\(\int \frac {c+d x^4}{x^2 (a+b x^4)^{17/4}} \, dx\) [189]

Optimal result

Integrand size = 22, antiderivative size = 181 \[ \int \frac {c+d x^4}{x^2 \left (a+b x^4\right )^{17/4}} \, dx=-\frac {c}{a x \left (a+b x^4\right )^{13/4}}-\frac {(14 b c-a d) x^3}{13 a^2 \left (a+b x^4\right )^{13/4}}-\frac {10 (14 b c-a d) x^3}{117 a^3 \left (a+b x^4\right )^{9/4}}-\frac {4 (14 b c-a d) x^3}{39 a^4 \left (a+b x^4\right )^{5/4}}+\frac {8 (14 b c-a d) \sqrt [4]{1+\frac {a}{b x^4}} x E\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{39 a^{9/2} \sqrt {b} \sqrt [4]{a+b x^4}} \] Output:

-c/a/x/(b*x^4+a)^(13/4)-1/13*(-a*d+14*b*c)*x^3/a^2/(b*x^4+a)^(13/4)-10/117 
*(-a*d+14*b*c)*x^3/a^3/(b*x^4+a)^(9/4)-4/39*(-a*d+14*b*c)*x^3/a^4/(b*x^4+a 
)^(5/4)+8/39*(-a*d+14*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^(9/2)/b^(1/2)/(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 = 10.04 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.46 \[ \int \frac {c+d x^4}{x^2 \left (a+b x^4\right )^{17/4}} \, dx=-\frac {c}{a x \left (a+b x^4\right )^{13/4}}-\frac {(14 b c-a d) x^3 \sqrt [4]{1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {17}{4},\frac {7}{4},-\frac {b x^4}{a}\right )}{3 a^5 \sqrt [4]{a+b x^4}} \] Input:

Integrate[(c + d*x^4)/(x^2*(a + b*x^4)^(17/4)),x]
 

Output:

-(c/(a*x*(a + b*x^4)^(13/4))) - ((14*b*c - a*d)*x^3*(1 + (b*x^4)/a)^(1/4)* 
Hypergeometric2F1[3/4, 17/4, 7/4, -((b*x^4)/a)])/(3*a^5*(a + b*x^4)^(1/4))
 

Rubi [A] (verified)

Time = 0.56 (sec) , antiderivative size = 176, normalized size of antiderivative = 0.97, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {955, 819, 819, 819, 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.

\(\displaystyle \int \frac {c+d x^4}{x^2 \left (a+b x^4\right )^{17/4}} \, dx\)

\(\Big \downarrow \) 955

\(\displaystyle -\frac {(14 b c-a d) \int \frac {x^2}{\left (b x^4+a\right )^{17/4}}dx}{a}-\frac {c}{a x \left (a+b x^4\right )^{13/4}}\)

\(\Big \downarrow \) 819

\(\displaystyle -\frac {(14 b c-a d) \left (\frac {10 \int \frac {x^2}{\left (b x^4+a\right )^{13/4}}dx}{13 a}+\frac {x^3}{13 a \left (a+b x^4\right )^{13/4}}\right )}{a}-\frac {c}{a x \left (a+b x^4\right )^{13/4}}\)

\(\Big \downarrow \) 819

\(\displaystyle -\frac {(14 b c-a d) \left (\frac {10 \left (\frac {2 \int \frac {x^2}{\left (b x^4+a\right )^{9/4}}dx}{3 a}+\frac {x^3}{9 a \left (a+b x^4\right )^{9/4}}\right )}{13 a}+\frac {x^3}{13 a \left (a+b x^4\right )^{13/4}}\right )}{a}-\frac {c}{a x \left (a+b x^4\right )^{13/4}}\)

\(\Big \downarrow \) 819

\(\displaystyle -\frac {(14 b c-a d) \left (\frac {10 \left (\frac {2 \left (\frac {2 \int \frac {x^2}{\left (b x^4+a\right )^{5/4}}dx}{5 a}+\frac {x^3}{5 a \left (a+b x^4\right )^{5/4}}\right )}{3 a}+\frac {x^3}{9 a \left (a+b x^4\right )^{9/4}}\right )}{13 a}+\frac {x^3}{13 a \left (a+b x^4\right )^{13/4}}\right )}{a}-\frac {c}{a x \left (a+b x^4\right )^{13/4}}\)

\(\Big \downarrow \) 813

\(\displaystyle -\frac {(14 b c-a d) \left (\frac {10 \left (\frac {2 \left (\frac {2 x \sqrt [4]{\frac {a}{b x^4}+1} \int \frac {1}{\left (\frac {a}{b x^4}+1\right )^{5/4} x^3}dx}{5 a b \sqrt [4]{a+b x^4}}+\frac {x^3}{5 a \left (a+b x^4\right )^{5/4}}\right )}{3 a}+\frac {x^3}{9 a \left (a+b x^4\right )^{9/4}}\right )}{13 a}+\frac {x^3}{13 a \left (a+b x^4\right )^{13/4}}\right )}{a}-\frac {c}{a x \left (a+b x^4\right )^{13/4}}\)

\(\Big \downarrow \) 858

\(\displaystyle -\frac {(14 b c-a d) \left (\frac {10 \left (\frac {2 \left (\frac {x^3}{5 a \left (a+b x^4\right )^{5/4}}-\frac {2 x \sqrt [4]{\frac {a}{b x^4}+1} \int \frac {1}{\left (\frac {a}{b x^4}+1\right )^{5/4} x}d\frac {1}{x}}{5 a b \sqrt [4]{a+b x^4}}\right )}{3 a}+\frac {x^3}{9 a \left (a+b x^4\right )^{9/4}}\right )}{13 a}+\frac {x^3}{13 a \left (a+b x^4\right )^{13/4}}\right )}{a}-\frac {c}{a x \left (a+b x^4\right )^{13/4}}\)

\(\Big \downarrow \) 807

\(\displaystyle -\frac {(14 b c-a d) \left (\frac {10 \left (\frac {2 \left (\frac {x^3}{5 a \left (a+b x^4\right )^{5/4}}-\frac {x \sqrt [4]{\frac {a}{b x^4}+1} \int \frac {1}{\left (\frac {a}{b x^2}+1\right )^{5/4}}d\frac {1}{x^2}}{5 a b \sqrt [4]{a+b x^4}}\right )}{3 a}+\frac {x^3}{9 a \left (a+b x^4\right )^{9/4}}\right )}{13 a}+\frac {x^3}{13 a \left (a+b x^4\right )^{13/4}}\right )}{a}-\frac {c}{a x \left (a+b x^4\right )^{13/4}}\)

\(\Big \downarrow \) 212

\(\displaystyle -\frac {(14 b c-a d) \left (\frac {10 \left (\frac {2 \left (\frac {x^3}{5 a \left (a+b x^4\right )^{5/4}}-\frac {2 x \sqrt [4]{\frac {a}{b x^4}+1} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {a}}{\sqrt {b} x^2}\right )\right |2\right )}{5 a^{3/2} \sqrt {b} \sqrt [4]{a+b x^4}}\right )}{3 a}+\frac {x^3}{9 a \left (a+b x^4\right )^{9/4}}\right )}{13 a}+\frac {x^3}{13 a \left (a+b x^4\right )^{13/4}}\right )}{a}-\frac {c}{a x \left (a+b x^4\right )^{13/4}}\)

Input:

Int[(c + d*x^4)/(x^2*(a + b*x^4)^(17/4)),x]
 

Output:

-(c/(a*x*(a + b*x^4)^(13/4))) - ((14*b*c - a*d)*(x^3/(13*a*(a + b*x^4)^(13 
/4)) + (10*(x^3/(9*a*(a + b*x^4)^(9/4)) + (2*(x^3/(5*a*(a + b*x^4)^(5/4)) 
- (2*(1 + a/(b*x^4))^(1/4)*x*EllipticE[ArcTan[Sqrt[a]/(Sqrt[b]*x^2)]/2, 2] 
)/(5*a^(3/2)*Sqrt[b]*(a + b*x^4)^(1/4))))/(3*a)))/(13*a)))/a
 

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 819

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

rule 955

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

Maple [F]

\[\int \frac {d \,x^{4}+c}{x^{2} \left (b \,x^{4}+a \right )^{\frac {17}{4}}}d x\]

Input:

int((d*x^4+c)/x^2/(b*x^4+a)^(17/4),x)
 

Output:

int((d*x^4+c)/x^2/(b*x^4+a)^(17/4),x)
 

Fricas [F]

\[ \int \frac {c+d x^4}{x^2 \left (a+b x^4\right )^{17/4}} \, dx=\int { \frac {d x^{4} + c}{{\left (b x^{4} + a\right )}^{\frac {17}{4}} x^{2}} \,d x } \] Input:

integrate((d*x^4+c)/x^2/(b*x^4+a)^(17/4),x, algorithm="fricas")
 

Output:

integral((b*x^4 + a)^(3/4)*(d*x^4 + c)/(b^5*x^22 + 5*a*b^4*x^18 + 10*a^2*b 
^3*x^14 + 10*a^3*b^2*x^10 + 5*a^4*b*x^6 + a^5*x^2), x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {c+d x^4}{x^2 \left (a+b x^4\right )^{17/4}} \, dx=\text {Timed out} \] Input:

integrate((d*x**4+c)/x**2/(b*x**4+a)**(17/4),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {c+d x^4}{x^2 \left (a+b x^4\right )^{17/4}} \, dx=\int { \frac {d x^{4} + c}{{\left (b x^{4} + a\right )}^{\frac {17}{4}} x^{2}} \,d x } \] Input:

integrate((d*x^4+c)/x^2/(b*x^4+a)^(17/4),x, algorithm="maxima")
 

Output:

integrate((d*x^4 + c)/((b*x^4 + a)^(17/4)*x^2), x)
 

Giac [F]

\[ \int \frac {c+d x^4}{x^2 \left (a+b x^4\right )^{17/4}} \, dx=\int { \frac {d x^{4} + c}{{\left (b x^{4} + a\right )}^{\frac {17}{4}} x^{2}} \,d x } \] Input:

integrate((d*x^4+c)/x^2/(b*x^4+a)^(17/4),x, algorithm="giac")
 

Output:

integrate((d*x^4 + c)/((b*x^4 + a)^(17/4)*x^2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {c+d x^4}{x^2 \left (a+b x^4\right )^{17/4}} \, dx=\int \frac {d\,x^4+c}{x^2\,{\left (b\,x^4+a\right )}^{17/4}} \,d x \] Input:

int((c + d*x^4)/(x^2*(a + b*x^4)^(17/4)),x)
 

Output:

int((c + d*x^4)/(x^2*(a + b*x^4)^(17/4)), x)
 

Reduce [F]

\[ \int \frac {c+d x^4}{x^2 \left (a+b x^4\right )^{17/4}} \, dx=\left (\int \frac {x^{2}}{\left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{4}+4 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{3} b \,x^{4}+6 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{2} b^{2} x^{8}+4 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a \,b^{3} x^{12}+\left (b \,x^{4}+a \right )^{\frac {1}{4}} b^{4} x^{16}}d x \right ) d +\left (\int \frac {1}{\left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{4} x^{2}+4 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{3} b \,x^{6}+6 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{2} b^{2} x^{10}+4 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a \,b^{3} x^{14}+\left (b \,x^{4}+a \right )^{\frac {1}{4}} b^{4} x^{18}}d x \right ) c \] Input:

int((d*x^4+c)/x^2/(b*x^4+a)^(17/4),x)
                                                                                    
                                                                                    
 

Output:

int(x**2/((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(1/((a + b*x**4)**(1/4)*a**4*x* 
*2 + 4*(a + b*x**4)**(1/4)*a**3*b*x**6 + 6*(a + b*x**4)**(1/4)*a**2*b**2*x 
**10 + 4*(a + b*x**4)**(1/4)*a*b**3*x**14 + (a + b*x**4)**(1/4)*b**4*x**18 
),x)*c