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

Optimal. Leaf size=64 \[ \frac{(a+3 b) (a-b) \tan ^{-1}(\sinh (c+d x))}{2 d}+\frac{(a-b)^2 \tanh (c+d x) \text{sech}(c+d x)}{2 d}+\frac{b^2 \sinh (c+d x)}{d} \]

[Out]

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

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Rubi [A]  time = 0.079862, antiderivative size = 64, normalized size of antiderivative = 1., 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 = {3190, 390, 385, 203} \[ \frac{(a+3 b) (a-b) \tan ^{-1}(\sinh (c+d x))}{2 d}+\frac{(a-b)^2 \tanh (c+d x) \text{sech}(c+d x)}{2 d}+\frac{b^2 \sinh (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a - b)*(a + 3*b)*ArcTan[Sinh[c + d*x]])/(2*d) + (b^2*Sinh[c + d*x])/d + ((a - b)^2*Sech[c + d*x]*Tanh[c + d*
x])/(2*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 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 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}^3(c+d x) \left (a+b \sinh ^2(c+d x)\right )^2 \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (a+b x^2\right )^2}{\left (1+x^2\right )^2} \, dx,x,\sinh (c+d x)\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \left (b^2+\frac{a^2-b^2+2 (a-b) b x^2}{\left (1+x^2\right )^2}\right ) \, dx,x,\sinh (c+d x)\right )}{d}\\ &=\frac{b^2 \sinh (c+d x)}{d}+\frac{\operatorname{Subst}\left (\int \frac{a^2-b^2+2 (a-b) b x^2}{\left (1+x^2\right )^2} \, dx,x,\sinh (c+d x)\right )}{d}\\ &=\frac{b^2 \sinh (c+d x)}{d}+\frac{(a-b)^2 \text{sech}(c+d x) \tanh (c+d x)}{2 d}+\frac{((a-b) (a+3 b)) \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\sinh (c+d x)\right )}{2 d}\\ &=\frac{(a-b) (a+3 b) \tan ^{-1}(\sinh (c+d x))}{2 d}+\frac{b^2 \sinh (c+d x)}{d}+\frac{(a-b)^2 \text{sech}(c+d x) \tanh (c+d x)}{2 d}\\ \end{align*}

Mathematica [C]  time = 7.36673, size = 253, normalized size = 3.95 \[ \frac{\text{csch}^3(c+d x) \left (-64 \sinh ^6(c+d x) \left (a+b \sinh ^2(c+d x)\right )^2 \text{HypergeometricPFQ}\left (\left \{\frac{3}{2},2,2,2\right \},\left \{1,1,\frac{9}{2}\right \},-\sinh ^2(c+d x)\right )-35 \left (a^2 \left (37 \sinh ^2(c+d x)+375\right )+2 a b \left (61 \sinh ^2(c+d x)+375\right ) \sinh ^2(c+d x)+b^2 \left (61 \sinh ^2(c+d x)+303\right ) \sinh ^4(c+d x)\right )+\frac{105 \tanh ^{-1}\left (\sqrt{-\sinh ^2(c+d x)}\right ) \left (a^2 \left (9 \sinh ^4(c+d x)+54 \sinh ^2(c+d x)+125\right )+2 a b \left (\sinh ^4(c+d x)+62 \sinh ^2(c+d x)+125\right ) \sinh ^2(c+d x)+b^2 \left (\sinh ^4(c+d x)+54 \sinh ^2(c+d x)+101\right ) \sinh ^4(c+d x)\right )}{\sqrt{-\sinh ^2(c+d x)}}\right )}{1680 d} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Csch[c + d*x]^3*(-64*HypergeometricPFQ[{3/2, 2, 2, 2}, {1, 1, 9/2}, -Sinh[c + d*x]^2]*Sinh[c + d*x]^6*(a + b*
Sinh[c + d*x]^2)^2 - 35*(a^2*(375 + 37*Sinh[c + d*x]^2) + b^2*Sinh[c + d*x]^4*(303 + 61*Sinh[c + d*x]^2) + 2*a
*b*Sinh[c + d*x]^2*(375 + 61*Sinh[c + d*x]^2)) + (105*ArcTanh[Sqrt[-Sinh[c + d*x]^2]]*(b^2*Sinh[c + d*x]^4*(10
1 + 54*Sinh[c + d*x]^2 + Sinh[c + d*x]^4) + 2*a*b*Sinh[c + d*x]^2*(125 + 62*Sinh[c + d*x]^2 + Sinh[c + d*x]^4)
+ a^2*(125 + 54*Sinh[c + d*x]^2 + 9*Sinh[c + d*x]^4)))/Sqrt[-Sinh[c + d*x]^2]))/(1680*d)

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Maple [B]  time = 0.047, size = 169, normalized size = 2.6 \begin{align*}{\frac{{a}^{2}{\rm sech} \left (dx+c\right )\tanh \left ( dx+c \right ) }{2\,d}}+{\frac{{a}^{2}\arctan \left ({{\rm e}^{dx+c}} \right ) }{d}}-2\,{\frac{ab\sinh \left ( dx+c \right ) }{d \left ( \cosh \left ( dx+c \right ) \right ) ^{2}}}+{\frac{ab{\rm sech} \left (dx+c\right )\tanh \left ( dx+c \right ) }{d}}+2\,{\frac{ab\arctan \left ({{\rm e}^{dx+c}} \right ) }{d}}+{\frac{{b}^{2} \left ( \sinh \left ( dx+c \right ) \right ) ^{3}}{d \left ( \cosh \left ( dx+c \right ) \right ) ^{2}}}+3\,{\frac{{b}^{2}\sinh \left ( dx+c \right ) }{d \left ( \cosh \left ( dx+c \right ) \right ) ^{2}}}-{\frac{3\,{b}^{2}{\rm sech} \left (dx+c\right )\tanh \left ( dx+c \right ) }{2\,d}}-3\,{\frac{{b}^{2}\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)^3*(a+b*sinh(d*x+c)^2)^2,x)

[Out]

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

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Maxima [B]  time = 1.89297, size = 316, normalized size = 4.94 \begin{align*} \frac{1}{2} \, b^{2}{\left (\frac{6 \, \arctan \left (e^{\left (-d x - c\right )}\right )}{d} - \frac{e^{\left (-d x - c\right )}}{d} + \frac{4 \, e^{\left (-2 \, d x - 2 \, c\right )} - e^{\left (-4 \, d x - 4 \, c\right )} + 1}{d{\left (e^{\left (-d x - c\right )} + 2 \, e^{\left (-3 \, d x - 3 \, c\right )} + e^{\left (-5 \, d x - 5 \, c\right )}\right )}}\right )} - 2 \, a b{\left (\frac{\arctan \left (e^{\left (-d x - c\right )}\right )}{d} + \frac{e^{\left (-d x - c\right )} - e^{\left (-3 \, d x - 3 \, c\right )}}{d{\left (2 \, e^{\left (-2 \, d x - 2 \, c\right )} + e^{\left (-4 \, d x - 4 \, c\right )} + 1\right )}}\right )} - a^{2}{\left (\frac{\arctan \left (e^{\left (-d x - c\right )}\right )}{d} - \frac{e^{\left (-d x - c\right )} - e^{\left (-3 \, d x - 3 \, c\right )}}{d{\left (2 \, e^{\left (-2 \, d x - 2 \, c\right )} + e^{\left (-4 \, d x - 4 \, c\right )} + 1\right )}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*b^2*(6*arctan(e^(-d*x - c))/d - e^(-d*x - c)/d + (4*e^(-2*d*x - 2*c) - e^(-4*d*x - 4*c) + 1)/(d*(e^(-d*x -
c) + 2*e^(-3*d*x - 3*c) + e^(-5*d*x - 5*c)))) - 2*a*b*(arctan(e^(-d*x - c))/d + (e^(-d*x - c) - e^(-3*d*x - 3
*c))/(d*(2*e^(-2*d*x - 2*c) + e^(-4*d*x - 4*c) + 1))) - a^2*(arctan(e^(-d*x - c))/d - (e^(-d*x - c) - e^(-3*d*
x - 3*c))/(d*(2*e^(-2*d*x - 2*c) + e^(-4*d*x - 4*c) + 1)))

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Fricas [B]  time = 1.62602, size = 1916, normalized size = 29.94 \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)^3*(a+b*sinh(d*x+c)^2)^2,x, algorithm="fricas")

[Out]

1/2*(b^2*cosh(d*x + c)^6 + 6*b^2*cosh(d*x + c)*sinh(d*x + c)^5 + b^2*sinh(d*x + c)^6 + (2*a^2 - 4*a*b + 3*b^2)
*cosh(d*x + c)^4 + (15*b^2*cosh(d*x + c)^2 + 2*a^2 - 4*a*b + 3*b^2)*sinh(d*x + c)^4 + 4*(5*b^2*cosh(d*x + c)^3
+ (2*a^2 - 4*a*b + 3*b^2)*cosh(d*x + c))*sinh(d*x + c)^3 - (2*a^2 - 4*a*b + 3*b^2)*cosh(d*x + c)^2 + (15*b^2*
cosh(d*x + c)^4 + 6*(2*a^2 - 4*a*b + 3*b^2)*cosh(d*x + c)^2 - 2*a^2 + 4*a*b - 3*b^2)*sinh(d*x + c)^2 - b^2 + 2
*((a^2 + 2*a*b - 3*b^2)*cosh(d*x + c)^5 + 5*(a^2 + 2*a*b - 3*b^2)*cosh(d*x + c)*sinh(d*x + c)^4 + (a^2 + 2*a*b
- 3*b^2)*sinh(d*x + c)^5 + 2*(a^2 + 2*a*b - 3*b^2)*cosh(d*x + c)^3 + 2*(5*(a^2 + 2*a*b - 3*b^2)*cosh(d*x + c)
^2 + a^2 + 2*a*b - 3*b^2)*sinh(d*x + c)^3 + 2*(5*(a^2 + 2*a*b - 3*b^2)*cosh(d*x + c)^3 + 3*(a^2 + 2*a*b - 3*b^
2)*cosh(d*x + c))*sinh(d*x + c)^2 + (a^2 + 2*a*b - 3*b^2)*cosh(d*x + c) + (5*(a^2 + 2*a*b - 3*b^2)*cosh(d*x +
c)^4 + 6*(a^2 + 2*a*b - 3*b^2)*cosh(d*x + c)^2 + a^2 + 2*a*b - 3*b^2)*sinh(d*x + c))*arctan(cosh(d*x + c) + si
nh(d*x + c)) + 2*(3*b^2*cosh(d*x + c)^5 + 2*(2*a^2 - 4*a*b + 3*b^2)*cosh(d*x + c)^3 - (2*a^2 - 4*a*b + 3*b^2)*
cosh(d*x + c))*sinh(d*x + c))/(d*cosh(d*x + c)^5 + 5*d*cosh(d*x + c)*sinh(d*x + c)^4 + d*sinh(d*x + c)^5 + 2*d
*cosh(d*x + c)^3 + 2*(5*d*cosh(d*x + c)^2 + d)*sinh(d*x + c)^3 + 2*(5*d*cosh(d*x + c)^3 + 3*d*cosh(d*x + c))*s
inh(d*x + c)^2 + d*cosh(d*x + c) + (5*d*cosh(d*x + c)^4 + 6*d*cosh(d*x + c)^2 + d)*sinh(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)**3*(a+b*sinh(d*x+c)**2)**2,x)

[Out]

Timed out

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Giac [B]  time = 1.2036, size = 225, normalized size = 3.52 \begin{align*} \frac{b^{2}{\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}}{2 \, d} + \frac{{\left (\pi + 2 \, \arctan \left (\frac{1}{2} \,{\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )} e^{\left (-d x - c\right )}\right )\right )}{\left (a^{2} + 2 \, a b - 3 \, b^{2}\right )}}{4 \, d} + \frac{a^{2}{\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )} - 2 \, a b{\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )} + b^{2}{\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}}{{\left ({\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{2} + 4\right )} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*b^2*(e^(d*x + c) - e^(-d*x - c))/d + 1/4*(pi + 2*arctan(1/2*(e^(2*d*x + 2*c) - 1)*e^(-d*x - c)))*(a^2 + 2*
a*b - 3*b^2)/d + (a^2*(e^(d*x + c) - e^(-d*x - c)) - 2*a*b*(e^(d*x + c) - e^(-d*x - c)) + b^2*(e^(d*x + c) - e
^(-d*x - c)))/(((e^(d*x + c) - e^(-d*x - c))^2 + 4)*d)