3.10 \(\int (c+d x)^2 \cosh ^2(a+b x) \, dx\)

Optimal. Leaf size=95 \[ \frac {d^2 \sinh (a+b x) \cosh (a+b x)}{4 b^3}-\frac {d (c+d x) \cosh ^2(a+b x)}{2 b^2}+\frac {(c+d x)^2 \sinh (a+b x) \cosh (a+b x)}{2 b}+\frac {d^2 x}{4 b^2}+\frac {(c+d x)^3}{6 d} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3311, 32, 2635, 8} \[ -\frac {d (c+d x) \cosh ^2(a+b x)}{2 b^2}+\frac {d^2 \sinh (a+b x) \cosh (a+b x)}{4 b^3}+\frac {(c+d x)^2 \sinh (a+b x) \cosh (a+b x)}{2 b}+\frac {d^2 x}{4 b^2}+\frac {(c+d x)^3}{6 d} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^2*Cosh[a + b*x]^2,x]

[Out]

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 2635

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

Rule 3311

Int[((c_.) + (d_.)*(x_))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(d*m*(c + d*x)^(m - 1)*(
b*Sin[e + f*x])^n)/(f^2*n^2), x] + (Dist[(b^2*(n - 1))/n, Int[(c + d*x)^m*(b*Sin[e + f*x])^(n - 2), x], x] - D
ist[(d^2*m*(m - 1))/(f^2*n^2), Int[(c + d*x)^(m - 2)*(b*Sin[e + f*x])^n, x], x] - Simp[(b*(c + d*x)^m*Cos[e +
f*x]*(b*Sin[e + f*x])^(n - 1))/(f*n), x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n, 1] && GtQ[m, 1]

Rubi steps

\begin {align*} \int (c+d x)^2 \cosh ^2(a+b x) \, dx &=-\frac {d (c+d x) \cosh ^2(a+b x)}{2 b^2}+\frac {(c+d x)^2 \cosh (a+b x) \sinh (a+b x)}{2 b}+\frac {1}{2} \int (c+d x)^2 \, dx+\frac {d^2 \int \cosh ^2(a+b x) \, dx}{2 b^2}\\ &=\frac {(c+d x)^3}{6 d}-\frac {d (c+d x) \cosh ^2(a+b x)}{2 b^2}+\frac {d^2 \cosh (a+b x) \sinh (a+b x)}{4 b^3}+\frac {(c+d x)^2 \cosh (a+b x) \sinh (a+b x)}{2 b}+\frac {d^2 \int 1 \, dx}{4 b^2}\\ &=\frac {d^2 x}{4 b^2}+\frac {(c+d x)^3}{6 d}-\frac {d (c+d x) \cosh ^2(a+b x)}{2 b^2}+\frac {d^2 \cosh (a+b x) \sinh (a+b x)}{4 b^3}+\frac {(c+d x)^2 \cosh (a+b x) \sinh (a+b x)}{2 b}\\ \end {align*}

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Mathematica [A]  time = 0.30, size = 75, normalized size = 0.79 \[ \frac {3 \sinh (2 (a+b x)) \left (2 b^2 (c+d x)^2+d^2\right )-6 b d (c+d x) \cosh (2 (a+b x))+4 b^3 x \left (3 c^2+3 c d x+d^2 x^2\right )}{24 b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.72, size = 123, normalized size = 1.29 \[ \frac {2 \, b^{3} d^{2} x^{3} + 6 \, b^{3} c d x^{2} + 6 \, b^{3} c^{2} x - 3 \, {\left (b d^{2} x + b c d\right )} \cosh \left (b x + a\right )^{2} + 3 \, {\left (2 \, b^{2} d^{2} x^{2} + 4 \, b^{2} c d x + 2 \, b^{2} c^{2} + d^{2}\right )} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) - 3 \, {\left (b d^{2} x + b c d\right )} \sinh \left (b x + a\right )^{2}}{12 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(2*b^3*d^2*x^3 + 6*b^3*c*d*x^2 + 6*b^3*c^2*x - 3*(b*d^2*x + b*c*d)*cosh(b*x + a)^2 + 3*(2*b^2*d^2*x^2 + 4
*b^2*c*d*x + 2*b^2*c^2 + d^2)*cosh(b*x + a)*sinh(b*x + a) - 3*(b*d^2*x + b*c*d)*sinh(b*x + a)^2)/b^3

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giac [A]  time = 0.12, size = 136, normalized size = 1.43 \[ \frac {1}{6} \, d^{2} x^{3} + \frac {1}{2} \, c d x^{2} + \frac {1}{2} \, c^{2} x + \frac {{\left (2 \, b^{2} d^{2} x^{2} + 4 \, b^{2} c d x + 2 \, b^{2} c^{2} - 2 \, b d^{2} x - 2 \, b c d + d^{2}\right )} e^{\left (2 \, b x + 2 \, a\right )}}{16 \, b^{3}} - \frac {{\left (2 \, b^{2} d^{2} x^{2} + 4 \, b^{2} c d x + 2 \, b^{2} c^{2} + 2 \, b d^{2} x + 2 \, b c d + d^{2}\right )} e^{\left (-2 \, b x - 2 \, a\right )}}{16 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*d^2*x^3 + 1/2*c*d*x^2 + 1/2*c^2*x + 1/16*(2*b^2*d^2*x^2 + 4*b^2*c*d*x + 2*b^2*c^2 - 2*b*d^2*x - 2*b*c*d +
d^2)*e^(2*b*x + 2*a)/b^3 - 1/16*(2*b^2*d^2*x^2 + 4*b^2*c*d*x + 2*b^2*c^2 + 2*b*d^2*x + 2*b*c*d + d^2)*e^(-2*b*
x - 2*a)/b^3

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maple [B]  time = 0.05, size = 262, normalized size = 2.76 \[ \frac {\frac {d^{2} \left (\frac {\left (b x +a \right )^{2} \cosh \left (b x +a \right ) \sinh \left (b x +a \right )}{2}+\frac {\left (b x +a \right )^{3}}{6}-\frac {\left (b x +a \right ) \left (\cosh ^{2}\left (b x +a \right )\right )}{2}+\frac {\cosh \left (b x +a \right ) \sinh \left (b x +a \right )}{4}+\frac {b x}{4}+\frac {a}{4}\right )}{b^{2}}-\frac {2 d^{2} a \left (\frac {\left (b x +a \right ) \cosh \left (b x +a \right ) \sinh \left (b x +a \right )}{2}+\frac {\left (b x +a \right )^{2}}{4}-\frac {\left (\cosh ^{2}\left (b x +a \right )\right )}{4}\right )}{b^{2}}+\frac {d^{2} a^{2} \left (\frac {\cosh \left (b x +a \right ) \sinh \left (b x +a \right )}{2}+\frac {b x}{2}+\frac {a}{2}\right )}{b^{2}}+\frac {2 c d \left (\frac {\left (b x +a \right ) \cosh \left (b x +a \right ) \sinh \left (b x +a \right )}{2}+\frac {\left (b x +a \right )^{2}}{4}-\frac {\left (\cosh ^{2}\left (b x +a \right )\right )}{4}\right )}{b}-\frac {2 c d a \left (\frac {\cosh \left (b x +a \right ) \sinh \left (b x +a \right )}{2}+\frac {b x}{2}+\frac {a}{2}\right )}{b}+c^{2} \left (\frac {\cosh \left (b x +a \right ) \sinh \left (b x +a \right )}{2}+\frac {b x}{2}+\frac {a}{2}\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.38, size = 165, normalized size = 1.74 \[ \frac {1}{8} \, {\left (4 \, x^{2} + \frac {{\left (2 \, b x e^{\left (2 \, a\right )} - e^{\left (2 \, a\right )}\right )} e^{\left (2 \, b x\right )}}{b^{2}} - \frac {{\left (2 \, b x + 1\right )} e^{\left (-2 \, b x - 2 \, a\right )}}{b^{2}}\right )} c d + \frac {1}{48} \, {\left (8 \, x^{3} + \frac {3 \, {\left (2 \, b^{2} x^{2} e^{\left (2 \, a\right )} - 2 \, b x e^{\left (2 \, a\right )} + e^{\left (2 \, a\right )}\right )} e^{\left (2 \, b x\right )}}{b^{3}} - \frac {3 \, {\left (2 \, b^{2} x^{2} + 2 \, b x + 1\right )} e^{\left (-2 \, b x - 2 \, a\right )}}{b^{3}}\right )} d^{2} + \frac {1}{8} \, c^{2} {\left (4 \, x + \frac {e^{\left (2 \, b x + 2 \, a\right )}}{b} - \frac {e^{\left (-2 \, b x - 2 \, a\right )}}{b}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(4*x^2 + (2*b*x*e^(2*a) - e^(2*a))*e^(2*b*x)/b^2 - (2*b*x + 1)*e^(-2*b*x - 2*a)/b^2)*c*d + 1/48*(8*x^3 + 3
*(2*b^2*x^2*e^(2*a) - 2*b*x*e^(2*a) + e^(2*a))*e^(2*b*x)/b^3 - 3*(2*b^2*x^2 + 2*b*x + 1)*e^(-2*b*x - 2*a)/b^3)
*d^2 + 1/8*c^2*(4*x + e^(2*b*x + 2*a)/b - e^(-2*b*x - 2*a)/b)

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mupad [B]  time = 0.98, size = 127, normalized size = 1.34 \[ \frac {c^2\,x}{2}+\frac {d^2\,x^3}{6}+\frac {c^2\,\mathrm {sinh}\left (2\,a+2\,b\,x\right )}{4\,b}+\frac {d^2\,\mathrm {sinh}\left (2\,a+2\,b\,x\right )}{8\,b^3}+\frac {c\,d\,x^2}{2}-\frac {d^2\,x\,\mathrm {cosh}\left (2\,a+2\,b\,x\right )}{4\,b^2}+\frac {d^2\,x^2\,\mathrm {sinh}\left (2\,a+2\,b\,x\right )}{4\,b}-\frac {c\,d\,\mathrm {cosh}\left (2\,a+2\,b\,x\right )}{4\,b^2}+\frac {c\,d\,x\,\mathrm {sinh}\left (2\,a+2\,b\,x\right )}{2\,b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(a + b*x)^2*(c + d*x)^2,x)

[Out]

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

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sympy [A]  time = 1.22, size = 264, normalized size = 2.78 \[ \begin {cases} - \frac {c^{2} x \sinh ^{2}{\left (a + b x \right )}}{2} + \frac {c^{2} x \cosh ^{2}{\left (a + b x \right )}}{2} - \frac {c d x^{2} \sinh ^{2}{\left (a + b x \right )}}{2} + \frac {c d x^{2} \cosh ^{2}{\left (a + b x \right )}}{2} - \frac {d^{2} x^{3} \sinh ^{2}{\left (a + b x \right )}}{6} + \frac {d^{2} x^{3} \cosh ^{2}{\left (a + b x \right )}}{6} + \frac {c^{2} \sinh {\left (a + b x \right )} \cosh {\left (a + b x \right )}}{2 b} + \frac {c d x \sinh {\left (a + b x \right )} \cosh {\left (a + b x \right )}}{b} + \frac {d^{2} x^{2} \sinh {\left (a + b x \right )} \cosh {\left (a + b x \right )}}{2 b} - \frac {c d \cosh ^{2}{\left (a + b x \right )}}{2 b^{2}} - \frac {d^{2} x \sinh ^{2}{\left (a + b x \right )}}{4 b^{2}} - \frac {d^{2} x \cosh ^{2}{\left (a + b x \right )}}{4 b^{2}} + \frac {d^{2} \sinh {\left (a + b x \right )} \cosh {\left (a + b x \right )}}{4 b^{3}} & \text {for}\: b \neq 0 \\\left (c^{2} x + c d x^{2} + \frac {d^{2} x^{3}}{3}\right ) \cosh ^{2}{\relax (a )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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