\(\int \frac {(c x)^{11/2}}{(a+b x^2)^{5/2}} \, dx\) [637]

Optimal result

Integrand size = 19, antiderivative size = 183 \[ \int \frac {(c x)^{11/2}}{\left (a+b x^2\right )^{5/2}} \, dx=-\frac {c (c x)^{9/2}}{3 b \left (a+b x^2\right )^{3/2}}-\frac {3 c^3 (c x)^{5/2}}{2 b^2 \sqrt {a+b x^2}}+\frac {5 c^5 \sqrt {c x} \sqrt {a+b x^2}}{2 b^3}-\frac {5 a^{3/4} c^{11/2} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a} \sqrt {c}}\right ),\frac {1}{2}\right )}{4 b^{13/4} \sqrt {a+b x^2}} \] Output:

-1/3*c*(c*x)^(9/2)/b/(b*x^2+a)^(3/2)-3/2*c^3*(c*x)^(5/2)/b^2/(b*x^2+a)^(1/ 
2)+5/2*c^5*(c*x)^(1/2)*(b*x^2+a)^(1/2)/b^3-5/4*a^(3/4)*c^(11/2)*(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))/b^(13/4)/(b*x^2+a)^(1/2 
)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 10.06 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.50 \[ \int \frac {(c x)^{11/2}}{\left (a+b x^2\right )^{5/2}} \, dx=\frac {c^5 \sqrt {c x} \left (15 a^2+21 a b x^2+4 b^2 x^4-15 a \left (a+b x^2\right ) \sqrt {1+\frac {b x^2}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-\frac {b x^2}{a}\right )\right )}{6 b^3 \left (a+b x^2\right )^{3/2}} \] Input:

Integrate[(c*x)^(11/2)/(a + b*x^2)^(5/2),x]
 

Output:

(c^5*Sqrt[c*x]*(15*a^2 + 21*a*b*x^2 + 4*b^2*x^4 - 15*a*(a + b*x^2)*Sqrt[1 
+ (b*x^2)/a]*Hypergeometric2F1[1/4, 1/2, 5/4, -((b*x^2)/a)]))/(6*b^3*(a + 
b*x^2)^(3/2))
 

Rubi [A] (verified)

Time = 0.26 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.16, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {252, 252, 262, 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.

\(\displaystyle \int \frac {(c x)^{11/2}}{\left (a+b x^2\right )^{5/2}} \, dx\)

\(\Big \downarrow \) 252

\(\displaystyle \frac {3 c^2 \int \frac {(c x)^{7/2}}{\left (b x^2+a\right )^{3/2}}dx}{2 b}-\frac {c (c x)^{9/2}}{3 b \left (a+b x^2\right )^{3/2}}\)

\(\Big \downarrow \) 252

\(\displaystyle \frac {3 c^2 \left (\frac {5 c^2 \int \frac {(c x)^{3/2}}{\sqrt {b x^2+a}}dx}{2 b}-\frac {c (c x)^{5/2}}{b \sqrt {a+b x^2}}\right )}{2 b}-\frac {c (c x)^{9/2}}{3 b \left (a+b x^2\right )^{3/2}}\)

\(\Big \downarrow \) 262

\(\displaystyle \frac {3 c^2 \left (\frac {5 c^2 \left (\frac {2 c \sqrt {c x} \sqrt {a+b x^2}}{3 b}-\frac {a c^2 \int \frac {1}{\sqrt {c x} \sqrt {b x^2+a}}dx}{3 b}\right )}{2 b}-\frac {c (c x)^{5/2}}{b \sqrt {a+b x^2}}\right )}{2 b}-\frac {c (c x)^{9/2}}{3 b \left (a+b x^2\right )^{3/2}}\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {3 c^2 \left (\frac {5 c^2 \left (\frac {2 c \sqrt {c x} \sqrt {a+b x^2}}{3 b}-\frac {2 a c \int \frac {1}{\sqrt {b x^2+a}}d\sqrt {c x}}{3 b}\right )}{2 b}-\frac {c (c x)^{5/2}}{b \sqrt {a+b x^2}}\right )}{2 b}-\frac {c (c x)^{9/2}}{3 b \left (a+b x^2\right )^{3/2}}\)

\(\Big \downarrow \) 761

\(\displaystyle \frac {3 c^2 \left (\frac {5 c^2 \left (\frac {2 c \sqrt {c x} \sqrt {a+b x^2}}{3 b}-\frac {a^{3/4} \sqrt {c} \left (\sqrt {a} c+\sqrt {b} c x\right ) \sqrt {\frac {a c^2+b c^2 x^2}{\left (\sqrt {a} c+\sqrt {b} c x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a} \sqrt {c}}\right ),\frac {1}{2}\right )}{3 b^{5/4} \sqrt {a+b x^2}}\right )}{2 b}-\frac {c (c x)^{5/2}}{b \sqrt {a+b x^2}}\right )}{2 b}-\frac {c (c x)^{9/2}}{3 b \left (a+b x^2\right )^{3/2}}\)

Input:

Int[(c*x)^(11/2)/(a + b*x^2)^(5/2),x]
 

Output:

-1/3*(c*(c*x)^(9/2))/(b*(a + b*x^2)^(3/2)) + (3*c^2*(-((c*(c*x)^(5/2))/(b* 
Sqrt[a + b*x^2])) + (5*c^2*((2*c*Sqrt[c*x]*Sqrt[a + b*x^2])/(3*b) - (a^(3/ 
4)*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])/(3*b^(5/4)*Sqrt[a + b*x^2])))/(2*b)))/(2*b)
 

Defintions of rubi rules used

rule 252

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]
 

rule 262

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) 
^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ 
(b*(m + 2*p + 1)))   Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b 
, c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c 
, 2, m, p, x]
 

rule 266

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]
 

rule 761

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]
 

Maple [A] (verified)

Time = 3.37 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.26

method	result	size
elliptic	\(\frac {\sqrt {c x}\, \sqrt {c x \left (b \,x^{2}+a \right )}\, \left (-\frac {a^{2} c^{5} \sqrt {b c \,x^{3}+a c x}}{3 b^{5} \left (x^{2}+\frac {a}{b}\right )^{2}}+\frac {13 c^{6} x a}{6 b^{3} \sqrt {\left (x^{2}+\frac {a}{b}\right ) b c x}}+\frac {2 c^{5} \sqrt {b c \,x^{3}+a c x}}{3 b^{3}}-\frac {5 a \,c^{6} \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{4 b^{4} \sqrt {b c \,x^{3}+a c x}}\right )}{c x \sqrt {b \,x^{2}+a}}\)	\(230\)
default	\(-\frac {\left (15 \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-a b}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right ) a b \,x^{2}+15 \sqrt {-a b}\, \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right ) a^{2}-8 b^{3} x^{5}-42 a \,b^{2} x^{3}-30 a^{2} b x \right ) c^{5} \sqrt {c x}}{12 x \,b^{4} \left (b \,x^{2}+a \right )^{\frac {3}{2}}}\)	\(233\)
risch	\(\frac {2 x \sqrt {b \,x^{2}+a}\, c^{6}}{3 b^{3} \sqrt {c x}}-\frac {a \left (\frac {7 \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b \sqrt {b c \,x^{3}+a c x}}-9 a \left (\frac {x}{a \sqrt {\left (x^{2}+\frac {a}{b}\right ) b c x}}+\frac {\sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{2 a b \sqrt {b c \,x^{3}+a c x}}\right )+3 a^{2} \left (\frac {\sqrt {b c \,x^{3}+a c x}}{3 a c \,b^{2} \left (x^{2}+\frac {a}{b}\right )^{2}}+\frac {5 x}{6 a^{2} \sqrt {\left (x^{2}+\frac {a}{b}\right ) b c x}}+\frac {5 \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{12 a^{2} b \sqrt {b c \,x^{3}+a c x}}\right )\right ) c^{6} \sqrt {c x \left (b \,x^{2}+a \right )}}{3 b^{3} \sqrt {c x}\, \sqrt {b \,x^{2}+a}}\)	\(483\)

Input:

int((c*x)^(11/2)/(b*x^2+a)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

1/c/x*(c*x)^(1/2)/(b*x^2+a)^(1/2)*(c*x*(b*x^2+a))^(1/2)*(-1/3*a^2*c^5/b^5* 
(b*c*x^3+a*c*x)^(1/2)/(x^2+a/b)^2+13/6/b^3*c^6*x*a/((x^2+a/b)*b*c*x)^(1/2) 
+2/3/b^3*c^5*(b*c*x^3+a*c*x)^(1/2)-5/4*a*c^6/b^4*(-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)*EllipticF(((x+1/b*(-a 
*b)^(1/2))*b/(-a*b)^(1/2))^(1/2),1/2*2^(1/2)))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.68 \[ \int \frac {(c x)^{11/2}}{\left (a+b x^2\right )^{5/2}} \, dx=-\frac {15 \, {\left (a b^{2} c^{5} x^{4} + 2 \, a^{2} b c^{5} x^{2} + a^{3} c^{5}\right )} \sqrt {b c} {\rm weierstrassPInverse}\left (-\frac {4 \, a}{b}, 0, x\right ) - {\left (4 \, b^{3} c^{5} x^{4} + 21 \, a b^{2} c^{5} x^{2} + 15 \, a^{2} b c^{5}\right )} \sqrt {b x^{2} + a} \sqrt {c x}}{6 \, {\left (b^{6} x^{4} + 2 \, a b^{5} x^{2} + a^{2} b^{4}\right )}} \] Input:

integrate((c*x)^(11/2)/(b*x^2+a)^(5/2),x, algorithm="fricas")
 

Output:

-1/6*(15*(a*b^2*c^5*x^4 + 2*a^2*b*c^5*x^2 + a^3*c^5)*sqrt(b*c)*weierstrass 
PInverse(-4*a/b, 0, x) - (4*b^3*c^5*x^4 + 21*a*b^2*c^5*x^2 + 15*a^2*b*c^5) 
*sqrt(b*x^2 + a)*sqrt(c*x))/(b^6*x^4 + 2*a*b^5*x^2 + a^2*b^4)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 102.88 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.24 \[ \int \frac {(c x)^{11/2}}{\left (a+b x^2\right )^{5/2}} \, dx=\frac {c^{\frac {11}{2}} x^{\frac {13}{2}} \Gamma \left (\frac {13}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{2}, \frac {13}{4} \\ \frac {17}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {5}{2}} \Gamma \left (\frac {17}{4}\right )} \] Input:

integrate((c*x)**(11/2)/(b*x**2+a)**(5/2),x)
 

Output:

c**(11/2)*x**(13/2)*gamma(13/4)*hyper((5/2, 13/4), (17/4,), b*x**2*exp_pol 
ar(I*pi)/a)/(2*a**(5/2)*gamma(17/4))
 

Maxima [F]

\[ \int \frac {(c x)^{11/2}}{\left (a+b x^2\right )^{5/2}} \, dx=\int { \frac {\left (c x\right )^{\frac {11}{2}}}{{\left (b x^{2} + a\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate((c*x)^(11/2)/(b*x^2+a)^(5/2),x, algorithm="maxima")
 

Output:

integrate((c*x)^(11/2)/(b*x^2 + a)^(5/2), x)
 

Giac [F]

\[ \int \frac {(c x)^{11/2}}{\left (a+b x^2\right )^{5/2}} \, dx=\int { \frac {\left (c x\right )^{\frac {11}{2}}}{{\left (b x^{2} + a\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate((c*x)^(11/2)/(b*x^2+a)^(5/2),x, algorithm="giac")
 

Output:

integrate((c*x)^(11/2)/(b*x^2 + a)^(5/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(c x)^{11/2}}{\left (a+b x^2\right )^{5/2}} \, dx=\int \frac {{\left (c\,x\right )}^{11/2}}{{\left (b\,x^2+a\right )}^{5/2}} \,d x \] Input:

int((c*x)^(11/2)/(a + b*x^2)^(5/2),x)
 

Output:

int((c*x)^(11/2)/(a + b*x^2)^(5/2), x)
 

Reduce [F]

\[ \int \frac {(c x)^{11/2}}{\left (a+b x^2\right )^{5/2}} \, dx=\frac {\sqrt {c}\, c^{5} \left (18 \sqrt {x}\, \sqrt {b \,x^{2}+a}\, a^{2}+18 \sqrt {x}\, \sqrt {b \,x^{2}+a}\, a b \,x^{2}+2 \sqrt {x}\, \sqrt {b \,x^{2}+a}\, b^{2} x^{4}-9 \left (\int \frac {\sqrt {x}\, \sqrt {b \,x^{2}+a}}{b^{3} x^{7}+3 a \,b^{2} x^{5}+3 a^{2} b \,x^{3}+a^{3} x}d x \right ) a^{5}-18 \left (\int \frac {\sqrt {x}\, \sqrt {b \,x^{2}+a}}{b^{3} x^{7}+3 a \,b^{2} x^{5}+3 a^{2} b \,x^{3}+a^{3} x}d x \right ) a^{4} b \,x^{2}-9 \left (\int \frac {\sqrt {x}\, \sqrt {b \,x^{2}+a}}{b^{3} x^{7}+3 a \,b^{2} x^{5}+3 a^{2} b \,x^{3}+a^{3} x}d x \right ) a^{3} b^{2} x^{4}\right )}{3 b^{3} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )} \] Input:

int((c*x)^(11/2)/(b*x^2+a)^(5/2),x)
 

Output:

(sqrt(c)*c**5*(18*sqrt(x)*sqrt(a + b*x**2)*a**2 + 18*sqrt(x)*sqrt(a + b*x* 
*2)*a*b*x**2 + 2*sqrt(x)*sqrt(a + b*x**2)*b**2*x**4 - 9*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)*a**5 
- 18*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)*a**4*b*x**2 - 9*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)*a**3*b**2*x**4))/(3*b**3*(a 
**2 + 2*a*b*x**2 + b**2*x**4))