3.1.65 \(\int \cosh ^4(c+d x) (a+b \text {sech}^2(c+d x))^3 \, dx\) [65]

Optimal. Leaf size=84 \[ \frac {3}{8} a \left (a^2+4 a b+8 b^2\right ) x+\frac {3 a^2 (a+4 b) \cosh (c+d x) \sinh (c+d x)}{8 d}+\frac {a^3 \cosh ^3(c+d x) \sinh (c+d x)}{4 d}+\frac {b^3 \tanh (c+d x)}{d} \]

[Out]

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

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Rubi [A]
time = 0.08, antiderivative size = 84, normalized size of antiderivative = 1.00, 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 = {4231, 398, 1171, 393, 212} \begin {gather*} \frac {a^3 \sinh (c+d x) \cosh ^3(c+d x)}{4 d}+\frac {3}{8} a x \left (a^2+4 a b+8 b^2\right )+\frac {3 a^2 (a+4 b) \sinh (c+d x) \cosh (c+d x)}{8 d}+\frac {b^3 \tanh (c+d x)}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 393

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 398

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 1171

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] &&
NeQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && LtQ[q, -1]

Rule 4231

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = Fre
eFactors[Tan[e + f*x], x]}, Dist[ff/f, Subst[Int[(1 + ff^2*x^2)^(m/2 - 1)*ExpandToSum[a + b*(1 + ff^2*x^2)^(n/
2), x]^p, x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[m/2] && IntegerQ[n/2]

Rubi steps

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

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Mathematica [A]
time = 0.29, size = 70, normalized size = 0.83 \begin {gather*} \frac {12 a \left (a^2+4 a b+8 b^2\right ) (c+d x)+8 a^2 (a+3 b) \sinh (2 (c+d x))+a^3 \sinh (4 (c+d x))+32 b^3 \tanh (c+d x)}{32 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 2.14, size = 147, normalized size = 1.75

method	result	size
risch	\(\frac {3 a^{3} x}{8}+\frac {3 a^{2} b x}{2}+3 a \,b^{2} x +\frac {a^{3} {\mathrm e}^{4 d x +4 c}}{64 d}+\frac {a^{3} {\mathrm e}^{2 d x +2 c}}{8 d}+\frac {3 a^{2} {\mathrm e}^{2 d x +2 c} b}{8 d}-\frac {a^{3} {\mathrm e}^{-2 d x -2 c}}{8 d}-\frac {3 a^{2} {\mathrm e}^{-2 d x -2 c} b}{8 d}-\frac {a^{3} {\mathrm e}^{-4 d x -4 c}}{64 d}-\frac {2 b^{3}}{d \left (1+{\mathrm e}^{2 d x +2 c}\right )}\)	\(147\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(d*x+c)^4*(a+b*sech(d*x+c)^2)^3,x,method=_RETURNVERBOSE)

[Out]

3/8*a^3*x+3/2*a^2*b*x+3*a*b^2*x+1/64*a^3/d*exp(4*d*x+4*c)+1/8*a^3/d*exp(2*d*x+2*c)+3/8*a^2/d*exp(2*d*x+2*c)*b-
1/8*a^3/d*exp(-2*d*x-2*c)-3/8*a^2/d*exp(-2*d*x-2*c)*b-1/64*a^3/d*exp(-4*d*x-4*c)-2*b^3/d/(1+exp(2*d*x+2*c))

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Maxima [A]
time = 0.28, size = 130, normalized size = 1.55 \begin {gather*} \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}{8} \, a^{2} 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 )} + 3 \, a b^{2} x + \frac {2 \, b^{3}}{d {\left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(d*x+c)^4*(a+b*sech(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) + 3/8*a^
2*b*(4*x + e^(2*d*x + 2*c)/d - e^(-2*d*x - 2*c)/d) + 3*a*b^2*x + 2*b^3/(d*(e^(-2*d*x - 2*c) + 1))

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Fricas [A]
time = 0.39, size = 153, normalized size = 1.82 \begin {gather*} \frac {a^{3} \sinh \left (d x + c\right )^{5} + {\left (10 \, a^{3} \cosh \left (d x + c\right )^{2} + 9 \, a^{3} + 24 \, a^{2} b\right )} \sinh \left (d x + c\right )^{3} - 8 \, {\left (8 \, b^{3} - 3 \, {\left (a^{3} + 4 \, a^{2} b + 8 \, a b^{2}\right )} d x\right )} \cosh \left (d x + c\right ) + {\left (5 \, a^{3} \cosh \left (d x + c\right )^{4} + 8 \, a^{3} + 24 \, a^{2} b + 64 \, b^{3} + 9 \, {\left (3 \, a^{3} + 8 \, a^{2} b\right )} \cosh \left (d x + c\right )^{2}\right )} \sinh \left (d x + c\right )}{64 \, d \cosh \left (d x + c\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: SystemError >> excessive stack use: stack is 4370 deep

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 177 vs. \(2 (78) = 156\).
time = 0.43, size = 177, normalized size = 2.11 \begin {gather*} \frac {a^{3} e^{\left (4 \, d x + 4 \, c\right )} + 8 \, a^{3} e^{\left (2 \, d x + 2 \, c\right )} + 24 \, a^{2} b e^{\left (2 \, d x + 2 \, c\right )} + 24 \, {\left (a^{3} + 4 \, a^{2} b + 8 \, a b^{2}\right )} {\left (d x + c\right )} - \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 )} + 72 \, a^{2} b e^{\left (4 \, d x + 4 \, c\right )} + 144 \, a b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 8 \, a^{3} e^{\left (2 \, d x + 2 \, c\right )} + 24 \, a^{2} b e^{\left (2 \, d x + 2 \, c\right )} + a^{3}\right )} e^{\left (-4 \, d x - 4 \, c\right )}}{64 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/64*(a^3*e^(4*d*x + 4*c) + 8*a^3*e^(2*d*x + 2*c) + 24*a^2*b*e^(2*d*x + 2*c) + 24*(a^3 + 4*a^2*b + 8*a*b^2)*(d
*x + c) - 128*b^3/(e^(2*d*x + 2*c) + 1) - (18*a^3*e^(4*d*x + 4*c) + 72*a^2*b*e^(4*d*x + 4*c) + 144*a*b^2*e^(4*
d*x + 4*c) + 8*a^3*e^(2*d*x + 2*c) + 24*a^2*b*e^(2*d*x + 2*c) + a^3)*e^(-4*d*x - 4*c))/d

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Mupad [B]
time = 1.53, size = 117, normalized size = 1.39 \begin {gather*} \frac {3\,a\,x\,\left (a^2+4\,a\,b+8\,b^2\right )}{8}-\frac {2\,b^3}{d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}+1\right )}-\frac {a^3\,{\mathrm {e}}^{-4\,c-4\,d\,x}}{64\,d}+\frac {a^3\,{\mathrm {e}}^{4\,c+4\,d\,x}}{64\,d}-\frac {a^2\,{\mathrm {e}}^{-2\,c-2\,d\,x}\,\left (a+3\,b\right )}{8\,d}+\frac {a^2\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (a+3\,b\right )}{8\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(3*a*x*(4*a*b + a^2 + 8*b^2))/8 - (2*b^3)/(d*(exp(2*c + 2*d*x) + 1)) - (a^3*exp(- 4*c - 4*d*x))/(64*d) + (a^3*
exp(4*c + 4*d*x))/(64*d) - (a^2*exp(- 2*c - 2*d*x)*(a + 3*b))/(8*d) + (a^2*exp(2*c + 2*d*x)*(a + 3*b))/(8*d)

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