\(\int \frac {\sqrt [4]{a-b x^2}}{(c x)^{13/2}} \, dx\) [994]

Optimal result

Integrand size = 20, antiderivative size = 159 \[ \int \frac {\sqrt [4]{a-b x^2}}{(c x)^{13/2}} \, dx=-\frac {2 \sqrt [4]{a-b x^2}}{11 c (c x)^{11/2}}+\frac {2 b \sqrt [4]{a-b x^2}}{77 a c^3 (c x)^{7/2}}+\frac {4 b^2 \sqrt [4]{a-b x^2}}{77 a^2 c^5 (c x)^{3/2}}+\frac {8 b^{7/2} \left (1-\frac {a}{b x^2}\right )^{3/4} (c x)^{3/2} \operatorname {EllipticF}\left (\frac {1}{2} \csc ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),2\right )}{77 a^{5/2} c^8 \left (a-b x^2\right )^{3/4}} \] Output:

-2/11*(-b*x^2+a)^(1/4)/c/(c*x)^(11/2)+2/77*b*(-b*x^2+a)^(1/4)/a/c^3/(c*x)^ 
(7/2)+4/77*b^2*(-b*x^2+a)^(1/4)/a^2/c^5/(c*x)^(3/2)+8/77*b^(7/2)*(1-a/b/x^ 
2)^(3/4)*(c*x)^(3/2)*InverseJacobiAM(1/2*arccsc(b^(1/2)*x/a^(1/2)),2^(1/2) 
)/a^(5/2)/c^8/(-b*x^2+a)^(3/4)
 

Mathematica [C] (verified)

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

Time = 10.01 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.36 \[ \int \frac {\sqrt [4]{a-b x^2}}{(c x)^{13/2}} \, dx=-\frac {2 x \sqrt [4]{a-b x^2} \operatorname {Hypergeometric2F1}\left (-\frac {11}{4},-\frac {1}{4},-\frac {7}{4},\frac {b x^2}{a}\right )}{11 (c x)^{13/2} \sqrt [4]{1-\frac {b x^2}{a}}} \] Input:

Integrate[(a - b*x^2)^(1/4)/(c*x)^(13/2),x]
 

Output:

(-2*x*(a - b*x^2)^(1/4)*Hypergeometric2F1[-11/4, -1/4, -7/4, (b*x^2)/a])/( 
11*(c*x)^(13/2)*(1 - (b*x^2)/a)^(1/4))
 

Rubi [A] (warning: unable to verify)

Time = 0.33 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.13, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {247, 264, 264, 266, 768, 858, 807, 230}

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.

\(\displaystyle \int \frac {\sqrt [4]{a-b x^2}}{(c x)^{13/2}} \, dx\)

\(\Big \downarrow \) 247

\(\displaystyle -\frac {b \int \frac {1}{(c x)^{9/2} \left (a-b x^2\right )^{3/4}}dx}{11 c^2}-\frac {2 \sqrt [4]{a-b x^2}}{11 c (c x)^{11/2}}\)

\(\Big \downarrow \) 264

\(\displaystyle -\frac {b \left (\frac {6 b \int \frac {1}{(c x)^{5/2} \left (a-b x^2\right )^{3/4}}dx}{7 a c^2}-\frac {2 \sqrt [4]{a-b x^2}}{7 a c (c x)^{7/2}}\right )}{11 c^2}-\frac {2 \sqrt [4]{a-b x^2}}{11 c (c x)^{11/2}}\)

\(\Big \downarrow \) 264

\(\displaystyle -\frac {b \left (\frac {6 b \left (\frac {2 b \int \frac {1}{\sqrt {c x} \left (a-b x^2\right )^{3/4}}dx}{3 a c^2}-\frac {2 \sqrt [4]{a-b x^2}}{3 a c (c x)^{3/2}}\right )}{7 a c^2}-\frac {2 \sqrt [4]{a-b x^2}}{7 a c (c x)^{7/2}}\right )}{11 c^2}-\frac {2 \sqrt [4]{a-b x^2}}{11 c (c x)^{11/2}}\)

\(\Big \downarrow \) 266

\(\displaystyle -\frac {b \left (\frac {6 b \left (\frac {4 b \int \frac {1}{\left (a-b x^2\right )^{3/4}}d\sqrt {c x}}{3 a c^3}-\frac {2 \sqrt [4]{a-b x^2}}{3 a c (c x)^{3/2}}\right )}{7 a c^2}-\frac {2 \sqrt [4]{a-b x^2}}{7 a c (c x)^{7/2}}\right )}{11 c^2}-\frac {2 \sqrt [4]{a-b x^2}}{11 c (c x)^{11/2}}\)

\(\Big \downarrow \) 768

\(\displaystyle -\frac {b \left (\frac {6 b \left (\frac {4 b (c x)^{3/2} \left (1-\frac {a}{b x^2}\right )^{3/4} \int \frac {1}{\left (1-\frac {a}{b x^2}\right )^{3/4} (c x)^{3/2}}d\sqrt {c x}}{3 a c^3 \left (a-b x^2\right )^{3/4}}-\frac {2 \sqrt [4]{a-b x^2}}{3 a c (c x)^{3/2}}\right )}{7 a c^2}-\frac {2 \sqrt [4]{a-b x^2}}{7 a c (c x)^{7/2}}\right )}{11 c^2}-\frac {2 \sqrt [4]{a-b x^2}}{11 c (c x)^{11/2}}\)

\(\Big \downarrow \) 858

\(\displaystyle -\frac {b \left (\frac {6 b \left (-\frac {4 b (c x)^{3/2} \left (1-\frac {a}{b x^2}\right )^{3/4} \int \frac {1}{\sqrt {c x} \left (1-\frac {a c^4 x^2}{b}\right )^{3/4}}d\frac {1}{\sqrt {c x}}}{3 a c^3 \left (a-b x^2\right )^{3/4}}-\frac {2 \sqrt [4]{a-b x^2}}{3 a c (c x)^{3/2}}\right )}{7 a c^2}-\frac {2 \sqrt [4]{a-b x^2}}{7 a c (c x)^{7/2}}\right )}{11 c^2}-\frac {2 \sqrt [4]{a-b x^2}}{11 c (c x)^{11/2}}\)

\(\Big \downarrow \) 807

\(\displaystyle -\frac {b \left (\frac {6 b \left (-\frac {2 b (c x)^{3/2} \left (1-\frac {a}{b x^2}\right )^{3/4} \int \frac {1}{\left (1-\frac {a c^3 x}{b}\right )^{3/4}}d(c x)}{3 a c^3 \left (a-b x^2\right )^{3/4}}-\frac {2 \sqrt [4]{a-b x^2}}{3 a c (c x)^{3/2}}\right )}{7 a c^2}-\frac {2 \sqrt [4]{a-b x^2}}{7 a c (c x)^{7/2}}\right )}{11 c^2}-\frac {2 \sqrt [4]{a-b x^2}}{11 c (c x)^{11/2}}\)

\(\Big \downarrow \) 230

\(\displaystyle -\frac {b \left (\frac {6 b \left (-\frac {4 b^{3/2} (c x)^{3/2} \left (1-\frac {a}{b x^2}\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arcsin \left (\frac {\sqrt {a} c^2 x}{\sqrt {b}}\right ),2\right )}{3 a^{3/2} c^4 \left (a-b x^2\right )^{3/4}}-\frac {2 \sqrt [4]{a-b x^2}}{3 a c (c x)^{3/2}}\right )}{7 a c^2}-\frac {2 \sqrt [4]{a-b x^2}}{7 a c (c x)^{7/2}}\right )}{11 c^2}-\frac {2 \sqrt [4]{a-b x^2}}{11 c (c x)^{11/2}}\)

Input:

Int[(a - b*x^2)^(1/4)/(c*x)^(13/2),x]
 

Output:

(-2*(a - b*x^2)^(1/4))/(11*c*(c*x)^(11/2)) - (b*((-2*(a - b*x^2)^(1/4))/(7 
*a*c*(c*x)^(7/2)) + (6*b*((-2*(a - b*x^2)^(1/4))/(3*a*c*(c*x)^(3/2)) - (4* 
b^(3/2)*(1 - a/(b*x^2))^(3/4)*(c*x)^(3/2)*EllipticF[ArcSin[(Sqrt[a]*c^2*x) 
/Sqrt[b]]/2, 2])/(3*a^(3/2)*c^4*(a - b*x^2)^(3/4))))/(7*a*c^2)))/(11*c^2)
 

Defintions of rubi rules used

rule 230

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

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 768

Int[((a_) + (b_.)*(x_)^4)^(-3/4), x_Symbol] :> Simp[x^3*((1 + a/(b*x^4))^(3 
/4)/(a + b*x^4)^(3/4))   Int[1/(x^3*(1 + a/(b*x^4))^(3/4)), x], x] /; FreeQ 
[{a, b}, x]
 

rule 807

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m 
+ 1, n]}, Simp[1/k   Subst[Int[x^((m + 1)/k - 1)*(a + b*x^(n/k))^p, x], x, 
x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]
 

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]

\[\int \frac {\left (-b \,x^{2}+a \right )^{\frac {1}{4}}}{\left (c x \right )^{\frac {13}{2}}}d x\]

Input:

int((-b*x^2+a)^(1/4)/(c*x)^(13/2),x)
 

Output:

int((-b*x^2+a)^(1/4)/(c*x)^(13/2),x)
 

Fricas [F]

\[ \int \frac {\sqrt [4]{a-b x^2}}{(c x)^{13/2}} \, dx=\int { \frac {{\left (-b x^{2} + a\right )}^{\frac {1}{4}}}{\left (c x\right )^{\frac {13}{2}}} \,d x } \] Input:

integrate((-b*x^2+a)^(1/4)/(c*x)^(13/2),x, algorithm="fricas")
 

Output:

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 139.76 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.25 \[ \int \frac {\sqrt [4]{a-b x^2}}{(c x)^{13/2}} \, dx=- \frac {i \sqrt [4]{b} e^{- \frac {i \pi }{4}} {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {5}{2} \\ \frac {7}{2} \end {matrix}\middle | {\frac {a}{b x^{2}}} \right )}}{5 c^{\frac {13}{2}} x^{5}} \] Input:

integrate((-b*x**2+a)**(1/4)/(c*x)**(13/2),x)
 

Output:

-I*b**(1/4)*exp(-I*pi/4)*hyper((-1/4, 5/2), (7/2,), a/(b*x**2))/(5*c**(13/ 
2)*x**5)
 

Maxima [F]

\[ \int \frac {\sqrt [4]{a-b x^2}}{(c x)^{13/2}} \, dx=\int { \frac {{\left (-b x^{2} + a\right )}^{\frac {1}{4}}}{\left (c x\right )^{\frac {13}{2}}} \,d x } \] Input:

integrate((-b*x^2+a)^(1/4)/(c*x)^(13/2),x, algorithm="maxima")
 

Output:

integrate((-b*x^2 + a)^(1/4)/(c*x)^(13/2), x)
 

Giac [F]

\[ \int \frac {\sqrt [4]{a-b x^2}}{(c x)^{13/2}} \, dx=\int { \frac {{\left (-b x^{2} + a\right )}^{\frac {1}{4}}}{\left (c x\right )^{\frac {13}{2}}} \,d x } \] Input:

integrate((-b*x^2+a)^(1/4)/(c*x)^(13/2),x, algorithm="giac")
 

Output:

integrate((-b*x^2 + a)^(1/4)/(c*x)^(13/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt [4]{a-b x^2}}{(c x)^{13/2}} \, dx=\int \frac {{\left (a-b\,x^2\right )}^{1/4}}{{\left (c\,x\right )}^{13/2}} \,d x \] Input:

int((a - b*x^2)^(1/4)/(c*x)^(13/2),x)
 

Output:

int((a - b*x^2)^(1/4)/(c*x)^(13/2), x)
 

Reduce [F]

\[ \int \frac {\sqrt [4]{a-b x^2}}{(c x)^{13/2}} \, dx=\frac {\sqrt {c}\, \left (-2 \left (-b \,x^{2}+a \right )^{\frac {1}{4}}-\sqrt {x}\, \left (\int \frac {\sqrt {x}\, \left (-b \,x^{2}+a \right )^{\frac {1}{4}}}{-b \,x^{9}+a \,x^{7}}d x \right ) a \,x^{5}\right )}{10 \sqrt {x}\, c^{7} x^{5}} \] Input:

int((-b*x^2+a)^(1/4)/(c*x)^(13/2),x)
 

Output:

(sqrt(c)*( - 2*(a - b*x**2)**(1/4) - sqrt(x)*int((sqrt(x)*(a - b*x**2)**(1 
/4))/(a*x**7 - b*x**9),x)*a*x**5))/(10*sqrt(x)*c**7*x**5)