3.18 \(\int (c+d x)^2 \cosh ^3(a+b x) \, dx\)

Optimal. Leaf size=123 \[ \frac {2 d^2 \sinh ^3(a+b x)}{27 b^3}+\frac {14 d^2 \sinh (a+b x)}{9 b^3}-\frac {2 d (c+d x) \cosh ^3(a+b x)}{9 b^2}-\frac {4 d (c+d x) \cosh (a+b x)}{3 b^2}+\frac {2 (c+d x)^2 \sinh (a+b x)}{3 b}+\frac {(c+d x)^2 \sinh (a+b x) \cosh ^2(a+b x)}{3 b} \]

[Out]

-4/3*d*(d*x+c)*cosh(b*x+a)/b^2-2/9*d*(d*x+c)*cosh(b*x+a)^3/b^2+14/9*d^2*sinh(b*x+a)/b^3+2/3*(d*x+c)^2*sinh(b*x
+a)/b+1/3*(d*x+c)^2*cosh(b*x+a)^2*sinh(b*x+a)/b+2/27*d^2*sinh(b*x+a)^3/b^3

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Rubi [A]  time = 0.10, antiderivative size = 123, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3311, 3296, 2637, 2633} \[ -\frac {2 d (c+d x) \cosh ^3(a+b x)}{9 b^2}-\frac {4 d (c+d x) \cosh (a+b x)}{3 b^2}+\frac {2 d^2 \sinh ^3(a+b x)}{27 b^3}+\frac {14 d^2 \sinh (a+b x)}{9 b^3}+\frac {2 (c+d x)^2 \sinh (a+b x)}{3 b}+\frac {(c+d x)^2 \sinh (a+b x) \cosh ^2(a+b x)}{3 b} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^2*Cosh[a + b*x]^3,x]

[Out]

(-4*d*(c + d*x)*Cosh[a + b*x])/(3*b^2) - (2*d*(c + d*x)*Cosh[a + b*x]^3)/(9*b^2) + (14*d^2*Sinh[a + b*x])/(9*b
^3) + (2*(c + d*x)^2*Sinh[a + b*x])/(3*b) + ((c + d*x)^2*Cosh[a + b*x]^2*Sinh[a + b*x])/(3*b) + (2*d^2*Sinh[a
+ b*x]^3)/(27*b^3)

Rule 2633

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3296

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> -Simp[((c + d*x)^m*Cos[e + f*x])/f, x] +
Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

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 ^3(a+b x) \, dx &=-\frac {2 d (c+d x) \cosh ^3(a+b x)}{9 b^2}+\frac {(c+d x)^2 \cosh ^2(a+b x) \sinh (a+b x)}{3 b}+\frac {2}{3} \int (c+d x)^2 \cosh (a+b x) \, dx+\frac {\left (2 d^2\right ) \int \cosh ^3(a+b x) \, dx}{9 b^2}\\ &=-\frac {2 d (c+d x) \cosh ^3(a+b x)}{9 b^2}+\frac {2 (c+d x)^2 \sinh (a+b x)}{3 b}+\frac {(c+d x)^2 \cosh ^2(a+b x) \sinh (a+b x)}{3 b}-\frac {(4 d) \int (c+d x) \sinh (a+b x) \, dx}{3 b}+\frac {\left (2 i d^2\right ) \operatorname {Subst}\left (\int \left (1-x^2\right ) \, dx,x,-i \sinh (a+b x)\right )}{9 b^3}\\ &=-\frac {4 d (c+d x) \cosh (a+b x)}{3 b^2}-\frac {2 d (c+d x) \cosh ^3(a+b x)}{9 b^2}+\frac {2 d^2 \sinh (a+b x)}{9 b^3}+\frac {2 (c+d x)^2 \sinh (a+b x)}{3 b}+\frac {(c+d x)^2 \cosh ^2(a+b x) \sinh (a+b x)}{3 b}+\frac {2 d^2 \sinh ^3(a+b x)}{27 b^3}+\frac {\left (4 d^2\right ) \int \cosh (a+b x) \, dx}{3 b^2}\\ &=-\frac {4 d (c+d x) \cosh (a+b x)}{3 b^2}-\frac {2 d (c+d x) \cosh ^3(a+b x)}{9 b^2}+\frac {14 d^2 \sinh (a+b x)}{9 b^3}+\frac {2 (c+d x)^2 \sinh (a+b x)}{3 b}+\frac {(c+d x)^2 \cosh ^2(a+b x) \sinh (a+b x)}{3 b}+\frac {2 d^2 \sinh ^3(a+b x)}{27 b^3}\\ \end {align*}

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Mathematica [A]  time = 0.55, size = 93, normalized size = 0.76 \[ \frac {2 \sinh (a+b x) \left (\cosh (2 (a+b x)) \left (9 b^2 (c+d x)^2+2 d^2\right )+45 b^2 (c+d x)^2+82 d^2\right )-162 b d (c+d x) \cosh (a+b x)-6 b d (c+d x) \cosh (3 (a+b x))}{108 b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)^2*Cosh[a + b*x]^3,x]

[Out]

(-162*b*d*(c + d*x)*Cosh[a + b*x] - 6*b*d*(c + d*x)*Cosh[3*(a + b*x)] + 2*(82*d^2 + 45*b^2*(c + d*x)^2 + (2*d^
2 + 9*b^2*(c + d*x)^2)*Cosh[2*(a + b*x)])*Sinh[a + b*x])/(108*b^3)

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fricas [A]  time = 0.54, size = 199, normalized size = 1.62 \[ -\frac {6 \, {\left (b d^{2} x + b c d\right )} \cosh \left (b x + a\right )^{3} + 18 \, {\left (b d^{2} x + b c d\right )} \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{2} - {\left (9 \, b^{2} d^{2} x^{2} + 18 \, b^{2} c d x + 9 \, b^{2} c^{2} + 2 \, d^{2}\right )} \sinh \left (b x + a\right )^{3} + 162 \, {\left (b d^{2} x + b c d\right )} \cosh \left (b x + a\right ) - 3 \, {\left (27 \, b^{2} d^{2} x^{2} + 54 \, b^{2} c d x + 27 \, b^{2} c^{2} + {\left (9 \, b^{2} d^{2} x^{2} + 18 \, b^{2} c d x + 9 \, b^{2} c^{2} + 2 \, d^{2}\right )} \cosh \left (b x + a\right )^{2} + 54 \, d^{2}\right )} \sinh \left (b x + a\right )}{108 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^2*cosh(b*x+a)^3,x, algorithm="fricas")

[Out]

-1/108*(6*(b*d^2*x + b*c*d)*cosh(b*x + a)^3 + 18*(b*d^2*x + b*c*d)*cosh(b*x + a)*sinh(b*x + a)^2 - (9*b^2*d^2*
x^2 + 18*b^2*c*d*x + 9*b^2*c^2 + 2*d^2)*sinh(b*x + a)^3 + 162*(b*d^2*x + b*c*d)*cosh(b*x + a) - 3*(27*b^2*d^2*
x^2 + 54*b^2*c*d*x + 27*b^2*c^2 + (9*b^2*d^2*x^2 + 18*b^2*c*d*x + 9*b^2*c^2 + 2*d^2)*cosh(b*x + a)^2 + 54*d^2)
*sinh(b*x + a))/b^3

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giac [B]  time = 0.16, size = 230, normalized size = 1.87 \[ \frac {{\left (9 \, b^{2} d^{2} x^{2} + 18 \, b^{2} c d x + 9 \, b^{2} c^{2} - 6 \, b d^{2} x - 6 \, b c d + 2 \, d^{2}\right )} e^{\left (3 \, b x + 3 \, a\right )}}{216 \, b^{3}} + \frac {3 \, {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2} - 2 \, b d^{2} x - 2 \, b c d + 2 \, d^{2}\right )} e^{\left (b x + a\right )}}{8 \, b^{3}} - \frac {3 \, {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2} + 2 \, b d^{2} x + 2 \, b c d + 2 \, d^{2}\right )} e^{\left (-b x - a\right )}}{8 \, b^{3}} - \frac {{\left (9 \, b^{2} d^{2} x^{2} + 18 \, b^{2} c d x + 9 \, b^{2} c^{2} + 6 \, b d^{2} x + 6 \, b c d + 2 \, d^{2}\right )} e^{\left (-3 \, b x - 3 \, a\right )}}{216 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^2*cosh(b*x+a)^3,x, algorithm="giac")

[Out]

1/216*(9*b^2*d^2*x^2 + 18*b^2*c*d*x + 9*b^2*c^2 - 6*b*d^2*x - 6*b*c*d + 2*d^2)*e^(3*b*x + 3*a)/b^3 + 3/8*(b^2*
d^2*x^2 + 2*b^2*c*d*x + b^2*c^2 - 2*b*d^2*x - 2*b*c*d + 2*d^2)*e^(b*x + a)/b^3 - 3/8*(b^2*d^2*x^2 + 2*b^2*c*d*
x + b^2*c^2 + 2*b*d^2*x + 2*b*c*d + 2*d^2)*e^(-b*x - a)/b^3 - 1/216*(9*b^2*d^2*x^2 + 18*b^2*c*d*x + 9*b^2*c^2
+ 6*b*d^2*x + 6*b*c*d + 2*d^2)*e^(-3*b*x - 3*a)/b^3

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maple [B]  time = 0.16, size = 302, normalized size = 2.46 \[ \frac {\frac {d^{2} \left (\frac {2 \left (b x +a \right )^{2} \sinh \left (b x +a \right )}{3}+\frac {\left (b x +a \right )^{2} \sinh \left (b x +a \right ) \left (\cosh ^{2}\left (b x +a \right )\right )}{3}-\frac {4 \left (b x +a \right ) \cosh \left (b x +a \right )}{3}+\frac {40 \sinh \left (b x +a \right )}{27}-\frac {2 \left (b x +a \right ) \left (\cosh ^{3}\left (b x +a \right )\right )}{9}+\frac {2 \left (\cosh ^{2}\left (b x +a \right )\right ) \sinh \left (b x +a \right )}{27}\right )}{b^{2}}-\frac {2 d^{2} a \left (\frac {2 \left (b x +a \right ) \sinh \left (b x +a \right )}{3}+\frac {\left (b x +a \right ) \sinh \left (b x +a \right ) \left (\cosh ^{2}\left (b x +a \right )\right )}{3}-\frac {2 \cosh \left (b x +a \right )}{3}-\frac {\left (\cosh ^{3}\left (b x +a \right )\right )}{9}\right )}{b^{2}}+\frac {d^{2} a^{2} \left (\frac {2}{3}+\frac {\left (\cosh ^{2}\left (b x +a \right )\right )}{3}\right ) \sinh \left (b x +a \right )}{b^{2}}+\frac {2 c d \left (\frac {2 \left (b x +a \right ) \sinh \left (b x +a \right )}{3}+\frac {\left (b x +a \right ) \sinh \left (b x +a \right ) \left (\cosh ^{2}\left (b x +a \right )\right )}{3}-\frac {2 \cosh \left (b x +a \right )}{3}-\frac {\left (\cosh ^{3}\left (b x +a \right )\right )}{9}\right )}{b}-\frac {2 c d a \left (\frac {2}{3}+\frac {\left (\cosh ^{2}\left (b x +a \right )\right )}{3}\right ) \sinh \left (b x +a \right )}{b}+c^{2} \left (\frac {2}{3}+\frac {\left (\cosh ^{2}\left (b x +a \right )\right )}{3}\right ) \sinh \left (b x +a \right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)^2*cosh(b*x+a)^3,x)

[Out]

1/b*(1/b^2*d^2*(2/3*(b*x+a)^2*sinh(b*x+a)+1/3*(b*x+a)^2*sinh(b*x+a)*cosh(b*x+a)^2-4/3*(b*x+a)*cosh(b*x+a)+40/2
7*sinh(b*x+a)-2/9*(b*x+a)*cosh(b*x+a)^3+2/27*cosh(b*x+a)^2*sinh(b*x+a))-2/b^2*d^2*a*(2/3*(b*x+a)*sinh(b*x+a)+1
/3*(b*x+a)*sinh(b*x+a)*cosh(b*x+a)^2-2/3*cosh(b*x+a)-1/9*cosh(b*x+a)^3)+1/b^2*d^2*a^2*(2/3+1/3*cosh(b*x+a)^2)*
sinh(b*x+a)+2/b*c*d*(2/3*(b*x+a)*sinh(b*x+a)+1/3*(b*x+a)*sinh(b*x+a)*cosh(b*x+a)^2-2/3*cosh(b*x+a)-1/9*cosh(b*
x+a)^3)-2/b*c*d*a*(2/3+1/3*cosh(b*x+a)^2)*sinh(b*x+a)+c^2*(2/3+1/3*cosh(b*x+a)^2)*sinh(b*x+a))

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maxima [B]  time = 0.42, size = 272, normalized size = 2.21 \[ \frac {1}{36} \, c d {\left (\frac {{\left (3 \, b x e^{\left (3 \, a\right )} - e^{\left (3 \, a\right )}\right )} e^{\left (3 \, b x\right )}}{b^{2}} + \frac {27 \, {\left (b x e^{a} - e^{a}\right )} e^{\left (b x\right )}}{b^{2}} - \frac {27 \, {\left (b x + 1\right )} e^{\left (-b x - a\right )}}{b^{2}} - \frac {{\left (3 \, b x + 1\right )} e^{\left (-3 \, b x - 3 \, a\right )}}{b^{2}}\right )} + \frac {1}{24} \, c^{2} {\left (\frac {e^{\left (3 \, b x + 3 \, a\right )}}{b} + \frac {9 \, e^{\left (b x + a\right )}}{b} - \frac {9 \, e^{\left (-b x - a\right )}}{b} - \frac {e^{\left (-3 \, b x - 3 \, a\right )}}{b}\right )} + \frac {1}{216} \, d^{2} {\left (\frac {{\left (9 \, b^{2} x^{2} e^{\left (3 \, a\right )} - 6 \, b x e^{\left (3 \, a\right )} + 2 \, e^{\left (3 \, a\right )}\right )} e^{\left (3 \, b x\right )}}{b^{3}} + \frac {81 \, {\left (b^{2} x^{2} e^{a} - 2 \, b x e^{a} + 2 \, e^{a}\right )} e^{\left (b x\right )}}{b^{3}} - \frac {81 \, {\left (b^{2} x^{2} + 2 \, b x + 2\right )} e^{\left (-b x - a\right )}}{b^{3}} - \frac {{\left (9 \, b^{2} x^{2} + 6 \, b x + 2\right )} e^{\left (-3 \, b x - 3 \, a\right )}}{b^{3}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^2*cosh(b*x+a)^3,x, algorithm="maxima")

[Out]

1/36*c*d*((3*b*x*e^(3*a) - e^(3*a))*e^(3*b*x)/b^2 + 27*(b*x*e^a - e^a)*e^(b*x)/b^2 - 27*(b*x + 1)*e^(-b*x - a)
/b^2 - (3*b*x + 1)*e^(-3*b*x - 3*a)/b^2) + 1/24*c^2*(e^(3*b*x + 3*a)/b + 9*e^(b*x + a)/b - 9*e^(-b*x - a)/b -
e^(-3*b*x - 3*a)/b) + 1/216*d^2*((9*b^2*x^2*e^(3*a) - 6*b*x*e^(3*a) + 2*e^(3*a))*e^(3*b*x)/b^3 + 81*(b^2*x^2*e
^a - 2*b*x*e^a + 2*e^a)*e^(b*x)/b^3 - 81*(b^2*x^2 + 2*b*x + 2)*e^(-b*x - a)/b^3 - (9*b^2*x^2 + 6*b*x + 2)*e^(-
3*b*x - 3*a)/b^3)

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mupad [B]  time = 1.22, size = 183, normalized size = 1.49 \[ \frac {\frac {3\,d^2\,\mathrm {sinh}\left (a+b\,x\right )}{2}+\frac {d^2\,\mathrm {sinh}\left (3\,a+3\,b\,x\right )}{54}+\frac {3\,b^2\,c^2\,\mathrm {sinh}\left (a+b\,x\right )}{4}+\frac {b^2\,c^2\,\mathrm {sinh}\left (3\,a+3\,b\,x\right )}{12}+\frac {3\,b^2\,d^2\,x^2\,\mathrm {sinh}\left (a+b\,x\right )}{4}-\frac {b\,c\,d\,\mathrm {cosh}\left (3\,a+3\,b\,x\right )}{18}-\frac {3\,b\,d^2\,x\,\mathrm {cosh}\left (a+b\,x\right )}{2}+\frac {b^2\,d^2\,x^2\,\mathrm {sinh}\left (3\,a+3\,b\,x\right )}{12}-\frac {b\,d^2\,x\,\mathrm {cosh}\left (3\,a+3\,b\,x\right )}{18}-\frac {3\,b\,c\,d\,\mathrm {cosh}\left (a+b\,x\right )}{2}+\frac {b^2\,c\,d\,x\,\mathrm {sinh}\left (3\,a+3\,b\,x\right )}{6}+\frac {3\,b^2\,c\,d\,x\,\mathrm {sinh}\left (a+b\,x\right )}{2}}{b^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(a + b*x)^3*(c + d*x)^2,x)

[Out]

((3*d^2*sinh(a + b*x))/2 + (d^2*sinh(3*a + 3*b*x))/54 + (3*b^2*c^2*sinh(a + b*x))/4 + (b^2*c^2*sinh(3*a + 3*b*
x))/12 + (3*b^2*d^2*x^2*sinh(a + b*x))/4 - (b*c*d*cosh(3*a + 3*b*x))/18 - (3*b*d^2*x*cosh(a + b*x))/2 + (b^2*d
^2*x^2*sinh(3*a + 3*b*x))/12 - (b*d^2*x*cosh(3*a + 3*b*x))/18 - (3*b*c*d*cosh(a + b*x))/2 + (b^2*c*d*x*sinh(3*
a + 3*b*x))/6 + (3*b^2*c*d*x*sinh(a + b*x))/2)/b^3

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sympy [A]  time = 2.25, size = 284, normalized size = 2.31 \[ \begin {cases} - \frac {2 c^{2} \sinh ^{3}{\left (a + b x \right )}}{3 b} + \frac {c^{2} \sinh {\left (a + b x \right )} \cosh ^{2}{\left (a + b x \right )}}{b} - \frac {4 c d x \sinh ^{3}{\left (a + b x \right )}}{3 b} + \frac {2 c d x \sinh {\left (a + b x \right )} \cosh ^{2}{\left (a + b x \right )}}{b} - \frac {2 d^{2} x^{2} \sinh ^{3}{\left (a + b x \right )}}{3 b} + \frac {d^{2} x^{2} \sinh {\left (a + b x \right )} \cosh ^{2}{\left (a + b x \right )}}{b} + \frac {4 c d \sinh ^{2}{\left (a + b x \right )} \cosh {\left (a + b x \right )}}{3 b^{2}} - \frac {14 c d \cosh ^{3}{\left (a + b x \right )}}{9 b^{2}} + \frac {4 d^{2} x \sinh ^{2}{\left (a + b x \right )} \cosh {\left (a + b x \right )}}{3 b^{2}} - \frac {14 d^{2} x \cosh ^{3}{\left (a + b x \right )}}{9 b^{2}} - \frac {40 d^{2} \sinh ^{3}{\left (a + b x \right )}}{27 b^{3}} + \frac {14 d^{2} \sinh {\left (a + b x \right )} \cosh ^{2}{\left (a + b x \right )}}{9 b^{3}} & \text {for}\: b \neq 0 \\\left (c^{2} x + c d x^{2} + \frac {d^{2} x^{3}}{3}\right ) \cosh ^{3}{\relax (a )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)**2*cosh(b*x+a)**3,x)

[Out]

Piecewise((-2*c**2*sinh(a + b*x)**3/(3*b) + c**2*sinh(a + b*x)*cosh(a + b*x)**2/b - 4*c*d*x*sinh(a + b*x)**3/(
3*b) + 2*c*d*x*sinh(a + b*x)*cosh(a + b*x)**2/b - 2*d**2*x**2*sinh(a + b*x)**3/(3*b) + d**2*x**2*sinh(a + b*x)
*cosh(a + b*x)**2/b + 4*c*d*sinh(a + b*x)**2*cosh(a + b*x)/(3*b**2) - 14*c*d*cosh(a + b*x)**3/(9*b**2) + 4*d**
2*x*sinh(a + b*x)**2*cosh(a + b*x)/(3*b**2) - 14*d**2*x*cosh(a + b*x)**3/(9*b**2) - 40*d**2*sinh(a + b*x)**3/(
27*b**3) + 14*d**2*sinh(a + b*x)*cosh(a + b*x)**2/(9*b**3), Ne(b, 0)), ((c**2*x + c*d*x**2 + d**2*x**3/3)*cosh
(a)**3, True))

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