3.3.98 \(\int \text {sech}^2(c+d x) (a+b \sinh ^2(c+d x))^2 \, dx\) [298]

Optimal. Leaf size=53 \[ \frac {1}{2} (4 a-3 b) b x+\frac {b^2 \cosh (c+d x) \sinh (c+d x)}{2 d}+\frac {(a-b)^2 \tanh (c+d x)}{d} \]

[Out]

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

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Rubi [A]
time = 0.06, antiderivative size = 53, 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 = {3270, 398, 393, 212} \begin {gather*} \frac {(a-b)^2 \tanh (c+d x)}{d}+\frac {1}{2} b x (4 a-3 b)+\frac {b^2 \sinh (c+d x) \cosh (c+d x)}{2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((4*a - 3*b)*b*x)/2 + (b^2*Cosh[c + d*x]*Sinh[c + d*x])/(2*d) + ((a - b)^2*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 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 )^2 \, dx &=\frac {\text {Subst}\left (\int \frac {\left (a-(a-b) x^2\right )^2}{\left (1-x^2\right )^2} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac {\text {Subst}\left (\int \left ((a-b)^2+\frac {(2 a-b) b-2 (a-b) b x^2}{\left (1-x^2\right )^2}\right ) \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac {(a-b)^2 \tanh (c+d x)}{d}+\frac {\text {Subst}\left (\int \frac {(2 a-b) b-2 (a-b) b x^2}{\left (1-x^2\right )^2} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac {b^2 \cosh (c+d x) \sinh (c+d x)}{2 d}+\frac {(a-b)^2 \tanh (c+d x)}{d}+\frac {((4 a-3 b) b) \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\tanh (c+d x)\right )}{2 d}\\ &=\frac {1}{2} (4 a-3 b) b x+\frac {b^2 \cosh (c+d x) \sinh (c+d x)}{2 d}+\frac {(a-b)^2 \tanh (c+d x)}{d}\\ \end {align*}

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Mathematica [A]
time = 0.23, size = 50, normalized size = 0.94 \begin {gather*} \frac {2 (4 a-3 b) b (c+d x)+b^2 \sinh (2 (c+d x))+4 (a-b)^2 \tanh (c+d x)}{4 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(108\) vs. \(2(49)=98\).
time = 1.59, size = 109, normalized size = 2.06

method	result	size
risch	\(2 a b x -\frac {3 b^{2} x}{2}+\frac {{\mathrm e}^{2 d x +2 c} b^{2}}{8 d}-\frac {{\mathrm e}^{-2 d x -2 c} b^{2}}{8 d}-\frac {2 a^{2}}{d \left (1+{\mathrm e}^{2 d x +2 c}\right )}+\frac {4 a b}{d \left (1+{\mathrm e}^{2 d x +2 c}\right )}-\frac {2 b^{2}}{d \left (1+{\mathrm e}^{2 d x +2 c}\right )}\)	\(109\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 119 vs. \(2 (49) = 98\).
time = 0.27, size = 119, normalized size = 2.25 \begin {gather*} 2 \, a b {\left (x + \frac {c}{d} - \frac {2}{d {\left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}}\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 )} + \frac {2 \, a^{2}}{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)^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.39, size = 97, normalized size = 1.83 \begin {gather*} \frac {b^{2} \sinh \left (d x + c\right )^{3} + 4 \, {\left ({\left (4 \, a b - 3 \, b^{2}\right )} d x - 2 \, a^{2} + 4 \, a b - 2 \, b^{2}\right )} \cosh \left (d x + c\right ) + {\left (3 \, b^{2} \cosh \left (d x + c\right )^{2} + 8 \, a^{2} - 16 \, a b + 9 \, b^{2}\right )} \sinh \left (d x + c\right )}{8 \, 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)^2,x, algorithm="fricas")

[Out]

1/8*(b^2*sinh(d*x + c)^3 + 4*((4*a*b - 3*b^2)*d*x - 2*a^2 + 4*a*b - 2*b^2)*cosh(d*x + c) + (3*b^2*cosh(d*x + c
)^2 + 8*a^2 - 16*a*b + 9*b^2)*sinh(d*x + c))/(d*cosh(d*x + c))

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a + b \sinh ^{2}{\left (c + d x \right )}\right )^{2} \operatorname {sech}^{2}{\left (c + d x \right )}\, dx \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)**2,x)

[Out]

Integral((a + b*sinh(c + d*x)**2)**2*sech(c + d*x)**2, x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 131 vs. \(2 (49) = 98\).
time = 0.43, size = 131, normalized size = 2.47 \begin {gather*} \frac {b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 4 \, {\left (4 \, a b - 3 \, b^{2}\right )} {\left (d x + c\right )} - \frac {4 \, a b e^{\left (4 \, d x + 4 \, c\right )} - 3 \, b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 16 \, a^{2} e^{\left (2 \, d x + 2 \, c\right )} - 28 \, a b e^{\left (2 \, d x + 2 \, c\right )} + 14 \, b^{2} e^{\left (2 \, d x + 2 \, c\right )} + b^{2}}{e^{\left (4 \, d x + 4 \, c\right )} + e^{\left (2 \, d x + 2 \, c\right )}}}{8 \, 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)^2,x, algorithm="giac")

[Out]

1/8*(b^2*e^(2*d*x + 2*c) + 4*(4*a*b - 3*b^2)*(d*x + c) - (4*a*b*e^(4*d*x + 4*c) - 3*b^2*e^(4*d*x + 4*c) + 16*a
^2*e^(2*d*x + 2*c) - 28*a*b*e^(2*d*x + 2*c) + 14*b^2*e^(2*d*x + 2*c) + b^2)/(e^(4*d*x + 4*c) + e^(2*d*x + 2*c)
))/d

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Mupad [B]
time = 0.88, size = 75, normalized size = 1.42 \begin {gather*} \frac {b\,x\,\left (4\,a-3\,b\right )}{2}-\frac {b^2\,{\mathrm {e}}^{-2\,c-2\,d\,x}}{8\,d}+\frac {b^2\,{\mathrm {e}}^{2\,c+2\,d\,x}}{8\,d}-\frac {2\,\left (a^2-2\,a\,b+b^2\right )}{d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}+1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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