\(\int \frac {c+d x^4}{x^3 (a+b x^4)^{9/4}} \, dx\) [163]

Optimal result

Integrand size = 22, antiderivative size = 122 \[ \int \frac {c+d x^4}{x^3 \left (a+b x^4\right )^{9/4}} \, dx=-\frac {c}{2 a x^2 \left (a+b x^4\right )^{5/4}}-\frac {(7 b c-2 a d) x^2}{10 a^2 \left (a+b x^4\right )^{5/4}}-\frac {3 (7 b c-2 a d) \sqrt [4]{1+\frac {b x^4}{a}} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{10 a^{5/2} \sqrt {b} \sqrt [4]{a+b x^4}} \] Output:

-1/2*c/a/x^2/(b*x^4+a)^(5/4)-1/10*(-2*a*d+7*b*c)*x^2/a^2/(b*x^4+a)^(5/4)-3 
/10*(-2*a*d+7*b*c)*(1+b*x^4/a)^(1/4)*EllipticE(sin(1/2*arctan(b^(1/2)*x^2/ 
a^(1/2))),2^(1/2))/a^(5/2)/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 = 10.05 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.97 \[ \int \frac {c+d x^4}{x^3 \left (a+b x^4\right )^{9/4}} \, dx=\frac {-2 \left (21 b^2 c x^8+a^2 \left (5 c-8 d x^4\right )+a b \left (28 c x^4-6 d x^8\right )\right )+3 (7 b c-2 a d) x^4 \left (a+b x^4\right ) \sqrt [4]{1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {3}{2},-\frac {b x^4}{a}\right )}{20 a^3 x^2 \left (a+b x^4\right )^{5/4}} \] Input:

Integrate[(c + d*x^4)/(x^3*(a + b*x^4)^(9/4)),x]
 

Output:

(-2*(21*b^2*c*x^8 + a^2*(5*c - 8*d*x^4) + a*b*(28*c*x^4 - 6*d*x^8)) + 3*(7 
*b*c - 2*a*d)*x^4*(a + b*x^4)*(1 + (b*x^4)/a)^(1/4)*Hypergeometric2F1[1/4, 
 1/2, 3/2, -((b*x^4)/a)])/(20*a^3*x^2*(a + b*x^4)^(5/4))
 

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.99, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {955, 807, 215, 213, 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.

\(\displaystyle \int \frac {c+d x^4}{x^3 \left (a+b x^4\right )^{9/4}} \, dx\)

\(\Big \downarrow \) 955

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

\(\Big \downarrow \) 807

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

\(\Big \downarrow \) 215

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

\(\Big \downarrow \) 213

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

\(\Big \downarrow \) 212

\(\displaystyle -\frac {(7 b c-2 a d) \left (\frac {6 \sqrt [4]{\frac {b x^4}{a}+1} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{5 a^{3/2} \sqrt {b} \sqrt [4]{a+b x^4}}+\frac {2 x^2}{5 a \left (a+b x^4\right )^{5/4}}\right )}{4 a}-\frac {c}{2 a x^2 \left (a+b x^4\right )^{5/4}}\)

Input:

Int[(c + d*x^4)/(x^3*(a + b*x^4)^(9/4)),x]
 

Output:

-1/2*c/(a*x^2*(a + b*x^4)^(5/4)) - ((7*b*c - 2*a*d)*((2*x^2)/(5*a*(a + b*x 
^4)^(5/4)) + (6*(1 + (b*x^4)/a)^(1/4)*EllipticE[ArcTan[(Sqrt[b]*x^2)/Sqrt[ 
a]]/2, 2])/(5*a^(3/2)*Sqrt[b]*(a + b*x^4)^(1/4))))/(4*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 213

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

rule 215

Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^2)^(p + 1) 
/(2*a*(p + 1))), x] + Simp[(2*p + 3)/(2*a*(p + 1))   Int[(a + b*x^2)^(p + 1 
), x], x] /; FreeQ[{a, b}, x] && LtQ[p, -1] && (IntegerQ[4*p] || IntegerQ[6 
*p])
 

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 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]

\[\int \frac {d \,x^{4}+c}{x^{3} \left (b \,x^{4}+a \right )^{\frac {9}{4}}}d x\]

Input:

int((d*x^4+c)/x^3/(b*x^4+a)^(9/4),x)
 

Output:

int((d*x^4+c)/x^3/(b*x^4+a)^(9/4),x)
 

Fricas [F]

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

integrate((d*x^4+c)/x^3/(b*x^4+a)^(9/4),x, algorithm="fricas")
 

Output:

integral((b*x^4 + a)^(3/4)*(d*x^4 + c)/(b^3*x^15 + 3*a*b^2*x^11 + 3*a^2*b* 
x^7 + a^3*x^3), x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 38.53 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.50 \[ \int \frac {c+d x^4}{x^3 \left (a+b x^4\right )^{9/4}} \, dx=- \frac {c {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {9}{4} \\ \frac {1}{2} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{2 a^{\frac {9}{4}} x^{2}} + \frac {d x^{2} {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {9}{4} \\ \frac {3}{2} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{2 a^{\frac {9}{4}}} \] Input:

integrate((d*x**4+c)/x**3/(b*x**4+a)**(9/4),x)
 

Output:

-c*hyper((-1/2, 9/4), (1/2,), b*x**4*exp_polar(I*pi)/a)/(2*a**(9/4)*x**2) 
+ d*x**2*hyper((1/2, 9/4), (3/2,), b*x**4*exp_polar(I*pi)/a)/(2*a**(9/4))
 

Maxima [F]

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

integrate((d*x^4+c)/x^3/(b*x^4+a)^(9/4),x, algorithm="maxima")
 

Output:

integrate((d*x^4 + c)/((b*x^4 + a)^(9/4)*x^3), x)
 

Giac [F]

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

integrate((d*x^4+c)/x^3/(b*x^4+a)^(9/4),x, algorithm="giac")
                                                                                    
                                                                                    
 

Output:

integrate((d*x^4 + c)/((b*x^4 + a)^(9/4)*x^3), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {c+d x^4}{x^3 \left (a+b x^4\right )^{9/4}} \, dx=\int \frac {d\,x^4+c}{x^3\,{\left (b\,x^4+a\right )}^{9/4}} \,d x \] Input:

int((c + d*x^4)/(x^3*(a + b*x^4)^(9/4)),x)
 

Output:

int((c + d*x^4)/(x^3*(a + b*x^4)^(9/4)), x)
 

Reduce [F]

\[ \int \frac {c+d x^4}{x^3 \left (a+b x^4\right )^{9/4}} \, dx=\left (\int \frac {x}{\left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{2}+2 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a b \,x^{4}+\left (b \,x^{4}+a \right )^{\frac {1}{4}} b^{2} x^{8}}d x \right ) d +\left (\int \frac {1}{\left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{2} x^{3}+2 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a b \,x^{7}+\left (b \,x^{4}+a \right )^{\frac {1}{4}} b^{2} x^{11}}d x \right ) c \] Input:

int((d*x^4+c)/x^3/(b*x^4+a)^(9/4),x)
 

Output:

int(x/((a + b*x**4)**(1/4)*a**2 + 2*(a + b*x**4)**(1/4)*a*b*x**4 + (a + b* 
x**4)**(1/4)*b**2*x**8),x)*d + int(1/((a + b*x**4)**(1/4)*a**2*x**3 + 2*(a 
 + b*x**4)**(1/4)*a*b*x**7 + (a + b*x**4)**(1/4)*b**2*x**11),x)*c