3.4.9 \(\int \text {sech}^2(c+d x) (a+b \sinh ^2(c+d x))^3 \, dx\) [309]

Optimal. Leaf size=92 \[ \frac {3}{8} b \left (8 a^2-12 a b+5 b^2\right ) x+\frac {3 (4 a-3 b) b^2 \cosh (c+d x) \sinh (c+d x)}{8 d}+\frac {b^3 \cosh ^3(c+d x) \sinh (c+d x)}{4 d}+\frac {(a-b)^3 \tanh (c+d x)}{d} \]

[Out]

3/8*b*(8*a^2-12*a*b+5*b^2)*x+3/8*(4*a-3*b)*b^2*cosh(d*x+c)*sinh(d*x+c)/d+1/4*b^3*cosh(d*x+c)^3*sinh(d*x+c)/d+(
a-b)^3*tanh(d*x+c)/d

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Rubi [A]
time = 0.09, antiderivative size = 92, 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 = {3270, 398, 1171, 393, 212} \begin {gather*} \frac {3}{8} b x \left (8 a^2-12 a b+5 b^2\right )+\frac {3 b^2 (4 a-3 b) \sinh (c+d x) \cosh (c+d x)}{8 d}+\frac {(a-b)^3 \tanh (c+d x)}{d}+\frac {b^3 \sinh (c+d x) \cosh ^3(c+d x)}{4 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*b*(8*a^2 - 12*a*b + 5*b^2)*x)/8 + (3*(4*a - 3*b)*b^2*Cosh[c + d*x]*Sinh[c + d*x])/(8*d) + (b^3*Cosh[c + d*x
]^3*Sinh[c + d*x])/(4*d) + ((a - 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 3270

Int[cos[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = FreeF
actors[Tan[e + f*x], x]}, Dist[ff/f, Subst[Int[(a + (a + b)*ff^2*x^2)^p/(1 + ff^2*x^2)^(m/2 + p + 1), x], x, T
an[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[m/2] && IntegerQ[p]

Rubi steps

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

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Mathematica [A]
time = 0.40, size = 78, normalized size = 0.85 \begin {gather*} \frac {12 b \left (8 a^2-12 a b+5 b^2\right ) (c+d x)+8 (3 a-2 b) b^2 \sinh (2 (c+d x))+b^3 \sinh (4 (c+d x))+32 (a-b)^3 \tanh (c+d x)}{32 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(12*b*(8*a^2 - 12*a*b + 5*b^2)*(c + d*x) + 8*(3*a - 2*b)*b^2*Sinh[2*(c + d*x)] + b^3*Sinh[4*(c + d*x)] + 32*(a
- 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. \(211\) vs. \(2(86)=172\).
time = 1.63, size = 212, normalized size = 2.30

method	result	size
risch	\(3 a^{2} b x -\frac {9 a \,b^{2} x}{2}+\frac {15 b^{3} x}{8}+\frac {b^{3} {\mathrm e}^{4 d x +4 c}}{64 d}+\frac {3 \,{\mathrm e}^{2 d x +2 c} a \,b^{2}}{8 d}-\frac {b^{3} {\mathrm e}^{2 d x +2 c}}{4 d}-\frac {3 \,{\mathrm e}^{-2 d x -2 c} a \,b^{2}}{8 d}+\frac {b^{3} {\mathrm e}^{-2 d x -2 c}}{4 d}-\frac {b^{3} {\mathrm e}^{-4 d x -4 c}}{64 d}-\frac {2 a^{3}}{d \left (1+{\mathrm e}^{2 d x +2 c}\right )}+\frac {6 a^{2} b}{d \left (1+{\mathrm e}^{2 d x +2 c}\right )}-\frac {6 a \,b^{2}}{d \left (1+{\mathrm e}^{2 d x +2 c}\right )}+\frac {2 b^{3}}{d \left (1+{\mathrm e}^{2 d x +2 c}\right )}\)	\(212\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*a^2*b*x-9/2*a*b^2*x+15/8*b^3*x+1/64*b^3/d*exp(4*d*x+4*c)+3/8/d*exp(2*d*x+2*c)*a*b^2-1/4*b^3/d*exp(2*d*x+2*c)
-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*b^3/d*exp(-4*d*x-4*c)-2/d/(1+exp(2*d*x+2*c))*a^3+6
/d/(1+exp(2*d*x+2*c))*a^2*b-6/d/(1+exp(2*d*x+2*c))*a*b^2+2/d/(1+exp(2*d*x+2*c))*b^3

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 215 vs. \(2 (86) = 172\).
time = 0.27, size = 215, normalized size = 2.34 \begin {gather*} 3 \, a^{2} b {\left (x + \frac {c}{d} - \frac {2}{d {\left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}}\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}{8} \, a 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 )} + \frac {2 \, a^{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(sech(d*x+c)^2*(a+b*sinh(d*x+c)^2)^3,x, algorithm="maxima")

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 178 vs. \(2 (86) = 172\).
time = 0.38, size = 178, normalized size = 1.93 \begin {gather*} \frac {b^{3} \sinh \left (d x + c\right )^{5} + {\left (10 \, b^{3} \cosh \left (d x + c\right )^{2} + 24 \, a b^{2} - 15 \, b^{3}\right )} \sinh \left (d x + c\right )^{3} - 8 \, {\left (8 \, a^{3} - 24 \, a^{2} b + 24 \, a b^{2} - 8 \, b^{3} - 3 \, {\left (8 \, a^{2} b - 12 \, a b^{2} + 5 \, b^{3}\right )} d x\right )} \cosh \left (d x + c\right ) + {\left (5 \, b^{3} \cosh \left (d x + c\right )^{4} + 64 \, a^{3} - 192 \, a^{2} b + 216 \, a b^{2} - 80 \, b^{3} + 9 \, {\left (8 \, 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(sech(d*x+c)^2*(a+b*sinh(d*x+c)^2)^3,x, algorithm="fricas")

[Out]

1/64*(b^3*sinh(d*x + c)^5 + (10*b^3*cosh(d*x + c)^2 + 24*a*b^2 - 15*b^3)*sinh(d*x + c)^3 - 8*(8*a^3 - 24*a^2*b
+ 24*a*b^2 - 8*b^3 - 3*(8*a^2*b - 12*a*b^2 + 5*b^3)*d*x)*cosh(d*x + c) + (5*b^3*cosh(d*x + c)^4 + 64*a^3 - 19
2*a^2*b + 216*a*b^2 - 80*b^3 + 9*(8*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(sech(d*x+c)**2*(a+b*sinh(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. 197 vs. \(2 (86) = 172\).
time = 0.46, size = 197, normalized size = 2.14 \begin {gather*} \frac {b^{3} e^{\left (4 \, d x + 4 \, 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 (8 \, a^{2} b - 12 \, a b^{2} + 5 \, b^{3}\right )} {\left (d x + c\right )} - {\left (144 \, a^{2} b e^{\left (4 \, d x + 4 \, c\right )} - 216 \, a b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 90 \, b^{3} e^{\left (4 \, d x + 4 \, 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 )} + b^{3}\right )} e^{\left (-4 \, d x - 4 \, c\right )} - \frac {128 \, {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )}}{e^{\left (2 \, d x + 2 \, c\right )} + 1}}{64 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/64*(b^3*e^(4*d*x + 4*c) + 24*a*b^2*e^(2*d*x + 2*c) - 16*b^3*e^(2*d*x + 2*c) + 24*(8*a^2*b - 12*a*b^2 + 5*b^3
)*(d*x + c) - (144*a^2*b*e^(4*d*x + 4*c) - 216*a*b^2*e^(4*d*x + 4*c) + 90*b^3*e^(4*d*x + 4*c) + 24*a*b^2*e^(2*
d*x + 2*c) - 16*b^3*e^(2*d*x + 2*c) + b^3)*e^(-4*d*x - 4*c) - 128*(a^3 - 3*a^2*b + 3*a*b^2 - b^3)/(e^(2*d*x +
2*c) + 1))/d

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Mupad [B]
time = 0.98, size = 141, normalized size = 1.53 \begin {gather*} \frac {b^3\,{\mathrm {e}}^{4\,c+4\,d\,x}}{64\,d}-\frac {b^3\,{\mathrm {e}}^{-4\,c-4\,d\,x}}{64\,d}-\frac {2\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )}{d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}+1\right )}+\frac {3\,b\,x\,\left (8\,a^2-12\,a\,b+5\,b^2\right )}{8}-\frac {b^2\,{\mathrm {e}}^{-2\,c-2\,d\,x}\,\left (3\,a-2\,b\right )}{8\,d}+\frac {b^2\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (3\,a-2\,b\right )}{8\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b^3*exp(4*c + 4*d*x))/(64*d) - (b^3*exp(- 4*c - 4*d*x))/(64*d) - (2*(3*a*b^2 - 3*a^2*b + a^3 - b^3))/(d*(exp(
2*c + 2*d*x) + 1)) + (3*b*x*(8*a^2 - 12*a*b + 5*b^2))/8 - (b^2*exp(- 2*c - 2*d*x)*(3*a - 2*b))/(8*d) + (b^2*ex
p(2*c + 2*d*x)*(3*a - 2*b))/(8*d)

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