Integrand size = 28, antiderivative size = 293 \[ \int \frac {x^{3/2} \left (A+B x^2\right )}{\sqrt {b x^2+c x^4}} \, dx=-\frac {2 (3 b B-5 A c) x^{3/2} \left (b+c x^2\right )}{5 c^{3/2} \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {b x^2+c x^4}}+\frac {2 B \sqrt {x} \sqrt {b x^2+c x^4}}{5 c}+\frac {2 \sqrt [4]{b} (3 b B-5 A c) x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{5 c^{7/4} \sqrt {b x^2+c x^4}}-\frac {\sqrt [4]{b} (3 b B-5 A c) x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{5 c^{7/4} \sqrt {b x^2+c x^4}} \]
-2/5*(-5*A*c+3*B*b)*x^(3/2)*(c*x^2+b)/c^(3/2)/(b^(1/2)+x*c^(1/2))/(c*x^4+b *x^2)^(1/2)+2/5*B*x^(1/2)*(c*x^4+b*x^2)^(1/2)/c+2/5*b^(1/4)*(-5*A*c+3*B*b) *x*(cos(2*arctan(c^(1/4)*x^(1/2)/b^(1/4)))^2)^(1/2)/cos(2*arctan(c^(1/4)*x ^(1/2)/b^(1/4)))*EllipticE(sin(2*arctan(c^(1/4)*x^(1/2)/b^(1/4))),1/2*2^(1 /2))*(b^(1/2)+x*c^(1/2))*((c*x^2+b)/(b^(1/2)+x*c^(1/2))^2)^(1/2)/c^(7/4)/( c*x^4+b*x^2)^(1/2)-1/5*b^(1/4)*(-5*A*c+3*B*b)*x*(cos(2*arctan(c^(1/4)*x^(1 /2)/b^(1/4)))^2)^(1/2)/cos(2*arctan(c^(1/4)*x^(1/2)/b^(1/4)))*EllipticF(si n(2*arctan(c^(1/4)*x^(1/2)/b^(1/4))),1/2*2^(1/2))*(b^(1/2)+x*c^(1/2))*((c* x^2+b)/(b^(1/2)+x*c^(1/2))^2)^(1/2)/c^(7/4)/(c*x^4+b*x^2)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.11 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.28 \[ \int \frac {x^{3/2} \left (A+B x^2\right )}{\sqrt {b x^2+c x^4}} \, dx=\frac {2 x^{5/2} \left (3 B \left (b+c x^2\right )+(-3 b B+5 A c) \sqrt {1+\frac {c x^2}{b}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-\frac {c x^2}{b}\right )\right )}{15 c \sqrt {x^2 \left (b+c x^2\right )}} \]
(2*x^(5/2)*(3*B*(b + c*x^2) + (-3*b*B + 5*A*c)*Sqrt[1 + (c*x^2)/b]*Hyperge ometric2F1[1/2, 3/4, 7/4, -((c*x^2)/b)]))/(15*c*Sqrt[x^2*(b + c*x^2)])
Time = 0.40 (sec) , antiderivative size = 287, normalized size of antiderivative = 0.98, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1945, 1431, 266, 834, 27, 761, 1510}
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.
| \(\displaystyle \int \frac {x^{3/2} \left (A+B x^2\right )}{\sqrt {b x^2+c x^4}} \, dx\) |
| \(\Big \downarrow \) 1945 |
| \(\displaystyle \frac {2 B \sqrt {x} \sqrt {b x^2+c x^4}}{5 c}-\frac {(3 b B-5 A c) \int \frac {x^{3/2}}{\sqrt {c x^4+b x^2}}dx}{5 c}\) |
| \(\Big \downarrow \) 1431 |
| \(\displaystyle \frac {2 B \sqrt {x} \sqrt {b x^2+c x^4}}{5 c}-\frac {x \sqrt {b+c x^2} (3 b B-5 A c) \int \frac {\sqrt {x}}{\sqrt {c x^2+b}}dx}{5 c \sqrt {b x^2+c x^4}}\) |
| \(\Big \downarrow \) 266 |
| \(\displaystyle \frac {2 B \sqrt {x} \sqrt {b x^2+c x^4}}{5 c}-\frac {2 x \sqrt {b+c x^2} (3 b B-5 A c) \int \frac {x}{\sqrt {c x^2+b}}d\sqrt {x}}{5 c \sqrt {b x^2+c x^4}}\) |
| \(\Big \downarrow \) 834 |
| \(\displaystyle \frac {2 B \sqrt {x} \sqrt {b x^2+c x^4}}{5 c}-\frac {2 x \sqrt {b+c x^2} (3 b B-5 A c) \left (\frac {\sqrt {b} \int \frac {1}{\sqrt {c x^2+b}}d\sqrt {x}}{\sqrt {c}}-\frac {\sqrt {b} \int \frac {\sqrt {b}-\sqrt {c} x}{\sqrt {b} \sqrt {c x^2+b}}d\sqrt {x}}{\sqrt {c}}\right )}{5 c \sqrt {b x^2+c x^4}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2 B \sqrt {x} \sqrt {b x^2+c x^4}}{5 c}-\frac {2 x \sqrt {b+c x^2} (3 b B-5 A c) \left (\frac {\sqrt {b} \int \frac {1}{\sqrt {c x^2+b}}d\sqrt {x}}{\sqrt {c}}-\frac {\int \frac {\sqrt {b}-\sqrt {c} x}{\sqrt {c x^2+b}}d\sqrt {x}}{\sqrt {c}}\right )}{5 c \sqrt {b x^2+c x^4}}\) |
| \(\Big \downarrow \) 761 |
| \(\displaystyle \frac {2 B \sqrt {x} \sqrt {b x^2+c x^4}}{5 c}-\frac {2 x \sqrt {b+c x^2} (3 b B-5 A c) \left (\frac {\sqrt [4]{b} \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{2 c^{3/4} \sqrt {b+c x^2}}-\frac {\int \frac {\sqrt {b}-\sqrt {c} x}{\sqrt {c x^2+b}}d\sqrt {x}}{\sqrt {c}}\right )}{5 c \sqrt {b x^2+c x^4}}\) |
| \(\Big \downarrow \) 1510 |
| \(\displaystyle \frac {2 B \sqrt {x} \sqrt {b x^2+c x^4}}{5 c}-\frac {2 x \sqrt {b+c x^2} (3 b B-5 A c) \left (\frac {\sqrt [4]{b} \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{2 c^{3/4} \sqrt {b+c x^2}}-\frac {\frac {\sqrt [4]{b} \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{\sqrt [4]{c} \sqrt {b+c x^2}}-\frac {\sqrt {x} \sqrt {b+c x^2}}{\sqrt {b}+\sqrt {c} x}}{\sqrt {c}}\right )}{5 c \sqrt {b x^2+c x^4}}\) |
(2*B*Sqrt[x]*Sqrt[b*x^2 + c*x^4])/(5*c) - (2*(3*b*B - 5*A*c)*x*Sqrt[b + c* x^2]*(-((-((Sqrt[x]*Sqrt[b + c*x^2])/(Sqrt[b] + Sqrt[c]*x)) + (b^(1/4)*(Sq rt[b] + Sqrt[c]*x)*Sqrt[(b + c*x^2)/(Sqrt[b] + Sqrt[c]*x)^2]*EllipticE[2*A rcTan[(c^(1/4)*Sqrt[x])/b^(1/4)], 1/2])/(c^(1/4)*Sqrt[b + c*x^2]))/Sqrt[c] ) + (b^(1/4)*(Sqrt[b] + Sqrt[c]*x)*Sqrt[(b + c*x^2)/(Sqrt[b] + Sqrt[c]*x)^ 2]*EllipticF[2*ArcTan[(c^(1/4)*Sqrt[x])/b^(1/4)], 1/2])/(2*c^(3/4)*Sqrt[b + c*x^2])))/(5*c*Sqrt[b*x^2 + c*x^4])
3.3.49.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 2]}, S imp[1/q Int[1/Sqrt[a + b*x^4], x], x] - Simp[1/q Int[(1 - q*x^2)/Sqrt[a + b*x^4], x], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]
Int[((d_.)*(x_))^(m_)*((b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp [(b*x^2 + c*x^4)^p/((d*x)^(2*p)*(b + c*x^2)^p) Int[(d*x)^(m + 2*p)*(b + c *x^2)^p, x], x] /; FreeQ[{b, c, d, m, p}, x] && !IntegerQ[p]
Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, Simp[(-d)*x*(Sqrt[a + c*x^4]/(a*(1 + q^2*x^2))), x] + Simp[d* (1 + q^2*x^2)*(Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]/(q*Sqrt[a + c*x^4]))*E llipticE[2*ArcTan[q*x], 1/2], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e }, x] && PosQ[c/a]
Int[((e_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(jn_.))^(p_)*((c_) + (d_.)*(x_)^(n_.)), x_Symbol] :> Simp[d*e^(j - 1)*(e*x)^(m - j + 1)*((a*x^j + b*x^(j + n))^(p + 1)/(b*(m + n + p*(j + n) + 1))), x] - Simp[(a*d*(m + j* p + 1) - b*c*(m + n + p*(j + n) + 1))/(b*(m + n + p*(j + n) + 1)) Int[(e* x)^m*(a*x^j + b*x^(j + n))^p, x], x] /; FreeQ[{a, b, c, d, e, j, m, n, p}, x] && EqQ[jn, j + n] && !IntegerQ[p] && NeQ[b*c - a*d, 0] && NeQ[m + n + p *(j + n) + 1, 0] && (GtQ[e, 0] || IntegerQ[j])
Time = 2.02 (sec) , antiderivative size = 225, normalized size of antiderivative = 0.77
| method | result | size |
| risch | \(\frac {2 B \,x^{\frac {5}{2}} \left (c \,x^{2}+b \right )}{5 c \sqrt {x^{2} \left (c \,x^{2}+b \right )}}+\frac {\left (5 A c -3 B b \right ) \sqrt {-b c}\, \sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {x c}{\sqrt {-b c}}}\, \left (-\frac {2 \sqrt {-b c}\, E\left (\sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )}{c}+\frac {\sqrt {-b c}\, F\left (\sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )}{c}\right ) \sqrt {x}\, \sqrt {x \left (c \,x^{2}+b \right )}}{5 c^{2} \sqrt {c \,x^{3}+b x}\, \sqrt {x^{2} \left (c \,x^{2}+b \right )}}\) | \(225\) |
| default | \(\frac {\sqrt {x}\, \left (10 A b c \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {x c}{\sqrt {-b c}}}\, E\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )-5 A b c \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {x c}{\sqrt {-b c}}}\, F\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )-6 B \,b^{2} \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {x c}{\sqrt {-b c}}}\, E\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )+3 B \,b^{2} \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {x c}{\sqrt {-b c}}}\, F\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )+2 B \,c^{2} x^{4}+2 B b c \,x^{2}\right )}{5 \sqrt {x^{4} c +b \,x^{2}}\, c^{2}}\) | \(378\) |
2/5*B/c*x^(5/2)*(c*x^2+b)/(x^2*(c*x^2+b))^(1/2)+1/5*(5*A*c-3*B*b)/c^2*(-b* c)^(1/2)*((x+1/c*(-b*c)^(1/2))*c/(-b*c)^(1/2))^(1/2)*(-2*(x-1/c*(-b*c)^(1/ 2))*c/(-b*c)^(1/2))^(1/2)*(-x*c/(-b*c)^(1/2))^(1/2)/(c*x^3+b*x)^(1/2)*(-2/ c*(-b*c)^(1/2)*EllipticE(((x+1/c*(-b*c)^(1/2))*c/(-b*c)^(1/2))^(1/2),1/2*2 ^(1/2))+1/c*(-b*c)^(1/2)*EllipticF(((x+1/c*(-b*c)^(1/2))*c/(-b*c)^(1/2))^( 1/2),1/2*2^(1/2)))*x^(1/2)/(x^2*(c*x^2+b))^(1/2)*(x*(c*x^2+b))^(1/2)
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.08 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.19 \[ \int \frac {x^{3/2} \left (A+B x^2\right )}{\sqrt {b x^2+c x^4}} \, dx=\frac {2 \, {\left (\sqrt {c x^{4} + b x^{2}} B c \sqrt {x} + {\left (3 \, B b - 5 \, A c\right )} \sqrt {c} {\rm weierstrassZeta}\left (-\frac {4 \, b}{c}, 0, {\rm weierstrassPInverse}\left (-\frac {4 \, b}{c}, 0, x\right )\right )\right )}}{5 \, c^{2}} \]
2/5*(sqrt(c*x^4 + b*x^2)*B*c*sqrt(x) + (3*B*b - 5*A*c)*sqrt(c)*weierstrass Zeta(-4*b/c, 0, weierstrassPInverse(-4*b/c, 0, x)))/c^2
\[ \int \frac {x^{3/2} \left (A+B x^2\right )}{\sqrt {b x^2+c x^4}} \, dx=\int \frac {x^{\frac {3}{2}} \left (A + B x^{2}\right )}{\sqrt {x^{2} \left (b + c x^{2}\right )}}\, dx \]
\[ \int \frac {x^{3/2} \left (A+B x^2\right )}{\sqrt {b x^2+c x^4}} \, dx=\int { \frac {{\left (B x^{2} + A\right )} x^{\frac {3}{2}}}{\sqrt {c x^{4} + b x^{2}}} \,d x } \]
\[ \int \frac {x^{3/2} \left (A+B x^2\right )}{\sqrt {b x^2+c x^4}} \, dx=\int { \frac {{\left (B x^{2} + A\right )} x^{\frac {3}{2}}}{\sqrt {c x^{4} + b x^{2}}} \,d x } \]
Timed out. \[ \int \frac {x^{3/2} \left (A+B x^2\right )}{\sqrt {b x^2+c x^4}} \, dx=\int \frac {x^{3/2}\,\left (B\,x^2+A\right )}{\sqrt {c\,x^4+b\,x^2}} \,d x \]