### 3.160 $$\int (a+b \tanh ^2(c+d x))^3 \, dx$$

Optimal. Leaf size=74 $-\frac{b \left (3 a^2+3 a b+b^2\right ) \tanh (c+d x)}{d}-\frac{b^2 (3 a+b) \tanh ^3(c+d x)}{3 d}+x (a+b)^3-\frac{b^3 \tanh ^5(c+d x)}{5 d}$

[Out]

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

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Rubi [A]  time = 0.0458704, antiderivative size = 74, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 14, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.214, Rules used = {3661, 390, 206} $-\frac{b \left (3 a^2+3 a b+b^2\right ) \tanh (c+d x)}{d}-\frac{b^2 (3 a+b) \tanh ^3(c+d x)}{3 d}+x (a+b)^3-\frac{b^3 \tanh ^5(c+d x)}{5 d}$

Antiderivative was successfully veriﬁed.

[In]

Int[(a + b*Tanh[c + d*x]^2)^3,x]

[Out]

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

Rule 3661

Int[((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[(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, e, f, n, p}, x] && (IntegersQ[n, p] || IGtQ[p, 0] || EqQ[n^2, 4] || EqQ[n^2, 16])

Rule 390

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Int[PolynomialDivide[(a + b*x^n)
^p, (c + d*x^n)^(-q), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && IGtQ[p, 0] && ILt
Q[q, 0] && GeQ[p, -q]

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

Mathematica [A]  time = 0.570454, size = 95, normalized size = 1.28 $\frac{\tanh (c+d x) \left (\frac{15 (a+b)^3 \tanh ^{-1}\left (\sqrt{\tanh ^2(c+d x)}\right )}{\sqrt{\tanh ^2(c+d x)}}-b \left (45 a^2+15 a b \left (\tanh ^2(c+d x)+3\right )+b^2 \left (3 \tanh ^4(c+d x)+5 \tanh ^2(c+d x)+15\right )\right )\right )}{15 d}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(Tanh[c + d*x]*((15*(a + b)^3*ArcTanh[Sqrt[Tanh[c + d*x]^2]])/Sqrt[Tanh[c + d*x]^2] - b*(45*a^2 + 15*a*b*(3 +
Tanh[c + d*x]^2) + b^2*(15 + 5*Tanh[c + d*x]^2 + 3*Tanh[c + d*x]^4))))/(15*d)

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Maple [B]  time = 0.004, size = 235, normalized size = 3.2 \begin{align*} -{\frac{{a}^{3}\ln \left ( \tanh \left ( dx+c \right ) -1 \right ) }{2\,d}}-{\frac{3\,\ln \left ( \tanh \left ( dx+c \right ) -1 \right ){a}^{2}b}{2\,d}}-{\frac{3\,\ln \left ( \tanh \left ( dx+c \right ) -1 \right ) a{b}^{2}}{2\,d}}-{\frac{\ln \left ( \tanh \left ( dx+c \right ) -1 \right ){b}^{3}}{2\,d}}-{\frac{{b}^{3} \left ( \tanh \left ( dx+c \right ) \right ) ^{5}}{5\,d}}-{\frac{{b}^{3} \left ( \tanh \left ( dx+c \right ) \right ) ^{3}}{3\,d}}-{\frac{{b}^{3}\tanh \left ( dx+c \right ) }{d}}-{\frac{a{b}^{2} \left ( \tanh \left ( dx+c \right ) \right ) ^{3}}{d}}-3\,{\frac{{a}^{2}b\tanh \left ( dx+c \right ) }{d}}+{\frac{\ln \left ( \tanh \left ( dx+c \right ) +1 \right ){a}^{3}}{2\,d}}+{\frac{3\,\ln \left ( \tanh \left ( dx+c \right ) +1 \right ){a}^{2}b}{2\,d}}+{\frac{3\,\ln \left ( \tanh \left ( dx+c \right ) +1 \right ) a{b}^{2}}{2\,d}}+{\frac{\ln \left ( \tanh \left ( dx+c \right ) +1 \right ){b}^{3}}{2\,d}}-3\,{\frac{a{b}^{2}\tanh \left ( dx+c \right ) }{d}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.10733, size = 323, normalized size = 4.36 \begin{align*} \frac{1}{15} \, b^{3}{\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 )} + a b^{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 )} + 3 \, a^{2} b{\left (x + \frac{c}{d} - \frac{2}{d{\left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}}\right )} + a^{3} x \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.0443, size = 1412, normalized size = 19.08 \begin{align*} \frac{{\left (45 \, a^{2} b + 60 \, a b^{2} + 23 \, b^{3} + 15 \,{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} d x\right )} \cosh \left (d x + c\right )^{5} + 5 \,{\left (45 \, a^{2} b + 60 \, a b^{2} + 23 \, b^{3} + 15 \,{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} d x\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{4} -{\left (45 \, a^{2} b + 60 \, a b^{2} + 23 \, b^{3}\right )} \sinh \left (d x + c\right )^{5} + 5 \,{\left (45 \, a^{2} b + 60 \, a b^{2} + 23 \, b^{3} + 15 \,{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} d x\right )} \cosh \left (d x + c\right )^{3} - 5 \,{\left (27 \, a^{2} b + 24 \, a b^{2} + 5 \, b^{3} + 2 \,{\left (45 \, a^{2} b + 60 \, a b^{2} + 23 \, b^{3}\right )} \cosh \left (d x + c\right )^{2}\right )} \sinh \left (d x + c\right )^{3} + 5 \,{\left (2 \,{\left (45 \, a^{2} b + 60 \, a b^{2} + 23 \, b^{3} + 15 \,{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} d x\right )} \cosh \left (d x + c\right )^{3} + 3 \,{\left (45 \, a^{2} b + 60 \, a b^{2} + 23 \, b^{3} + 15 \,{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} d x\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} + 10 \,{\left (45 \, a^{2} b + 60 \, a b^{2} + 23 \, b^{3} + 15 \,{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} d x\right )} \cosh \left (d x + c\right ) - 5 \,{\left ({\left (45 \, a^{2} b + 60 \, a b^{2} + 23 \, b^{3}\right )} \cosh \left (d x + c\right )^{4} + 18 \, a^{2} b + 12 \, a b^{2} + 10 \, b^{3} + 3 \,{\left (27 \, a^{2} b + 24 \, a b^{2} + 5 \, b^{3}\right )} \cosh \left (d x + c\right )^{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 )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*((45*a^2*b + 60*a*b^2 + 23*b^3 + 15*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*x)*cosh(d*x + c)^5 + 5*(45*a^2*b +
60*a*b^2 + 23*b^3 + 15*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*x)*cosh(d*x + c)*sinh(d*x + c)^4 - (45*a^2*b + 60*a*b
^2 + 23*b^3)*sinh(d*x + c)^5 + 5*(45*a^2*b + 60*a*b^2 + 23*b^3 + 15*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*x)*cosh(
d*x + c)^3 - 5*(27*a^2*b + 24*a*b^2 + 5*b^3 + 2*(45*a^2*b + 60*a*b^2 + 23*b^3)*cosh(d*x + c)^2)*sinh(d*x + c)^
3 + 5*(2*(45*a^2*b + 60*a*b^2 + 23*b^3 + 15*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*x)*cosh(d*x + c)^3 + 3*(45*a^2*b
+ 60*a*b^2 + 23*b^3 + 15*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*x)*cosh(d*x + c))*sinh(d*x + c)^2 + 10*(45*a^2*b +
60*a*b^2 + 23*b^3 + 15*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*x)*cosh(d*x + c) - 5*((45*a^2*b + 60*a*b^2 + 23*b^3)
*cosh(d*x + c)^4 + 18*a^2*b + 12*a*b^2 + 10*b^3 + 3*(27*a^2*b + 24*a*b^2 + 5*b^3)*cosh(d*x + c)^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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Sympy [A]  time = 0.904005, size = 126, normalized size = 1.7 \begin{align*} \begin{cases} a^{3} x + 3 a^{2} b x - \frac{3 a^{2} b \tanh{\left (c + d x \right )}}{d} + 3 a b^{2} x - \frac{a b^{2} \tanh ^{3}{\left (c + d x \right )}}{d} - \frac{3 a b^{2} \tanh{\left (c + d x \right )}}{d} + b^{3} x - \frac{b^{3} \tanh ^{5}{\left (c + d x \right )}}{5 d} - \frac{b^{3} \tanh ^{3}{\left (c + d x \right )}}{3 d} - \frac{b^{3} \tanh{\left (c + d x \right )}}{d} & \text{for}\: d \neq 0 \\x \left (a + b \tanh ^{2}{\left (c \right )}\right )^{3} & \text{otherwise} \end{cases} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a**3*x + 3*a**2*b*x - 3*a**2*b*tanh(c + d*x)/d + 3*a*b**2*x - a*b**2*tanh(c + d*x)**3/d - 3*a*b**2*
tanh(c + d*x)/d + b**3*x - b**3*tanh(c + d*x)**5/(5*d) - b**3*tanh(c + d*x)**3/(3*d) - b**3*tanh(c + d*x)/d, N
e(d, 0)), (x*(a + b*tanh(c)**2)**3, True))

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Giac [B]  time = 1.18731, size = 325, normalized size = 4.39 \begin{align*} \frac{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )}{\left (d x + c\right )}}{d} + \frac{2 \,{\left (45 \, a^{2} b e^{\left (8 \, d x + 8 \, c\right )} + 90 \, a b^{2} e^{\left (8 \, d x + 8 \, c\right )} + 45 \, b^{3} e^{\left (8 \, d x + 8 \, c\right )} + 180 \, a^{2} b e^{\left (6 \, d x + 6 \, c\right )} + 270 \, a b^{2} e^{\left (6 \, d x + 6 \, c\right )} + 90 \, b^{3} e^{\left (6 \, d x + 6 \, c\right )} + 270 \, a^{2} b e^{\left (4 \, d x + 4 \, c\right )} + 330 \, a b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 140 \, b^{3} e^{\left (4 \, d x + 4 \, c\right )} + 180 \, a^{2} b e^{\left (2 \, d x + 2 \, c\right )} + 210 \, a b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 70 \, b^{3} e^{\left (2 \, d x + 2 \, c\right )} + 45 \, a^{2} b + 60 \, a b^{2} + 23 \, b^{3}\right )}}{15 \, d{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{5}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*(d*x + c)/d + 2/15*(45*a^2*b*e^(8*d*x + 8*c) + 90*a*b^2*e^(8*d*x + 8*c) + 45*b
^3*e^(8*d*x + 8*c) + 180*a^2*b*e^(6*d*x + 6*c) + 270*a*b^2*e^(6*d*x + 6*c) + 90*b^3*e^(6*d*x + 6*c) + 270*a^2*
b*e^(4*d*x + 4*c) + 330*a*b^2*e^(4*d*x + 4*c) + 140*b^3*e^(4*d*x + 4*c) + 180*a^2*b*e^(2*d*x + 2*c) + 210*a*b^
2*e^(2*d*x + 2*c) + 70*b^3*e^(2*d*x + 2*c) + 45*a^2*b + 60*a*b^2 + 23*b^3)/(d*(e^(2*d*x + 2*c) + 1)^5)