3.49 \(\int \sinh ^4(c+d x) (a+b \tanh ^3(c+d x)) \, dx\)

Optimal. Leaf size=132 \[ -\frac {3 (a+8 b) \log (1-\tanh (c+d x))}{16 d}+\frac {3 (a-8 b) \log (\tanh (c+d x)+1)}{16 d}+\frac {\sinh ^4(c+d x) (a \tanh (c+d x)+b)}{4 d}-\frac {\sinh ^2(c+d x) \tanh (c+d x) (a+8 b \tanh (c+d x))}{8 d}-\frac {3 a \tanh (c+d x)}{8 d}-\frac {3 b \tanh ^2(c+d x)}{2 d} \]

[Out]

-3/16*(a+8*b)*ln(1-tanh(d*x+c))/d+3/16*(a-8*b)*ln(1+tanh(d*x+c))/d-3/8*a*tanh(d*x+c)/d-3/2*b*tanh(d*x+c)^2/d+1
/4*sinh(d*x+c)^4*(b+a*tanh(d*x+c))/d-1/8*sinh(d*x+c)^2*tanh(d*x+c)*(a+8*b*tanh(d*x+c))/d

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Rubi [A]  time = 0.17, antiderivative size = 132, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3663, 1804, 801, 633, 31} \[ -\frac {3 (a+8 b) \log (1-\tanh (c+d x))}{16 d}+\frac {3 (a-8 b) \log (\tanh (c+d x)+1)}{16 d}+\frac {\sinh ^4(c+d x) (a \tanh (c+d x)+b)}{4 d}-\frac {\sinh ^2(c+d x) \tanh (c+d x) (a+8 b \tanh (c+d x))}{8 d}-\frac {3 a \tanh (c+d x)}{8 d}-\frac {3 b \tanh ^2(c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[c + d*x]^4*(a + b*Tanh[c + d*x]^3),x]

[Out]

(-3*(a + 8*b)*Log[1 - Tanh[c + d*x]])/(16*d) + (3*(a - 8*b)*Log[1 + Tanh[c + d*x]])/(16*d) - (3*a*Tanh[c + d*x
])/(8*d) - (3*b*Tanh[c + d*x]^2)/(2*d) + (Sinh[c + d*x]^4*(b + a*Tanh[c + d*x]))/(4*d) - (Sinh[c + d*x]^2*Tanh
[c + d*x]*(a + 8*b*Tanh[c + d*x]))/(8*d)

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 633

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> With[{q = Rt[-(a*c), 2]}, Dist[e/2 + (c*d)/(2*q),
Int[1/(-q + c*x), x], x] + Dist[e/2 - (c*d)/(2*q), Int[1/(q + c*x), x], x]] /; FreeQ[{a, c, d, e}, x] && NiceS
qrtQ[-(a*c)]

Rule 801

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
(d + e*x)^m*(f + g*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

Rule 1804

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, a + b*x
^2, x], f = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x
], x, 1]}, Simp[((c*x)^m*(a + b*x^2)^(p + 1)*(a*g - b*f*x))/(2*a*b*(p + 1)), x] + Dist[c/(2*a*b*(p + 1)), Int[
(c*x)^(m - 1)*(a + b*x^2)^(p + 1)*ExpandToSum[2*a*b*(p + 1)*x*Q - a*g*m + b*f*(m + 2*p + 3)*x, x], x], x]] /;
FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && LtQ[p, -1] && GtQ[m, 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 ^4(c+d x) \left (a+b \tanh ^3(c+d x)\right ) \, dx &=\frac {\operatorname {Subst}\left (\int \frac {x^4 \left (a+b x^3\right )}{\left (1-x^2\right )^3} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac {\sinh ^4(c+d x) (b+a \tanh (c+d x))}{4 d}+\frac {\operatorname {Subst}\left (\int \frac {x^3 \left (-4 b-a x-4 b x^2\right )}{\left (1-x^2\right )^2} \, dx,x,\tanh (c+d x)\right )}{4 d}\\ &=\frac {\sinh ^4(c+d x) (b+a \tanh (c+d x))}{4 d}-\frac {\sinh ^2(c+d x) \tanh (c+d x) (a+8 b \tanh (c+d x))}{8 d}+\frac {\operatorname {Subst}\left (\int \frac {x^2 (3 a+24 b x)}{1-x^2} \, dx,x,\tanh (c+d x)\right )}{8 d}\\ &=\frac {\sinh ^4(c+d x) (b+a \tanh (c+d x))}{4 d}-\frac {\sinh ^2(c+d x) \tanh (c+d x) (a+8 b \tanh (c+d x))}{8 d}+\frac {\operatorname {Subst}\left (\int \left (-3 a-24 b x+\frac {3 (a+8 b x)}{1-x^2}\right ) \, dx,x,\tanh (c+d x)\right )}{8 d}\\ &=-\frac {3 a \tanh (c+d x)}{8 d}-\frac {3 b \tanh ^2(c+d x)}{2 d}+\frac {\sinh ^4(c+d x) (b+a \tanh (c+d x))}{4 d}-\frac {\sinh ^2(c+d x) \tanh (c+d x) (a+8 b \tanh (c+d x))}{8 d}+\frac {3 \operatorname {Subst}\left (\int \frac {a+8 b x}{1-x^2} \, dx,x,\tanh (c+d x)\right )}{8 d}\\ &=-\frac {3 a \tanh (c+d x)}{8 d}-\frac {3 b \tanh ^2(c+d x)}{2 d}+\frac {\sinh ^4(c+d x) (b+a \tanh (c+d x))}{4 d}-\frac {\sinh ^2(c+d x) \tanh (c+d x) (a+8 b \tanh (c+d x))}{8 d}-\frac {(3 (a-8 b)) \operatorname {Subst}\left (\int \frac {1}{-1-x} \, dx,x,\tanh (c+d x)\right )}{16 d}+\frac {(3 (a+8 b)) \operatorname {Subst}\left (\int \frac {1}{1-x} \, dx,x,\tanh (c+d x)\right )}{16 d}\\ &=-\frac {3 (a+8 b) \log (1-\tanh (c+d x))}{16 d}+\frac {3 (a-8 b) \log (1+\tanh (c+d x))}{16 d}-\frac {3 a \tanh (c+d x)}{8 d}-\frac {3 b \tanh ^2(c+d x)}{2 d}+\frac {\sinh ^4(c+d x) (b+a \tanh (c+d x))}{4 d}-\frac {\sinh ^2(c+d x) \tanh (c+d x) (a+8 b \tanh (c+d x))}{8 d}\\ \end {align*}

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Mathematica [A]  time = 0.20, size = 92, normalized size = 0.70 \[ \frac {3 a (c+d x)}{8 d}-\frac {a \sinh (2 (c+d x))}{4 d}+\frac {a \sinh (4 (c+d x))}{32 d}+\frac {b \left (\sinh ^4(c+d x)-4 \sinh ^2(c+d x)+2 \text {sech}^2(c+d x)+12 \log (\cosh (c+d x))\right )}{4 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[c + d*x]^4*(a + b*Tanh[c + d*x]^3),x]

[Out]

(3*a*(c + d*x))/(8*d) + (b*(12*Log[Cosh[c + d*x]] + 2*Sech[c + d*x]^2 - 4*Sinh[c + d*x]^2 + Sinh[c + d*x]^4))/
(4*d) - (a*Sinh[2*(c + d*x)])/(4*d) + (a*Sinh[4*(c + d*x)])/(32*d)

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fricas [B]  time = 0.57, size = 1530, normalized size = 11.59 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(d*x+c)^4*(a+b*tanh(d*x+c)^3),x, algorithm="fricas")

[Out]

1/64*((a + b)*cosh(d*x + c)^12 + 12*(a + b)*cosh(d*x + c)*sinh(d*x + c)^11 + (a + b)*sinh(d*x + c)^12 - 6*(a +
3*b)*cosh(d*x + c)^10 + 6*(11*(a + b)*cosh(d*x + c)^2 - a - 3*b)*sinh(d*x + c)^10 + 20*(11*(a + b)*cosh(d*x +
c)^3 - 3*(a + 3*b)*cosh(d*x + c))*sinh(d*x + c)^9 + 3*(8*(a - 8*b)*d*x - 5*a - 13*b)*cosh(d*x + c)^8 + 3*(165
*(a + b)*cosh(d*x + c)^4 + 8*(a - 8*b)*d*x - 90*(a + 3*b)*cosh(d*x + c)^2 - 5*a - 13*b)*sinh(d*x + c)^8 + 24*(
33*(a + b)*cosh(d*x + c)^5 - 30*(a + 3*b)*cosh(d*x + c)^3 + (8*(a - 8*b)*d*x - 5*a - 13*b)*cosh(d*x + c))*sinh
(d*x + c)^7 + 8*(6*(a - 8*b)*d*x + 11*b)*cosh(d*x + c)^6 + 4*(231*(a + b)*cosh(d*x + c)^6 - 315*(a + 3*b)*cosh
(d*x + c)^4 + 12*(a - 8*b)*d*x + 21*(8*(a - 8*b)*d*x - 5*a - 13*b)*cosh(d*x + c)^2 + 22*b)*sinh(d*x + c)^6 + 2
4*(33*(a + b)*cosh(d*x + c)^7 - 63*(a + 3*b)*cosh(d*x + c)^5 + 7*(8*(a - 8*b)*d*x - 5*a - 13*b)*cosh(d*x + c)^
3 + 2*(6*(a - 8*b)*d*x + 11*b)*cosh(d*x + c))*sinh(d*x + c)^5 + 3*(8*(a - 8*b)*d*x + 5*a - 13*b)*cosh(d*x + c)
^4 + 3*(165*(a + b)*cosh(d*x + c)^8 - 420*(a + 3*b)*cosh(d*x + c)^6 + 70*(8*(a - 8*b)*d*x - 5*a - 13*b)*cosh(d
*x + c)^4 + 8*(a - 8*b)*d*x + 40*(6*(a - 8*b)*d*x + 11*b)*cosh(d*x + c)^2 + 5*a - 13*b)*sinh(d*x + c)^4 + 4*(5
5*(a + b)*cosh(d*x + c)^9 - 180*(a + 3*b)*cosh(d*x + c)^7 + 42*(8*(a - 8*b)*d*x - 5*a - 13*b)*cosh(d*x + c)^5
+ 40*(6*(a - 8*b)*d*x + 11*b)*cosh(d*x + c)^3 + 3*(8*(a - 8*b)*d*x + 5*a - 13*b)*cosh(d*x + c))*sinh(d*x + c)^
3 + 6*(a - 3*b)*cosh(d*x + c)^2 + 6*(11*(a + b)*cosh(d*x + c)^10 - 45*(a + 3*b)*cosh(d*x + c)^8 + 14*(8*(a - 8
*b)*d*x - 5*a - 13*b)*cosh(d*x + c)^6 + 20*(6*(a - 8*b)*d*x + 11*b)*cosh(d*x + c)^4 + 3*(8*(a - 8*b)*d*x + 5*a
- 13*b)*cosh(d*x + c)^2 + a - 3*b)*sinh(d*x + c)^2 + 192*(b*cosh(d*x + c)^8 + 8*b*cosh(d*x + c)*sinh(d*x + c)
^7 + b*sinh(d*x + c)^8 + 2*b*cosh(d*x + c)^6 + 2*(14*b*cosh(d*x + c)^2 + b)*sinh(d*x + c)^6 + 4*(14*b*cosh(d*x
+ c)^3 + 3*b*cosh(d*x + c))*sinh(d*x + c)^5 + b*cosh(d*x + c)^4 + (70*b*cosh(d*x + c)^4 + 30*b*cosh(d*x + c)^
2 + b)*sinh(d*x + c)^4 + 4*(14*b*cosh(d*x + c)^5 + 10*b*cosh(d*x + c)^3 + b*cosh(d*x + c))*sinh(d*x + c)^3 + 2
*(14*b*cosh(d*x + c)^6 + 15*b*cosh(d*x + c)^4 + 3*b*cosh(d*x + c)^2)*sinh(d*x + c)^2 + 4*(2*b*cosh(d*x + c)^7
+ 3*b*cosh(d*x + c)^5 + b*cosh(d*x + c)^3)*sinh(d*x + c))*log(2*cosh(d*x + c)/(cosh(d*x + c) - sinh(d*x + c)))
+ 12*((a + b)*cosh(d*x + c)^11 - 5*(a + 3*b)*cosh(d*x + c)^9 + 2*(8*(a - 8*b)*d*x - 5*a - 13*b)*cosh(d*x + c)
^7 + 4*(6*(a - 8*b)*d*x + 11*b)*cosh(d*x + c)^5 + (8*(a - 8*b)*d*x + 5*a - 13*b)*cosh(d*x + c)^3 + (a - 3*b)*c
osh(d*x + c))*sinh(d*x + c) - a + b)/(d*cosh(d*x + c)^8 + 8*d*cosh(d*x + c)*sinh(d*x + c)^7 + d*sinh(d*x + c)^
8 + 2*d*cosh(d*x + c)^6 + 2*(14*d*cosh(d*x + c)^2 + d)*sinh(d*x + c)^6 + 4*(14*d*cosh(d*x + c)^3 + 3*d*cosh(d*
x + c))*sinh(d*x + c)^5 + d*cosh(d*x + c)^4 + (70*d*cosh(d*x + c)^4 + 30*d*cosh(d*x + c)^2 + d)*sinh(d*x + c)^
4 + 4*(14*d*cosh(d*x + c)^5 + 10*d*cosh(d*x + c)^3 + d*cosh(d*x + c))*sinh(d*x + c)^3 + 2*(14*d*cosh(d*x + c)^
6 + 15*d*cosh(d*x + c)^4 + 3*d*cosh(d*x + c)^2)*sinh(d*x + c)^2 + 4*(2*d*cosh(d*x + c)^7 + 3*d*cosh(d*x + c)^5
+ d*cosh(d*x + c)^3)*sinh(d*x + c))

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giac [A]  time = 0.25, size = 206, normalized size = 1.56 \[ \frac {24 \, {\left (a - 8 \, b\right )} d x + {\left (a e^{\left (4 \, d x + 24 \, c\right )} + b e^{\left (4 \, d x + 24 \, c\right )} - 8 \, a e^{\left (2 \, d x + 22 \, c\right )} - 20 \, b e^{\left (2 \, d x + 22 \, c\right )}\right )} e^{\left (-20 \, c\right )} + 192 \, b \log \left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right ) - \frac {{\left (9 \, a e^{\left (8 \, d x + 8 \, c\right )} + 72 \, b e^{\left (8 \, d x + 8 \, c\right )} + 10 \, a e^{\left (6 \, d x + 6 \, c\right )} + 36 \, b e^{\left (6 \, d x + 6 \, c\right )} - 6 \, a e^{\left (4 \, d x + 4 \, c\right )} + 111 \, b e^{\left (4 \, d x + 4 \, c\right )} - 6 \, a e^{\left (2 \, d x + 2 \, c\right )} + 18 \, b e^{\left (2 \, d x + 2 \, c\right )} + a - b\right )} e^{\left (-4 \, c\right )}}{{\left (e^{\left (2 \, d x\right )} + e^{\left (4 \, d x + 2 \, c\right )}\right )}^{2}}}{64 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(d*x+c)^4*(a+b*tanh(d*x+c)^3),x, algorithm="giac")

[Out]

1/64*(24*(a - 8*b)*d*x + (a*e^(4*d*x + 24*c) + b*e^(4*d*x + 24*c) - 8*a*e^(2*d*x + 22*c) - 20*b*e^(2*d*x + 22*
c))*e^(-20*c) + 192*b*log(e^(2*d*x + 2*c) + 1) - (9*a*e^(8*d*x + 8*c) + 72*b*e^(8*d*x + 8*c) + 10*a*e^(6*d*x +
6*c) + 36*b*e^(6*d*x + 6*c) - 6*a*e^(4*d*x + 4*c) + 111*b*e^(4*d*x + 4*c) - 6*a*e^(2*d*x + 2*c) + 18*b*e^(2*d
*x + 2*c) + a - b)*e^(-4*c)/(e^(2*d*x) + e^(4*d*x + 2*c))^2)/d

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maple [A]  time = 0.30, size = 122, normalized size = 0.92 \[ \frac {a \cosh \left (d x +c \right ) \left (\sinh ^{3}\left (d x +c \right )\right )}{4 d}-\frac {3 a \cosh \left (d x +c \right ) \sinh \left (d x +c \right )}{8 d}+\frac {3 a x}{8}+\frac {3 a c}{8 d}+\frac {b \left (\sinh ^{6}\left (d x +c \right )\right )}{4 d \cosh \left (d x +c \right )^{2}}-\frac {3 b \left (\sinh ^{4}\left (d x +c \right )\right )}{4 d \cosh \left (d x +c \right )^{2}}+\frac {3 b \ln \left (\cosh \left (d x +c \right )\right )}{d}-\frac {3 b \left (\tanh ^{2}\left (d x +c \right )\right )}{2 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(d*x+c)^4*(a+b*tanh(d*x+c)^3),x)

[Out]

1/4*a*cosh(d*x+c)*sinh(d*x+c)^3/d-3/8*a*cosh(d*x+c)*sinh(d*x+c)/d+3/8*a*x+3/8/d*a*c+1/4/d*b*sinh(d*x+c)^6/cosh
(d*x+c)^2-3/4/d*b*sinh(d*x+c)^4/cosh(d*x+c)^2+3*b*ln(cosh(d*x+c))/d-3/2*b*tanh(d*x+c)^2/d

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maxima [A]  time = 0.42, size = 194, normalized size = 1.47 \[ \frac {1}{64} \, a {\left (24 \, x + \frac {e^{\left (4 \, d x + 4 \, c\right )}}{d} - \frac {8 \, e^{\left (2 \, d x + 2 \, c\right )}}{d} + \frac {8 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d} - \frac {e^{\left (-4 \, d x - 4 \, c\right )}}{d}\right )} + \frac {1}{64} \, b {\left (\frac {192 \, {\left (d x + c\right )}}{d} - \frac {20 \, e^{\left (-2 \, d x - 2 \, c\right )} - e^{\left (-4 \, d x - 4 \, c\right )}}{d} + \frac {192 \, \log \left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}{d} - \frac {18 \, e^{\left (-2 \, d x - 2 \, c\right )} + 39 \, e^{\left (-4 \, d x - 4 \, c\right )} - 108 \, e^{\left (-6 \, d x - 6 \, c\right )} - 1}{d {\left (e^{\left (-4 \, d x - 4 \, c\right )} + 2 \, e^{\left (-6 \, d x - 6 \, c\right )} + e^{\left (-8 \, d x - 8 \, c\right )}\right )}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(d*x+c)^4*(a+b*tanh(d*x+c)^3),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*(
192*(d*x + c)/d - (20*e^(-2*d*x - 2*c) - e^(-4*d*x - 4*c))/d + 192*log(e^(-2*d*x - 2*c) + 1)/d - (18*e^(-2*d*x
- 2*c) + 39*e^(-4*d*x - 4*c) - 108*e^(-6*d*x - 6*c) - 1)/(d*(e^(-4*d*x - 4*c) + 2*e^(-6*d*x - 6*c) + e^(-8*d*
x - 8*c))))

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mupad [B]  time = 0.27, size = 156, normalized size = 1.18 \[ x\,\left (\frac {3\,a}{8}-3\,b\right )+\frac {2\,b}{d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}+1\right )}+\frac {{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (a+b\right )}{64\,d}-\frac {2\,b}{d\,\left (2\,{\mathrm {e}}^{2\,c+2\,d\,x}+{\mathrm {e}}^{4\,c+4\,d\,x}+1\right )}-\frac {{\mathrm {e}}^{-4\,c-4\,d\,x}\,\left (a-b\right )}{64\,d}+\frac {3\,b\,\ln \left ({\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}+1\right )}{d}+\frac {{\mathrm {e}}^{-2\,c-2\,d\,x}\,\left (2\,a-5\,b\right )}{16\,d}-\frac {{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (2\,a+5\,b\right )}{16\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(c + d*x)^4*(a + b*tanh(c + d*x)^3),x)

[Out]

x*((3*a)/8 - 3*b) + (2*b)/(d*(exp(2*c + 2*d*x) + 1)) + (exp(4*c + 4*d*x)*(a + b))/(64*d) - (2*b)/(d*(2*exp(2*c
+ 2*d*x) + exp(4*c + 4*d*x) + 1)) - (exp(- 4*c - 4*d*x)*(a - b))/(64*d) + (3*b*log(exp(2*c)*exp(2*d*x) + 1))/
d + (exp(- 2*c - 2*d*x)*(2*a - 5*b))/(16*d) - (exp(2*c + 2*d*x)*(2*a + 5*b))/(16*d)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \tanh ^{3}{\left (c + d x \right )}\right ) \sinh ^{4}{\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(d*x+c)**4*(a+b*tanh(d*x+c)**3),x)

[Out]

Integral((a + b*tanh(c + d*x)**3)*sinh(c + d*x)**4, x)

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