\(\int \sqrt [4]{x^{4 (-1+n)} (a+b x^n)} \, dx\) [450]

Optimal result

Integrand size = 19, antiderivative size = 44 \[ \int \sqrt [4]{x^{4 (-1+n)} \left (a+b x^n\right )} \, dx=\frac {4 x^{5 (1-n)} \left (a x^{-4 (1-n)}+b x^{-4+5 n}\right )^{5/4}}{5 b n} \]

[Out]

4/5*x^(5-5*n)*(a/(x^(4-4*n))+b*x^(-4+5*n))^(5/4)/b/n

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2004, 2025} \[ \int \sqrt [4]{x^{4 (-1+n)} \left (a+b x^n\right )} \, dx=\frac {4 x^{5 (1-n)} \left (a x^{-4 (1-n)}+b x^{5 n-4}\right )^{5/4}}{5 b n} \]

[In]

Int[(x^(4*(-1 + n))*(a + b*x^n))^(1/4),x]

[Out]

(4*x^(5*(1 - n))*(a/x^(4*(1 - n)) + b*x^(-4 + 5*n))^(5/4))/(5*b*n)

Rule 2004

Int[(u_)^(p_), x_Symbol] :> Int[ExpandToSum[u, x]^p, x] /; FreeQ[p, x] && GeneralizedBinomialQ[u, x] &&  !Gene
ralizedBinomialMatchQ[u, x]

Rule 2025

Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(a*x^j + b*x^n)^(p + 1)/(b*(n - j)*(p + 1)*x
^(n - 1)), x] /; FreeQ[{a, b, j, n, p}, x] &&  !IntegerQ[p] && NeQ[n, j] && EqQ[j*p - n + j + 1, 0]

Rubi steps \begin{align*} \text {integral}& = \int \sqrt [4]{a x^{4 (-1+n)}+b x^{4 (-1+n)+n}} \, dx \\ & = \frac {4 x^{5 (1-n)} \left (a x^{-4 (1-n)}+b x^{-4+5 n}\right )^{5/4}}{5 b n} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.82 \[ \int \sqrt [4]{x^{4 (-1+n)} \left (a+b x^n\right )} \, dx=\frac {4 x^{5-5 n} \left (x^{-4+4 n} \left (a+b x^n\right )\right )^{5/4}}{5 b n} \]

[In]

Integrate[(x^(4*(-1 + n))*(a + b*x^n))^(1/4),x]

[Out]

(4*x^(5 - 5*n)*(x^(-4 + 4*n)*(a + b*x^n))^(5/4))/(5*b*n)

Maple [A] (verified)

Time = 1.83 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.91

method	result	size
risch	\(\frac {4 \left (\frac {x^{4 n} \left (a +b \,x^{n}\right )}{x^{4}}\right )^{\frac {1}{4}} x \,x^{-n} \left (a +b \,x^{n}\right )}{5 b n}\)	\(40\)

[In]

int((x^(-4+4*n)*(a+b*x^n))^(1/4),x,method=_RETURNVERBOSE)

[Out]

4/5*(1/x^4*(x^n)^4*(a+b*x^n))^(1/4)*x/(x^n)*(a+b*x^n)/b/n

Fricas [A] (verification not implemented)

none

Time = 0.69 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.00 \[ \int \sqrt [4]{x^{4 (-1+n)} \left (a+b x^n\right )} \, dx=\frac {4 \, {\left (b x x^{n} + a x\right )} \left (\frac {b x^{5 \, n} + a x^{4 \, n}}{x^{4}}\right )^{\frac {1}{4}}}{5 \, b n x^{n}} \]

[In]

integrate((x^(-4+4*n)*(a+b*x^n))^(1/4),x, algorithm="fricas")

[Out]

4/5*(b*x*x^n + a*x)*((b*x^(5*n) + a*x^(4*n))/x^4)^(1/4)/(b*n*x^n)

Sympy [F(-1)]

Timed out. \[ \int \sqrt [4]{x^{4 (-1+n)} \left (a+b x^n\right )} \, dx=\text {Timed out} \]

[In]

integrate((x**(-4+4*n)*(a+b*x**n))**(1/4),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.39 \[ \int \sqrt [4]{x^{4 (-1+n)} \left (a+b x^n\right )} \, dx=\frac {4 \, {\left (b x^{n} + a\right )}^{\frac {5}{4}}}{5 \, b n} \]

[In]

integrate((x^(-4+4*n)*(a+b*x^n))^(1/4),x, algorithm="maxima")

[Out]

4/5*(b*x^n + a)^(5/4)/(b*n)

Giac [F]

\[ \int \sqrt [4]{x^{4 (-1+n)} \left (a+b x^n\right )} \, dx=\int { \left ({\left (b x^{n} + a\right )} x^{4 \, n - 4}\right )^{\frac {1}{4}} \,d x } \]

[In]

integrate((x^(-4+4*n)*(a+b*x^n))^(1/4),x, algorithm="giac")

[Out]

integrate(((b*x^n + a)*x^(4*n - 4))^(1/4), x)

Mupad [F(-1)]

Timed out. \[ \int \sqrt [4]{x^{4 (-1+n)} \left (a+b x^n\right )} \, dx=\int {\left (x^{4\,n-4}\,\left (a+b\,x^n\right )\right )}^{1/4} \,d x \]

[In]

int((x^(4*n - 4)*(a + b*x^n))^(1/4),x)

[Out]

int((x^(4*n - 4)*(a + b*x^n))^(1/4), x)