Integrand size = 19, antiderivative size = 125 \[ \int \frac {(c x)^{3/2}}{\left (a+b x^2\right )^{3/2}} \, dx=-\frac {c \sqrt {c x}}{b \sqrt {a+b x^2}}+\frac {c^{3/2} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a} \sqrt {c}}\right ),\frac {1}{2}\right )}{2 \sqrt [4]{a} b^{5/4} \sqrt {a+b x^2}} \] Output:
-c*(c*x)^(1/2)/b/(b*x^2+a)^(1/2)+1/2*c^(3/2)*(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)/b^(5/4)/(b*x^2+a)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.02 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.47 \[ \int \frac {(c x)^{3/2}}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {c \sqrt {c x} \left (-1+\sqrt {1+\frac {b x^2}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-\frac {b x^2}{a}\right )\right )}{b \sqrt {a+b x^2}} \] Input:
Integrate[(c*x)^(3/2)/(a + b*x^2)^(3/2),x]
Output:
(c*Sqrt[c*x]*(-1 + Sqrt[1 + (b*x^2)/a]*Hypergeometric2F1[1/4, 1/2, 5/4, -( (b*x^2)/a)]))/(b*Sqrt[a + b*x^2])
Time = 0.22 (sec) , antiderivative size = 138, 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 = {252, 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.
\(\displaystyle \int \frac {(c x)^{3/2}}{\left (a+b x^2\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 252 |
\(\displaystyle \frac {c^2 \int \frac {1}{\sqrt {c x} \sqrt {b x^2+a}}dx}{2 b}-\frac {c \sqrt {c x}}{b \sqrt {a+b x^2}}\) |
\(\Big \downarrow \) 266 |
\(\displaystyle \frac {c \int \frac {1}{\sqrt {b x^2+a}}d\sqrt {c x}}{b}-\frac {c \sqrt {c x}}{b \sqrt {a+b x^2}}\) |
\(\Big \downarrow \) 761 |
\(\displaystyle \frac {\sqrt {c} \left (\sqrt {a} c+\sqrt {b} c x\right ) \sqrt {\frac {a c^2+b c^2 x^2}{\left (\sqrt {a} c+\sqrt {b} c x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a} \sqrt {c}}\right ),\frac {1}{2}\right )}{2 \sqrt [4]{a} b^{5/4} \sqrt {a+b x^2}}-\frac {c \sqrt {c x}}{b \sqrt {a+b x^2}}\) |
Input:
Int[(c*x)^(3/2)/(a + b*x^2)^(3/2),x]
Output:
-((c*Sqrt[c*x])/(b*Sqrt[a + b*x^2])) + (Sqrt[c]*(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^(1/4)*b^(5/4)*Sqrt[a + b *x^2])
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x )^(m - 1)*((a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] - Simp[c^2*((m - 1)/(2*b* (p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c }, x] && LtQ[p, -1] && GtQ[m, 1] && !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomi alQ[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 = 0.44 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.92
method | result | size |
default | \(\frac {c \sqrt {c x}\, \left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {-a b}-2 b x \right )}{2 x \sqrt {b \,x^{2}+a}\, b^{2}}\) | \(115\) |
elliptic | \(\frac {\sqrt {c x}\, \sqrt {c x \left (b \,x^{2}+a \right )}\, \left (-\frac {c^{2} x}{b \sqrt {\left (x^{2}+\frac {a}{b}\right ) b c x}}+\frac {c^{2} \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{2 b^{2} \sqrt {b c \,x^{3}+a c x}}\right )}{c x \sqrt {b \,x^{2}+a}}\) | \(172\) |
Input:
int((c*x)^(3/2)/(b*x^2+a)^(3/2),x,method=_RETURNVERBOSE)
Output:
1/2*c/x*(c*x)^(1/2)*(((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)*(-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)-2*b*x)/(b* x^2+a)^(1/2)/b^2
Time = 0.07 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.49 \[ \int \frac {(c x)^{3/2}}{\left (a+b x^2\right )^{3/2}} \, dx=-\frac {\sqrt {b x^{2} + a} \sqrt {c x} b c - {\left (b c x^{2} + a c\right )} \sqrt {b c} {\rm weierstrassPInverse}\left (-\frac {4 \, a}{b}, 0, x\right )}{b^{3} x^{2} + a b^{2}} \] Input:
integrate((c*x)^(3/2)/(b*x^2+a)^(3/2),x, algorithm="fricas")
Output:
-(sqrt(b*x^2 + a)*sqrt(c*x)*b*c - (b*c*x^2 + a*c)*sqrt(b*c)*weierstrassPIn verse(-4*a/b, 0, x))/(b^3*x^2 + a*b^2)
Result contains complex when optimal does not.
Time = 1.10 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.35 \[ \int \frac {(c x)^{3/2}}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {c^{\frac {3}{2}} x^{\frac {5}{2}} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{4}, \frac {3}{2} \\ \frac {9}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {3}{2}} \Gamma \left (\frac {9}{4}\right )} \] Input:
integrate((c*x)**(3/2)/(b*x**2+a)**(3/2),x)
Output:
c**(3/2)*x**(5/2)*gamma(5/4)*hyper((5/4, 3/2), (9/4,), b*x**2*exp_polar(I* pi)/a)/(2*a**(3/2)*gamma(9/4))
\[ \int \frac {(c x)^{3/2}}{\left (a+b x^2\right )^{3/2}} \, dx=\int { \frac {\left (c x\right )^{\frac {3}{2}}}{{\left (b x^{2} + a\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate((c*x)^(3/2)/(b*x^2+a)^(3/2),x, algorithm="maxima")
Output:
integrate((c*x)^(3/2)/(b*x^2 + a)^(3/2), x)
\[ \int \frac {(c x)^{3/2}}{\left (a+b x^2\right )^{3/2}} \, dx=\int { \frac {\left (c x\right )^{\frac {3}{2}}}{{\left (b x^{2} + a\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate((c*x)^(3/2)/(b*x^2+a)^(3/2),x, algorithm="giac")
Output:
integrate((c*x)^(3/2)/(b*x^2 + a)^(3/2), x)
Timed out. \[ \int \frac {(c x)^{3/2}}{\left (a+b x^2\right )^{3/2}} \, dx=\int \frac {{\left (c\,x\right )}^{3/2}}{{\left (b\,x^2+a\right )}^{3/2}} \,d x \] Input:
int((c*x)^(3/2)/(a + b*x^2)^(3/2),x)
Output:
int((c*x)^(3/2)/(a + b*x^2)^(3/2), x)
\[ \int \frac {(c x)^{3/2}}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {\sqrt {c}\, c \left (-2 \sqrt {x}\, \sqrt {b \,x^{2}+a}+\left (\int \frac {\sqrt {x}\, \sqrt {b \,x^{2}+a}}{b^{2} x^{5}+2 a b \,x^{3}+a^{2} x}d x \right ) a^{2}+\left (\int \frac {\sqrt {x}\, \sqrt {b \,x^{2}+a}}{b^{2} x^{5}+2 a b \,x^{3}+a^{2} x}d x \right ) a b \,x^{2}\right )}{b \left (b \,x^{2}+a \right )} \] Input:
int((c*x)^(3/2)/(b*x^2+a)^(3/2),x)
Output:
(sqrt(c)*c*( - 2*sqrt(x)*sqrt(a + b*x**2) + int((sqrt(x)*sqrt(a + b*x**2)) /(a**2*x + 2*a*b*x**3 + b**2*x**5),x)*a**2 + int((sqrt(x)*sqrt(a + b*x**2) )/(a**2*x + 2*a*b*x**3 + b**2*x**5),x)*a*b*x**2))/(b*(a + b*x**2))