\(\int \frac {x^{19/2}}{(a x+b x^3)^{9/2}} \, dx\) [80]

Optimal result

Integrand size = 19, antiderivative size = 76 \[ \int \frac {x^{19/2}}{\left (a x+b x^3\right )^{9/2}} \, dx=-\frac {x^{15/2}}{7 b \left (a x+b x^3\right )^{7/2}}-\frac {4 x^{9/2}}{35 b^2 \left (a x+b x^3\right )^{5/2}}-\frac {8 x^{3/2}}{105 b^3 \left (a x+b x^3\right )^{3/2}} \]

[Out]

-1/7*x^(15/2)/b/(b*x^3+a*x)^(7/2)-4/35*x^(9/2)/b^2/(b*x^3+a*x)^(5/2)-8/105*x^(3/2)/b^3/(b*x^3+a*x)^(3/2)

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2040, 2039} \[ \int \frac {x^{19/2}}{\left (a x+b x^3\right )^{9/2}} \, dx=-\frac {8 x^{3/2}}{105 b^3 \left (a x+b x^3\right )^{3/2}}-\frac {4 x^{9/2}}{35 b^2 \left (a x+b x^3\right )^{5/2}}-\frac {x^{15/2}}{7 b \left (a x+b x^3\right )^{7/2}} \]

[In]

Int[x^(19/2)/(a*x + b*x^3)^(9/2),x]

[Out]

-1/7*x^(15/2)/(b*(a*x + b*x^3)^(7/2)) - (4*x^(9/2))/(35*b^2*(a*x + b*x^3)^(5/2)) - (8*x^(3/2))/(105*b^3*(a*x +
b*x^3)^(3/2))

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^{15/2}}{7 b \left (a x+b x^3\right )^{7/2}}+\frac {4 \int \frac {x^{13/2}}{\left (a x+b x^3\right )^{7/2}} \, dx}{7 b} \\ & = -\frac {x^{15/2}}{7 b \left (a x+b x^3\right )^{7/2}}-\frac {4 x^{9/2}}{35 b^2 \left (a x+b x^3\right )^{5/2}}+\frac {8 \int \frac {x^{7/2}}{\left (a x+b x^3\right )^{5/2}} \, dx}{35 b^2} \\ & = -\frac {x^{15/2}}{7 b \left (a x+b x^3\right )^{7/2}}-\frac {4 x^{9/2}}{35 b^2 \left (a x+b x^3\right )^{5/2}}-\frac {8 x^{3/2}}{105 b^3 \left (a x+b x^3\right )^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.61 \[ \int \frac {x^{19/2}}{\left (a x+b x^3\right )^{9/2}} \, dx=\frac {x^{7/2} \left (-8 a^2-28 a b x^2-35 b^2 x^4\right )}{105 b^3 \left (x \left (a+b x^2\right )\right )^{7/2}} \]

[In]

Integrate[x^(19/2)/(a*x + b*x^3)^(9/2),x]

[Out]

(x^(7/2)*(-8*a^2 - 28*a*b*x^2 - 35*b^2*x^4))/(105*b^3*(x*(a + b*x^2))^(7/2))

Maple [A] (verified)

Time = 2.03 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.63

method	result	size
gosper	\(-\frac {\left (b \,x^{2}+a \right ) \left (35 b^{2} x^{4}+28 a b \,x^{2}+8 a^{2}\right ) x^{\frac {9}{2}}}{105 b^{3} \left (b \,x^{3}+a x \right )^{\frac {9}{2}}}\)	\(48\)
default	\(-\frac {\sqrt {x \left (b \,x^{2}+a \right )}\, \left (35 b^{2} x^{4}+28 a b \,x^{2}+8 a^{2}\right )}{105 \sqrt {x}\, \left (b \,x^{2}+a \right )^{4} b^{3}}\)	\(50\)

[In]

int(x^(19/2)/(b*x^3+a*x)^(9/2),x,method=_RETURNVERBOSE)

[Out]

-1/105*(b*x^2+a)*(35*b^2*x^4+28*a*b*x^2+8*a^2)*x^(9/2)/b^3/(b*x^3+a*x)^(9/2)

Fricas [A] (verification not implemented)

none

Time = 0.57 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.13 \[ \int \frac {x^{19/2}}{\left (a x+b x^3\right )^{9/2}} \, dx=-\frac {{\left (35 \, b^{2} x^{4} + 28 \, a b x^{2} + 8 \, a^{2}\right )} \sqrt {b x^{3} + a x} \sqrt {x}}{105 \, {\left (b^{7} x^{9} + 4 \, a b^{6} x^{7} + 6 \, a^{2} b^{5} x^{5} + 4 \, a^{3} b^{4} x^{3} + a^{4} b^{3} x\right )}} \]

[In]

integrate(x^(19/2)/(b*x^3+a*x)^(9/2),x, algorithm="fricas")

[Out]

-1/105*(35*b^2*x^4 + 28*a*b*x^2 + 8*a^2)*sqrt(b*x^3 + a*x)*sqrt(x)/(b^7*x^9 + 4*a*b^6*x^7 + 6*a^2*b^5*x^5 + 4*
a^3*b^4*x^3 + a^4*b^3*x)

Sympy [F(-1)]

Timed out. \[ \int \frac {x^{19/2}}{\left (a x+b x^3\right )^{9/2}} \, dx=\text {Timed out} \]

[In]

integrate(x**(19/2)/(b*x**3+a*x)**(9/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {x^{19/2}}{\left (a x+b x^3\right )^{9/2}} \, dx=\int { \frac {x^{\frac {19}{2}}}{{\left (b x^{3} + a x\right )}^{\frac {9}{2}}} \,d x } \]

[In]

integrate(x^(19/2)/(b*x^3+a*x)^(9/2),x, algorithm="maxima")

[Out]

integrate(x^(19/2)/(b*x^3 + a*x)^(9/2), x)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.66 \[ \int \frac {x^{19/2}}{\left (a x+b x^3\right )^{9/2}} \, dx=\frac {8}{105 \, a^{\frac {3}{2}} b^{3}} - \frac {35 \, {\left (b x^{2} + a\right )}^{2} - 42 \, {\left (b x^{2} + a\right )} a + 15 \, a^{2}}{105 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} b^{3}} \]

[In]

integrate(x^(19/2)/(b*x^3+a*x)^(9/2),x, algorithm="giac")

[Out]

8/105/(a^(3/2)*b^3) - 1/105*(35*(b*x^2 + a)^2 - 42*(b*x^2 + a)*a + 15*a^2)/((b*x^2 + a)^(7/2)*b^3)

Mupad [F(-1)]

Timed out. \[ \int \frac {x^{19/2}}{\left (a x+b x^3\right )^{9/2}} \, dx=\int \frac {x^{19/2}}{{\left (b\,x^3+a\,x\right )}^{9/2}} \,d x \]

[In]

int(x^(19/2)/(a*x + b*x^3)^(9/2),x)

[Out]

int(x^(19/2)/(a*x + b*x^3)^(9/2), x)