\(\int \frac {\sqrt [4]{c x}}{(a+b x^2)^{5/2}} \, dx\) [695]

Optimal result

Integrand size = 19, antiderivative size = 1029 \[ \int \frac {\sqrt [4]{c x}}{\left (a+b x^2\right )^{5/2}} \, dx =\text {Too large to display} \] Output:

1/3*(c*x)^(5/4)/a/c/(b*x^2+a)^(3/2)+7/12*(c*x)^(5/4)/a^2/c/(b*x^2+a)^(1/2) 
+7/24*(2+2^(1/2))^(1/2)*(c*x)^(3/4)*(-(b*x^2+a)/a^(1/2)/b^(1/2)/x)^(1/2)*( 
(a^(1/4)*c^(1/2)+b^(1/4)*(c*x)^(1/2))^2/a^(1/4)/b^(1/4)/c^(1/2)/(c*x)^(1/2 
))^(1/2)*EllipticE(1/2*(-a^(1/4)*c^(1/2)*(2^(1/2)+2^(1/2)*b^(1/2)*x/a^(1/2 
)-2*b^(1/4)*(c*x)^(1/2)/a^(1/4)/c^(1/2))/b^(1/4)/(c*x)^(1/2))^(1/2),(-2+2* 
2^(1/2))^(1/2))/a^(3/2)/b^(1/4)/(b*x^2+a)^(1/2)/(a^(1/4)*c^(1/2)+b^(1/4)*( 
c*x)^(1/2))-7/24*(2+2^(1/2))^(1/2)*(c*x)^(3/4)*(-(b*x^2+a)/a^(1/2)/b^(1/2) 
/x)^(1/2)*(-(a^(1/4)*c^(1/2)-b^(1/4)*(c*x)^(1/2))^2/a^(1/4)/b^(1/4)/c^(1/2 
)/(c*x)^(1/2))^(1/2)*EllipticE(1/2*(a^(1/4)*c^(1/2)*(2^(1/2)+2^(1/2)*b^(1/ 
2)*x/a^(1/2)+2*b^(1/4)*(c*x)^(1/2)/a^(1/4)/c^(1/2))/b^(1/4)/(c*x)^(1/2))^( 
1/2),(-2+2*2^(1/2))^(1/2))/a^(3/2)/b^(1/4)/(b*x^2+a)^(1/2)/(a^(1/4)*c^(1/2 
)-b^(1/4)*(c*x)^(1/2))-7/24*(c*x)^(3/4)*(-(b*x^2+a)/a^(1/2)/b^(1/2)/x)^(1/ 
2)*((a^(1/4)*c^(1/2)+b^(1/4)*(c*x)^(1/2))^2/a^(1/4)/b^(1/4)/c^(1/2)/(c*x)^ 
(1/2))^(1/2)*EllipticF(1/2*(-a^(1/4)*c^(1/2)*(2^(1/2)+2^(1/2)*b^(1/2)*x/a^ 
(1/2)-2*b^(1/4)*(c*x)^(1/2)/a^(1/4)/c^(1/2))/b^(1/4)/(c*x)^(1/2))^(1/2),(- 
2+2*2^(1/2))^(1/2))/(2+2^(1/2))^(1/2)/a^(3/2)/b^(1/4)/(b*x^2+a)^(1/2)/(a^( 
1/4)*c^(1/2)+b^(1/4)*(c*x)^(1/2))+7/24*(c*x)^(3/4)*(-(b*x^2+a)/a^(1/2)/b^( 
1/2)/x)^(1/2)*(-(a^(1/4)*c^(1/2)-b^(1/4)*(c*x)^(1/2))^2/a^(1/4)/b^(1/4)/c^ 
(1/2)/(c*x)^(1/2))^(1/2)*EllipticF(1/2*(a^(1/4)*c^(1/2)*(2^(1/2)+2^(1/2)*b 
^(1/2)*x/a^(1/2)+2*b^(1/4)*(c*x)^(1/2)/a^(1/4)/c^(1/2))/b^(1/4)/(c*x)^(...
 

Mathematica [C] (verified)

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

Time = 10.02 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.06 \[ \int \frac {\sqrt [4]{c x}}{\left (a+b x^2\right )^{5/2}} \, dx=\frac {4 x \sqrt [4]{c x} \sqrt {1+\frac {b x^2}{a}} \operatorname {Hypergeometric2F1}\left (\frac {5}{8},\frac {5}{2},\frac {13}{8},-\frac {b x^2}{a}\right )}{5 a^2 \sqrt {a+b x^2}} \] Input:

Integrate[(c*x)^(1/4)/(a + b*x^2)^(5/2),x]
 

Output:

(4*x*(c*x)^(1/4)*Sqrt[1 + (b*x^2)/a]*Hypergeometric2F1[5/8, 5/2, 13/8, -(( 
b*x^2)/a)])/(5*a^2*Sqrt[a + b*x^2])
 

Rubi [C] (verified)

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

Time = 0.25 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.12, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {253, 253, 266, 889, 888}

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.

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

\(\Big \downarrow \) 253

\(\displaystyle \frac {7 \int \frac {\sqrt [4]{c x}}{\left (b x^2+a\right )^{3/2}}dx}{12 a}+\frac {(c x)^{5/4}}{3 a c \left (a+b x^2\right )^{3/2}}\)

\(\Big \downarrow \) 253

\(\displaystyle \frac {7 \left (\frac {(c x)^{5/4}}{a c \sqrt {a+b x^2}}-\frac {\int \frac {\sqrt [4]{c x}}{\sqrt {b x^2+a}}dx}{4 a}\right )}{12 a}+\frac {(c x)^{5/4}}{3 a c \left (a+b x^2\right )^{3/2}}\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {7 \left (\frac {(c x)^{5/4}}{a c \sqrt {a+b x^2}}-\frac {\int \frac {c x}{\sqrt {b x^2+a}}d\sqrt [4]{c x}}{a c}\right )}{12 a}+\frac {(c x)^{5/4}}{3 a c \left (a+b x^2\right )^{3/2}}\)

\(\Big \downarrow \) 889

\(\displaystyle \frac {7 \left (\frac {(c x)^{5/4}}{a c \sqrt {a+b x^2}}-\frac {\sqrt {\frac {b x^2}{a}+1} \int \frac {c x}{\sqrt {\frac {b x^2}{a}+1}}d\sqrt [4]{c x}}{a c \sqrt {a+b x^2}}\right )}{12 a}+\frac {(c x)^{5/4}}{3 a c \left (a+b x^2\right )^{3/2}}\)

\(\Big \downarrow \) 888

\(\displaystyle \frac {7 \left (\frac {(c x)^{5/4}}{a c \sqrt {a+b x^2}}-\frac {(c x)^{5/4} \sqrt {\frac {b x^2}{a}+1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{8},\frac {13}{8},-\frac {b x^2}{a}\right )}{5 a c \sqrt {a+b x^2}}\right )}{12 a}+\frac {(c x)^{5/4}}{3 a c \left (a+b x^2\right )^{3/2}}\)

Input:

Int[(c*x)^(1/4)/(a + b*x^2)^(5/2),x]
 

Output:

(c*x)^(5/4)/(3*a*c*(a + b*x^2)^(3/2)) + (7*((c*x)^(5/4)/(a*c*Sqrt[a + b*x^ 
2]) - ((c*x)^(5/4)*Sqrt[1 + (b*x^2)/a]*Hypergeometric2F1[1/2, 5/8, 13/8, - 
((b*x^2)/a)])/(5*a*c*Sqrt[a + b*x^2])))/(12*a)
 

Defintions of rubi rules used

rule 253

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-(c*x 
)^(m + 1))*((a + b*x^2)^(p + 1)/(2*a*c*(p + 1))), x] + Simp[(m + 2*p + 3)/( 
2*a*(p + 1))   Int[(c*x)^m*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, m 
}, x] && LtQ[p, -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 888

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p 
*((c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/n, (m + 1)/n + 1 
, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] && (ILt 
Q[p, 0] || GtQ[a, 0])
 

rule 889

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^I 
ntPart[p]*((a + b*x^n)^FracPart[p]/(1 + b*(x^n/a))^FracPart[p])   Int[(c*x) 
^m*(1 + b*(x^n/a))^p, x], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0 
] &&  !(ILtQ[p, 0] || GtQ[a, 0])
 

Maple [F]

\[\int \frac {\left (c x \right )^{\frac {1}{4}}}{\left (b \,x^{2}+a \right )^{\frac {5}{2}}}d x\]

Input:

int((c*x)^(1/4)/(b*x^2+a)^(5/2),x)
 

Output:

int((c*x)^(1/4)/(b*x^2+a)^(5/2),x)
 

Fricas [F]

\[ \int \frac {\sqrt [4]{c x}}{\left (a+b x^2\right )^{5/2}} \, dx=\int { \frac {\left (c x\right )^{\frac {1}{4}}}{{\left (b x^{2} + a\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate((c*x)^(1/4)/(b*x^2+a)^(5/2),x, algorithm="fricas")
 

Output:

integral(sqrt(b*x^2 + a)*(c*x)^(1/4)/(b^3*x^6 + 3*a*b^2*x^4 + 3*a^2*b*x^2 
+ a^3), x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.88 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.04 \[ \int \frac {\sqrt [4]{c x}}{\left (a+b x^2\right )^{5/2}} \, dx=\frac {\sqrt [4]{c} x^{\frac {5}{4}} \Gamma \left (\frac {5}{8}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{8}, \frac {5}{2} \\ \frac {13}{8} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {5}{2}} \Gamma \left (\frac {13}{8}\right )} \] Input:

integrate((c*x)**(1/4)/(b*x**2+a)**(5/2),x)
 

Output:

c**(1/4)*x**(5/4)*gamma(5/8)*hyper((5/8, 5/2), (13/8,), b*x**2*exp_polar(I 
*pi)/a)/(2*a**(5/2)*gamma(13/8))
 

Maxima [F]

\[ \int \frac {\sqrt [4]{c x}}{\left (a+b x^2\right )^{5/2}} \, dx=\int { \frac {\left (c x\right )^{\frac {1}{4}}}{{\left (b x^{2} + a\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate((c*x)^(1/4)/(b*x^2+a)^(5/2),x, algorithm="maxima")
 

Output:

integrate((c*x)^(1/4)/(b*x^2 + a)^(5/2), x)
 

Giac [F]

\[ \int \frac {\sqrt [4]{c x}}{\left (a+b x^2\right )^{5/2}} \, dx=\int { \frac {\left (c x\right )^{\frac {1}{4}}}{{\left (b x^{2} + a\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate((c*x)^(1/4)/(b*x^2+a)^(5/2),x, algorithm="giac")
 

Output:

integrate((c*x)^(1/4)/(b*x^2 + a)^(5/2), x)
                                                                                    
                                                                                    
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt [4]{c x}}{\left (a+b x^2\right )^{5/2}} \, dx=\int \frac {{\left (c\,x\right )}^{1/4}}{{\left (b\,x^2+a\right )}^{5/2}} \,d x \] Input:

int((c*x)^(1/4)/(a + b*x^2)^(5/2),x)
 

Output:

int((c*x)^(1/4)/(a + b*x^2)^(5/2), x)
 

Reduce [F]

\[ \int \frac {\sqrt [4]{c x}}{\left (a+b x^2\right )^{5/2}} \, dx=c^{\frac {1}{4}} \left (\int \frac {x^{\frac {1}{4}} \sqrt {b \,x^{2}+a}}{b^{3} x^{6}+3 a \,b^{2} x^{4}+3 a^{2} b \,x^{2}+a^{3}}d x \right ) \] Input:

int((c*x)^(1/4)/(b*x^2+a)^(5/2),x)
 

Output:

c**(1/4)*int((x**(1/4)*sqrt(a + b*x**2))/(a**3 + 3*a**2*b*x**2 + 3*a*b**2* 
x**4 + b**3*x**6),x)