\(\int \frac {(c x)^{9/2}}{\sqrt [4]{a+b x^2}} \, dx\) [1033]

Optimal result

Integrand size = 19, antiderivative size = 155 \[ \int \frac {(c x)^{9/2}}{\sqrt [4]{a+b x^2}} \, dx=\frac {7 a^2 c^3 (c x)^{3/2}}{20 b^2 \sqrt [4]{a+b x^2}}-\frac {7 a c^3 (c x)^{3/2} \left (a+b x^2\right )^{3/4}}{30 b^2}+\frac {c (c x)^{7/2} \left (a+b x^2\right )^{3/4}}{5 b}+\frac {7 a^{5/2} c^4 \sqrt [4]{1+\frac {a}{b x^2}} \sqrt {c x} E\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{20 b^{5/2} \sqrt [4]{a+b x^2}} \] Output:

7/20*a^2*c^3*(c*x)^(3/2)/b^2/(b*x^2+a)^(1/4)-7/30*a*c^3*(c*x)^(3/2)*(b*x^2 
+a)^(3/4)/b^2+1/5*c*(c*x)^(7/2)*(b*x^2+a)^(3/4)/b+7/20*a^(5/2)*c^4*(1+a/b/ 
x^2)^(1/4)*(c*x)^(1/2)*EllipticE(sin(1/2*arccot(b^(1/2)*x/a^(1/2))),2^(1/2 
))/b^(5/2)/(b*x^2+a)^(1/4)
                                                                                    
                                                                                    
 

Mathematica [C] (verified)

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

Time = 10.02 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.56 \[ \int \frac {(c x)^{9/2}}{\sqrt [4]{a+b x^2}} \, dx=\frac {c^3 (c x)^{3/2} \left (-7 a^2-a b x^2+6 b^2 x^4+7 a^2 \sqrt [4]{1+\frac {b x^2}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {3}{4},\frac {7}{4},-\frac {b x^2}{a}\right )\right )}{30 b^2 \sqrt [4]{a+b x^2}} \] Input:

Integrate[(c*x)^(9/2)/(a + b*x^2)^(1/4),x]
 

Output:

(c^3*(c*x)^(3/2)*(-7*a^2 - a*b*x^2 + 6*b^2*x^4 + 7*a^2*(1 + (b*x^2)/a)^(1/ 
4)*Hypergeometric2F1[1/4, 3/4, 7/4, -((b*x^2)/a)]))/(30*b^2*(a + b*x^2)^(1 
/4))
 

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.04, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {262, 262, 255, 249, 858, 212}

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.

\(\displaystyle \int \frac {(c x)^{9/2}}{\sqrt [4]{a+b x^2}} \, dx\)

\(\Big \downarrow \) 262

\(\displaystyle \frac {c (c x)^{7/2} \left (a+b x^2\right )^{3/4}}{5 b}-\frac {7 a c^2 \int \frac {(c x)^{5/2}}{\sqrt [4]{b x^2+a}}dx}{10 b}\)

\(\Big \downarrow \) 262

\(\displaystyle \frac {c (c x)^{7/2} \left (a+b x^2\right )^{3/4}}{5 b}-\frac {7 a c^2 \left (\frac {c (c x)^{3/2} \left (a+b x^2\right )^{3/4}}{3 b}-\frac {a c^2 \int \frac {\sqrt {c x}}{\sqrt [4]{b x^2+a}}dx}{2 b}\right )}{10 b}\)

\(\Big \downarrow \) 255

\(\displaystyle \frac {c (c x)^{7/2} \left (a+b x^2\right )^{3/4}}{5 b}-\frac {7 a c^2 \left (\frac {c (c x)^{3/2} \left (a+b x^2\right )^{3/4}}{3 b}-\frac {a c^2 \left (\frac {x \sqrt {c x}}{\sqrt [4]{a+b x^2}}-\frac {1}{2} a \int \frac {\sqrt {c x}}{\left (b x^2+a\right )^{5/4}}dx\right )}{2 b}\right )}{10 b}\)

\(\Big \downarrow \) 249

\(\displaystyle \frac {c (c x)^{7/2} \left (a+b x^2\right )^{3/4}}{5 b}-\frac {7 a c^2 \left (\frac {c (c x)^{3/2} \left (a+b x^2\right )^{3/4}}{3 b}-\frac {a c^2 \left (\frac {x \sqrt {c x}}{\sqrt [4]{a+b x^2}}-\frac {a \sqrt {c x} \sqrt [4]{\frac {a}{b x^2}+1} \int \frac {1}{\left (\frac {a}{b x^2}+1\right )^{5/4} x^2}dx}{2 b \sqrt [4]{a+b x^2}}\right )}{2 b}\right )}{10 b}\)

\(\Big \downarrow \) 858

\(\displaystyle \frac {c (c x)^{7/2} \left (a+b x^2\right )^{3/4}}{5 b}-\frac {7 a c^2 \left (\frac {c (c x)^{3/2} \left (a+b x^2\right )^{3/4}}{3 b}-\frac {a c^2 \left (\frac {a \sqrt {c x} \sqrt [4]{\frac {a}{b x^2}+1} \int \frac {1}{\left (\frac {a}{b x^2}+1\right )^{5/4}}d\frac {1}{x}}{2 b \sqrt [4]{a+b x^2}}+\frac {x \sqrt {c x}}{\sqrt [4]{a+b x^2}}\right )}{2 b}\right )}{10 b}\)

\(\Big \downarrow \) 212

\(\displaystyle \frac {c (c x)^{7/2} \left (a+b x^2\right )^{3/4}}{5 b}-\frac {7 a c^2 \left (\frac {c (c x)^{3/2} \left (a+b x^2\right )^{3/4}}{3 b}-\frac {a c^2 \left (\frac {\sqrt {a} \sqrt {c x} \sqrt [4]{\frac {a}{b x^2}+1} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {a}}{\sqrt {b} x}\right )\right |2\right )}{\sqrt {b} \sqrt [4]{a+b x^2}}+\frac {x \sqrt {c x}}{\sqrt [4]{a+b x^2}}\right )}{2 b}\right )}{10 b}\)

Input:

Int[(c*x)^(9/2)/(a + b*x^2)^(1/4),x]
 

Output:

(c*(c*x)^(7/2)*(a + b*x^2)^(3/4))/(5*b) - (7*a*c^2*((c*(c*x)^(3/2)*(a + b* 
x^2)^(3/4))/(3*b) - (a*c^2*((x*Sqrt[c*x])/(a + b*x^2)^(1/4) + (Sqrt[a]*(1 
+ a/(b*x^2))^(1/4)*Sqrt[c*x]*EllipticE[ArcTan[Sqrt[a]/(Sqrt[b]*x)]/2, 2])/ 
(Sqrt[b]*(a + b*x^2)^(1/4))))/(2*b)))/(10*b)
 

Defintions of rubi rules used

rule 212

Int[((a_) + (b_.)*(x_)^2)^(-5/4), x_Symbol] :> Simp[(2/(a^(5/4)*Rt[b/a, 2]) 
)*EllipticE[(1/2)*ArcTan[Rt[b/a, 2]*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a 
, 0] && PosQ[b/a]
 

rule 249

Int[Sqrt[(c_.)*(x_)]/((a_) + (b_.)*(x_)^2)^(5/4), x_Symbol] :> Simp[Sqrt[c* 
x]*((1 + a/(b*x^2))^(1/4)/(b*(a + b*x^2)^(1/4)))   Int[1/(x^2*(1 + a/(b*x^2 
))^(5/4)), x], x] /; FreeQ[{a, b, c}, x] && PosQ[b/a]
 

rule 255

Int[Sqrt[(c_)*(x_)]/((a_) + (b_.)*(x_)^2)^(1/4), x_Symbol] :> Simp[x*(Sqrt[ 
c*x]/(a + b*x^2)^(1/4)), x] - Simp[a/2   Int[Sqrt[c*x]/(a + b*x^2)^(5/4), x 
], x] /; FreeQ[{a, b, c}, x] && 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 858

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + 
b/x^n)^p/x^(m + 2), x], x, 1/x] /; FreeQ[{a, b, p}, x] && ILtQ[n, 0] && Int 
egerQ[m]
 

Maple [F]

\[\int \frac {\left (c x \right )^{\frac {9}{2}}}{\left (b \,x^{2}+a \right )^{\frac {1}{4}}}d x\]

Input:

int((c*x)^(9/2)/(b*x^2+a)^(1/4),x)
 

Output:

int((c*x)^(9/2)/(b*x^2+a)^(1/4),x)
 

Fricas [F]

\[ \int \frac {(c x)^{9/2}}{\sqrt [4]{a+b x^2}} \, dx=\int { \frac {\left (c x\right )^{\frac {9}{2}}}{{\left (b x^{2} + a\right )}^{\frac {1}{4}}} \,d x } \] Input:

integrate((c*x)^(9/2)/(b*x^2+a)^(1/4),x, algorithm="fricas")
 

Output:

integral(sqrt(c*x)*c^4*x^4/(b*x^2 + a)^(1/4), x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 27.61 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.28 \[ \int \frac {(c x)^{9/2}}{\sqrt [4]{a+b x^2}} \, dx=\frac {c^{\frac {9}{2}} x^{\frac {11}{2}} \Gamma \left (\frac {11}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {11}{4} \\ \frac {15}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \sqrt [4]{a} \Gamma \left (\frac {15}{4}\right )} \] Input:

integrate((c*x)**(9/2)/(b*x**2+a)**(1/4),x)
 

Output:

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

Maxima [F]

\[ \int \frac {(c x)^{9/2}}{\sqrt [4]{a+b x^2}} \, dx=\int { \frac {\left (c x\right )^{\frac {9}{2}}}{{\left (b x^{2} + a\right )}^{\frac {1}{4}}} \,d x } \] Input:

integrate((c*x)^(9/2)/(b*x^2+a)^(1/4),x, algorithm="maxima")
 

Output:

integrate((c*x)^(9/2)/(b*x^2 + a)^(1/4), x)
 

Giac [F]

\[ \int \frac {(c x)^{9/2}}{\sqrt [4]{a+b x^2}} \, dx=\int { \frac {\left (c x\right )^{\frac {9}{2}}}{{\left (b x^{2} + a\right )}^{\frac {1}{4}}} \,d x } \] Input:

integrate((c*x)^(9/2)/(b*x^2+a)^(1/4),x, algorithm="giac")
                                                                                    
                                                                                    
 

Output:

integrate((c*x)^(9/2)/(b*x^2 + a)^(1/4), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(c x)^{9/2}}{\sqrt [4]{a+b x^2}} \, dx=\int \frac {{\left (c\,x\right )}^{9/2}}{{\left (b\,x^2+a\right )}^{1/4}} \,d x \] Input:

int((c*x)^(9/2)/(a + b*x^2)^(1/4),x)
 

Output:

int((c*x)^(9/2)/(a + b*x^2)^(1/4), x)
 

Reduce [F]

\[ \int \frac {(c x)^{9/2}}{\sqrt [4]{a+b x^2}} \, dx=\sqrt {c}\, \left (\int \frac {\sqrt {x}\, x^{4}}{\left (b \,x^{2}+a \right )^{\frac {1}{4}}}d x \right ) c^{4} \] Input:

int((c*x)^(9/2)/(b*x^2+a)^(1/4),x)
 

Output:

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