\(\int \frac {x^{10} (c+d x^4)}{(a+b x^4)^{17/4}} \, dx\) [186]

Optimal result

Integrand size = 22, antiderivative size = 163 \[ \int \frac {x^{10} \left (c+d x^4\right )}{\left (a+b x^4\right )^{17/4}} \, dx=-\frac {(b c-a d) x^7}{13 b^2 \left (a+b x^4\right )^{13/4}}-\frac {(7 b c-20 a d) x^3}{117 b^3 \left (a+b x^4\right )^{9/4}}+\frac {(7 b c-59 a d) x^3}{195 a b^3 \left (a+b x^4\right )^{5/4}}-\frac {7 (2 b c+11 a d) \sqrt [4]{1+\frac {a}{b x^4}} x E\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{195 a^{3/2} b^{7/2} \sqrt [4]{a+b x^4}} \] Output:

-1/13*(-a*d+b*c)*x^7/b^2/(b*x^4+a)^(13/4)-1/117*(-20*a*d+7*b*c)*x^3/b^3/(b 
*x^4+a)^(9/4)+1/195*(-59*a*d+7*b*c)*x^3/a/b^3/(b*x^4+a)^(5/4)-7/195*(11*a* 
d+2*b*c)*(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^(7/2)/(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 = 10.12 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.74 \[ \int \frac {x^{10} \left (c+d x^4\right )}{\left (a+b x^4\right )^{17/4}} \, dx=\frac {x^3 \left (-a^2 \left (77 a^2 d+20 b^2 x^4 \left (c+3 d x^4\right )+2 a b \left (7 c+55 d x^4\right )\right )+7 (2 b c+11 a d) \left (a+b x^4\right )^3 \sqrt [4]{1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {17}{4},\frac {7}{4},-\frac {b x^4}{a}\right )\right )}{120 a^2 b^3 \left (a+b x^4\right )^{13/4}} \] Input:

Integrate[(x^10*(c + d*x^4))/(a + b*x^4)^(17/4),x]
 

Output:

(x^3*(-(a^2*(77*a^2*d + 20*b^2*x^4*(c + 3*d*x^4) + 2*a*b*(7*c + 55*d*x^4)) 
) + 7*(2*b*c + 11*a*d)*(a + b*x^4)^3*(1 + (b*x^4)/a)^(1/4)*Hypergeometric2 
F1[3/4, 17/4, 7/4, -((b*x^4)/a)]))/(120*a^2*b^3*(a + b*x^4)^(13/4))
 

Rubi [A] (verified)

Time = 0.53 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.01, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {957, 817, 817, 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.

\(\displaystyle \int \frac {x^{10} \left (c+d x^4\right )}{\left (a+b x^4\right )^{17/4}} \, dx\)

\(\Big \downarrow \) 957

\(\displaystyle \frac {(11 a d+2 b c) \int \frac {x^{10}}{\left (b x^4+a\right )^{13/4}}dx}{13 a b}+\frac {x^{11} (b c-a d)}{13 a b \left (a+b x^4\right )^{13/4}}\)

\(\Big \downarrow \) 817

\(\displaystyle \frac {(11 a d+2 b c) \left (\frac {7 \int \frac {x^6}{\left (b x^4+a\right )^{9/4}}dx}{9 b}-\frac {x^7}{9 b \left (a+b x^4\right )^{9/4}}\right )}{13 a b}+\frac {x^{11} (b c-a d)}{13 a b \left (a+b x^4\right )^{13/4}}\)

\(\Big \downarrow \) 817

\(\displaystyle \frac {(11 a d+2 b c) \left (\frac {7 \left (\frac {3 \int \frac {x^2}{\left (b x^4+a\right )^{5/4}}dx}{5 b}-\frac {x^3}{5 b \left (a+b x^4\right )^{5/4}}\right )}{9 b}-\frac {x^7}{9 b \left (a+b x^4\right )^{9/4}}\right )}{13 a b}+\frac {x^{11} (b c-a d)}{13 a b \left (a+b x^4\right )^{13/4}}\)

\(\Big \downarrow \) 813

\(\displaystyle \frac {(11 a d+2 b c) \left (\frac {7 \left (\frac {3 x \sqrt [4]{\frac {a}{b x^4}+1} \int \frac {1}{\left (\frac {a}{b x^4}+1\right )^{5/4} x^3}dx}{5 b^2 \sqrt [4]{a+b x^4}}-\frac {x^3}{5 b \left (a+b x^4\right )^{5/4}}\right )}{9 b}-\frac {x^7}{9 b \left (a+b x^4\right )^{9/4}}\right )}{13 a b}+\frac {x^{11} (b c-a d)}{13 a b \left (a+b x^4\right )^{13/4}}\)

\(\Big \downarrow \) 858

\(\displaystyle \frac {(11 a d+2 b c) \left (\frac {7 \left (-\frac {3 x \sqrt [4]{\frac {a}{b x^4}+1} \int \frac {1}{\left (\frac {a}{b x^4}+1\right )^{5/4} x}d\frac {1}{x}}{5 b^2 \sqrt [4]{a+b x^4}}-\frac {x^3}{5 b \left (a+b x^4\right )^{5/4}}\right )}{9 b}-\frac {x^7}{9 b \left (a+b x^4\right )^{9/4}}\right )}{13 a b}+\frac {x^{11} (b c-a d)}{13 a b \left (a+b x^4\right )^{13/4}}\)

\(\Big \downarrow \) 807

\(\displaystyle \frac {(11 a d+2 b c) \left (\frac {7 \left (-\frac {3 x \sqrt [4]{\frac {a}{b x^4}+1} \int \frac {1}{\left (\frac {a}{b x^2}+1\right )^{5/4}}d\frac {1}{x^2}}{10 b^2 \sqrt [4]{a+b x^4}}-\frac {x^3}{5 b \left (a+b x^4\right )^{5/4}}\right )}{9 b}-\frac {x^7}{9 b \left (a+b x^4\right )^{9/4}}\right )}{13 a b}+\frac {x^{11} (b c-a d)}{13 a b \left (a+b x^4\right )^{13/4}}\)

\(\Big \downarrow \) 212

\(\displaystyle \frac {(11 a d+2 b c) \left (\frac {7 \left (-\frac {3 x \sqrt [4]{\frac {a}{b x^4}+1} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {a}}{\sqrt {b} x^2}\right )\right |2\right )}{5 \sqrt {a} b^{3/2} \sqrt [4]{a+b x^4}}-\frac {x^3}{5 b \left (a+b x^4\right )^{5/4}}\right )}{9 b}-\frac {x^7}{9 b \left (a+b x^4\right )^{9/4}}\right )}{13 a b}+\frac {x^{11} (b c-a d)}{13 a b \left (a+b x^4\right )^{13/4}}\)

Input:

Int[(x^10*(c + d*x^4))/(a + b*x^4)^(17/4),x]
 

Output:

((b*c - a*d)*x^11)/(13*a*b*(a + b*x^4)^(13/4)) + ((2*b*c + 11*a*d)*(-1/9*x 
^7/(b*(a + b*x^4)^(9/4)) + (7*(-1/5*x^3/(b*(a + b*x^4)^(5/4)) - (3*(1 + a/ 
(b*x^4))^(1/4)*x*EllipticE[ArcTan[Sqrt[a]/(Sqrt[b]*x^2)]/2, 2])/(5*Sqrt[a] 
*b^(3/2)*(a + b*x^4)^(1/4))))/(9*b)))/(13*a*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 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 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 817

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

rule 957

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n 
_)), x_Symbol] :> Simp[(-(b*c - a*d))*(e*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a 
*b*e*n*(p + 1))), x] - Simp[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*b*n* 
(p + 1))   Int[(e*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, 
 m, n}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && (( !IntegerQ[p + 1/2] && N 
eQ[p, -5/4]) ||  !RationalQ[m] || (IGtQ[n, 0] && ILtQ[p + 1/2, 0] && LeQ[-1 
, m, (-n)*(p + 1)]))
 

Maple [F]

\[\int \frac {x^{10} \left (d \,x^{4}+c \right )}{\left (b \,x^{4}+a \right )^{\frac {17}{4}}}d x\]

Input:

int(x^10*(d*x^4+c)/(b*x^4+a)^(17/4),x)
 

Output:

int(x^10*(d*x^4+c)/(b*x^4+a)^(17/4),x)
 

Fricas [F]

\[ \int \frac {x^{10} \left (c+d x^4\right )}{\left (a+b x^4\right )^{17/4}} \, dx=\int { \frac {{\left (d x^{4} + c\right )} x^{10}}{{\left (b x^{4} + a\right )}^{\frac {17}{4}}} \,d x } \] Input:

integrate(x^10*(d*x^4+c)/(b*x^4+a)^(17/4),x, algorithm="fricas")
 

Output:

integral((d*x^14 + c*x^10)*(b*x^4 + a)^(3/4)/(b^5*x^20 + 5*a*b^4*x^16 + 10 
*a^2*b^3*x^12 + 10*a^3*b^2*x^8 + 5*a^4*b*x^4 + a^5), x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {x^{10} \left (c+d x^4\right )}{\left (a+b x^4\right )^{17/4}} \, dx=\text {Timed out} \] Input:

integrate(x**10*(d*x**4+c)/(b*x**4+a)**(17/4),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {x^{10} \left (c+d x^4\right )}{\left (a+b x^4\right )^{17/4}} \, dx=\int { \frac {{\left (d x^{4} + c\right )} x^{10}}{{\left (b x^{4} + a\right )}^{\frac {17}{4}}} \,d x } \] Input:

integrate(x^10*(d*x^4+c)/(b*x^4+a)^(17/4),x, algorithm="maxima")
 

Output:

integrate((d*x^4 + c)*x^10/(b*x^4 + a)^(17/4), x)
 

Giac [F]

\[ \int \frac {x^{10} \left (c+d x^4\right )}{\left (a+b x^4\right )^{17/4}} \, dx=\int { \frac {{\left (d x^{4} + c\right )} x^{10}}{{\left (b x^{4} + a\right )}^{\frac {17}{4}}} \,d x } \] Input:

integrate(x^10*(d*x^4+c)/(b*x^4+a)^(17/4),x, algorithm="giac")
 

Output:

integrate((d*x^4 + c)*x^10/(b*x^4 + a)^(17/4), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^{10} \left (c+d x^4\right )}{\left (a+b x^4\right )^{17/4}} \, dx=\int \frac {x^{10}\,\left (d\,x^4+c\right )}{{\left (b\,x^4+a\right )}^{17/4}} \,d x \] Input:

int((x^10*(c + d*x^4))/(a + b*x^4)^(17/4),x)
 

Output:

int((x^10*(c + d*x^4))/(a + b*x^4)^(17/4), x)
 

Reduce [F]

\[ \int \frac {x^{10} \left (c+d x^4\right )}{\left (a+b x^4\right )^{17/4}} \, dx=\left (\int \frac {x^{14}}{\left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{4}+4 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{3} b \,x^{4}+6 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{2} b^{2} x^{8}+4 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a \,b^{3} x^{12}+\left (b \,x^{4}+a \right )^{\frac {1}{4}} b^{4} x^{16}}d x \right ) d +\left (\int \frac {x^{10}}{\left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{4}+4 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{3} b \,x^{4}+6 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{2} b^{2} x^{8}+4 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a \,b^{3} x^{12}+\left (b \,x^{4}+a \right )^{\frac {1}{4}} b^{4} x^{16}}d x \right ) c \] Input:

int(x^10*(d*x^4+c)/(b*x^4+a)^(17/4),x)
                                                                                    
                                                                                    
 

Output:

int(x**14/((a + b*x**4)**(1/4)*a**4 + 4*(a + b*x**4)**(1/4)*a**3*b*x**4 + 
6*(a + b*x**4)**(1/4)*a**2*b**2*x**8 + 4*(a + b*x**4)**(1/4)*a*b**3*x**12 
+ (a + b*x**4)**(1/4)*b**4*x**16),x)*d + int(x**10/((a + b*x**4)**(1/4)*a* 
*4 + 4*(a + b*x**4)**(1/4)*a**3*b*x**4 + 6*(a + b*x**4)**(1/4)*a**2*b**2*x 
**8 + 4*(a + b*x**4)**(1/4)*a*b**3*x**12 + (a + b*x**4)**(1/4)*b**4*x**16) 
,x)*c