3.3.95 \(\int \cosh ^2(c+d x) (a+b \sinh ^2(c+d x))^2 \, dx\) [295]

Optimal. Leaf size=119 \[ \frac {1}{16} \left (8 a^2-4 a b+b^2\right ) x+\frac {\left (8 a^2-4 a b+b^2\right ) \cosh (c+d x) \sinh (c+d x)}{16 d}+\frac {(8 a-3 b) b \cosh ^3(c+d x) \sinh (c+d x)}{24 d}+\frac {b \cosh ^5(c+d x) \sinh (c+d x) \left (a-(a-b) \tanh ^2(c+d x)\right )}{6 d} \]

[Out]

1/16*(8*a^2-4*a*b+b^2)*x+1/16*(8*a^2-4*a*b+b^2)*cosh(d*x+c)*sinh(d*x+c)/d+1/24*(8*a-3*b)*b*cosh(d*x+c)^3*sinh(
d*x+c)/d+1/6*b*cosh(d*x+c)^5*sinh(d*x+c)*(a-(a-b)*tanh(d*x+c)^2)/d

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Rubi [A]
time = 0.09, antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3270, 424, 393, 205, 212} \begin {gather*} \frac {\left (8 a^2-4 a b+b^2\right ) \sinh (c+d x) \cosh (c+d x)}{16 d}+\frac {1}{16} x \left (8 a^2-4 a b+b^2\right )+\frac {b (8 a-3 b) \sinh (c+d x) \cosh ^3(c+d x)}{24 d}+\frac {b \sinh (c+d x) \cosh ^5(c+d x) \left (a-(a-b) \tanh ^2(c+d x)\right )}{6 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((8*a^2 - 4*a*b + b^2)*x)/16 + ((8*a^2 - 4*a*b + b^2)*Cosh[c + d*x]*Sinh[c + d*x])/(16*d) + ((8*a - 3*b)*b*Cos
h[c + d*x]^3*Sinh[c + d*x])/(24*d) + (b*Cosh[c + d*x]^5*Sinh[c + d*x]*(a - (a - b)*Tanh[c + d*x]^2))/(6*d)

Rule 205

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^n)^(p + 1)/(a*n*(p + 1))), x] + Dist[(n*(p
+ 1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (
IntegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[
p])

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 424

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

Rule 3270

Int[cos[(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/f, Subst[Int[(a + (a + b)*ff^2*x^2)^p/(1 + ff^2*x^2)^(m/2 + p + 1), x], x, T
an[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[m/2] && IntegerQ[p]

Rubi steps

\begin {align*} \int \cosh ^2(c+d x) \left (a+b \sinh ^2(c+d x)\right )^2 \, dx &=\frac {\text {Subst}\left (\int \frac {\left (a-(a-b) x^2\right )^2}{\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) \left (a-(a-b) \tanh ^2(c+d x)\right )}{6 d}-\frac {\text {Subst}\left (\int \frac {-a (6 a-b)+3 (a-b) (2 a-b) x^2}{\left (1-x^2\right )^3} \, dx,x,\tanh (c+d x)\right )}{6 d}\\ &=\frac {(8 a-3 b) b \cosh ^3(c+d x) \sinh (c+d x)}{24 d}+\frac {b \cosh ^5(c+d x) \sinh (c+d x) \left (a-(a-b) \tanh ^2(c+d x)\right )}{6 d}+\frac {\left (8 a^2-4 a b+b^2\right ) \text {Subst}\left (\int \frac {1}{\left (1-x^2\right )^2} \, dx,x,\tanh (c+d x)\right )}{8 d}\\ &=\frac {\left (8 a^2-4 a b+b^2\right ) \cosh (c+d x) \sinh (c+d x)}{16 d}+\frac {(8 a-3 b) b \cosh ^3(c+d x) \sinh (c+d x)}{24 d}+\frac {b \cosh ^5(c+d x) \sinh (c+d x) \left (a-(a-b) \tanh ^2(c+d x)\right )}{6 d}+\frac {\left (8 a^2-4 a b+b^2\right ) \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\tanh (c+d x)\right )}{16 d}\\ &=\frac {1}{16} \left (8 a^2-4 a b+b^2\right ) x+\frac {\left (8 a^2-4 a b+b^2\right ) \cosh (c+d x) \sinh (c+d x)}{16 d}+\frac {(8 a-3 b) b \cosh ^3(c+d x) \sinh (c+d x)}{24 d}+\frac {b \cosh ^5(c+d x) \sinh (c+d x) \left (a-(a-b) \tanh ^2(c+d x)\right )}{6 d}\\ \end {align*}

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Mathematica [A]
time = 0.21, size = 79, normalized size = 0.66 \begin {gather*} \frac {12 \left (8 a^2-4 a b+b^2\right ) (c+d x)+3 \left (16 a^2-b^2\right ) \sinh (2 (c+d x))+3 (4 a-b) b \sinh (4 (c+d x))+b^2 \sinh (6 (c+d x))}{192 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(12*(8*a^2 - 4*a*b + b^2)*(c + d*x) + 3*(16*a^2 - b^2)*Sinh[2*(c + d*x)] + 3*(4*a - b)*b*Sinh[4*(c + d*x)] + b
^2*Sinh[6*(c + d*x)])/(192*d)

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Maple [A]
time = 1.25, size = 134, normalized size = 1.13

method	result	size
derivativedivides	\(\frac {b^{2} \left (\frac {\left (\sinh ^{3}\left (d x +c \right )\right ) \left (\cosh ^{3}\left (d x +c \right )\right )}{6}-\frac {\sinh \left (d x +c \right ) \left (\cosh ^{3}\left (d x +c \right )\right )}{8}+\frac {\cosh \left (d x +c \right ) \sinh \left (d x +c \right )}{16}+\frac {d x}{16}+\frac {c}{16}\right )+2 a b \left (\frac {\sinh \left (d x +c \right ) \left (\cosh ^{3}\left (d x +c \right )\right )}{4}-\frac {\cosh \left (d x +c \right ) \sinh \left (d x +c \right )}{8}-\frac {d x}{8}-\frac {c}{8}\right )+a^{2} \left (\frac {\cosh \left (d x +c \right ) \sinh \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(134\)
default	\(\frac {b^{2} \left (\frac {\left (\sinh ^{3}\left (d x +c \right )\right ) \left (\cosh ^{3}\left (d x +c \right )\right )}{6}-\frac {\sinh \left (d x +c \right ) \left (\cosh ^{3}\left (d x +c \right )\right )}{8}+\frac {\cosh \left (d x +c \right ) \sinh \left (d x +c \right )}{16}+\frac {d x}{16}+\frac {c}{16}\right )+2 a b \left (\frac {\sinh \left (d x +c \right ) \left (\cosh ^{3}\left (d x +c \right )\right )}{4}-\frac {\cosh \left (d x +c \right ) \sinh \left (d x +c \right )}{8}-\frac {d x}{8}-\frac {c}{8}\right )+a^{2} \left (\frac {\cosh \left (d x +c \right ) \sinh \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(134\)
risch	\(\frac {a^{2} x}{2}-\frac {a b x}{4}+\frac {b^{2} x}{16}+\frac {b^{2} {\mathrm e}^{6 d x +6 c}}{384 d}+\frac {{\mathrm e}^{4 d x +4 c} a b}{32 d}-\frac {{\mathrm e}^{4 d x +4 c} b^{2}}{128 d}+\frac {{\mathrm e}^{2 d x +2 c} a^{2}}{8 d}-\frac {{\mathrm e}^{2 d x +2 c} b^{2}}{128 d}-\frac {{\mathrm e}^{-2 d x -2 c} a^{2}}{8 d}+\frac {{\mathrm e}^{-2 d x -2 c} b^{2}}{128 d}-\frac {{\mathrm e}^{-4 d x -4 c} a b}{32 d}+\frac {{\mathrm e}^{-4 d x -4 c} b^{2}}{128 d}-\frac {b^{2} {\mathrm e}^{-6 d x -6 c}}{384 d}\)	\(187\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(b^2*(1/6*sinh(d*x+c)^3*cosh(d*x+c)^3-1/8*sinh(d*x+c)*cosh(d*x+c)^3+1/16*cosh(d*x+c)*sinh(d*x+c)+1/16*d*x+
1/16*c)+2*a*b*(1/4*sinh(d*x+c)*cosh(d*x+c)^3-1/8*cosh(d*x+c)*sinh(d*x+c)-1/8*d*x-1/8*c)+a^2*(1/2*cosh(d*x+c)*s
inh(d*x+c)+1/2*d*x+1/2*c))

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Maxima [A]
time = 0.27, size = 171, normalized size = 1.44 \begin {gather*} \frac {1}{8} \, a^{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 )} - \frac {1}{384} \, b^{2} {\left (\frac {{\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, e^{\left (-4 \, d x - 4 \, c\right )} - 1\right )} e^{\left (6 \, d x + 6 \, c\right )}}{d} - \frac {24 \, {\left (d x + c\right )}}{d} - \frac {3 \, e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, e^{\left (-4 \, d x - 4 \, c\right )} - e^{\left (-6 \, d x - 6 \, c\right )}}{d}\right )} - \frac {1}{32} \, a b {\left (\frac {8 \, {\left (d x + c\right )}}{d} - \frac {e^{\left (4 \, d x + 4 \, c\right )}}{d} + \frac {e^{\left (-4 \, d x - 4 \, c\right )}}{d}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*a^2*(4*x + e^(2*d*x + 2*c)/d - e^(-2*d*x - 2*c)/d) - 1/384*b^2*((3*e^(-2*d*x - 2*c) + 3*e^(-4*d*x - 4*c) -
1)*e^(6*d*x + 6*c)/d - 24*(d*x + c)/d - (3*e^(-2*d*x - 2*c) + 3*e^(-4*d*x - 4*c) - e^(-6*d*x - 6*c))/d) - 1/3
2*a*b*(8*(d*x + c)/d - e^(4*d*x + 4*c)/d + e^(-4*d*x - 4*c)/d)

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Fricas [A]
time = 0.38, size = 143, normalized size = 1.20 \begin {gather*} \frac {3 \, b^{2} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{5} + 2 \, {\left (5 \, b^{2} \cosh \left (d x + c\right )^{3} + 3 \, {\left (4 \, a b - b^{2}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{3} + 6 \, {\left (8 \, a^{2} - 4 \, a b + b^{2}\right )} d x + 3 \, {\left (b^{2} \cosh \left (d x + c\right )^{5} + 2 \, {\left (4 \, a b - b^{2}\right )} \cosh \left (d x + c\right )^{3} + {\left (16 \, a^{2} - b^{2}\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(cosh(d*x+c)^2*(a+b*sinh(d*x+c)^2)^2,x, algorithm="fricas")

[Out]

1/96*(3*b^2*cosh(d*x + c)*sinh(d*x + c)^5 + 2*(5*b^2*cosh(d*x + c)^3 + 3*(4*a*b - b^2)*cosh(d*x + c))*sinh(d*x
+ c)^3 + 6*(8*a^2 - 4*a*b + b^2)*d*x + 3*(b^2*cosh(d*x + c)^5 + 2*(4*a*b - b^2)*cosh(d*x + c)^3 + (16*a^2 - 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. 314 vs. \(2 (105) = 210\).
time = 0.49, size = 314, normalized size = 2.64 \begin {gather*} \begin {cases} - \frac {a^{2} x \sinh ^{2}{\left (c + d x \right )}}{2} + \frac {a^{2} x \cosh ^{2}{\left (c + d x \right )}}{2} + \frac {a^{2} \sinh {\left (c + d x \right )} \cosh {\left (c + d x \right )}}{2 d} - \frac {a b x \sinh ^{4}{\left (c + d x \right )}}{4} + \frac {a b x \sinh ^{2}{\left (c + d x \right )} \cosh ^{2}{\left (c + d x \right )}}{2} - \frac {a b x \cosh ^{4}{\left (c + d x \right )}}{4} + \frac {a b \sinh ^{3}{\left (c + d x \right )} \cosh {\left (c + d x \right )}}{4 d} + \frac {a b \sinh {\left (c + d x \right )} \cosh ^{3}{\left (c + d x \right )}}{4 d} - \frac {b^{2} x \sinh ^{6}{\left (c + d x \right )}}{16} + \frac {3 b^{2} x \sinh ^{4}{\left (c + d x \right )} \cosh ^{2}{\left (c + d x \right )}}{16} - \frac {3 b^{2} x \sinh ^{2}{\left (c + d x \right )} \cosh ^{4}{\left (c + d x \right )}}{16} + \frac {b^{2} x \cosh ^{6}{\left (c + d x \right )}}{16} + \frac {b^{2} \sinh ^{5}{\left (c + d x \right )} \cosh {\left (c + d x \right )}}{16 d} + \frac {b^{2} \sinh ^{3}{\left (c + d x \right )} \cosh ^{3}{\left (c + d x \right )}}{6 d} - \frac {b^{2} \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 ^{2}{\left (c \right )}\right )^{2} \cosh ^{2}{\left (c \right )} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.42, size = 149, normalized size = 1.25 \begin {gather*} \frac {1}{16} \, {\left (8 \, a^{2} - 4 \, a b + b^{2}\right )} x + \frac {b^{2} e^{\left (6 \, d x + 6 \, c\right )}}{384 \, d} - \frac {b^{2} e^{\left (-6 \, d x - 6 \, c\right )}}{384 \, d} + \frac {{\left (4 \, a b - b^{2}\right )} e^{\left (4 \, d x + 4 \, c\right )}}{128 \, d} + \frac {{\left (16 \, a^{2} - b^{2}\right )} e^{\left (2 \, d x + 2 \, c\right )}}{128 \, d} - \frac {{\left (16 \, a^{2} - b^{2}\right )} e^{\left (-2 \, d x - 2 \, c\right )}}{128 \, d} - \frac {{\left (4 \, a b - b^{2}\right )} e^{\left (-4 \, d x - 4 \, c\right )}}{128 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.22, size = 95, normalized size = 0.80 \begin {gather*} \frac {12\,a^2\,\mathrm {sinh}\left (2\,c+2\,d\,x\right )-\frac {3\,b^2\,\mathrm {sinh}\left (2\,c+2\,d\,x\right )}{4}-\frac {3\,b^2\,\mathrm {sinh}\left (4\,c+4\,d\,x\right )}{4}+\frac {b^2\,\mathrm {sinh}\left (6\,c+6\,d\,x\right )}{4}+3\,a\,b\,\mathrm {sinh}\left (4\,c+4\,d\,x\right )+24\,a^2\,d\,x+3\,b^2\,d\,x-12\,a\,b\,d\,x}{48\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(12*a^2*sinh(2*c + 2*d*x) - (3*b^2*sinh(2*c + 2*d*x))/4 - (3*b^2*sinh(4*c + 4*d*x))/4 + (b^2*sinh(6*c + 6*d*x)
)/4 + 3*a*b*sinh(4*c + 4*d*x) + 24*a^2*d*x + 3*b^2*d*x - 12*a*b*d*x)/(48*d)

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