Integrand size = 19, antiderivative size = 117 \[ \int \frac {1}{(c x)^{17/2} \left (a+b x^3\right )^{3/2}} \, dx=\frac {2}{3 a c (c x)^{15/2} \sqrt {a+b x^3}}-\frac {4 \sqrt {a+b x^3}}{5 a^2 c (c x)^{15/2}}+\frac {16 b \sqrt {a+b x^3}}{15 a^3 c^4 (c x)^{9/2}}-\frac {32 b^2 \sqrt {a+b x^3}}{15 a^4 c^7 (c x)^{3/2}} \] Output:
2/3/a/c/(c*x)^(15/2)/(b*x^3+a)^(1/2)-4/5*(b*x^3+a)^(1/2)/a^2/c/(c*x)^(15/2 )+16/15*b*(b*x^3+a)^(1/2)/a^3/c^4/(c*x)^(9/2)-32/15*b^2*(b*x^3+a)^(1/2)/a^ 4/c^7/(c*x)^(3/2)
Time = 2.17 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.48 \[ \int \frac {1}{(c x)^{17/2} \left (a+b x^3\right )^{3/2}} \, dx=-\frac {2 x \left (a^3-2 a^2 b x^3+8 a b^2 x^6+16 b^3 x^9\right )}{15 a^4 (c x)^{17/2} \sqrt {a+b x^3}} \] Input:
Integrate[1/((c*x)^(17/2)*(a + b*x^3)^(3/2)),x]
Output:
(-2*x*(a^3 - 2*a^2*b*x^3 + 8*a*b^2*x^6 + 16*b^3*x^9))/(15*a^4*(c*x)^(17/2) *Sqrt[a + b*x^3])
Time = 0.41 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.07, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {805, 805, 805, 796}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {1}{(c x)^{17/2} \left (a+b x^3\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 805 |
\(\displaystyle \frac {6 \int \frac {1}{(c x)^{17/2} \sqrt {b x^3+a}}dx}{a}+\frac {2}{3 a c (c x)^{15/2} \sqrt {a+b x^3}}\) |
\(\Big \downarrow \) 805 |
\(\displaystyle \frac {6 \left (-\frac {4 \int \frac {\sqrt {b x^3+a}}{(c x)^{17/2}}dx}{a}-\frac {2 \sqrt {a+b x^3}}{3 a c (c x)^{15/2}}\right )}{a}+\frac {2}{3 a c (c x)^{15/2} \sqrt {a+b x^3}}\) |
\(\Big \downarrow \) 805 |
\(\displaystyle \frac {6 \left (-\frac {4 \left (-\frac {2 \int \frac {\left (b x^3+a\right )^{3/2}}{(c x)^{17/2}}dx}{3 a}-\frac {2 \left (a+b x^3\right )^{3/2}}{9 a c (c x)^{15/2}}\right )}{a}-\frac {2 \sqrt {a+b x^3}}{3 a c (c x)^{15/2}}\right )}{a}+\frac {2}{3 a c (c x)^{15/2} \sqrt {a+b x^3}}\) |
\(\Big \downarrow \) 796 |
\(\displaystyle \frac {6 \left (-\frac {4 \left (\frac {4 \left (a+b x^3\right )^{5/2}}{45 a^2 c (c x)^{15/2}}-\frac {2 \left (a+b x^3\right )^{3/2}}{9 a c (c x)^{15/2}}\right )}{a}-\frac {2 \sqrt {a+b x^3}}{3 a c (c x)^{15/2}}\right )}{a}+\frac {2}{3 a c (c x)^{15/2} \sqrt {a+b x^3}}\) |
Input:
Int[1/((c*x)^(17/2)*(a + b*x^3)^(3/2)),x]
Output:
2/(3*a*c*(c*x)^(15/2)*Sqrt[a + b*x^3]) + (6*((-2*Sqrt[a + b*x^3])/(3*a*c*( c*x)^(15/2)) - (4*((-2*(a + b*x^3)^(3/2))/(9*a*c*(c*x)^(15/2)) + (4*(a + b *x^3)^(5/2))/(45*a^2*c*(c*x)^(15/2))))/a))/a
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]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-( c*x)^(m + 1))*((a + b*x^n)^(p + 1)/(a*c*n*(p + 1))), x] + Simp[(m + n*(p + 1) + 1)/(a*n*(p + 1)) Int[(c*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a , b, c, m, n, p}, x] && ILtQ[Simplify[(m + 1)/n + p + 1], 0] && NeQ[p, -1]
Time = 0.66 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.44
method | result | size |
gosper | \(-\frac {2 x \left (16 b^{3} x^{9}+8 a \,b^{2} x^{6}-2 a^{2} b \,x^{3}+a^{3}\right )}{15 \sqrt {b \,x^{3}+a}\, a^{4} \left (c x \right )^{\frac {17}{2}}}\) | \(51\) |
orering | \(-\frac {2 x \left (16 b^{3} x^{9}+8 a \,b^{2} x^{6}-2 a^{2} b \,x^{3}+a^{3}\right )}{15 \sqrt {b \,x^{3}+a}\, a^{4} \left (c x \right )^{\frac {17}{2}}}\) | \(51\) |
default | \(-\frac {2 \left (16 b^{3} x^{9}+8 a \,b^{2} x^{6}-2 a^{2} b \,x^{3}+a^{3}\right )}{15 x^{7} \sqrt {b \,x^{3}+a}\, a^{4} c^{8} \sqrt {c x}}\) | \(56\) |
risch | \(-\frac {2 \sqrt {b \,x^{3}+a}\, \left (11 b^{2} x^{6}-3 a b \,x^{3}+a^{2}\right )}{15 a^{4} x^{7} c^{8} \sqrt {c x}}-\frac {2 b^{3} x^{2}}{3 a^{4} c^{8} \sqrt {c x}\, \sqrt {b \,x^{3}+a}}\) | \(74\) |
elliptic | \(\frac {\sqrt {c x \left (b \,x^{3}+a \right )}\, \left (-\frac {2 \sqrt {b c \,x^{4}+a c x}}{15 a^{2} c^{9} x^{8}}+\frac {2 b \sqrt {b c \,x^{4}+a c x}}{5 c^{9} a^{3} x^{5}}-\frac {22 b^{2} \sqrt {b c \,x^{4}+a c x}}{15 a^{4} c^{9} x^{2}}-\frac {2 b^{3} x^{2}}{3 c^{8} a^{4} \sqrt {\left (x^{3}+\frac {a}{b}\right ) b c x}}\right )}{\sqrt {c x}\, \sqrt {b \,x^{3}+a}}\) | \(134\) |
Input:
int(1/(c*x)^(17/2)/(b*x^3+a)^(3/2),x,method=_RETURNVERBOSE)
Output:
-2/15*x*(16*b^3*x^9+8*a*b^2*x^6-2*a^2*b*x^3+a^3)/(b*x^3+a)^(1/2)/a^4/(c*x) ^(17/2)
Time = 0.09 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.60 \[ \int \frac {1}{(c x)^{17/2} \left (a+b x^3\right )^{3/2}} \, dx=-\frac {2 \, {\left (16 \, b^{3} x^{9} + 8 \, a b^{2} x^{6} - 2 \, a^{2} b x^{3} + a^{3}\right )} \sqrt {b x^{3} + a} \sqrt {c x}}{15 \, {\left (a^{4} b c^{9} x^{11} + a^{5} c^{9} x^{8}\right )}} \] Input:
integrate(1/(c*x)^(17/2)/(b*x^3+a)^(3/2),x, algorithm="fricas")
Output:
-2/15*(16*b^3*x^9 + 8*a*b^2*x^6 - 2*a^2*b*x^3 + a^3)*sqrt(b*x^3 + a)*sqrt( c*x)/(a^4*b*c^9*x^11 + a^5*c^9*x^8)
Timed out. \[ \int \frac {1}{(c x)^{17/2} \left (a+b x^3\right )^{3/2}} \, dx=\text {Timed out} \] Input:
integrate(1/(c*x)**(17/2)/(b*x**3+a)**(3/2),x)
Output:
Timed out
\[ \int \frac {1}{(c x)^{17/2} \left (a+b x^3\right )^{3/2}} \, dx=\int { \frac {1}{{\left (b x^{3} + a\right )}^{\frac {3}{2}} \left (c x\right )^{\frac {17}{2}}} \,d x } \] Input:
integrate(1/(c*x)^(17/2)/(b*x^3+a)^(3/2),x, algorithm="maxima")
Output:
integrate(1/((b*x^3 + a)^(3/2)*(c*x)^(17/2)), x)
Error detected during grading. Assigning place holder grade for now.
Time = 0.22 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.17 \[ \int \frac {1}{(c x)^{17/2} \left (a+b x^3\right )^{3/2}} \, dx=\mathit {Recursive} a \mathit {assumption} \geq -\frac {2 \, {\left (\frac {15 \, \sqrt {b c + \frac {a c}{x^{3}}} b^{2}}{a c^{7}} - \frac {11 \, \sqrt {b c} b^{2}}{a c^{7}} - \frac {5 \, {\left (b c + \frac {a c}{x^{3}}\right )}^{\frac {3}{2}} b c - {\left (b c + \frac {a c}{x^{3}}\right )}^{\frac {5}{2}}}{a c^{9}}\right )}}{15 \, a^{3} {\left | c \right |}^{2}} - \frac {b \mathit {ignored}}{c^{3} t_{\mathit {nostep}}^{6}} - \frac {2 \, \sqrt {c x} b^{3} x}{3 \, \sqrt {b c^{4} x^{3} + a c^{4}} a^{4} c^{7}} \] Input:
integrate(1/(c*x)^(17/2)/(b*x^3+a)^(3/2),x, algorithm="giac")
Output:
Recursive*a*assumption >= -2/15*(15*sqrt(b*c + a*c/x^3)*b^2/(a*c^7) - 11*s qrt(b*c)*b^2/(a*c^7) - (5*(b*c + a*c/x^3)^(3/2)*b*c - (b*c + a*c/x^3)^(5/2 ))/(a*c^9))/(a^3*abs(c)^2) - b*ignored/(c^3*t_nostep^6) - 2/3*sqrt(c*x)*b^ 3*x/(sqrt(b*c^4*x^3 + a*c^4)*a^4*c^7)
Time = 0.53 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.73 \[ \int \frac {1}{(c x)^{17/2} \left (a+b x^3\right )^{3/2}} \, dx=-\frac {\sqrt {b\,x^3+a}\,\left (\frac {2}{15\,a\,b\,c^8}-\frac {4\,x^3}{15\,a^2\,c^8}+\frac {16\,b\,x^6}{15\,a^3\,c^8}+\frac {32\,b^2\,x^9}{15\,a^4\,c^8}\right )}{x^{10}\,\sqrt {c\,x}+\frac {a\,x^7\,\sqrt {c\,x}}{b}} \] Input:
int(1/((c*x)^(17/2)*(a + b*x^3)^(3/2)),x)
Output:
-((a + b*x^3)^(1/2)*(2/(15*a*b*c^8) - (4*x^3)/(15*a^2*c^8) + (16*b*x^6)/(1 5*a^3*c^8) + (32*b^2*x^9)/(15*a^4*c^8)))/(x^10*(c*x)^(1/2) + (a*x^7*(c*x)^ (1/2))/b)
Time = 0.31 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.56 \[ \int \frac {1}{(c x)^{17/2} \left (a+b x^3\right )^{3/2}} \, dx=\frac {2 \sqrt {c}\, \sqrt {b \,x^{3}+a}\, \left (-16 b^{3} x^{9}-8 a \,b^{2} x^{6}+2 a^{2} b \,x^{3}-a^{3}\right )}{15 \sqrt {x}\, a^{4} c^{9} x^{7} \left (b \,x^{3}+a \right )} \] Input:
int(1/(c*x)^(17/2)/(b*x^3+a)^(3/2),x)
Output:
(2*sqrt(c)*sqrt(a + b*x**3)*( - a**3 + 2*a**2*b*x**3 - 8*a*b**2*x**6 - 16* b**3*x**9))/(15*sqrt(x)*a**4*c**9*x**7*(a + b*x**3))