\(\int x^4 (\frac {c}{\sqrt {a+b x^2}})^{3/2} \, dx\) [55]

Optimal result

Integrand size = 21, antiderivative size = 132 \[ \int x^4 \left (\frac {c}{\sqrt {a+b x^2}}\right )^{3/2} \, dx=-\frac {4 a x \left (\frac {c}{\sqrt {a+b x^2}}\right )^{3/2} \left (a+b x^2\right )}{7 b^2}+\frac {2 x^3 \left (\frac {c}{\sqrt {a+b x^2}}\right )^{3/2} \left (a+b x^2\right )}{7 b}+\frac {8 a^{5/2} \left (\frac {c}{\sqrt {a+b x^2}}\right )^{3/2} \left (1+\frac {b x^2}{a}\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),2\right )}{7 b^{5/2}} \] Output:

-4/7*a*x*(c/(b*x^2+a)^(1/2))^(3/2)*(b*x^2+a)/b^2+2/7*x^3*(c/(b*x^2+a)^(1/2 
))^(3/2)*(b*x^2+a)/b+8/7*a^(5/2)*(c/(b*x^2+a)^(1/2))^(3/2)*(1+b*x^2/a)^(3/ 
4)*InverseJacobiAM(1/2*arctan(b^(1/2)*x/a^(1/2)),2^(1/2))/b^(5/2)
 

Mathematica [C] (verified)

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

Time = 6.37 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.63 \[ \int x^4 \left (\frac {c}{\sqrt {a+b x^2}}\right )^{3/2} \, dx=\frac {2 x \left (\frac {c}{\sqrt {a+b x^2}}\right )^{3/2} \left (-2 a^2-a b x^2+b^2 x^4+2 a^2 \left (1+\frac {b x^2}{a}\right )^{3/4} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {3}{2},-\frac {b x^2}{a}\right )\right )}{7 b^2} \] Input:

Integrate[x^4*(c/Sqrt[a + b*x^2])^(3/2),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.93, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {2045, 262, 262, 229}

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.

\(\displaystyle \int x^4 \left (\frac {c}{\sqrt {a+b x^2}}\right )^{3/2} \, dx\)

\(\Big \downarrow \) 2045

\(\displaystyle \left (\frac {b x^2}{a}+1\right )^{3/4} \left (\frac {c}{\sqrt {a+b x^2}}\right )^{3/2} \int \frac {x^4}{\left (\frac {b x^2}{a}+1\right )^{3/4}}dx\)

\(\Big \downarrow \) 262

\(\displaystyle \left (\frac {b x^2}{a}+1\right )^{3/4} \left (\frac {c}{\sqrt {a+b x^2}}\right )^{3/2} \left (\frac {2 a x^3 \sqrt [4]{\frac {b x^2}{a}+1}}{7 b}-\frac {6 a \int \frac {x^2}{\left (\frac {b x^2}{a}+1\right )^{3/4}}dx}{7 b}\right )\)

\(\Big \downarrow \) 262

\(\displaystyle \left (\frac {b x^2}{a}+1\right )^{3/4} \left (\frac {c}{\sqrt {a+b x^2}}\right )^{3/2} \left (\frac {2 a x^3 \sqrt [4]{\frac {b x^2}{a}+1}}{7 b}-\frac {6 a \left (\frac {2 a x \sqrt [4]{\frac {b x^2}{a}+1}}{3 b}-\frac {2 a \int \frac {1}{\left (\frac {b x^2}{a}+1\right )^{3/4}}dx}{3 b}\right )}{7 b}\right )\)

\(\Big \downarrow \) 229

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

Input:

Int[x^4*(c/Sqrt[a + b*x^2])^(3/2),x]
 

Output:

(c/Sqrt[a + b*x^2])^(3/2)*(1 + (b*x^2)/a)^(3/4)*((2*a*x^3*(1 + (b*x^2)/a)^ 
(1/4))/(7*b) - (6*a*((2*a*x*(1 + (b*x^2)/a)^(1/4))/(3*b) - (4*a^(3/2)*Elli 
pticF[ArcTan[(Sqrt[b]*x)/Sqrt[a]]/2, 2])/(3*b^(3/2))))/(7*b))
 

Defintions of rubi rules used

rule 229

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[(2/(a^(3/4)*Rt[b/a, 2]) 
)*EllipticF[(1/2)*ArcTan[Rt[b/a, 2]*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a 
, 0] && PosQ[b/a]
 

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 2045

Int[(u_.)*((c_.)*((a_) + (b_.)*(x_)^(n_.))^(q_))^(p_), x_Symbol] :> Simp[Si 
mp[(c*(a + b*x^n)^q)^p/(1 + b*(x^n/a))^(p*q)]   Int[u*(1 + b*(x^n/a))^(p*q) 
, x], x] /; FreeQ[{a, b, c, n, p, q}, x] &&  !GeQ[a, 0]
 

Maple [F]

\[\int x^{4} {\left (\frac {c}{\sqrt {b \,x^{2}+a}}\right )}^{\frac {3}{2}}d x\]

Input:

int(x^4*(c/(b*x^2+a)^(1/2))^(3/2),x)
 

Output:

int(x^4*(c/(b*x^2+a)^(1/2))^(3/2),x)
 

Fricas [F]

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

integrate(x^4*(c/(b*x^2+a)^(1/2))^(3/2),x, algorithm="fricas")
 

Output:

integral(c*x^4*sqrt(c/sqrt(b*x^2 + a))/sqrt(b*x^2 + a), x)
 

Sympy [F]

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

integrate(x**4*(c/(b*x**2+a)**(1/2))**(3/2),x)
 

Output:

Integral(x**4*(c/sqrt(a + b*x**2))**(3/2), x)
 

Maxima [F]

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

integrate(x^4*(c/(b*x^2+a)^(1/2))^(3/2),x, algorithm="maxima")
 

Output:

integrate(x^4*(c/sqrt(b*x^2 + a))^(3/2), x)
 

Giac [F]

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

integrate(x^4*(c/(b*x^2+a)^(1/2))^(3/2),x, algorithm="giac")
 

Output:

integrate(x^4*(c/sqrt(b*x^2 + a))^(3/2), x)
 

Mupad [F(-1)]

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

int(x^4*(c/(a + b*x^2)^(1/2))^(3/2),x)
 

Output:

int(x^4*(c/(a + b*x^2)^(1/2))^(3/2), x)
 

Reduce [F]

\[ \int x^4 \left (\frac {c}{\sqrt {a+b x^2}}\right )^{3/2} \, dx=\frac {2 \sqrt {c}\, c \left (-2 \left (b \,x^{2}+a \right )^{\frac {1}{4}} a x +\left (b \,x^{2}+a \right )^{\frac {1}{4}} b \,x^{3}+2 \left (\int \frac {1}{\left (b \,x^{2}+a \right )^{\frac {3}{4}}}d x \right ) a^{2}\right )}{7 b^{2}} \] Input:

int(x^4*(c/(b*x^2+a)^(1/2))^(3/2),x)
 

Output:

(2*sqrt(c)*c*( - 2*(a + b*x**2)**(1/4)*a*x + (a + b*x**2)**(1/4)*b*x**3 + 
2*int((a + b*x**2)**(1/4)/(a + b*x**2),x)*a**2))/(7*b**2)