\(\int x^{2+m} \cosh (a+b x) \, dx\) [82]

Optimal result

Integrand size = 12, antiderivative size = 59 \[ \int x^{2+m} \cosh (a+b x) \, dx=\frac {e^a x^m (-b x)^{-m} \Gamma (3+m,-b x)}{2 b^3}-\frac {e^{-a} x^m (b x)^{-m} \Gamma (3+m,b x)}{2 b^3} \]

[Out]

1/2*exp(a)*x^m*GAMMA(3+m,-b*x)/b^3/((-b*x)^m)-1/2*x^m*GAMMA(3+m,b*x)/b^3/exp(a)/((b*x)^m)

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 59, 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 = {3388, 2212} \[ \int x^{2+m} \cosh (a+b x) \, dx=\frac {e^a x^m (-b x)^{-m} \Gamma (m+3,-b x)}{2 b^3}-\frac {e^{-a} x^m (b x)^{-m} \Gamma (m+3,b x)}{2 b^3} \]

[In]

Int[x^(2 + m)*Cosh[a + b*x],x]

[Out]

(E^a*x^m*Gamma[3 + m, -(b*x)])/(2*b^3*(-(b*x))^m) - (x^m*Gamma[3 + m, b*x])/(2*b^3*E^a*(b*x)^m)

Rule 2212

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[(-F^(g*(e - c*(f/d))))*((c
+ d*x)^FracPart[m]/(d*((-f)*g*(Log[F]/d))^(IntPart[m] + 1)*((-f)*g*Log[F]*((c + d*x)/d))^FracPart[m]))*Gamma[m
+ 1, ((-f)*g*(Log[F]/d))*(c + d*x)], x] /; FreeQ[{F, c, d, e, f, g, m}, x] &&  !IntegerQ[m]

Rule 3388

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + Pi*(k_.) + (f_.)*(x_)], x_Symbol] :> Dist[I/2, Int[(c + d*x)^m/(E^(
I*k*Pi)*E^(I*(e + f*x))), x], x] - Dist[I/2, Int[(c + d*x)^m*E^(I*k*Pi)*E^(I*(e + f*x)), x], x] /; FreeQ[{c, d
, e, f, m}, x] && IntegerQ[2*k]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \int e^{-i (i a+i b x)} x^{2+m} \, dx+\frac {1}{2} \int e^{i (i a+i b x)} x^{2+m} \, dx \\ & = \frac {e^a x^m (-b x)^{-m} \Gamma (3+m,-b x)}{2 b^3}-\frac {e^{-a} x^m (b x)^{-m} \Gamma (3+m,b x)}{2 b^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.92 \[ \int x^{2+m} \cosh (a+b x) \, dx=\frac {e^{-a} x^m \left (e^{2 a} (-b x)^{-m} \Gamma (3+m,-b x)-(b x)^{-m} \Gamma (3+m,b x)\right )}{2 b^3} \]

[In]

Integrate[x^(2 + m)*Cosh[a + b*x],x]

[Out]

(x^m*((E^(2*a)*Gamma[3 + m, -(b*x)])/(-(b*x))^m - Gamma[3 + m, b*x]/(b*x)^m))/(2*b^3*E^a)

Maple [C] (verified)

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

Time = 0.09 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.24

method	result	size
meijerg	\(\frac {x^{3+m} \operatorname {hypergeom}\left (\left [\frac {3}{2}+\frac {m}{2}\right ], \left [\frac {1}{2}, \frac {5}{2}+\frac {m}{2}\right ], \frac {x^{2} b^{2}}{4}\right ) \cosh \left (a \right )}{3+m}+\frac {b \,x^{4+m} \operatorname {hypergeom}\left (\left [2+\frac {m}{2}\right ], \left [\frac {3}{2}, 3+\frac {m}{2}\right ], \frac {x^{2} b^{2}}{4}\right ) \sinh \left (a \right )}{4+m}\)	\(73\)

[In]

int(x^(2+m)*cosh(b*x+a),x,method=_RETURNVERBOSE)

[Out]

1/(3+m)*x^(3+m)*hypergeom([3/2+1/2*m],[1/2,5/2+1/2*m],1/4*x^2*b^2)*cosh(a)+b/(4+m)*x^(4+m)*hypergeom([2+1/2*m]
,[3/2,3+1/2*m],1/4*x^2*b^2)*sinh(a)

Fricas [A] (verification not implemented)

none

Time = 0.08 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.46 \[ \int x^{2+m} \cosh (a+b x) \, dx=-\frac {\cosh \left ({\left (m + 2\right )} \log \left (b\right ) + a\right ) \Gamma \left (m + 3, b x\right ) - \cosh \left ({\left (m + 2\right )} \log \left (-b\right ) - a\right ) \Gamma \left (m + 3, -b x\right ) + \Gamma \left (m + 3, -b x\right ) \sinh \left ({\left (m + 2\right )} \log \left (-b\right ) - a\right ) - \Gamma \left (m + 3, b x\right ) \sinh \left ({\left (m + 2\right )} \log \left (b\right ) + a\right )}{2 \, b} \]

[In]

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

[Out]

-1/2*(cosh((m + 2)*log(b) + a)*gamma(m + 3, b*x) - cosh((m + 2)*log(-b) - a)*gamma(m + 3, -b*x) + gamma(m + 3,
-b*x)*sinh((m + 2)*log(-b) - a) - gamma(m + 3, b*x)*sinh((m + 2)*log(b) + a))/b

Sympy [F(-2)]

Exception generated. \[ \int x^{2+m} \cosh (a+b x) \, dx=\text {Exception raised: TypeError} \]

[In]

integrate(x**(2+m)*cosh(b*x+a),x)

[Out]

Exception raised: TypeError >> cannot determine truth value of Relational

Maxima [A] (verification not implemented)

none

Time = 0.08 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.93 \[ \int x^{2+m} \cosh (a+b x) \, dx=-\frac {1}{2} \, \left (b x\right )^{-m - 3} x^{m + 3} e^{\left (-a\right )} \Gamma \left (m + 3, b x\right ) - \frac {1}{2} \, \left (-b x\right )^{-m - 3} x^{m + 3} e^{a} \Gamma \left (m + 3, -b x\right ) \]

[In]

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

[Out]

-1/2*(b*x)^(-m - 3)*x^(m + 3)*e^(-a)*gamma(m + 3, b*x) - 1/2*(-b*x)^(-m - 3)*x^(m + 3)*e^a*gamma(m + 3, -b*x)

Giac [F]

\[ \int x^{2+m} \cosh (a+b x) \, dx=\int { x^{m + 2} \cosh \left (b x + a\right ) \,d x } \]

[In]

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

[Out]

integrate(x^(m + 2)*cosh(b*x + a), x)

Mupad [F(-1)]

Timed out. \[ \int x^{2+m} \cosh (a+b x) \, dx=\int x^{m+2}\,\mathrm {cosh}\left (a+b\,x\right ) \,d x \]

[In]

int(x^(m + 2)*cosh(a + b*x),x)

[Out]

int(x^(m + 2)*cosh(a + b*x), x)