\(\int \frac {(c x)^{9/2}}{\sqrt {a+b x^2}} \, dx\) [623]

Optimal result

Integrand size = 19, antiderivative size = 304 \[ \int \frac {(c x)^{9/2}}{\sqrt {a+b x^2}} \, dx=-\frac {14 a c^3 (c x)^{3/2} \sqrt {a+b x^2}}{45 b^2}+\frac {2 c (c x)^{7/2} \sqrt {a+b x^2}}{9 b}+\frac {14 a^2 c^4 \sqrt {c x} \sqrt {a+b x^2}}{15 b^{5/2} \left (\sqrt {a}+\sqrt {b} x\right )}-\frac {14 a^{9/4} c^{9/2} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a} \sqrt {c}}\right )|\frac {1}{2}\right )}{15 b^{11/4} \sqrt {a+b x^2}}+\frac {7 a^{9/4} c^{9/2} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a} \sqrt {c}}\right ),\frac {1}{2}\right )}{15 b^{11/4} \sqrt {a+b x^2}} \] Output:

-14/45*a*c^3*(c*x)^(3/2)*(b*x^2+a)^(1/2)/b^2+2/9*c*(c*x)^(7/2)*(b*x^2+a)^( 
1/2)/b+14/15*a^2*c^4*(c*x)^(1/2)*(b*x^2+a)^(1/2)/b^(5/2)/(a^(1/2)+b^(1/2)* 
x)-14/15*a^(9/4)*c^(9/2)*(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))/b^(11/4)/(b*x^2+a)^(1/2)+7/15*a^(9/4)*c^(9/2)*(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^(11/4)/(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 = 10.04 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.29 \[ \int \frac {(c x)^{9/2}}{\sqrt {a+b x^2}} \, dx=\frac {2 c^3 (c x)^{3/2} \left (-7 a^2-2 a b x^2+5 b^2 x^4+7 a^2 \sqrt {1+\frac {b x^2}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-\frac {b x^2}{a}\right )\right )}{45 b^2 \sqrt {a+b x^2}} \] Input:

Integrate[(c*x)^(9/2)/Sqrt[a + b*x^2],x]
 

Output:

(2*c^3*(c*x)^(3/2)*(-7*a^2 - 2*a*b*x^2 + 5*b^2*x^4 + 7*a^2*Sqrt[1 + (b*x^2 
)/a]*Hypergeometric2F1[1/2, 3/4, 7/4, -((b*x^2)/a)]))/(45*b^2*Sqrt[a + b*x 
^2])
 

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 347, normalized size of antiderivative = 1.14, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {262, 262, 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.

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

\(\Big \downarrow \) 262

\(\displaystyle \frac {2 c (c x)^{7/2} \sqrt {a+b x^2}}{9 b}-\frac {7 a c^2 \int \frac {(c x)^{5/2}}{\sqrt {b x^2+a}}dx}{9 b}\)

\(\Big \downarrow \) 262

\(\displaystyle \frac {2 c (c x)^{7/2} \sqrt {a+b x^2}}{9 b}-\frac {7 a c^2 \left (\frac {2 c (c x)^{3/2} \sqrt {a+b x^2}}{5 b}-\frac {3 a c^2 \int \frac {\sqrt {c x}}{\sqrt {b x^2+a}}dx}{5 b}\right )}{9 b}\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {2 c (c x)^{7/2} \sqrt {a+b x^2}}{9 b}-\frac {7 a c^2 \left (\frac {2 c (c x)^{3/2} \sqrt {a+b x^2}}{5 b}-\frac {6 a c \int \frac {c x}{\sqrt {b x^2+a}}d\sqrt {c x}}{5 b}\right )}{9 b}\)

\(\Big \downarrow \) 834

\(\displaystyle \frac {2 c (c x)^{7/2} \sqrt {a+b x^2}}{9 b}-\frac {7 a c^2 \left (\frac {2 c (c x)^{3/2} \sqrt {a+b x^2}}{5 b}-\frac {6 a c \left (\frac {\sqrt {a} c \int \frac {1}{\sqrt {b x^2+a}}d\sqrt {c x}}{\sqrt {b}}-\frac {\sqrt {a} c \int \frac {\sqrt {a} c-\sqrt {b} c x}{\sqrt {a} c \sqrt {b x^2+a}}d\sqrt {c x}}{\sqrt {b}}\right )}{5 b}\right )}{9 b}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 c (c x)^{7/2} \sqrt {a+b x^2}}{9 b}-\frac {7 a c^2 \left (\frac {2 c (c x)^{3/2} \sqrt {a+b x^2}}{5 b}-\frac {6 a c \left (\frac {\sqrt {a} c \int \frac {1}{\sqrt {b x^2+a}}d\sqrt {c x}}{\sqrt {b}}-\frac {\int \frac {\sqrt {a} c-\sqrt {b} c x}{\sqrt {b x^2+a}}d\sqrt {c x}}{\sqrt {b}}\right )}{5 b}\right )}{9 b}\)

\(\Big \downarrow \) 761

\(\displaystyle \frac {2 c (c x)^{7/2} \sqrt {a+b x^2}}{9 b}-\frac {7 a c^2 \left (\frac {2 c (c x)^{3/2} \sqrt {a+b x^2}}{5 b}-\frac {6 a c \left (\frac {\sqrt [4]{a} \sqrt {c} \left (\sqrt {a} c+\sqrt {b} c x\right ) \sqrt {\frac {a c^2+b c^2 x^2}{\left (\sqrt {a} c+\sqrt {b} c x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a} \sqrt {c}}\right ),\frac {1}{2}\right )}{2 b^{3/4} \sqrt {a+b x^2}}-\frac {\int \frac {\sqrt {a} c-\sqrt {b} c x}{\sqrt {b x^2+a}}d\sqrt {c x}}{\sqrt {b}}\right )}{5 b}\right )}{9 b}\)

\(\Big \downarrow \) 1510

\(\displaystyle \frac {2 c (c x)^{7/2} \sqrt {a+b x^2}}{9 b}-\frac {7 a c^2 \left (\frac {2 c (c x)^{3/2} \sqrt {a+b x^2}}{5 b}-\frac {6 a c \left (\frac {\sqrt [4]{a} \sqrt {c} \left (\sqrt {a} c+\sqrt {b} c x\right ) \sqrt {\frac {a c^2+b c^2 x^2}{\left (\sqrt {a} c+\sqrt {b} c x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a} \sqrt {c}}\right ),\frac {1}{2}\right )}{2 b^{3/4} \sqrt {a+b x^2}}-\frac {\frac {\sqrt [4]{a} \sqrt {c} \left (\sqrt {a} c+\sqrt {b} c x\right ) \sqrt {\frac {a c^2+b c^2 x^2}{\left (\sqrt {a} c+\sqrt {b} c x\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a} \sqrt {c}}\right )|\frac {1}{2}\right )}{\sqrt [4]{b} \sqrt {a+b x^2}}-\frac {c^2 \sqrt {c x} \sqrt {a+b x^2}}{\sqrt {a} c+\sqrt {b} c x}}{\sqrt {b}}\right )}{5 b}\right )}{9 b}\)

Input:

Int[(c*x)^(9/2)/Sqrt[a + b*x^2],x]
 

Output:

(2*c*(c*x)^(7/2)*Sqrt[a + b*x^2])/(9*b) - (7*a*c^2*((2*c*(c*x)^(3/2)*Sqrt[ 
a + b*x^2])/(5*b) - (6*a*c*(-((-((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)*Sqr 
t[c*x])/(a^(1/4)*Sqrt[c])], 1/2])/(b^(1/4)*Sqrt[a + b*x^2]))/Sqrt[b]) + (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]*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])))/(5*b)))/(9*b)
 

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 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]
 

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 = 1.04 (sec) , antiderivative size = 221, normalized size of antiderivative = 0.73

method	result	size
default	\(\frac {c^{4} \sqrt {c x}\, \left (10 b^{3} x^{6}+42 \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \operatorname {EllipticE}\left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right ) a^{3}-21 \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right ) a^{3}-4 a \,b^{2} x^{4}-14 a^{2} b \,x^{2}\right )}{45 x \sqrt {b \,x^{2}+a}\, b^{3}}\)	\(221\)
risch	\(-\frac {2 x^{2} \left (-5 b \,x^{2}+7 a \right ) \sqrt {b \,x^{2}+a}\, c^{5}}{45 b^{2} \sqrt {c x}}+\frac {7 a^{2} \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \left (-\frac {2 \sqrt {-a b}\, \operatorname {EllipticE}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}+\frac {\sqrt {-a b}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}\right ) c^{5} \sqrt {c x \left (b \,x^{2}+a \right )}}{15 b^{3} \sqrt {b c \,x^{3}+a c x}\, \sqrt {c x}\, \sqrt {b \,x^{2}+a}}\)	\(229\)
elliptic	\(\frac {\sqrt {c x}\, \sqrt {c x \left (b \,x^{2}+a \right )}\, \left (\frac {2 c^{4} x^{3} \sqrt {b c \,x^{3}+a c x}}{9 b}-\frac {14 c^{4} a x \sqrt {b c \,x^{3}+a c x}}{45 b^{2}}+\frac {7 c^{5} a^{2} \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \left (-\frac {2 \sqrt {-a b}\, \operatorname {EllipticE}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}+\frac {\sqrt {-a b}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}\right )}{15 b^{3} \sqrt {b c \,x^{3}+a c x}}\right )}{c x \sqrt {b \,x^{2}+a}}\)	\(248\)

Input:

int((c*x)^(9/2)/(b*x^2+a)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

1/45*c^4/x*(c*x)^(1/2)/(b*x^2+a)^(1/2)/b^3*(10*b^3*x^6+42*((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^3-21*((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^3-4*a*b^2*x^4-14*a^2* 
b*x^2)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.24 \[ \int \frac {(c x)^{9/2}}{\sqrt {a+b x^2}} \, dx=-\frac {2 \, {\left (21 \, \sqrt {b c} a^{2} c^{4} {\rm weierstrassZeta}\left (-\frac {4 \, a}{b}, 0, {\rm weierstrassPInverse}\left (-\frac {4 \, a}{b}, 0, x\right )\right ) - {\left (5 \, b^{2} c^{4} x^{3} - 7 \, a b c^{4} x\right )} \sqrt {b x^{2} + a} \sqrt {c x}\right )}}{45 \, b^{3}} \] Input:

integrate((c*x)^(9/2)/(b*x^2+a)^(1/2),x, algorithm="fricas")
 

Output:

-2/45*(21*sqrt(b*c)*a^2*c^4*weierstrassZeta(-4*a/b, 0, weierstrassPInverse 
(-4*a/b, 0, x)) - (5*b^2*c^4*x^3 - 7*a*b*c^4*x)*sqrt(b*x^2 + a)*sqrt(c*x)) 
/b^3
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 30.42 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.14 \[ \int \frac {(c x)^{9/2}}{\sqrt {a+b x^2}} \, dx=\frac {c^{\frac {9}{2}} x^{\frac {11}{2}} \Gamma \left (\frac {11}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {11}{4} \\ \frac {15}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \sqrt {a} \Gamma \left (\frac {15}{4}\right )} \] Input:

integrate((c*x)**(9/2)/(b*x**2+a)**(1/2),x)
 

Output:

c**(9/2)*x**(11/2)*gamma(11/4)*hyper((1/2, 11/4), (15/4,), b*x**2*exp_pola 
r(I*pi)/a)/(2*sqrt(a)*gamma(15/4))
 

Maxima [F]

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

integrate((c*x)^(9/2)/(b*x^2+a)^(1/2),x, algorithm="maxima")
 

Output:

integrate((c*x)^(9/2)/sqrt(b*x^2 + a), x)
 

Giac [F]

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

integrate((c*x)^(9/2)/(b*x^2+a)^(1/2),x, algorithm="giac")
 

Output:

integrate((c*x)^(9/2)/sqrt(b*x^2 + a), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(c x)^{9/2}}{\sqrt {a+b x^2}} \, dx=\int \frac {{\left (c\,x\right )}^{9/2}}{\sqrt {b\,x^2+a}} \,d x \] Input:

int((c*x)^(9/2)/(a + b*x^2)^(1/2),x)
 

Output:

int((c*x)^(9/2)/(a + b*x^2)^(1/2), x)
                                                                                    
                                                                                    
 

Reduce [F]

\[ \int \frac {(c x)^{9/2}}{\sqrt {a+b x^2}} \, dx=\frac {\sqrt {c}\, c^{4} \left (-14 \sqrt {x}\, \sqrt {b \,x^{2}+a}\, a x +10 \sqrt {x}\, \sqrt {b \,x^{2}+a}\, b \,x^{3}+21 \left (\int \frac {\sqrt {x}\, \sqrt {b \,x^{2}+a}}{b \,x^{2}+a}d x \right ) a^{2}\right )}{45 b^{2}} \] Input:

int((c*x)^(9/2)/(b*x^2+a)^(1/2),x)
 

Output:

(sqrt(c)*c**4*( - 14*sqrt(x)*sqrt(a + b*x**2)*a*x + 10*sqrt(x)*sqrt(a + b* 
x**2)*b*x**3 + 21*int((sqrt(x)*sqrt(a + b*x**2))/(a + b*x**2),x)*a**2))/(4 
5*b**2)