Integrand size = 17, antiderivative size = 163 \[ \int \frac {\left (a x+b x^3\right )^{3/2}}{x^8} \, dx=-\frac {12 b \sqrt {a x+b x^3}}{77 x^4}-\frac {8 b^2 \sqrt {a x+b x^3}}{77 a x^2}-\frac {2 \left (a x+b x^3\right )^{3/2}}{11 x^7}-\frac {4 b^{11/4} \sqrt {x} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{77 a^{5/4} \sqrt {a x+b x^3}} \]
-2/11*(b*x^3+a*x)^(3/2)/x^7-12/77*b*(b*x^3+a*x)^(1/2)/x^4-8/77*b^2*(b*x^3+ a*x)^(1/2)/a/x^2-4/77*b^(11/4)*(cos(2*arctan(b^(1/4)*x^(1/2)/a^(1/4)))^2)^ (1/2)/cos(2*arctan(b^(1/4)*x^(1/2)/a^(1/4)))*EllipticF(sin(2*arctan(b^(1/4 )*x^(1/2)/a^(1/4))),1/2*2^(1/2))*(a^(1/2)+x*b^(1/2))*x^(1/2)*((b*x^2+a)/(a ^(1/2)+x*b^(1/2))^2)^(1/2)/a^(5/4)/(b*x^3+a*x)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.02 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.33 \[ \int \frac {\left (a x+b x^3\right )^{3/2}}{x^8} \, dx=-\frac {2 a \sqrt {x \left (a+b x^2\right )} \operatorname {Hypergeometric2F1}\left (-\frac {11}{4},-\frac {3}{2},-\frac {7}{4},-\frac {b x^2}{a}\right )}{11 x^6 \sqrt {1+\frac {b x^2}{a}}} \]
(-2*a*Sqrt[x*(a + b*x^2)]*Hypergeometric2F1[-11/4, -3/2, -7/4, -((b*x^2)/a )])/(11*x^6*Sqrt[1 + (b*x^2)/a])
Time = 0.33 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.05, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {1926, 1926, 1931, 1917, 266, 761}
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.
\(\displaystyle \int \frac {\left (a x+b x^3\right )^{3/2}}{x^8} \, dx\) |
\(\Big \downarrow \) 1926 |
\(\displaystyle \frac {6}{11} b \int \frac {\sqrt {b x^3+a x}}{x^5}dx-\frac {2 \left (a x+b x^3\right )^{3/2}}{11 x^7}\) |
\(\Big \downarrow \) 1926 |
\(\displaystyle \frac {6}{11} b \left (\frac {2}{7} b \int \frac {1}{x^2 \sqrt {b x^3+a x}}dx-\frac {2 \sqrt {a x+b x^3}}{7 x^4}\right )-\frac {2 \left (a x+b x^3\right )^{3/2}}{11 x^7}\) |
\(\Big \downarrow \) 1931 |
\(\displaystyle \frac {6}{11} b \left (\frac {2}{7} b \left (-\frac {b \int \frac {1}{\sqrt {b x^3+a x}}dx}{3 a}-\frac {2 \sqrt {a x+b x^3}}{3 a x^2}\right )-\frac {2 \sqrt {a x+b x^3}}{7 x^4}\right )-\frac {2 \left (a x+b x^3\right )^{3/2}}{11 x^7}\) |
\(\Big \downarrow \) 1917 |
\(\displaystyle \frac {6}{11} b \left (\frac {2}{7} b \left (-\frac {b \sqrt {x} \sqrt {a+b x^2} \int \frac {1}{\sqrt {x} \sqrt {b x^2+a}}dx}{3 a \sqrt {a x+b x^3}}-\frac {2 \sqrt {a x+b x^3}}{3 a x^2}\right )-\frac {2 \sqrt {a x+b x^3}}{7 x^4}\right )-\frac {2 \left (a x+b x^3\right )^{3/2}}{11 x^7}\) |
\(\Big \downarrow \) 266 |
\(\displaystyle \frac {6}{11} b \left (\frac {2}{7} b \left (-\frac {2 b \sqrt {x} \sqrt {a+b x^2} \int \frac {1}{\sqrt {b x^2+a}}d\sqrt {x}}{3 a \sqrt {a x+b x^3}}-\frac {2 \sqrt {a x+b x^3}}{3 a x^2}\right )-\frac {2 \sqrt {a x+b x^3}}{7 x^4}\right )-\frac {2 \left (a x+b x^3\right )^{3/2}}{11 x^7}\) |
\(\Big \downarrow \) 761 |
\(\displaystyle \frac {6}{11} b \left (\frac {2}{7} b \left (-\frac {b^{3/4} \sqrt {x} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{3 a^{5/4} \sqrt {a x+b x^3}}-\frac {2 \sqrt {a x+b x^3}}{3 a x^2}\right )-\frac {2 \sqrt {a x+b x^3}}{7 x^4}\right )-\frac {2 \left (a x+b x^3\right )^{3/2}}{11 x^7}\) |
(-2*(a*x + b*x^3)^(3/2))/(11*x^7) + (6*b*((-2*Sqrt[a*x + b*x^3])/(7*x^4) + (2*b*((-2*Sqrt[a*x + b*x^3])/(3*a*x^2) - (b^(3/4)*Sqrt[x]*(Sqrt[a] + Sqrt [b]*x)*Sqrt[(a + b*x^2)/(Sqrt[a] + Sqrt[b]*x)^2]*EllipticF[2*ArcTan[(b^(1/ 4)*Sqrt[x])/a^(1/4)], 1/2])/(3*a^(5/4)*Sqrt[a*x + b*x^3])))/7))/11
3.1.56.3.1 Defintions of rubi rules used
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De nominator[m]}, Simp[k/c Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) ^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I ntBinomialQ[a, b, c, 2, m, p, x]
Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[( 1 + q^2*x^2)*(Sqrt[(a + b*x^4)/(a*(1 + q^2*x^2)^2)]/(2*q*Sqrt[a + b*x^4]))* EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]
Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(a*x^j + b*x^n)^FracPart[p]/(x^(j*FracPart[p])*(a + b*x^(n - j))^FracPart[p]) Int[ x^(j*p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, j, n, p}, x] && !Integ erQ[p] && NeQ[n, j] && PosQ[n - j]
Int[((c_.)*(x_))^(m_)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a*x^j + b*x^n)^p/(c*(m + j*p + 1))), x] - Simp[b*p *((n - j)/(c^n*(m + j*p + 1))) Int[(c*x)^(m + n)*(a*x^j + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && !IntegerQ[p] && LtQ[0, j, n] && (Integer sQ[j, n] || GtQ[c, 0]) && GtQ[p, 0] && LtQ[m + j*p + 1, 0]
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] - Simp[b*((m + n*p + n - j + 1)/(a*c^(n - j)*(m + j*p + 1))) I nt[(c*x)^(m + n - j)*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, p}, x] && !IntegerQ[p] && LtQ[0, j, n] && (IntegersQ[j, n] || GtQ[c, 0]) && LtQ[ m + j*p + 1, 0]
Time = 2.28 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.98
method | result | size |
risch | \(-\frac {2 \left (b \,x^{2}+a \right ) \left (4 b^{2} x^{4}+13 a b \,x^{2}+7 a^{2}\right )}{77 x^{5} \sqrt {x \left (b \,x^{2}+a \right )}\, a}-\frac {4 b^{2} \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, F\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{77 a \sqrt {b \,x^{3}+a x}}\) | \(160\) |
default | \(-\frac {2 a \sqrt {b \,x^{3}+a x}}{11 x^{6}}-\frac {26 b \sqrt {b \,x^{3}+a x}}{77 x^{4}}-\frac {8 b^{2} \sqrt {b \,x^{3}+a x}}{77 a \,x^{2}}-\frac {4 b^{2} \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, F\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{77 a \sqrt {b \,x^{3}+a x}}\) | \(169\) |
elliptic | \(-\frac {2 a \sqrt {b \,x^{3}+a x}}{11 x^{6}}-\frac {26 b \sqrt {b \,x^{3}+a x}}{77 x^{4}}-\frac {8 b^{2} \sqrt {b \,x^{3}+a x}}{77 a \,x^{2}}-\frac {4 b^{2} \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, F\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{77 a \sqrt {b \,x^{3}+a x}}\) | \(169\) |
-2/77*(b*x^2+a)*(4*b^2*x^4+13*a*b*x^2+7*a^2)/x^5/(x*(b*x^2+a))^(1/2)/a-4/7 7*b^2/a*(-a*b)^(1/2)*((x+(-a*b)^(1/2)/b)/(-a*b)^(1/2)*b)^(1/2)*(-2*(x-(-a* b)^(1/2)/b)/(-a*b)^(1/2)*b)^(1/2)*(-x/(-a*b)^(1/2)*b)^(1/2)/(b*x^3+a*x)^(1 /2)*EllipticF(((x+(-a*b)^(1/2)/b)/(-a*b)^(1/2)*b)^(1/2),1/2*2^(1/2))
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.13 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.36 \[ \int \frac {\left (a x+b x^3\right )^{3/2}}{x^8} \, dx=-\frac {2 \, {\left (4 \, b^{\frac {5}{2}} x^{6} {\rm weierstrassPInverse}\left (-\frac {4 \, a}{b}, 0, x\right ) + {\left (4 \, b^{2} x^{4} + 13 \, a b x^{2} + 7 \, a^{2}\right )} \sqrt {b x^{3} + a x}\right )}}{77 \, a x^{6}} \]
-2/77*(4*b^(5/2)*x^6*weierstrassPInverse(-4*a/b, 0, x) + (4*b^2*x^4 + 13*a *b*x^2 + 7*a^2)*sqrt(b*x^3 + a*x))/(a*x^6)
\[ \int \frac {\left (a x+b x^3\right )^{3/2}}{x^8} \, dx=\int \frac {\left (x \left (a + b x^{2}\right )\right )^{\frac {3}{2}}}{x^{8}}\, dx \]
\[ \int \frac {\left (a x+b x^3\right )^{3/2}}{x^8} \, dx=\int { \frac {{\left (b x^{3} + a x\right )}^{\frac {3}{2}}}{x^{8}} \,d x } \]
\[ \int \frac {\left (a x+b x^3\right )^{3/2}}{x^8} \, dx=\int { \frac {{\left (b x^{3} + a x\right )}^{\frac {3}{2}}}{x^{8}} \,d x } \]
Timed out. \[ \int \frac {\left (a x+b x^3\right )^{3/2}}{x^8} \, dx=\int \frac {{\left (b\,x^3+a\,x\right )}^{3/2}}{x^8} \,d x \]