\(\int \frac {x^{27/2}}{(a x+b x^3)^{9/2}} \, dx\) [76]

Optimal result

Integrand size = 19, antiderivative size = 126 \[ \int \frac {x^{27/2}}{\left (a x+b x^3\right )^{9/2}} \, dx=-\frac {x^{23/2}}{7 b \left (a x+b x^3\right )^{7/2}}-\frac {8 x^{17/2}}{35 b^2 \left (a x+b x^3\right )^{5/2}}-\frac {16 x^{11/2}}{35 b^3 \left (a x+b x^3\right )^{3/2}}-\frac {64 x^{5/2}}{35 b^4 \sqrt {a x+b x^3}}+\frac {128 \sqrt {a x+b x^3}}{35 b^5 \sqrt {x}} \]

[Out]

-1/7*x^(23/2)/b/(b*x^3+a*x)^(7/2)-8/35*x^(17/2)/b^2/(b*x^3+a*x)^(5/2)-16/35*x^(11/2)/b^3/(b*x^3+a*x)^(3/2)-64/
35*x^(5/2)/b^4/(b*x^3+a*x)^(1/2)+128/35*(b*x^3+a*x)^(1/2)/b^5/x^(1/2)

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2040, 2039} \[ \int \frac {x^{27/2}}{\left (a x+b x^3\right )^{9/2}} \, dx=\frac {128 \sqrt {a x+b x^3}}{35 b^5 \sqrt {x}}-\frac {64 x^{5/2}}{35 b^4 \sqrt {a x+b x^3}}-\frac {16 x^{11/2}}{35 b^3 \left (a x+b x^3\right )^{3/2}}-\frac {8 x^{17/2}}{35 b^2 \left (a x+b x^3\right )^{5/2}}-\frac {x^{23/2}}{7 b \left (a x+b x^3\right )^{7/2}} \]

[In]

Int[x^(27/2)/(a*x + b*x^3)^(9/2),x]

[Out]

-1/7*x^(23/2)/(b*(a*x + b*x^3)^(7/2)) - (8*x^(17/2))/(35*b^2*(a*x + b*x^3)^(5/2)) - (16*x^(11/2))/(35*b^3*(a*x
+ b*x^3)^(3/2)) - (64*x^(5/2))/(35*b^4*Sqrt[a*x + b*x^3]) + (128*Sqrt[a*x + b*x^3])/(35*b^5*Sqrt[x])

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^{23/2}}{7 b \left (a x+b x^3\right )^{7/2}}+\frac {8 \int \frac {x^{21/2}}{\left (a x+b x^3\right )^{7/2}} \, dx}{7 b} \\ & = -\frac {x^{23/2}}{7 b \left (a x+b x^3\right )^{7/2}}-\frac {8 x^{17/2}}{35 b^2 \left (a x+b x^3\right )^{5/2}}+\frac {48 \int \frac {x^{15/2}}{\left (a x+b x^3\right )^{5/2}} \, dx}{35 b^2} \\ & = -\frac {x^{23/2}}{7 b \left (a x+b x^3\right )^{7/2}}-\frac {8 x^{17/2}}{35 b^2 \left (a x+b x^3\right )^{5/2}}-\frac {16 x^{11/2}}{35 b^3 \left (a x+b x^3\right )^{3/2}}+\frac {64 \int \frac {x^{9/2}}{\left (a x+b x^3\right )^{3/2}} \, dx}{35 b^3} \\ & = -\frac {x^{23/2}}{7 b \left (a x+b x^3\right )^{7/2}}-\frac {8 x^{17/2}}{35 b^2 \left (a x+b x^3\right )^{5/2}}-\frac {16 x^{11/2}}{35 b^3 \left (a x+b x^3\right )^{3/2}}-\frac {64 x^{5/2}}{35 b^4 \sqrt {a x+b x^3}}+\frac {128 \int \frac {x^{3/2}}{\sqrt {a x+b x^3}} \, dx}{35 b^4} \\ & = -\frac {x^{23/2}}{7 b \left (a x+b x^3\right )^{7/2}}-\frac {8 x^{17/2}}{35 b^2 \left (a x+b x^3\right )^{5/2}}-\frac {16 x^{11/2}}{35 b^3 \left (a x+b x^3\right )^{3/2}}-\frac {64 x^{5/2}}{35 b^4 \sqrt {a x+b x^3}}+\frac {128 \sqrt {a x+b x^3}}{35 b^5 \sqrt {x}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.54 \[ \int \frac {x^{27/2}}{\left (a x+b x^3\right )^{9/2}} \, dx=\frac {x^{7/2} \left (128 a^4+448 a^3 b x^2+560 a^2 b^2 x^4+280 a b^3 x^6+35 b^4 x^8\right )}{35 b^5 \left (x \left (a+b x^2\right )\right )^{7/2}} \]

[In]

Integrate[x^(27/2)/(a*x + b*x^3)^(9/2),x]

[Out]

(x^(7/2)*(128*a^4 + 448*a^3*b*x^2 + 560*a^2*b^2*x^4 + 280*a*b^3*x^6 + 35*b^4*x^8))/(35*b^5*(x*(a + b*x^2))^(7/
2))

Maple [A] (verified)

Time = 2.06 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.56

method	result	size
gosper	\(\frac {\left (b \,x^{2}+a \right ) \left (35 x^{8} b^{4}+280 a \,b^{3} x^{6}+560 a^{2} x^{4} b^{2}+448 a^{3} b \,x^{2}+128 a^{4}\right ) x^{\frac {9}{2}}}{35 b^{5} \left (b \,x^{3}+a x \right )^{\frac {9}{2}}}\)	\(70\)
default	\(\frac {\sqrt {x \left (b \,x^{2}+a \right )}\, \left (35 x^{8} b^{4}+280 a \,b^{3} x^{6}+560 a^{2} x^{4} b^{2}+448 a^{3} b \,x^{2}+128 a^{4}\right )}{35 \sqrt {x}\, \left (b \,x^{2}+a \right )^{4} b^{5}}\)	\(72\)
risch	\(\frac {\left (b \,x^{2}+a \right ) \sqrt {x}}{b^{5} \sqrt {x \left (b \,x^{2}+a \right )}}+\frac {\left (b \,x^{2}+a \right ) \left (140 b^{3} x^{6}+350 a \,b^{2} x^{4}+308 a^{2} b \,x^{2}+93 a^{3}\right ) a \sqrt {x}}{35 b^{5} \left (x^{8} b^{4}+4 a \,b^{3} x^{6}+6 a^{2} x^{4} b^{2}+4 a^{3} b \,x^{2}+a^{4}\right ) \sqrt {x \left (b \,x^{2}+a \right )}}\)	\(128\)

[In]

int(x^(27/2)/(b*x^3+a*x)^(9/2),x,method=_RETURNVERBOSE)

[Out]

1/35*(b*x^2+a)*(35*b^4*x^8+280*a*b^3*x^6+560*a^2*b^2*x^4+448*a^3*b*x^2+128*a^4)*x^(9/2)/b^5/(b*x^3+a*x)^(9/2)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.86 \[ \int \frac {x^{27/2}}{\left (a x+b x^3\right )^{9/2}} \, dx=\frac {{\left (35 \, b^{4} x^{8} + 280 \, a b^{3} x^{6} + 560 \, a^{2} b^{2} x^{4} + 448 \, a^{3} b x^{2} + 128 \, a^{4}\right )} \sqrt {b x^{3} + a x} \sqrt {x}}{35 \, {\left (b^{9} x^{9} + 4 \, a b^{8} x^{7} + 6 \, a^{2} b^{7} x^{5} + 4 \, a^{3} b^{6} x^{3} + a^{4} b^{5} x\right )}} \]

[In]

integrate(x^(27/2)/(b*x^3+a*x)^(9/2),x, algorithm="fricas")

[Out]

1/35*(35*b^4*x^8 + 280*a*b^3*x^6 + 560*a^2*b^2*x^4 + 448*a^3*b*x^2 + 128*a^4)*sqrt(b*x^3 + a*x)*sqrt(x)/(b^9*x
^9 + 4*a*b^8*x^7 + 6*a^2*b^7*x^5 + 4*a^3*b^6*x^3 + a^4*b^5*x)

Sympy [F(-1)]

Timed out. \[ \int \frac {x^{27/2}}{\left (a x+b x^3\right )^{9/2}} \, dx=\text {Timed out} \]

[In]

integrate(x**(27/2)/(b*x**3+a*x)**(9/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {x^{27/2}}{\left (a x+b x^3\right )^{9/2}} \, dx=\int { \frac {x^{\frac {27}{2}}}{{\left (b x^{3} + a x\right )}^{\frac {9}{2}}} \,d x } \]

[In]

integrate(x^(27/2)/(b*x^3+a*x)^(9/2),x, algorithm="maxima")

[Out]

integrate(x^(27/2)/(b*x^3 + a*x)^(9/2), x)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.63 \[ \int \frac {x^{27/2}}{\left (a x+b x^3\right )^{9/2}} \, dx=\frac {\sqrt {b x^{2} + a}}{b^{5}} - \frac {128 \, \sqrt {a}}{35 \, b^{5}} + \frac {140 \, {\left (b x^{2} + a\right )}^{3} a - 70 \, {\left (b x^{2} + a\right )}^{2} a^{2} + 28 \, {\left (b x^{2} + a\right )} a^{3} - 5 \, a^{4}}{35 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} b^{5}} \]

[In]

integrate(x^(27/2)/(b*x^3+a*x)^(9/2),x, algorithm="giac")

[Out]

sqrt(b*x^2 + a)/b^5 - 128/35*sqrt(a)/b^5 + 1/35*(140*(b*x^2 + a)^3*a - 70*(b*x^2 + a)^2*a^2 + 28*(b*x^2 + a)*a
^3 - 5*a^4)/((b*x^2 + a)^(7/2)*b^5)

Mupad [F(-1)]

Timed out. \[ \int \frac {x^{27/2}}{\left (a x+b x^3\right )^{9/2}} \, dx=\int \frac {x^{27/2}}{{\left (b\,x^3+a\,x\right )}^{9/2}} \,d x \]

[In]

int(x^(27/2)/(a*x + b*x^3)^(9/2),x)

[Out]

int(x^(27/2)/(a*x + b*x^3)^(9/2), x)