Integrand size = 17, antiderivative size = 161 \[ \int \frac {x^7}{\left (a x+b x^3\right )^{3/2}} \, dx=-\frac {x^5}{b \sqrt {a x+b x^3}}-\frac {15 a \sqrt {a x+b x^3}}{7 b^3}+\frac {9 x^2 \sqrt {a x+b x^3}}{7 b^2}+\frac {15 a^{7/4} \sqrt {x} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{14 b^{13/4} \sqrt {a x+b x^3}} \]
-x^5/b/(b*x^3+a*x)^(1/2)-15/7*a*(b*x^3+a*x)^(1/2)/b^3+9/7*x^2*(b*x^3+a*x)^ (1/2)/b^2+15/14*a^(7/4)*(cos(2*arctan(b^(1/4)*x^(1/2)/a^(1/4)))^2)^(1/2)/c os(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)/b^(13/4)/(b*x^3+a*x)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.05 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.50 \[ \int \frac {x^7}{\left (a x+b x^3\right )^{3/2}} \, dx=\frac {x \left (-15 a^2-6 a b x^2+2 b^2 x^4+15 a^2 \sqrt {1+\frac {b x^2}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-\frac {b x^2}{a}\right )\right )}{7 b^3 \sqrt {x \left (a+b x^2\right )}} \]
(x*(-15*a^2 - 6*a*b*x^2 + 2*b^2*x^4 + 15*a^2*Sqrt[1 + (b*x^2)/a]*Hypergeom etric2F1[1/4, 1/2, 5/4, -((b*x^2)/a)]))/(7*b^3*Sqrt[x*(a + b*x^2)])
Time = 0.34 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.10, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {1928, 1930, 1930, 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.
| \(\displaystyle \int \frac {x^7}{\left (a x+b x^3\right )^{3/2}} \, dx\) |
| \(\Big \downarrow \) 1928 |
| \(\displaystyle \frac {9 \int \frac {x^4}{\sqrt {b x^3+a x}}dx}{2 b}-\frac {x^5}{b \sqrt {a x+b x^3}}\) |
| \(\Big \downarrow \) 1930 |
| \(\displaystyle \frac {9 \left (\frac {2 x^2 \sqrt {a x+b x^3}}{7 b}-\frac {5 a \int \frac {x^2}{\sqrt {b x^3+a x}}dx}{7 b}\right )}{2 b}-\frac {x^5}{b \sqrt {a x+b x^3}}\) |
| \(\Big \downarrow \) 1930 |
| \(\displaystyle \frac {9 \left (\frac {2 x^2 \sqrt {a x+b x^3}}{7 b}-\frac {5 a \left (\frac {2 \sqrt {a x+b x^3}}{3 b}-\frac {a \int \frac {1}{\sqrt {b x^3+a x}}dx}{3 b}\right )}{7 b}\right )}{2 b}-\frac {x^5}{b \sqrt {a x+b x^3}}\) |
| \(\Big \downarrow \) 1917 |
| \(\displaystyle \frac {9 \left (\frac {2 x^2 \sqrt {a x+b x^3}}{7 b}-\frac {5 a \left (\frac {2 \sqrt {a x+b x^3}}{3 b}-\frac {a \sqrt {x} \sqrt {a+b x^2} \int \frac {1}{\sqrt {x} \sqrt {b x^2+a}}dx}{3 b \sqrt {a x+b x^3}}\right )}{7 b}\right )}{2 b}-\frac {x^5}{b \sqrt {a x+b x^3}}\) |
| \(\Big \downarrow \) 266 |
| \(\displaystyle \frac {9 \left (\frac {2 x^2 \sqrt {a x+b x^3}}{7 b}-\frac {5 a \left (\frac {2 \sqrt {a x+b x^3}}{3 b}-\frac {2 a \sqrt {x} \sqrt {a+b x^2} \int \frac {1}{\sqrt {b x^2+a}}d\sqrt {x}}{3 b \sqrt {a x+b x^3}}\right )}{7 b}\right )}{2 b}-\frac {x^5}{b \sqrt {a x+b x^3}}\) |
| \(\Big \downarrow \) 761 |
| \(\displaystyle \frac {9 \left (\frac {2 x^2 \sqrt {a x+b x^3}}{7 b}-\frac {5 a \left (\frac {2 \sqrt {a x+b x^3}}{3 b}-\frac {a^{3/4} \sqrt {x} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{3 b^{5/4} \sqrt {a x+b x^3}}\right )}{7 b}\right )}{2 b}-\frac {x^5}{b \sqrt {a x+b x^3}}\) |
-(x^5/(b*Sqrt[a*x + b*x^3])) + (9*((2*x^2*Sqrt[a*x + b*x^3])/(7*b) - (5*a* ((2*Sqrt[a*x + b*x^3])/(3*b) - (a^(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*b^(5/4)*Sqrt[a*x + b*x^3])))/(7*b)))/(2*b)
3.1.65.3.1 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^(n - 1)*(c*x)^(m - n + 1)*((a*x^j + b*x^n)^(p + 1)/(b*(n - j)*( p + 1))), x] - Simp[c^n*((m + j*p - n + j + 1)/(b*(n - j)*(p + 1))) Int[( c*x)^(m - n)*(a*x^j + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c}, x] && !In tegerQ[p] && LtQ[0, j, n] && (IntegersQ[j, n] || GtQ[c, 0]) && LtQ[p, -1] & & GtQ[m + j*p + 1, n - j]
Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol ] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a*x^j + b*x^n)^(p + 1)/(b*(m + n*p + 1))), x] - Simp[a*c^(n - j)*((m + j*p - n + j + 1)/(b*(m + n*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]) && Gt Q[m + j*p - n + j + 1, 0] && NeQ[m + n*p + 1, 0]
Time = 2.71 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.07
| method | result | size |
| default | \(-\frac {x \,a^{2}}{b^{3} \sqrt {\left (x^{2}+\frac {a}{b}\right ) b x}}+\frac {2 x^{2} \sqrt {b \,x^{3}+a x}}{7 b^{2}}-\frac {8 a \sqrt {b \,x^{3}+a x}}{7 b^{3}}+\frac {15 a^{2} \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, F\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{14 b^{4} \sqrt {b \,x^{3}+a x}}\) | \(172\) |
| elliptic | \(-\frac {x \,a^{2}}{b^{3} \sqrt {\left (x^{2}+\frac {a}{b}\right ) b x}}+\frac {2 x^{2} \sqrt {b \,x^{3}+a x}}{7 b^{2}}-\frac {8 a \sqrt {b \,x^{3}+a x}}{7 b^{3}}+\frac {15 a^{2} \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, F\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{14 b^{4} \sqrt {b \,x^{3}+a x}}\) | \(172\) |
| risch | \(-\frac {2 \left (-b \,x^{2}+4 a \right ) \left (b \,x^{2}+a \right ) x}{7 b^{3} \sqrt {x \left (b \,x^{2}+a \right )}}+\frac {a^{2} \left (\frac {11 \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, F\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b \sqrt {b \,x^{3}+a x}}-7 a \left (\frac {x}{a \sqrt {\left (x^{2}+\frac {a}{b}\right ) b x}}+\frac {\sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, F\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{2 a b \sqrt {b \,x^{3}+a x}}\right )\right )}{7 b^{3}}\) | \(287\) |
-1/b^3*x*a^2/((x^2+a/b)*b*x)^(1/2)+2/7*x^2*(b*x^3+a*x)^(1/2)/b^2-8/7*a*(b* x^3+a*x)^(1/2)/b^3+15/14*a^2/b^4*(-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 = 0.09 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.50 \[ \int \frac {x^7}{\left (a x+b x^3\right )^{3/2}} \, dx=\frac {15 \, {\left (a^{2} b x^{2} + a^{3}\right )} \sqrt {b} {\rm weierstrassPInverse}\left (-\frac {4 \, a}{b}, 0, x\right ) + {\left (2 \, b^{3} x^{4} - 6 \, a b^{2} x^{2} - 15 \, a^{2} b\right )} \sqrt {b x^{3} + a x}}{7 \, {\left (b^{5} x^{2} + a b^{4}\right )}} \]
1/7*(15*(a^2*b*x^2 + a^3)*sqrt(b)*weierstrassPInverse(-4*a/b, 0, x) + (2*b ^3*x^4 - 6*a*b^2*x^2 - 15*a^2*b)*sqrt(b*x^3 + a*x))/(b^5*x^2 + a*b^4)
\[ \int \frac {x^7}{\left (a x+b x^3\right )^{3/2}} \, dx=\int \frac {x^{7}}{\left (x \left (a + b x^{2}\right )\right )^{\frac {3}{2}}}\, dx \]
\[ \int \frac {x^7}{\left (a x+b x^3\right )^{3/2}} \, dx=\int { \frac {x^{7}}{{\left (b x^{3} + a x\right )}^{\frac {3}{2}}} \,d x } \]
\[ \int \frac {x^7}{\left (a x+b x^3\right )^{3/2}} \, dx=\int { \frac {x^{7}}{{\left (b x^{3} + a x\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {x^7}{\left (a x+b x^3\right )^{3/2}} \, dx=\int \frac {x^7}{{\left (b\,x^3+a\,x\right )}^{3/2}} \,d x \]