\(\int x (a+b x^3) \cosh (c+d x) \, dx\) [81]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [B] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 15, antiderivative size = 94 \[ \int x \left (a+b x^3\right ) \cosh (c+d x) \, dx=-\frac {a \cosh (c+d x)}{d^2}-\frac {24 b x \cosh (c+d x)}{d^4}-\frac {4 b x^3 \cosh (c+d x)}{d^2}+\frac {24 b \sinh (c+d x)}{d^5}+\frac {a x \sinh (c+d x)}{d}+\frac {12 b x^2 \sinh (c+d x)}{d^3}+\frac {b x^4 \sinh (c+d x)}{d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {5395, 3377, 2718, 2717} \[ \int x \left (a+b x^3\right ) \cosh (c+d x) \, dx=-\frac {a \cosh (c+d x)}{d^2}+\frac {a x \sinh (c+d x)}{d}+\frac {24 b \sinh (c+d x)}{d^5}-\frac {24 b x \cosh (c+d x)}{d^4}+\frac {12 b x^2 \sinh (c+d x)}{d^3}-\frac {4 b x^3 \cosh (c+d x)}{d^2}+\frac {b x^4 \sinh (c+d x)}{d} \]

[In]

Int[x*(a + b*x^3)*Cosh[c + d*x],x]

[Out]

-((a*Cosh[c + d*x])/d^2) - (24*b*x*Cosh[c + d*x])/d^4 - (4*b*x^3*Cosh[c + d*x])/d^2 + (24*b*Sinh[c + d*x])/d^5
 + (a*x*Sinh[c + d*x])/d + (12*b*x^2*Sinh[c + d*x])/d^3 + (b*x^4*Sinh[c + d*x])/d

Rule 2717

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

Rule 2718

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

Rule 3377

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 5395

Int[Cosh[(c_.) + (d_.)*(x_)]*((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegra
nd[Cosh[c + d*x], (e*x)^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && IGtQ[p, 0]

Rubi steps \begin{align*} \text {integral}& = \int \left (a x \cosh (c+d x)+b x^4 \cosh (c+d x)\right ) \, dx \\ & = a \int x \cosh (c+d x) \, dx+b \int x^4 \cosh (c+d x) \, dx \\ & = \frac {a x \sinh (c+d x)}{d}+\frac {b x^4 \sinh (c+d x)}{d}-\frac {a \int \sinh (c+d x) \, dx}{d}-\frac {(4 b) \int x^3 \sinh (c+d x) \, dx}{d} \\ & = -\frac {a \cosh (c+d x)}{d^2}-\frac {4 b x^3 \cosh (c+d x)}{d^2}+\frac {a x \sinh (c+d x)}{d}+\frac {b x^4 \sinh (c+d x)}{d}+\frac {(12 b) \int x^2 \cosh (c+d x) \, dx}{d^2} \\ & = -\frac {a \cosh (c+d x)}{d^2}-\frac {4 b x^3 \cosh (c+d x)}{d^2}+\frac {a x \sinh (c+d x)}{d}+\frac {12 b x^2 \sinh (c+d x)}{d^3}+\frac {b x^4 \sinh (c+d x)}{d}-\frac {(24 b) \int x \sinh (c+d x) \, dx}{d^3} \\ & = -\frac {a \cosh (c+d x)}{d^2}-\frac {24 b x \cosh (c+d x)}{d^4}-\frac {4 b x^3 \cosh (c+d x)}{d^2}+\frac {a x \sinh (c+d x)}{d}+\frac {12 b x^2 \sinh (c+d x)}{d^3}+\frac {b x^4 \sinh (c+d x)}{d}+\frac {(24 b) \int \cosh (c+d x) \, dx}{d^4} \\ & = -\frac {a \cosh (c+d x)}{d^2}-\frac {24 b x \cosh (c+d x)}{d^4}-\frac {4 b x^3 \cosh (c+d x)}{d^2}+\frac {24 b \sinh (c+d x)}{d^5}+\frac {a x \sinh (c+d x)}{d}+\frac {12 b x^2 \sinh (c+d x)}{d^3}+\frac {b x^4 \sinh (c+d x)}{d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.70 \[ \int x \left (a+b x^3\right ) \cosh (c+d x) \, dx=\frac {-d \left (a d^2+4 b x \left (6+d^2 x^2\right )\right ) \cosh (c+d x)+\left (a d^4 x+b \left (24+12 d^2 x^2+d^4 x^4\right )\right ) \sinh (c+d x)}{d^5} \]

[In]

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

[Out]

(-(d*(a*d^2 + 4*b*x*(6 + d^2*x^2))*Cosh[c + d*x]) + (a*d^4*x + b*(24 + 12*d^2*x^2 + d^4*x^4))*Sinh[c + d*x])/d
^5

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.11

method result size
parallelrisch \(\frac {4 d x b \left (x^{2} d^{2}+6\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+2 \left (\left (-b \,x^{4}-a x \right ) d^{4}-12 b \,d^{2} x^{2}-24 b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+2 \left (\left (2 b \,x^{3}+a \right ) d^{2}+12 b x \right ) d}{d^{5} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )}\) \(104\)
risch \(\frac {\left (b \,x^{4} d^{4}-4 b \,d^{3} x^{3}+a \,d^{4} x +12 b \,d^{2} x^{2}-d^{3} a -24 d x b +24 b \right ) {\mathrm e}^{d x +c}}{2 d^{5}}-\frac {\left (b \,x^{4} d^{4}+4 b \,d^{3} x^{3}+a \,d^{4} x +12 b \,d^{2} x^{2}+d^{3} a +24 d x b +24 b \right ) {\mathrm e}^{-d x -c}}{2 d^{5}}\) \(120\)
parts \(\frac {b \,x^{4} \sinh \left (d x +c \right )}{d}+\frac {a x \sinh \left (d x +c \right )}{d}-\frac {-\frac {4 b \,c^{3} \cosh \left (d x +c \right )}{d^{3}}+\frac {12 b \,c^{2} \left (\left (d x +c \right ) \cosh \left (d x +c \right )-\sinh \left (d x +c \right )\right )}{d^{3}}-\frac {12 b c \left (\left (d x +c \right )^{2} \cosh \left (d x +c \right )-2 \left (d x +c \right ) \sinh \left (d x +c \right )+2 \cosh \left (d x +c \right )\right )}{d^{3}}+\frac {4 b \left (\left (d x +c \right )^{3} \cosh \left (d x +c \right )-3 \left (d x +c \right )^{2} \sinh \left (d x +c \right )+6 \left (d x +c \right ) \cosh \left (d x +c \right )-6 \sinh \left (d x +c \right )\right )}{d^{3}}+a \cosh \left (d x +c \right )}{d^{2}}\) \(187\)
meijerg \(-\frac {16 i b \cosh \left (c \right ) \sqrt {\pi }\, \left (-\frac {i x d \left (\frac {5 x^{2} d^{2}}{2}+15\right ) \cosh \left (d x \right )}{10 \sqrt {\pi }}+\frac {i \left (\frac {5}{8} d^{4} x^{4}+\frac {15}{2} x^{2} d^{2}+15\right ) \sinh \left (d x \right )}{10 \sqrt {\pi }}\right )}{d^{5}}-\frac {16 b \sinh \left (c \right ) \sqrt {\pi }\, \left (\frac {3}{2 \sqrt {\pi }}-\frac {\left (\frac {3}{8} d^{4} x^{4}+\frac {9}{2} x^{2} d^{2}+9\right ) \cosh \left (d x \right )}{6 \sqrt {\pi }}+\frac {x d \left (\frac {3 x^{2} d^{2}}{2}+9\right ) \sinh \left (d x \right )}{6 \sqrt {\pi }}\right )}{d^{5}}-\frac {2 a \cosh \left (c \right ) \sqrt {\pi }\, \left (-\frac {1}{2 \sqrt {\pi }}+\frac {\cosh \left (d x \right )}{2 \sqrt {\pi }}-\frac {d x \sinh \left (d x \right )}{2 \sqrt {\pi }}\right )}{d^{2}}+\frac {a \sinh \left (c \right ) \left (\cosh \left (d x \right ) x d -\sinh \left (d x \right )\right )}{d^{2}}\) \(188\)
derivativedivides \(\frac {-\frac {4 b \,c^{3} \left (\left (d x +c \right ) \sinh \left (d x +c \right )-\cosh \left (d x +c \right )\right )}{d^{3}}+\frac {6 b \,c^{2} \left (\left (d x +c \right )^{2} \sinh \left (d x +c \right )-2 \left (d x +c \right ) \cosh \left (d x +c \right )+2 \sinh \left (d x +c \right )\right )}{d^{3}}-\frac {4 b c \left (\left (d x +c \right )^{3} \sinh \left (d x +c \right )-3 \left (d x +c \right )^{2} \cosh \left (d x +c \right )+6 \left (d x +c \right ) \sinh \left (d x +c \right )-6 \cosh \left (d x +c \right )\right )}{d^{3}}+\frac {b \left (\left (d x +c \right )^{4} \sinh \left (d x +c \right )-4 \left (d x +c \right )^{3} \cosh \left (d x +c \right )+12 \left (d x +c \right )^{2} \sinh \left (d x +c \right )-24 \left (d x +c \right ) \cosh \left (d x +c \right )+24 \sinh \left (d x +c \right )\right )}{d^{3}}+a \left (\left (d x +c \right ) \sinh \left (d x +c \right )-\cosh \left (d x +c \right )\right )+\frac {b \,c^{4} \sinh \left (d x +c \right )}{d^{3}}-c a \sinh \left (d x +c \right )}{d^{2}}\) \(257\)
default \(\frac {-\frac {4 b \,c^{3} \left (\left (d x +c \right ) \sinh \left (d x +c \right )-\cosh \left (d x +c \right )\right )}{d^{3}}+\frac {6 b \,c^{2} \left (\left (d x +c \right )^{2} \sinh \left (d x +c \right )-2 \left (d x +c \right ) \cosh \left (d x +c \right )+2 \sinh \left (d x +c \right )\right )}{d^{3}}-\frac {4 b c \left (\left (d x +c \right )^{3} \sinh \left (d x +c \right )-3 \left (d x +c \right )^{2} \cosh \left (d x +c \right )+6 \left (d x +c \right ) \sinh \left (d x +c \right )-6 \cosh \left (d x +c \right )\right )}{d^{3}}+\frac {b \left (\left (d x +c \right )^{4} \sinh \left (d x +c \right )-4 \left (d x +c \right )^{3} \cosh \left (d x +c \right )+12 \left (d x +c \right )^{2} \sinh \left (d x +c \right )-24 \left (d x +c \right ) \cosh \left (d x +c \right )+24 \sinh \left (d x +c \right )\right )}{d^{3}}+a \left (\left (d x +c \right ) \sinh \left (d x +c \right )-\cosh \left (d x +c \right )\right )+\frac {b \,c^{4} \sinh \left (d x +c \right )}{d^{3}}-c a \sinh \left (d x +c \right )}{d^{2}}\) \(257\)

[In]

int(x*(b*x^3+a)*cosh(d*x+c),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.72 \[ \int x \left (a+b x^3\right ) \cosh (c+d x) \, dx=-\frac {{\left (4 \, b d^{3} x^{3} + a d^{3} + 24 \, b d x\right )} \cosh \left (d x + c\right ) - {\left (b d^{4} x^{4} + a d^{4} x + 12 \, b d^{2} x^{2} + 24 \, b\right )} \sinh \left (d x + c\right )}{d^{5}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.32 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.23 \[ \int x \left (a+b x^3\right ) \cosh (c+d x) \, dx=\begin {cases} \frac {a x \sinh {\left (c + d x \right )}}{d} - \frac {a \cosh {\left (c + d x \right )}}{d^{2}} + \frac {b x^{4} \sinh {\left (c + d x \right )}}{d} - \frac {4 b x^{3} \cosh {\left (c + d x \right )}}{d^{2}} + \frac {12 b x^{2} \sinh {\left (c + d x \right )}}{d^{3}} - \frac {24 b x \cosh {\left (c + d x \right )}}{d^{4}} + \frac {24 b \sinh {\left (c + d x \right )}}{d^{5}} & \text {for}\: d \neq 0 \\\left (\frac {a x^{2}}{2} + \frac {b x^{5}}{5}\right ) \cosh {\left (c \right )} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((a*x*sinh(c + d*x)/d - a*cosh(c + d*x)/d**2 + b*x**4*sinh(c + d*x)/d - 4*b*x**3*cosh(c + d*x)/d**2 +
 12*b*x**2*sinh(c + d*x)/d**3 - 24*b*x*cosh(c + d*x)/d**4 + 24*b*sinh(c + d*x)/d**5, Ne(d, 0)), ((a*x**2/2 + b
*x**5/5)*cosh(c), True))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 196 vs. \(2 (94) = 188\).

Time = 0.20 (sec) , antiderivative size = 196, normalized size of antiderivative = 2.09 \[ \int x \left (a+b x^3\right ) \cosh (c+d x) \, dx=-\frac {1}{20} \, d {\left (\frac {5 \, {\left (d^{2} x^{2} e^{c} - 2 \, d x e^{c} + 2 \, e^{c}\right )} a e^{\left (d x\right )}}{d^{3}} + \frac {5 \, {\left (d^{2} x^{2} + 2 \, d x + 2\right )} a e^{\left (-d x - c\right )}}{d^{3}} + \frac {2 \, {\left (d^{5} x^{5} e^{c} - 5 \, d^{4} x^{4} e^{c} + 20 \, d^{3} x^{3} e^{c} - 60 \, d^{2} x^{2} e^{c} + 120 \, d x e^{c} - 120 \, e^{c}\right )} b e^{\left (d x\right )}}{d^{6}} + \frac {2 \, {\left (d^{5} x^{5} + 5 \, d^{4} x^{4} + 20 \, d^{3} x^{3} + 60 \, d^{2} x^{2} + 120 \, d x + 120\right )} b e^{\left (-d x - c\right )}}{d^{6}}\right )} + \frac {1}{10} \, {\left (2 \, b x^{5} + 5 \, a x^{2}\right )} \cosh \left (d x + c\right ) \]

[In]

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

[Out]

-1/20*d*(5*(d^2*x^2*e^c - 2*d*x*e^c + 2*e^c)*a*e^(d*x)/d^3 + 5*(d^2*x^2 + 2*d*x + 2)*a*e^(-d*x - c)/d^3 + 2*(d
^5*x^5*e^c - 5*d^4*x^4*e^c + 20*d^3*x^3*e^c - 60*d^2*x^2*e^c + 120*d*x*e^c - 120*e^c)*b*e^(d*x)/d^6 + 2*(d^5*x
^5 + 5*d^4*x^4 + 20*d^3*x^3 + 60*d^2*x^2 + 120*d*x + 120)*b*e^(-d*x - c)/d^6) + 1/10*(2*b*x^5 + 5*a*x^2)*cosh(
d*x + c)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.27 \[ \int x \left (a+b x^3\right ) \cosh (c+d x) \, dx=\frac {{\left (b d^{4} x^{4} - 4 \, b d^{3} x^{3} + a d^{4} x + 12 \, b d^{2} x^{2} - a d^{3} - 24 \, b d x + 24 \, b\right )} e^{\left (d x + c\right )}}{2 \, d^{5}} - \frac {{\left (b d^{4} x^{4} + 4 \, b d^{3} x^{3} + a d^{4} x + 12 \, b d^{2} x^{2} + a d^{3} + 24 \, b d x + 24 \, b\right )} e^{\left (-d x - c\right )}}{2 \, d^{5}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 1.58 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.98 \[ \int x \left (a+b x^3\right ) \cosh (c+d x) \, dx=\frac {b\,x^4\,\mathrm {sinh}\left (c+d\,x\right )+a\,x\,\mathrm {sinh}\left (c+d\,x\right )}{d}-\frac {a\,\mathrm {cosh}\left (c+d\,x\right )+4\,b\,x^3\,\mathrm {cosh}\left (c+d\,x\right )}{d^2}+\frac {24\,b\,\mathrm {sinh}\left (c+d\,x\right )}{d^5}-\frac {24\,b\,x\,\mathrm {cosh}\left (c+d\,x\right )}{d^4}+\frac {12\,b\,x^2\,\mathrm {sinh}\left (c+d\,x\right )}{d^3} \]

[In]

int(x*cosh(c + d*x)*(a + b*x^3),x)

[Out]

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