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]

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

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Rubi [A]  time = 0.05, antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, 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 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 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 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 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]

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*}

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Mathematica [A]  time = 0.21, 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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fricas [A]  time = 0.65, size = 71, normalized size = 1.61 \[ \frac {{\left (a + b\right )} \sinh \left (d x + c\right )^{3} - 4 \, {\left ({\left (a + 3 \, b\right )} d x + 2 \, b\right )} \cosh \left (d x + c\right ) + {\left (3 \, {\left (a + b\right )} \cosh \left (d x + c\right )^{2} + a + 9 \, b\right )} \sinh \left (d x + c\right )}{8 \, d \cosh \left (d x + c\right )} \]

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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giac [B]  time = 0.18, size = 109, normalized size = 2.48 \[ -\frac {4 \, {\left (a + 3 \, b\right )} d x - {\left (a e^{\left (2 \, d x + 8 \, c\right )} + b e^{\left (2 \, d x + 8 \, c\right )}\right )} e^{\left (-6 \, c\right )} - \frac {{\left (a e^{\left (4 \, d x + 4 \, c\right )} + 3 \, b e^{\left (4 \, d x + 4 \, c\right )} - 14 \, b e^{\left (2 \, d x + 2 \, c\right )} - a - b\right )} e^{\left (-2 \, c\right )}}{e^{\left (2 \, d x\right )} + e^{\left (4 \, d x + 2 \, c\right )}}}{8 \, d} \]

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

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maple [A]  time = 0.18, size = 66, normalized size = 1.50 \[ \frac {a \left (\frac {\cosh \left (d x +c \right ) \sinh \left (d x +c \right )}{2}-\frac {d x}{2}-\frac {c}{2}\right )+b \left (\frac {\sinh ^{3}\left (d x +c \right )}{2 \cosh \left (d x +c \right )}-\frac {3 d x}{2}-\frac {3 c}{2}+\frac {3 \tanh \left (d x +c \right )}{2}\right )}{d} \]

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 = 1.38, size = 101, normalized size = 2.30 \[ -\frac {1}{8} \, a {\left (4 \, x - \frac {e^{\left (2 \, d x + 2 \, c\right )}}{d} + \frac {e^{\left (-2 \, d x - 2 \, c\right )}}{d}\right )} - \frac {1}{8} \, b {\left (\frac {12 \, {\left (d x + c\right )}}{d} + \frac {e^{\left (-2 \, d x - 2 \, c\right )}}{d} - \frac {17 \, e^{\left (-2 \, d x - 2 \, c\right )} + 1}{d {\left (e^{\left (-2 \, d x - 2 \, c\right )} + e^{\left (-4 \, d x - 4 \, c\right )}\right )}}\right )} \]

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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mupad [B]  time = 0.15, size = 64, normalized size = 1.45 \[ \frac {{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (a+b\right )}{8\,d}-\frac {2\,b}{d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}+1\right )}-\frac {{\mathrm {e}}^{-2\,c-2\,d\,x}\,\left (a+b\right )}{8\,d}-x\,\left (\frac {a}{2}+\frac {3\,b}{2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(exp(2*c + 2*d*x)*(a + b))/(8*d) - (2*b)/(d*(exp(2*c + 2*d*x) + 1)) - (exp(- 2*c - 2*d*x)*(a + b))/(8*d) - x*(
a/2 + (3*b)/2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \tanh ^{2}{\left (c + d x \right )}\right ) \sinh ^{2}{\left (c + d x \right )}\, dx \]

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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