3.3 \(\int \sinh ^2(c+d x) (a+b \tanh ^2(c+d x)) \, dx\)

Optimal. Leaf size=44 \[ \frac{(a+b) \sinh (c+d x) \cosh (c+d x)}{2 d}-\frac{1}{2} x (a+3 b)+\frac{b \tanh (c+d x)}{d} \]

[Out]

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

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Rubi [A]  time = 0.0507793, antiderivative size = 44, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {3663, 455, 388, 206} \[ \frac{(a+b) \sinh (c+d x) \cosh (c+d x)}{2 d}-\frac{1}{2} x (a+3 b)+\frac{b \tanh (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3663

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

Rule 455

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[((-a)^(m/2 - 1)*(b*c - a*d)*
x*(a + b*x^2)^(p + 1))/(2*b^(m/2 + 1)*(p + 1)), x] + Dist[1/(2*b^(m/2 + 1)*(p + 1)), Int[(a + b*x^2)^(p + 1)*E
xpandToSum[2*b*(p + 1)*x^2*Together[(b^(m/2)*x^(m - 2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d))/(a + b*x^2)]
- (-a)^(m/2 - 1)*(b*c - a*d), x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && IGtQ[
m/2, 0] && (IntegerQ[p] || EqQ[m + 2*p + 1, 0])

Rule 388

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(d*x*(a + b*x^n)^(p + 1))/(b*(n*
(p + 1) + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(b*(n*(p + 1) + 1)), Int[(a + b*x^n)^p, x], x] /; FreeQ[{
a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && NeQ[n*(p + 1) + 1, 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 \sinh ^2(c+d x) \left (a+b \tanh ^2(c+d x)\right ) \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^2 \left (a+b x^2\right )}{\left (1-x^2\right )^2} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac{(a+b) \cosh (c+d x) \sinh (c+d x)}{2 d}-\frac{\operatorname{Subst}\left (\int \frac{a+b+2 b x^2}{1-x^2} \, dx,x,\tanh (c+d x)\right )}{2 d}\\ &=\frac{(a+b) \cosh (c+d x) \sinh (c+d x)}{2 d}+\frac{b \tanh (c+d x)}{d}-\frac{(a+3 b) \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\tanh (c+d x)\right )}{2 d}\\ &=-\frac{1}{2} (a+3 b) x+\frac{(a+b) \cosh (c+d x) \sinh (c+d x)}{2 d}+\frac{b \tanh (c+d x)}{d}\\ \end{align*}

Mathematica [A]  time = 0.189214, size = 41, normalized size = 0.93 \[ \frac{-2 (a+3 b) (c+d x)+(a+b) \sinh (2 (c+d x))+4 b \tanh (c+d x)}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(a + 3*b)*(c + d*x) + (a + b)*Sinh[2*(c + d*x)] + 4*b*Tanh[c + d*x])/(4*d)

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Maple [A]  time = 0.033, size = 66, normalized size = 1.5 \begin{align*}{\frac{1}{d} \left ( a \left ({\frac{\cosh \left ( dx+c \right ) \sinh \left ( dx+c \right ) }{2}}-{\frac{dx}{2}}-{\frac{c}{2}} \right ) +b \left ({\frac{ \left ( \sinh \left ( dx+c \right ) \right ) ^{3}}{2\,\cosh \left ( dx+c \right ) }}-{\frac{3\,dx}{2}}-{\frac{3\,c}{2}}+{\frac{3\,\tanh \left ( dx+c \right ) }{2}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.12602, size = 136, normalized size = 3.09 \begin{align*} -\frac{1}{8} \, a{\left (4 \, x - \frac{e^{\left (2 \, d x + 2 \, c\right )}}{d} + \frac{e^{\left (-2 \, d x - 2 \, c\right )}}{d}\right )} - \frac{1}{8} \, b{\left (\frac{12 \,{\left (d x + c\right )}}{d} + \frac{e^{\left (-2 \, d x - 2 \, c\right )}}{d} - \frac{17 \, e^{\left (-2 \, d x - 2 \, c\right )} + 1}{d{\left (e^{\left (-2 \, d x - 2 \, c\right )} + e^{\left (-4 \, d x - 4 \, c\right )}\right )}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*a*(4*x - e^(2*d*x + 2*c)/d + e^(-2*d*x - 2*c)/d) - 1/8*b*(12*(d*x + c)/d + e^(-2*d*x - 2*c)/d - (17*e^(-2
*d*x - 2*c) + 1)/(d*(e^(-2*d*x - 2*c) + e^(-4*d*x - 4*c))))

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Fricas [A]  time = 1.95945, size = 193, normalized size = 4.39 \begin{align*} \frac{{\left (a + b\right )} \sinh \left (d x + c\right )^{3} - 4 \,{\left ({\left (a + 3 \, b\right )} d x + 2 \, b\right )} \cosh \left (d x + c\right ) +{\left (3 \,{\left (a + b\right )} \cosh \left (d x + c\right )^{2} + a + 9 \, b\right )} \sinh \left (d x + c\right )}{8 \, d \cosh \left (d x + c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*((a + b)*sinh(d*x + c)^3 - 4*((a + 3*b)*d*x + 2*b)*cosh(d*x + c) + (3*(a + b)*cosh(d*x + c)^2 + a + 9*b)*s
inh(d*x + c))/(d*cosh(d*x + c))

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

[Out]

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

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Giac [B]  time = 1.2702, size = 147, normalized size = 3.34 \begin{align*} -\frac{4 \,{\left (a + 3 \, b\right )} d x -{\left (a e^{\left (2 \, d x + 8 \, c\right )} + b e^{\left (2 \, d x + 8 \, c\right )}\right )} e^{\left (-6 \, c\right )} - \frac{{\left (a e^{\left (4 \, d x + 4 \, c\right )} + 3 \, b e^{\left (4 \, d x + 4 \, c\right )} - 14 \, b e^{\left (2 \, d x + 2 \, c\right )} - a - b\right )} e^{\left (-2 \, c\right )}}{e^{\left (2 \, d x\right )} + e^{\left (4 \, d x + 2 \, c\right )}}}{8 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(4*(a + 3*b)*d*x - (a*e^(2*d*x + 8*c) + b*e^(2*d*x + 8*c))*e^(-6*c) - (a*e^(4*d*x + 4*c) + 3*b*e^(4*d*x +
4*c) - 14*b*e^(2*d*x + 2*c) - a - b)*e^(-2*c)/(e^(2*d*x) + e^(4*d*x + 2*c)))/d