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

Optimal. Leaf size=76 \[ \frac{b^2 \tan ^{-1}\left (\frac{\sqrt{a} \sinh (c+d x)}{\sqrt{a+b}}\right )}{a^{5/2} d \sqrt{a+b}}+\frac{(a-b) \sinh (c+d x)}{a^2 d}+\frac{\sinh ^3(c+d x)}{3 a d} \]

[Out]

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

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Rubi [A]  time = 0.09035, antiderivative size = 76, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {4147, 390, 205} \[ \frac{b^2 \tan ^{-1}\left (\frac{\sqrt{a} \sinh (c+d x)}{\sqrt{a+b}}\right )}{a^{5/2} d \sqrt{a+b}}+\frac{(a-b) \sinh (c+d x)}{a^2 d}+\frac{\sinh ^3(c+d x)}{3 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4147

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

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 205

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

Rubi steps

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

Mathematica [A]  time = 0.287139, size = 79, normalized size = 1.04 \[ \frac{a^{3/2} \sinh (3 (c+d x))-\frac{12 b^2 \tan ^{-1}\left (\frac{\sqrt{a+b} \text{csch}(c+d x)}{\sqrt{a}}\right )}{\sqrt{a+b}}+3 \sqrt{a} (3 a-4 b) \sinh (c+d x)}{12 a^{5/2} d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-12*b^2*ArcTan[(Sqrt[a + b]*Csch[c + d*x])/Sqrt[a]])/Sqrt[a + b] + 3*Sqrt[a]*(3*a - 4*b)*Sinh[c + d*x] + a^(
3/2)*Sinh[3*(c + d*x)])/(12*a^(5/2)*d)

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Maple [B]  time = 0.083, size = 256, normalized size = 3.4 \begin{align*} -{\frac{1}{3\,da} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-3}}+{\frac{1}{2\,da} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-2}}-{\frac{1}{da} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}+{\frac{b}{d{a}^{2}} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}+{\frac{{b}^{2}}{d}\arctan \left ({\frac{1}{2} \left ( 2\,\tanh \left ( 1/2\,dx+c/2 \right ) \sqrt{a+b}+2\,\sqrt{b} \right ){\frac{1}{\sqrt{a}}}} \right ){a}^{-{\frac{5}{2}}}{\frac{1}{\sqrt{a+b}}}}+{\frac{{b}^{2}}{d}\arctan \left ({\frac{1}{2} \left ( 2\,\tanh \left ( 1/2\,dx+c/2 \right ) \sqrt{a+b}-2\,\sqrt{b} \right ){\frac{1}{\sqrt{a}}}} \right ){a}^{-{\frac{5}{2}}}{\frac{1}{\sqrt{a+b}}}}-{\frac{1}{3\,da} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-3}}-{\frac{1}{2\,da} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-2}}-{\frac{1}{da} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}+{\frac{b}{d{a}^{2}} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3/d/a/(tanh(1/2*d*x+1/2*c)+1)^3+1/2/d/a/(tanh(1/2*d*x+1/2*c)+1)^2-1/d/a/(tanh(1/2*d*x+1/2*c)+1)+1/d/a^2/(ta
nh(1/2*d*x+1/2*c)+1)*b+1/d/a^(5/2)*b^2/(a+b)^(1/2)*arctan(1/2*(2*tanh(1/2*d*x+1/2*c)*(a+b)^(1/2)+2*b^(1/2))/a^
(1/2))+1/d/a^(5/2)*b^2/(a+b)^(1/2)*arctan(1/2*(2*tanh(1/2*d*x+1/2*c)*(a+b)^(1/2)-2*b^(1/2))/a^(1/2))-1/3/d/a/(
tanh(1/2*d*x+1/2*c)-1)^3-1/2/d/a/(tanh(1/2*d*x+1/2*c)-1)^2-1/d/a/(tanh(1/2*d*x+1/2*c)-1)+1/d/a^2/(tanh(1/2*d*x
+1/2*c)-1)*b

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{{\left (3 \,{\left (3 \, a e^{\left (4 \, c\right )} - 4 \, b e^{\left (4 \, c\right )}\right )} e^{\left (4 \, d x\right )} - 3 \,{\left (3 \, a e^{\left (2 \, c\right )} - 4 \, b e^{\left (2 \, c\right )}\right )} e^{\left (2 \, d x\right )} + a e^{\left (6 \, d x + 6 \, c\right )} - a\right )} e^{\left (-3 \, d x - 3 \, c\right )}}{24 \, a^{2} d} + \frac{1}{8} \, \int \frac{16 \,{\left (b^{2} e^{\left (3 \, d x + 3 \, c\right )} + b^{2} e^{\left (d x + c\right )}\right )}}{a^{3} e^{\left (4 \, d x + 4 \, c\right )} + a^{3} + 2 \,{\left (a^{3} e^{\left (2 \, c\right )} + 2 \, a^{2} b e^{\left (2 \, c\right )}\right )} e^{\left (2 \, d x\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(3*(3*a*e^(4*c) - 4*b*e^(4*c))*e^(4*d*x) - 3*(3*a*e^(2*c) - 4*b*e^(2*c))*e^(2*d*x) + a*e^(6*d*x + 6*c) -
a)*e^(-3*d*x - 3*c)/(a^2*d) + 1/8*integrate(16*(b^2*e^(3*d*x + 3*c) + b^2*e^(d*x + c))/(a^3*e^(4*d*x + 4*c) +
a^3 + 2*(a^3*e^(2*c) + 2*a^2*b*e^(2*c))*e^(2*d*x)), x)

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Fricas [B]  time = 2.79247, size = 3991, normalized size = 52.51 \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)^3/(a+b*sech(d*x+c)^2),x, algorithm="fricas")

[Out]

[1/24*((a^3 + a^2*b)*cosh(d*x + c)^6 + 6*(a^3 + a^2*b)*cosh(d*x + c)*sinh(d*x + c)^5 + (a^3 + a^2*b)*sinh(d*x
+ c)^6 + 3*(3*a^3 - a^2*b - 4*a*b^2)*cosh(d*x + c)^4 + 3*(3*a^3 - a^2*b - 4*a*b^2 + 5*(a^3 + a^2*b)*cosh(d*x +
c)^2)*sinh(d*x + c)^4 + 4*(5*(a^3 + a^2*b)*cosh(d*x + c)^3 + 3*(3*a^3 - a^2*b - 4*a*b^2)*cosh(d*x + c))*sinh(
d*x + c)^3 - a^3 - a^2*b - 3*(3*a^3 - a^2*b - 4*a*b^2)*cosh(d*x + c)^2 + 3*(5*(a^3 + a^2*b)*cosh(d*x + c)^4 -
3*a^3 + a^2*b + 4*a*b^2 + 6*(3*a^3 - a^2*b - 4*a*b^2)*cosh(d*x + c)^2)*sinh(d*x + c)^2 - 12*(b^2*cosh(d*x + c)
^3 + 3*b^2*cosh(d*x + c)^2*sinh(d*x + c) + 3*b^2*cosh(d*x + c)*sinh(d*x + c)^2 + b^2*sinh(d*x + c)^3)*sqrt(-a^
2 - a*b)*log((a*cosh(d*x + c)^4 + 4*a*cosh(d*x + c)*sinh(d*x + c)^3 + a*sinh(d*x + c)^4 - 2*(3*a + 2*b)*cosh(d
*x + c)^2 + 2*(3*a*cosh(d*x + c)^2 - 3*a - 2*b)*sinh(d*x + c)^2 + 4*(a*cosh(d*x + c)^3 - (3*a + 2*b)*cosh(d*x
+ c))*sinh(d*x + c) - 4*(cosh(d*x + c)^3 + 3*cosh(d*x + c)*sinh(d*x + c)^2 + sinh(d*x + c)^3 + (3*cosh(d*x + c
)^2 - 1)*sinh(d*x + c) - cosh(d*x + c))*sqrt(-a^2 - a*b) + a)/(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)) + 6*((a^3 + a^2*b)*cosh(d*x + c)^5 + 2*(3
*a^3 - a^2*b - 4*a*b^2)*cosh(d*x + c)^3 - (3*a^3 - a^2*b - 4*a*b^2)*cosh(d*x + c))*sinh(d*x + c))/((a^4 + a^3*
b)*d*cosh(d*x + c)^3 + 3*(a^4 + a^3*b)*d*cosh(d*x + c)^2*sinh(d*x + c) + 3*(a^4 + a^3*b)*d*cosh(d*x + c)*sinh(
d*x + c)^2 + (a^4 + a^3*b)*d*sinh(d*x + c)^3), 1/24*((a^3 + a^2*b)*cosh(d*x + c)^6 + 6*(a^3 + a^2*b)*cosh(d*x
+ c)*sinh(d*x + c)^5 + (a^3 + a^2*b)*sinh(d*x + c)^6 + 3*(3*a^3 - a^2*b - 4*a*b^2)*cosh(d*x + c)^4 + 3*(3*a^3
- a^2*b - 4*a*b^2 + 5*(a^3 + a^2*b)*cosh(d*x + c)^2)*sinh(d*x + c)^4 + 4*(5*(a^3 + a^2*b)*cosh(d*x + c)^3 + 3*
(3*a^3 - a^2*b - 4*a*b^2)*cosh(d*x + c))*sinh(d*x + c)^3 - a^3 - a^2*b - 3*(3*a^3 - a^2*b - 4*a*b^2)*cosh(d*x
+ c)^2 + 3*(5*(a^3 + a^2*b)*cosh(d*x + c)^4 - 3*a^3 + a^2*b + 4*a*b^2 + 6*(3*a^3 - a^2*b - 4*a*b^2)*cosh(d*x +
c)^2)*sinh(d*x + c)^2 + 24*(b^2*cosh(d*x + c)^3 + 3*b^2*cosh(d*x + c)^2*sinh(d*x + c) + 3*b^2*cosh(d*x + c)*s
inh(d*x + c)^2 + b^2*sinh(d*x + c)^3)*sqrt(a^2 + a*b)*arctan(1/2*(a*cosh(d*x + c)^3 + 3*a*cosh(d*x + c)*sinh(d
*x + c)^2 + a*sinh(d*x + c)^3 + (3*a + 4*b)*cosh(d*x + c) + (3*a*cosh(d*x + c)^2 + 3*a + 4*b)*sinh(d*x + c))/s
qrt(a^2 + a*b)) + 24*(b^2*cosh(d*x + c)^3 + 3*b^2*cosh(d*x + c)^2*sinh(d*x + c) + 3*b^2*cosh(d*x + c)*sinh(d*x
+ c)^2 + b^2*sinh(d*x + c)^3)*sqrt(a^2 + a*b)*arctan(1/2*sqrt(a^2 + a*b)*(cosh(d*x + c) + sinh(d*x + c))/(a +
b)) + 6*((a^3 + a^2*b)*cosh(d*x + c)^5 + 2*(3*a^3 - a^2*b - 4*a*b^2)*cosh(d*x + c)^3 - (3*a^3 - a^2*b - 4*a*b
^2)*cosh(d*x + c))*sinh(d*x + c))/((a^4 + a^3*b)*d*cosh(d*x + c)^3 + 3*(a^4 + a^3*b)*d*cosh(d*x + c)^2*sinh(d*
x + c) + 3*(a^4 + a^3*b)*d*cosh(d*x + c)*sinh(d*x + c)^2 + (a^4 + a^3*b)*d*sinh(d*x + c)^3)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cosh ^{3}{\left (c + d x \right )}}{a + b \operatorname{sech}^{2}{\left (c + d x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError