\(\int \frac {(a+b x^4)^{3/4}}{x^{10}} \, dx\) [500]

Optimal result

Integrand size = 15, antiderivative size = 126 \[ \int \frac {\left (a+b x^4\right )^{3/4}}{x^{10}} \, dx=\frac {2 b^2}{15 a x \sqrt [4]{a+b x^4}}-\frac {\left (a+b x^4\right )^{3/4}}{9 x^9}-\frac {b \left (a+b x^4\right )^{3/4}}{15 a x^5}-\frac {2 b^{5/2} \sqrt [4]{1+\frac {a}{b x^4}} x E\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{15 a^{3/2} \sqrt [4]{a+b x^4}} \] Output:

2/15*b^2/a/x/(b*x^4+a)^(1/4)-1/9*(b*x^4+a)^(3/4)/x^9-1/15*b*(b*x^4+a)^(3/4 
)/a/x^5-2/15*b^(5/2)*(1+a/b/x^4)^(1/4)*x*EllipticE(sin(1/2*arccot(b^(1/2)* 
x^2/a^(1/2))),2^(1/2))/a^(3/2)/(b*x^4+a)^(1/4)
 

Mathematica [C] (verified)

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

Time = 10.01 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.40 \[ \int \frac {\left (a+b x^4\right )^{3/4}}{x^{10}} \, dx=-\frac {\left (a+b x^4\right )^{3/4} \operatorname {Hypergeometric2F1}\left (-\frac {9}{4},-\frac {3}{4},-\frac {5}{4},-\frac {b x^4}{a}\right )}{9 x^9 \left (1+\frac {b x^4}{a}\right )^{3/4}} \] Input:

Integrate[(a + b*x^4)^(3/4)/x^10,x]
 

Output:

-1/9*((a + b*x^4)^(3/4)*Hypergeometric2F1[-9/4, -3/4, -5/4, -((b*x^4)/a)]) 
/(x^9*(1 + (b*x^4)/a)^(3/4))
 

Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.02, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.467, Rules used = {809, 847, 841, 813, 858, 807, 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.

\(\displaystyle \int \frac {\left (a+b x^4\right )^{3/4}}{x^{10}} \, dx\)

\(\Big \downarrow \) 809

\(\displaystyle \frac {1}{3} b \int \frac {1}{x^6 \sqrt [4]{b x^4+a}}dx-\frac {\left (a+b x^4\right )^{3/4}}{9 x^9}\)

\(\Big \downarrow \) 847

\(\displaystyle \frac {1}{3} b \left (-\frac {2 b \int \frac {1}{x^2 \sqrt [4]{b x^4+a}}dx}{5 a}-\frac {\left (a+b x^4\right )^{3/4}}{5 a x^5}\right )-\frac {\left (a+b x^4\right )^{3/4}}{9 x^9}\)

\(\Big \downarrow \) 841

\(\displaystyle \frac {1}{3} b \left (-\frac {2 b \left (-b \int \frac {x^2}{\left (b x^4+a\right )^{5/4}}dx-\frac {1}{x \sqrt [4]{a+b x^4}}\right )}{5 a}-\frac {\left (a+b x^4\right )^{3/4}}{5 a x^5}\right )-\frac {\left (a+b x^4\right )^{3/4}}{9 x^9}\)

\(\Big \downarrow \) 813

\(\displaystyle \frac {1}{3} b \left (-\frac {2 b \left (-\frac {x \sqrt [4]{\frac {a}{b x^4}+1} \int \frac {1}{\left (\frac {a}{b x^4}+1\right )^{5/4} x^3}dx}{\sqrt [4]{a+b x^4}}-\frac {1}{x \sqrt [4]{a+b x^4}}\right )}{5 a}-\frac {\left (a+b x^4\right )^{3/4}}{5 a x^5}\right )-\frac {\left (a+b x^4\right )^{3/4}}{9 x^9}\)

\(\Big \downarrow \) 858

\(\displaystyle \frac {1}{3} b \left (-\frac {2 b \left (\frac {x \sqrt [4]{\frac {a}{b x^4}+1} \int \frac {1}{\left (\frac {a}{b x^4}+1\right )^{5/4} x}d\frac {1}{x}}{\sqrt [4]{a+b x^4}}-\frac {1}{x \sqrt [4]{a+b x^4}}\right )}{5 a}-\frac {\left (a+b x^4\right )^{3/4}}{5 a x^5}\right )-\frac {\left (a+b x^4\right )^{3/4}}{9 x^9}\)

\(\Big \downarrow \) 807

\(\displaystyle \frac {1}{3} b \left (-\frac {2 b \left (\frac {x \sqrt [4]{\frac {a}{b x^4}+1} \int \frac {1}{\left (\frac {a}{b x^2}+1\right )^{5/4}}d\frac {1}{x^2}}{2 \sqrt [4]{a+b x^4}}-\frac {1}{x \sqrt [4]{a+b x^4}}\right )}{5 a}-\frac {\left (a+b x^4\right )^{3/4}}{5 a x^5}\right )-\frac {\left (a+b x^4\right )^{3/4}}{9 x^9}\)

\(\Big \downarrow \) 212

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

Input:

Int[(a + b*x^4)^(3/4)/x^10,x]
 

Output:

-1/9*(a + b*x^4)^(3/4)/x^9 + (b*(-1/5*(a + b*x^4)^(3/4)/(a*x^5) - (2*b*(-( 
1/(x*(a + b*x^4)^(1/4))) + (Sqrt[b]*(1 + a/(b*x^4))^(1/4)*x*EllipticE[ArcT 
an[Sqrt[a]/(Sqrt[b]*x^2)]/2, 2])/(Sqrt[a]*(a + b*x^4)^(1/4))))/(5*a)))/3
 

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 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 809

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

rule 813

Int[(x_)^2/((a_) + (b_.)*(x_)^4)^(5/4), x_Symbol] :> Simp[x*((1 + a/(b*x^4) 
)^(1/4)/(b*(a + b*x^4)^(1/4)))   Int[1/(x^3*(1 + a/(b*x^4))^(5/4)), x], x] 
/; FreeQ[{a, b}, x] && PosQ[b/a]
 

rule 841

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

rule 847

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

\[\int \frac {\left (b \,x^{4}+a \right )^{\frac {3}{4}}}{x^{10}}d x\]

Input:

int((b*x^4+a)^(3/4)/x^10,x)
 

Output:

int((b*x^4+a)^(3/4)/x^10,x)
 

Fricas [F]

\[ \int \frac {\left (a+b x^4\right )^{3/4}}{x^{10}} \, dx=\int { \frac {{\left (b x^{4} + a\right )}^{\frac {3}{4}}}{x^{10}} \,d x } \] Input:

integrate((b*x^4+a)^(3/4)/x^10,x, algorithm="fricas")
 

Output:

integral((b*x^4 + a)^(3/4)/x^10, x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.77 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.25 \[ \int \frac {\left (a+b x^4\right )^{3/4}}{x^{10}} \, dx=- \frac {b^{\frac {3}{4}} {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, \frac {3}{2} \\ \frac {5}{2} \end {matrix}\middle | {\frac {a e^{i \pi }}{b x^{4}}} \right )}}{6 x^{6}} \] Input:

integrate((b*x**4+a)**(3/4)/x**10,x)
 

Output:

-b**(3/4)*hyper((-3/4, 3/2), (5/2,), a*exp_polar(I*pi)/(b*x**4))/(6*x**6)
 

Maxima [F]

\[ \int \frac {\left (a+b x^4\right )^{3/4}}{x^{10}} \, dx=\int { \frac {{\left (b x^{4} + a\right )}^{\frac {3}{4}}}{x^{10}} \,d x } \] Input:

integrate((b*x^4+a)^(3/4)/x^10,x, algorithm="maxima")
 

Output:

integrate((b*x^4 + a)^(3/4)/x^10, x)
 

Giac [F]

\[ \int \frac {\left (a+b x^4\right )^{3/4}}{x^{10}} \, dx=\int { \frac {{\left (b x^{4} + a\right )}^{\frac {3}{4}}}{x^{10}} \,d x } \] Input:

integrate((b*x^4+a)^(3/4)/x^10,x, algorithm="giac")
 

Output:

integrate((b*x^4 + a)^(3/4)/x^10, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^4\right )^{3/4}}{x^{10}} \, dx=\int \frac {{\left (b\,x^4+a\right )}^{3/4}}{x^{10}} \,d x \] Input:

int((a + b*x^4)^(3/4)/x^10,x)
 

Output:

int((a + b*x^4)^(3/4)/x^10, x)
 

Reduce [F]

\[ \int \frac {\left (a+b x^4\right )^{3/4}}{x^{10}} \, dx=\frac {-\left (b \,x^{4}+a \right )^{\frac {3}{4}}-3 \left (\int \frac {\left (b \,x^{4}+a \right )^{\frac {3}{4}}}{b \,x^{14}+a \,x^{10}}d x \right ) a \,x^{9}}{6 x^{9}} \] Input:

int((b*x^4+a)^(3/4)/x^10,x)
 

Output:

( - (a + b*x**4)**(3/4) - 3*int((a + b*x**4)**(3/4)/(a*x**10 + b*x**14),x) 
*a*x**9)/(6*x**9)