\(\int \frac {x^{23/2}}{(a x+b x^3)^{9/2}} \, dx\) [78]

Optimal result

Integrand size = 19, antiderivative size = 101 \[ \int \frac {x^{23/2}}{\left (a x+b x^3\right )^{9/2}} \, dx=-\frac {x^{19/2}}{7 b \left (a x+b x^3\right )^{7/2}}-\frac {6 x^{13/2}}{35 b^2 \left (a x+b x^3\right )^{5/2}}-\frac {8 x^{7/2}}{35 b^3 \left (a x+b x^3\right )^{3/2}}-\frac {16 \sqrt {x}}{35 b^4 \sqrt {a x+b x^3}} \]

[Out]

-1/7*x^(19/2)/b/(b*x^3+a*x)^(7/2)-6/35*x^(13/2)/b^2/(b*x^3+a*x)^(5/2)-8/35*x^(7/2)/b^3/(b*x^3+a*x)^(3/2)-16/35
*x^(1/2)/b^4/(b*x^3+a*x)^(1/2)

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2040, 2039} \[ \int \frac {x^{23/2}}{\left (a x+b x^3\right )^{9/2}} \, dx=-\frac {16 \sqrt {x}}{35 b^4 \sqrt {a x+b x^3}}-\frac {8 x^{7/2}}{35 b^3 \left (a x+b x^3\right )^{3/2}}-\frac {6 x^{13/2}}{35 b^2 \left (a x+b x^3\right )^{5/2}}-\frac {x^{19/2}}{7 b \left (a x+b x^3\right )^{7/2}} \]

[In]

Int[x^(23/2)/(a*x + b*x^3)^(9/2),x]

[Out]

-1/7*x^(19/2)/(b*(a*x + b*x^3)^(7/2)) - (6*x^(13/2))/(35*b^2*(a*x + b*x^3)^(5/2)) - (8*x^(7/2))/(35*b^3*(a*x +
b*x^3)^(3/2)) - (16*Sqrt[x])/(35*b^4*Sqrt[a*x + b*x^3])

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^{19/2}}{7 b \left (a x+b x^3\right )^{7/2}}+\frac {6 \int \frac {x^{17/2}}{\left (a x+b x^3\right )^{7/2}} \, dx}{7 b} \\ & = -\frac {x^{19/2}}{7 b \left (a x+b x^3\right )^{7/2}}-\frac {6 x^{13/2}}{35 b^2 \left (a x+b x^3\right )^{5/2}}+\frac {24 \int \frac {x^{11/2}}{\left (a x+b x^3\right )^{5/2}} \, dx}{35 b^2} \\ & = -\frac {x^{19/2}}{7 b \left (a x+b x^3\right )^{7/2}}-\frac {6 x^{13/2}}{35 b^2 \left (a x+b x^3\right )^{5/2}}-\frac {8 x^{7/2}}{35 b^3 \left (a x+b x^3\right )^{3/2}}+\frac {16 \int \frac {x^{5/2}}{\left (a x+b x^3\right )^{3/2}} \, dx}{35 b^3} \\ & = -\frac {x^{19/2}}{7 b \left (a x+b x^3\right )^{7/2}}-\frac {6 x^{13/2}}{35 b^2 \left (a x+b x^3\right )^{5/2}}-\frac {8 x^{7/2}}{35 b^3 \left (a x+b x^3\right )^{3/2}}-\frac {16 \sqrt {x}}{35 b^4 \sqrt {a x+b x^3}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.63 \[ \int \frac {x^{23/2}}{\left (a x+b x^3\right )^{9/2}} \, dx=\frac {x^{9/2} \left (a+b x^2\right ) \left (-16 a^3-56 a^2 b x^2-70 a b^2 x^4-35 b^3 x^6\right )}{35 b^4 \left (x \left (a+b x^2\right )\right )^{9/2}} \]

[In]

Integrate[x^(23/2)/(a*x + b*x^3)^(9/2),x]

[Out]

(x^(9/2)*(a + b*x^2)*(-16*a^3 - 56*a^2*b*x^2 - 70*a*b^2*x^4 - 35*b^3*x^6))/(35*b^4*(x*(a + b*x^2))^(9/2))

Maple [A] (verified)

Time = 2.00 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.58

method	result	size
gosper	\(-\frac {\left (b \,x^{2}+a \right ) \left (35 b^{3} x^{6}+70 a \,b^{2} x^{4}+56 a^{2} b \,x^{2}+16 a^{3}\right ) x^{\frac {9}{2}}}{35 b^{4} \left (b \,x^{3}+a x \right )^{\frac {9}{2}}}\)	\(59\)
default	\(-\frac {\sqrt {x \left (b \,x^{2}+a \right )}\, \left (35 b^{3} x^{6}+70 a \,b^{2} x^{4}+56 a^{2} b \,x^{2}+16 a^{3}\right )}{35 \sqrt {x}\, \left (b \,x^{2}+a \right )^{4} b^{4}}\)	\(61\)

[In]

int(x^(23/2)/(b*x^3+a*x)^(9/2),x,method=_RETURNVERBOSE)

[Out]

-1/35*(b*x^2+a)*(35*b^3*x^6+70*a*b^2*x^4+56*a^2*b*x^2+16*a^3)*x^(9/2)/b^4/(b*x^3+a*x)^(9/2)

Fricas [A] (verification not implemented)

none

Time = 0.42 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.96 \[ \int \frac {x^{23/2}}{\left (a x+b x^3\right )^{9/2}} \, dx=-\frac {{\left (35 \, b^{3} x^{6} + 70 \, a b^{2} x^{4} + 56 \, a^{2} b x^{2} + 16 \, a^{3}\right )} \sqrt {b x^{3} + a x} \sqrt {x}}{35 \, {\left (b^{8} x^{9} + 4 \, a b^{7} x^{7} + 6 \, a^{2} b^{6} x^{5} + 4 \, a^{3} b^{5} x^{3} + a^{4} b^{4} x\right )}} \]

[In]

integrate(x^(23/2)/(b*x^3+a*x)^(9/2),x, algorithm="fricas")

[Out]

-1/35*(35*b^3*x^6 + 70*a*b^2*x^4 + 56*a^2*b*x^2 + 16*a^3)*sqrt(b*x^3 + a*x)*sqrt(x)/(b^8*x^9 + 4*a*b^7*x^7 + 6
*a^2*b^6*x^5 + 4*a^3*b^5*x^3 + a^4*b^4*x)

Sympy [F(-1)]

Timed out. \[ \int \frac {x^{23/2}}{\left (a x+b x^3\right )^{9/2}} \, dx=\text {Timed out} \]

[In]

integrate(x**(23/2)/(b*x**3+a*x)**(9/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {x^{23/2}}{\left (a x+b x^3\right )^{9/2}} \, dx=\int { \frac {x^{\frac {23}{2}}}{{\left (b x^{3} + a x\right )}^{\frac {9}{2}}} \,d x } \]

[In]

integrate(x^(23/2)/(b*x^3+a*x)^(9/2),x, algorithm="maxima")

[Out]

integrate(x^(23/2)/(b*x^3 + a*x)^(9/2), x)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.63 \[ \int \frac {x^{23/2}}{\left (a x+b x^3\right )^{9/2}} \, dx=\frac {16}{35 \, \sqrt {a} b^{4}} - \frac {35 \, {\left (b x^{2} + a\right )}^{3} - 35 \, {\left (b x^{2} + a\right )}^{2} a + 21 \, {\left (b x^{2} + a\right )} a^{2} - 5 \, a^{3}}{35 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} b^{4}} \]

[In]

integrate(x^(23/2)/(b*x^3+a*x)^(9/2),x, algorithm="giac")

[Out]

16/35/(sqrt(a)*b^4) - 1/35*(35*(b*x^2 + a)^3 - 35*(b*x^2 + a)^2*a + 21*(b*x^2 + a)*a^2 - 5*a^3)/((b*x^2 + a)^(
7/2)*b^4)

Mupad [F(-1)]

Timed out. \[ \int \frac {x^{23/2}}{\left (a x+b x^3\right )^{9/2}} \, dx=\int \frac {x^{23/2}}{{\left (b\,x^3+a\,x\right )}^{9/2}} \,d x \]

[In]

int(x^(23/2)/(a*x + b*x^3)^(9/2),x)

[Out]

int(x^(23/2)/(a*x + b*x^3)^(9/2), x)