Integrand size = 26, antiderivative size = 342 \[ \int \frac {A+B x^2}{(e x)^{7/2} \sqrt {a+b x^2}} \, dx=-\frac {2 A \sqrt {a+b x^2}}{5 a e (e x)^{5/2}}+\frac {2 (3 A b-5 a B) \sqrt {a+b x^2}}{5 a^2 e^3 \sqrt {e x}}-\frac {2 \sqrt {b} (3 A b-5 a B) \sqrt {e x} \sqrt {a+b x^2}}{5 a^2 e^4 \left (\sqrt {a}+\sqrt {b} x\right )}+\frac {2 \sqrt [4]{b} (3 A b-5 a B) \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt [4]{a} \sqrt {e}}\right )|\frac {1}{2}\right )}{5 a^{7/4} e^{7/2} \sqrt {a+b x^2}}-\frac {\sqrt [4]{b} (3 A b-5 a B) \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt [4]{a} \sqrt {e}}\right ),\frac {1}{2}\right )}{5 a^{7/4} e^{7/2} \sqrt {a+b x^2}} \]
-2/5*A*(b*x^2+a)^(1/2)/a/e/(e*x)^(5/2)+2/5*(3*A*b-5*B*a)*(b*x^2+a)^(1/2)/a ^2/e^3/(e*x)^(1/2)-2/5*(3*A*b-5*B*a)*b^(1/2)*(e*x)^(1/2)*(b*x^2+a)^(1/2)/a ^2/e^4/(a^(1/2)+x*b^(1/2))+2/5*b^(1/4)*(3*A*b-5*B*a)*(cos(2*arctan(b^(1/4) *(e*x)^(1/2)/a^(1/4)/e^(1/2)))^2)^(1/2)/cos(2*arctan(b^(1/4)*(e*x)^(1/2)/a ^(1/4)/e^(1/2)))*EllipticE(sin(2*arctan(b^(1/4)*(e*x)^(1/2)/a^(1/4)/e^(1/2 ))),1/2*2^(1/2))*(a^(1/2)+x*b^(1/2))*((b*x^2+a)/(a^(1/2)+x*b^(1/2))^2)^(1/ 2)/a^(7/4)/e^(7/2)/(b*x^2+a)^(1/2)-1/5*b^(1/4)*(3*A*b-5*B*a)*(cos(2*arctan (b^(1/4)*(e*x)^(1/2)/a^(1/4)/e^(1/2)))^2)^(1/2)/cos(2*arctan(b^(1/4)*(e*x) ^(1/2)/a^(1/4)/e^(1/2)))*EllipticF(sin(2*arctan(b^(1/4)*(e*x)^(1/2)/a^(1/4 )/e^(1/2))),1/2*2^(1/2))*(a^(1/2)+x*b^(1/2))*((b*x^2+a)/(a^(1/2)+x*b^(1/2) )^2)^(1/2)/a^(7/4)/e^(7/2)/(b*x^2+a)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.03 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.24 \[ \int \frac {A+B x^2}{(e x)^{7/2} \sqrt {a+b x^2}} \, dx=-\frac {2 x \left (A \left (a+b x^2\right )+(-3 A b+5 a B) x^2 \sqrt {1+\frac {b x^2}{a}} \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{2},\frac {3}{4},-\frac {b x^2}{a}\right )\right )}{5 a (e x)^{7/2} \sqrt {a+b x^2}} \]
(-2*x*(A*(a + b*x^2) + (-3*A*b + 5*a*B)*x^2*Sqrt[1 + (b*x^2)/a]*Hypergeome tric2F1[-1/4, 1/2, 3/4, -((b*x^2)/a)]))/(5*a*(e*x)^(7/2)*Sqrt[a + b*x^2])
Time = 0.42 (sec) , antiderivative size = 358, normalized size of antiderivative = 1.05, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {359, 264, 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.
| \(\displaystyle \int \frac {A+B x^2}{(e x)^{7/2} \sqrt {a+b x^2}} \, dx\) |
| \(\Big \downarrow \) 359 |
| \(\displaystyle -\frac {(3 A b-5 a B) \int \frac {1}{(e x)^{3/2} \sqrt {b x^2+a}}dx}{5 a e^2}-\frac {2 A \sqrt {a+b x^2}}{5 a e (e x)^{5/2}}\) |
| \(\Big \downarrow \) 264 |
| \(\displaystyle -\frac {(3 A b-5 a B) \left (\frac {b \int \frac {\sqrt {e x}}{\sqrt {b x^2+a}}dx}{a e^2}-\frac {2 \sqrt {a+b x^2}}{a e \sqrt {e x}}\right )}{5 a e^2}-\frac {2 A \sqrt {a+b x^2}}{5 a e (e x)^{5/2}}\) |
| \(\Big \downarrow \) 266 |
| \(\displaystyle -\frac {(3 A b-5 a B) \left (\frac {2 b \int \frac {e x}{\sqrt {b x^2+a}}d\sqrt {e x}}{a e^3}-\frac {2 \sqrt {a+b x^2}}{a e \sqrt {e x}}\right )}{5 a e^2}-\frac {2 A \sqrt {a+b x^2}}{5 a e (e x)^{5/2}}\) |
| \(\Big \downarrow \) 834 |
| \(\displaystyle -\frac {(3 A b-5 a B) \left (\frac {2 b \left (\frac {\sqrt {a} e \int \frac {1}{\sqrt {b x^2+a}}d\sqrt {e x}}{\sqrt {b}}-\frac {\sqrt {a} e \int \frac {\sqrt {a} e-\sqrt {b} e x}{\sqrt {a} e \sqrt {b x^2+a}}d\sqrt {e x}}{\sqrt {b}}\right )}{a e^3}-\frac {2 \sqrt {a+b x^2}}{a e \sqrt {e x}}\right )}{5 a e^2}-\frac {2 A \sqrt {a+b x^2}}{5 a e (e x)^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {(3 A b-5 a B) \left (\frac {2 b \left (\frac {\sqrt {a} e \int \frac {1}{\sqrt {b x^2+a}}d\sqrt {e x}}{\sqrt {b}}-\frac {\int \frac {\sqrt {a} e-\sqrt {b} e x}{\sqrt {b x^2+a}}d\sqrt {e x}}{\sqrt {b}}\right )}{a e^3}-\frac {2 \sqrt {a+b x^2}}{a e \sqrt {e x}}\right )}{5 a e^2}-\frac {2 A \sqrt {a+b x^2}}{5 a e (e x)^{5/2}}\) |
| \(\Big \downarrow \) 761 |
| \(\displaystyle -\frac {(3 A b-5 a B) \left (\frac {2 b \left (\frac {\sqrt [4]{a} \sqrt {e} \left (\sqrt {a} e+\sqrt {b} e x\right ) \sqrt {\frac {a e^2+b e^2 x^2}{\left (\sqrt {a} e+\sqrt {b} e x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt [4]{a} \sqrt {e}}\right ),\frac {1}{2}\right )}{2 b^{3/4} \sqrt {a+b x^2}}-\frac {\int \frac {\sqrt {a} e-\sqrt {b} e x}{\sqrt {b x^2+a}}d\sqrt {e x}}{\sqrt {b}}\right )}{a e^3}-\frac {2 \sqrt {a+b x^2}}{a e \sqrt {e x}}\right )}{5 a e^2}-\frac {2 A \sqrt {a+b x^2}}{5 a e (e x)^{5/2}}\) |
| \(\Big \downarrow \) 1510 |
| \(\displaystyle -\frac {(3 A b-5 a B) \left (\frac {2 b \left (\frac {\sqrt [4]{a} \sqrt {e} \left (\sqrt {a} e+\sqrt {b} e x\right ) \sqrt {\frac {a e^2+b e^2 x^2}{\left (\sqrt {a} e+\sqrt {b} e x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt [4]{a} \sqrt {e}}\right ),\frac {1}{2}\right )}{2 b^{3/4} \sqrt {a+b x^2}}-\frac {\frac {\sqrt [4]{a} \sqrt {e} \left (\sqrt {a} e+\sqrt {b} e x\right ) \sqrt {\frac {a e^2+b e^2 x^2}{\left (\sqrt {a} e+\sqrt {b} e x\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt [4]{a} \sqrt {e}}\right )|\frac {1}{2}\right )}{\sqrt [4]{b} \sqrt {a+b x^2}}-\frac {e^2 \sqrt {e x} \sqrt {a+b x^2}}{\sqrt {a} e+\sqrt {b} e x}}{\sqrt {b}}\right )}{a e^3}-\frac {2 \sqrt {a+b x^2}}{a e \sqrt {e x}}\right )}{5 a e^2}-\frac {2 A \sqrt {a+b x^2}}{5 a e (e x)^{5/2}}\) |
(-2*A*Sqrt[a + b*x^2])/(5*a*e*(e*x)^(5/2)) - ((3*A*b - 5*a*B)*((-2*Sqrt[a + b*x^2])/(a*e*Sqrt[e*x]) + (2*b*(-((-((e^2*Sqrt[e*x]*Sqrt[a + b*x^2])/(Sq rt[a]*e + Sqrt[b]*e*x)) + (a^(1/4)*Sqrt[e]*(Sqrt[a]*e + Sqrt[b]*e*x)*Sqrt[ (a*e^2 + b*e^2*x^2)/(Sqrt[a]*e + Sqrt[b]*e*x)^2]*EllipticE[2*ArcTan[(b^(1/ 4)*Sqrt[e*x])/(a^(1/4)*Sqrt[e])], 1/2])/(b^(1/4)*Sqrt[a + b*x^2]))/Sqrt[b] ) + (a^(1/4)*Sqrt[e]*(Sqrt[a]*e + Sqrt[b]*e*x)*Sqrt[(a*e^2 + b*e^2*x^2)/(S qrt[a]*e + Sqrt[b]*e*x)^2]*EllipticF[2*ArcTan[(b^(1/4)*Sqrt[e*x])/(a^(1/4) *Sqrt[e])], 1/2])/(2*b^(3/4)*Sqrt[a + b*x^2])))/(a*e^3)))/(5*a*e^2)
3.9.6.3.1 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] :> Simp[(c*x)^( m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + 2*p + 3)/(a*c ^2*(m + 1))) Int[(c*x)^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, p }, x] && LtQ[m, -1] && IntBinomialQ[a, b, c, 2, m, p, 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[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x _Symbol] :> Simp[c*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*e*(m + 1))), x] + Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(a*e^2*(m + 1)) Int[(e*x)^(m + 2)* (a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] && !ILtQ[p, -1]
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_) + (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]
Time = 3.08 (sec) , antiderivative size = 243, normalized size of antiderivative = 0.71
| method | result | size |
| risch | \(-\frac {2 \sqrt {b \,x^{2}+a}\, \left (-3 A b \,x^{2}+5 B a \,x^{2}+A a \right )}{5 a^{2} x^{2} e^{3} \sqrt {e x}}-\frac {\left (3 A b -5 B a \right ) \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, \left (-\frac {2 \sqrt {-a b}\, E\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}+\frac {\sqrt {-a b}\, F\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}\right ) \sqrt {\left (b \,x^{2}+a \right ) e x}}{5 a^{2} \sqrt {b e \,x^{3}+a e x}\, e^{3} \sqrt {e x}\, \sqrt {b \,x^{2}+a}}\) | \(243\) |
| elliptic | \(\frac {\sqrt {\left (b \,x^{2}+a \right ) e x}\, \left (-\frac {2 A \sqrt {b e \,x^{3}+a e x}}{5 e^{4} a \,x^{3}}+\frac {2 \left (b e \,x^{2}+a e \right ) \left (3 A b -5 B a \right )}{5 e^{4} a^{2} \sqrt {x \left (b e \,x^{2}+a e \right )}}-\frac {\left (3 A b -5 B a \right ) \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, \left (-\frac {2 \sqrt {-a b}\, E\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}+\frac {\sqrt {-a b}\, F\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}\right )}{5 a^{2} e^{3} \sqrt {b e \,x^{3}+a e x}}\right )}{\sqrt {e x}\, \sqrt {b \,x^{2}+a}}\) | \(267\) |
| default | \(-\frac {6 A \sqrt {2}\, \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, E\left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right ) a b \,x^{2}-3 A \sqrt {2}\, \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, F\left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right ) a b \,x^{2}-10 B \sqrt {2}\, \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, E\left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right ) a^{2} x^{2}+5 B \sqrt {2}\, \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, F\left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right ) a^{2} x^{2}-6 A \,b^{2} x^{4}+10 B a b \,x^{4}-4 a A b \,x^{2}+10 a^{2} B \,x^{2}+2 a^{2} A}{5 x^{2} \sqrt {b \,x^{2}+a}\, e^{3} \sqrt {e x}\, a^{2}}\) | \(417\) |
-2/5*(b*x^2+a)^(1/2)*(-3*A*b*x^2+5*B*a*x^2+A*a)/a^2/x^2/e^3/(e*x)^(1/2)-1/ 5*(3*A*b-5*B*a)/a^2*(-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 *e*x^3+a*e*x)^(1/2)*(-2*(-a*b)^(1/2)/b*EllipticE(((x+(-a*b)^(1/2)/b)/(-a*b )^(1/2)*b)^(1/2),1/2*2^(1/2))+(-a*b)^(1/2)/b*EllipticF(((x+(-a*b)^(1/2)/b) /(-a*b)^(1/2)*b)^(1/2),1/2*2^(1/2)))/e^3*((b*x^2+a)*e*x)^(1/2)/(e*x)^(1/2) /(b*x^2+a)^(1/2)
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.09 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.23 \[ \int \frac {A+B x^2}{(e x)^{7/2} \sqrt {a+b x^2}} \, dx=-\frac {2 \, {\left ({\left (5 \, B a - 3 \, A b\right )} \sqrt {b e} x^{3} {\rm weierstrassZeta}\left (-\frac {4 \, a}{b}, 0, {\rm weierstrassPInverse}\left (-\frac {4 \, a}{b}, 0, x\right )\right ) + {\left ({\left (5 \, B a - 3 \, A b\right )} x^{2} + A a\right )} \sqrt {b x^{2} + a} \sqrt {e x}\right )}}{5 \, a^{2} e^{4} x^{3}} \]
-2/5*((5*B*a - 3*A*b)*sqrt(b*e)*x^3*weierstrassZeta(-4*a/b, 0, weierstrass PInverse(-4*a/b, 0, x)) + ((5*B*a - 3*A*b)*x^2 + A*a)*sqrt(b*x^2 + a)*sqrt (e*x))/(a^2*e^4*x^3)
Result contains complex when optimal does not.
Time = 15.98 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.30 \[ \int \frac {A+B x^2}{(e x)^{7/2} \sqrt {a+b x^2}} \, dx=\frac {A \Gamma \left (- \frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{4}, \frac {1}{2} \\ - \frac {1}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \sqrt {a} e^{\frac {7}{2}} x^{\frac {5}{2}} \Gamma \left (- \frac {1}{4}\right )} + \frac {B \Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {1}{2} \\ \frac {3}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \sqrt {a} e^{\frac {7}{2}} \sqrt {x} \Gamma \left (\frac {3}{4}\right )} \]
A*gamma(-5/4)*hyper((-5/4, 1/2), (-1/4,), b*x**2*exp_polar(I*pi)/a)/(2*sqr t(a)*e**(7/2)*x**(5/2)*gamma(-1/4)) + B*gamma(-1/4)*hyper((-1/4, 1/2), (3/ 4,), b*x**2*exp_polar(I*pi)/a)/(2*sqrt(a)*e**(7/2)*sqrt(x)*gamma(3/4))
\[ \int \frac {A+B x^2}{(e x)^{7/2} \sqrt {a+b x^2}} \, dx=\int { \frac {B x^{2} + A}{\sqrt {b x^{2} + a} \left (e x\right )^{\frac {7}{2}}} \,d x } \]
\[ \int \frac {A+B x^2}{(e x)^{7/2} \sqrt {a+b x^2}} \, dx=\int { \frac {B x^{2} + A}{\sqrt {b x^{2} + a} \left (e x\right )^{\frac {7}{2}}} \,d x } \]
Timed out. \[ \int \frac {A+B x^2}{(e x)^{7/2} \sqrt {a+b x^2}} \, dx=\int \frac {B\,x^2+A}{{\left (e\,x\right )}^{7/2}\,\sqrt {b\,x^2+a}} \,d x \]