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

Optimal. Leaf size=71 \[ -\frac{(a+b) \cosh (c+d x)}{a^2 d}+\frac{\sqrt{b} (a+b) \tan ^{-1}\left (\frac{\sqrt{a} \cosh (c+d x)}{\sqrt{b}}\right )}{a^{5/2} d}+\frac{\cosh ^3(c+d x)}{3 a d} \]

[Out]

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

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Rubi [A]  time = 0.10468, antiderivative size = 71, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {4133, 459, 321, 205} \[ -\frac{(a+b) \cosh (c+d x)}{a^2 d}+\frac{\sqrt{b} (a+b) \tan ^{-1}\left (\frac{\sqrt{a} \cosh (c+d x)}{\sqrt{b}}\right )}{a^{5/2} d}+\frac{\cosh ^3(c+d x)}{3 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4133

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

Rule 459

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(d*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(b*e*(m + n*(p + 1) + 1)), x] - Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(b*(m +
n*(p + 1) + 1)), Int[(e*x)^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0]
&& NeQ[m + n*(p + 1) + 1, 0]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

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

Mathematica [C]  time = 2.14591, size = 372, normalized size = 5.24 \[ \frac{(a \cosh (2 (c+d x))+a+2 b) \left (3 \left (a^2+8 a b+8 b^2\right ) \tan ^{-1}\left (\frac{\sinh (c) \tanh \left (\frac{d x}{2}\right ) \left (\sqrt{a}-i \sqrt{a+b} \sqrt{(\cosh (c)-\sinh (c))^2}\right )+\cosh (c) \left (\sqrt{a}-i \sqrt{a+b} \sqrt{(\cosh (c)-\sinh (c))^2} \tanh \left (\frac{d x}{2}\right )\right )}{\sqrt{b}}\right )+3 \left (a^2+8 a b+8 b^2\right ) \tan ^{-1}\left (\frac{\sinh (c) \tanh \left (\frac{d x}{2}\right ) \left (\sqrt{a}+i \sqrt{a+b} \sqrt{(\cosh (c)-\sinh (c))^2}\right )+\cosh (c) \left (\sqrt{a}+i \sqrt{a+b} \sqrt{(\cosh (c)-\sinh (c))^2} \tanh \left (\frac{d x}{2}\right )\right )}{\sqrt{b}}\right )+2 a^{3/2} \sqrt{b} \cosh (3 (c+d x))-3 a^2 \left (\tan ^{-1}\left (\frac{\sqrt{a}-i \sqrt{a+b} \tanh \left (\frac{1}{2} (c+d x)\right )}{\sqrt{b}}\right )+\tan ^{-1}\left (\frac{\sqrt{a}+i \sqrt{a+b} \tanh \left (\frac{1}{2} (c+d x)\right )}{\sqrt{b}}\right )\right )-6 \sqrt{a} \sqrt{b} (3 a+4 b) \cosh (c+d x)\right )}{48 a^{5/2} \sqrt{b} d \left (a \cosh ^2(c+d x)+b\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + 2*b + a*Cosh[2*(c + d*x)])*(3*(a^2 + 8*a*b + 8*b^2)*ArcTan[((Sqrt[a] - I*Sqrt[a + b]*Sqrt[(Cosh[c] - Sin
h[c])^2])*Sinh[c]*Tanh[(d*x)/2] + Cosh[c]*(Sqrt[a] - I*Sqrt[a + b]*Sqrt[(Cosh[c] - Sinh[c])^2]*Tanh[(d*x)/2]))
/Sqrt[b]] + 3*(a^2 + 8*a*b + 8*b^2)*ArcTan[((Sqrt[a] + I*Sqrt[a + b]*Sqrt[(Cosh[c] - Sinh[c])^2])*Sinh[c]*Tanh
[(d*x)/2] + Cosh[c]*(Sqrt[a] + I*Sqrt[a + b]*Sqrt[(Cosh[c] - Sinh[c])^2]*Tanh[(d*x)/2]))/Sqrt[b]] - 3*a^2*(Arc
Tan[(Sqrt[a] - I*Sqrt[a + b]*Tanh[(c + d*x)/2])/Sqrt[b]] + ArcTan[(Sqrt[a] + I*Sqrt[a + b]*Tanh[(c + d*x)/2])/
Sqrt[b]]) - 6*Sqrt[a]*Sqrt[b]*(3*a + 4*b)*Cosh[c + d*x] + 2*a^(3/2)*Sqrt[b]*Cosh[3*(c + d*x)]))/(48*a^(5/2)*Sq
rt[b]*d*(b + a*Cosh[c + d*x]^2))

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Maple [B]  time = 0.061, size = 261, normalized size = 3.7 \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}{2\,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}{da}\arctan \left ({\frac{1}{4} \left ( 2\, \left ( a+b \right ) \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+2\,a-2\,b \right ){\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}+{\frac{{b}^{2}}{d{a}^{2}}\arctan \left ({\frac{1}{4} \left ( 2\, \left ( a+b \right ) \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+2\,a-2\,b \right ){\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}-{\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}{2\,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(sinh(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)^2-1/2/d/a/(tanh(1/2*d*x+1/2*c)+1)-1/d/a^2/(t
anh(1/2*d*x+1/2*c)+1)*b+1/d*b/a/(a*b)^(1/2)*arctan(1/4*(2*(a+b)*tanh(1/2*d*x+1/2*c)^2+2*a-2*b)/(a*b)^(1/2))+1/
d*b^2/a^2/(a*b)^(1/2)*arctan(1/4*(2*(a+b)*tanh(1/2*d*x+1/2*c)^2+2*a-2*b)/(a*b)^(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+1/2/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 ({\left (a b e^{\left (3 \, c\right )} + b^{2} e^{\left (3 \, c\right )}\right )} e^{\left (3 \, d x\right )} -{\left (a b e^{c} + b^{2} e^{c}\right )} e^{\left (d x\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(sinh(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*((a*b*e^(3*c) + b^2*e^(3*c))*e^(3*d*x) - (a*b*e^c + b^2*e^c)*e
^(d*x))/(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 = 2.95373, size = 3343, normalized size = 47.08 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/24*(a*cosh(d*x + c)^6 + 6*a*cosh(d*x + c)*sinh(d*x + c)^5 + a*sinh(d*x + c)^6 - 3*(3*a + 4*b)*cosh(d*x + c)
^4 + 3*(5*a*cosh(d*x + c)^2 - 3*a - 4*b)*sinh(d*x + c)^4 + 4*(5*a*cosh(d*x + c)^3 - 3*(3*a + 4*b)*cosh(d*x + c
))*sinh(d*x + c)^3 - 3*(3*a + 4*b)*cosh(d*x + c)^2 + 3*(5*a*cosh(d*x + c)^4 - 6*(3*a + 4*b)*cosh(d*x + c)^2 -
3*a - 4*b)*sinh(d*x + c)^2 + 12*((a + b)*cosh(d*x + c)^3 + 3*(a + b)*cosh(d*x + c)^2*sinh(d*x + c) + 3*(a + b)
*cosh(d*x + c)*sinh(d*x + c)^2 + (a + b)*sinh(d*x + c)^3)*sqrt(-b/a)*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*(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) + 4*(a*cosh(d*x + c)^3 + 3*a*cosh(d*
x + c)*sinh(d*x + c)^2 + a*sinh(d*x + c)^3 + a*cosh(d*x + c) + (3*a*cosh(d*x + c)^2 + a)*sinh(d*x + c))*sqrt(-
b/a) + 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))*sin
h(d*x + c) + a)) + 6*(a*cosh(d*x + c)^5 - 2*(3*a + 4*b)*cosh(d*x + c)^3 - (3*a + 4*b)*cosh(d*x + c))*sinh(d*x
+ c) + a)/(a^2*d*cosh(d*x + c)^3 + 3*a^2*d*cosh(d*x + c)^2*sinh(d*x + c) + 3*a^2*d*cosh(d*x + c)*sinh(d*x + c)
^2 + a^2*d*sinh(d*x + c)^3), 1/24*(a*cosh(d*x + c)^6 + 6*a*cosh(d*x + c)*sinh(d*x + c)^5 + a*sinh(d*x + c)^6 -
3*(3*a + 4*b)*cosh(d*x + c)^4 + 3*(5*a*cosh(d*x + c)^2 - 3*a - 4*b)*sinh(d*x + c)^4 + 4*(5*a*cosh(d*x + c)^3
- 3*(3*a + 4*b)*cosh(d*x + c))*sinh(d*x + c)^3 - 3*(3*a + 4*b)*cosh(d*x + c)^2 + 3*(5*a*cosh(d*x + c)^4 - 6*(3
*a + 4*b)*cosh(d*x + c)^2 - 3*a - 4*b)*sinh(d*x + c)^2 - 24*((a + b)*cosh(d*x + c)^3 + 3*(a + b)*cosh(d*x + c)
^2*sinh(d*x + c) + 3*(a + b)*cosh(d*x + c)*sinh(d*x + c)^2 + (a + b)*sinh(d*x + c)^3)*sqrt(b/a)*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 + (a + 4*b)*cosh(d*x + c) + (3*a*cosh(
d*x + c)^2 + a + 4*b)*sinh(d*x + c))*sqrt(b/a)/b) + 24*((a + b)*cosh(d*x + c)^3 + 3*(a + b)*cosh(d*x + c)^2*si
nh(d*x + c) + 3*(a + b)*cosh(d*x + c)*sinh(d*x + c)^2 + (a + b)*sinh(d*x + c)^3)*sqrt(b/a)*arctan(1/2*(a*cosh(
d*x + c) + a*sinh(d*x + c))*sqrt(b/a)/b) + 6*(a*cosh(d*x + c)^5 - 2*(3*a + 4*b)*cosh(d*x + c)^3 - (3*a + 4*b)*
cosh(d*x + c))*sinh(d*x + c) + a)/(a^2*d*cosh(d*x + c)^3 + 3*a^2*d*cosh(d*x + c)^2*sinh(d*x + c) + 3*a^2*d*cos
h(d*x + c)*sinh(d*x + c)^2 + a^2*d*sinh(d*x + c)^3)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sinh ^{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(sinh(d*x+c)**3/(a+b*sech(d*x+c)**2),x)

[Out]

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

[Out]

Exception raised: TypeError