\(\int \frac {\sqrt {a+b x^2}}{(c x)^{13/2}} \, dx\) [591]

Optimal result

Integrand size = 19, antiderivative size = 186 \[ \int \frac {\sqrt {a+b x^2}}{(c x)^{13/2}} \, dx=-\frac {2 \sqrt {a+b x^2}}{11 c (c x)^{11/2}}-\frac {4 b \sqrt {a+b x^2}}{77 a c^3 (c x)^{7/2}}+\frac {20 b^2 \sqrt {a+b x^2}}{231 a^2 c^5 (c x)^{3/2}}+\frac {10 b^{11/4} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a} \sqrt {c}}\right ),\frac {1}{2}\right )}{231 a^{9/4} c^{13/2} \sqrt {a+b x^2}} \] Output:

-2/11*(b*x^2+a)^(1/2)/c/(c*x)^(11/2)-4/77*b*(b*x^2+a)^(1/2)/a/c^3/(c*x)^(7 
/2)+20/231*b^2*(b*x^2+a)^(1/2)/a^2/c^5/(c*x)^(3/2)+10/231*b^(11/4)*(a^(1/2 
)+b^(1/2)*x)*((b*x^2+a)/(a^(1/2)+b^(1/2)*x)^2)^(1/2)*InverseJacobiAM(2*arc 
tan(b^(1/4)*(c*x)^(1/2)/a^(1/4)/c^(1/2)),1/2*2^(1/2))/a^(9/4)/c^(13/2)/(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 = 10.01 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.30 \[ \int \frac {\sqrt {a+b x^2}}{(c x)^{13/2}} \, dx=-\frac {2 x \sqrt {a+b x^2} \operatorname {Hypergeometric2F1}\left (-\frac {11}{4},-\frac {1}{2},-\frac {7}{4},-\frac {b x^2}{a}\right )}{11 (c x)^{13/2} \sqrt {1+\frac {b x^2}{a}}} \] Input:

Integrate[Sqrt[a + b*x^2]/(c*x)^(13/2),x]
 

Output:

(-2*x*Sqrt[a + b*x^2]*Hypergeometric2F1[-11/4, -1/2, -7/4, -((b*x^2)/a)])/ 
(11*(c*x)^(13/2)*Sqrt[1 + (b*x^2)/a])
 

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 216, 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 = {247, 264, 264, 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.

\(\displaystyle \int \frac {\sqrt {a+b x^2}}{(c x)^{13/2}} \, dx\)

\(\Big \downarrow \) 247

\(\displaystyle \frac {2 b \int \frac {1}{(c x)^{9/2} \sqrt {b x^2+a}}dx}{11 c^2}-\frac {2 \sqrt {a+b x^2}}{11 c (c x)^{11/2}}\)

\(\Big \downarrow \) 264

\(\displaystyle \frac {2 b \left (-\frac {5 b \int \frac {1}{(c x)^{5/2} \sqrt {b x^2+a}}dx}{7 a c^2}-\frac {2 \sqrt {a+b x^2}}{7 a c (c x)^{7/2}}\right )}{11 c^2}-\frac {2 \sqrt {a+b x^2}}{11 c (c x)^{11/2}}\)

\(\Big \downarrow \) 264

\(\displaystyle \frac {2 b \left (-\frac {5 b \left (-\frac {b \int \frac {1}{\sqrt {c x} \sqrt {b x^2+a}}dx}{3 a c^2}-\frac {2 \sqrt {a+b x^2}}{3 a c (c x)^{3/2}}\right )}{7 a c^2}-\frac {2 \sqrt {a+b x^2}}{7 a c (c x)^{7/2}}\right )}{11 c^2}-\frac {2 \sqrt {a+b x^2}}{11 c (c x)^{11/2}}\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {2 b \left (-\frac {5 b \left (-\frac {2 b \int \frac {1}{\sqrt {b x^2+a}}d\sqrt {c x}}{3 a c^3}-\frac {2 \sqrt {a+b x^2}}{3 a c (c x)^{3/2}}\right )}{7 a c^2}-\frac {2 \sqrt {a+b x^2}}{7 a c (c x)^{7/2}}\right )}{11 c^2}-\frac {2 \sqrt {a+b x^2}}{11 c (c x)^{11/2}}\)

\(\Big \downarrow \) 761

\(\displaystyle \frac {2 b \left (-\frac {5 b \left (-\frac {b^{3/4} \left (\sqrt {a} c+\sqrt {b} c x\right ) \sqrt {\frac {a c^2+b c^2 x^2}{\left (\sqrt {a} c+\sqrt {b} c x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a} \sqrt {c}}\right ),\frac {1}{2}\right )}{3 a^{5/4} c^{7/2} \sqrt {a+b x^2}}-\frac {2 \sqrt {a+b x^2}}{3 a c (c x)^{3/2}}\right )}{7 a c^2}-\frac {2 \sqrt {a+b x^2}}{7 a c (c x)^{7/2}}\right )}{11 c^2}-\frac {2 \sqrt {a+b x^2}}{11 c (c x)^{11/2}}\)

Input:

Int[Sqrt[a + b*x^2]/(c*x)^(13/2),x]
 

Output:

(-2*Sqrt[a + b*x^2])/(11*c*(c*x)^(11/2)) + (2*b*((-2*Sqrt[a + b*x^2])/(7*a 
*c*(c*x)^(7/2)) - (5*b*((-2*Sqrt[a + b*x^2])/(3*a*c*(c*x)^(3/2)) - (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*ArcTan[(b^(1/4)*Sqrt[c*x])/(a^(1/4)*Sqrt[c])], 1/2])/( 
3*a^(5/4)*c^(7/2)*Sqrt[a + b*x^2])))/(7*a*c^2)))/(11*c^2)
 

Defintions of rubi rules used

rule 247

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]
 

rule 264

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^( 
m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + 2*p + 3)/(a*c 
^2*(m + 1)))   Int[(c*x)^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, p 
}, x] && LtQ[m, -1] && 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 = 0.73 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.81

method	result	size
default	\(\frac {\frac {10 \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {-a b}\, b^{2} x^{5}}{231}+\frac {20 b^{3} x^{6}}{231}+\frac {8 a \,b^{2} x^{4}}{231}-\frac {18 a^{2} b \,x^{2}}{77}-\frac {2 a^{3}}{11}}{\sqrt {b \,x^{2}+a}\, x^{5} c^{6} \sqrt {c x}\, a^{2}}\)	\(151\)
risch	\(-\frac {2 \sqrt {b \,x^{2}+a}\, \left (-10 b^{2} x^{4}+6 a b \,x^{2}+21 a^{2}\right )}{231 x^{5} a^{2} c^{6} \sqrt {c x}}+\frac {10 b^{2} \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {c x \left (b \,x^{2}+a \right )}}{231 a^{2} \sqrt {b c \,x^{3}+a c x}\, c^{6} \sqrt {c x}\, \sqrt {b \,x^{2}+a}}\)	\(190\)
elliptic	\(\frac {\sqrt {c x \left (b \,x^{2}+a \right )}\, \left (-\frac {2 \sqrt {b c \,x^{3}+a c x}}{11 c^{7} x^{6}}-\frac {4 b \sqrt {b c \,x^{3}+a c x}}{77 a \,c^{7} x^{4}}+\frac {20 b^{2} \sqrt {b c \,x^{3}+a c x}}{231 a^{2} c^{7} x^{2}}+\frac {10 b^{2} \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{231 a^{2} c^{6} \sqrt {b c \,x^{3}+a c x}}\right )}{\sqrt {c x}\, \sqrt {b \,x^{2}+a}}\)	\(218\)

Input:

int((b*x^2+a)^(1/2)/(c*x)^(13/2),x,method=_RETURNVERBOSE)
 

Output:

2/231/(b*x^2+a)^(1/2)/x^5*(5*((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)*Elli 
pticF(((b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2),1/2*2^(1/2))*(-a*b)^(1/2)*b^ 
2*x^5+10*b^3*x^6+4*a*b^2*x^4-27*a^2*b*x^2-21*a^3)/c^6/(c*x)^(1/2)/a^2
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.38 \[ \int \frac {\sqrt {a+b x^2}}{(c x)^{13/2}} \, dx=\frac {2 \, {\left (10 \, \sqrt {b c} b^{2} x^{6} {\rm weierstrassPInverse}\left (-\frac {4 \, a}{b}, 0, x\right ) + {\left (10 \, b^{2} x^{4} - 6 \, a b x^{2} - 21 \, a^{2}\right )} \sqrt {b x^{2} + a} \sqrt {c x}\right )}}{231 \, a^{2} c^{7} x^{6}} \] Input:

integrate((b*x^2+a)^(1/2)/(c*x)^(13/2),x, algorithm="fricas")
 

Output:

2/231*(10*sqrt(b*c)*b^2*x^6*weierstrassPInverse(-4*a/b, 0, x) + (10*b^2*x^ 
4 - 6*a*b*x^2 - 21*a^2)*sqrt(b*x^2 + a)*sqrt(c*x))/(a^2*c^7*x^6)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 157.36 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.28 \[ \int \frac {\sqrt {a+b x^2}}{(c x)^{13/2}} \, dx=\frac {\sqrt {a} \Gamma \left (- \frac {11}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {11}{4}, - \frac {1}{2} \\ - \frac {7}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 c^{\frac {13}{2}} x^{\frac {11}{2}} \Gamma \left (- \frac {7}{4}\right )} \] Input:

integrate((b*x**2+a)**(1/2)/(c*x)**(13/2),x)
 

Output:

sqrt(a)*gamma(-11/4)*hyper((-11/4, -1/2), (-7/4,), b*x**2*exp_polar(I*pi)/ 
a)/(2*c**(13/2)*x**(11/2)*gamma(-7/4))
 

Maxima [F]

\[ \int \frac {\sqrt {a+b x^2}}{(c x)^{13/2}} \, dx=\int { \frac {\sqrt {b x^{2} + a}}{\left (c x\right )^{\frac {13}{2}}} \,d x } \] Input:

integrate((b*x^2+a)^(1/2)/(c*x)^(13/2),x, algorithm="maxima")
 

Output:

integrate(sqrt(b*x^2 + a)/(c*x)^(13/2), x)
 

Giac [F]

\[ \int \frac {\sqrt {a+b x^2}}{(c x)^{13/2}} \, dx=\int { \frac {\sqrt {b x^{2} + a}}{\left (c x\right )^{\frac {13}{2}}} \,d x } \] Input:

integrate((b*x^2+a)^(1/2)/(c*x)^(13/2),x, algorithm="giac")
 

Output:

integrate(sqrt(b*x^2 + a)/(c*x)^(13/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {a+b x^2}}{(c x)^{13/2}} \, dx=\int \frac {\sqrt {b\,x^2+a}}{{\left (c\,x\right )}^{13/2}} \,d x \] Input:

int((a + b*x^2)^(1/2)/(c*x)^(13/2),x)
 

Output:

int((a + b*x^2)^(1/2)/(c*x)^(13/2), x)
 

Reduce [F]

\[ \int \frac {\sqrt {a+b x^2}}{(c x)^{13/2}} \, dx=-\frac {2 \sqrt {c}\, \left (\sqrt {b \,x^{2}+a}+\sqrt {x}\, \left (\int \frac {\sqrt {x}\, \sqrt {b \,x^{2}+a}}{b \,x^{9}+a \,x^{7}}d x \right ) a \,x^{5}\right )}{9 \sqrt {x}\, c^{7} x^{5}} \] Input:

int((b*x^2+a)^(1/2)/(c*x)^(13/2),x)
 

Output:

( - 2*sqrt(c)*(sqrt(a + b*x**2) + sqrt(x)*int((sqrt(x)*sqrt(a + b*x**2))/( 
a*x**7 + b*x**9),x)*a*x**5))/(9*sqrt(x)*c**7*x**5)