3.1.97 \(\int (a+b \sinh (c+d x))^4 \, dx\) [97]

Optimal. Leaf size=137 \[ \frac {1}{8} \left (8 a^4-24 a^2 b^2+3 b^4\right ) x+\frac {a b \left (19 a^2-16 b^2\right ) \cosh (c+d x)}{6 d}+\frac {b^2 \left (26 a^2-9 b^2\right ) \cosh (c+d x) \sinh (c+d x)}{24 d}+\frac {7 a b \cosh (c+d x) (a+b \sinh (c+d x))^2}{12 d}+\frac {b \cosh (c+d x) (a+b \sinh (c+d x))^3}{4 d} \]

[Out]

1/8*(8*a^4-24*a^2*b^2+3*b^4)*x+1/6*a*b*(19*a^2-16*b^2)*cosh(d*x+c)/d+1/24*b^2*(26*a^2-9*b^2)*cosh(d*x+c)*sinh(
d*x+c)/d+7/12*a*b*cosh(d*x+c)*(a+b*sinh(d*x+c))^2/d+1/4*b*cosh(d*x+c)*(a+b*sinh(d*x+c))^3/d

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Rubi [A]
time = 0.11, antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2735, 2832, 2813} \begin {gather*} \frac {a b \left (19 a^2-16 b^2\right ) \cosh (c+d x)}{6 d}+\frac {b^2 \left (26 a^2-9 b^2\right ) \sinh (c+d x) \cosh (c+d x)}{24 d}+\frac {1}{8} x \left (8 a^4-24 a^2 b^2+3 b^4\right )+\frac {b \cosh (c+d x) (a+b \sinh (c+d x))^3}{4 d}+\frac {7 a b \cosh (c+d x) (a+b \sinh (c+d x))^2}{12 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*Sinh[c + d*x])^4,x]

[Out]

((8*a^4 - 24*a^2*b^2 + 3*b^4)*x)/8 + (a*b*(19*a^2 - 16*b^2)*Cosh[c + d*x])/(6*d) + (b^2*(26*a^2 - 9*b^2)*Cosh[
c + d*x]*Sinh[c + d*x])/(24*d) + (7*a*b*Cosh[c + d*x]*(a + b*Sinh[c + d*x])^2)/(12*d) + (b*Cosh[c + d*x]*(a +
b*Sinh[c + d*x])^3)/(4*d)

Rule 2735

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((a + b*Sin[c + d*x])^(n
- 1)/(d*n)), x] + Dist[1/n, Int[(a + b*Sin[c + d*x])^(n - 2)*Simp[a^2*n + b^2*(n - 1) + a*b*(2*n - 1)*Sin[c +
d*x], x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[n, 1] && IntegerQ[2*n]

Rule 2813

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(2*a*c +
b*d)*(x/2), x] + (-Simp[(b*c + a*d)*(Cos[e + f*x]/f), x] - Simp[b*d*Cos[e + f*x]*(Sin[e + f*x]/(2*f)), x]) /;
FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]

Rule 2832

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-d
)*Cos[e + f*x]*((a + b*Sin[e + f*x])^m/(f*(m + 1))), x] + Dist[1/(m + 1), Int[(a + b*Sin[e + f*x])^(m - 1)*Sim
p[b*d*m + a*c*(m + 1) + (a*d*m + b*c*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[
b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && GtQ[m, 0] && IntegerQ[2*m]

Rubi steps

\begin {align*} \int (a+b \sinh (c+d x))^4 \, dx &=\frac {b \cosh (c+d x) (a+b \sinh (c+d x))^3}{4 d}+\frac {1}{4} \int (a+b \sinh (c+d x))^2 \left (4 a^2-3 b^2+7 a b \sinh (c+d x)\right ) \, dx\\ &=\frac {7 a b \cosh (c+d x) (a+b \sinh (c+d x))^2}{12 d}+\frac {b \cosh (c+d x) (a+b \sinh (c+d x))^3}{4 d}+\frac {1}{12} \int (a+b \sinh (c+d x)) \left (a \left (12 a^2-23 b^2\right )+b \left (26 a^2-9 b^2\right ) \sinh (c+d x)\right ) \, dx\\ &=\frac {1}{8} \left (8 a^4-24 a^2 b^2+3 b^4\right ) x+\frac {a b \left (19 a^2-16 b^2\right ) \cosh (c+d x)}{6 d}+\frac {b^2 \left (26 a^2-9 b^2\right ) \cosh (c+d x) \sinh (c+d x)}{24 d}+\frac {7 a b \cosh (c+d x) (a+b \sinh (c+d x))^2}{12 d}+\frac {b \cosh (c+d x) (a+b \sinh (c+d x))^3}{4 d}\\ \end {align*}

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Mathematica [A]
time = 0.26, size = 108, normalized size = 0.79 \begin {gather*} \frac {96 a b \left (4 a^2-3 b^2\right ) \cosh (c+d x)+32 a b^3 \cosh (3 (c+d x))+3 \left (4 \left (8 a^4-24 a^2 b^2+3 b^4\right ) (c+d x)+8 \left (6 a^2 b^2-b^4\right ) \sinh (2 (c+d x))+b^4 \sinh (4 (c+d x))\right )}{96 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sinh[c + d*x])^4,x]

[Out]

(96*a*b*(4*a^2 - 3*b^2)*Cosh[c + d*x] + 32*a*b^3*Cosh[3*(c + d*x)] + 3*(4*(8*a^4 - 24*a^2*b^2 + 3*b^4)*(c + d*
x) + 8*(6*a^2*b^2 - b^4)*Sinh[2*(c + d*x)] + b^4*Sinh[4*(c + d*x)]))/(96*d)

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Maple [A]
time = 0.86, size = 108, normalized size = 0.79

method	result	size
default	\(x \,a^{4}+\frac {\left (-\frac {1}{2} b^{4}+3 a^{2} b^{2}\right ) \sinh \left (2 d x +2 c \right )}{2 d}+\frac {\left (4 a^{3} b -3 a \,b^{3}\right ) \cosh \left (d x +c \right )}{d}+\frac {a \,b^{3} \cosh \left (3 d x +3 c \right )}{3 d}+\frac {3 x \,b^{4}}{8}-3 x \,a^{2} b^{2}+\frac {b^{4} \sinh \left (4 d x +4 c \right )}{32 d}\)	\(108\)
risch	\(x \,a^{4}+\frac {3 x \,b^{4}}{8}-3 x \,a^{2} b^{2}+\frac {b^{4} {\mathrm e}^{4 d x +4 c}}{64 d}+\frac {a \,b^{3} {\mathrm e}^{3 d x +3 c}}{6 d}+\frac {3 b^{2} {\mathrm e}^{2 d x +2 c} a^{2}}{4 d}-\frac {b^{4} {\mathrm e}^{2 d x +2 c}}{8 d}+\frac {2 a^{3} b \,{\mathrm e}^{d x +c}}{d}-\frac {3 a \,b^{3} {\mathrm e}^{d x +c}}{2 d}+\frac {2 a^{3} b \,{\mathrm e}^{-d x -c}}{d}-\frac {3 a \,b^{3} {\mathrm e}^{-d x -c}}{2 d}-\frac {3 b^{2} {\mathrm e}^{-2 d x -2 c} a^{2}}{4 d}+\frac {b^{4} {\mathrm e}^{-2 d x -2 c}}{8 d}+\frac {a \,b^{3} {\mathrm e}^{-3 d x -3 c}}{6 d}-\frac {b^{4} {\mathrm e}^{-4 d x -4 c}}{64 d}\)	\(232\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*a^4+1/2*(-1/2*b^4+3*a^2*b^2)*sinh(2*d*x+2*c)/d+(4*a^3*b-3*a*b^3)/d*cosh(d*x+c)+1/3*a*b^3/d*cosh(3*d*x+3*c)+3
/8*x*b^4-3*x*a^2*b^2+1/32*b^4*sinh(4*d*x+4*c)/d

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Maxima [A]
time = 0.34, size = 182, normalized size = 1.33 \begin {gather*} \frac {1}{64} \, b^{4} {\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 {3}{4} \, a^{2} b^{2} {\left (4 \, x - \frac {e^{\left (2 \, d x + 2 \, c\right )}}{d} + \frac {e^{\left (-2 \, d x - 2 \, c\right )}}{d}\right )} + a^{4} x + \frac {1}{6} \, a b^{3} {\left (\frac {e^{\left (3 \, d x + 3 \, c\right )}}{d} - \frac {9 \, e^{\left (d x + c\right )}}{d} - \frac {9 \, e^{\left (-d x - c\right )}}{d} + \frac {e^{\left (-3 \, d x - 3 \, c\right )}}{d}\right )} + \frac {4 \, a^{3} b \cosh \left (d x + c\right )}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/64*b^4*(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) - 3/4*a^
2*b^2*(4*x - e^(2*d*x + 2*c)/d + e^(-2*d*x - 2*c)/d) + a^4*x + 1/6*a*b^3*(e^(3*d*x + 3*c)/d - 9*e^(d*x + c)/d
- 9*e^(-d*x - c)/d + e^(-3*d*x - 3*c)/d) + 4*a^3*b*cosh(d*x + c)/d

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Fricas [A]
time = 0.34, size = 146, normalized size = 1.07 \begin {gather*} \frac {3 \, b^{4} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + 8 \, a b^{3} \cosh \left (d x + c\right )^{3} + 24 \, a b^{3} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{2} + 3 \, {\left (8 \, a^{4} - 24 \, a^{2} b^{2} + 3 \, b^{4}\right )} d x + 24 \, {\left (4 \, a^{3} b - 3 \, a b^{3}\right )} \cosh \left (d x + c\right ) + 3 \, {\left (b^{4} \cosh \left (d x + c\right )^{3} + 4 \, {\left (6 \, a^{2} b^{2} - b^{4}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )}{24 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(3*b^4*cosh(d*x + c)*sinh(d*x + c)^3 + 8*a*b^3*cosh(d*x + c)^3 + 24*a*b^3*cosh(d*x + c)*sinh(d*x + c)^2 +
3*(8*a^4 - 24*a^2*b^2 + 3*b^4)*d*x + 24*(4*a^3*b - 3*a*b^3)*cosh(d*x + c) + 3*(b^4*cosh(d*x + c)^3 + 4*(6*a^2
*b^2 - b^4)*cosh(d*x + c))*sinh(d*x + c))/d

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Sympy [A]
time = 0.20, size = 240, normalized size = 1.75 \begin {gather*} \begin {cases} a^{4} x + \frac {4 a^{3} b \cosh {\left (c + d x \right )}}{d} + 3 a^{2} b^{2} x \sinh ^{2}{\left (c + d x \right )} - 3 a^{2} b^{2} x \cosh ^{2}{\left (c + d x \right )} + \frac {3 a^{2} b^{2} \sinh {\left (c + d x \right )} \cosh {\left (c + d x \right )}}{d} + \frac {4 a b^{3} \sinh ^{2}{\left (c + d x \right )} \cosh {\left (c + d x \right )}}{d} - \frac {8 a b^{3} \cosh ^{3}{\left (c + d x \right )}}{3 d} + \frac {3 b^{4} x \sinh ^{4}{\left (c + d x \right )}}{8} - \frac {3 b^{4} x \sinh ^{2}{\left (c + d x \right )} \cosh ^{2}{\left (c + d x \right )}}{4} + \frac {3 b^{4} x \cosh ^{4}{\left (c + d x \right )}}{8} + \frac {5 b^{4} \sinh ^{3}{\left (c + d x \right )} \cosh {\left (c + d x \right )}}{8 d} - \frac {3 b^{4} \sinh {\left (c + d x \right )} \cosh ^{3}{\left (c + d x \right )}}{8 d} & \text {for}\: d \neq 0 \\x \left (a + b \sinh {\left (c \right )}\right )^{4} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a**4*x + 4*a**3*b*cosh(c + d*x)/d + 3*a**2*b**2*x*sinh(c + d*x)**2 - 3*a**2*b**2*x*cosh(c + d*x)**2
+ 3*a**2*b**2*sinh(c + d*x)*cosh(c + d*x)/d + 4*a*b**3*sinh(c + d*x)**2*cosh(c + d*x)/d - 8*a*b**3*cosh(c + d
*x)**3/(3*d) + 3*b**4*x*sinh(c + d*x)**4/8 - 3*b**4*x*sinh(c + d*x)**2*cosh(c + d*x)**2/4 + 3*b**4*x*cosh(c +
d*x)**4/8 + 5*b**4*sinh(c + d*x)**3*cosh(c + d*x)/(8*d) - 3*b**4*sinh(c + d*x)*cosh(c + d*x)**3/(8*d), Ne(d, 0
)), (x*(a + b*sinh(c))**4, True))

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Giac [A]
time = 0.45, size = 200, normalized size = 1.46 \begin {gather*} \frac {b^{4} e^{\left (4 \, d x + 4 \, c\right )}}{64 \, d} + \frac {a b^{3} e^{\left (3 \, d x + 3 \, c\right )}}{6 \, d} + \frac {a b^{3} e^{\left (-3 \, d x - 3 \, c\right )}}{6 \, d} - \frac {b^{4} e^{\left (-4 \, d x - 4 \, c\right )}}{64 \, d} + \frac {1}{8} \, {\left (8 \, a^{4} - 24 \, a^{2} b^{2} + 3 \, b^{4}\right )} x + \frac {{\left (6 \, a^{2} b^{2} - b^{4}\right )} e^{\left (2 \, d x + 2 \, c\right )}}{8 \, d} + \frac {{\left (4 \, a^{3} b - 3 \, a b^{3}\right )} e^{\left (d x + c\right )}}{2 \, d} + \frac {{\left (4 \, a^{3} b - 3 \, a b^{3}\right )} e^{\left (-d x - c\right )}}{2 \, d} - \frac {{\left (6 \, a^{2} b^{2} - b^{4}\right )} e^{\left (-2 \, d x - 2 \, c\right )}}{8 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/64*b^4*e^(4*d*x + 4*c)/d + 1/6*a*b^3*e^(3*d*x + 3*c)/d + 1/6*a*b^3*e^(-3*d*x - 3*c)/d - 1/64*b^4*e^(-4*d*x -
4*c)/d + 1/8*(8*a^4 - 24*a^2*b^2 + 3*b^4)*x + 1/8*(6*a^2*b^2 - b^4)*e^(2*d*x + 2*c)/d + 1/2*(4*a^3*b - 3*a*b^
3)*e^(d*x + c)/d + 1/2*(4*a^3*b - 3*a*b^3)*e^(-d*x - c)/d - 1/8*(6*a^2*b^2 - b^4)*e^(-2*d*x - 2*c)/d

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Mupad [B]
time = 0.34, size = 114, normalized size = 0.83 \begin {gather*} \frac {\frac {3\,b^4\,\mathrm {sinh}\left (4\,c+4\,d\,x\right )}{4}-6\,b^4\,\mathrm {sinh}\left (2\,c+2\,d\,x\right )+8\,a\,b^3\,\mathrm {cosh}\left (3\,c+3\,d\,x\right )+36\,a^2\,b^2\,\mathrm {sinh}\left (2\,c+2\,d\,x\right )-72\,a\,b^3\,\mathrm {cosh}\left (c+d\,x\right )+96\,a^3\,b\,\mathrm {cosh}\left (c+d\,x\right )+24\,a^4\,d\,x+9\,b^4\,d\,x-72\,a^2\,b^2\,d\,x}{24\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*sinh(c + d*x))^4,x)

[Out]

((3*b^4*sinh(4*c + 4*d*x))/4 - 6*b^4*sinh(2*c + 2*d*x) + 8*a*b^3*cosh(3*c + 3*d*x) + 36*a^2*b^2*sinh(2*c + 2*d
*x) - 72*a*b^3*cosh(c + d*x) + 96*a^3*b*cosh(c + d*x) + 24*a^4*d*x + 9*b^4*d*x - 72*a^2*b^2*d*x)/(24*d)

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