\(\int x^{-1+3 n} (a+b x^n) \, dx\) [2519]

Optimal result

Integrand size = 15, antiderivative size = 27 \[ \int x^{-1+3 n} \left (a+b x^n\right ) \, dx=\frac {a x^{3 n}}{3 n}+\frac {b x^{4 n}}{4 n} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {14} \[ \int x^{-1+3 n} \left (a+b x^n\right ) \, dx=\frac {a x^{3 n}}{3 n}+\frac {b x^{4 n}}{4 n} \]

[In]

Int[x^(-1 + 3*n)*(a + b*x^n),x]

[Out]

(a*x^(3*n))/(3*n) + (b*x^(4*n))/(4*n)

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81 \[ \int x^{-1+3 n} \left (a+b x^n\right ) \, dx=\frac {x^{3 n} \left (4 a+3 b x^n\right )}{12 n} \]

[In]

Integrate[x^(-1 + 3*n)*(a + b*x^n),x]

[Out]

(x^(3*n)*(4*a + 3*b*x^n))/(12*n)

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89

method	result	size
risch	\(\frac {a \,x^{3 n}}{3 n}+\frac {b \,x^{4 n}}{4 n}\)	\(24\)
norman	\(\frac {a \,{\mathrm e}^{3 n \ln \left (x \right )}}{3 n}+\frac {b \,{\mathrm e}^{4 n \ln \left (x \right )}}{4 n}\)	\(28\)
parallelrisch	\(\frac {3 x \,x^{n} x^{-1+3 n} b +4 x \,x^{-1+3 n} a}{12 n}\)	\(32\)

[In]

int(x^(-1+3*n)*(a+b*x^n),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81 \[ \int x^{-1+3 n} \left (a+b x^n\right ) \, dx=\frac {3 \, b x^{4 \, n} + 4 \, a x^{3 \, n}}{12 \, n} \]

[In]

integrate(x^(-1+3*n)*(a+b*x^n),x, algorithm="fricas")

[Out]

1/12*(3*b*x^(4*n) + 4*a*x^(3*n))/n

Sympy [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.33 \[ \int x^{-1+3 n} \left (a+b x^n\right ) \, dx=\begin {cases} \frac {a x x^{3 n - 1}}{3 n} + \frac {b x x^{n} x^{3 n - 1}}{4 n} & \text {for}\: n \neq 0 \\\left (a + b\right ) \log {\left (x \right )} & \text {otherwise} \end {cases} \]

[In]

integrate(x**(-1+3*n)*(a+b*x**n),x)

[Out]

Piecewise((a*x*x**(3*n - 1)/(3*n) + b*x*x**n*x**(3*n - 1)/(4*n), Ne(n, 0)), ((a + b)*log(x), True))

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85 \[ \int x^{-1+3 n} \left (a+b x^n\right ) \, dx=\frac {b x^{4 \, n}}{4 \, n} + \frac {a x^{3 \, n}}{3 \, n} \]

[In]

integrate(x^(-1+3*n)*(a+b*x^n),x, algorithm="maxima")

[Out]

1/4*b*x^(4*n)/n + 1/3*a*x^(3*n)/n

Giac [F]

\[ \int x^{-1+3 n} \left (a+b x^n\right ) \, dx=\int { {\left (b x^{n} + a\right )} x^{3 \, n - 1} \,d x } \]

[In]

integrate(x^(-1+3*n)*(a+b*x^n),x, algorithm="giac")

[Out]

integrate((b*x^n + a)*x^(3*n - 1), x)

Mupad [B] (verification not implemented)

Time = 5.96 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.74 \[ \int x^{-1+3 n} \left (a+b x^n\right ) \, dx=\frac {x^{3\,n}\,\left (4\,a+3\,b\,x^n\right )}{12\,n} \]

[In]

int(x^(3*n - 1)*(a + b*x^n),x)

[Out]

(x^(3*n)*(4*a + 3*b*x^n))/(12*n)