3.1.97 \(\int \cosh ^4(c+d x) (a+b \tanh ^2(c+d x))^3 \, dx\) [97]

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

[Out]

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

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Rubi [A]
time = 0.09, antiderivative size = 91, 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 = {3756, 398, 1171, 393, 212} \begin {gather*} \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} \end {gather*}

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 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 3756

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 ^4(c+d x) \left (a+b \tanh ^2(c+d x)\right )^3 \, dx &=\frac {\text {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 {\text {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 {\text {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 {\text {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 ) \text {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*}

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Mathematica [A]
time = 0.50, size = 81, normalized size = 0.89 \begin {gather*} \frac {12 \left (a^3-a^2 b+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} \end {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. \(292\) vs. \(2(85)=170\).
time = 1.86, size = 293, normalized size = 3.22

method	result	size
risch	\(\frac {3 a^{3} x}{8}-\frac {3 a^{2} b x}{8}+\frac {9 a \,b^{2} x}{8}+\frac {15 b^{3} x}{8}+\frac {{\mathrm e}^{4 d x +4 c} a^{3}}{64 d}+\frac {3 \,{\mathrm e}^{4 d x +4 c} a^{2} b}{64 d}+\frac {3 b^{2} {\mathrm e}^{4 d x +4 c} a}{64 d}+\frac {b^{3} {\mathrm e}^{4 d x +4 c}}{64 d}+\frac {{\mathrm e}^{2 d x +2 c} a^{3}}{8 d}-\frac {3 \,{\mathrm e}^{2 d x +2 c} a \,b^{2}}{8 d}-\frac {b^{3} {\mathrm e}^{2 d x +2 c}}{4 d}-\frac {{\mathrm e}^{-2 d x -2 c} a^{3}}{8 d}+\frac {3 \,{\mathrm e}^{-2 d x -2 c} a \,b^{2}}{8 d}+\frac {b^{3} {\mathrm e}^{-2 d x -2 c}}{4 d}-\frac {{\mathrm e}^{-4 d x -4 c} a^{3}}{64 d}-\frac {3 \,{\mathrm e}^{-4 d x -4 c} a^{2} b}{64 d}-\frac {3 b^{2} {\mathrm e}^{-4 d x -4 c} a}{64 d}-\frac {b^{3} {\mathrm e}^{-4 d x -4 c}}{64 d}+\frac {2 b^{3}}{d \left (1+{\mathrm e}^{2 d x +2 c}\right )}\)	\(293\)

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,method=_RETURNVERBOSE)

[Out]

3/8*a^3*x-3/8*a^2*b*x+9/8*a*b^2*x+15/8*b^3*x+1/64/d*exp(4*d*x+4*c)*a^3+3/64/d*exp(4*d*x+4*c)*a^2*b+3/64*b^2/d*
exp(4*d*x+4*c)*a+1/64*b^3/d*exp(4*d*x+4*c)+1/8/d*exp(2*d*x+2*c)*a^3-3/8/d*exp(2*d*x+2*c)*a*b^2-1/4*b^3/d*exp(2
*d*x+2*c)-1/8/d*exp(-2*d*x-2*c)*a^3+3/8/d*exp(-2*d*x-2*c)*a*b^2+1/4*b^3/d*exp(-2*d*x-2*c)-1/64/d*exp(-4*d*x-4*
c)*a^3-3/64/d*exp(-4*d*x-4*c)*a^2*b-3/64*b^2/d*exp(-4*d*x-4*c)*a-1/64*b^3/d*exp(-4*d*x-4*c)+2*b^3/d/(1+exp(2*d
*x+2*c))

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 267 vs. \(2 (85) = 170\).
time = 0.27, size = 267, normalized size = 2.93 \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}{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 {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. 227 vs. \(2 (85) = 170\).
time = 0.37, size = 227, normalized size = 2.49 \begin {gather*} \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 {gather*}

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)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. 282 vs. \(2 (85) = 170\).
time = 0.62, size = 282, normalized size = 3.10 \begin {gather*} \frac {a^{3} e^{\left (4 \, d x + 4 \, c\right )} + 3 \, a^{2} b e^{\left (4 \, d x + 4 \, c\right )} + 3 \, a b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 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 )} + 24 \, {\left (a^{3} - a^{2} b + 3 \, a b^{2} + 5 \, b^{3}\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 )} - 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 )}}{64 \, d} \end {gather*}

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*(a^3*e^(4*d*x + 4*c) + 3*a^2*b*e^(4*d*x + 4*c) + 3*a*b^2*e^(4*d*x + 4*c) + 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) + 24*(a^3 - a^2*b + 3*a*b^2 + 5*b^3)*(d*x +
c) + 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))/d

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Mupad [B]
time = 1.41, size = 133, normalized size = 1.46 \begin {gather*} x\,\left (\frac {3\,a^3}{8}-\frac {3\,a^2\,b}{8}+\frac {9\,a\,b^2}{8}+\frac {15\,b^3}{8}\right )+\frac {2\,b^3}{d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}+1\right )}-\frac {{\mathrm {e}}^{-4\,c-4\,d\,x}\,{\left (a+b\right )}^3}{64\,d}+\frac {{\mathrm {e}}^{4\,c+4\,d\,x}\,{\left (a+b\right )}^3}{64\,d}-\frac {{\mathrm {e}}^{-2\,c-2\,d\,x}\,{\left (a+b\right )}^2\,\left (a-2\,b\right )}{8\,d}+\frac {{\mathrm {e}}^{2\,c+2\,d\,x}\,{\left (a+b\right )}^2\,\left (a-2\,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*tanh(c + d*x)^2)^3,x)

[Out]

x*((9*a*b^2)/8 - (3*a^2*b)/8 + (3*a^3)/8 + (15*b^3)/8) + (2*b^3)/(d*(exp(2*c + 2*d*x) + 1)) - (exp(- 4*c - 4*d
*x)*(a + b)^3)/(64*d) + (exp(4*c + 4*d*x)*(a + b)^3)/(64*d) - (exp(- 2*c - 2*d*x)*(a + b)^2*(a - 2*b))/(8*d) +
(exp(2*c + 2*d*x)*(a + b)^2*(a - 2*b))/(8*d)

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