tanh \left (d x + c\right )^{3}}{d} + \frac {2 \, a^{3}}{d {\left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(d*x+c)^2*(a+b*tanh(d*x+c)^2)^3,x, algorithm="maxima")

[Out]

1/7*b^3*tanh(d*x + c)^7/d + 3/5*a*b^2*tanh(d*x + c)^5/d + a^2*b*tanh(d*x + c)^3/d + 2*a^3/(d*(e^(-2*d*x - 2*c)
+ 1))

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mupad [B]  time = 1.33, size = 1050, normalized size = 15.67 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*tanh(c + d*x)^2)^3/cosh(c + d*x)^2,x)

[Out]

- ((2*(3*a*b^2 - 3*a^2*b + 5*a^3 - 5*b^3))/(35*d) + (2*exp(6*c + 6*d*x)*(a + b)^3)/(7*d) - (6*exp(2*c + 2*d*x)
*(a*b^2 + a^2*b - 5*a^3 - 5*b^3))/(35*d) + (6*exp(4*c + 4*d*x)*(a + b)^2*(a - b))/(7*d))/(4*exp(2*c + 2*d*x) +
6*exp(4*c + 4*d*x) + 4*exp(6*c + 6*d*x) + exp(8*c + 8*d*x) + 1) - ((2*(a + b)^3)/(7*d) + (2*exp(12*c + 12*d*x
)*(a + b)^3)/(7*d) - (6*exp(4*c + 4*d*x)*(a*b^2 + a^2*b - 5*a^3 - 5*b^3))/(7*d) - (6*exp(8*c + 8*d*x)*(a*b^2 +
a^2*b - 5*a^3 - 5*b^3))/(7*d) + (8*exp(6*c + 6*d*x)*(3*a*b^2 - 3*a^2*b + 5*a^3 - 5*b^3))/(7*d) + (12*exp(2*c
+ 2*d*x)*(a + b)^2*(a - b))/(7*d) + (12*exp(10*c + 10*d*x)*(a + b)^2*(a - b))/(7*d))/(7*exp(2*c + 2*d*x) + 21*
exp(4*c + 4*d*x) + 35*exp(6*c + 6*d*x) + 35*exp(8*c + 8*d*x) + 21*exp(10*c + 10*d*x) + 7*exp(12*c + 12*d*x) +
exp(14*c + 14*d*x) + 1) - ((2*exp(4*c + 4*d*x)*(a + b)^3)/(7*d) - (2*(a*b^2 + a^2*b - 5*a^3 - 5*b^3))/(35*d) +
(4*exp(2*c + 2*d*x)*(a + b)^2*(a - b))/(7*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)^2*(a - b))/(7*d) + (2*exp(2*c + 2*d*x)*(a + b)^3)/(7*d))/(2*exp(2*c + 2*d*x) + exp(4*c + 4*d*x
) + 1) - ((2*exp(8*c + 8*d*x)*(a + b)^3)/(7*d) - (2*(a*b^2 + a^2*b - 5*a^3 - 5*b^3))/(35*d) - (12*exp(4*c + 4*
d*x)*(a*b^2 + a^2*b - 5*a^3 - 5*b^3))/(35*d) + (8*exp(2*c + 2*d*x)*(3*a*b^2 - 3*a^2*b + 5*a^3 - 5*b^3))/(35*d)
+ (8*exp(6*c + 6*d*x)*(a + b)^2*(a - b))/(7*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)^2*(a - b))/(7*d) + (2*exp(10*c + 10*d*x)*(a +
b)^3)/(7*d) - (2*exp(2*c + 2*d*x)*(a*b^2 + a^2*b - 5*a^3 - 5*b^3))/(7*d) - (4*exp(6*c + 6*d*x)*(a*b^2 + a^2*b
- 5*a^3 - 5*b^3))/(7*d) + (4*exp(4*c + 4*d*x)*(3*a*b^2 - 3*a^2*b + 5*a^3 - 5*b^3))/(7*d) + (10*exp(8*c + 8*d*x
)*(a + b)^2*(a - b))/(7*d))/(6*exp(2*c + 2*d*x) + 15*exp(4*c + 4*d*x) + 20*exp(6*c + 6*d*x) + 15*exp(8*c + 8*d
*x) + 6*exp(10*c + 10*d*x) + exp(12*c + 12*d*x) + 1) - (2*(a + b)^3)/(7*d*(exp(2*c + 2*d*x) + 1))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \tanh ^{2}{\left (c + d x \right )}\right )^{3} \operatorname {sech}^{2}{\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(d*x+c)**2*(a+b*tanh(d*x+c)**2)**3,x)

[Out]

Integral((a + b*tanh(c + d*x)**2)**3*sech(c + d*x)**2, x)

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