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

Optimal. Leaf size=94 \[ \frac {2 a \sinh (c+d x)}{d^3}-\frac {2 a x \cosh (c+d x)}{d^2}+\frac {a x^2 \sinh (c+d x)}{d}-\frac {6 b \cosh (c+d x)}{d^4}+\frac {6 b x \sinh (c+d x)}{d^3}-\frac {3 b x^2 \cosh (c+d x)}{d^2}+\frac {b x^3 \sinh (c+d x)}{d} \]

[Out]

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

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Rubi [A]  time = 0.22, antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {6742, 3296, 2637, 2638} \[ \frac {2 a \sinh (c+d x)}{d^3}-\frac {2 a x \cosh (c+d x)}{d^2}+\frac {a x^2 \sinh (c+d x)}{d}-\frac {3 b x^2 \cosh (c+d x)}{d^2}+\frac {6 b x \sinh (c+d x)}{d^3}-\frac {6 b \cosh (c+d x)}{d^4}+\frac {b x^3 \sinh (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Int[x^2*(a + b*x)*Cosh[c + d*x],x]

[Out]

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

Rule 2637

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

Rule 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[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 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

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

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Mathematica [A]  time = 0.14, size = 65, normalized size = 0.69 \[ \frac {d \left (a \left (d^2 x^2+2\right )+b x \left (d^2 x^2+6\right )\right ) \sinh (c+d x)-\left (2 a d^2 x+3 b \left (d^2 x^2+2\right )\right ) \cosh (c+d x)}{d^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.00, size = 67, normalized size = 0.71 \[ -\frac {{\left (3 \, b d^{2} x^{2} + 2 \, a d^{2} x + 6 \, b\right )} \cosh \left (d x + c\right ) - {\left (b d^{3} x^{3} + a d^{3} x^{2} + 6 \, b d x + 2 \, a d\right )} \sinh \left (d x + c\right )}{d^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.11, size = 116, normalized size = 1.23 \[ \frac {{\left (b d^{3} x^{3} + a d^{3} x^{2} - 3 \, b d^{2} x^{2} - 2 \, a d^{2} x + 6 \, b d x + 2 \, a d - 6 \, b\right )} e^{\left (d x + c\right )}}{2 \, d^{4}} - \frac {{\left (b d^{3} x^{3} + a d^{3} x^{2} + 3 \, b d^{2} x^{2} + 2 \, a d^{2} x + 6 \, b d x + 2 \, a d + 6 \, b\right )} e^{\left (-d x - c\right )}}{2 \, d^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.04, size = 224, normalized size = 2.38 \[ \frac {\frac {b \left (\left (d x +c \right )^{3} \sinh \left (d x +c \right )-3 \left (d x +c \right )^{2} \cosh \left (d x +c \right )+6 \left (d x +c \right ) \sinh \left (d x +c \right )-6 \cosh \left (d x +c \right )\right )}{d}-\frac {3 b c \left (\left (d x +c \right )^{2} \sinh \left (d x +c \right )-2 \left (d x +c \right ) \cosh \left (d x +c \right )+2 \sinh \left (d x +c \right )\right )}{d}+\frac {3 b \,c^{2} \left (\left (d x +c \right ) \sinh \left (d x +c \right )-\cosh \left (d x +c \right )\right )}{d}-\frac {b \,c^{3} \sinh \left (d x +c \right )}{d}+a \left (\left (d x +c \right )^{2} \sinh \left (d x +c \right )-2 \left (d x +c \right ) \cosh \left (d x +c \right )+2 \sinh \left (d x +c \right )\right )-2 a c \left (\left (d x +c \right ) \sinh \left (d x +c \right )-\cosh \left (d x +c \right )\right )+a \,c^{2} \sinh \left (d x +c \right )}{d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.41, size = 196, normalized size = 2.09 \[ -\frac {1}{24} \, d {\left (\frac {4 \, {\left (d^{3} x^{3} e^{c} - 3 \, d^{2} x^{2} e^{c} + 6 \, d x e^{c} - 6 \, e^{c}\right )} a e^{\left (d x\right )}}{d^{4}} + \frac {4 \, {\left (d^{3} x^{3} + 3 \, d^{2} x^{2} + 6 \, d x + 6\right )} a e^{\left (-d x - c\right )}}{d^{4}} + \frac {3 \, {\left (d^{4} x^{4} e^{c} - 4 \, d^{3} x^{3} e^{c} + 12 \, d^{2} x^{2} e^{c} - 24 \, d x e^{c} + 24 \, e^{c}\right )} b e^{\left (d x\right )}}{d^{5}} + \frac {3 \, {\left (d^{4} x^{4} + 4 \, d^{3} x^{3} + 12 \, d^{2} x^{2} + 24 \, d x + 24\right )} b e^{\left (-d x - c\right )}}{d^{5}}\right )} + \frac {1}{12} \, {\left (3 \, b x^{4} + 4 \, a x^{3}\right )} \cosh \left (d x + c\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*d*(4*(d^3*x^3*e^c - 3*d^2*x^2*e^c + 6*d*x*e^c - 6*e^c)*a*e^(d*x)/d^4 + 4*(d^3*x^3 + 3*d^2*x^2 + 6*d*x +
6)*a*e^(-d*x - c)/d^4 + 3*(d^4*x^4*e^c - 4*d^3*x^3*e^c + 12*d^2*x^2*e^c - 24*d*x*e^c + 24*e^c)*b*e^(d*x)/d^5 +
3*(d^4*x^4 + 4*d^3*x^3 + 12*d^2*x^2 + 24*d*x + 24)*b*e^(-d*x - c)/d^5) + 1/12*(3*b*x^4 + 4*a*x^3)*cosh(d*x +
c)

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mupad [B]  time = 0.96, size = 92, normalized size = 0.98 \[ \frac {2\,a\,\mathrm {sinh}\left (c+d\,x\right )+6\,b\,x\,\mathrm {sinh}\left (c+d\,x\right )}{d^3}-\frac {2\,a\,x\,\mathrm {cosh}\left (c+d\,x\right )+3\,b\,x^2\,\mathrm {cosh}\left (c+d\,x\right )}{d^2}+\frac {a\,x^2\,\mathrm {sinh}\left (c+d\,x\right )+b\,x^3\,\mathrm {sinh}\left (c+d\,x\right )}{d}-\frac {6\,b\,\mathrm {cosh}\left (c+d\,x\right )}{d^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*cosh(c + d*x)*(a + b*x),x)

[Out]

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

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sympy [A]  time = 0.94, size = 117, normalized size = 1.24 \[ \begin {cases} \frac {a x^{2} \sinh {\left (c + d x \right )}}{d} - \frac {2 a x \cosh {\left (c + d x \right )}}{d^{2}} + \frac {2 a \sinh {\left (c + d x \right )}}{d^{3}} + \frac {b x^{3} \sinh {\left (c + d x \right )}}{d} - \frac {3 b x^{2} \cosh {\left (c + d x \right )}}{d^{2}} + \frac {6 b x \sinh {\left (c + d x \right )}}{d^{3}} - \frac {6 b \cosh {\left (c + d x \right )}}{d^{4}} & \text {for}\: d \neq 0 \\\left (\frac {a x^{3}}{3} + \frac {b x^{4}}{4}\right ) \cosh {\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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