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

Optimal. Leaf size=134 \[ -\frac {a^2 \cosh (c+d x)}{d^2}+\frac {a^2 x \sinh (c+d x)}{d}+\frac {4 a b \sinh (c+d x)}{d^3}-\frac {4 a b x \cosh (c+d x)}{d^2}+\frac {2 a b x^2 \sinh (c+d x)}{d}-\frac {6 b^2 \cosh (c+d x)}{d^4}+\frac {6 b^2 x \sinh (c+d x)}{d^3}-\frac {3 b^2 x^2 \cosh (c+d x)}{d^2}+\frac {b^2 x^3 \sinh (c+d x)}{d} \]

[Out]

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

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Rubi [A]  time = 0.21, antiderivative size = 134, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {6742, 3296, 2638, 2637} \[ -\frac {a^2 \cosh (c+d x)}{d^2}+\frac {a^2 x \sinh (c+d x)}{d}+\frac {4 a b \sinh (c+d x)}{d^3}-\frac {4 a b x \cosh (c+d x)}{d^2}+\frac {2 a b x^2 \sinh (c+d x)}{d}-\frac {3 b^2 x^2 \cosh (c+d x)}{d^2}+\frac {6 b^2 x \sinh (c+d x)}{d^3}-\frac {6 b^2 \cosh (c+d x)}{d^4}+\frac {b^2 x^3 \sinh (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Int[x*(a + b*x)^2*Cosh[c + d*x],x]

[Out]

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

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Mathematica [A]  time = 0.22, size = 87, normalized size = 0.65 \[ \frac {d \left (a^2 d^2 x+2 a b \left (d^2 x^2+2\right )+b^2 x \left (d^2 x^2+6\right )\right ) \sinh (c+d x)-\left (a^2 d^2+4 a b d^2 x+3 b^2 \left (d^2 x^2+2\right )\right ) \cosh (c+d x)}{d^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.53, size = 95, normalized size = 0.71 \[ -\frac {{\left (3 \, b^{2} d^{2} x^{2} + 4 \, a b d^{2} x + a^{2} d^{2} + 6 \, b^{2}\right )} \cosh \left (d x + c\right ) - {\left (b^{2} d^{3} x^{3} + 2 \, a b d^{3} x^{2} + 4 \, a b d + {\left (a^{2} d^{3} + 6 \, b^{2} d\right )} x\right )} \sinh \left (d x + c\right )}{d^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.12, size = 171, normalized size = 1.28 \[ \frac {{\left (b^{2} d^{3} x^{3} + 2 \, a b d^{3} x^{2} + a^{2} d^{3} x - 3 \, b^{2} d^{2} x^{2} - 4 \, a b d^{2} x - a^{2} d^{2} + 6 \, b^{2} d x + 4 \, a b d - 6 \, b^{2}\right )} e^{\left (d x + c\right )}}{2 \, d^{4}} - \frac {{\left (b^{2} d^{3} x^{3} + 2 \, a b d^{3} x^{2} + a^{2} d^{3} x + 3 \, b^{2} d^{2} x^{2} + 4 \, a b d^{2} x + a^{2} d^{2} + 6 \, b^{2} d x + 4 \, a b d + 6 \, b^{2}\right )} e^{\left (-d x - c\right )}}{2 \, d^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.04, size = 283, normalized size = 2.11 \[ \frac {\frac {b^{2} \left (\left (d x +c \right )^{3} \sinh \left (d x +c \right )-3 \left (d x +c \right )^{2} \cosh \left (d x +c \right )+6 \left (d x +c \right ) \sinh \left (d x +c \right )-6 \cosh \left (d x +c \right )\right )}{d^{2}}-\frac {3 b^{2} c \left (\left (d x +c \right )^{2} \sinh \left (d x +c \right )-2 \left (d x +c \right ) \cosh \left (d x +c \right )+2 \sinh \left (d x +c \right )\right )}{d^{2}}+\frac {2 b a \left (\left (d x +c \right )^{2} \sinh \left (d x +c \right )-2 \left (d x +c \right ) \cosh \left (d x +c \right )+2 \sinh \left (d x +c \right )\right )}{d}+\frac {3 b^{2} c^{2} \left (\left (d x +c \right ) \sinh \left (d x +c \right )-\cosh \left (d x +c \right )\right )}{d^{2}}-\frac {4 b c a \left (\left (d x +c \right ) \sinh \left (d x +c \right )-\cosh \left (d x +c \right )\right )}{d}+a^{2} \left (\left (d x +c \right ) \sinh \left (d x +c \right )-\cosh \left (d x +c \right )\right )-\frac {b^{2} c^{3} \sinh \left (d x +c \right )}{d^{2}}+\frac {2 b \,c^{2} a \sinh \left (d x +c \right )}{d}-c \,a^{2} \sinh \left (d x +c \right )}{d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.33, size = 275, normalized size = 2.05 \[ -\frac {1}{24} \, d {\left (\frac {6 \, {\left (d^{2} x^{2} e^{c} - 2 \, d x e^{c} + 2 \, e^{c}\right )} a^{2} e^{\left (d x\right )}}{d^{3}} + \frac {6 \, {\left (d^{2} x^{2} + 2 \, d x + 2\right )} a^{2} e^{\left (-d x - c\right )}}{d^{3}} + \frac {8 \, {\left (d^{3} x^{3} e^{c} - 3 \, d^{2} x^{2} e^{c} + 6 \, d x e^{c} - 6 \, e^{c}\right )} a b e^{\left (d x\right )}}{d^{4}} + \frac {8 \, {\left (d^{3} x^{3} + 3 \, d^{2} x^{2} + 6 \, d x + 6\right )} a b e^{\left (-d x - c\right )}}{d^{4}} + \frac {3 \, {\left (d^{4} x^{4} e^{c} - 4 \, d^{3} x^{3} e^{c} + 12 \, d^{2} x^{2} e^{c} - 24 \, d x e^{c} + 24 \, e^{c}\right )} b^{2} e^{\left (d x\right )}}{d^{5}} + \frac {3 \, {\left (d^{4} x^{4} + 4 \, d^{3} x^{3} + 12 \, d^{2} x^{2} + 24 \, d x + 24\right )} b^{2} e^{\left (-d x - c\right )}}{d^{5}}\right )} + \frac {1}{12} \, {\left (3 \, b^{2} x^{4} + 8 \, a b x^{3} + 6 \, a^{2} x^{2}\right )} \cosh \left (d x + c\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*d*(6*(d^2*x^2*e^c - 2*d*x*e^c + 2*e^c)*a^2*e^(d*x)/d^3 + 6*(d^2*x^2 + 2*d*x + 2)*a^2*e^(-d*x - c)/d^3 +
8*(d^3*x^3*e^c - 3*d^2*x^2*e^c + 6*d*x*e^c - 6*e^c)*a*b*e^(d*x)/d^4 + 8*(d^3*x^3 + 3*d^2*x^2 + 6*d*x + 6)*a*b*
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^2*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^2*e^(-d*x - c)/d^5) + 1/12*(3*b^2*x^4 + 8*a*b*x^3 + 6*a^2*x
^2)*cosh(d*x + c)

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mupad [B]  time = 0.92, size = 125, normalized size = 0.93 \[ \frac {b^2\,x^3\,\mathrm {sinh}\left (c+d\,x\right )}{d}-\frac {3\,b^2\,x^2\,\mathrm {cosh}\left (c+d\,x\right )}{d^2}-\frac {\mathrm {cosh}\left (c+d\,x\right )\,\left (a^2\,d^2+6\,b^2\right )}{d^4}+\frac {4\,a\,b\,\mathrm {sinh}\left (c+d\,x\right )}{d^3}+\frac {x\,\mathrm {sinh}\left (c+d\,x\right )\,\left (a^2\,d^2+6\,b^2\right )}{d^3}+\frac {2\,a\,b\,x^2\,\mathrm {sinh}\left (c+d\,x\right )}{d}-\frac {4\,a\,b\,x\,\mathrm {cosh}\left (c+d\,x\right )}{d^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*cosh(c + d*x)*(a + b*x)^2,x)

[Out]

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

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sympy [A]  time = 1.15, size = 172, normalized size = 1.28 \[ \begin {cases} \frac {a^{2} x \sinh {\left (c + d x \right )}}{d} - \frac {a^{2} \cosh {\left (c + d x \right )}}{d^{2}} + \frac {2 a b x^{2} \sinh {\left (c + d x \right )}}{d} - \frac {4 a b x \cosh {\left (c + d x \right )}}{d^{2}} + \frac {4 a b \sinh {\left (c + d x \right )}}{d^{3}} + \frac {b^{2} x^{3} \sinh {\left (c + d x \right )}}{d} - \frac {3 b^{2} x^{2} \cosh {\left (c + d x \right )}}{d^{2}} + \frac {6 b^{2} x \sinh {\left (c + d x \right )}}{d^{3}} - \frac {6 b^{2} \cosh {\left (c + d x \right )}}{d^{4}} & \text {for}\: d \neq 0 \\\left (\frac {a^{2} x^{2}}{2} + \frac {2 a b x^{3}}{3} + \frac {b^{2} x^{4}}{4}\right ) \cosh {\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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