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

Optimal. Leaf size=51 \[ \frac {(a+b)^2 \sinh (c+d x) \cosh (c+d x)}{2 d}+\frac {1}{2} x (a-3 b) (a+b)+\frac {b^2 \tanh (c+d x)}{d} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3675, 390, 385, 206} \[ \frac {(a+b)^2 \sinh (c+d x) \cosh (c+d x)}{2 d}+\frac {1}{2} x (a-3 b) (a+b)+\frac {b^2 \tanh (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a - 3*b)*(a + b)*x)/2 + ((a + b)^2*Cosh[c + d*x]*Sinh[c + d*x])/(2*d) + (b^2*Tanh[c + d*x])/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 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 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 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 \cosh ^2(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (a+b x^2\right )^2}{\left (1-x^2\right )^2} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac {\operatorname {Subst}\left (\int \left (b^2+\frac {a^2-b^2+2 b (a+b) x^2}{\left (1-x^2\right )^2}\right ) \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac {b^2 \tanh (c+d x)}{d}+\frac {\operatorname {Subst}\left (\int \frac {a^2-b^2+2 b (a+b) x^2}{\left (1-x^2\right )^2} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac {(a+b)^2 \cosh (c+d x) \sinh (c+d x)}{2 d}+\frac {b^2 \tanh (c+d x)}{d}+\frac {((a-3 b) (a+b)) \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\tanh (c+d x)\right )}{2 d}\\ &=\frac {1}{2} (a-3 b) (a+b) x+\frac {(a+b)^2 \cosh (c+d x) \sinh (c+d x)}{2 d}+\frac {b^2 \tanh (c+d x)}{d}\\ \end {align*}

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Mathematica [A]  time = 0.41, size = 54, normalized size = 1.06 \[ \frac {(a-3 b) (a+b) (c+d x)}{2 d}+\frac {(a+b)^2 \sinh (2 (c+d x))}{4 d}+\frac {b^2 \tanh (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.43, size = 105, normalized size = 2.06 \[ \frac {{\left (a^{2} + 2 \, a b + b^{2}\right )} \sinh \left (d x + c\right )^{3} + 4 \, {\left ({\left (a^{2} - 2 \, a b - 3 \, b^{2}\right )} d x - 2 \, b^{2}\right )} \cosh \left (d x + c\right ) + {\left (3 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \cosh \left (d x + c\right )^{2} + a^{2} + 2 \, a b + 9 \, b^{2}\right )} \sinh \left (d x + c\right )}{8 \, d \cosh \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*((a^2 + 2*a*b + b^2)*sinh(d*x + c)^3 + 4*((a^2 - 2*a*b - 3*b^2)*d*x - 2*b^2)*cosh(d*x + c) + (3*(a^2 + 2*a
*b + b^2)*cosh(d*x + c)^2 + a^2 + 2*a*b + 9*b^2)*sinh(d*x + c))/(d*cosh(d*x + c))

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giac [B]  time = 0.28, size = 170, normalized size = 3.33 \[ \frac {4 \, {\left (a^{2} - 2 \, a b - 3 \, b^{2}\right )} d x + {\left (a^{2} e^{\left (2 \, d x + 8 \, c\right )} + 2 \, a b e^{\left (2 \, d x + 8 \, c\right )} + b^{2} e^{\left (2 \, d x + 8 \, c\right )}\right )} e^{\left (-6 \, c\right )} - \frac {{\left (a^{2} e^{\left (4 \, d x + 4 \, c\right )} - 2 \, a b e^{\left (4 \, d x + 4 \, c\right )} - 3 \, b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 2 \, a^{2} e^{\left (2 \, d x + 2 \, c\right )} + 14 \, b^{2} e^{\left (2 \, d x + 2 \, c\right )} + a^{2} + 2 \, a b + b^{2}\right )} e^{\left (-2 \, c\right )}}{e^{\left (2 \, d x\right )} + e^{\left (4 \, d x + 2 \, c\right )}}}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(4*(a^2 - 2*a*b - 3*b^2)*d*x + (a^2*e^(2*d*x + 8*c) + 2*a*b*e^(2*d*x + 8*c) + b^2*e^(2*d*x + 8*c))*e^(-6*c
) - (a^2*e^(4*d*x + 4*c) - 2*a*b*e^(4*d*x + 4*c) - 3*b^2*e^(4*d*x + 4*c) + 2*a^2*e^(2*d*x + 2*c) + 14*b^2*e^(2
*d*x + 2*c) + a^2 + 2*a*b + b^2)*e^(-2*c)/(e^(2*d*x) + e^(4*d*x + 2*c)))/d

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maple [B]  time = 0.21, size = 96, normalized size = 1.88 \[ \frac {a^{2} \left (\frac {\cosh \left (d x +c \right ) \sinh \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+2 a b \left (\frac {\cosh \left (d x +c \right ) \sinh \left (d x +c \right )}{2}-\frac {d x}{2}-\frac {c}{2}\right )+b^{2} \left (\frac {\sinh ^{3}\left (d x +c \right )}{2 \cosh \left (d x +c \right )}-\frac {3 d x}{2}-\frac {3 c}{2}+\frac {3 \tanh \left (d x +c \right )}{2}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.34, size = 140, normalized size = 2.75 \[ \frac {1}{8} \, a^{2} {\left (4 \, x + \frac {e^{\left (2 \, d x + 2 \, c\right )}}{d} - \frac {e^{\left (-2 \, d x - 2 \, c\right )}}{d}\right )} - \frac {1}{4} \, a b {\left (4 \, x - \frac {e^{\left (2 \, d x + 2 \, c\right )}}{d} + \frac {e^{\left (-2 \, d x - 2 \, c\right )}}{d}\right )} - \frac {1}{8} \, b^{2} {\left (\frac {12 \, {\left (d x + c\right )}}{d} + \frac {e^{\left (-2 \, d x - 2 \, c\right )}}{d} - \frac {17 \, e^{\left (-2 \, d x - 2 \, c\right )} + 1}{d {\left (e^{\left (-2 \, d x - 2 \, c\right )} + e^{\left (-4 \, d x - 4 \, c\right )}\right )}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*a^2*(4*x + e^(2*d*x + 2*c)/d - e^(-2*d*x - 2*c)/d) - 1/4*a*b*(4*x - e^(2*d*x + 2*c)/d + e^(-2*d*x - 2*c)/d
) - 1/8*b^2*(12*(d*x + c)/d + e^(-2*d*x - 2*c)/d - (17*e^(-2*d*x - 2*c) + 1)/(d*(e^(-2*d*x - 2*c) + e^(-4*d*x
- 4*c))))

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mupad [B]  time = 1.29, size = 77, normalized size = 1.51 \[ \frac {{\mathrm {e}}^{2\,c+2\,d\,x}\,{\left (a+b\right )}^2}{8\,d}-\frac {2\,b^2}{d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}+1\right )}-\frac {{\mathrm {e}}^{-2\,c-2\,d\,x}\,{\left (a+b\right )}^2}{8\,d}-x\,\left (-\frac {a^2}{2}+a\,b+\frac {3\,b^2}{2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(exp(2*c + 2*d*x)*(a + b)^2)/(8*d) - (2*b^2)/(d*(exp(2*c + 2*d*x) + 1)) - (exp(- 2*c - 2*d*x)*(a + b)^2)/(8*d)
- x*(a*b - a^2/2 + (3*b^2)/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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