3.1 \(\int (a+b \text{sech}^2(c+d x)) \sinh ^4(c+d x) \, dx\)

Optimal. Leaf size=70 \[ -\frac{(5 a-4 b) \sinh (c+d x) \cosh (c+d x)}{8 d}+\frac{3}{8} x (a-4 b)+\frac{a \sinh (c+d x) \cosh ^3(c+d x)}{4 d}+\frac{b \tanh (c+d x)}{d} \]

[Out]

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

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Rubi [A]  time = 0.0808843, antiderivative size = 70, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {4132, 455, 1157, 388, 206} \[ -\frac{(5 a-4 b) \sinh (c+d x) \cosh (c+d x)}{8 d}+\frac{3}{8} x (a-4 b)+\frac{a \sinh (c+d x) \cosh ^3(c+d x)}{4 d}+\frac{b \tanh (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4132

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]

Rule 455

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 1157

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

Rule 388

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 206

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

Rubi steps

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

Mathematica [A]  time = 0.313628, size = 54, normalized size = 0.77 \[ \frac{12 (a-4 b) (c+d x)-8 (a-b) \sinh (2 (c+d x))+a \sinh (4 (c+d x))+32 b \tanh (c+d x)}{32 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(12*(a - 4*b)*(c + d*x) - 8*(a - b)*Sinh[2*(c + d*x)] + a*Sinh[4*(c + d*x)] + 32*b*Tanh[c + d*x])/(32*d)

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Maple [A]  time = 0.032, size = 78, normalized size = 1.1 \begin{align*}{\frac{1}{d} \left ( a \left ( \left ({\frac{ \left ( \sinh \left ( dx+c \right ) \right ) ^{3}}{4}}-{\frac{3\,\sinh \left ( dx+c \right ) }{8}} \right ) \cosh \left ( dx+c \right ) +{\frac{3\,dx}{8}}+{\frac{3\,c}{8}} \right ) +b \left ({\frac{ \left ( \sinh \left ( dx+c \right ) \right ) ^{3}}{2\,\cosh \left ( dx+c \right ) }}-{\frac{3\,dx}{2}}-{\frac{3\,c}{2}}+{\frac{3\,\tanh \left ( dx+c \right ) }{2}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sech(d*x+c)^2)*sinh(d*x+c)^4,x)

[Out]

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

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Maxima [B]  time = 1.01564, size = 174, normalized size = 2.49 \begin{align*} \frac{1}{64} \, a{\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}{8} \, b{\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 )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sech(d*x+c)^2)*sinh(d*x+c)^4,x, algorithm="maxima")

[Out]

1/64*a*(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) - 1/8*b*(1
2*(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))))

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Fricas [A]  time = 2.55299, size = 300, normalized size = 4.29 \begin{align*} \frac{a \sinh \left (d x + c\right )^{5} +{\left (10 \, a \cosh \left (d x + c\right )^{2} - 7 \, a + 8 \, b\right )} \sinh \left (d x + c\right )^{3} + 8 \,{\left (3 \,{\left (a - 4 \, b\right )} d x - 8 \, b\right )} \cosh \left (d x + c\right ) +{\left (5 \, a \cosh \left (d x + c\right )^{4} - 3 \,{\left (7 \, a - 8 \, b\right )} \cosh \left (d x + c\right )^{2} - 8 \, a + 72 \, b\right )} \sinh \left (d x + c\right )}{64 \, d \cosh \left (d x + c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/64*(a*sinh(d*x + c)^5 + (10*a*cosh(d*x + c)^2 - 7*a + 8*b)*sinh(d*x + c)^3 + 8*(3*(a - 4*b)*d*x - 8*b)*cosh(
d*x + c) + (5*a*cosh(d*x + c)^4 - 3*(7*a - 8*b)*cosh(d*x + c)^2 - 8*a + 72*b)*sinh(d*x + c))/(d*cosh(d*x + c))

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [B]  time = 1.16455, size = 193, normalized size = 2.76 \begin{align*} \frac{3 \,{\left (d x + c\right )}{\left (a - 4 \, b\right )}}{8 \, d} - \frac{{\left (18 \, a e^{\left (4 \, d x + 4 \, c\right )} - 72 \, b e^{\left (4 \, d x + 4 \, c\right )} - 8 \, a e^{\left (2 \, d x + 2 \, c\right )} + 8 \, b e^{\left (2 \, d x + 2 \, c\right )} + a\right )} e^{\left (-4 \, d x - 4 \, c\right )}}{64 \, d} + \frac{a d e^{\left (4 \, d x + 4 \, c\right )} - 8 \, a d e^{\left (2 \, d x + 2 \, c\right )} + 8 \, b d e^{\left (2 \, d x + 2 \, c\right )}}{64 \, d^{2}} - \frac{2 \, b}{d{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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