3.1.10 \(\int \sinh ^4(c+d x) (a+b \sinh ^2(c+d x))^2 \, dx\) [10]

Optimal. Leaf size=146 \[ \frac {1}{128} \left (48 a^2-80 a b+35 b^2\right ) x-\frac {\left (80 a^2-176 a b+93 b^2\right ) \cosh (c+d x) \sinh (c+d x)}{128 d}+\frac {\left (48 a^2-208 a b+139 b^2\right ) \cosh ^3(c+d x) \sinh (c+d x)}{192 d}+\frac {(16 a-13 b) b \cosh ^5(c+d x) \sinh (c+d x)}{48 d}+\frac {b^2 \cosh ^3(c+d x) \sinh ^5(c+d x)}{8 d} \]

[Out]

1/128*(48*a^2-80*a*b+35*b^2)*x-1/128*(80*a^2-176*a*b+93*b^2)*cosh(d*x+c)*sinh(d*x+c)/d+1/192*(48*a^2-208*a*b+1
39*b^2)*cosh(d*x+c)^3*sinh(d*x+c)/d+1/48*(16*a-13*b)*b*cosh(d*x+c)^5*sinh(d*x+c)/d+1/8*b^2*cosh(d*x+c)^3*sinh(
d*x+c)^5/d

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Rubi [A]
time = 0.13, antiderivative size = 146, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3266, 474, 466, 1171, 393, 212} \begin {gather*} \frac {\left (48 a^2-208 a b+139 b^2\right ) \sinh (c+d x) \cosh ^3(c+d x)}{192 d}-\frac {\left (80 a^2-176 a b+93 b^2\right ) \sinh (c+d x) \cosh (c+d x)}{128 d}+\frac {1}{128} x \left (48 a^2-80 a b+35 b^2\right )+\frac {b (16 a-13 b) \sinh (c+d x) \cosh ^5(c+d x)}{48 d}+\frac {b^2 \sinh ^5(c+d x) \cosh ^3(c+d x)}{8 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((48*a^2 - 80*a*b + 35*b^2)*x)/128 - ((80*a^2 - 176*a*b + 93*b^2)*Cosh[c + d*x]*Sinh[c + d*x])/(128*d) + ((48*
a^2 - 208*a*b + 139*b^2)*Cosh[c + d*x]^3*Sinh[c + d*x])/(192*d) + ((16*a - 13*b)*b*Cosh[c + d*x]^5*Sinh[c + d*
x])/(48*d) + (b^2*Cosh[c + d*x]^3*Sinh[c + d*x]^5)/(8*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 393

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(-(b*c - a*d))*x*((a + b*x^n)^(p
+ 1)/(a*b*n*(p + 1))), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x]
/; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/n + p, 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 474

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

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 3266

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

Rubi steps

\begin {align*} \int \sinh ^4(c+d x) \left (a+b \sinh ^2(c+d x)\right )^2 \, dx &=\frac {\text {Subst}\left (\int \frac {x^4 \left (a-(a-b) x^2\right )^2}{\left (1-x^2\right )^5} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac {b^2 \cosh ^3(c+d x) \sinh ^5(c+d x)}{8 d}-\frac {\text {Subst}\left (\int \frac {x^4 \left (-8 a^2+5 b^2+8 (a-b)^2 x^2\right )}{\left (1-x^2\right )^4} \, dx,x,\tanh (c+d x)\right )}{8 d}\\ &=\frac {(16 a-13 b) b \cosh ^5(c+d x) \sinh (c+d x)}{48 d}+\frac {b^2 \cosh ^3(c+d x) \sinh ^5(c+d x)}{8 d}-\frac {\text {Subst}\left (\int \frac {(16 a-13 b) b+6 (16 a-13 b) b x^2-48 (a-b)^2 x^4}{\left (1-x^2\right )^3} \, dx,x,\tanh (c+d x)\right )}{48 d}\\ &=\frac {\left (48 a^2-208 a b+139 b^2\right ) \cosh ^3(c+d x) \sinh (c+d x)}{192 d}+\frac {(16 a-13 b) b \cosh ^5(c+d x) \sinh (c+d x)}{48 d}+\frac {b^2 \cosh ^3(c+d x) \sinh ^5(c+d x)}{8 d}+\frac {\text {Subst}\left (\int \frac {-3 \left (16 a^2-48 a b+29 b^2\right )-192 (a-b)^2 x^2}{\left (1-x^2\right )^2} \, dx,x,\tanh (c+d x)\right )}{192 d}\\ &=-\frac {\left (80 a^2-176 a b+93 b^2\right ) \cosh (c+d x) \sinh (c+d x)}{128 d}+\frac {\left (48 a^2-208 a b+139 b^2\right ) \cosh ^3(c+d x) \sinh (c+d x)}{192 d}+\frac {(16 a-13 b) b \cosh ^5(c+d x) \sinh (c+d x)}{48 d}+\frac {b^2 \cosh ^3(c+d x) \sinh ^5(c+d x)}{8 d}+\frac {\left (48 a^2-80 a b+35 b^2\right ) \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\tanh (c+d x)\right )}{128 d}\\ &=\frac {1}{128} \left (48 a^2-80 a b+35 b^2\right ) x-\frac {\left (80 a^2-176 a b+93 b^2\right ) \cosh (c+d x) \sinh (c+d x)}{128 d}+\frac {\left (48 a^2-208 a b+139 b^2\right ) \cosh ^3(c+d x) \sinh (c+d x)}{192 d}+\frac {(16 a-13 b) b \cosh ^5(c+d x) \sinh (c+d x)}{48 d}+\frac {b^2 \cosh ^3(c+d x) \sinh ^5(c+d x)}{8 d}\\ \end {align*}

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Mathematica [A]
time = 0.14, size = 133, normalized size = 0.91 \begin {gather*} \frac {1152 a^2 c-1920 a b c+840 b^2 c+1152 a^2 d x-1920 a b d x+840 b^2 d x-96 \left (8 a^2-15 a b+7 b^2\right ) \sinh (2 (c+d x))+24 \left (4 a^2-12 a b+7 b^2\right ) \sinh (4 (c+d x))+32 a b \sinh (6 (c+d x))-32 b^2 \sinh (6 (c+d x))+3 b^2 \sinh (8 (c+d x))}{3072 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(1152*a^2*c - 1920*a*b*c + 840*b^2*c + 1152*a^2*d*x - 1920*a*b*d*x + 840*b^2*d*x - 96*(8*a^2 - 15*a*b + 7*b^2)
*Sinh[2*(c + d*x)] + 24*(4*a^2 - 12*a*b + 7*b^2)*Sinh[4*(c + d*x)] + 32*a*b*Sinh[6*(c + d*x)] - 32*b^2*Sinh[6*
(c + d*x)] + 3*b^2*Sinh[8*(c + d*x)])/(3072*d)

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Maple [A]
time = 1.34, size = 118, normalized size = 0.81

method	result	size
default	\(\frac {\left (-\frac {1}{16} b^{2}+\frac {1}{16} a b \right ) \sinh \left (6 d x +6 c \right )}{6 d}+\frac {\left (-\frac {7}{16} b^{2}+\frac {15}{16} a b -\frac {1}{2} a^{2}\right ) \sinh \left (2 d x +2 c \right )}{2 d}+\frac {\left (\frac {7}{32} b^{2}-\frac {3}{8} a b +\frac {1}{8} a^{2}\right ) \sinh \left (4 d x +4 c \right )}{4 d}+\frac {3 a^{2} x}{8}+\frac {35 b^{2} x}{128}-\frac {5 a b x}{8}+\frac {b^{2} \sinh \left (8 d x +8 c \right )}{1024 d}\)	\(118\)
risch	\(\frac {35 b^{2} x}{128}-\frac {5 a b x}{8}+\frac {3 a^{2} x}{8}+\frac {b^{2} {\mathrm e}^{8 d x +8 c}}{2048 d}+\frac {b \,{\mathrm e}^{6 d x +6 c} a}{192 d}-\frac {b^{2} {\mathrm e}^{6 d x +6 c}}{192 d}+\frac {{\mathrm e}^{4 d x +4 c} a^{2}}{64 d}-\frac {3 \,{\mathrm e}^{4 d x +4 c} a b}{64 d}+\frac {7 \,{\mathrm e}^{4 d x +4 c} b^{2}}{256 d}-\frac {{\mathrm e}^{2 d x +2 c} a^{2}}{8 d}+\frac {15 \,{\mathrm e}^{2 d x +2 c} a b}{64 d}-\frac {7 \,{\mathrm e}^{2 d x +2 c} b^{2}}{64 d}+\frac {{\mathrm e}^{-2 d x -2 c} a^{2}}{8 d}-\frac {15 \,{\mathrm e}^{-2 d x -2 c} a b}{64 d}+\frac {7 \,{\mathrm e}^{-2 d x -2 c} b^{2}}{64 d}-\frac {{\mathrm e}^{-4 d x -4 c} a^{2}}{64 d}+\frac {3 \,{\mathrm e}^{-4 d x -4 c} a b}{64 d}-\frac {7 \,{\mathrm e}^{-4 d x -4 c} b^{2}}{256 d}-\frac {b \,{\mathrm e}^{-6 d x -6 c} a}{192 d}+\frac {b^{2} {\mathrm e}^{-6 d x -6 c}}{192 d}-\frac {b^{2} {\mathrm e}^{-8 d x -8 c}}{2048 d}\)	\(319\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(-1/16*b^2+1/16*a*b)/d*sinh(6*d*x+6*c)+1/2*(-7/16*b^2+15/16*a*b-1/2*a^2)*sinh(2*d*x+2*c)/d+1/4*(7/32*b^2-3
/8*a*b+1/8*a^2)/d*sinh(4*d*x+4*c)+3/8*a^2*x+35/128*b^2*x-5/8*a*b*x+1/1024*b^2/d*sinh(8*d*x+8*c)

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Maxima [A]
time = 0.27, size = 267, normalized size = 1.83 \begin {gather*} \frac {1}{64} \, a^{2} {\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}{6144} \, b^{2} {\left (\frac {{\left (32 \, e^{\left (-2 \, d x - 2 \, c\right )} - 168 \, e^{\left (-4 \, d x - 4 \, c\right )} + 672 \, e^{\left (-6 \, d x - 6 \, c\right )} - 3\right )} e^{\left (8 \, d x + 8 \, c\right )}}{d} - \frac {1680 \, {\left (d x + c\right )}}{d} - \frac {672 \, e^{\left (-2 \, d x - 2 \, c\right )} - 168 \, e^{\left (-4 \, d x - 4 \, c\right )} + 32 \, e^{\left (-6 \, d x - 6 \, c\right )} - 3 \, e^{\left (-8 \, d x - 8 \, c\right )}}{d}\right )} - \frac {1}{192} \, a b {\left (\frac {{\left (9 \, e^{\left (-2 \, d x - 2 \, c\right )} - 45 \, e^{\left (-4 \, d x - 4 \, c\right )} - 1\right )} e^{\left (6 \, d x + 6 \, c\right )}}{d} + \frac {120 \, {\left (d x + c\right )}}{d} + \frac {45 \, e^{\left (-2 \, d x - 2 \, c\right )} - 9 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )}}{d}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/64*a^2*(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) - 1/6144
*b^2*((32*e^(-2*d*x - 2*c) - 168*e^(-4*d*x - 4*c) + 672*e^(-6*d*x - 6*c) - 3)*e^(8*d*x + 8*c)/d - 1680*(d*x +
c)/d - (672*e^(-2*d*x - 2*c) - 168*e^(-4*d*x - 4*c) + 32*e^(-6*d*x - 6*c) - 3*e^(-8*d*x - 8*c))/d) - 1/192*a*b
*((9*e^(-2*d*x - 2*c) - 45*e^(-4*d*x - 4*c) - 1)*e^(6*d*x + 6*c)/d + 120*(d*x + c)/d + (45*e^(-2*d*x - 2*c) -
9*e^(-4*d*x - 4*c) + e^(-6*d*x - 6*c))/d)

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Fricas [A]
time = 0.39, size = 238, normalized size = 1.63 \begin {gather*} \frac {3 \, b^{2} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{7} + 3 \, {\left (7 \, b^{2} \cosh \left (d x + c\right )^{3} + 8 \, {\left (a b - b^{2}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{5} + {\left (21 \, b^{2} \cosh \left (d x + c\right )^{5} + 80 \, {\left (a b - b^{2}\right )} \cosh \left (d x + c\right )^{3} + 12 \, {\left (4 \, a^{2} - 12 \, a b + 7 \, b^{2}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{3} + 3 \, {\left (48 \, a^{2} - 80 \, a b + 35 \, b^{2}\right )} d x + 3 \, {\left (b^{2} \cosh \left (d x + c\right )^{7} + 8 \, {\left (a b - b^{2}\right )} \cosh \left (d x + c\right )^{5} + 4 \, {\left (4 \, a^{2} - 12 \, a b + 7 \, b^{2}\right )} \cosh \left (d x + c\right )^{3} - 8 \, {\left (8 \, a^{2} - 15 \, a b + 7 \, b^{2}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )}{384 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/384*(3*b^2*cosh(d*x + c)*sinh(d*x + c)^7 + 3*(7*b^2*cosh(d*x + c)^3 + 8*(a*b - b^2)*cosh(d*x + c))*sinh(d*x
+ c)^5 + (21*b^2*cosh(d*x + c)^5 + 80*(a*b - b^2)*cosh(d*x + c)^3 + 12*(4*a^2 - 12*a*b + 7*b^2)*cosh(d*x + c))
*sinh(d*x + c)^3 + 3*(48*a^2 - 80*a*b + 35*b^2)*d*x + 3*(b^2*cosh(d*x + c)^7 + 8*(a*b - b^2)*cosh(d*x + c)^5 +
4*(4*a^2 - 12*a*b + 7*b^2)*cosh(d*x + c)^3 - 8*(8*a^2 - 15*a*b + 7*b^2)*cosh(d*x + c))*sinh(d*x + c))/d

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 490 vs. \(2 (143) = 286\).
time = 1.04, size = 490, normalized size = 3.36 \begin {gather*} \begin {cases} \frac {3 a^{2} x \sinh ^{4}{\left (c + d x \right )}}{8} - \frac {3 a^{2} x \sinh ^{2}{\left (c + d x \right )} \cosh ^{2}{\left (c + d x \right )}}{4} + \frac {3 a^{2} x \cosh ^{4}{\left (c + d x \right )}}{8} + \frac {5 a^{2} \sinh ^{3}{\left (c + d x \right )} \cosh {\left (c + d x \right )}}{8 d} - \frac {3 a^{2} \sinh {\left (c + d x \right )} \cosh ^{3}{\left (c + d x \right )}}{8 d} + \frac {5 a b x \sinh ^{6}{\left (c + d x \right )}}{8} - \frac {15 a b x \sinh ^{4}{\left (c + d x \right )} \cosh ^{2}{\left (c + d x \right )}}{8} + \frac {15 a b x \sinh ^{2}{\left (c + d x \right )} \cosh ^{4}{\left (c + d x \right )}}{8} - \frac {5 a b x \cosh ^{6}{\left (c + d x \right )}}{8} + \frac {11 a b \sinh ^{5}{\left (c + d x \right )} \cosh {\left (c + d x \right )}}{8 d} - \frac {5 a b \sinh ^{3}{\left (c + d x \right )} \cosh ^{3}{\left (c + d x \right )}}{3 d} + \frac {5 a b \sinh {\left (c + d x \right )} \cosh ^{5}{\left (c + d x \right )}}{8 d} + \frac {35 b^{2} x \sinh ^{8}{\left (c + d x \right )}}{128} - \frac {35 b^{2} x \sinh ^{6}{\left (c + d x \right )} \cosh ^{2}{\left (c + d x \right )}}{32} + \frac {105 b^{2} x \sinh ^{4}{\left (c + d x \right )} \cosh ^{4}{\left (c + d x \right )}}{64} - \frac {35 b^{2} x \sinh ^{2}{\left (c + d x \right )} \cosh ^{6}{\left (c + d x \right )}}{32} + \frac {35 b^{2} x \cosh ^{8}{\left (c + d x \right )}}{128} + \frac {93 b^{2} \sinh ^{7}{\left (c + d x \right )} \cosh {\left (c + d x \right )}}{128 d} - \frac {511 b^{2} \sinh ^{5}{\left (c + d x \right )} \cosh ^{3}{\left (c + d x \right )}}{384 d} + \frac {385 b^{2} \sinh ^{3}{\left (c + d x \right )} \cosh ^{5}{\left (c + d x \right )}}{384 d} - \frac {35 b^{2} \sinh {\left (c + d x \right )} \cosh ^{7}{\left (c + d x \right )}}{128 d} & \text {for}\: d \neq 0 \\x \left (a + b \sinh ^{2}{\left (c \right )}\right )^{2} \sinh ^{4}{\left (c \right )} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((3*a**2*x*sinh(c + d*x)**4/8 - 3*a**2*x*sinh(c + d*x)**2*cosh(c + d*x)**2/4 + 3*a**2*x*cosh(c + d*x)
**4/8 + 5*a**2*sinh(c + d*x)**3*cosh(c + d*x)/(8*d) - 3*a**2*sinh(c + d*x)*cosh(c + d*x)**3/(8*d) + 5*a*b*x*si
nh(c + d*x)**6/8 - 15*a*b*x*sinh(c + d*x)**4*cosh(c + d*x)**2/8 + 15*a*b*x*sinh(c + d*x)**2*cosh(c + d*x)**4/8
- 5*a*b*x*cosh(c + d*x)**6/8 + 11*a*b*sinh(c + d*x)**5*cosh(c + d*x)/(8*d) - 5*a*b*sinh(c + d*x)**3*cosh(c +
d*x)**3/(3*d) + 5*a*b*sinh(c + d*x)*cosh(c + d*x)**5/(8*d) + 35*b**2*x*sinh(c + d*x)**8/128 - 35*b**2*x*sinh(c
+ d*x)**6*cosh(c + d*x)**2/32 + 105*b**2*x*sinh(c + d*x)**4*cosh(c + d*x)**4/64 - 35*b**2*x*sinh(c + d*x)**2*
cosh(c + d*x)**6/32 + 35*b**2*x*cosh(c + d*x)**8/128 + 93*b**2*sinh(c + d*x)**7*cosh(c + d*x)/(128*d) - 511*b*
*2*sinh(c + d*x)**5*cosh(c + d*x)**3/(384*d) + 385*b**2*sinh(c + d*x)**3*cosh(c + d*x)**5/(384*d) - 35*b**2*si
nh(c + d*x)*cosh(c + d*x)**7/(128*d), Ne(d, 0)), (x*(a + b*sinh(c)**2)**2*sinh(c)**4, True))

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Giac [A]
time = 0.47, size = 215, normalized size = 1.47 \begin {gather*} \frac {1}{128} \, {\left (48 \, a^{2} - 80 \, a b + 35 \, b^{2}\right )} x + \frac {b^{2} e^{\left (8 \, d x + 8 \, c\right )}}{2048 \, d} - \frac {b^{2} e^{\left (-8 \, d x - 8 \, c\right )}}{2048 \, d} + \frac {{\left (a b - b^{2}\right )} e^{\left (6 \, d x + 6 \, c\right )}}{192 \, d} + \frac {{\left (4 \, a^{2} - 12 \, a b + 7 \, b^{2}\right )} e^{\left (4 \, d x + 4 \, c\right )}}{256 \, d} - \frac {{\left (8 \, a^{2} - 15 \, a b + 7 \, b^{2}\right )} e^{\left (2 \, d x + 2 \, c\right )}}{64 \, d} + \frac {{\left (8 \, a^{2} - 15 \, a b + 7 \, b^{2}\right )} e^{\left (-2 \, d x - 2 \, c\right )}}{64 \, d} - \frac {{\left (4 \, a^{2} - 12 \, a b + 7 \, b^{2}\right )} e^{\left (-4 \, d x - 4 \, c\right )}}{256 \, d} - \frac {{\left (a b - b^{2}\right )} e^{\left (-6 \, d x - 6 \, c\right )}}{192 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/128*(48*a^2 - 80*a*b + 35*b^2)*x + 1/2048*b^2*e^(8*d*x + 8*c)/d - 1/2048*b^2*e^(-8*d*x - 8*c)/d + 1/192*(a*b
- b^2)*e^(6*d*x + 6*c)/d + 1/256*(4*a^2 - 12*a*b + 7*b^2)*e^(4*d*x + 4*c)/d - 1/64*(8*a^2 - 15*a*b + 7*b^2)*e
^(2*d*x + 2*c)/d + 1/64*(8*a^2 - 15*a*b + 7*b^2)*e^(-2*d*x - 2*c)/d - 1/256*(4*a^2 - 12*a*b + 7*b^2)*e^(-4*d*x
- 4*c)/d - 1/192*(a*b - b^2)*e^(-6*d*x - 6*c)/d

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Mupad [B]
time = 0.90, size = 149, normalized size = 1.02 \begin {gather*} \frac {12\,a^2\,\mathrm {sinh}\left (4\,c+4\,d\,x\right )-96\,a^2\,\mathrm {sinh}\left (2\,c+2\,d\,x\right )-84\,b^2\,\mathrm {sinh}\left (2\,c+2\,d\,x\right )+21\,b^2\,\mathrm {sinh}\left (4\,c+4\,d\,x\right )-4\,b^2\,\mathrm {sinh}\left (6\,c+6\,d\,x\right )+\frac {3\,b^2\,\mathrm {sinh}\left (8\,c+8\,d\,x\right )}{8}+180\,a\,b\,\mathrm {sinh}\left (2\,c+2\,d\,x\right )-36\,a\,b\,\mathrm {sinh}\left (4\,c+4\,d\,x\right )+4\,a\,b\,\mathrm {sinh}\left (6\,c+6\,d\,x\right )+144\,a^2\,d\,x+105\,b^2\,d\,x-240\,a\,b\,d\,x}{384\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(12*a^2*sinh(4*c + 4*d*x) - 96*a^2*sinh(2*c + 2*d*x) - 84*b^2*sinh(2*c + 2*d*x) + 21*b^2*sinh(4*c + 4*d*x) - 4
*b^2*sinh(6*c + 6*d*x) + (3*b^2*sinh(8*c + 8*d*x))/8 + 180*a*b*sinh(2*c + 2*d*x) - 36*a*b*sinh(4*c + 4*d*x) +
4*a*b*sinh(6*c + 6*d*x) + 144*a^2*d*x + 105*b^2*d*x - 240*a*b*d*x)/(384*d)

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