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

Optimal. Leaf size=43 \[ -\frac {1}{2} (a-2 b) x+\frac {a \cosh (c+d x) \sinh (c+d x)}{2 d}-\frac {b \tanh (c+d x)}{d} \]

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {4217, 466, 396, 212} \begin {gather*} -\frac {1}{2} x (a-2 b)+\frac {a \sinh (c+d x) \cosh (c+d x)}{2 d}-\frac {b \tanh (c+d x)}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*((a - 2*b)*x) + (a*Cosh[c + d*x]*Sinh[c + d*x])/(2*d) - (b*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 396

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[d*x*((a + b*x^n)^(p + 1)/(b*(n*(
p + 1) + 1))), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(b*(n*(p + 1) + 1)), Int[(a + b*x^n)^p, x], x] /; FreeQ[{
a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && NeQ[n*(p + 1) + 1, 0]

Rule 466

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[(-a)^(m/2 - 1)*(b*c - a*d)*x
*((a + b*x^2)^(p + 1)/(2*b^(m/2 + 1)*(p + 1))), x] + Dist[1/(2*b^(m/2 + 1)*(p + 1)), Int[(a + b*x^2)^(p + 1)*E
xpandToSum[2*b*(p + 1)*x^2*Together[(b^(m/2)*x^(m - 2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d))/(a + b*x^2)]
- (-a)^(m/2 - 1)*(b*c - a*d), x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && IGtQ[
m/2, 0] && (IntegerQ[p] || EqQ[m + 2*p + 1, 0])

Rule 4217

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

Rubi steps

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

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Mathematica [A]
time = 0.14, size = 57, normalized size = 1.33 \begin {gather*} \frac {a (-c-d x)}{2 d}+\frac {b \tanh ^{-1}(\tanh (c+d x))}{d}+\frac {a \sinh (2 (c+d x))}{4 d}-\frac {b \tanh (c+d x)}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 1.00, size = 45, normalized size = 1.05

method	result	size
derivativedivides	\(\frac {a \left (\frac {\cosh \left (d x +c \right ) \sinh \left (d x +c \right )}{2}-\frac {d x}{2}-\frac {c}{2}\right )+b \left (d x +c -\tanh \left (d x +c \right )\right )}{d}\)	\(45\)
default	\(\frac {a \left (\frac {\cosh \left (d x +c \right ) \sinh \left (d x +c \right )}{2}-\frac {d x}{2}-\frac {c}{2}\right )+b \left (d x +c -\tanh \left (d x +c \right )\right )}{d}\)	\(45\)
risch	\(-\frac {a x}{2}+b x +\frac {{\mathrm e}^{2 d x +2 c} a}{8 d}-\frac {{\mathrm e}^{-2 d x -2 c} a}{8 d}+\frac {2 b}{d \left (1+{\mathrm e}^{2 d x +2 c}\right )}\)	\(58\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.26, size = 62, normalized size = 1.44 \begin {gather*} -\frac {1}{8} \, a {\left (4 \, x - \frac {e^{\left (2 \, d x + 2 \, c\right )}}{d} + \frac {e^{\left (-2 \, d x - 2 \, c\right )}}{d}\right )} + b {\left (x + \frac {c}{d} - \frac {2}{d {\left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.43, size = 67, normalized size = 1.56 \begin {gather*} \frac {a \sinh \left (d x + c\right )^{3} - 4 \, {\left ({\left (a - 2 \, b\right )} d x - 2 \, b\right )} \cosh \left (d x + c\right ) + {\left (3 \, a \cosh \left (d x + c\right )^{2} + a - 8 \, b\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((a+b*sech(d*x+c)^2)*sinh(d*x+c)^2,x, algorithm="fricas")

[Out]

1/8*(a*sinh(d*x + c)^3 - 4*((a - 2*b)*d*x - 2*b)*cosh(d*x + c) + (3*a*cosh(d*x + c)^2 + a - 8*b)*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 \operatorname {sech}^{2}{\left (c + d x \right )}\right ) \sinh ^{2}{\left (c + d x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 92 vs. \(2 (39) = 78\).
time = 0.40, size = 92, normalized size = 2.14 \begin {gather*} -\frac {4 \, {\left (d x + c\right )} {\left (a - 2 \, b\right )} - a e^{\left (2 \, d x + 2 \, c\right )} - \frac {a e^{\left (4 \, d x + 4 \, c\right )} - 2 \, b e^{\left (4 \, d x + 4 \, c\right )} + 14 \, b e^{\left (2 \, d x + 2 \, c\right )} - a}{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((a+b*sech(d*x+c)^2)*sinh(d*x+c)^2,x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 1.43, size = 55, normalized size = 1.28 \begin {gather*} \frac {a\,\mathrm {cosh}\left (c+d\,x\right )\,\mathrm {sinh}\left (c+d\,x\right )}{2\,d}-\frac {\frac {a\,d\,x}{2}-b\,d\,x}{d}-\frac {b\,\mathrm {sinh}\left (c+d\,x\right )}{d\,\mathrm {cosh}\left (c+d\,x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a*cosh(c + d*x)*sinh(c + d*x))/(2*d) - ((a*d*x)/2 - b*d*x)/d - (b*sinh(c + d*x))/(d*cosh(c + d*x))

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