### 3.97 $$\int \cosh ^4(c+d x) (a+b \tanh ^2(c+d x))^3 \, dx$$

Optimal. Leaf size=91 $\frac{3}{8} x (a+b) \left (a^2-2 a b+5 b^2\right )+\frac{(a+b)^3 \sinh (c+d x) \cosh ^3(c+d x)}{4 d}+\frac{3 (a-3 b) (a+b)^2 \sinh (c+d x) \cosh (c+d x)}{8 d}-\frac{b^3 \tanh (c+d x)}{d}$

[Out]

(3*(a + b)*(a^2 - 2*a*b + 5*b^2)*x)/8 + (3*(a - 3*b)*(a + b)^2*Cosh[c + d*x]*Sinh[c + d*x])/(8*d) + ((a + b)^3
*Cosh[c + d*x]^3*Sinh[c + d*x])/(4*d) - (b^3*Tanh[c + d*x])/d

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Rubi [A]  time = 0.130354, antiderivative size = 91, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 23, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.217, Rules used = {3675, 390, 1157, 385, 206} $\frac{3}{8} x (a+b) \left (a^2-2 a b+5 b^2\right )+\frac{(a+b)^3 \sinh (c+d x) \cosh ^3(c+d x)}{4 d}+\frac{3 (a-3 b) (a+b)^2 \sinh (c+d x) \cosh (c+d x)}{8 d}-\frac{b^3 \tanh (c+d x)}{d}$

Antiderivative was successfully veriﬁed.

[In]

Int[Cosh[c + d*x]^4*(a + b*Tanh[c + d*x]^2)^3,x]

[Out]

(3*(a + b)*(a^2 - 2*a*b + 5*b^2)*x)/8 + (3*(a - 3*b)*(a + b)^2*Cosh[c + d*x]*Sinh[c + d*x])/(8*d) + ((a + b)^3
*Cosh[c + d*x]^3*Sinh[c + d*x])/(4*d) - (b^3*Tanh[c + d*x])/d

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])

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 1157

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> With[{Qx = PolynomialQ
uotient[(a + b*x^2 + c*x^4)^p, d + e*x^2, x], R = Coeff[PolynomialRemainder[(a + b*x^2 + c*x^4)^p, d + e*x^2,
x], x, 0]}, -Simp[(R*x*(d + e*x^2)^(q + 1))/(2*d*(q + 1)), x] + Dist[1/(2*d*(q + 1)), Int[(d + e*x^2)^(q + 1)*
ExpandToSum[2*d*(q + 1)*Qx + R*(2*q + 3), x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && N
eQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && LtQ[q, -1]

Rule 385

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> -Simp[((b*c - a*d)*x*(a + b*x^n)^(p +
1))/(a*b*n*(p + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /
; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/n + p, 0])

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 \cosh ^4(c+d x) \left (a+b \tanh ^2(c+d x)\right )^3 \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (a+b x^2\right )^3}{\left (1-x^2\right )^3} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \left (-b^3+\frac{a^3+b^3+3 b \left (a^2-b^2\right ) x^2+3 b^2 (a+b) x^4}{\left (1-x^2\right )^3}\right ) \, dx,x,\tanh (c+d x)\right )}{d}\\ &=-\frac{b^3 \tanh (c+d x)}{d}+\frac{\operatorname{Subst}\left (\int \frac{a^3+b^3+3 b \left (a^2-b^2\right ) x^2+3 b^2 (a+b) x^4}{\left (1-x^2\right )^3} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac{(a+b)^3 \cosh ^3(c+d x) \sinh (c+d x)}{4 d}-\frac{b^3 \tanh (c+d x)}{d}-\frac{\operatorname{Subst}\left (\int \frac{-3 (a-b)^2 (a+b)+12 b^2 (a+b) x^2}{\left (1-x^2\right )^2} \, dx,x,\tanh (c+d x)\right )}{4 d}\\ &=\frac{3 (a-3 b) (a+b)^2 \cosh (c+d x) \sinh (c+d x)}{8 d}+\frac{(a+b)^3 \cosh ^3(c+d x) \sinh (c+d x)}{4 d}-\frac{b^3 \tanh (c+d x)}{d}+\frac{\left (3 (a+b) \left (a^2-2 a b+5 b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\tanh (c+d x)\right )}{8 d}\\ &=\frac{3}{8} (a+b) \left (a^2-2 a b+5 b^2\right ) x+\frac{3 (a-3 b) (a+b)^2 \cosh (c+d x) \sinh (c+d x)}{8 d}+\frac{(a+b)^3 \cosh ^3(c+d x) \sinh (c+d x)}{4 d}-\frac{b^3 \tanh (c+d x)}{d}\\ \end{align*}

Mathematica [A]  time = 0.668448, size = 81, normalized size = 0.89 $\frac{12 \left (-a^2 b+a^3+3 a b^2+5 b^3\right ) (c+d x)+8 (a-2 b) (a+b)^2 \sinh (2 (c+d x))+(a+b)^3 \sinh (4 (c+d x))-32 b^3 \tanh (c+d x)}{32 d}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[Cosh[c + d*x]^4*(a + b*Tanh[c + d*x]^2)^3,x]

[Out]

(12*(a^3 - a^2*b + 3*a*b^2 + 5*b^3)*(c + d*x) + 8*(a - 2*b)*(a + b)^2*Sinh[2*(c + d*x)] + (a + b)^3*Sinh[4*(c
+ d*x)] - 32*b^3*Tanh[c + d*x])/(32*d)

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Maple [B]  time = 0.042, size = 184, normalized size = 2. \begin{align*}{\frac{1}{d} \left ({a}^{3} \left ( \left ({\frac{ \left ( \cosh \left ( dx+c \right ) \right ) ^{3}}{4}}+{\frac{3\,\cosh \left ( dx+c \right ) }{8}} \right ) \sinh \left ( dx+c \right ) +{\frac{3\,dx}{8}}+{\frac{3\,c}{8}} \right ) +3\,{a}^{2}b \left ( 1/4\,\sinh \left ( dx+c \right ) \left ( \cosh \left ( dx+c \right ) \right ) ^{3}-1/8\,\cosh \left ( dx+c \right ) \sinh \left ( dx+c \right ) -1/8\,dx-c/8 \right ) +3\,a{b}^{2} \left ( \left ( 1/4\, \left ( \sinh \left ( dx+c \right ) \right ) ^{3}-3/8\,\sinh \left ( dx+c \right ) \right ) \cosh \left ( dx+c \right ) +3/8\,dx+3/8\,c \right ) +{b}^{3} \left ({\frac{ \left ( \sinh \left ( dx+c \right ) \right ) ^{5}}{4\,\cosh \left ( dx+c \right ) }}-{\frac{5\, \left ( \sinh \left ( dx+c \right ) \right ) ^{3}}{8\,\cosh \left ( dx+c \right ) }}+{\frac{15\,dx}{8}}+{\frac{15\,c}{8}}-{\frac{15\,\tanh \left ( dx+c \right ) }{8}} \right ) \right ) } \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(d*x+c)^4*(a+b*tanh(d*x+c)^2)^3,x)

[Out]

1/d*(a^3*((1/4*cosh(d*x+c)^3+3/8*cosh(d*x+c))*sinh(d*x+c)+3/8*d*x+3/8*c)+3*a^2*b*(1/4*sinh(d*x+c)*cosh(d*x+c)^
3-1/8*cosh(d*x+c)*sinh(d*x+c)-1/8*d*x-1/8*c)+3*a*b^2*((1/4*sinh(d*x+c)^3-3/8*sinh(d*x+c))*cosh(d*x+c)+3/8*d*x+
3/8*c)+b^3*(1/4*sinh(d*x+c)^5/cosh(d*x+c)-5/8*sinh(d*x+c)^3/cosh(d*x+c)+15/8*d*x+15/8*c-15/8*tanh(d*x+c)))

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Maxima [B]  time = 1.16712, size = 360, normalized size = 3.96 \begin{align*} \frac{1}{64} \, a^{3}{\left (24 \, x + \frac{e^{\left (4 \, d x + 4 \, c\right )}}{d} + \frac{8 \, e^{\left (2 \, d x + 2 \, c\right )}}{d} - \frac{8 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d} - \frac{e^{\left (-4 \, d x - 4 \, c\right )}}{d}\right )} + \frac{3}{64} \, a b^{2}{\left (24 \, x + \frac{e^{\left (4 \, d x + 4 \, c\right )}}{d} - \frac{8 \, e^{\left (2 \, d x + 2 \, c\right )}}{d} + \frac{8 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d} - \frac{e^{\left (-4 \, d x - 4 \, c\right )}}{d}\right )} + \frac{1}{64} \, b^{3}{\left (\frac{120 \,{\left (d x + c\right )}}{d} + \frac{16 \, e^{\left (-2 \, d x - 2 \, c\right )} - e^{\left (-4 \, d x - 4 \, c\right )}}{d} - \frac{15 \, e^{\left (-2 \, d x - 2 \, c\right )} + 144 \, e^{\left (-4 \, d x - 4 \, c\right )} - 1}{d{\left (e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )}\right )}}\right )} - \frac{3}{64} \, a^{2} b{\left (\frac{8 \,{\left (d x + c\right )}}{d} - \frac{e^{\left (4 \, d x + 4 \, c\right )}}{d} + \frac{e^{\left (-4 \, d x - 4 \, c\right )}}{d}\right )} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(d*x+c)^4*(a+b*tanh(d*x+c)^2)^3,x, algorithm="maxima")

[Out]

1/64*a^3*(24*x + e^(4*d*x + 4*c)/d + 8*e^(2*d*x + 2*c)/d - 8*e^(-2*d*x - 2*c)/d - e^(-4*d*x - 4*c)/d) + 3/64*a
*b^2*(24*x + e^(4*d*x + 4*c)/d - 8*e^(2*d*x + 2*c)/d + 8*e^(-2*d*x - 2*c)/d - e^(-4*d*x - 4*c)/d) + 1/64*b^3*(
120*(d*x + c)/d + (16*e^(-2*d*x - 2*c) - e^(-4*d*x - 4*c))/d - (15*e^(-2*d*x - 2*c) + 144*e^(-4*d*x - 4*c) - 1
)/(d*(e^(-4*d*x - 4*c) + e^(-6*d*x - 6*c)))) - 3/64*a^2*b*(8*(d*x + c)/d - e^(4*d*x + 4*c)/d + e^(-4*d*x - 4*c
)/d)

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Fricas [B]  time = 2.0398, size = 539, normalized size = 5.92 \begin{align*} \frac{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \sinh \left (d x + c\right )^{5} +{\left (9 \, a^{3} + 3 \, a^{2} b - 21 \, a b^{2} - 15 \, b^{3} + 10 \,{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \cosh \left (d x + c\right )^{2}\right )} \sinh \left (d x + c\right )^{3} + 8 \,{\left (8 \, b^{3} + 3 \,{\left (a^{3} - a^{2} b + 3 \, a b^{2} + 5 \, b^{3}\right )} d x\right )} \cosh \left (d x + c\right ) +{\left (5 \,{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \cosh \left (d x + c\right )^{4} + 8 \, a^{3} - 24 \, a b^{2} - 80 \, b^{3} + 9 \,{\left (3 \, a^{3} + a^{2} b - 7 \, a b^{2} - 5 \, b^{3}\right )} \cosh \left (d x + c\right )^{2}\right )} \sinh \left (d x + c\right )}{64 \, d \cosh \left (d x + c\right )} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(d*x+c)^4*(a+b*tanh(d*x+c)^2)^3,x, algorithm="fricas")

[Out]

1/64*((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*sinh(d*x + c)^5 + (9*a^3 + 3*a^2*b - 21*a*b^2 - 15*b^3 + 10*(a^3 + 3*a^2
*b + 3*a*b^2 + b^3)*cosh(d*x + c)^2)*sinh(d*x + c)^3 + 8*(8*b^3 + 3*(a^3 - a^2*b + 3*a*b^2 + 5*b^3)*d*x)*cosh(
d*x + c) + (5*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cosh(d*x + c)^4 + 8*a^3 - 24*a*b^2 - 80*b^3 + 9*(3*a^3 + a^2*b -
7*a*b^2 - 5*b^3)*cosh(d*x + c)^2)*sinh(d*x + c))/(d*cosh(d*x + c))

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(d*x+c)**4*(a+b*tanh(d*x+c)**2)**3,x)

[Out]

Timed out

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Giac [B]  time = 2.67663, size = 385, normalized size = 4.23 \begin{align*} \frac{24 \,{\left (a^{3} - a^{2} b + 3 \, a b^{2} + 5 \, b^{3}\right )} d x + \frac{128 \, b^{3}}{e^{\left (2 \, d x + 2 \, c\right )} + 1} -{\left (18 \, a^{3} e^{\left (4 \, d x + 4 \, c\right )} - 18 \, a^{2} b e^{\left (4 \, d x + 4 \, c\right )} + 54 \, a b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 90 \, b^{3} e^{\left (4 \, d x + 4 \, c\right )} + 8 \, a^{3} e^{\left (2 \, d x + 2 \, c\right )} - 24 \, a b^{2} e^{\left (2 \, d x + 2 \, c\right )} - 16 \, b^{3} e^{\left (2 \, d x + 2 \, c\right )} + a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} e^{\left (-4 \, d x - 4 \, c\right )} +{\left (a^{3} e^{\left (4 \, d x + 20 \, c\right )} + 3 \, a^{2} b e^{\left (4 \, d x + 20 \, c\right )} + 3 \, a b^{2} e^{\left (4 \, d x + 20 \, c\right )} + b^{3} e^{\left (4 \, d x + 20 \, c\right )} + 8 \, a^{3} e^{\left (2 \, d x + 18 \, c\right )} - 24 \, a b^{2} e^{\left (2 \, d x + 18 \, c\right )} - 16 \, b^{3} e^{\left (2 \, d x + 18 \, c\right )}\right )} e^{\left (-16 \, c\right )}}{64 \, d} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(d*x+c)^4*(a+b*tanh(d*x+c)^2)^3,x, algorithm="giac")

[Out]

1/64*(24*(a^3 - a^2*b + 3*a*b^2 + 5*b^3)*d*x + 128*b^3/(e^(2*d*x + 2*c) + 1) - (18*a^3*e^(4*d*x + 4*c) - 18*a^
2*b*e^(4*d*x + 4*c) + 54*a*b^2*e^(4*d*x + 4*c) + 90*b^3*e^(4*d*x + 4*c) + 8*a^3*e^(2*d*x + 2*c) - 24*a*b^2*e^(
2*d*x + 2*c) - 16*b^3*e^(2*d*x + 2*c) + a^3 + 3*a^2*b + 3*a*b^2 + b^3)*e^(-4*d*x - 4*c) + (a^3*e^(4*d*x + 20*c
) + 3*a^2*b*e^(4*d*x + 20*c) + 3*a*b^2*e^(4*d*x + 20*c) + b^3*e^(4*d*x + 20*c) + 8*a^3*e^(2*d*x + 18*c) - 24*a
*b^2*e^(2*d*x + 18*c) - 16*b^3*e^(2*d*x + 18*c))*e^(-16*c))/d