3.1 \(\int \sinh ^4(c+d x) (a+b \tanh ^2(c+d x)) \, dx\)

Optimal. Leaf size=73 \[ \frac{(a+b) \sinh (c+d x) \cosh ^3(c+d x)}{4 d}-\frac{(5 a+9 b) \sinh (c+d x) \cosh (c+d x)}{8 d}+\frac{3}{8} x (a+5 b)-\frac{b \tanh (c+d x)}{d} \]

[Out]

(3*(a + 5*b)*x)/8 - ((5*a + 9*b)*Cosh[c + d*x]*Sinh[c + d*x])/(8*d) + ((a + b)*Cosh[c + d*x]^3*Sinh[c + d*x])/
(4*d) - (b*Tanh[c + d*x])/d

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Rubi [A]  time = 0.0757167, antiderivative size = 73, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {3663, 455, 1157, 388, 206} \[ \frac{(a+b) \sinh (c+d x) \cosh ^3(c+d x)}{4 d}-\frac{(5 a+9 b) \sinh (c+d x) \cosh (c+d x)}{8 d}+\frac{3}{8} x (a+5 b)-\frac{b \tanh (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[c + d*x]^4*(a + b*Tanh[c + d*x]^2),x]

[Out]

(3*(a + 5*b)*x)/8 - ((5*a + 9*b)*Cosh[c + d*x]*Sinh[c + d*x])/(8*d) + ((a + b)*Cosh[c + d*x]^3*Sinh[c + d*x])/
(4*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 1157

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> With[{Qx = PolynomialQ
uotient[(a + b*x^2 + c*x^4)^p, d + e*x^2, x], R = Coeff[PolynomialRemainder[(a + b*x^2 + c*x^4)^p, d + e*x^2,
x], x, 0]}, -Simp[(R*x*(d + e*x^2)^(q + 1))/(2*d*(q + 1)), x] + Dist[1/(2*d*(q + 1)), Int[(d + e*x^2)^(q + 1)*
ExpandToSum[2*d*(q + 1)*Qx + R*(2*q + 3), x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && N
eQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && LtQ[q, -1]

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 ^4(c+d x) \left (a+b \tanh ^2(c+d x)\right ) \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^4 \left (a+b x^2\right )}{\left (1-x^2\right )^3} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac{(a+b) \cosh ^3(c+d x) \sinh (c+d x)}{4 d}+\frac{\operatorname{Subst}\left (\int \frac{-a-b-4 (a+b) x^2-4 b x^4}{\left (1-x^2\right )^2} \, dx,x,\tanh (c+d x)\right )}{4 d}\\ &=-\frac{(5 a+9 b) \cosh (c+d x) \sinh (c+d x)}{8 d}+\frac{(a+b) \cosh ^3(c+d x) \sinh (c+d x)}{4 d}-\frac{\operatorname{Subst}\left (\int \frac{-3 a-7 b-8 b x^2}{1-x^2} \, dx,x,\tanh (c+d x)\right )}{8 d}\\ &=-\frac{(5 a+9 b) \cosh (c+d x) \sinh (c+d x)}{8 d}+\frac{(a+b) \cosh ^3(c+d x) \sinh (c+d x)}{4 d}-\frac{b \tanh (c+d x)}{d}+\frac{(3 (a+5 b)) \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\tanh (c+d x)\right )}{8 d}\\ &=\frac{3}{8} (a+5 b) x-\frac{(5 a+9 b) \cosh (c+d x) \sinh (c+d x)}{8 d}+\frac{(a+b) \cosh ^3(c+d x) \sinh (c+d x)}{4 d}-\frac{b \tanh (c+d x)}{d}\\ \end{align*}

Mathematica [A]  time = 0.340941, size = 56, normalized size = 0.77 \[ \frac{12 (a+5 b) (c+d x)-8 (a+2 b) \sinh (2 (c+d x))+(a+b) \sinh (4 (c+d x))-32 b \tanh (c+d x)}{32 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[c + d*x]^4*(a + b*Tanh[c + d*x]^2),x]

[Out]

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

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Maple [A]  time = 0.041, size = 96, normalized size = 1.3 \begin{align*}{\frac{1}{d} \left ( a \left ( \left ({\frac{ \left ( \sinh \left ( dx+c \right ) \right ) ^{3}}{4}}-{\frac{3\,\sinh \left ( dx+c \right ) }{8}} \right ) \cosh \left ( dx+c \right ) +{\frac{3\,dx}{8}}+{\frac{3\,c}{8}} \right ) +b \left ({\frac{ \left ( \sinh \left ( dx+c \right ) \right ) ^{5}}{4\,\cosh \left ( dx+c \right ) }}-{\frac{5\, \left ( \sinh \left ( dx+c \right ) \right ) ^{3}}{8\,\cosh \left ( dx+c \right ) }}+{\frac{15\,dx}{8}}+{\frac{15\,c}{8}}-{\frac{15\,\tanh \left ( dx+c \right ) }{8}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(d*x+c)^4*(a+b*tanh(d*x+c)^2),x)

[Out]

1/d*(a*((1/4*sinh(d*x+c)^3-3/8*sinh(d*x+c))*cosh(d*x+c)+3/8*d*x+3/8*c)+b*(1/4*sinh(d*x+c)^5/cosh(d*x+c)-5/8*si
nh(d*x+c)^3/cosh(d*x+c)+15/8*d*x+15/8*c-15/8*tanh(d*x+c)))

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Maxima [B]  time = 1.20197, size = 208, normalized size = 2.85 \begin{align*} \frac{1}{64} \, a{\left (24 \, x + \frac{e^{\left (4 \, d x + 4 \, c\right )}}{d} - \frac{8 \, e^{\left (2 \, d x + 2 \, c\right )}}{d} + \frac{8 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d} - \frac{e^{\left (-4 \, d x - 4 \, c\right )}}{d}\right )} + \frac{1}{64} \, b{\left (\frac{120 \,{\left (d x + c\right )}}{d} + \frac{16 \, e^{\left (-2 \, d x - 2 \, c\right )} - e^{\left (-4 \, d x - 4 \, c\right )}}{d} - \frac{15 \, e^{\left (-2 \, d x - 2 \, c\right )} + 144 \, e^{\left (-4 \, d x - 4 \, c\right )} - 1}{d{\left (e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )}\right )}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(d*x+c)^4*(a+b*tanh(d*x+c)^2),x, algorithm="maxima")

[Out]

1/64*a*(24*x + e^(4*d*x + 4*c)/d - 8*e^(2*d*x + 2*c)/d + 8*e^(-2*d*x - 2*c)/d - e^(-4*d*x - 4*c)/d) + 1/64*b*(
120*(d*x + c)/d + (16*e^(-2*d*x - 2*c) - e^(-4*d*x - 4*c))/d - (15*e^(-2*d*x - 2*c) + 144*e^(-4*d*x - 4*c) - 1
)/(d*(e^(-4*d*x - 4*c) + e^(-6*d*x - 6*c))))

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Fricas [A]  time = 1.84961, size = 327, normalized size = 4.48 \begin{align*} \frac{{\left (a + b\right )} \sinh \left (d x + c\right )^{5} +{\left (10 \,{\left (a + b\right )} \cosh \left (d x + c\right )^{2} - 7 \, a - 15 \, b\right )} \sinh \left (d x + c\right )^{3} + 8 \,{\left (3 \,{\left (a + 5 \, b\right )} d x + 8 \, b\right )} \cosh \left (d x + c\right ) +{\left (5 \,{\left (a + b\right )} \cosh \left (d x + c\right )^{4} - 3 \,{\left (7 \, a + 15 \, b\right )} \cosh \left (d x + c\right )^{2} - 8 \, a - 80 \, b\right )} \sinh \left (d x + c\right )}{64 \, 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)^4*(a+b*tanh(d*x+c)^2),x, algorithm="fricas")

[Out]

1/64*((a + b)*sinh(d*x + c)^5 + (10*(a + b)*cosh(d*x + c)^2 - 7*a - 15*b)*sinh(d*x + c)^3 + 8*(3*(a + 5*b)*d*x
+ 8*b)*cosh(d*x + c) + (5*(a + b)*cosh(d*x + c)^4 - 3*(7*a + 15*b)*cosh(d*x + c)^2 - 8*a - 80*b)*sinh(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 ^{4}{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(d*x+c)**4*(a+b*tanh(d*x+c)**2),x)

[Out]

Integral((a + b*tanh(c + d*x)**2)*sinh(c + d*x)**4, x)

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Giac [B]  time = 1.32276, size = 196, normalized size = 2.68 \begin{align*} \frac{24 \,{\left (a + 5 \, b\right )} d x -{\left (18 \, a e^{\left (4 \, d x + 4 \, c\right )} + 90 \, b e^{\left (4 \, d x + 4 \, c\right )} - 8 \, a e^{\left (2 \, d x + 2 \, c\right )} - 16 \, b e^{\left (2 \, d x + 2 \, c\right )} + a + b\right )} e^{\left (-4 \, d x - 4 \, c\right )} +{\left (a e^{\left (4 \, d x + 20 \, c\right )} + b e^{\left (4 \, d x + 20 \, c\right )} - 8 \, a e^{\left (2 \, d x + 18 \, c\right )} - 16 \, b e^{\left (2 \, d x + 18 \, c\right )}\right )} e^{\left (-16 \, c\right )} + \frac{128 \, b}{e^{\left (2 \, d x + 2 \, c\right )} + 1}}{64 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(d*x+c)^4*(a+b*tanh(d*x+c)^2),x, algorithm="giac")

[Out]

1/64*(24*(a + 5*b)*d*x - (18*a*e^(4*d*x + 4*c) + 90*b*e^(4*d*x + 4*c) - 8*a*e^(2*d*x + 2*c) - 16*b*e^(2*d*x +
2*c) + a + b)*e^(-4*d*x - 4*c) + (a*e^(4*d*x + 20*c) + b*e^(4*d*x + 20*c) - 8*a*e^(2*d*x + 18*c) - 16*b*e^(2*d
*x + 18*c))*e^(-16*c) + 128*b/(e^(2*d*x + 2*c) + 1))/d