3.65 \(\int \cosh ^4(c+d x) (a+b \text{sech}^2(c+d x))^3 \, dx\)

Optimal. Leaf size=84 \[ \frac{3}{8} a x \left (a^2+4 a b+8 b^2\right )+\frac{3 a^2 (a+4 b) \sinh (c+d x) \cosh (c+d x)}{8 d}+\frac{a^3 \sinh (c+d x) \cosh ^3(c+d x)}{4 d}+\frac{b^3 \tanh (c+d x)}{d} \]

[Out]

(3*a*(a^2 + 4*a*b + 8*b^2)*x)/8 + (3*a^2*(a + 4*b)*Cosh[c + d*x]*Sinh[c + d*x])/(8*d) + (a^3*Cosh[c + d*x]^3*S
inh[c + d*x])/(4*d) + (b^3*Tanh[c + d*x])/d

________________________________________________________________________________________

Rubi [A]  time = 0.111196, antiderivative size = 84, 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 = {4146, 390, 1157, 385, 206} \[ \frac{3}{8} a x \left (a^2+4 a b+8 b^2\right )+\frac{3 a^2 (a+4 b) \sinh (c+d x) \cosh (c+d x)}{8 d}+\frac{a^3 \sinh (c+d x) \cosh ^3(c+d x)}{4 d}+\frac{b^3 \tanh (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Int[Cosh[c + d*x]^4*(a + b*Sech[c + d*x]^2)^3,x]

[Out]

(3*a*(a^2 + 4*a*b + 8*b^2)*x)/8 + (3*a^2*(a + 4*b)*Cosh[c + d*x]*Sinh[c + d*x])/(8*d) + (a^3*Cosh[c + d*x]^3*S
inh[c + d*x])/(4*d) + (b^3*Tanh[c + d*x])/d

Rule 4146

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = Fre
eFactors[Tan[e + f*x], x]}, Dist[ff/f, Subst[Int[(1 + ff^2*x^2)^(m/2 - 1)*ExpandToSum[a + b*(1 + ff^2*x^2)^(n/
2), x]^p, x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[m/2] && IntegerQ[n/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 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 \cosh ^4(c+d x) \left (a+b \text{sech}^2(c+d x)\right )^3 \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (a+b-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 (b^3+\frac{a \left (a^2+3 a b+3 b^2\right )-3 a b (a+2 b) x^2+3 a b^2 x^4}{\left (1-x^2\right )^3}\right ) \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac{b^3 \tanh (c+d x)}{d}+\frac{\operatorname{Subst}\left (\int \frac{a \left (a^2+3 a b+3 b^2\right )-3 a b (a+2 b) x^2+3 a b^2 x^4}{\left (1-x^2\right )^3} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac{a^3 \cosh ^3(c+d x) \sinh (c+d x)}{4 d}+\frac{b^3 \tanh (c+d x)}{d}-\frac{\operatorname{Subst}\left (\int \frac{-3 a (a+2 b)^2+12 a b^2 x^2}{\left (1-x^2\right )^2} \, dx,x,\tanh (c+d x)\right )}{4 d}\\ &=\frac{3 a^2 (a+4 b) \cosh (c+d x) \sinh (c+d x)}{8 d}+\frac{a^3 \cosh ^3(c+d x) \sinh (c+d x)}{4 d}+\frac{b^3 \tanh (c+d x)}{d}+\frac{\left (3 a \left (a^2+4 a b+8 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} a \left (a^2+4 a b+8 b^2\right ) x+\frac{3 a^2 (a+4 b) \cosh (c+d x) \sinh (c+d x)}{8 d}+\frac{a^3 \cosh ^3(c+d x) \sinh (c+d x)}{4 d}+\frac{b^3 \tanh (c+d x)}{d}\\ \end{align*}

Mathematica [A]  time = 0.408999, size = 70, normalized size = 0.83 \[ \frac{12 a \left (a^2+4 a b+8 b^2\right ) (c+d x)+8 a^2 (a+3 b) \sinh (2 (c+d x))+a^3 \sinh (4 (c+d x))+32 b^3 \tanh (c+d x)}{32 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cosh[c + d*x]^4*(a + b*Sech[c + d*x]^2)^3,x]

[Out]

(12*a*(a^2 + 4*a*b + 8*b^2)*(c + d*x) + 8*a^2*(a + 3*b)*Sinh[2*(c + d*x)] + a^3*Sinh[4*(c + d*x)] + 32*b^3*Tan
h[c + d*x])/(32*d)

________________________________________________________________________________________

Maple [A]  time = 0.038, size = 93, normalized size = 1.1 \begin{align*}{\frac{1}{d} \left ({a}^{3} \left ( \left ({\frac{ \left ( \cosh \left ( dx+c \right ) \right ) ^{3}}{4}}+{\frac{3\,\cosh \left ( dx+c \right ) }{8}} \right ) \sinh \left ( dx+c \right ) +{\frac{3\,dx}{8}}+{\frac{3\,c}{8}} \right ) +3\,{a}^{2}b \left ( 1/2\,\cosh \left ( dx+c \right ) \sinh \left ( dx+c \right ) +1/2\,dx+c/2 \right ) +3\,a{b}^{2} \left ( dx+c \right ) +{b}^{3}\tanh \left ( dx+c \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(d*x+c)^4*(a+b*sech(d*x+c)^2)^3,x)

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.10589, size = 176, normalized size = 2.1 \begin{align*} \frac{1}{64} \, a^{3}{\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{3}{8} \, a^{2} b{\left (4 \, x + \frac{e^{\left (2 \, d x + 2 \, c\right )}}{d} - \frac{e^{\left (-2 \, d x - 2 \, c\right )}}{d}\right )} + 3 \, a b^{2} x + \frac{2 \, b^{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(cosh(d*x+c)^4*(a+b*sech(d*x+c)^2)^3,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 2.08693, size = 370, normalized size = 4.4 \begin{align*} \frac{a^{3} \sinh \left (d x + c\right )^{5} +{\left (10 \, a^{3} \cosh \left (d x + c\right )^{2} + 9 \, a^{3} + 24 \, a^{2} b\right )} \sinh \left (d x + c\right )^{3} - 8 \,{\left (8 \, b^{3} - 3 \,{\left (a^{3} + 4 \, a^{2} b + 8 \, a b^{2}\right )} d x\right )} \cosh \left (d x + c\right ) +{\left (5 \, a^{3} \cosh \left (d x + c\right )^{4} + 8 \, a^{3} + 24 \, a^{2} b + 64 \, b^{3} + 9 \,{\left (3 \, a^{3} + 8 \, a^{2} b\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(cosh(d*x+c)^4*(a+b*sech(d*x+c)^2)^3,x, algorithm="fricas")

[Out]

1/64*(a^3*sinh(d*x + c)^5 + (10*a^3*cosh(d*x + c)^2 + 9*a^3 + 24*a^2*b)*sinh(d*x + c)^3 - 8*(8*b^3 - 3*(a^3 +
4*a^2*b + 8*a*b^2)*d*x)*cosh(d*x + c) + (5*a^3*cosh(d*x + c)^4 + 8*a^3 + 24*a^2*b + 64*b^3 + 9*(3*a^3 + 8*a^2*
b)*cosh(d*x + c)^2)*sinh(d*x + c))/(d*cosh(d*x + c))

________________________________________________________________________________________

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(cosh(d*x+c)**4*(a+b*sech(d*x+c)**2)**3,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [B]  time = 1.19182, size = 257, normalized size = 3.06 \begin{align*} \frac{3 \,{\left (a^{3} + 4 \, a^{2} b + 8 \, a b^{2}\right )}{\left (d x + c\right )}}{8 \, d} - \frac{2 \, b^{3}}{d{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}} - \frac{{\left (18 \, a^{3} e^{\left (4 \, d x + 4 \, c\right )} + 72 \, a^{2} b e^{\left (4 \, d x + 4 \, c\right )} + 144 \, a b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 8 \, a^{3} e^{\left (2 \, d x + 2 \, c\right )} + 24 \, a^{2} b e^{\left (2 \, d x + 2 \, c\right )} + a^{3}\right )} e^{\left (-4 \, d x - 4 \, c\right )}}{64 \, d} + \frac{a^{3} d e^{\left (4 \, d x + 4 \, c\right )} + 8 \, a^{3} d e^{\left (2 \, d x + 2 \, c\right )} + 24 \, a^{2} b d e^{\left (2 \, d x + 2 \, c\right )}}{64 \, d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(d*x+c)^4*(a+b*sech(d*x+c)^2)^3,x, algorithm="giac")

[Out]

3/8*(a^3 + 4*a^2*b + 8*a*b^2)*(d*x + c)/d - 2*b^3/(d*(e^(2*d*x + 2*c) + 1)) - 1/64*(18*a^3*e^(4*d*x + 4*c) + 7
2*a^2*b*e^(4*d*x + 4*c) + 144*a*b^2*e^(4*d*x + 4*c) + 8*a^3*e^(2*d*x + 2*c) + 24*a^2*b*e^(2*d*x + 2*c) + a^3)*
e^(-4*d*x - 4*c)/d + 1/64*(a^3*d*e^(4*d*x + 4*c) + 8*a^3*d*e^(2*d*x + 2*c) + 24*a^2*b*d*e^(2*d*x + 2*c))/d^2