\(\int (a+b e^x)^4 \, dx\) [697]

Optimal result

Integrand size = 9, antiderivative size = 53 \[ \int \left (a+b e^x\right )^4 \, dx=4 a^3 b e^x+3 a^2 b^2 e^{2 x}+\frac {4}{3} a b^3 e^{3 x}+\frac {1}{4} b^4 e^{4 x}+a^4 x \]

[Out]

4*a^3*b*exp(x)+3*a^2*b^2*exp(2*x)+4/3*a*b^3*exp(3*x)+1/4*b^4*exp(4*x)+a^4*x

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {2320, 45} \[ \int \left (a+b e^x\right )^4 \, dx=a^4 x+4 a^3 b e^x+3 a^2 b^2 e^{2 x}+\frac {4}{3} a b^3 e^{3 x}+\frac {1}{4} b^4 e^{4 x} \]

[In]

Int[(a + b*E^x)^4,x]

[Out]

4*a^3*b*E^x + 3*a^2*b^2*E^(2*x) + (4*a*b^3*E^(3*x))/3 + (b^4*E^(4*x))/4 + a^4*x

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2320

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rubi steps \begin{align*} \text {integral}& = \text {Subst}\left (\int \frac {(a+b x)^4}{x} \, dx,x,e^x\right ) \\ & = \text {Subst}\left (\int \left (4 a^3 b+\frac {a^4}{x}+6 a^2 b^2 x+4 a b^3 x^2+b^4 x^3\right ) \, dx,x,e^x\right ) \\ & = 4 a^3 b e^x+3 a^2 b^2 e^{2 x}+\frac {4}{3} a b^3 e^{3 x}+\frac {1}{4} b^4 e^{4 x}+a^4 x \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.00 \[ \int \left (a+b e^x\right )^4 \, dx=\frac {1}{12} b e^x \left (48 a^3+36 a^2 b e^x+16 a b^2 e^{2 x}+3 b^3 e^{3 x}\right )+a^4 \log \left (e^x\right ) \]

[In]

Integrate[(a + b*E^x)^4,x]

[Out]

(b*E^x*(48*a^3 + 36*a^2*b*E^x + 16*a*b^2*E^(2*x) + 3*b^3*E^(3*x)))/12 + a^4*Log[E^x]

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.87

method	result	size
norman	\(4 a^{3} b \,{\mathrm e}^{x}+3 a^{2} b^{2} {\mathrm e}^{2 x}+\frac {4 a \,b^{3} {\mathrm e}^{3 x}}{3}+\frac {b^{4} {\mathrm e}^{4 x}}{4}+a^{4} x\)	\(46\)
risch	\(4 a^{3} b \,{\mathrm e}^{x}+3 a^{2} b^{2} {\mathrm e}^{2 x}+\frac {4 a \,b^{3} {\mathrm e}^{3 x}}{3}+\frac {b^{4} {\mathrm e}^{4 x}}{4}+a^{4} x\)	\(46\)
parallelrisch	\(4 a^{3} b \,{\mathrm e}^{x}+3 a^{2} b^{2} {\mathrm e}^{2 x}+\frac {4 a \,b^{3} {\mathrm e}^{3 x}}{3}+\frac {b^{4} {\mathrm e}^{4 x}}{4}+a^{4} x\)	\(46\)
parts	\(4 a^{3} b \,{\mathrm e}^{x}+3 a^{2} b^{2} {\mathrm e}^{2 x}+\frac {4 a \,b^{3} {\mathrm e}^{3 x}}{3}+\frac {b^{4} {\mathrm e}^{4 x}}{4}+a^{4} x\)	\(46\)
derivativedivides	\(\frac {b^{4} {\mathrm e}^{4 x}}{4}+\frac {4 a \,b^{3} {\mathrm e}^{3 x}}{3}+3 a^{2} b^{2} {\mathrm e}^{2 x}+4 a^{3} b \,{\mathrm e}^{x}+a^{4} \ln \left ({\mathrm e}^{x}\right )\)	\(48\)
default	\(\frac {b^{4} {\mathrm e}^{4 x}}{4}+\frac {4 a \,b^{3} {\mathrm e}^{3 x}}{3}+3 a^{2} b^{2} {\mathrm e}^{2 x}+4 a^{3} b \,{\mathrm e}^{x}+a^{4} \ln \left ({\mathrm e}^{x}\right )\)	\(48\)

[In]

int((a+b*exp(x))^4,x,method=_RETURNVERBOSE)

[Out]

a^4*x+1/4*b^4*exp(x)^4+4/3*a*b^3*exp(x)^3+3*a^2*b^2*exp(x)^2+4*a^3*b*exp(x)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.85 \[ \int \left (a+b e^x\right )^4 \, dx=a^{4} x + \frac {1}{4} \, b^{4} e^{\left (4 \, x\right )} + \frac {4}{3} \, a b^{3} e^{\left (3 \, x\right )} + 3 \, a^{2} b^{2} e^{\left (2 \, x\right )} + 4 \, a^{3} b e^{x} \]

[In]

integrate((a+b*exp(x))^4,x, algorithm="fricas")

[Out]

a^4*x + 1/4*b^4*e^(4*x) + 4/3*a*b^3*e^(3*x) + 3*a^2*b^2*e^(2*x) + 4*a^3*b*e^x

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.96 \[ \int \left (a+b e^x\right )^4 \, dx=a^{4} x + 4 a^{3} b e^{x} + 3 a^{2} b^{2} e^{2 x} + \frac {4 a b^{3} e^{3 x}}{3} + \frac {b^{4} e^{4 x}}{4} \]

[In]

integrate((a+b*exp(x))**4,x)

[Out]

a**4*x + 4*a**3*b*exp(x) + 3*a**2*b**2*exp(2*x) + 4*a*b**3*exp(3*x)/3 + b**4*exp(4*x)/4

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.85 \[ \int \left (a+b e^x\right )^4 \, dx=a^{4} x + \frac {1}{4} \, b^{4} e^{\left (4 \, x\right )} + \frac {4}{3} \, a b^{3} e^{\left (3 \, x\right )} + 3 \, a^{2} b^{2} e^{\left (2 \, x\right )} + 4 \, a^{3} b e^{x} \]

[In]

integrate((a+b*exp(x))^4,x, algorithm="maxima")

[Out]

a^4*x + 1/4*b^4*e^(4*x) + 4/3*a*b^3*e^(3*x) + 3*a^2*b^2*e^(2*x) + 4*a^3*b*e^x

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.85 \[ \int \left (a+b e^x\right )^4 \, dx=a^{4} x + \frac {1}{4} \, b^{4} e^{\left (4 \, x\right )} + \frac {4}{3} \, a b^{3} e^{\left (3 \, x\right )} + 3 \, a^{2} b^{2} e^{\left (2 \, x\right )} + 4 \, a^{3} b e^{x} \]

[In]

integrate((a+b*exp(x))^4,x, algorithm="giac")

[Out]

a^4*x + 1/4*b^4*e^(4*x) + 4/3*a*b^3*e^(3*x) + 3*a^2*b^2*e^(2*x) + 4*a^3*b*e^x

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.85 \[ \int \left (a+b e^x\right )^4 \, dx=x\,a^4+4\,{\mathrm {e}}^x\,a^3\,b+3\,{\mathrm {e}}^{2\,x}\,a^2\,b^2+\frac {4\,{\mathrm {e}}^{3\,x}\,a\,b^3}{3}+\frac {{\mathrm {e}}^{4\,x}\,b^4}{4} \]

[In]

int((a + b*exp(x))^4,x)

[Out]

(b^4*exp(4*x))/4 + a^4*x + 3*a^2*b^2*exp(2*x) + 4*a^3*b*exp(x) + (4*a*b^3*exp(3*x))/3