\(\int \frac {x^7}{(a+b x^4)^{5/4}} \, dx\) [1142]

Optimal result

Integrand size = 15, antiderivative size = 35 \[ \int \frac {x^7}{\left (a+b x^4\right )^{5/4}} \, dx=\frac {a}{b^2 \sqrt [4]{a+b x^4}}+\frac {\left (a+b x^4\right )^{3/4}}{3 b^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {272, 45} \[ \int \frac {x^7}{\left (a+b x^4\right )^{5/4}} \, dx=\frac {a}{b^2 \sqrt [4]{a+b x^4}}+\frac {\left (a+b x^4\right )^{3/4}}{3 b^2} \]

[In]

Int[x^7/(a + b*x^4)^(5/4),x]

[Out]

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

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 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{4} \text {Subst}\left (\int \frac {x}{(a+b x)^{5/4}} \, dx,x,x^4\right ) \\ & = \frac {1}{4} \text {Subst}\left (\int \left (-\frac {a}{b (a+b x)^{5/4}}+\frac {1}{b \sqrt [4]{a+b x}}\right ) \, dx,x,x^4\right ) \\ & = \frac {a}{b^2 \sqrt [4]{a+b x^4}}+\frac {\left (a+b x^4\right )^{3/4}}{3 b^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.77 \[ \int \frac {x^7}{\left (a+b x^4\right )^{5/4}} \, dx=\frac {4 a+b x^4}{3 b^2 \sqrt [4]{a+b x^4}} \]

[In]

Integrate[x^7/(a + b*x^4)^(5/4),x]

[Out]

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

Maple [A] (verified)

Time = 4.30 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.69

method	result	size
gosper	\(\frac {b \,x^{4}+4 a}{3 \left (b \,x^{4}+a \right )^{\frac {1}{4}} b^{2}}\)	\(24\)
trager	\(\frac {b \,x^{4}+4 a}{3 \left (b \,x^{4}+a \right )^{\frac {1}{4}} b^{2}}\)	\(24\)
pseudoelliptic	\(\frac {b \,x^{4}+4 a}{3 \left (b \,x^{4}+a \right )^{\frac {1}{4}} b^{2}}\)	\(24\)
risch	\(\frac {a}{b^{2} \left (b \,x^{4}+a \right )^{\frac {1}{4}}}+\frac {\left (b \,x^{4}+a \right )^{\frac {3}{4}}}{3 b^{2}}\)	\(30\)

[In]

int(x^7/(b*x^4+a)^(5/4),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.00 \[ \int \frac {x^7}{\left (a+b x^4\right )^{5/4}} \, dx=\frac {{\left (b x^{4} + 4 \, a\right )} {\left (b x^{4} + a\right )}^{\frac {3}{4}}}{3 \, {\left (b^{3} x^{4} + a b^{2}\right )}} \]

[In]

integrate(x^7/(b*x^4+a)^(5/4),x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.26 \[ \int \frac {x^7}{\left (a+b x^4\right )^{5/4}} \, dx=\begin {cases} \frac {4 a}{3 b^{2} \sqrt [4]{a + b x^{4}}} + \frac {x^{4}}{3 b \sqrt [4]{a + b x^{4}}} & \text {for}\: b \neq 0 \\\frac {x^{8}}{8 a^{\frac {5}{4}}} & \text {otherwise} \end {cases} \]

[In]

integrate(x**7/(b*x**4+a)**(5/4),x)

[Out]

Piecewise((4*a/(3*b**2*(a + b*x**4)**(1/4)) + x**4/(3*b*(a + b*x**4)**(1/4)), Ne(b, 0)), (x**8/(8*a**(5/4)), T
rue))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.83 \[ \int \frac {x^7}{\left (a+b x^4\right )^{5/4}} \, dx=\frac {{\left (b x^{4} + a\right )}^{\frac {3}{4}}}{3 \, b^{2}} + \frac {a}{{\left (b x^{4} + a\right )}^{\frac {1}{4}} b^{2}} \]

[In]

integrate(x^7/(b*x^4+a)^(5/4),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.97 \[ \int \frac {x^7}{\left (a+b x^4\right )^{5/4}} \, dx=\frac {\frac {{\left (b x^{4} + a\right )}^{\frac {3}{4}}}{b} + \frac {3 \, a}{{\left (b x^{4} + a\right )}^{\frac {1}{4}} b}}{3 \, b} \]

[In]

integrate(x^7/(b*x^4+a)^(5/4),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.77 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.66 \[ \int \frac {x^7}{\left (a+b x^4\right )^{5/4}} \, dx=\frac {b\,x^4+4\,a}{3\,b^2\,{\left (b\,x^4+a\right )}^{1/4}} \]

[In]

int(x^7/(a + b*x^4)^(5/4),x)

[Out]

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