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

Optimal. Leaf size=117 \[ -\frac{b^{5/2} \tanh ^{-1}\left (\frac{\sqrt{b} \tanh (c+d x)}{\sqrt{a+b}}\right )}{a^3 d \sqrt{a+b}}+\frac{x \left (3 a^2-4 a b+8 b^2\right )}{8 a^3}+\frac{(3 a-4 b) \sinh (c+d x) \cosh (c+d x)}{8 a^2 d}+\frac{\sinh (c+d x) \cosh ^3(c+d x)}{4 a d} \]

[Out]

((3*a^2 - 4*a*b + 8*b^2)*x)/(8*a^3) - (b^(5/2)*ArcTanh[(Sqrt[b]*Tanh[c + d*x])/Sqrt[a + b]])/(a^3*Sqrt[a + b]*
d) + ((3*a - 4*b)*Cosh[c + d*x]*Sinh[c + d*x])/(8*a^2*d) + (Cosh[c + d*x]^3*Sinh[c + d*x])/(4*a*d)

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Rubi [A]  time = 0.189282, antiderivative size = 117, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {4146, 414, 527, 522, 206, 208} \[ -\frac{b^{5/2} \tanh ^{-1}\left (\frac{\sqrt{b} \tanh (c+d x)}{\sqrt{a+b}}\right )}{a^3 d \sqrt{a+b}}+\frac{x \left (3 a^2-4 a b+8 b^2\right )}{8 a^3}+\frac{(3 a-4 b) \sinh (c+d x) \cosh (c+d x)}{8 a^2 d}+\frac{\sinh (c+d x) \cosh ^3(c+d x)}{4 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((3*a^2 - 4*a*b + 8*b^2)*x)/(8*a^3) - (b^(5/2)*ArcTanh[(Sqrt[b]*Tanh[c + d*x])/Sqrt[a + b]])/(a^3*Sqrt[a + b]*
d) + ((3*a - 4*b)*Cosh[c + d*x]*Sinh[c + d*x])/(8*a^2*d) + (Cosh[c + d*x]^3*Sinh[c + d*x])/(4*a*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 414

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(b*x*(a + b*x^n)^(p + 1)*(
c + d*x^n)^(q + 1))/(a*n*(p + 1)*(b*c - a*d)), x] + Dist[1/(a*n*(p + 1)*(b*c - a*d)), Int[(a + b*x^n)^(p + 1)*
(c + d*x^n)^q*Simp[b*c + n*(p + 1)*(b*c - a*d) + d*b*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d,
n, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &&  !( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomial
Q[a, b, c, d, n, p, q, x]

Rule 527

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> -Simp[
((b*e - a*f)*x*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*n*(b*c - a*d)*(p + 1)), x] + Dist[1/(a*n*(b*c - a*d
)*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(b*e - a*f) + e*n*(b*c - a*d)*(p + 1) + d*(b*e - a*f)
*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, q}, x] && LtQ[p, -1]

Rule 522

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[(b*e - a*f
)/(b*c - a*d), Int[1/(a + b*x^n), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[1/(c + d*x^n), x], x] /; FreeQ[{a
, b, c, d, e, f, n}, x]

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])

Rule 208

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

Rubi steps

\begin{align*} \int \frac{\cosh ^4(c+d x)}{a+b \text{sech}^2(c+d x)} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{\left (1-x^2\right )^3 \left (a+b-b x^2\right )} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac{\cosh ^3(c+d x) \sinh (c+d x)}{4 a d}+\frac{\operatorname{Subst}\left (\int \frac{3 a-b-3 b x^2}{\left (1-x^2\right )^2 \left (a+b-b x^2\right )} \, dx,x,\tanh (c+d x)\right )}{4 a d}\\ &=\frac{(3 a-4 b) \cosh (c+d x) \sinh (c+d x)}{8 a^2 d}+\frac{\cosh ^3(c+d x) \sinh (c+d x)}{4 a d}+\frac{\operatorname{Subst}\left (\int \frac{3 a^2-a b+4 b^2-(3 a-4 b) b x^2}{\left (1-x^2\right ) \left (a+b-b x^2\right )} \, dx,x,\tanh (c+d x)\right )}{8 a^2 d}\\ &=\frac{(3 a-4 b) \cosh (c+d x) \sinh (c+d x)}{8 a^2 d}+\frac{\cosh ^3(c+d x) \sinh (c+d x)}{4 a d}-\frac{b^3 \operatorname{Subst}\left (\int \frac{1}{a+b-b x^2} \, dx,x,\tanh (c+d x)\right )}{a^3 d}+\frac{\left (3 a^2-4 a b+8 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\tanh (c+d x)\right )}{8 a^3 d}\\ &=\frac{\left (3 a^2-4 a b+8 b^2\right ) x}{8 a^3}-\frac{b^{5/2} \tanh ^{-1}\left (\frac{\sqrt{b} \tanh (c+d x)}{\sqrt{a+b}}\right )}{a^3 \sqrt{a+b} d}+\frac{(3 a-4 b) \cosh (c+d x) \sinh (c+d x)}{8 a^2 d}+\frac{\cosh ^3(c+d x) \sinh (c+d x)}{4 a d}\\ \end{align*}

Mathematica [A]  time = 0.521615, size = 95, normalized size = 0.81 \[ \frac{4 \left (3 a^2-4 a b+8 b^2\right ) (c+d x)+a^2 \sinh (4 (c+d x))-\frac{32 b^{5/2} \tanh ^{-1}\left (\frac{\sqrt{b} \tanh (c+d x)}{\sqrt{a+b}}\right )}{\sqrt{a+b}}+8 a (a-b) \sinh (2 (c+d x))}{32 a^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*(3*a^2 - 4*a*b + 8*b^2)*(c + d*x) - (32*b^(5/2)*ArcTanh[(Sqrt[b]*Tanh[c + d*x])/Sqrt[a + b]])/Sqrt[a + b] +
8*a*(a - b)*Sinh[2*(c + d*x)] + a^2*Sinh[4*(c + d*x)])/(32*a^3*d)

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Maple [B]  time = 0.084, size = 493, normalized size = 4.2 \begin{align*} \text{result too large to display} \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),x)

[Out]

-1/4/d/a/(tanh(1/2*d*x+1/2*c)+1)^4+1/2/d/a/(tanh(1/2*d*x+1/2*c)+1)^3+5/8/d/a/(tanh(1/2*d*x+1/2*c)+1)-1/2/d/a^2
/(tanh(1/2*d*x+1/2*c)+1)*b-7/8/d/a/(tanh(1/2*d*x+1/2*c)+1)^2+1/2/d/a^2/(tanh(1/2*d*x+1/2*c)+1)^2*b+3/8/d/a*ln(
tanh(1/2*d*x+1/2*c)+1)-1/2/d/a^2*ln(tanh(1/2*d*x+1/2*c)+1)*b+1/d/a^3*ln(tanh(1/2*d*x+1/2*c)+1)*b^2-1/2/d*b^(5/
2)/a^3/(a+b)^(1/2)*ln((a+b)^(1/2)*tanh(1/2*d*x+1/2*c)^2+2*tanh(1/2*d*x+1/2*c)*b^(1/2)+(a+b)^(1/2))+1/2/d*b^(5/
2)/a^3/(a+b)^(1/2)*ln((a+b)^(1/2)*tanh(1/2*d*x+1/2*c)^2-2*tanh(1/2*d*x+1/2*c)*b^(1/2)+(a+b)^(1/2))+1/4/d/a/(ta
nh(1/2*d*x+1/2*c)-1)^4+1/2/d/a/(tanh(1/2*d*x+1/2*c)-1)^3+7/8/d/a/(tanh(1/2*d*x+1/2*c)-1)^2-1/2/d/a^2/(tanh(1/2
*d*x+1/2*c)-1)^2*b+5/8/d/a/(tanh(1/2*d*x+1/2*c)-1)-1/2/d/a^2/(tanh(1/2*d*x+1/2*c)-1)*b-3/8/d/a*ln(tanh(1/2*d*x
+1/2*c)-1)+1/2/d/a^2*ln(tanh(1/2*d*x+1/2*c)-1)*b-1/d/a^3*ln(tanh(1/2*d*x+1/2*c)-1)*b^2

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \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),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.75175, size = 4313, normalized size = 36.86 \begin{align*} \text{result too large to display} \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),x, algorithm="fricas")

[Out]

[1/64*(a^2*cosh(d*x + c)^8 + 8*a^2*cosh(d*x + c)*sinh(d*x + c)^7 + a^2*sinh(d*x + c)^8 + 8*(3*a^2 - 4*a*b + 8*
b^2)*d*x*cosh(d*x + c)^4 + 8*(a^2 - a*b)*cosh(d*x + c)^6 + 4*(7*a^2*cosh(d*x + c)^2 + 2*a^2 - 2*a*b)*sinh(d*x
+ c)^6 + 8*(7*a^2*cosh(d*x + c)^3 + 6*(a^2 - a*b)*cosh(d*x + c))*sinh(d*x + c)^5 + 2*(35*a^2*cosh(d*x + c)^4 +
4*(3*a^2 - 4*a*b + 8*b^2)*d*x + 60*(a^2 - a*b)*cosh(d*x + c)^2)*sinh(d*x + c)^4 + 8*(7*a^2*cosh(d*x + c)^5 +
4*(3*a^2 - 4*a*b + 8*b^2)*d*x*cosh(d*x + c) + 20*(a^2 - a*b)*cosh(d*x + c)^3)*sinh(d*x + c)^3 - 8*(a^2 - a*b)*
cosh(d*x + c)^2 + 4*(7*a^2*cosh(d*x + c)^6 + 12*(3*a^2 - 4*a*b + 8*b^2)*d*x*cosh(d*x + c)^2 + 30*(a^2 - a*b)*c
osh(d*x + c)^4 - 2*a^2 + 2*a*b)*sinh(d*x + c)^2 + 32*(b^2*cosh(d*x + c)^4 + 4*b^2*cosh(d*x + c)^3*sinh(d*x + c
) + 6*b^2*cosh(d*x + c)^2*sinh(d*x + c)^2 + 4*b^2*cosh(d*x + c)*sinh(d*x + c)^3 + b^2*sinh(d*x + c)^4)*sqrt(b/
(a + b))*log((a^2*cosh(d*x + c)^4 + 4*a^2*cosh(d*x + c)*sinh(d*x + c)^3 + a^2*sinh(d*x + c)^4 + 2*(a^2 + 2*a*b
)*cosh(d*x + c)^2 + 2*(3*a^2*cosh(d*x + c)^2 + a^2 + 2*a*b)*sinh(d*x + c)^2 + a^2 + 8*a*b + 8*b^2 + 4*(a^2*cos
h(d*x + c)^3 + (a^2 + 2*a*b)*cosh(d*x + c))*sinh(d*x + c) + 4*((a^2 + a*b)*cosh(d*x + c)^2 + 2*(a^2 + a*b)*cos
h(d*x + c)*sinh(d*x + c) + (a^2 + a*b)*sinh(d*x + c)^2 + a^2 + 3*a*b + 2*b^2)*sqrt(b/(a + b)))/(a*cosh(d*x + c
)^4 + 4*a*cosh(d*x + c)*sinh(d*x + c)^3 + a*sinh(d*x + c)^4 + 2*(a + 2*b)*cosh(d*x + c)^2 + 2*(3*a*cosh(d*x +
c)^2 + a + 2*b)*sinh(d*x + c)^2 + 4*(a*cosh(d*x + c)^3 + (a + 2*b)*cosh(d*x + c))*sinh(d*x + c) + a)) - a^2 +
8*(a^2*cosh(d*x + c)^7 + 4*(3*a^2 - 4*a*b + 8*b^2)*d*x*cosh(d*x + c)^3 + 6*(a^2 - a*b)*cosh(d*x + c)^5 - 2*(a^
2 - a*b)*cosh(d*x + c))*sinh(d*x + c))/(a^3*d*cosh(d*x + c)^4 + 4*a^3*d*cosh(d*x + c)^3*sinh(d*x + c) + 6*a^3*
d*cosh(d*x + c)^2*sinh(d*x + c)^2 + 4*a^3*d*cosh(d*x + c)*sinh(d*x + c)^3 + a^3*d*sinh(d*x + c)^4), 1/64*(a^2*
cosh(d*x + c)^8 + 8*a^2*cosh(d*x + c)*sinh(d*x + c)^7 + a^2*sinh(d*x + c)^8 + 8*(3*a^2 - 4*a*b + 8*b^2)*d*x*co
sh(d*x + c)^4 + 8*(a^2 - a*b)*cosh(d*x + c)^6 + 4*(7*a^2*cosh(d*x + c)^2 + 2*a^2 - 2*a*b)*sinh(d*x + c)^6 + 8*
(7*a^2*cosh(d*x + c)^3 + 6*(a^2 - a*b)*cosh(d*x + c))*sinh(d*x + c)^5 + 2*(35*a^2*cosh(d*x + c)^4 + 4*(3*a^2 -
4*a*b + 8*b^2)*d*x + 60*(a^2 - a*b)*cosh(d*x + c)^2)*sinh(d*x + c)^4 + 8*(7*a^2*cosh(d*x + c)^5 + 4*(3*a^2 -
4*a*b + 8*b^2)*d*x*cosh(d*x + c) + 20*(a^2 - a*b)*cosh(d*x + c)^3)*sinh(d*x + c)^3 - 8*(a^2 - a*b)*cosh(d*x +
c)^2 + 4*(7*a^2*cosh(d*x + c)^6 + 12*(3*a^2 - 4*a*b + 8*b^2)*d*x*cosh(d*x + c)^2 + 30*(a^2 - a*b)*cosh(d*x + c
)^4 - 2*a^2 + 2*a*b)*sinh(d*x + c)^2 - 64*(b^2*cosh(d*x + c)^4 + 4*b^2*cosh(d*x + c)^3*sinh(d*x + c) + 6*b^2*c
osh(d*x + c)^2*sinh(d*x + c)^2 + 4*b^2*cosh(d*x + c)*sinh(d*x + c)^3 + b^2*sinh(d*x + c)^4)*sqrt(-b/(a + b))*a
rctan(1/2*(a*cosh(d*x + c)^2 + 2*a*cosh(d*x + c)*sinh(d*x + c) + a*sinh(d*x + c)^2 + a + 2*b)*sqrt(-b/(a + b))
/b) - a^2 + 8*(a^2*cosh(d*x + c)^7 + 4*(3*a^2 - 4*a*b + 8*b^2)*d*x*cosh(d*x + c)^3 + 6*(a^2 - a*b)*cosh(d*x +
c)^5 - 2*(a^2 - a*b)*cosh(d*x + c))*sinh(d*x + c))/(a^3*d*cosh(d*x + c)^4 + 4*a^3*d*cosh(d*x + c)^3*sinh(d*x +
c) + 6*a^3*d*cosh(d*x + c)^2*sinh(d*x + c)^2 + 4*a^3*d*cosh(d*x + c)*sinh(d*x + c)^3 + a^3*d*sinh(d*x + c)^4)
]

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

[Out]

Timed out

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Giac [B]  time = 1.18464, size = 296, normalized size = 2.53 \begin{align*} -\frac{b^{3} \arctan \left (\frac{a e^{\left (2 \, d x + 2 \, c\right )} + a + 2 \, b}{2 \, \sqrt{-a b - b^{2}}}\right )}{\sqrt{-a b - b^{2}} a^{3} d} + \frac{{\left (3 \, a^{2} - 4 \, a b + 8 \, b^{2}\right )}{\left (d x + c\right )}}{8 \, a^{3} d} - \frac{{\left (18 \, a^{2} e^{\left (4 \, d x + 4 \, c\right )} - 24 \, a b e^{\left (4 \, d x + 4 \, c\right )} + 48 \, b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 8 \, a^{2} e^{\left (2 \, d x + 2 \, c\right )} - 8 \, a b e^{\left (2 \, d x + 2 \, c\right )} + a^{2}\right )} e^{\left (-4 \, d x - 4 \, c\right )}}{64 \, a^{3} d} + \frac{a d e^{\left (4 \, d x + 4 \, c\right )} + 8 \, a d e^{\left (2 \, d x + 2 \, c\right )} - 8 \, b d e^{\left (2 \, d x + 2 \, c\right )}}{64 \, a^{2} 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),x, algorithm="giac")

[Out]

-b^3*arctan(1/2*(a*e^(2*d*x + 2*c) + a + 2*b)/sqrt(-a*b - b^2))/(sqrt(-a*b - b^2)*a^3*d) + 1/8*(3*a^2 - 4*a*b
+ 8*b^2)*(d*x + c)/(a^3*d) - 1/64*(18*a^2*e^(4*d*x + 4*c) - 24*a*b*e^(4*d*x + 4*c) + 48*b^2*e^(4*d*x + 4*c) +
8*a^2*e^(2*d*x + 2*c) - 8*a*b*e^(2*d*x + 2*c) + a^2)*e^(-4*d*x - 4*c)/(a^3*d) + 1/64*(a*d*e^(4*d*x + 4*c) + 8*
a*d*e^(2*d*x + 2*c) - 8*b*d*e^(2*d*x + 2*c))/(a^2*d^2)