\(\int \frac {\sqrt {a+b x^2}}{(c x)^{7/2}} \, dx\) [596]

Optimal result

Integrand size = 19, antiderivative size = 303 \[ \int \frac {\sqrt {a+b x^2}}{(c x)^{7/2}} \, dx=-\frac {2 \sqrt {a+b x^2}}{5 c (c x)^{5/2}}-\frac {4 b \sqrt {a+b x^2}}{5 a c^3 \sqrt {c x}}+\frac {4 b^{3/2} \sqrt {c x} \sqrt {a+b x^2}}{5 a c^4 \left (\sqrt {a}+\sqrt {b} x\right )}-\frac {4 b^{5/4} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a} \sqrt {c}}\right )|\frac {1}{2}\right )}{5 a^{3/4} c^{7/2} \sqrt {a+b x^2}}+\frac {2 b^{5/4} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a} \sqrt {c}}\right ),\frac {1}{2}\right )}{5 a^{3/4} c^{7/2} \sqrt {a+b x^2}} \] Output:

-2/5*(b*x^2+a)^(1/2)/c/(c*x)^(5/2)-4/5*b*(b*x^2+a)^(1/2)/a/c^3/(c*x)^(1/2) 
+4/5*b^(3/2)*(c*x)^(1/2)*(b*x^2+a)^(1/2)/a/c^4/(a^(1/2)+b^(1/2)*x)-4/5*b^( 
5/4)*(a^(1/2)+b^(1/2)*x)*((b*x^2+a)/(a^(1/2)+b^(1/2)*x)^2)^(1/2)*EllipticE 
(sin(2*arctan(b^(1/4)*(c*x)^(1/2)/a^(1/4)/c^(1/2))),1/2*2^(1/2))/a^(3/4)/c 
^(7/2)/(b*x^2+a)^(1/2)+2/5*b^(5/4)*(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^(3/4)/c^(7/2)/(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 = 0.18 \[ \int \frac {\sqrt {a+b x^2}}{(c x)^{7/2}} \, dx=-\frac {2 x \sqrt {a+b x^2} \operatorname {Hypergeometric2F1}\left (-\frac {5}{4},-\frac {1}{2},-\frac {1}{4},-\frac {b x^2}{a}\right )}{5 (c x)^{7/2} \sqrt {1+\frac {b x^2}{a}}} \] Input:

Integrate[Sqrt[a + b*x^2]/(c*x)^(7/2),x]
 

Output:

(-2*x*Sqrt[a + b*x^2]*Hypergeometric2F1[-5/4, -1/2, -1/4, -((b*x^2)/a)])/( 
5*(c*x)^(7/2)*Sqrt[1 + (b*x^2)/a])
 

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 343, normalized size of antiderivative = 1.13, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {247, 264, 266, 834, 27, 761, 1510}

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.

\(\displaystyle \int \frac {\sqrt {a+b x^2}}{(c x)^{7/2}} \, dx\)

\(\Big \downarrow \) 247

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

\(\Big \downarrow \) 264

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 834

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 761

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

\(\Big \downarrow \) 1510

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

Input:

Int[Sqrt[a + b*x^2]/(c*x)^(7/2),x]
 

Output:

(-2*Sqrt[a + b*x^2])/(5*c*(c*x)^(5/2)) + (2*b*((-2*Sqrt[a + b*x^2])/(a*c*S 
qrt[c*x]) + (2*b*(-((-((c^2*Sqrt[c*x]*Sqrt[a + b*x^2])/(Sqrt[a]*c + Sqrt[b 
]*c*x)) + (a^(1/4)*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]*EllipticE[2*ArcTan[(b^(1/4)*Sqrt[c*x])/(a 
^(1/4)*Sqrt[c])], 1/2])/(b^(1/4)*Sqrt[a + b*x^2]))/Sqrt[b]) + (a^(1/4)*Sqr 
t[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*b^(3/4)*Sqrt[a + b*x^2])))/(a*c^3)))/(5*c^2)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

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]
 

rule 834

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 2]}, S 
imp[1/q   Int[1/Sqrt[a + b*x^4], x], x] - Simp[1/q   Int[(1 - q*x^2)/Sqrt[a 
 + b*x^4], x], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]
 

rule 1510

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = 
 Rt[c/a, 4]}, Simp[(-d)*x*(Sqrt[a + c*x^4]/(a*(1 + q^2*x^2))), x] + Simp[d* 
(1 + q^2*x^2)*(Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]/(q*Sqrt[a + c*x^4]))*E 
llipticE[2*ArcTan[q*x], 1/2], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e 
}, x] && PosQ[c/a]
 

Maple [A] (verified)

Time = 0.56 (sec) , antiderivative size = 219, normalized size of antiderivative = 0.72

method	result	size
default	\(\frac {\frac {4 \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \operatorname {EllipticE}\left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right ) a b \,x^{2}}{5}-\frac {2 \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right ) a b \,x^{2}}{5}-\frac {4 b^{2} x^{4}}{5}-\frac {6 a b \,x^{2}}{5}-\frac {2 a^{2}}{5}}{x^{2} \sqrt {b \,x^{2}+a}\, c^{3} \sqrt {c x}\, a}\)	\(219\)
risch	\(-\frac {2 \sqrt {b \,x^{2}+a}\, \left (2 b \,x^{2}+a \right )}{5 x^{2} a \,c^{3} \sqrt {c x}}+\frac {2 b \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \left (-\frac {2 \sqrt {-a b}\, \operatorname {EllipticE}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}+\frac {\sqrt {-a b}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}\right ) \sqrt {c x \left (b \,x^{2}+a \right )}}{5 a \sqrt {b c \,x^{3}+a c x}\, c^{3} \sqrt {c x}\, \sqrt {b \,x^{2}+a}}\)	\(225\)
elliptic	\(\frac {\sqrt {c x \left (b \,x^{2}+a \right )}\, \left (-\frac {2 \sqrt {b c \,x^{3}+a c x}}{5 c^{4} x^{3}}-\frac {4 \left (x^{2} b c +a c \right ) b}{5 a \,c^{4} \sqrt {x \left (x^{2} b c +a c \right )}}+\frac {2 b \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \left (-\frac {2 \sqrt {-a b}\, \operatorname {EllipticE}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}+\frac {\sqrt {-a b}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}\right )}{5 a \,c^{3} \sqrt {b c \,x^{3}+a c x}}\right )}{\sqrt {c x}\, \sqrt {b \,x^{2}+a}}\)	\(247\)

Input:

int((b*x^2+a)^(1/2)/(c*x)^(7/2),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.21 \[ \int \frac {\sqrt {a+b x^2}}{(c x)^{7/2}} \, dx=-\frac {2 \, {\left (2 \, \sqrt {b c} b x^{3} {\rm weierstrassZeta}\left (-\frac {4 \, a}{b}, 0, {\rm weierstrassPInverse}\left (-\frac {4 \, a}{b}, 0, x\right )\right ) + {\left (2 \, b x^{2} + a\right )} \sqrt {b x^{2} + a} \sqrt {c x}\right )}}{5 \, a c^{4} x^{3}} \] Input:

integrate((b*x^2+a)^(1/2)/(c*x)^(7/2),x, algorithm="fricas")
 

Output:

-2/5*(2*sqrt(b*c)*b*x^3*weierstrassZeta(-4*a/b, 0, weierstrassPInverse(-4* 
a/b, 0, x)) + (2*b*x^2 + a)*sqrt(b*x^2 + a)*sqrt(c*x))/(a*c^4*x^3)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 5.31 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.17 \[ \int \frac {\sqrt {a+b x^2}}{(c x)^{7/2}} \, dx=\frac {\sqrt {a} \Gamma \left (- \frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{4}, - \frac {1}{2} \\ - \frac {1}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 c^{\frac {7}{2}} x^{\frac {5}{2}} \Gamma \left (- \frac {1}{4}\right )} \] Input:

integrate((b*x**2+a)**(1/2)/(c*x)**(7/2),x)
 

Output:

sqrt(a)*gamma(-5/4)*hyper((-5/4, -1/2), (-1/4,), b*x**2*exp_polar(I*pi)/a) 
/(2*c**(7/2)*x**(5/2)*gamma(-1/4))
 

Maxima [F]

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

integrate((b*x^2+a)^(1/2)/(c*x)^(7/2),x, algorithm="maxima")
 

Output:

integrate(sqrt(b*x^2 + a)/(c*x)^(7/2), x)
 

Giac [F]

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

integrate((b*x^2+a)^(1/2)/(c*x)^(7/2),x, algorithm="giac")
 

Output:

integrate(sqrt(b*x^2 + a)/(c*x)^(7/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {a+b x^2}}{(c x)^{7/2}} \, dx=\int \frac {\sqrt {b\,x^2+a}}{{\left (c\,x\right )}^{7/2}} \,d x \] Input:

int((a + b*x^2)^(1/2)/(c*x)^(7/2),x)
 

Output:

int((a + b*x^2)^(1/2)/(c*x)^(7/2), x)
 

Reduce [F]

\[ \int \frac {\sqrt {a+b x^2}}{(c x)^{7/2}} \, dx=-\frac {2 \sqrt {c}\, \left (\sqrt {b \,x^{2}+a}+\sqrt {x}\, \left (\int \frac {\sqrt {x}\, \sqrt {b \,x^{2}+a}}{b \,x^{6}+a \,x^{4}}d x \right ) a \,x^{2}\right )}{3 \sqrt {x}\, c^{4} x^{2}} \] Input:

int((b*x^2+a)^(1/2)/(c*x)^(7/2),x)
 

Output:

( - 2*sqrt(c)*(sqrt(a + b*x**2) + sqrt(x)*int((sqrt(x)*sqrt(a + b*x**2))/( 
a*x**4 + b*x**6),x)*a*x**2))/(3*sqrt(x)*c**4*x**2)