\(\int x^m \cosh (a+b x^n) \, dx\) [47]

Optimal result

Integrand size = 12, antiderivative size = 89 \[ \int x^m \cosh \left (a+b x^n\right ) \, dx=-\frac {e^a x^{1+m} \left (-b x^n\right )^{-\frac {1+m}{n}} \Gamma \left (\frac {1+m}{n},-b x^n\right )}{2 n}-\frac {e^{-a} x^{1+m} \left (b x^n\right )^{-\frac {1+m}{n}} \Gamma \left (\frac {1+m}{n},b x^n\right )}{2 n} \]

[Out]

-1/2*exp(a)*x^(1+m)*GAMMA((1+m)/n,-b*x^n)/n/((-b*x^n)^((1+m)/n))-1/2*x^(1+m)*GAMMA((1+m)/n,b*x^n)/exp(a)/n/((b
*x^n)^((1+m)/n))

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {5469, 2250} \[ \int x^m \cosh \left (a+b x^n\right ) \, dx=-\frac {e^a x^{m+1} \left (-b x^n\right )^{-\frac {m+1}{n}} \Gamma \left (\frac {m+1}{n},-b x^n\right )}{2 n}-\frac {e^{-a} x^{m+1} \left (b x^n\right )^{-\frac {m+1}{n}} \Gamma \left (\frac {m+1}{n},b x^n\right )}{2 n} \]

[In]

Int[x^m*Cosh[a + b*x^n],x]

[Out]

-1/2*(E^a*x^(1 + m)*Gamma[(1 + m)/n, -(b*x^n)])/(n*(-(b*x^n))^((1 + m)/n)) - (x^(1 + m)*Gamma[(1 + m)/n, b*x^n
])/(2*E^a*n*(b*x^n)^((1 + m)/n))

Rule 2250

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[(-F^a)*((e +
f*x)^(m + 1)/(f*n*((-b)*(c + d*x)^n*Log[F])^((m + 1)/n)))*Gamma[(m + 1)/n, (-b)*(c + d*x)^n*Log[F]], x] /; Fre
eQ[{F, a, b, c, d, e, f, m, n}, x] && EqQ[d*e - c*f, 0]

Rule 5469

Int[Cosh[(c_.) + (d_.)*(x_)^(n_)]*((e_.)*(x_))^(m_.), x_Symbol] :> Dist[1/2, Int[(e*x)^m*E^(c + d*x^n), x], x]
+ Dist[1/2, Int[(e*x)^m*E^(-c - d*x^n), x], x] /; FreeQ[{c, d, e, m, n}, x]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \int e^{-a-b x^n} x^m \, dx+\frac {1}{2} \int e^{a+b x^n} x^m \, dx \\ & = -\frac {e^a x^{1+m} \left (-b x^n\right )^{-\frac {1+m}{n}} \Gamma \left (\frac {1+m}{n},-b x^n\right )}{2 n}-\frac {e^{-a} x^{1+m} \left (b x^n\right )^{-\frac {1+m}{n}} \Gamma \left (\frac {1+m}{n},b x^n\right )}{2 n} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.94 \[ \int x^m \cosh \left (a+b x^n\right ) \, dx=-\frac {e^a x^{1+m} \left (-b x^n\right )^{-\frac {1+m}{n}} \Gamma \left (\frac {1+m}{n},-b x^n\right )+e^{-a} x^{1+m} \left (b x^n\right )^{-\frac {1+m}{n}} \Gamma \left (\frac {1+m}{n},b x^n\right )}{2 n} \]

[In]

Integrate[x^m*Cosh[a + b*x^n],x]

[Out]

-1/2*((E^a*x^(1 + m)*Gamma[(1 + m)/n, -(b*x^n)])/(-(b*x^n))^((1 + m)/n) + (x^(1 + m)*Gamma[(1 + m)/n, b*x^n])/
(E^a*(b*x^n)^((1 + m)/n)))/n

Maple [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4.

Time = 0.30 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.24

method	result	size
meijerg	\(\frac {x^{1+m} \operatorname {hypergeom}\left (\left [\frac {m}{2 n}+\frac {1}{2 n}\right ], \left [\frac {1}{2}, 1+\frac {m}{2 n}+\frac {1}{2 n}\right ], \frac {x^{2 n} b^{2}}{4}\right ) \cosh \left (a \right )}{1+m}+\frac {x^{n +m +1} b \operatorname {hypergeom}\left (\left [\frac {1}{2}+\frac {m}{2 n}+\frac {1}{2 n}\right ], \left [\frac {3}{2}, \frac {3}{2}+\frac {m}{2 n}+\frac {1}{2 n}\right ], \frac {x^{2 n} b^{2}}{4}\right ) \sinh \left (a \right )}{n +m +1}\)	\(110\)

[In]

int(x^m*cosh(a+b*x^n),x,method=_RETURNVERBOSE)

[Out]

1/(1+m)*x^(1+m)*hypergeom([1/2/n*m+1/2/n],[1/2,1+1/2/n*m+1/2/n],1/4*x^(2*n)*b^2)*cosh(a)+1/(n+m+1)*x^(n+m+1)*b
*hypergeom([1/2+1/2/n*m+1/2/n],[3/2,3/2+1/2/n*m+1/2/n],1/4*x^(2*n)*b^2)*sinh(a)

Fricas [F]

\[ \int x^m \cosh \left (a+b x^n\right ) \, dx=\int { x^{m} \cosh \left (b x^{n} + a\right ) \,d x } \]

[In]

integrate(x^m*cosh(a+b*x^n),x, algorithm="fricas")

[Out]

integral(x^m*cosh(b*x^n + a), x)

Sympy [F]

\[ \int x^m \cosh \left (a+b x^n\right ) \, dx=\int x^{m} \cosh {\left (a + b x^{n} \right )}\, dx \]

[In]

integrate(x**m*cosh(a+b*x**n),x)

[Out]

Integral(x**m*cosh(a + b*x**n), x)

Maxima [A] (verification not implemented)

none

Time = 0.11 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.96 \[ \int x^m \cosh \left (a+b x^n\right ) \, dx=-\frac {x^{m + 1} e^{\left (-a\right )} \Gamma \left (\frac {m + 1}{n}, b x^{n}\right )}{2 \, \left (b x^{n}\right )^{\frac {m + 1}{n}} n} - \frac {x^{m + 1} e^{a} \Gamma \left (\frac {m + 1}{n}, -b x^{n}\right )}{2 \, \left (-b x^{n}\right )^{\frac {m + 1}{n}} n} \]

[In]

integrate(x^m*cosh(a+b*x^n),x, algorithm="maxima")

[Out]

-1/2*x^(m + 1)*e^(-a)*gamma((m + 1)/n, b*x^n)/((b*x^n)^((m + 1)/n)*n) - 1/2*x^(m + 1)*e^a*gamma((m + 1)/n, -b*
x^n)/((-b*x^n)^((m + 1)/n)*n)

Giac [F]

\[ \int x^m \cosh \left (a+b x^n\right ) \, dx=\int { x^{m} \cosh \left (b x^{n} + a\right ) \,d x } \]

[In]

integrate(x^m*cosh(a+b*x^n),x, algorithm="giac")

[Out]

integrate(x^m*cosh(b*x^n + a), x)

Mupad [F(-1)]

Timed out. \[ \int x^m \cosh \left (a+b x^n\right ) \, dx=\int x^m\,\mathrm {cosh}\left (a+b\,x^n\right ) \,d x \]

[In]

int(x^m*cosh(a + b*x^n),x)

[Out]

int(x^m*cosh(a + b*x^n), x)