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

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

[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

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Rubi [A]  time = 0.1291, antiderivative size = 92, normalized size of antiderivative = 1., 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 = {3191, 390, 1157, 385, 206} \[ \frac{3}{8} b x \left (8 a^2-12 a b+5 b^2\right )+\frac{3 b^2 (4 a-3 b) \sinh (c+d x) \cosh (c+d x)}{8 d}+\frac{(a-b)^3 \tanh (c+d x)}{d}+\frac{b^3 \sinh (c+d x) \cosh ^3(c+d x)}{4 d} \]

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 3191

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]

Rule 390

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 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 385

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

Mathematica [A]  time = 0.510685, size = 78, normalized size = 0.85 \[ \frac{12 b \left (8 a^2-12 a b+5 b^2\right ) (c+d x)+8 b^2 (3 a-2 b) \sinh (2 (c+d x))+32 (a-b)^3 \tanh (c+d x)+b^3 \sinh (4 (c+d x))}{32 d} \]

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 [A]  time = 0.039, size = 131, normalized size = 1.4 \begin{align*}{\frac{1}{d} \left ({a}^{3}\tanh \left ( dx+c \right ) +3\,{a}^{2}b \left ( dx+c-\tanh \left ( dx+c \right ) \right ) +3\,a{b}^{2} \left ( 1/2\,{\frac{ \left ( \sinh \left ( dx+c \right ) \right ) ^{3}}{\cosh \left ( dx+c \right ) }}-3/2\,dx-3/2\,c+3/2\,\tanh \left ( dx+c \right ) \right ) +{b}^{3} \left ({\frac{ \left ( \sinh \left ( dx+c \right ) \right ) ^{5}}{4\,\cosh \left ( dx+c \right ) }}-{\frac{5\, \left ( \sinh \left ( dx+c \right ) \right ) ^{3}}{8\,\cosh \left ( dx+c \right ) }}+{\frac{15\,dx}{8}}+{\frac{15\,c}{8}}-{\frac{15\,\tanh \left ( dx+c \right ) }{8}} \right ) \right ) } \end{align*}

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)

[Out]

1/d*(a^3*tanh(d*x+c)+3*a^2*b*(d*x+c-tanh(d*x+c))+3*a*b^2*(1/2*sinh(d*x+c)^3/cosh(d*x+c)-3/2*d*x-3/2*c+3/2*tanh
(d*x+c))+b^3*(1/4*sinh(d*x+c)^5/cosh(d*x+c)-5/8*sinh(d*x+c)^3/cosh(d*x+c)+15/8*d*x+15/8*c-15/8*tanh(d*x+c)))

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Maxima [B]  time = 1.09097, size = 290, normalized size = 3.15 \begin{align*} 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{align*}

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]  time = 1.55093, size = 435, normalized size = 4.73 \begin{align*} \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{align*}

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)]  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(sech(d*x+c)**2*(a+b*sinh(d*x+c)**2)**3,x)

[Out]

Timed out

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Giac [B]  time = 1.30828, size = 284, normalized size = 3.09 \begin{align*} \frac{3 \,{\left (8 \, a^{2} b - 12 \, a b^{2} + 5 \, b^{3}\right )}{\left (d x + c\right )}}{8 \, d} - \frac{{\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 )}}{64 \, d} + \frac{b^{3} d e^{\left (4 \, d x + 4 \, c\right )} + 24 \, a b^{2} d e^{\left (2 \, d x + 2 \, c\right )} - 16 \, b^{3} d e^{\left (2 \, d x + 2 \, c\right )}}{64 \, d^{2}} - \frac{2 \,{\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )}}{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(sech(d*x+c)^2*(a+b*sinh(d*x+c)^2)^3,x, algorithm="giac")

[Out]

3/8*(8*a^2*b - 12*a*b^2 + 5*b^3)*(d*x + c)/d - 1/64*(144*a^2*b*e^(4*d*x + 4*c) - 216*a*b^2*e^(4*d*x + 4*c) + 9
0*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)/d + 1/64*(b^
3*d*e^(4*d*x + 4*c) + 24*a*b^2*d*e^(2*d*x + 2*c) - 16*b^3*d*e^(2*d*x + 2*c))/d^2 - 2*(a^3 - 3*a^2*b + 3*a*b^2
- b^3)/(d*(e^(2*d*x + 2*c) + 1))