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

Optimal. Leaf size=55 \[ \frac {b (2 a-b) \sinh (c+d x)}{d}+\frac {(a-b)^2 \tan ^{-1}(\sinh (c+d x))}{d}+\frac {b^2 \sinh ^3(c+d x)}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 55, normalized size of antiderivative = 1.00, 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} \[ \frac {b (2 a-b) \sinh (c+d x)}{d}+\frac {(a-b)^2 \tan ^{-1}(\sinh (c+d x))}{d}+\frac {b^2 \sinh ^3(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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])

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 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]

Rubi steps

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

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Mathematica [A]  time = 0.29, size = 70, normalized size = 1.27 \[ \frac {\sinh (c+d x) \left (b \left (6 a+b \left (\sinh ^2(c+d x)-3\right )\right )+\frac {3 (a-b)^2 \tanh ^{-1}\left (\sqrt {-\sinh ^2(c+d x)}\right )}{\sqrt {-\sinh ^2(c+d x)}}\right )}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sinh[c + d*x]*((3*(a - b)^2*ArcTanh[Sqrt[-Sinh[c + d*x]^2]])/Sqrt[-Sinh[c + d*x]^2] + b*(6*a + b*(-3 + Sinh[c
+ d*x]^2))))/(3*d)

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fricas [B]  time = 0.91, size = 446, normalized size = 8.11 \[ \frac {b^{2} \cosh \left (d x + c\right )^{6} + 6 \, b^{2} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{5} + b^{2} \sinh \left (d x + c\right )^{6} + 3 \, {\left (8 \, a b - 5 \, b^{2}\right )} \cosh \left (d x + c\right )^{4} + 3 \, {\left (5 \, b^{2} \cosh \left (d x + c\right )^{2} + 8 \, a b - 5 \, b^{2}\right )} \sinh \left (d x + c\right )^{4} + 4 \, {\left (5 \, b^{2} \cosh \left (d x + c\right )^{3} + 3 \, {\left (8 \, a b - 5 \, b^{2}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{3} - 3 \, {\left (8 \, a b - 5 \, b^{2}\right )} \cosh \left (d x + c\right )^{2} + 3 \, {\left (5 \, b^{2} \cosh \left (d x + c\right )^{4} + 6 \, {\left (8 \, a b - 5 \, b^{2}\right )} \cosh \left (d x + c\right )^{2} - 8 \, a b + 5 \, b^{2}\right )} \sinh \left (d x + c\right )^{2} - b^{2} + 48 \, {\left ({\left (a^{2} - 2 \, a b + b^{2}\right )} \cosh \left (d x + c\right )^{3} + 3 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \cosh \left (d x + c\right )^{2} \sinh \left (d x + c\right ) + 3 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{2} + {\left (a^{2} - 2 \, a b + b^{2}\right )} \sinh \left (d x + c\right )^{3}\right )} \arctan \left (\cosh \left (d x + c\right ) + \sinh \left (d x + c\right )\right ) + 6 \, {\left (b^{2} \cosh \left (d x + c\right )^{5} + 2 \, {\left (8 \, a b - 5 \, b^{2}\right )} \cosh \left (d x + c\right )^{3} - {\left (8 \, a b - 5 \, b^{2}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )}{24 \, {\left (d \cosh \left (d x + c\right )^{3} + 3 \, d \cosh \left (d x + c\right )^{2} \sinh \left (d x + c\right ) + 3 \, d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{2} + d \sinh \left (d x + c\right )^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(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 + 3*(8*a*b - 5*b^2)*cosh
(d*x + c)^4 + 3*(5*b^2*cosh(d*x + c)^2 + 8*a*b - 5*b^2)*sinh(d*x + c)^4 + 4*(5*b^2*cosh(d*x + c)^3 + 3*(8*a*b
- 5*b^2)*cosh(d*x + c))*sinh(d*x + c)^3 - 3*(8*a*b - 5*b^2)*cosh(d*x + c)^2 + 3*(5*b^2*cosh(d*x + c)^4 + 6*(8*
a*b - 5*b^2)*cosh(d*x + c)^2 - 8*a*b + 5*b^2)*sinh(d*x + c)^2 - b^2 + 48*((a^2 - 2*a*b + b^2)*cosh(d*x + c)^3
+ 3*(a^2 - 2*a*b + b^2)*cosh(d*x + c)^2*sinh(d*x + c) + 3*(a^2 - 2*a*b + b^2)*cosh(d*x + c)*sinh(d*x + c)^2 +
(a^2 - 2*a*b + b^2)*sinh(d*x + c)^3)*arctan(cosh(d*x + c) + sinh(d*x + c)) + 6*(b^2*cosh(d*x + c)^5 + 2*(8*a*b
- 5*b^2)*cosh(d*x + c)^3 - (8*a*b - 5*b^2)*cosh(d*x + c))*sinh(d*x + c))/(d*cosh(d*x + c)^3 + 3*d*cosh(d*x +
c)^2*sinh(d*x + c) + 3*d*cosh(d*x + c)*sinh(d*x + c)^2 + d*sinh(d*x + c)^3)

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giac [A]  time = 0.16, size = 102, normalized size = 1.85 \[ \frac {b^{2} e^{\left (3 \, d x + 3 \, c\right )} + 24 \, a b e^{\left (d x + c\right )} - 15 \, b^{2} e^{\left (d x + c\right )} + 48 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \arctan \left (e^{\left (d x + c\right )}\right ) - {\left (24 \, a b e^{\left (2 \, d x + 2 \, c\right )} - 15 \, b^{2} e^{\left (2 \, d x + 2 \, c\right )} + b^{2}\right )} e^{\left (-3 \, d x - 3 \, c\right )}}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(b^2*e^(3*d*x + 3*c) + 24*a*b*e^(d*x + c) - 15*b^2*e^(d*x + c) + 48*(a^2 - 2*a*b + b^2)*arctan(e^(d*x + c
)) - (24*a*b*e^(2*d*x + 2*c) - 15*b^2*e^(2*d*x + 2*c) + b^2)*e^(-3*d*x - 3*c))/d

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maple [A]  time = 0.09, size = 89, normalized size = 1.62 \[ \frac {2 a^{2} \arctan \left ({\mathrm e}^{d x +c}\right )}{d}+\frac {2 a b \sinh \left (d x +c \right )}{d}-\frac {4 a b \arctan \left ({\mathrm e}^{d x +c}\right )}{d}+\frac {b^{2} \left (\sinh ^{3}\left (d x +c \right )\right )}{3 d}-\frac {b^{2} \sinh \left (d x +c \right )}{d}+\frac {2 b^{2} \arctan \left ({\mathrm e}^{d x +c}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.67, size = 133, normalized size = 2.42 \[ -\frac {1}{24} \, b^{2} {\left (\frac {{\left (15 \, e^{\left (-2 \, d x - 2 \, c\right )} - 1\right )} e^{\left (3 \, d x + 3 \, c\right )}}{d} - \frac {15 \, e^{\left (-d x - c\right )} - e^{\left (-3 \, d x - 3 \, c\right )}}{d} + \frac {48 \, \arctan \left (e^{\left (-d x - c\right )}\right )}{d}\right )} + a b {\left (\frac {4 \, \arctan \left (e^{\left (-d x - c\right )}\right )}{d} + \frac {e^{\left (d x + c\right )}}{d} - \frac {e^{\left (-d x - c\right )}}{d}\right )} + \frac {a^{2} \arctan \left (\sinh \left (d x + c\right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*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) + a*b*(4*arctan(e^(-d*x - c))/d + e^(d*x + c)/d - e^(-d*x - c)/d) + a^2*arctan(sinh(d*x + c))/d

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mupad [B]  time = 0.17, size = 182, normalized size = 3.31 \[ \frac {b^2\,{\mathrm {e}}^{3\,c+3\,d\,x}}{24\,d}-\frac {b^2\,{\mathrm {e}}^{-3\,c-3\,d\,x}}{24\,d}-\frac {{\mathrm {e}}^{-c-d\,x}\,\left (8\,a\,b-5\,b^2\right )}{8\,d}+\frac {2\,\mathrm {atan}\left (\frac {{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\,\left (a^2\,\sqrt {d^2}+b^2\,\sqrt {d^2}-2\,a\,b\,\sqrt {d^2}\right )}{d\,\sqrt {a^4-4\,a^3\,b+6\,a^2\,b^2-4\,a\,b^3+b^4}}\right )\,\sqrt {a^4-4\,a^3\,b+6\,a^2\,b^2-4\,a\,b^3+b^4}}{\sqrt {d^2}}+\frac {b\,{\mathrm {e}}^{c+d\,x}\,\left (8\,a-5\,b\right )}{8\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b^2*exp(3*c + 3*d*x))/(24*d) - (b^2*exp(- 3*c - 3*d*x))/(24*d) - (exp(- c - d*x)*(8*a*b - 5*b^2))/(8*d) + (2*
atan((exp(d*x)*exp(c)*(a^2*(d^2)^(1/2) + b^2*(d^2)^(1/2) - 2*a*b*(d^2)^(1/2)))/(d*(a^4 - 4*a^3*b - 4*a*b^3 + b
^4 + 6*a^2*b^2)^(1/2)))*(a^4 - 4*a^3*b - 4*a*b^3 + b^4 + 6*a^2*b^2)^(1/2))/(d^2)^(1/2) + (b*exp(c + d*x)*(8*a
- 5*b))/(8*d)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \sinh ^{2}{\left (c + d x \right )}\right )^{2} \operatorname {sech}{\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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