\(\int \frac {x^{17/2}}{(a x+b x^3)^{9/2}} \, dx\) [81]

Optimal result

Integrand size = 19, antiderivative size = 51 \[ \int \frac {x^{17/2}}{\left (a x+b x^3\right )^{9/2}} \, dx=\frac {x^{17/2}}{7 a \left (a x+b x^3\right )^{7/2}}+\frac {2 x^{15/2}}{35 a^2 \left (a x+b x^3\right )^{5/2}} \]

[Out]

1/7*x^(17/2)/a/(b*x^3+a*x)^(7/2)+2/35*x^(15/2)/a^2/(b*x^3+a*x)^(5/2)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 51, 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 = {2040, 2039} \[ \int \frac {x^{17/2}}{\left (a x+b x^3\right )^{9/2}} \, dx=\frac {2 x^{15/2}}{35 a^2 \left (a x+b x^3\right )^{5/2}}+\frac {x^{17/2}}{7 a \left (a x+b x^3\right )^{7/2}} \]

[In]

Int[x^(17/2)/(a*x + b*x^3)^(9/2),x]

[Out]

x^(17/2)/(7*a*(a*x + b*x^3)^(7/2)) + (2*x^(15/2))/(35*a^2*(a*x + b*x^3)^(5/2))

Rule 2039

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

Rule 2040

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(-c^(j - 1))*(c*x)^(m - j
+ 1)*((a*x^j + b*x^n)^(p + 1)/(a*(n - j)*(p + 1))), x] + Dist[c^j*((m + n*p + n - j + 1)/(a*(n - j)*(p + 1)))
, Int[(c*x)^(m - j)*(a*x^j + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, j, m, n}, x] &&  !IntegerQ[p] && NeQ[n,
j] && ILtQ[Simplify[(m + n*p + n - j + 1)/(n - j)], 0] && LtQ[p, -1] && (IntegerQ[j] || GtQ[c, 0])

Rubi steps \begin{align*} \text {integral}& = \frac {x^{17/2}}{7 a \left (a x+b x^3\right )^{7/2}}+\frac {2 \int \frac {x^{15/2}}{\left (a x+b x^3\right )^{7/2}} \, dx}{7 a} \\ & = \frac {x^{17/2}}{7 a \left (a x+b x^3\right )^{7/2}}+\frac {2 x^{15/2}}{35 a^2 \left (a x+b x^3\right )^{5/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.75 \[ \int \frac {x^{17/2}}{\left (a x+b x^3\right )^{9/2}} \, dx=\frac {x^{7/2} \left (7 a x^5+2 b x^7\right )}{35 a^2 \left (x \left (a+b x^2\right )\right )^{7/2}} \]

[In]

Integrate[x^(17/2)/(a*x + b*x^3)^(9/2),x]

[Out]

(x^(7/2)*(7*a*x^5 + 2*b*x^7))/(35*a^2*(x*(a + b*x^2))^(7/2))

Maple [A] (verified)

Time = 2.05 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.73

method	result	size
gosper	\(\frac {\left (b \,x^{2}+a \right ) x^{\frac {19}{2}} \left (2 b \,x^{2}+7 a \right )}{35 a^{2} \left (b \,x^{3}+a x \right )^{\frac {9}{2}}}\)	\(37\)
default	\(\frac {x^{\frac {9}{2}} \sqrt {x \left (b \,x^{2}+a \right )}\, \left (2 b \,x^{2}+7 a \right )}{35 a^{2} \left (b \,x^{2}+a \right )^{4}}\)	\(39\)

[In]

int(x^(17/2)/(b*x^3+a*x)^(9/2),x,method=_RETURNVERBOSE)

[Out]

1/35*(b*x^2+a)*x^(19/2)*(2*b*x^2+7*a)/a^2/(b*x^3+a*x)^(9/2)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.49 \[ \int \frac {x^{17/2}}{\left (a x+b x^3\right )^{9/2}} \, dx=\frac {{\left (2 \, b x^{6} + 7 \, a x^{4}\right )} \sqrt {b x^{3} + a x} \sqrt {x}}{35 \, {\left (a^{2} b^{4} x^{8} + 4 \, a^{3} b^{3} x^{6} + 6 \, a^{4} b^{2} x^{4} + 4 \, a^{5} b x^{2} + a^{6}\right )}} \]

[In]

integrate(x^(17/2)/(b*x^3+a*x)^(9/2),x, algorithm="fricas")

[Out]

1/35*(2*b*x^6 + 7*a*x^4)*sqrt(b*x^3 + a*x)*sqrt(x)/(a^2*b^4*x^8 + 4*a^3*b^3*x^6 + 6*a^4*b^2*x^4 + 4*a^5*b*x^2
+ a^6)

Sympy [F(-1)]

Timed out. \[ \int \frac {x^{17/2}}{\left (a x+b x^3\right )^{9/2}} \, dx=\text {Timed out} \]

[In]

integrate(x**(17/2)/(b*x**3+a*x)**(9/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {x^{17/2}}{\left (a x+b x^3\right )^{9/2}} \, dx=\int { \frac {x^{\frac {17}{2}}}{{\left (b x^{3} + a x\right )}^{\frac {9}{2}}} \,d x } \]

[In]

integrate(x^(17/2)/(b*x^3+a*x)^(9/2),x, algorithm="maxima")

[Out]

integrate(x^(17/2)/(b*x^3 + a*x)^(9/2), x)

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.57 \[ \int \frac {x^{17/2}}{\left (a x+b x^3\right )^{9/2}} \, dx=\frac {x^{5} {\left (\frac {2 \, b x^{2}}{a^{2}} + \frac {7}{a}\right )}}{35 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}}} \]

[In]

integrate(x^(17/2)/(b*x^3+a*x)^(9/2),x, algorithm="giac")

[Out]

1/35*x^5*(2*b*x^2/a^2 + 7/a)/(b*x^2 + a)^(7/2)

Mupad [F(-1)]

Timed out. \[ \int \frac {x^{17/2}}{\left (a x+b x^3\right )^{9/2}} \, dx=\int \frac {x^{17/2}}{{\left (b\,x^3+a\,x\right )}^{9/2}} \,d x \]

[In]

int(x^(17/2)/(a*x + b*x^3)^(9/2),x)

[Out]

int(x^(17/2)/(a*x + b*x^3)^(9/2), x)