Integrand size = 19, antiderivative size = 126 \[ \int \frac {\sqrt {a+b x^2}}{(c x)^{5/2}} \, dx=-\frac {2 \sqrt {a+b x^2}}{3 c (c x)^{3/2}}+\frac {2 b^{3/4} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a} \sqrt {c}}\right ),\frac {1}{2}\right )}{3 \sqrt [4]{a} c^{5/2} \sqrt {a+b x^2}} \] Output:
-2/3*(b*x^2+a)^(1/2)/c/(c*x)^(3/2)+2/3*b^(3/4)*(a^(1/2)+b^(1/2)*x)*((b*x^2 +a)/(a^(1/2)+b^(1/2)*x)^2)^(1/2)*InverseJacobiAM(2*arctan(b^(1/4)*(c*x)^(1 /2)/a^(1/4)/c^(1/2)),1/2*2^(1/2))/a^(1/4)/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 = 56, normalized size of antiderivative = 0.44 \[ \int \frac {\sqrt {a+b x^2}}{(c x)^{5/2}} \, dx=-\frac {2 x \sqrt {a+b x^2} \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},-\frac {1}{2},\frac {1}{4},-\frac {b x^2}{a}\right )}{3 (c x)^{5/2} \sqrt {1+\frac {b x^2}{a}}} \] Input:
Integrate[Sqrt[a + b*x^2]/(c*x)^(5/2),x]
Output:
(-2*x*Sqrt[a + b*x^2]*Hypergeometric2F1[-3/4, -1/2, 1/4, -((b*x^2)/a)])/(3 *(c*x)^(5/2)*Sqrt[1 + (b*x^2)/a])
Time = 0.21 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.10, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {247, 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.
\(\displaystyle \int \frac {\sqrt {a+b x^2}}{(c x)^{5/2}} \, dx\) |
\(\Big \downarrow \) 247 |
\(\displaystyle \frac {2 b \int \frac {1}{\sqrt {c x} \sqrt {b x^2+a}}dx}{3 c^2}-\frac {2 \sqrt {a+b x^2}}{3 c (c x)^{3/2}}\) |
\(\Big \downarrow \) 266 |
\(\displaystyle \frac {4 b \int \frac {1}{\sqrt {b x^2+a}}d\sqrt {c x}}{3 c^3}-\frac {2 \sqrt {a+b x^2}}{3 c (c x)^{3/2}}\) |
\(\Big \downarrow \) 761 |
\(\displaystyle \frac {2 b^{3/4} \left (\sqrt {a} c+\sqrt {b} c x\right ) \sqrt {\frac {a c^2+b c^2 x^2}{\left (\sqrt {a} c+\sqrt {b} c x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a} \sqrt {c}}\right ),\frac {1}{2}\right )}{3 \sqrt [4]{a} c^{7/2} \sqrt {a+b x^2}}-\frac {2 \sqrt {a+b x^2}}{3 c (c x)^{3/2}}\) |
Input:
Int[Sqrt[a + b*x^2]/(c*x)^(5/2),x]
Output:
(-2*Sqrt[a + b*x^2])/(3*c*(c*x)^(3/2)) + (2*b^(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*ArcT an[(b^(1/4)*Sqrt[c*x])/(a^(1/4)*Sqrt[c])], 1/2])/(3*a^(1/4)*c^(7/2)*Sqrt[a + b*x^2])
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] :> 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 = 0.46 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.95
method | result | size |
default | \(\frac {\frac {2 \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {-a b}\, x}{3}-\frac {2 b \,x^{2}}{3}-\frac {2 a}{3}}{\sqrt {b \,x^{2}+a}\, x \,c^{2} \sqrt {c x}}\) | \(120\) |
risch | \(-\frac {2 \sqrt {b \,x^{2}+a}}{3 x \,c^{2} \sqrt {c x}}+\frac {2 \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {c x \left (b \,x^{2}+a \right )}}{3 \sqrt {b c \,x^{3}+a c x}\, c^{2} \sqrt {c x}\, \sqrt {b \,x^{2}+a}}\) | \(160\) |
elliptic | \(\frac {\sqrt {c x \left (b \,x^{2}+a \right )}\, \left (-\frac {2 \sqrt {b c \,x^{3}+a c x}}{3 c^{3} x^{2}}+\frac {2 \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{3 c^{2} \sqrt {b c \,x^{3}+a c x}}\right )}{\sqrt {c x}\, \sqrt {b \,x^{2}+a}}\) | \(160\) |
Input:
int((b*x^2+a)^(1/2)/(c*x)^(5/2),x,method=_RETURNVERBOSE)
Output:
2/3/(b*x^2+a)^(1/2)/x*(((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)*(-b/(-a*b)^(1/2)*x)^(1/2)*EllipticF( ((b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2),1/2*2^(1/2))*(-a*b)^(1/2)*x-b*x^2- a)/c^2/(c*x)^(1/2)
Time = 0.07 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.35 \[ \int \frac {\sqrt {a+b x^2}}{(c x)^{5/2}} \, dx=\frac {2 \, {\left (2 \, \sqrt {b c} x^{2} {\rm weierstrassPInverse}\left (-\frac {4 \, a}{b}, 0, x\right ) - \sqrt {b x^{2} + a} \sqrt {c x}\right )}}{3 \, c^{3} x^{2}} \] Input:
integrate((b*x^2+a)^(1/2)/(c*x)^(5/2),x, algorithm="fricas")
Output:
2/3*(2*sqrt(b*c)*x^2*weierstrassPInverse(-4*a/b, 0, x) - sqrt(b*x^2 + a)*s qrt(c*x))/(c^3*x^2)
Result contains complex when optimal does not.
Time = 1.48 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.39 \[ \int \frac {\sqrt {a+b x^2}}{(c x)^{5/2}} \, dx=\frac {\sqrt {a} \Gamma \left (- \frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, - \frac {1}{2} \\ \frac {1}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 c^{\frac {5}{2}} x^{\frac {3}{2}} \Gamma \left (\frac {1}{4}\right )} \] Input:
integrate((b*x**2+a)**(1/2)/(c*x)**(5/2),x)
Output:
sqrt(a)*gamma(-3/4)*hyper((-3/4, -1/2), (1/4,), b*x**2*exp_polar(I*pi)/a)/ (2*c**(5/2)*x**(3/2)*gamma(1/4))
\[ \int \frac {\sqrt {a+b x^2}}{(c x)^{5/2}} \, dx=\int { \frac {\sqrt {b x^{2} + a}}{\left (c x\right )^{\frac {5}{2}}} \,d x } \] Input:
integrate((b*x^2+a)^(1/2)/(c*x)^(5/2),x, algorithm="maxima")
Output:
integrate(sqrt(b*x^2 + a)/(c*x)^(5/2), x)
\[ \int \frac {\sqrt {a+b x^2}}{(c x)^{5/2}} \, dx=\int { \frac {\sqrt {b x^{2} + a}}{\left (c x\right )^{\frac {5}{2}}} \,d x } \] Input:
integrate((b*x^2+a)^(1/2)/(c*x)^(5/2),x, algorithm="giac")
Output:
integrate(sqrt(b*x^2 + a)/(c*x)^(5/2), x)
Timed out. \[ \int \frac {\sqrt {a+b x^2}}{(c x)^{5/2}} \, dx=\int \frac {\sqrt {b\,x^2+a}}{{\left (c\,x\right )}^{5/2}} \,d x \] Input:
int((a + b*x^2)^(1/2)/(c*x)^(5/2),x)
Output:
int((a + b*x^2)^(1/2)/(c*x)^(5/2), x)
\[ \int \frac {\sqrt {a+b x^2}}{(c x)^{5/2}} \, dx=-\frac {2 \sqrt {c}\, \left (\sqrt {b \,x^{2}+a}+\sqrt {x}\, \left (\int \frac {\sqrt {x}\, \sqrt {b \,x^{2}+a}}{b \,x^{5}+a \,x^{3}}d x \right ) a x \right )}{\sqrt {x}\, c^{3} x} \] Input:
int((b*x^2+a)^(1/2)/(c*x)^(5/2),x)
Output:
( - 2*sqrt(c)*(sqrt(a + b*x**2) + sqrt(x)*int((sqrt(x)*sqrt(a + b*x**2))/( a*x**3 + b*x**5),x)*a*x))/(sqrt(x)*c**3*x)