Integrand size = 19, antiderivative size = 152 \[ \int \frac {\left (a+b x^2\right )^{3/2}}{(c x)^{5/2}} \, dx=\frac {4 b \sqrt {c x} \sqrt {a+b x^2}}{3 c^3}-\frac {2 \left (a+b x^2\right )^{3/2}}{3 c (c x)^{3/2}}+\frac {4 a^{3/4} b^{3/4} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a} \sqrt {c}}\right ),\frac {1}{2}\right )}{3 c^{5/2} \sqrt {a+b x^2}} \]
-2/3*(b*x^2+a)^(3/2)/c/(c*x)^(3/2)+4/3*b*(c*x)^(1/2)*(b*x^2+a)^(1/2)/c^3+4 /3*a^(3/4)*b^(3/4)*(cos(2*arctan(b^(1/4)*(c*x)^(1/2)/a^(1/4)/c^(1/2)))^2)^ (1/2)/cos(2*arctan(b^(1/4)*(c*x)^(1/2)/a^(1/4)/c^(1/2)))*EllipticF(sin(2*a rctan(b^(1/4)*(c*x)^(1/2)/a^(1/4)/c^(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)/c^(5/2)/(b*x^2+a)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.01 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.38 \[ \int \frac {\left (a+b x^2\right )^{3/2}}{(c x)^{5/2}} \, dx=-\frac {2 a x \sqrt {a+b x^2} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-\frac {3}{4},\frac {1}{4},-\frac {b x^2}{a}\right )}{3 (c x)^{5/2} \sqrt {1+\frac {b x^2}{a}}} \]
(-2*a*x*Sqrt[a + b*x^2]*Hypergeometric2F1[-3/2, -3/4, 1/4, -((b*x^2)/a)])/ (3*(c*x)^(5/2)*Sqrt[1 + (b*x^2)/a])
Time = 0.22 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.12, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {247, 248, 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.
\(\displaystyle \int \frac {\left (a+b x^2\right )^{3/2}}{(c x)^{5/2}} \, dx\) |
\(\Big \downarrow \) 247 |
\(\displaystyle \frac {2 b \int \frac {\sqrt {b x^2+a}}{\sqrt {c x}}dx}{c^2}-\frac {2 \left (a+b x^2\right )^{3/2}}{3 c (c x)^{3/2}}\) |
\(\Big \downarrow \) 248 |
\(\displaystyle \frac {2 b \left (\frac {2}{3} a \int \frac {1}{\sqrt {c x} \sqrt {b x^2+a}}dx+\frac {2 \sqrt {c x} \sqrt {a+b x^2}}{3 c}\right )}{c^2}-\frac {2 \left (a+b x^2\right )^{3/2}}{3 c (c x)^{3/2}}\) |
\(\Big \downarrow \) 266 |
\(\displaystyle \frac {2 b \left (\frac {4 a \int \frac {1}{\sqrt {b x^2+a}}d\sqrt {c x}}{3 c}+\frac {2 \sqrt {c x} \sqrt {a+b x^2}}{3 c}\right )}{c^2}-\frac {2 \left (a+b x^2\right )^{3/2}}{3 c (c x)^{3/2}}\) |
\(\Big \downarrow \) 761 |
\(\displaystyle \frac {2 b \left (\frac {2 a^{3/4} \left (\sqrt {a} c+\sqrt {b} c x\right ) \sqrt {\frac {a c^2+b c^2 x^2}{\left (\sqrt {a} c+\sqrt {b} c x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a} \sqrt {c}}\right ),\frac {1}{2}\right )}{3 \sqrt [4]{b} c^{3/2} \sqrt {a+b x^2}}+\frac {2 \sqrt {c x} \sqrt {a+b x^2}}{3 c}\right )}{c^2}-\frac {2 \left (a+b x^2\right )^{3/2}}{3 c (c x)^{3/2}}\) |
(-2*(a + b*x^2)^(3/2))/(3*c*(c*x)^(3/2)) + (2*b*((2*Sqrt[c*x]*Sqrt[a + b*x ^2])/(3*c) + (2*a^(3/4)*(Sqrt[a]*c + Sqrt[b]*c*x)*Sqrt[(a*c^2 + b*c^2*x^2) /(Sqrt[a]*c + Sqrt[b]*c*x)^2]*EllipticF[2*ArcTan[(b^(1/4)*Sqrt[c*x])/(a^(1 /4)*Sqrt[c])], 1/2])/(3*b^(1/4)*c^(3/2)*Sqrt[a + b*x^2])))/c^2
3.7.3.3.1 Defintions of rubi rules used
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^ (m + 1)*((a + b*x^2)^p/(c*(m + 1))), x] - Simp[2*b*(p/(c^2*(m + 1))) Int[ (c*x)^(m + 2)*(a + b*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && GtQ[p, 0] && LtQ[m, -1] && !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomialQ[a, b, c, 2, m, p, x]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^ (m + 1)*((a + b*x^2)^p/(c*(m + 2*p + 1))), x] + Simp[2*a*(p/(m + 2*p + 1)) Int[(c*x)^m*(a + b*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, m}, x] && GtQ[ p, 0] && NeQ[m + 2*p + 1, 0] && 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[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]
Time = 1.99 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.82
method | result | size |
default | \(\frac {\frac {4 \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, F\left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {-a b}\, a x}{3}+\frac {2 b^{2} x^{4}}{3}-\frac {2 a^{2}}{3}}{\sqrt {b \,x^{2}+a}\, x \,c^{2} \sqrt {c x}}\) | \(125\) |
risch | \(-\frac {2 \sqrt {b \,x^{2}+a}\, \left (-b \,x^{2}+a \right )}{3 x \,c^{2} \sqrt {c x}}+\frac {4 a \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, F\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {c x \left (b \,x^{2}+a \right )}}{3 \sqrt {b c \,x^{3}+a c x}\, c^{2} \sqrt {c x}\, \sqrt {b \,x^{2}+a}}\) | \(169\) |
elliptic | \(\frac {\sqrt {c x \left (b \,x^{2}+a \right )}\, \left (-\frac {2 a \sqrt {b c \,x^{3}+a c x}}{3 c^{3} x^{2}}+\frac {2 b \sqrt {b c \,x^{3}+a c x}}{3 c^{3}}+\frac {4 a \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, F\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{3 c^{2} \sqrt {b c \,x^{3}+a c x}}\right )}{\sqrt {c x}\, \sqrt {b \,x^{2}+a}}\) | \(181\) |
2/3/(b*x^2+a)^(1/2)/x*(2*((b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2)*2^(1/2)*( (-b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2)*(-x/(-a*b)^(1/2)*b)^(1/2)*Elliptic F(((b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2),1/2*2^(1/2))*(-a*b)^(1/2)*a*x+b^ 2*x^4-a^2)/c^2/(c*x)^(1/2)
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.09 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.35 \[ \int \frac {\left (a+b x^2\right )^{3/2}}{(c x)^{5/2}} \, dx=\frac {2 \, {\left (4 \, \sqrt {b c} a x^{2} {\rm weierstrassPInverse}\left (-\frac {4 \, a}{b}, 0, x\right ) + \sqrt {b x^{2} + a} {\left (b x^{2} - a\right )} \sqrt {c x}\right )}}{3 \, c^{3} x^{2}} \]
2/3*(4*sqrt(b*c)*a*x^2*weierstrassPInverse(-4*a/b, 0, x) + sqrt(b*x^2 + a) *(b*x^2 - a)*sqrt(c*x))/(c^3*x^2)
Result contains complex when optimal does not.
Time = 1.94 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.32 \[ \int \frac {\left (a+b x^2\right )^{3/2}}{(c x)^{5/2}} \, dx=\frac {a^{\frac {3}{2}} \Gamma \left (- \frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{2}, - \frac {3}{4} \\ \frac {1}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 c^{\frac {5}{2}} x^{\frac {3}{2}} \Gamma \left (\frac {1}{4}\right )} \]
a**(3/2)*gamma(-3/4)*hyper((-3/2, -3/4), (1/4,), b*x**2*exp_polar(I*pi)/a) /(2*c**(5/2)*x**(3/2)*gamma(1/4))
\[ \int \frac {\left (a+b x^2\right )^{3/2}}{(c x)^{5/2}} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}}}{\left (c x\right )^{\frac {5}{2}}} \,d x } \]
\[ \int \frac {\left (a+b x^2\right )^{3/2}}{(c x)^{5/2}} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}}}{\left (c x\right )^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {\left (a+b x^2\right )^{3/2}}{(c x)^{5/2}} \, dx=\int \frac {{\left (b\,x^2+a\right )}^{3/2}}{{\left (c\,x\right )}^{5/2}} \,d x \]