\(\int \sinh ^4(c+d x) (a+b \tanh ^2(c+d x)) \, dx\) [1]

Optimal result

Integrand size = 21, antiderivative size = 73 \[ \int \sinh ^4(c+d x) \left (a+b \tanh ^2(c+d x)\right ) \, dx=\frac {3}{8} (a+5 b) x-\frac {(5 a+9 b) \cosh (c+d x) \sinh (c+d x)}{8 d}+\frac {(a+b) \cosh ^3(c+d x) \sinh (c+d x)}{4 d}-\frac {b \tanh (c+d x)}{d} \]

[Out]

3/8*(a+5*b)*x-1/8*(5*a+9*b)*cosh(d*x+c)*sinh(d*x+c)/d+1/4*(a+b)*cosh(d*x+c)^3*sinh(d*x+c)/d-b*tanh(d*x+c)/d

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3744, 466, 1171, 396, 212} \[ \int \sinh ^4(c+d x) \left (a+b \tanh ^2(c+d x)\right ) \, dx=\frac {(a+b) \sinh (c+d x) \cosh ^3(c+d x)}{4 d}-\frac {(5 a+9 b) \sinh (c+d x) \cosh (c+d x)}{8 d}+\frac {3}{8} x (a+5 b)-\frac {b \tanh (c+d x)}{d} \]

[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 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 396

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 466

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 1171

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] &&
NeQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && LtQ[q, -1]

Rule 3744

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*} \text {integral}& = \frac {\text {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 {\text {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 {\text {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)) \text {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] (verified)

Time = 0.69 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.77 \[ \int \sinh ^4(c+d x) \left (a+b \tanh ^2(c+d x)\right ) \, dx=\frac {12 (a+5 b) (c+d x)-8 (a+2 b) \sinh (2 (c+d x))+(a+b) \sinh (4 (c+d x))-32 b \tanh (c+d x)}{32 d} \]

[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)

Maple [A] (verified)

Time = 1.47 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.32

method	result	size
derivativedivides	\(\frac {a \left (\left (\frac {\sinh \left (d x +c \right )^{3}}{4}-\frac {3 \sinh \left (d x +c \right )}{8}\right ) \cosh \left (d x +c \right )+\frac {3 d x}{8}+\frac {3 c}{8}\right )+b \left (\frac {\sinh \left (d x +c \right )^{5}}{4 \cosh \left (d x +c \right )}-\frac {5 \sinh \left (d x +c \right )^{3}}{8 \cosh \left (d x +c \right )}+\frac {15 d x}{8}+\frac {15 c}{8}-\frac {15 \tanh \left (d x +c \right )}{8}\right )}{d}\)	\(96\)
default	\(\frac {a \left (\left (\frac {\sinh \left (d x +c \right )^{3}}{4}-\frac {3 \sinh \left (d x +c \right )}{8}\right ) \cosh \left (d x +c \right )+\frac {3 d x}{8}+\frac {3 c}{8}\right )+b \left (\frac {\sinh \left (d x +c \right )^{5}}{4 \cosh \left (d x +c \right )}-\frac {5 \sinh \left (d x +c \right )^{3}}{8 \cosh \left (d x +c \right )}+\frac {15 d x}{8}+\frac {15 c}{8}-\frac {15 \tanh \left (d x +c \right )}{8}\right )}{d}\)	\(96\)
risch	\(\frac {3 a x}{8}+\frac {15 b x}{8}+\frac {{\mathrm e}^{4 d x +4 c} a}{64 d}+\frac {{\mathrm e}^{4 d x +4 c} b}{64 d}-\frac {{\mathrm e}^{2 d x +2 c} a}{8 d}-\frac {{\mathrm e}^{2 d x +2 c} b}{4 d}+\frac {{\mathrm e}^{-2 d x -2 c} a}{8 d}+\frac {{\mathrm e}^{-2 d x -2 c} b}{4 d}-\frac {{\mathrm e}^{-4 d x -4 c} a}{64 d}-\frac {{\mathrm e}^{-4 d x -4 c} b}{64 d}+\frac {2 b}{d \left ({\mathrm e}^{2 d x +2 c}+1\right )}\)	\(149\)

[In]

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

[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)))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.64 \[ \int \sinh ^4(c+d x) \left (a+b \tanh ^2(c+d x)\right ) \, dx=\frac {{\left (a + b\right )} \sinh \left (d x + c\right )^{5} + {\left (10 \, {\left (a + b\right )} \cosh \left (d x + c\right )^{2} - 7 \, a - 15 \, b\right )} \sinh \left (d x + c\right )^{3} + 8 \, {\left (3 \, {\left (a + 5 \, b\right )} d x + 8 \, b\right )} \cosh \left (d x + c\right ) + {\left (5 \, {\left (a + b\right )} \cosh \left (d x + c\right )^{4} - 3 \, {\left (7 \, a + 15 \, b\right )} \cosh \left (d x + c\right )^{2} - 8 \, a - 80 \, b\right )} \sinh \left (d x + c\right )}{64 \, d \cosh \left (d x + c\right )} \]

[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))

Sympy [F]

\[ \int \sinh ^4(c+d x) \left (a+b \tanh ^2(c+d x)\right ) \, dx=\int \left (a + b \tanh ^{2}{\left (c + d x \right )}\right ) \sinh ^{4}{\left (c + d x \right )}\, dx \]

[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)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 154 vs. \(2 (67) = 134\).

Time = 0.20 (sec) , antiderivative size = 154, normalized size of antiderivative = 2.11 \[ \int \sinh ^4(c+d x) \left (a+b \tanh ^2(c+d x)\right ) \, dx=\frac {1}{64} \, a {\left (24 \, x + \frac {e^{\left (4 \, d x + 4 \, c\right )}}{d} - \frac {8 \, e^{\left (2 \, d x + 2 \, c\right )}}{d} + \frac {8 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d} - \frac {e^{\left (-4 \, d x - 4 \, c\right )}}{d}\right )} + \frac {1}{64} \, b {\left (\frac {120 \, {\left (d x + c\right )}}{d} + \frac {16 \, e^{\left (-2 \, d x - 2 \, c\right )} - e^{\left (-4 \, d x - 4 \, c\right )}}{d} - \frac {15 \, e^{\left (-2 \, d x - 2 \, c\right )} + 144 \, e^{\left (-4 \, d x - 4 \, c\right )} - 1}{d {\left (e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )}\right )}}\right )} \]

[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))))

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 142 vs. \(2 (67) = 134\).

Time = 0.32 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.95 \[ \int \sinh ^4(c+d x) \left (a+b \tanh ^2(c+d x)\right ) \, dx=\frac {24 \, {\left (d x + c\right )} {\left (a + 5 \, b\right )} + a e^{\left (4 \, d x + 4 \, c\right )} + b e^{\left (4 \, d x + 4 \, c\right )} - 8 \, a e^{\left (2 \, d x + 2 \, c\right )} - 16 \, b e^{\left (2 \, d x + 2 \, c\right )} - {\left (18 \, a e^{\left (4 \, d x + 4 \, c\right )} + 90 \, b e^{\left (4 \, d x + 4 \, c\right )} - 8 \, a e^{\left (2 \, d x + 2 \, c\right )} - 16 \, b e^{\left (2 \, d x + 2 \, c\right )} + a + b\right )} e^{\left (-4 \, d x - 4 \, c\right )} + \frac {128 \, b}{e^{\left (2 \, d x + 2 \, c\right )} + 1}}{64 \, d} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.38 \[ \int \sinh ^4(c+d x) \left (a+b \tanh ^2(c+d x)\right ) \, dx=x\,\left (\frac {3\,a}{8}+\frac {15\,b}{8}\right )+\frac {2\,b}{d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}+1\right )}-\frac {{\mathrm {e}}^{-4\,c-4\,d\,x}\,\left (a+b\right )}{64\,d}+\frac {{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (a+b\right )}{64\,d}+\frac {{\mathrm {e}}^{-2\,c-2\,d\,x}\,\left (a+2\,b\right )}{8\,d}-\frac {{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (a+2\,b\right )}{8\,d} \]

[In]

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

[Out]

x*((3*a)/8 + (15*b)/8) + (2*b)/(d*(exp(2*c + 2*d*x) + 1)) - (exp(- 4*c - 4*d*x)*(a + b))/(64*d) + (exp(4*c + 4
*d*x)*(a + b))/(64*d) + (exp(- 2*c - 2*d*x)*(a + 2*b))/(8*d) - (exp(2*c + 2*d*x)*(a + 2*b))/(8*d)