\(\int \frac {1}{x^{3/2} (a x+b x^3)^{9/2}} \, dx\) [91]

Optimal result

Integrand size = 19, antiderivative size = 180 \[ \int \frac {1}{x^{3/2} \left (a x+b x^3\right )^{9/2}} \, dx=\frac {1}{7 a x^{3/2} \left (a x+b x^3\right )^{7/2}}+\frac {12}{35 a^2 x^{5/2} \left (a x+b x^3\right )^{5/2}}+\frac {8}{7 a^3 x^{7/2} \left (a x+b x^3\right )^{3/2}}+\frac {64}{7 a^4 x^{9/2} \sqrt {a x+b x^3}}-\frac {384 \sqrt {a x+b x^3}}{35 a^5 x^{11/2}}+\frac {512 b \sqrt {a x+b x^3}}{35 a^6 x^{7/2}}-\frac {1024 b^2 \sqrt {a x+b x^3}}{35 a^7 x^{3/2}} \]

[Out]

1/7/a/x^(3/2)/(b*x^3+a*x)^(7/2)+12/35/a^2/x^(5/2)/(b*x^3+a*x)^(5/2)+8/7/a^3/x^(7/2)/(b*x^3+a*x)^(3/2)+64/7/a^4
/x^(9/2)/(b*x^3+a*x)^(1/2)-384/35*(b*x^3+a*x)^(1/2)/a^5/x^(11/2)+512/35*b*(b*x^3+a*x)^(1/2)/a^6/x^(7/2)-1024/3
5*b^2*(b*x^3+a*x)^(1/2)/a^7/x^(3/2)

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {2040, 2041, 2039} \[ \int \frac {1}{x^{3/2} \left (a x+b x^3\right )^{9/2}} \, dx=-\frac {1024 b^2 \sqrt {a x+b x^3}}{35 a^7 x^{3/2}}+\frac {512 b \sqrt {a x+b x^3}}{35 a^6 x^{7/2}}-\frac {384 \sqrt {a x+b x^3}}{35 a^5 x^{11/2}}+\frac {64}{7 a^4 x^{9/2} \sqrt {a x+b x^3}}+\frac {8}{7 a^3 x^{7/2} \left (a x+b x^3\right )^{3/2}}+\frac {12}{35 a^2 x^{5/2} \left (a x+b x^3\right )^{5/2}}+\frac {1}{7 a x^{3/2} \left (a x+b x^3\right )^{7/2}} \]

[In]

Int[1/(x^(3/2)*(a*x + b*x^3)^(9/2)),x]

[Out]

1/(7*a*x^(3/2)*(a*x + b*x^3)^(7/2)) + 12/(35*a^2*x^(5/2)*(a*x + b*x^3)^(5/2)) + 8/(7*a^3*x^(7/2)*(a*x + b*x^3)
^(3/2)) + 64/(7*a^4*x^(9/2)*Sqrt[a*x + b*x^3]) - (384*Sqrt[a*x + b*x^3])/(35*a^5*x^(11/2)) + (512*b*Sqrt[a*x +
b*x^3])/(35*a^6*x^(7/2)) - (1024*b^2*Sqrt[a*x + b*x^3])/(35*a^7*x^(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])

Rule 2041

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*(m + j*p + 1))), x] - Dist[b*((m + n*p + n - j + 1)/(a*c^(n - j)*(m + j*p + 1))
), Int[(c*x)^(m + n - j)*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[
n, j] && ILtQ[Simplify[(m + n*p + n - j + 1)/(n - j)], 0] && NeQ[m + j*p + 1, 0] && (IntegersQ[j, n] || GtQ[c,
0])

Rubi steps \begin{align*} \text {integral}& = \frac {1}{7 a x^{3/2} \left (a x+b x^3\right )^{7/2}}+\frac {12 \int \frac {1}{x^{5/2} \left (a x+b x^3\right )^{7/2}} \, dx}{7 a} \\ & = \frac {1}{7 a x^{3/2} \left (a x+b x^3\right )^{7/2}}+\frac {12}{35 a^2 x^{5/2} \left (a x+b x^3\right )^{5/2}}+\frac {24 \int \frac {1}{x^{7/2} \left (a x+b x^3\right )^{5/2}} \, dx}{7 a^2} \\ & = \frac {1}{7 a x^{3/2} \left (a x+b x^3\right )^{7/2}}+\frac {12}{35 a^2 x^{5/2} \left (a x+b x^3\right )^{5/2}}+\frac {8}{7 a^3 x^{7/2} \left (a x+b x^3\right )^{3/2}}+\frac {64 \int \frac {1}{x^{9/2} \left (a x+b x^3\right )^{3/2}} \, dx}{7 a^3} \\ & = \frac {1}{7 a x^{3/2} \left (a x+b x^3\right )^{7/2}}+\frac {12}{35 a^2 x^{5/2} \left (a x+b x^3\right )^{5/2}}+\frac {8}{7 a^3 x^{7/2} \left (a x+b x^3\right )^{3/2}}+\frac {64}{7 a^4 x^{9/2} \sqrt {a x+b x^3}}+\frac {384 \int \frac {1}{x^{11/2} \sqrt {a x+b x^3}} \, dx}{7 a^4} \\ & = \frac {1}{7 a x^{3/2} \left (a x+b x^3\right )^{7/2}}+\frac {12}{35 a^2 x^{5/2} \left (a x+b x^3\right )^{5/2}}+\frac {8}{7 a^3 x^{7/2} \left (a x+b x^3\right )^{3/2}}+\frac {64}{7 a^4 x^{9/2} \sqrt {a x+b x^3}}-\frac {384 \sqrt {a x+b x^3}}{35 a^5 x^{11/2}}-\frac {(1536 b) \int \frac {1}{x^{7/2} \sqrt {a x+b x^3}} \, dx}{35 a^5} \\ & = \frac {1}{7 a x^{3/2} \left (a x+b x^3\right )^{7/2}}+\frac {12}{35 a^2 x^{5/2} \left (a x+b x^3\right )^{5/2}}+\frac {8}{7 a^3 x^{7/2} \left (a x+b x^3\right )^{3/2}}+\frac {64}{7 a^4 x^{9/2} \sqrt {a x+b x^3}}-\frac {384 \sqrt {a x+b x^3}}{35 a^5 x^{11/2}}+\frac {512 b \sqrt {a x+b x^3}}{35 a^6 x^{7/2}}+\frac {\left (1024 b^2\right ) \int \frac {1}{x^{3/2} \sqrt {a x+b x^3}} \, dx}{35 a^6} \\ & = \frac {1}{7 a x^{3/2} \left (a x+b x^3\right )^{7/2}}+\frac {12}{35 a^2 x^{5/2} \left (a x+b x^3\right )^{5/2}}+\frac {8}{7 a^3 x^{7/2} \left (a x+b x^3\right )^{3/2}}+\frac {64}{7 a^4 x^{9/2} \sqrt {a x+b x^3}}-\frac {384 \sqrt {a x+b x^3}}{35 a^5 x^{11/2}}+\frac {512 b \sqrt {a x+b x^3}}{35 a^6 x^{7/2}}-\frac {1024 b^2 \sqrt {a x+b x^3}}{35 a^7 x^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.50 \[ \int \frac {1}{x^{3/2} \left (a x+b x^3\right )^{9/2}} \, dx=\frac {-7 a^6+28 a^5 b x^2-280 a^4 b^2 x^4-2240 a^3 b^3 x^6-4480 a^2 b^4 x^8-3584 a b^5 x^{10}-1024 b^6 x^{12}}{35 a^7 x^{3/2} \left (x \left (a+b x^2\right )\right )^{7/2}} \]

[In]

Integrate[1/(x^(3/2)*(a*x + b*x^3)^(9/2)),x]

[Out]

(-7*a^6 + 28*a^5*b*x^2 - 280*a^4*b^2*x^4 - 2240*a^3*b^3*x^6 - 4480*a^2*b^4*x^8 - 3584*a*b^5*x^10 - 1024*b^6*x^
12)/(35*a^7*x^(3/2)*(x*(a + b*x^2))^(7/2))

Maple [A] (verified)

Time = 2.06 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.51

method	result	size
gosper	\(-\frac {\left (b \,x^{2}+a \right ) \left (1024 b^{6} x^{12}+3584 a \,b^{5} x^{10}+4480 a^{2} b^{4} x^{8}+2240 a^{3} b^{3} x^{6}+280 a^{4} b^{2} x^{4}-28 a^{5} b \,x^{2}+7 a^{6}\right )}{35 \sqrt {x}\, a^{7} \left (b \,x^{3}+a x \right )^{\frac {9}{2}}}\)	\(92\)
default	\(-\frac {\sqrt {x \left (b \,x^{2}+a \right )}\, \left (1024 b^{6} x^{12}+3584 a \,b^{5} x^{10}+4480 a^{2} b^{4} x^{8}+2240 a^{3} b^{3} x^{6}+280 a^{4} b^{2} x^{4}-28 a^{5} b \,x^{2}+7 a^{6}\right )}{35 x^{\frac {11}{2}} \left (b \,x^{2}+a \right )^{4} a^{7}}\)	\(94\)
risch	\(-\frac {\left (b \,x^{2}+a \right ) \left (66 b^{2} x^{4}-8 a b \,x^{2}+a^{2}\right )}{5 a^{7} x^{\frac {9}{2}} \sqrt {x \left (b \,x^{2}+a \right )}}-\frac {\left (b \,x^{2}+a \right ) x^{\frac {3}{2}} \left (562 b^{3} x^{6}+1792 a \,b^{2} x^{4}+1925 a^{2} b \,x^{2}+700 a^{3}\right ) b^{3}}{35 a^{7} \left (x^{8} b^{4}+4 a \,b^{3} x^{6}+6 a^{2} x^{4} b^{2}+4 a^{3} b \,x^{2}+a^{4}\right ) \sqrt {x \left (b \,x^{2}+a \right )}}\)	\(150\)

[In]

int(1/x^(3/2)/(b*x^3+a*x)^(9/2),x,method=_RETURNVERBOSE)

[Out]

-1/35*(b*x^2+a)*(1024*b^6*x^12+3584*a*b^5*x^10+4480*a^2*b^4*x^8+2240*a^3*b^3*x^6+280*a^4*b^2*x^4-28*a^5*b*x^2+
7*a^6)/x^(1/2)/a^7/(b*x^3+a*x)^(9/2)

Fricas [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.73 \[ \int \frac {1}{x^{3/2} \left (a x+b x^3\right )^{9/2}} \, dx=-\frac {{\left (1024 \, b^{6} x^{12} + 3584 \, a b^{5} x^{10} + 4480 \, a^{2} b^{4} x^{8} + 2240 \, a^{3} b^{3} x^{6} + 280 \, a^{4} b^{2} x^{4} - 28 \, a^{5} b x^{2} + 7 \, a^{6}\right )} \sqrt {b x^{3} + a x} \sqrt {x}}{35 \, {\left (a^{7} b^{4} x^{14} + 4 \, a^{8} b^{3} x^{12} + 6 \, a^{9} b^{2} x^{10} + 4 \, a^{10} b x^{8} + a^{11} x^{6}\right )}} \]

[In]

integrate(1/x^(3/2)/(b*x^3+a*x)^(9/2),x, algorithm="fricas")

[Out]

-1/35*(1024*b^6*x^12 + 3584*a*b^5*x^10 + 4480*a^2*b^4*x^8 + 2240*a^3*b^3*x^6 + 280*a^4*b^2*x^4 - 28*a^5*b*x^2
+ 7*a^6)*sqrt(b*x^3 + a*x)*sqrt(x)/(a^7*b^4*x^14 + 4*a^8*b^3*x^12 + 6*a^9*b^2*x^10 + 4*a^10*b*x^8 + a^11*x^6)

Sympy [F]

\[ \int \frac {1}{x^{3/2} \left (a x+b x^3\right )^{9/2}} \, dx=\int \frac {1}{x^{\frac {3}{2}} \left (x \left (a + b x^{2}\right )\right )^{\frac {9}{2}}}\, dx \]

[In]

integrate(1/x**(3/2)/(b*x**3+a*x)**(9/2),x)

[Out]

Integral(1/(x**(3/2)*(x*(a + b*x**2))**(9/2)), x)

Maxima [F]

\[ \int \frac {1}{x^{3/2} \left (a x+b x^3\right )^{9/2}} \, dx=\int { \frac {1}{{\left (b x^{3} + a x\right )}^{\frac {9}{2}} x^{\frac {3}{2}}} \,d x } \]

[In]

integrate(1/x^(3/2)/(b*x^3+a*x)^(9/2),x, algorithm="maxima")

[Out]

integrate(1/((b*x^3 + a*x)^(9/2)*x^(3/2)), x)

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.12 \[ \int \frac {1}{x^{3/2} \left (a x+b x^3\right )^{9/2}} \, dx=-\frac {{\left ({\left (2 \, x^{2} {\left (\frac {281 \, b^{6} x^{2}}{a^{7}} + \frac {896 \, b^{5}}{a^{6}}\right )} + \frac {1925 \, b^{4}}{a^{5}}\right )} x^{2} + \frac {700 \, b^{3}}{a^{4}}\right )} x}{35 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}}} + \frac {4 \, {\left (25 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{8} b^{\frac {5}{2}} - 120 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{6} a b^{\frac {5}{2}} + 210 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{4} a^{2} b^{\frac {5}{2}} - 140 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} a^{3} b^{\frac {5}{2}} + 33 \, a^{4} b^{\frac {5}{2}}\right )}}{5 \, {\left ({\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} - a\right )}^{5} a^{6}} \]

[In]

integrate(1/x^(3/2)/(b*x^3+a*x)^(9/2),x, algorithm="giac")

[Out]

-1/35*((2*x^2*(281*b^6*x^2/a^7 + 896*b^5/a^6) + 1925*b^4/a^5)*x^2 + 700*b^3/a^4)*x/(b*x^2 + a)^(7/2) + 4/5*(25
*(sqrt(b)*x - sqrt(b*x^2 + a))^8*b^(5/2) - 120*(sqrt(b)*x - sqrt(b*x^2 + a))^6*a*b^(5/2) + 210*(sqrt(b)*x - sq
rt(b*x^2 + a))^4*a^2*b^(5/2) - 140*(sqrt(b)*x - sqrt(b*x^2 + a))^2*a^3*b^(5/2) + 33*a^4*b^(5/2))/(((sqrt(b)*x
- sqrt(b*x^2 + a))^2 - a)^5*a^6)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x^{3/2} \left (a x+b x^3\right )^{9/2}} \, dx=\int \frac {1}{x^{3/2}\,{\left (b\,x^3+a\,x\right )}^{9/2}} \,d x \]

[In]

int(1/(x^(3/2)*(a*x + b*x^3)^(9/2)),x)

[Out]

int(1/(x^(3/2)*(a*x + b*x^3)^(9/2)), x)