Integrand size = 14, antiderivative size = 66 \[ \int \left (a+b x^3\right ) \cosh (c+d x) \, dx=-\frac {6 b \cosh (c+d x)}{d^4}-\frac {3 b x^2 \cosh (c+d x)}{d^2}+\frac {a \sinh (c+d x)}{d}+\frac {6 b x \sinh (c+d x)}{d^3}+\frac {b x^3 \sinh (c+d x)}{d} \]
-6*b*cosh(d*x+c)/d^4-3*b*x^2*cosh(d*x+c)/d^2+a*sinh(d*x+c)/d+6*b*x*sinh(d* x+c)/d^3+b*x^3*sinh(d*x+c)/d
Time = 0.05 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.74 \[ \int \left (a+b x^3\right ) \cosh (c+d x) \, dx=\frac {-3 b \left (2+d^2 x^2\right ) \cosh (c+d x)+d \left (a d^2+b x \left (6+d^2 x^2\right )\right ) \sinh (c+d x)}{d^4} \]
Time = 0.30 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {5800, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \left (a+b x^3\right ) \cosh (c+d x) \, dx\) |
\(\Big \downarrow \) 5800 |
\(\displaystyle \int \left (a \cosh (c+d x)+b x^3 \cosh (c+d x)\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {a \sinh (c+d x)}{d}-\frac {6 b \cosh (c+d x)}{d^4}+\frac {6 b x \sinh (c+d x)}{d^3}-\frac {3 b x^2 \cosh (c+d x)}{d^2}+\frac {b x^3 \sinh (c+d x)}{d}\) |
(-6*b*Cosh[c + d*x])/d^4 - (3*b*x^2*Cosh[c + d*x])/d^2 + (a*Sinh[c + d*x]) /d + (6*b*x*Sinh[c + d*x])/d^3 + (b*x^3*Sinh[c + d*x])/d
3.1.82.3.1 Defintions of rubi rules used
Int[Cosh[(c_.) + (d_.)*(x_)]*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> I nt[ExpandIntegrand[Cosh[c + d*x], (a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, n}, x] && IGtQ[p, 0]
Time = 0.07 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.23
method | result | size |
parallelrisch | \(\frac {3 b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} x^{2} d^{2}-2 d \left (\left (b \,x^{3}+a \right ) d^{2}+6 b x \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+3 b \,d^{2} x^{2}+12 b}{d^{4} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )}\) | \(81\) |
risch | \(\frac {\left (b \,d^{3} x^{3}-3 b \,d^{2} x^{2}+d^{3} a +6 d x b -6 b \right ) {\mathrm e}^{d x +c}}{2 d^{4}}-\frac {\left (b \,d^{3} x^{3}+3 b \,d^{2} x^{2}+d^{3} a +6 d x b +6 b \right ) {\mathrm e}^{-d x -c}}{2 d^{4}}\) | \(89\) |
parts | \(\frac {b \,x^{3} \sinh \left (d x +c \right )}{d}+\frac {a \sinh \left (d x +c \right )}{d}-\frac {3 b \left (c^{2} \cosh \left (d x +c \right )-2 c \left (\left (d x +c \right ) \cosh \left (d x +c \right )-\sinh \left (d x +c \right )\right )+\left (d x +c \right )^{2} \cosh \left (d x +c \right )-2 \left (d x +c \right ) \sinh \left (d x +c \right )+2 \cosh \left (d x +c \right )\right )}{d^{4}}\) | \(103\) |
meijerg | \(\frac {8 b \cosh \left (c \right ) \sqrt {\pi }\, \left (\frac {3}{4 \sqrt {\pi }}-\frac {\left (\frac {3 x^{2} d^{2}}{2}+3\right ) \cosh \left (d x \right )}{4 \sqrt {\pi }}+\frac {d x \left (\frac {x^{2} d^{2}}{2}+3\right ) \sinh \left (d x \right )}{4 \sqrt {\pi }}\right )}{d^{4}}-\frac {8 i b \sinh \left (c \right ) \sqrt {\pi }\, \left (\frac {i x d \left (\frac {5 x^{2} d^{2}}{2}+15\right ) \cosh \left (d x \right )}{20 \sqrt {\pi }}-\frac {i \left (\frac {15 x^{2} d^{2}}{2}+15\right ) \sinh \left (d x \right )}{20 \sqrt {\pi }}\right )}{d^{4}}+\frac {a \cosh \left (c \right ) \sinh \left (d x \right )}{d}-\frac {a \sinh \left (c \right ) \sqrt {\pi }\, \left (\frac {1}{\sqrt {\pi }}-\frac {\cosh \left (d x \right )}{\sqrt {\pi }}\right )}{d}\) | \(149\) |
derivativedivides | \(\frac {-\frac {b \,c^{3} \sinh \left (d x +c \right )}{d^{3}}+\frac {3 b \,c^{2} \left (\left (d x +c \right ) \sinh \left (d x +c \right )-\cosh \left (d x +c \right )\right )}{d^{3}}-\frac {3 b c \left (\left (d x +c \right )^{2} \sinh \left (d x +c \right )-2 \left (d x +c \right ) \cosh \left (d x +c \right )+2 \sinh \left (d x +c \right )\right )}{d^{3}}+\frac {b \left (\left (d x +c \right )^{3} \sinh \left (d x +c \right )-3 \left (d x +c \right )^{2} \cosh \left (d x +c \right )+6 \left (d x +c \right ) \sinh \left (d x +c \right )-6 \cosh \left (d x +c \right )\right )}{d^{3}}+a \sinh \left (d x +c \right )}{d}\) | \(158\) |
default | \(\frac {-\frac {b \,c^{3} \sinh \left (d x +c \right )}{d^{3}}+\frac {3 b \,c^{2} \left (\left (d x +c \right ) \sinh \left (d x +c \right )-\cosh \left (d x +c \right )\right )}{d^{3}}-\frac {3 b c \left (\left (d x +c \right )^{2} \sinh \left (d x +c \right )-2 \left (d x +c \right ) \cosh \left (d x +c \right )+2 \sinh \left (d x +c \right )\right )}{d^{3}}+\frac {b \left (\left (d x +c \right )^{3} \sinh \left (d x +c \right )-3 \left (d x +c \right )^{2} \cosh \left (d x +c \right )+6 \left (d x +c \right ) \sinh \left (d x +c \right )-6 \cosh \left (d x +c \right )\right )}{d^{3}}+a \sinh \left (d x +c \right )}{d}\) | \(158\) |
(3*b*tanh(1/2*d*x+1/2*c)^2*x^2*d^2-2*d*((b*x^3+a)*d^2+6*b*x)*tanh(1/2*d*x+ 1/2*c)+3*b*d^2*x^2+12*b)/d^4/(tanh(1/2*d*x+1/2*c)^2-1)
Time = 0.24 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.80 \[ \int \left (a+b x^3\right ) \cosh (c+d x) \, dx=-\frac {3 \, {\left (b d^{2} x^{2} + 2 \, b\right )} \cosh \left (d x + c\right ) - {\left (b d^{3} x^{3} + a d^{3} + 6 \, b d x\right )} \sinh \left (d x + c\right )}{d^{4}} \]
Time = 0.24 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.24 \[ \int \left (a+b x^3\right ) \cosh (c+d x) \, dx=\begin {cases} \frac {a \sinh {\left (c + d x \right )}}{d} + \frac {b x^{3} \sinh {\left (c + d x \right )}}{d} - \frac {3 b x^{2} \cosh {\left (c + d x \right )}}{d^{2}} + \frac {6 b x \sinh {\left (c + d x \right )}}{d^{3}} - \frac {6 b \cosh {\left (c + d x \right )}}{d^{4}} & \text {for}\: d \neq 0 \\\left (a x + \frac {b x^{4}}{4}\right ) \cosh {\left (c \right )} & \text {otherwise} \end {cases} \]
Piecewise((a*sinh(c + d*x)/d + 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 + b*x**4/4)*cosh(c), True))
Time = 0.17 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.58 \[ \int \left (a+b x^3\right ) \cosh (c+d x) \, dx=\frac {a e^{\left (d x + c\right )}}{2 \, d} - \frac {a e^{\left (-d x - c\right )}}{2 \, d} + \frac {{\left (d^{3} x^{3} e^{c} - 3 \, d^{2} x^{2} e^{c} + 6 \, d x e^{c} - 6 \, e^{c}\right )} b e^{\left (d x\right )}}{2 \, d^{4}} - \frac {{\left (d^{3} x^{3} + 3 \, d^{2} x^{2} + 6 \, d x + 6\right )} b e^{\left (-d x - c\right )}}{2 \, d^{4}} \]
1/2*a*e^(d*x + c)/d - 1/2*a*e^(-d*x - c)/d + 1/2*(d^3*x^3*e^c - 3*d^2*x^2* e^c + 6*d*x*e^c - 6*e^c)*b*e^(d*x)/d^4 - 1/2*(d^3*x^3 + 3*d^2*x^2 + 6*d*x + 6)*b*e^(-d*x - c)/d^4
Time = 0.26 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.33 \[ \int \left (a+b x^3\right ) \cosh (c+d x) \, dx=\frac {{\left (b d^{3} x^{3} - 3 \, b d^{2} x^{2} + a d^{3} + 6 \, b d x - 6 \, b\right )} e^{\left (d x + c\right )}}{2 \, d^{4}} - \frac {{\left (b d^{3} x^{3} + 3 \, b d^{2} x^{2} + a d^{3} + 6 \, b d x + 6 \, b\right )} e^{\left (-d x - c\right )}}{2 \, d^{4}} \]
1/2*(b*d^3*x^3 - 3*b*d^2*x^2 + a*d^3 + 6*b*d*x - 6*b)*e^(d*x + c)/d^4 - 1/ 2*(b*d^3*x^3 + 3*b*d^2*x^2 + a*d^3 + 6*b*d*x + 6*b)*e^(-d*x - c)/d^4
Time = 0.10 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.98 \[ \int \left (a+b x^3\right ) \cosh (c+d x) \, dx=\frac {a\,\mathrm {sinh}\left (c+d\,x\right )+b\,x^3\,\mathrm {sinh}\left (c+d\,x\right )}{d}-\frac {6\,b\,\mathrm {cosh}\left (c+d\,x\right )}{d^4}+\frac {6\,b\,x\,\mathrm {sinh}\left (c+d\,x\right )}{d^3}-\frac {3\,b\,x^2\,\mathrm {cosh}\left (c+d\,x\right )}{d^2} \]