\(\int \frac {1}{x^9 (a+b x^{3/2})^{2/3}} \, dx\) [2275]

Optimal result

Integrand size = 17, antiderivative size = 158 \[ \int \frac {1}{x^9 \left (a+b x^{3/2}\right )^{2/3}} \, dx=-\frac {\sqrt [3]{a+b x^{3/2}}}{8 a x^8}+\frac {15 b \sqrt [3]{a+b x^{3/2}}}{104 a^2 x^{13/2}}-\frac {9 b^2 \sqrt [3]{a+b x^{3/2}}}{52 a^3 x^5}+\frac {81 b^3 \sqrt [3]{a+b x^{3/2}}}{364 a^4 x^{7/2}}-\frac {243 b^4 \sqrt [3]{a+b x^{3/2}}}{728 a^5 x^2}+\frac {729 b^5 \sqrt [3]{a+b x^{3/2}}}{728 a^6 \sqrt {x}} \]

[Out]

-1/8*(a+b*x^(3/2))^(1/3)/a/x^8+15/104*b*(a+b*x^(3/2))^(1/3)/a^2/x^(13/2)-9/52*b^2*(a+b*x^(3/2))^(1/3)/a^3/x^5+
81/364*b^3*(a+b*x^(3/2))^(1/3)/a^4/x^(7/2)-243/728*b^4*(a+b*x^(3/2))^(1/3)/a^5/x^2+729/728*b^5*(a+b*x^(3/2))^(
1/3)/a^6/x^(1/2)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {277, 270} \[ \int \frac {1}{x^9 \left (a+b x^{3/2}\right )^{2/3}} \, dx=\frac {729 b^5 \sqrt [3]{a+b x^{3/2}}}{728 a^6 \sqrt {x}}-\frac {243 b^4 \sqrt [3]{a+b x^{3/2}}}{728 a^5 x^2}+\frac {81 b^3 \sqrt [3]{a+b x^{3/2}}}{364 a^4 x^{7/2}}-\frac {9 b^2 \sqrt [3]{a+b x^{3/2}}}{52 a^3 x^5}+\frac {15 b \sqrt [3]{a+b x^{3/2}}}{104 a^2 x^{13/2}}-\frac {\sqrt [3]{a+b x^{3/2}}}{8 a x^8} \]

[In]

Int[1/(x^9*(a + b*x^(3/2))^(2/3)),x]

[Out]

-1/8*(a + b*x^(3/2))^(1/3)/(a*x^8) + (15*b*(a + b*x^(3/2))^(1/3))/(104*a^2*x^(13/2)) - (9*b^2*(a + b*x^(3/2))^
(1/3))/(52*a^3*x^5) + (81*b^3*(a + b*x^(3/2))^(1/3))/(364*a^4*x^(7/2)) - (243*b^4*(a + b*x^(3/2))^(1/3))/(728*
a^5*x^2) + (729*b^5*(a + b*x^(3/2))^(1/3))/(728*a^6*Sqrt[x])

Rule 270

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

Rule 277

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

Rubi steps \begin{align*} \text {integral}& = -\frac {\sqrt [3]{a+b x^{3/2}}}{8 a x^8}-\frac {(15 b) \int \frac {1}{x^{15/2} \left (a+b x^{3/2}\right )^{2/3}} \, dx}{16 a} \\ & = -\frac {\sqrt [3]{a+b x^{3/2}}}{8 a x^8}+\frac {15 b \sqrt [3]{a+b x^{3/2}}}{104 a^2 x^{13/2}}+\frac {\left (45 b^2\right ) \int \frac {1}{x^6 \left (a+b x^{3/2}\right )^{2/3}} \, dx}{52 a^2} \\ & = -\frac {\sqrt [3]{a+b x^{3/2}}}{8 a x^8}+\frac {15 b \sqrt [3]{a+b x^{3/2}}}{104 a^2 x^{13/2}}-\frac {9 b^2 \sqrt [3]{a+b x^{3/2}}}{52 a^3 x^5}-\frac {\left (81 b^3\right ) \int \frac {1}{x^{9/2} \left (a+b x^{3/2}\right )^{2/3}} \, dx}{104 a^3} \\ & = -\frac {\sqrt [3]{a+b x^{3/2}}}{8 a x^8}+\frac {15 b \sqrt [3]{a+b x^{3/2}}}{104 a^2 x^{13/2}}-\frac {9 b^2 \sqrt [3]{a+b x^{3/2}}}{52 a^3 x^5}+\frac {81 b^3 \sqrt [3]{a+b x^{3/2}}}{364 a^4 x^{7/2}}+\frac {\left (243 b^4\right ) \int \frac {1}{x^3 \left (a+b x^{3/2}\right )^{2/3}} \, dx}{364 a^4} \\ & = -\frac {\sqrt [3]{a+b x^{3/2}}}{8 a x^8}+\frac {15 b \sqrt [3]{a+b x^{3/2}}}{104 a^2 x^{13/2}}-\frac {9 b^2 \sqrt [3]{a+b x^{3/2}}}{52 a^3 x^5}+\frac {81 b^3 \sqrt [3]{a+b x^{3/2}}}{364 a^4 x^{7/2}}-\frac {243 b^4 \sqrt [3]{a+b x^{3/2}}}{728 a^5 x^2}-\frac {\left (729 b^5\right ) \int \frac {1}{x^{3/2} \left (a+b x^{3/2}\right )^{2/3}} \, dx}{1456 a^5} \\ & = -\frac {\sqrt [3]{a+b x^{3/2}}}{8 a x^8}+\frac {15 b \sqrt [3]{a+b x^{3/2}}}{104 a^2 x^{13/2}}-\frac {9 b^2 \sqrt [3]{a+b x^{3/2}}}{52 a^3 x^5}+\frac {81 b^3 \sqrt [3]{a+b x^{3/2}}}{364 a^4 x^{7/2}}-\frac {243 b^4 \sqrt [3]{a+b x^{3/2}}}{728 a^5 x^2}+\frac {729 b^5 \sqrt [3]{a+b x^{3/2}}}{728 a^6 \sqrt {x}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.42 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.53 \[ \int \frac {1}{x^9 \left (a+b x^{3/2}\right )^{2/3}} \, dx=\frac {\sqrt [3]{a+b x^{3/2}} \left (-91 a^5+105 a^4 b x^{3/2}-126 a^3 b^2 x^3+162 a^2 b^3 x^{9/2}-243 a b^4 x^6+729 b^5 x^{15/2}\right )}{728 a^6 x^8} \]

[In]

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

[Out]

((a + b*x^(3/2))^(1/3)*(-91*a^5 + 105*a^4*b*x^(3/2) - 126*a^3*b^2*x^3 + 162*a^2*b^3*x^(9/2) - 243*a*b^4*x^6 +
729*b^5*x^(15/2)))/(728*a^6*x^8)

Maple [F]

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.54 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.47 \[ \int \frac {1}{x^9 \left (a+b x^{3/2}\right )^{2/3}} \, dx=-\frac {{\left (243 \, a b^{4} x^{6} + 126 \, a^{3} b^{2} x^{3} + 91 \, a^{5} - 3 \, {\left (243 \, b^{5} x^{7} + 54 \, a^{2} b^{3} x^{4} + 35 \, a^{4} b x\right )} \sqrt {x}\right )} {\left (b x^{\frac {3}{2}} + a\right )}^{\frac {1}{3}}}{728 \, a^{6} x^{8}} \]

[In]

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

[Out]

-1/728*(243*a*b^4*x^6 + 126*a^3*b^2*x^3 + 91*a^5 - 3*(243*b^5*x^7 + 54*a^2*b^3*x^4 + 35*a^4*b*x)*sqrt(x))*(b*x
^(3/2) + a)^(1/3)/(a^6*x^8)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1554 vs. \(2 (148) = 296\).

Time = 32.84 (sec) , antiderivative size = 1554, normalized size of antiderivative = 9.84 \[ \int \frac {1}{x^9 \left (a+b x^{3/2}\right )^{2/3}} \, dx=\text {Too large to display} \]

[In]

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

[Out]

-7280*a**10*b**(76/3)*x**30*(a/(b*x**(3/2)) + 1)**(1/3)*gamma(-16/3)/(729*a**11*b**25*x**(75/2)*gamma(2/3) + 3
645*a**10*b**26*x**39*gamma(2/3) + 7290*a**9*b**27*x**(81/2)*gamma(2/3) + 7290*a**8*b**28*x**42*gamma(2/3) + 3
645*a**7*b**29*x**(87/2)*gamma(2/3) + 729*a**6*b**30*x**45*gamma(2/3)) - 28000*a**9*b**(79/3)*x**(63/2)*(a/(b*
x**(3/2)) + 1)**(1/3)*gamma(-16/3)/(729*a**11*b**25*x**(75/2)*gamma(2/3) + 3645*a**10*b**26*x**39*gamma(2/3) +
7290*a**9*b**27*x**(81/2)*gamma(2/3) + 7290*a**8*b**28*x**42*gamma(2/3) + 3645*a**7*b**29*x**(87/2)*gamma(2/3
) + 729*a**6*b**30*x**45*gamma(2/3)) - 40880*a**8*b**(82/3)*x**33*(a/(b*x**(3/2)) + 1)**(1/3)*gamma(-16/3)/(72
9*a**11*b**25*x**(75/2)*gamma(2/3) + 3645*a**10*b**26*x**39*gamma(2/3) + 7290*a**9*b**27*x**(81/2)*gamma(2/3)
+ 7290*a**8*b**28*x**42*gamma(2/3) + 3645*a**7*b**29*x**(87/2)*gamma(2/3) + 729*a**6*b**30*x**45*gamma(2/3)) -
26240*a**7*b**(85/3)*x**(69/2)*(a/(b*x**(3/2)) + 1)**(1/3)*gamma(-16/3)/(729*a**11*b**25*x**(75/2)*gamma(2/3)
+ 3645*a**10*b**26*x**39*gamma(2/3) + 7290*a**9*b**27*x**(81/2)*gamma(2/3) + 7290*a**8*b**28*x**42*gamma(2/3)
+ 3645*a**7*b**29*x**(87/2)*gamma(2/3) + 729*a**6*b**30*x**45*gamma(2/3)) - 7840*a**6*b**(88/3)*x**36*(a/(b*x
**(3/2)) + 1)**(1/3)*gamma(-16/3)/(729*a**11*b**25*x**(75/2)*gamma(2/3) + 3645*a**10*b**26*x**39*gamma(2/3) +
7290*a**9*b**27*x**(81/2)*gamma(2/3) + 7290*a**8*b**28*x**42*gamma(2/3) + 3645*a**7*b**29*x**(87/2)*gamma(2/3)
+ 729*a**6*b**30*x**45*gamma(2/3)) + 24640*a**5*b**(91/3)*x**(75/2)*(a/(b*x**(3/2)) + 1)**(1/3)*gamma(-16/3)/
(729*a**11*b**25*x**(75/2)*gamma(2/3) + 3645*a**10*b**26*x**39*gamma(2/3) + 7290*a**9*b**27*x**(81/2)*gamma(2/
3) + 7290*a**8*b**28*x**42*gamma(2/3) + 3645*a**7*b**29*x**(87/2)*gamma(2/3) + 729*a**6*b**30*x**45*gamma(2/3)
) + 184800*a**4*b**(94/3)*x**39*(a/(b*x**(3/2)) + 1)**(1/3)*gamma(-16/3)/(729*a**11*b**25*x**(75/2)*gamma(2/3)
+ 3645*a**10*b**26*x**39*gamma(2/3) + 7290*a**9*b**27*x**(81/2)*gamma(2/3) + 7290*a**8*b**28*x**42*gamma(2/3)
+ 3645*a**7*b**29*x**(87/2)*gamma(2/3) + 729*a**6*b**30*x**45*gamma(2/3)) + 443520*a**3*b**(97/3)*x**(81/2)*(
a/(b*x**(3/2)) + 1)**(1/3)*gamma(-16/3)/(729*a**11*b**25*x**(75/2)*gamma(2/3) + 3645*a**10*b**26*x**39*gamma(2
/3) + 7290*a**9*b**27*x**(81/2)*gamma(2/3) + 7290*a**8*b**28*x**42*gamma(2/3) + 3645*a**7*b**29*x**(87/2)*gamm
a(2/3) + 729*a**6*b**30*x**45*gamma(2/3)) + 498960*a**2*b**(100/3)*x**42*(a/(b*x**(3/2)) + 1)**(1/3)*gamma(-16
/3)/(729*a**11*b**25*x**(75/2)*gamma(2/3) + 3645*a**10*b**26*x**39*gamma(2/3) + 7290*a**9*b**27*x**(81/2)*gamm
a(2/3) + 7290*a**8*b**28*x**42*gamma(2/3) + 3645*a**7*b**29*x**(87/2)*gamma(2/3) + 729*a**6*b**30*x**45*gamma(
2/3)) + 272160*a*b**(103/3)*x**(87/2)*(a/(b*x**(3/2)) + 1)**(1/3)*gamma(-16/3)/(729*a**11*b**25*x**(75/2)*gamm
a(2/3) + 3645*a**10*b**26*x**39*gamma(2/3) + 7290*a**9*b**27*x**(81/2)*gamma(2/3) + 7290*a**8*b**28*x**42*gamm
a(2/3) + 3645*a**7*b**29*x**(87/2)*gamma(2/3) + 729*a**6*b**30*x**45*gamma(2/3)) + 58320*b**(106/3)*x**45*(a/(
b*x**(3/2)) + 1)**(1/3)*gamma(-16/3)/(729*a**11*b**25*x**(75/2)*gamma(2/3) + 3645*a**10*b**26*x**39*gamma(2/3)
+ 7290*a**9*b**27*x**(81/2)*gamma(2/3) + 7290*a**8*b**28*x**42*gamma(2/3) + 3645*a**7*b**29*x**(87/2)*gamma(2
/3) + 729*a**6*b**30*x**45*gamma(2/3))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.65 \[ \int \frac {1}{x^9 \left (a+b x^{3/2}\right )^{2/3}} \, dx=\frac {\frac {1456 \, {\left (b x^{\frac {3}{2}} + a\right )}^{\frac {1}{3}} b^{5}}{\sqrt {x}} - \frac {1820 \, {\left (b x^{\frac {3}{2}} + a\right )}^{\frac {4}{3}} b^{4}}{x^{2}} + \frac {2080 \, {\left (b x^{\frac {3}{2}} + a\right )}^{\frac {7}{3}} b^{3}}{x^{\frac {7}{2}}} - \frac {1456 \, {\left (b x^{\frac {3}{2}} + a\right )}^{\frac {10}{3}} b^{2}}{x^{5}} + \frac {560 \, {\left (b x^{\frac {3}{2}} + a\right )}^{\frac {13}{3}} b}{x^{\frac {13}{2}}} - \frac {91 \, {\left (b x^{\frac {3}{2}} + a\right )}^{\frac {16}{3}}}{x^{8}}}{728 \, a^{6}} \]

[In]

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

[Out]

1/728*(1456*(b*x^(3/2) + a)^(1/3)*b^5/sqrt(x) - 1820*(b*x^(3/2) + a)^(4/3)*b^4/x^2 + 2080*(b*x^(3/2) + a)^(7/3
)*b^3/x^(7/2) - 1456*(b*x^(3/2) + a)^(10/3)*b^2/x^5 + 560*(b*x^(3/2) + a)^(13/3)*b/x^(13/2) - 91*(b*x^(3/2) +
a)^(16/3)/x^8)/a^6

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

int(1/(x^9*(a + b*x^(3/2))^(2/3)),x)

[Out]

int(1/(x^9*(a + b*x^(3/2))^(2/3)), x)