3.2.86 \(\int \sinh ^2(c+d x) (a+b \sinh ^4(c+d x)) \, dx\) [186]

Optimal. Leaf size=83 \[ -\frac {1}{16} (8 a+5 b) x+\frac {(8 a+11 b) \cosh (c+d x) \sinh (c+d x)}{16 d}-\frac {13 b \cosh ^3(c+d x) \sinh (c+d x)}{24 d}+\frac {b \cosh ^5(c+d x) \sinh (c+d x)}{6 d} \]

[Out]

-1/16*(8*a+5*b)*x+1/16*(8*a+11*b)*cosh(d*x+c)*sinh(d*x+c)/d-13/24*b*cosh(d*x+c)^3*sinh(d*x+c)/d+1/6*b*cosh(d*x
+c)^5*sinh(d*x+c)/d

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Rubi [A]
time = 0.07, antiderivative size = 83, normalized size of antiderivative = 1.00, 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 = {3296, 1271, 1171, 393, 212} \begin {gather*} \frac {(8 a+11 b) \sinh (c+d x) \cosh (c+d x)}{16 d}-\frac {1}{16} x (8 a+5 b)+\frac {b \sinh (c+d x) \cosh ^5(c+d x)}{6 d}-\frac {13 b \sinh (c+d x) \cosh ^3(c+d x)}{24 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/16*((8*a + 5*b)*x) + ((8*a + 11*b)*Cosh[c + d*x]*Sinh[c + d*x])/(16*d) - (13*b*Cosh[c + d*x]^3*Sinh[c + d*x
])/(24*d) + (b*Cosh[c + d*x]^5*Sinh[c + d*x])/(6*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 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 1271

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Simp[(-d)^(
m/2 - 1)*(c*d^2 - b*d*e + a*e^2)^p*x*((d + e*x^2)^(q + 1)/(2*e^(2*p + m/2)*(q + 1))), x] + Dist[1/(2*e^(2*p +
m/2)*(q + 1)), Int[(d + e*x^2)^(q + 1)*ExpandToSum[Together[(1/(d + e*x^2))*(2*e^(2*p + m/2)*(q + 1)*x^m*(a +
b*x^2 + c*x^4)^p - (-d)^(m/2 - 1)*(c*d^2 - b*d*e + a*e^2)^p*(d + e*(2*q + 3)*x^2))], x], x], x] /; FreeQ[{a, b
, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && ILtQ[q, -1] && IGtQ[m/2, 0]

Rule 3296

Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^4)^(p_.), x_Symbol] :> With[{ff = FreeF
actors[Tan[e + f*x], x]}, Dist[ff^(m + 1)/f, Subst[Int[x^m*((a + 2*a*ff^2*x^2 + (a + b)*ff^4*x^4)^p/(1 + ff^2*
x^2)^(m/2 + 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 ^2(c+d x) \left (a+b \sinh ^4(c+d x)\right ) \, dx &=\frac {\text {Subst}\left (\int \frac {x^2 \left (a-2 a x^2+(a+b) x^4\right )}{\left (1-x^2\right )^4} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac {b \cosh ^5(c+d x) \sinh (c+d x)}{6 d}+\frac {\text {Subst}\left (\int \frac {-b+6 (a-b) x^2-6 (a+b) x^4}{\left (1-x^2\right )^3} \, dx,x,\tanh (c+d x)\right )}{6 d}\\ &=-\frac {13 b \cosh ^3(c+d x) \sinh (c+d x)}{24 d}+\frac {b \cosh ^5(c+d x) \sinh (c+d x)}{6 d}-\frac {\text {Subst}\left (\int \frac {-9 b-24 (a+b) x^2}{\left (1-x^2\right )^2} \, dx,x,\tanh (c+d x)\right )}{24 d}\\ &=\frac {(8 a+11 b) \cosh (c+d x) \sinh (c+d x)}{16 d}-\frac {13 b \cosh ^3(c+d x) \sinh (c+d x)}{24 d}+\frac {b \cosh ^5(c+d x) \sinh (c+d x)}{6 d}-\frac {(8 a+5 b) \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\tanh (c+d x)\right )}{16 d}\\ &=-\frac {1}{16} (8 a+5 b) x+\frac {(8 a+11 b) \cosh (c+d x) \sinh (c+d x)}{16 d}-\frac {13 b \cosh ^3(c+d x) \sinh (c+d x)}{24 d}+\frac {b \cosh ^5(c+d x) \sinh (c+d x)}{6 d}\\ \end {align*}

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Mathematica [A]
time = 0.07, size = 63, normalized size = 0.76 \begin {gather*} \frac {-96 a c-60 b c-96 a d x-60 b d x+(48 a+45 b) \sinh (2 (c+d x))-9 b \sinh (4 (c+d x))+b \sinh (6 (c+d x))}{192 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-96*a*c - 60*b*c - 96*a*d*x - 60*b*d*x + (48*a + 45*b)*Sinh[2*(c + d*x)] - 9*b*Sinh[4*(c + d*x)] + b*Sinh[6*(
c + d*x)])/(192*d)

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Maple [A]
time = 1.03, size = 61, normalized size = 0.73

method	result	size
default	\(\frac {\left (\frac {15 b}{32}+\frac {a}{2}\right ) \sinh \left (2 d x +2 c \right )}{2 d}-\frac {a x}{2}-\frac {5 b x}{16}-\frac {3 b \sinh \left (4 d x +4 c \right )}{64 d}+\frac {b \sinh \left (6 d x +6 c \right )}{192 d}\)	\(61\)
risch	\(-\frac {5 b x}{16}-\frac {a x}{2}+\frac {b \,{\mathrm e}^{6 d x +6 c}}{384 d}-\frac {3 \,{\mathrm e}^{4 d x +4 c} b}{128 d}+\frac {15 b \,{\mathrm e}^{2 d x +2 c}}{128 d}+\frac {{\mathrm e}^{2 d x +2 c} a}{8 d}-\frac {15 \,{\mathrm e}^{-2 d x -2 c} b}{128 d}-\frac {{\mathrm e}^{-2 d x -2 c} a}{8 d}+\frac {3 \,{\mathrm e}^{-4 d x -4 c} b}{128 d}-\frac {b \,{\mathrm e}^{-6 d x -6 c}}{384 d}\)	\(130\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(15/32*b+1/2*a)*sinh(2*d*x+2*c)/d-1/2*a*x-5/16*b*x-3/64*b*sinh(4*d*x+4*c)/d+1/192*b*sinh(6*d*x+6*c)/d

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Maxima [A]
time = 0.27, size = 122, normalized size = 1.47 \begin {gather*} -\frac {1}{8} \, a {\left (4 \, x - \frac {e^{\left (2 \, d x + 2 \, c\right )}}{d} + \frac {e^{\left (-2 \, d x - 2 \, c\right )}}{d}\right )} - \frac {1}{384} \, 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)^2*(a+b*sinh(d*x+c)^4),x, algorithm="maxima")

[Out]

-1/8*a*(4*x - e^(2*d*x + 2*c)/d + e^(-2*d*x - 2*c)/d) - 1/384*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 = 0.42, size = 109, normalized size = 1.31 \begin {gather*} \frac {3 \, b \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{5} + 2 \, {\left (5 \, b \cosh \left (d x + c\right )^{3} - 9 \, b \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{3} - 6 \, {\left (8 \, a + 5 \, b\right )} d x + 3 \, {\left (b \cosh \left (d x + c\right )^{5} - 6 \, b \cosh \left (d x + c\right )^{3} + {\left (16 \, a + 15 \, b\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )}{96 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/96*(3*b*cosh(d*x + c)*sinh(d*x + c)^5 + 2*(5*b*cosh(d*x + c)^3 - 9*b*cosh(d*x + c))*sinh(d*x + c)^3 - 6*(8*a
+ 5*b)*d*x + 3*(b*cosh(d*x + c)^5 - 6*b*cosh(d*x + c)^3 + (16*a + 15*b)*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. 206 vs. \(2 (76) = 152\).
time = 0.46, size = 206, normalized size = 2.48 \begin {gather*} \begin {cases} \frac {a x \sinh ^{2}{\left (c + d x \right )}}{2} - \frac {a x \cosh ^{2}{\left (c + d x \right )}}{2} + \frac {a \sinh {\left (c + d x \right )} \cosh {\left (c + d x \right )}}{2 d} + \frac {5 b x \sinh ^{6}{\left (c + d x \right )}}{16} - \frac {15 b x \sinh ^{4}{\left (c + d x \right )} \cosh ^{2}{\left (c + d x \right )}}{16} + \frac {15 b x \sinh ^{2}{\left (c + d x \right )} \cosh ^{4}{\left (c + d x \right )}}{16} - \frac {5 b x \cosh ^{6}{\left (c + d x \right )}}{16} + \frac {11 b \sinh ^{5}{\left (c + d x \right )} \cosh {\left (c + d x \right )}}{16 d} - \frac {5 b \sinh ^{3}{\left (c + d x \right )} \cosh ^{3}{\left (c + d x \right )}}{6 d} + \frac {5 b \sinh {\left (c + d x \right )} \cosh ^{5}{\left (c + d x \right )}}{16 d} & \text {for}\: d \neq 0 \\x \left (a + b \sinh ^{4}{\left (c \right )}\right ) \sinh ^{2}{\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)**2*(a+b*sinh(d*x+c)**4),x)

[Out]

Piecewise((a*x*sinh(c + d*x)**2/2 - a*x*cosh(c + d*x)**2/2 + a*sinh(c + d*x)*cosh(c + d*x)/(2*d) + 5*b*x*sinh(
c + d*x)**6/16 - 15*b*x*sinh(c + d*x)**4*cosh(c + d*x)**2/16 + 15*b*x*sinh(c + d*x)**2*cosh(c + d*x)**4/16 - 5
*b*x*cosh(c + d*x)**6/16 + 11*b*sinh(c + d*x)**5*cosh(c + d*x)/(16*d) - 5*b*sinh(c + d*x)**3*cosh(c + d*x)**3/
(6*d) + 5*b*sinh(c + d*x)*cosh(c + d*x)**5/(16*d), Ne(d, 0)), (x*(a + b*sinh(c)**4)*sinh(c)**2, True))

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Giac [A]
time = 0.42, size = 113, normalized size = 1.36 \begin {gather*} -\frac {1}{16} \, {\left (8 \, a + 5 \, b\right )} x + \frac {b e^{\left (6 \, d x + 6 \, c\right )}}{384 \, d} - \frac {3 \, b e^{\left (4 \, d x + 4 \, c\right )}}{128 \, d} + \frac {{\left (16 \, a + 15 \, b\right )} e^{\left (2 \, d x + 2 \, c\right )}}{128 \, d} - \frac {{\left (16 \, a + 15 \, b\right )} e^{\left (-2 \, d x - 2 \, c\right )}}{128 \, d} + \frac {3 \, b e^{\left (-4 \, d x - 4 \, c\right )}}{128 \, d} - \frac {b e^{\left (-6 \, d x - 6 \, c\right )}}{384 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(8*a + 5*b)*x + 1/384*b*e^(6*d*x + 6*c)/d - 3/128*b*e^(4*d*x + 4*c)/d + 1/128*(16*a + 15*b)*e^(2*d*x + 2
*c)/d - 1/128*(16*a + 15*b)*e^(-2*d*x - 2*c)/d + 3/128*b*e^(-4*d*x - 4*c)/d - 1/384*b*e^(-6*d*x - 6*c)/d

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Mupad [B]
time = 0.16, size = 64, normalized size = 0.77 \begin {gather*} \frac {12\,a\,\mathrm {sinh}\left (2\,c+2\,d\,x\right )+\frac {45\,b\,\mathrm {sinh}\left (2\,c+2\,d\,x\right )}{4}-\frac {9\,b\,\mathrm {sinh}\left (4\,c+4\,d\,x\right )}{4}+\frac {b\,\mathrm {sinh}\left (6\,c+6\,d\,x\right )}{4}-24\,a\,d\,x-15\,b\,d\,x}{48\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(12*a*sinh(2*c + 2*d*x) + (45*b*sinh(2*c + 2*d*x))/4 - (9*b*sinh(4*c + 4*d*x))/4 + (b*sinh(6*c + 6*d*x))/4 - 2
4*a*d*x - 15*b*d*x)/(48*d)

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