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

Optimal. Leaf size=86 \[ \frac{b \left (3 a^2-3 a b+b^2\right ) \sinh (c+d x)}{d}+\frac{b^2 (3 a-b) \sinh ^3(c+d x)}{3 d}+\frac{(a-b)^3 \tan ^{-1}(\sinh (c+d x))}{d}+\frac{b^3 \sinh ^5(c+d x)}{5 d} \]

[Out]

((a - b)^3*ArcTan[Sinh[c + d*x]])/d + (b*(3*a^2 - 3*a*b + b^2)*Sinh[c + d*x])/d + ((3*a - b)*b^2*Sinh[c + d*x]
^3)/(3*d) + (b^3*Sinh[c + d*x]^5)/(5*d)

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Rubi [A]  time = 0.0755837, antiderivative size = 86, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {3190, 390, 203} \[ \frac{b \left (3 a^2-3 a b+b^2\right ) \sinh (c+d x)}{d}+\frac{b^2 (3 a-b) \sinh ^3(c+d x)}{3 d}+\frac{(a-b)^3 \tan ^{-1}(\sinh (c+d x))}{d}+\frac{b^3 \sinh ^5(c+d x)}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a - b)^3*ArcTan[Sinh[c + d*x]])/d + (b*(3*a^2 - 3*a*b + b^2)*Sinh[c + d*x])/d + ((3*a - b)*b^2*Sinh[c + d*x]
^3)/(3*d) + (b^3*Sinh[c + d*x]^5)/(5*d)

Rule 3190

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

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 203

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

Rubi steps

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

Mathematica [A]  time = 0.530513, size = 100, normalized size = 1.16 \[ \frac{\sinh (c+d x) \left (b \left (45 a^2+15 a b \left (\sinh ^2(c+d x)-3\right )+b^2 \left (3 \sinh ^4(c+d x)-5 \sinh ^2(c+d x)+15\right )\right )+\frac{15 (a-b)^3 \tanh ^{-1}\left (\sqrt{-\sinh ^2(c+d x)}\right )}{\sqrt{-\sinh ^2(c+d x)}}\right )}{15 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sinh[c + d*x]*((15*(a - b)^3*ArcTanh[Sqrt[-Sinh[c + d*x]^2]])/Sqrt[-Sinh[c + d*x]^2] + b*(45*a^2 + 15*a*b*(-3
+ Sinh[c + d*x]^2) + b^2*(15 - 5*Sinh[c + d*x]^2 + 3*Sinh[c + d*x]^4))))/(15*d)

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Maple [A]  time = 0.056, size = 155, normalized size = 1.8 \begin{align*} 2\,{\frac{{a}^{3}\arctan \left ({{\rm e}^{dx+c}} \right ) }{d}}+3\,{\frac{{a}^{2}b\sinh \left ( dx+c \right ) }{d}}-6\,{\frac{{a}^{2}b\arctan \left ({{\rm e}^{dx+c}} \right ) }{d}}+{\frac{a{b}^{2} \left ( \sinh \left ( dx+c \right ) \right ) ^{3}}{d}}-3\,{\frac{a{b}^{2}\sinh \left ( dx+c \right ) }{d}}+6\,{\frac{a{b}^{2}\arctan \left ({{\rm e}^{dx+c}} \right ) }{d}}+{\frac{{b}^{3} \left ( \sinh \left ( dx+c \right ) \right ) ^{5}}{5\,d}}-{\frac{{b}^{3} \left ( \sinh \left ( dx+c \right ) \right ) ^{3}}{3\,d}}+{\frac{{b}^{3}\sinh \left ( dx+c \right ) }{d}}-2\,{\frac{{b}^{3}\arctan \left ({{\rm e}^{dx+c}} \right ) }{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/d*a^3*arctan(exp(d*x+c))+3/d*a^2*b*sinh(d*x+c)-6/d*a^2*b*arctan(exp(d*x+c))+1/d*a*b^2*sinh(d*x+c)^3-3/d*a*b^
2*sinh(d*x+c)+6/d*a*b^2*arctan(exp(d*x+c))+1/5*b^3*sinh(d*x+c)^5/d-1/3*b^3*sinh(d*x+c)^3/d+b^3*sinh(d*x+c)/d-2
/d*b^3*arctan(exp(d*x+c))

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Maxima [B]  time = 1.69462, size = 315, normalized size = 3.66 \begin{align*} -\frac{1}{480} \, b^{3}{\left (\frac{{\left (35 \, e^{\left (-2 \, d x - 2 \, c\right )} - 330 \, e^{\left (-4 \, d x - 4 \, c\right )} - 3\right )} e^{\left (5 \, d x + 5 \, c\right )}}{d} + \frac{330 \, e^{\left (-d x - c\right )} - 35 \, e^{\left (-3 \, d x - 3 \, c\right )} + 3 \, e^{\left (-5 \, d x - 5 \, c\right )}}{d} - \frac{960 \, \arctan \left (e^{\left (-d x - c\right )}\right )}{d}\right )} - \frac{1}{8} \, a b^{2}{\left (\frac{{\left (15 \, e^{\left (-2 \, d x - 2 \, c\right )} - 1\right )} e^{\left (3 \, d x + 3 \, c\right )}}{d} - \frac{15 \, e^{\left (-d x - c\right )} - e^{\left (-3 \, d x - 3 \, c\right )}}{d} + \frac{48 \, \arctan \left (e^{\left (-d x - c\right )}\right )}{d}\right )} + \frac{3}{2} \, a^{2} b{\left (\frac{4 \, \arctan \left (e^{\left (-d x - c\right )}\right )}{d} + \frac{e^{\left (d x + c\right )}}{d} - \frac{e^{\left (-d x - c\right )}}{d}\right )} + \frac{a^{3} \arctan \left (\sinh \left (d x + c\right )\right )}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/480*b^3*((35*e^(-2*d*x - 2*c) - 330*e^(-4*d*x - 4*c) - 3)*e^(5*d*x + 5*c)/d + (330*e^(-d*x - c) - 35*e^(-3*
d*x - 3*c) + 3*e^(-5*d*x - 5*c))/d - 960*arctan(e^(-d*x - c))/d) - 1/8*a*b^2*((15*e^(-2*d*x - 2*c) - 1)*e^(3*d
*x + 3*c)/d - (15*e^(-d*x - c) - e^(-3*d*x - 3*c))/d + 48*arctan(e^(-d*x - c))/d) + 3/2*a^2*b*(4*arctan(e^(-d*
x - c))/d + e^(d*x + c)/d - e^(-d*x - c)/d) + a^3*arctan(sinh(d*x + c))/d

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Fricas [B]  time = 1.62245, size = 2795, normalized size = 32.5 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/480*(3*b^3*cosh(d*x + c)^10 + 30*b^3*cosh(d*x + c)*sinh(d*x + c)^9 + 3*b^3*sinh(d*x + c)^10 + 5*(12*a*b^2 -
7*b^3)*cosh(d*x + c)^8 + 5*(27*b^3*cosh(d*x + c)^2 + 12*a*b^2 - 7*b^3)*sinh(d*x + c)^8 + 40*(9*b^3*cosh(d*x +
c)^3 + (12*a*b^2 - 7*b^3)*cosh(d*x + c))*sinh(d*x + c)^7 + 30*(24*a^2*b - 30*a*b^2 + 11*b^3)*cosh(d*x + c)^6 +
10*(63*b^3*cosh(d*x + c)^4 + 72*a^2*b - 90*a*b^2 + 33*b^3 + 14*(12*a*b^2 - 7*b^3)*cosh(d*x + c)^2)*sinh(d*x +
c)^6 + 4*(189*b^3*cosh(d*x + c)^5 + 70*(12*a*b^2 - 7*b^3)*cosh(d*x + c)^3 + 45*(24*a^2*b - 30*a*b^2 + 11*b^3)
*cosh(d*x + c))*sinh(d*x + c)^5 - 30*(24*a^2*b - 30*a*b^2 + 11*b^3)*cosh(d*x + c)^4 + 10*(63*b^3*cosh(d*x + c)
^6 + 35*(12*a*b^2 - 7*b^3)*cosh(d*x + c)^4 - 72*a^2*b + 90*a*b^2 - 33*b^3 + 45*(24*a^2*b - 30*a*b^2 + 11*b^3)*
cosh(d*x + c)^2)*sinh(d*x + c)^4 + 40*(9*b^3*cosh(d*x + c)^7 + 7*(12*a*b^2 - 7*b^3)*cosh(d*x + c)^5 + 15*(24*a
^2*b - 30*a*b^2 + 11*b^3)*cosh(d*x + c)^3 - 3*(24*a^2*b - 30*a*b^2 + 11*b^3)*cosh(d*x + c))*sinh(d*x + c)^3 -
3*b^3 - 5*(12*a*b^2 - 7*b^3)*cosh(d*x + c)^2 + 5*(27*b^3*cosh(d*x + c)^8 + 28*(12*a*b^2 - 7*b^3)*cosh(d*x + c)
^6 + 90*(24*a^2*b - 30*a*b^2 + 11*b^3)*cosh(d*x + c)^4 - 12*a*b^2 + 7*b^3 - 36*(24*a^2*b - 30*a*b^2 + 11*b^3)*
cosh(d*x + c)^2)*sinh(d*x + c)^2 + 960*((a^3 - 3*a^2*b + 3*a*b^2 - b^3)*cosh(d*x + c)^5 + 5*(a^3 - 3*a^2*b + 3
*a*b^2 - b^3)*cosh(d*x + c)^4*sinh(d*x + c) + 10*(a^3 - 3*a^2*b + 3*a*b^2 - b^3)*cosh(d*x + c)^3*sinh(d*x + c)
^2 + 10*(a^3 - 3*a^2*b + 3*a*b^2 - b^3)*cosh(d*x + c)^2*sinh(d*x + c)^3 + 5*(a^3 - 3*a^2*b + 3*a*b^2 - b^3)*co
sh(d*x + c)*sinh(d*x + c)^4 + (a^3 - 3*a^2*b + 3*a*b^2 - b^3)*sinh(d*x + c)^5)*arctan(cosh(d*x + c) + sinh(d*x
+ c)) + 10*(3*b^3*cosh(d*x + c)^9 + 4*(12*a*b^2 - 7*b^3)*cosh(d*x + c)^7 + 18*(24*a^2*b - 30*a*b^2 + 11*b^3)*
cosh(d*x + c)^5 - 12*(24*a^2*b - 30*a*b^2 + 11*b^3)*cosh(d*x + c)^3 - (12*a*b^2 - 7*b^3)*cosh(d*x + c))*sinh(d
*x + c))/(d*cosh(d*x + c)^5 + 5*d*cosh(d*x + c)^4*sinh(d*x + c) + 10*d*cosh(d*x + c)^3*sinh(d*x + c)^2 + 10*d*
cosh(d*x + c)^2*sinh(d*x + c)^3 + 5*d*cosh(d*x + c)*sinh(d*x + c)^4 + d*sinh(d*x + c)^5)

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

[Out]

Timed out

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Giac [B]  time = 1.279, size = 309, normalized size = 3.59 \begin{align*} \frac{2 \,{\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} \arctan \left (e^{\left (d x + c\right )}\right )}{d} - \frac{{\left (720 \, a^{2} b e^{\left (4 \, d x + 4 \, c\right )} - 900 \, a b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 330 \, b^{3} e^{\left (4 \, d x + 4 \, c\right )} + 60 \, a b^{2} e^{\left (2 \, d x + 2 \, c\right )} - 35 \, b^{3} e^{\left (2 \, d x + 2 \, c\right )} + 3 \, b^{3}\right )} e^{\left (-5 \, d x - 5 \, c\right )}}{480 \, d} + \frac{3 \, b^{3} d^{4} e^{\left (5 \, d x + 5 \, c\right )} + 60 \, a b^{2} d^{4} e^{\left (3 \, d x + 3 \, c\right )} - 35 \, b^{3} d^{4} e^{\left (3 \, d x + 3 \, c\right )} + 720 \, a^{2} b d^{4} e^{\left (d x + c\right )} - 900 \, a b^{2} d^{4} e^{\left (d x + c\right )} + 330 \, b^{3} d^{4} e^{\left (d x + c\right )}}{480 \, d^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(a^3 - 3*a^2*b + 3*a*b^2 - b^3)*arctan(e^(d*x + c))/d - 1/480*(720*a^2*b*e^(4*d*x + 4*c) - 900*a*b^2*e^(4*d*
x + 4*c) + 330*b^3*e^(4*d*x + 4*c) + 60*a*b^2*e^(2*d*x + 2*c) - 35*b^3*e^(2*d*x + 2*c) + 3*b^3)*e^(-5*d*x - 5*
c)/d + 1/480*(3*b^3*d^4*e^(5*d*x + 5*c) + 60*a*b^2*d^4*e^(3*d*x + 3*c) - 35*b^3*d^4*e^(3*d*x + 3*c) + 720*a^2*
b*d^4*e^(d*x + c) - 900*a*b^2*d^4*e^(d*x + c) + 330*b^3*d^4*e^(d*x + c))/d^5