Integrand size = 19, antiderivative size = 157 \[ \int \frac {1}{\sqrt {c x} \left (a+b x^2\right )^{5/2}} \, dx=\frac {\sqrt {c x}}{3 a c \left (a+b x^2\right )^{3/2}}+\frac {5 \sqrt {c x}}{6 a^2 c \sqrt {a+b x^2}}+\frac {5 \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a} \sqrt {c}}\right ),\frac {1}{2}\right )}{12 a^{9/4} \sqrt [4]{b} \sqrt {c} \sqrt {a+b x^2}} \] Output:
1/3*(c*x)^(1/2)/a/c/(b*x^2+a)^(3/2)+5/6*(c*x)^(1/2)/a^2/c/(b*x^2+a)^(1/2)+ 5/12*(a^(1/2)+b^(1/2)*x)*((b*x^2+a)/(a^(1/2)+b^(1/2)*x)^2)^(1/2)*InverseJa cobiAM(2*arctan(b^(1/4)*(c*x)^(1/2)/a^(1/4)/c^(1/2)),1/2*2^(1/2))/a^(9/4)/ b^(1/4)/c^(1/2)/(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 = 79, normalized size of antiderivative = 0.50 \[ \int \frac {1}{\sqrt {c x} \left (a+b x^2\right )^{5/2}} \, dx=\frac {7 a x+5 b x^3+5 x \left (a+b x^2\right ) \sqrt {1+\frac {b x^2}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-\frac {b x^2}{a}\right )}{6 a^2 \sqrt {c x} \left (a+b x^2\right )^{3/2}} \] Input:
Integrate[1/(Sqrt[c*x]*(a + b*x^2)^(5/2)),x]
Output:
(7*a*x + 5*b*x^3 + 5*x*(a + b*x^2)*Sqrt[1 + (b*x^2)/a]*Hypergeometric2F1[1 /4, 1/2, 5/4, -((b*x^2)/a)])/(6*a^2*Sqrt[c*x]*(a + b*x^2)^(3/2))
Time = 0.25 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.11, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {253, 253, 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.
\(\displaystyle \int \frac {1}{\sqrt {c x} \left (a+b x^2\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 253 |
\(\displaystyle \frac {5 \int \frac {1}{\sqrt {c x} \left (b x^2+a\right )^{3/2}}dx}{6 a}+\frac {\sqrt {c x}}{3 a c \left (a+b x^2\right )^{3/2}}\) |
\(\Big \downarrow \) 253 |
\(\displaystyle \frac {5 \left (\frac {\int \frac {1}{\sqrt {c x} \sqrt {b x^2+a}}dx}{2 a}+\frac {\sqrt {c x}}{a c \sqrt {a+b x^2}}\right )}{6 a}+\frac {\sqrt {c x}}{3 a c \left (a+b x^2\right )^{3/2}}\) |
\(\Big \downarrow \) 266 |
\(\displaystyle \frac {5 \left (\frac {\int \frac {1}{\sqrt {b x^2+a}}d\sqrt {c x}}{a c}+\frac {\sqrt {c x}}{a c \sqrt {a+b x^2}}\right )}{6 a}+\frac {\sqrt {c x}}{3 a c \left (a+b x^2\right )^{3/2}}\) |
\(\Big \downarrow \) 761 |
\(\displaystyle \frac {5 \left (\frac {\left (\sqrt {a} c+\sqrt {b} c x\right ) \sqrt {\frac {a c^2+b c^2 x^2}{\left (\sqrt {a} c+\sqrt {b} c x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a} \sqrt {c}}\right ),\frac {1}{2}\right )}{2 a^{5/4} \sqrt [4]{b} c^{3/2} \sqrt {a+b x^2}}+\frac {\sqrt {c x}}{a c \sqrt {a+b x^2}}\right )}{6 a}+\frac {\sqrt {c x}}{3 a c \left (a+b x^2\right )^{3/2}}\) |
Input:
Int[1/(Sqrt[c*x]*(a + b*x^2)^(5/2)),x]
Output:
Sqrt[c*x]/(3*a*c*(a + b*x^2)^(3/2)) + (5*(Sqrt[c*x]/(a*c*Sqrt[a + b*x^2]) + ((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])/ (2*a^(5/4)*b^(1/4)*c^(3/2)*Sqrt[a + b*x^2])))/(6*a)
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-(c*x )^(m + 1))*((a + b*x^2)^(p + 1)/(2*a*c*(p + 1))), x] + Simp[(m + 2*p + 3)/( 2*a*(p + 1)) Int[(c*x)^m*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, m }, x] && LtQ[p, -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[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 = 2.11 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.26
method | result | size |
elliptic | \(\frac {\sqrt {c x \left (b \,x^{2}+a \right )}\, \left (\frac {\sqrt {b c \,x^{3}+a c x}}{3 a c \,b^{2} \left (x^{2}+\frac {a}{b}\right )^{2}}+\frac {5 x}{6 a^{2} \sqrt {\left (x^{2}+\frac {a}{b}\right ) b c x}}+\frac {5 \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{12 a^{2} b \sqrt {b c \,x^{3}+a c x}}\right )}{\sqrt {c x}\, \sqrt {b \,x^{2}+a}}\) | \(198\) |
default | \(\frac {5 \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {-a b}\, b \,x^{2}+5 \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {-a b}\, a +10 b^{2} x^{3}+14 a b x}{12 \sqrt {c x}\, a^{2} b \left (b \,x^{2}+a \right )^{\frac {3}{2}}}\) | \(216\) |
Input:
int(1/(c*x)^(1/2)/(b*x^2+a)^(5/2),x,method=_RETURNVERBOSE)
Output:
(c*x*(b*x^2+a))^(1/2)/(c*x)^(1/2)/(b*x^2+a)^(1/2)*(1/3/a/c/b^2*(b*c*x^3+a* c*x)^(1/2)/(x^2+a/b)^2+5/6*x/a^2/((x^2+a/b)*b*c*x)^(1/2)+5/12/a^2/b*(-a*b) ^(1/2)*((x+1/b*(-a*b)^(1/2))*b/(-a*b)^(1/2))^(1/2)*(-2*(x-1/b*(-a*b)^(1/2) )*b/(-a*b)^(1/2))^(1/2)*(-b/(-a*b)^(1/2)*x)^(1/2)/(b*c*x^3+a*c*x)^(1/2)*El lipticF(((x+1/b*(-a*b)^(1/2))*b/(-a*b)^(1/2))^(1/2),1/2*2^(1/2)))
Time = 0.08 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.62 \[ \int \frac {1}{\sqrt {c x} \left (a+b x^2\right )^{5/2}} \, dx=\frac {5 \, {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )} \sqrt {b c} {\rm weierstrassPInverse}\left (-\frac {4 \, a}{b}, 0, x\right ) + {\left (5 \, b^{2} x^{2} + 7 \, a b\right )} \sqrt {b x^{2} + a} \sqrt {c x}}{6 \, {\left (a^{2} b^{3} c x^{4} + 2 \, a^{3} b^{2} c x^{2} + a^{4} b c\right )}} \] Input:
integrate(1/(c*x)^(1/2)/(b*x^2+a)^(5/2),x, algorithm="fricas")
Output:
1/6*(5*(b^2*x^4 + 2*a*b*x^2 + a^2)*sqrt(b*c)*weierstrassPInverse(-4*a/b, 0 , x) + (5*b^2*x^2 + 7*a*b)*sqrt(b*x^2 + a)*sqrt(c*x))/(a^2*b^3*c*x^4 + 2*a ^3*b^2*c*x^2 + a^4*b*c)
Result contains complex when optimal does not.
Time = 2.36 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.28 \[ \int \frac {1}{\sqrt {c x} \left (a+b x^2\right )^{5/2}} \, dx=\frac {\sqrt {x} \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {5}{2} \\ \frac {5}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {5}{2}} \sqrt {c} \Gamma \left (\frac {5}{4}\right )} \] Input:
integrate(1/(c*x)**(1/2)/(b*x**2+a)**(5/2),x)
Output:
sqrt(x)*gamma(1/4)*hyper((1/4, 5/2), (5/4,), b*x**2*exp_polar(I*pi)/a)/(2* a**(5/2)*sqrt(c)*gamma(5/4))
\[ \int \frac {1}{\sqrt {c x} \left (a+b x^2\right )^{5/2}} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )}^{\frac {5}{2}} \sqrt {c x}} \,d x } \] Input:
integrate(1/(c*x)^(1/2)/(b*x^2+a)^(5/2),x, algorithm="maxima")
Output:
integrate(1/((b*x^2 + a)^(5/2)*sqrt(c*x)), x)
\[ \int \frac {1}{\sqrt {c x} \left (a+b x^2\right )^{5/2}} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )}^{\frac {5}{2}} \sqrt {c x}} \,d x } \] Input:
integrate(1/(c*x)^(1/2)/(b*x^2+a)^(5/2),x, algorithm="giac")
Output:
integrate(1/((b*x^2 + a)^(5/2)*sqrt(c*x)), x)
Timed out. \[ \int \frac {1}{\sqrt {c x} \left (a+b x^2\right )^{5/2}} \, dx=\int \frac {1}{\sqrt {c\,x}\,{\left (b\,x^2+a\right )}^{5/2}} \,d x \] Input:
int(1/((c*x)^(1/2)*(a + b*x^2)^(5/2)),x)
Output:
int(1/((c*x)^(1/2)*(a + b*x^2)^(5/2)), x)
\[ \int \frac {1}{\sqrt {c x} \left (a+b x^2\right )^{5/2}} \, dx=\frac {\sqrt {c}\, \left (\int \frac {\sqrt {x}\, \sqrt {b \,x^{2}+a}}{b^{3} x^{7}+3 a \,b^{2} x^{5}+3 a^{2} b \,x^{3}+a^{3} x}d x \right )}{c} \] Input:
int(1/(c*x)^(1/2)/(b*x^2+a)^(5/2),x)
Output:
(sqrt(c)*int((sqrt(x)*sqrt(a + b*x**2))/(a**3*x + 3*a**2*b*x**3 + 3*a*b**2 *x**5 + b**3*x**7),x))/c