3.144 \(\int \tanh ^4(c+d x) (a+b \tanh ^2(c+d x))^2 \, dx\)

Optimal. Leaf size=83 \[ -\frac{b (2 a+b) \tanh ^5(c+d x)}{5 d}-\frac{(a+b)^2 \tanh ^3(c+d x)}{3 d}-\frac{(a+b)^2 \tanh (c+d x)}{d}+x (a+b)^2-\frac{b^2 \tanh ^7(c+d x)}{7 d} \]

[Out]

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

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Rubi [A]  time = 0.0788679, antiderivative size = 83, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {3670, 461, 206} \[ -\frac{b (2 a+b) \tanh ^5(c+d x)}{5 d}-\frac{(a+b)^2 \tanh ^3(c+d x)}{3 d}-\frac{(a+b)^2 \tanh (c+d x)}{d}+x (a+b)^2-\frac{b^2 \tanh ^7(c+d x)}{7 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3670

Int[((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol]
:> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist[(c*ff)/f, Subst[Int[(((d*ff*x)/c)^m*(a + b*(ff*x)^n)^p)/(c^
2 + ff^2*x^2), x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && (IGtQ[p, 0] || EqQ
[n, 2] || EqQ[n, 4] || (IntegerQ[p] && RationalQ[n]))

Rule 461

Int[(((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_))/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Int[ExpandIntegr
and[((e*x)^m*(a + b*x^n)^p)/(c + d*x^n), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && IGtQ[n
, 0] && IGtQ[p, 0] && (IntegerQ[m] || IGtQ[2*(m + 1), 0] ||  !RationalQ[m])

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 \tanh ^4(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^4 \left (a+b x^2\right )^2}{1-x^2} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \left (-(a+b)^2-(a+b)^2 x^2-b (2 a+b) x^4-b^2 x^6+\frac{a^2+2 a b+b^2}{1-x^2}\right ) \, dx,x,\tanh (c+d x)\right )}{d}\\ &=-\frac{(a+b)^2 \tanh (c+d x)}{d}-\frac{(a+b)^2 \tanh ^3(c+d x)}{3 d}-\frac{b (2 a+b) \tanh ^5(c+d x)}{5 d}-\frac{b^2 \tanh ^7(c+d x)}{7 d}+\frac{(a+b)^2 \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=(a+b)^2 x-\frac{(a+b)^2 \tanh (c+d x)}{d}-\frac{(a+b)^2 \tanh ^3(c+d x)}{3 d}-\frac{b (2 a+b) \tanh ^5(c+d x)}{5 d}-\frac{b^2 \tanh ^7(c+d x)}{7 d}\\ \end{align*}

Mathematica [B]  time = 0.0767448, size = 190, normalized size = 2.29 \[ -\frac{a^2 \tanh ^3(c+d x)}{3 d}+\frac{a^2 \tanh ^{-1}(\tanh (c+d x))}{d}-\frac{a^2 \tanh (c+d x)}{d}-\frac{2 a b \tanh ^5(c+d x)}{5 d}-\frac{2 a b \tanh ^3(c+d x)}{3 d}+\frac{2 a b \tanh ^{-1}(\tanh (c+d x))}{d}-\frac{2 a b \tanh (c+d x)}{d}-\frac{b^2 \tanh ^7(c+d x)}{7 d}-\frac{b^2 \tanh ^5(c+d x)}{5 d}-\frac{b^2 \tanh ^3(c+d x)}{3 d}+\frac{b^2 \tanh ^{-1}(\tanh (c+d x))}{d}-\frac{b^2 \tanh (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.006, size = 236, normalized size = 2.8 \begin{align*} -{\frac{{a}^{2}\ln \left ( \tanh \left ( dx+c \right ) -1 \right ) }{2\,d}}-{\frac{\ln \left ( \tanh \left ( dx+c \right ) -1 \right ) ab}{d}}-{\frac{\ln \left ( \tanh \left ( dx+c \right ) -1 \right ){b}^{2}}{2\,d}}-{\frac{{b}^{2} \left ( \tanh \left ( dx+c \right ) \right ) ^{3}}{3\,d}}-{\frac{{b}^{2} \left ( \tanh \left ( dx+c \right ) \right ) ^{5}}{5\,d}}-{\frac{ \left ( \tanh \left ( dx+c \right ) \right ) ^{3}{a}^{2}}{3\,d}}-{\frac{{b}^{2} \left ( \tanh \left ( dx+c \right ) \right ) ^{7}}{7\,d}}-2\,{\frac{ab\tanh \left ( dx+c \right ) }{d}}-{\frac{2\, \left ( \tanh \left ( dx+c \right ) \right ) ^{5}ab}{5\,d}}-{\frac{2\,ab \left ( \tanh \left ( dx+c \right ) \right ) ^{3}}{3\,d}}+{\frac{\ln \left ( \tanh \left ( dx+c \right ) +1 \right ){a}^{2}}{2\,d}}+{\frac{\ln \left ( \tanh \left ( dx+c \right ) +1 \right ) ab}{d}}+{\frac{\ln \left ( \tanh \left ( dx+c \right ) +1 \right ){b}^{2}}{2\,d}}-{\frac{{a}^{2}\tanh \left ( dx+c \right ) }{d}}-{\frac{{b}^{2}\tanh \left ( dx+c \right ) }{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.16148, size = 498, normalized size = 6. \begin{align*} \frac{1}{105} \, b^{2}{\left (105 \, x + \frac{105 \, c}{d} - \frac{8 \,{\left (203 \, e^{\left (-2 \, d x - 2 \, c\right )} + 609 \, e^{\left (-4 \, d x - 4 \, c\right )} + 770 \, e^{\left (-6 \, d x - 6 \, c\right )} + 770 \, e^{\left (-8 \, d x - 8 \, c\right )} + 315 \, e^{\left (-10 \, d x - 10 \, c\right )} + 105 \, e^{\left (-12 \, d x - 12 \, c\right )} + 44\right )}}{d{\left (7 \, e^{\left (-2 \, d x - 2 \, c\right )} + 21 \, e^{\left (-4 \, d x - 4 \, c\right )} + 35 \, e^{\left (-6 \, d x - 6 \, c\right )} + 35 \, e^{\left (-8 \, d x - 8 \, c\right )} + 21 \, e^{\left (-10 \, d x - 10 \, c\right )} + 7 \, e^{\left (-12 \, d x - 12 \, c\right )} + e^{\left (-14 \, d x - 14 \, c\right )} + 1\right )}}\right )} + \frac{2}{15} \, a b{\left (15 \, x + \frac{15 \, c}{d} - \frac{2 \,{\left (70 \, e^{\left (-2 \, d x - 2 \, c\right )} + 140 \, e^{\left (-4 \, d x - 4 \, c\right )} + 90 \, e^{\left (-6 \, d x - 6 \, c\right )} + 45 \, e^{\left (-8 \, d x - 8 \, c\right )} + 23\right )}}{d{\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} + 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} + 1\right )}}\right )} + \frac{1}{3} \, a^{2}{\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(tanh(d*x+c)^4*(a+b*tanh(d*x+c)^2)^2,x, algorithm="maxima")

[Out]

1/105*b^2*(105*x + 105*c/d - 8*(203*e^(-2*d*x - 2*c) + 609*e^(-4*d*x - 4*c) + 770*e^(-6*d*x - 6*c) + 770*e^(-8
*d*x - 8*c) + 315*e^(-10*d*x - 10*c) + 105*e^(-12*d*x - 12*c) + 44)/(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 - 1
4*c) + 1))) + 2/15*a*b*(15*x + 15*c/d - 2*(70*e^(-2*d*x - 2*c) + 140*e^(-4*d*x - 4*c) + 90*e^(-6*d*x - 6*c) +
45*e^(-8*d*x - 8*c) + 23)/(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))) + 1/3*a^2*(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 = 1.97528, size = 2140, normalized size = 25.78 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/105*((105*(a^2 + 2*a*b + b^2)*d*x + 140*a^2 + 322*a*b + 176*b^2)*cosh(d*x + c)^7 + 7*(105*(a^2 + 2*a*b + b^2
)*d*x + 140*a^2 + 322*a*b + 176*b^2)*cosh(d*x + c)*sinh(d*x + c)^6 - 2*(70*a^2 + 161*a*b + 88*b^2)*sinh(d*x +
c)^7 + 7*(105*(a^2 + 2*a*b + b^2)*d*x + 140*a^2 + 322*a*b + 176*b^2)*cosh(d*x + c)^5 - 14*(3*(70*a^2 + 161*a*b
+ 88*b^2)*cosh(d*x + c)^2 + 40*a^2 + 71*a*b + 28*b^2)*sinh(d*x + c)^5 + 35*((105*(a^2 + 2*a*b + b^2)*d*x + 14
0*a^2 + 322*a*b + 176*b^2)*cosh(d*x + c)^3 + (105*(a^2 + 2*a*b + b^2)*d*x + 140*a^2 + 322*a*b + 176*b^2)*cosh(
d*x + c))*sinh(d*x + c)^4 + 21*(105*(a^2 + 2*a*b + b^2)*d*x + 140*a^2 + 322*a*b + 176*b^2)*cosh(d*x + c)^3 - 1
4*(5*(70*a^2 + 161*a*b + 88*b^2)*cosh(d*x + c)^4 + 10*(40*a^2 + 71*a*b + 28*b^2)*cosh(d*x + c)^2 + 60*a^2 + 12
3*a*b + 84*b^2)*sinh(d*x + c)^3 + 7*(3*(105*(a^2 + 2*a*b + b^2)*d*x + 140*a^2 + 322*a*b + 176*b^2)*cosh(d*x +
c)^5 + 10*(105*(a^2 + 2*a*b + b^2)*d*x + 140*a^2 + 322*a*b + 176*b^2)*cosh(d*x + c)^3 + 9*(105*(a^2 + 2*a*b +
b^2)*d*x + 140*a^2 + 322*a*b + 176*b^2)*cosh(d*x + c))*sinh(d*x + c)^2 + 35*(105*(a^2 + 2*a*b + b^2)*d*x + 140
*a^2 + 322*a*b + 176*b^2)*cosh(d*x + c) - 14*((70*a^2 + 161*a*b + 88*b^2)*cosh(d*x + c)^6 + 5*(40*a^2 + 71*a*b
+ 28*b^2)*cosh(d*x + c)^4 + 9*(20*a^2 + 41*a*b + 28*b^2)*cosh(d*x + c)^2 + 30*a^2 + 75*a*b)*sinh(d*x + c))/(d
*cosh(d*x + c)^7 + 7*d*cosh(d*x + c)*sinh(d*x + c)^6 + 7*d*cosh(d*x + c)^5 + 35*(d*cosh(d*x + c)^3 + d*cosh(d*
x + c))*sinh(d*x + c)^4 + 21*d*cosh(d*x + c)^3 + 7*(3*d*cosh(d*x + c)^5 + 10*d*cosh(d*x + c)^3 + 9*d*cosh(d*x
+ c))*sinh(d*x + c)^2 + 35*d*cosh(d*x + c))

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Sympy [A]  time = 1.51795, size = 165, normalized size = 1.99 \begin{align*} \begin{cases} a^{2} x - \frac{a^{2} \tanh ^{3}{\left (c + d x \right )}}{3 d} - \frac{a^{2} \tanh{\left (c + d x \right )}}{d} + 2 a b x - \frac{2 a b \tanh ^{5}{\left (c + d x \right )}}{5 d} - \frac{2 a b \tanh ^{3}{\left (c + d x \right )}}{3 d} - \frac{2 a b \tanh{\left (c + d x \right )}}{d} + b^{2} x - \frac{b^{2} \tanh ^{7}{\left (c + d x \right )}}{7 d} - \frac{b^{2} \tanh ^{5}{\left (c + d x \right )}}{5 d} - \frac{b^{2} \tanh ^{3}{\left (c + d x \right )}}{3 d} - \frac{b^{2} \tanh{\left (c + d x \right )}}{d} & \text{for}\: d \neq 0 \\x \left (a + b \tanh ^{2}{\left (c \right )}\right )^{2} \tanh ^{4}{\left (c \right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a**2*x - a**2*tanh(c + d*x)**3/(3*d) - a**2*tanh(c + d*x)/d + 2*a*b*x - 2*a*b*tanh(c + d*x)**5/(5*d
) - 2*a*b*tanh(c + d*x)**3/(3*d) - 2*a*b*tanh(c + d*x)/d + b**2*x - b**2*tanh(c + d*x)**7/(7*d) - b**2*tanh(c
+ d*x)**5/(5*d) - b**2*tanh(c + d*x)**3/(3*d) - b**2*tanh(c + d*x)/d, Ne(d, 0)), (x*(a + b*tanh(c)**2)**2*tanh
(c)**4, True))

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Giac [B]  time = 1.31415, size = 405, normalized size = 4.88 \begin{align*} \frac{{\left (a^{2} + 2 \, a b + b^{2}\right )}{\left (d x + c\right )}}{d} + \frac{4 \,{\left (105 \, a^{2} e^{\left (12 \, d x + 12 \, c\right )} + 315 \, a b e^{\left (12 \, d x + 12 \, c\right )} + 210 \, b^{2} e^{\left (12 \, d x + 12 \, c\right )} + 525 \, a^{2} e^{\left (10 \, d x + 10 \, c\right )} + 1260 \, a b e^{\left (10 \, d x + 10 \, c\right )} + 630 \, b^{2} e^{\left (10 \, d x + 10 \, c\right )} + 1120 \, a^{2} e^{\left (8 \, d x + 8 \, c\right )} + 2555 \, a b e^{\left (8 \, d x + 8 \, c\right )} + 1540 \, b^{2} e^{\left (8 \, d x + 8 \, c\right )} + 1330 \, a^{2} e^{\left (6 \, d x + 6 \, c\right )} + 3080 \, a b e^{\left (6 \, d x + 6 \, c\right )} + 1540 \, b^{2} e^{\left (6 \, d x + 6 \, c\right )} + 945 \, a^{2} e^{\left (4 \, d x + 4 \, c\right )} + 2121 \, a b e^{\left (4 \, d x + 4 \, c\right )} + 1218 \, b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 385 \, a^{2} e^{\left (2 \, d x + 2 \, c\right )} + 812 \, a b e^{\left (2 \, d x + 2 \, c\right )} + 406 \, b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 70 \, a^{2} + 161 \, a b + 88 \, 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(tanh(d*x+c)^4*(a+b*tanh(d*x+c)^2)^2,x, algorithm="giac")

[Out]

(a^2 + 2*a*b + b^2)*(d*x + c)/d + 4/105*(105*a^2*e^(12*d*x + 12*c) + 315*a*b*e^(12*d*x + 12*c) + 210*b^2*e^(12
*d*x + 12*c) + 525*a^2*e^(10*d*x + 10*c) + 1260*a*b*e^(10*d*x + 10*c) + 630*b^2*e^(10*d*x + 10*c) + 1120*a^2*e
^(8*d*x + 8*c) + 2555*a*b*e^(8*d*x + 8*c) + 1540*b^2*e^(8*d*x + 8*c) + 1330*a^2*e^(6*d*x + 6*c) + 3080*a*b*e^(
6*d*x + 6*c) + 1540*b^2*e^(6*d*x + 6*c) + 945*a^2*e^(4*d*x + 4*c) + 2121*a*b*e^(4*d*x + 4*c) + 1218*b^2*e^(4*d
*x + 4*c) + 385*a^2*e^(2*d*x + 2*c) + 812*a*b*e^(2*d*x + 2*c) + 406*b^2*e^(2*d*x + 2*c) + 70*a^2 + 161*a*b + 8
8*b^2)/(d*(e^(2*d*x + 2*c) + 1)^7)