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

Optimal. Leaf size=63 \[ -\frac {b (2 a+b) \tanh ^3(c+d x)}{3 d}-\frac {(a+b)^2 \tanh (c+d x)}{d}+x (a+b)^2-\frac {b^2 \tanh ^5(c+d x)}{5 d} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3670, 461, 206} \[ -\frac {b (2 a+b) \tanh ^3(c+d x)}{3 d}-\frac {(a+b)^2 \tanh (c+d x)}{d}+x (a+b)^2-\frac {b^2 \tanh ^5(c+d x)}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [B]  time = 0.05, size = 137, normalized size = 2.17 \[ \frac {a^2 \tanh ^{-1}(\tanh (c+d x))}{d}-\frac {a^2 \tanh (c+d x)}{d}+\frac {2 a b \tanh ^{-1}(\tanh (c+d x))}{d}-\frac {2 a b \tanh ^3(c+d x)}{3 d}-\frac {2 a b \tanh (c+d x)}{d}+\frac {b^2 \tanh ^{-1}(\tanh (c+d x))}{d}-\frac {b^2 \tanh ^5(c+d x)}{5 d}-\frac {b^2 \tanh ^3(c+d x)}{3 d}-\frac {b^2 \tanh (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

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fricas [B]  time = 0.43, size = 483, normalized size = 7.67 \[ \frac {{\left (15 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} d x + 15 \, a^{2} + 40 \, a b + 23 \, b^{2}\right )} \cosh \left (d x + c\right )^{5} + 5 \, {\left (15 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} d x + 15 \, a^{2} + 40 \, a b + 23 \, b^{2}\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{4} - {\left (15 \, a^{2} + 40 \, a b + 23 \, b^{2}\right )} \sinh \left (d x + c\right )^{5} + 5 \, {\left (15 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} d x + 15 \, a^{2} + 40 \, a b + 23 \, b^{2}\right )} \cosh \left (d x + c\right )^{3} - 5 \, {\left (2 \, {\left (15 \, a^{2} + 40 \, a b + 23 \, b^{2}\right )} \cosh \left (d x + c\right )^{2} + 9 \, a^{2} + 16 \, a b + 5 \, b^{2}\right )} \sinh \left (d x + c\right )^{3} + 5 \, {\left (2 \, {\left (15 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} d x + 15 \, a^{2} + 40 \, a b + 23 \, b^{2}\right )} \cosh \left (d x + c\right )^{3} + 3 \, {\left (15 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} d x + 15 \, a^{2} + 40 \, a b + 23 \, b^{2}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} + 10 \, {\left (15 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} d x + 15 \, a^{2} + 40 \, a b + 23 \, b^{2}\right )} \cosh \left (d x + c\right ) - 5 \, {\left ({\left (15 \, a^{2} + 40 \, a b + 23 \, b^{2}\right )} \cosh \left (d x + c\right )^{4} + 3 \, {\left (9 \, a^{2} + 16 \, a b + 5 \, b^{2}\right )} \cosh \left (d x + c\right )^{2} + 6 \, a^{2} + 8 \, a b + 10 \, b^{2}\right )} \sinh \left (d x + c\right )}{15 \, {\left (d \cosh \left (d x + c\right )^{5} + 5 \, d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{4} + 5 \, d \cosh \left (d x + c\right )^{3} + 5 \, {\left (2 \, d \cosh \left (d x + c\right )^{3} + 3 \, d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} + 10 \, d \cosh \left (d x + c\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*((15*(a^2 + 2*a*b + b^2)*d*x + 15*a^2 + 40*a*b + 23*b^2)*cosh(d*x + c)^5 + 5*(15*(a^2 + 2*a*b + b^2)*d*x
+ 15*a^2 + 40*a*b + 23*b^2)*cosh(d*x + c)*sinh(d*x + c)^4 - (15*a^2 + 40*a*b + 23*b^2)*sinh(d*x + c)^5 + 5*(15
*(a^2 + 2*a*b + b^2)*d*x + 15*a^2 + 40*a*b + 23*b^2)*cosh(d*x + c)^3 - 5*(2*(15*a^2 + 40*a*b + 23*b^2)*cosh(d*
x + c)^2 + 9*a^2 + 16*a*b + 5*b^2)*sinh(d*x + c)^3 + 5*(2*(15*(a^2 + 2*a*b + b^2)*d*x + 15*a^2 + 40*a*b + 23*b
^2)*cosh(d*x + c)^3 + 3*(15*(a^2 + 2*a*b + b^2)*d*x + 15*a^2 + 40*a*b + 23*b^2)*cosh(d*x + c))*sinh(d*x + c)^2
+ 10*(15*(a^2 + 2*a*b + b^2)*d*x + 15*a^2 + 40*a*b + 23*b^2)*cosh(d*x + c) - 5*((15*a^2 + 40*a*b + 23*b^2)*co
sh(d*x + c)^4 + 3*(9*a^2 + 16*a*b + 5*b^2)*cosh(d*x + c)^2 + 6*a^2 + 8*a*b + 10*b^2)*sinh(d*x + c))/(d*cosh(d*
x + c)^5 + 5*d*cosh(d*x + c)*sinh(d*x + c)^4 + 5*d*cosh(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 = 0.26, size = 218, normalized size = 3.46 \[ \frac {15 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} {\left (d x + c\right )} + \frac {2 \, {\left (15 \, a^{2} e^{\left (8 \, d x + 8 \, c\right )} + 60 \, a b e^{\left (8 \, d x + 8 \, c\right )} + 45 \, b^{2} e^{\left (8 \, d x + 8 \, c\right )} + 60 \, a^{2} e^{\left (6 \, d x + 6 \, c\right )} + 180 \, a b e^{\left (6 \, d x + 6 \, c\right )} + 90 \, b^{2} e^{\left (6 \, d x + 6 \, c\right )} + 90 \, a^{2} e^{\left (4 \, d x + 4 \, c\right )} + 220 \, a b e^{\left (4 \, d x + 4 \, c\right )} + 140 \, b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 60 \, a^{2} e^{\left (2 \, d x + 2 \, c\right )} + 140 \, a b e^{\left (2 \, d x + 2 \, c\right )} + 70 \, b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 15 \, a^{2} + 40 \, a b + 23 \, b^{2}\right )}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{5}}}{15 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(15*(a^2 + 2*a*b + b^2)*(d*x + c) + 2*(15*a^2*e^(8*d*x + 8*c) + 60*a*b*e^(8*d*x + 8*c) + 45*b^2*e^(8*d*x
+ 8*c) + 60*a^2*e^(6*d*x + 6*c) + 180*a*b*e^(6*d*x + 6*c) + 90*b^2*e^(6*d*x + 6*c) + 90*a^2*e^(4*d*x + 4*c) +
220*a*b*e^(4*d*x + 4*c) + 140*b^2*e^(4*d*x + 4*c) + 60*a^2*e^(2*d*x + 2*c) + 140*a*b*e^(2*d*x + 2*c) + 70*b^2*
e^(2*d*x + 2*c) + 15*a^2 + 40*a*b + 23*b^2)/(e^(2*d*x + 2*c) + 1)^5)/d

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maple [B]  time = 0.02, size = 189, normalized size = 3.00 \[ -\frac {a^{2} \tanh \left (d x +c \right )}{d}-\frac {\ln \left (\tanh \left (d x +c \right )-1\right ) a^{2}}{2 d}-\frac {\ln \left (\tanh \left (d x +c \right )-1\right ) a b}{d}-\frac {\ln \left (\tanh \left (d x +c \right )-1\right ) b^{2}}{2 d}-\frac {b^{2} \left (\tanh ^{3}\left (d x +c \right )\right )}{3 d}-\frac {b^{2} \left (\tanh ^{5}\left (d x +c \right )\right )}{5 d}+\frac {\ln \left (1+\tanh \left (d x +c \right )\right ) a^{2}}{2 d}+\frac {\ln \left (1+\tanh \left (d x +c \right )\right ) a b}{d}+\frac {\ln \left (1+\tanh \left (d x +c \right )\right ) b^{2}}{2 d}-\frac {2 a b \left (\tanh ^{3}\left (d x +c \right )\right )}{3 d}-\frac {2 a b \tanh \left (d x +c \right )}{d}-\frac {b^{2} \tanh \left (d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.34, size = 231, normalized size = 3.67 \[ \frac {1}{15} \, b^{2} {\left (15 \, x + \frac {15 \, c}{d} - \frac {2 \, {\left (70 \, e^{\left (-2 \, d x - 2 \, c\right )} + 140 \, e^{\left (-4 \, d x - 4 \, c\right )} + 90 \, e^{\left (-6 \, d x - 6 \, c\right )} + 45 \, e^{\left (-8 \, d x - 8 \, c\right )} + 23\right )}}{d {\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} + 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} + 1\right )}}\right )} + \frac {2}{3} \, a b {\left (3 \, x + \frac {3 \, c}{d} - \frac {4 \, {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + 2\right )}}{d {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} + 1\right )}}\right )} + a^{2} {\left (x + \frac {c}{d} - \frac {2}{d {\left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*b^2*(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))) + 2/3*a*b*(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))) + a^2*(x + c/d - 2/(d*(e^(-2*d*x - 2*c) + 1)))

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mupad [B]  time = 1.30, size = 67, normalized size = 1.06 \[ x\,\left (a^2+2\,a\,b+b^2\right )-\frac {\mathrm {tanh}\left (c+d\,x\right )\,{\left (a+b\right )}^2}{d}-\frac {{\mathrm {tanh}\left (c+d\,x\right )}^3\,\left (b^2+2\,a\,b\right )}{3\,d}-\frac {b^2\,{\mathrm {tanh}\left (c+d\,x\right )}^5}{5\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.64, size = 117, normalized size = 1.86 \[ \begin {cases} a^{2} x - \frac {a^{2} \tanh {\left (c + d x \right )}}{d} + 2 a b x - \frac {2 a b \tanh ^{3}{\left (c + d x \right )}}{3 d} - \frac {2 a b \tanh {\left (c + d x \right )}}{d} + b^{2} x - \frac {b^{2} \tanh ^{5}{\left (c + d x \right )}}{5 d} - \frac {b^{2} \tanh ^{3}{\left (c + d x \right )}}{3 d} - \frac {b^{2} \tanh {\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (a + b \tanh ^{2}{\relax (c )}\right )^{2} \tanh ^{2}{\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a**2*x - a**2*tanh(c + d*x)/d + 2*a*b*x - 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)**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)**2, True))

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