\(\int \frac {c+d x^4}{x^7 (a+b x^4)^{5/4}} \, dx\) [124]

Optimal result

Integrand size = 22, antiderivative size = 122 \[ \int \frac {c+d x^4}{x^7 \left (a+b x^4\right )^{5/4}} \, dx=-\frac {c}{6 a x^6 \sqrt [4]{a+b x^4}}+\frac {7 b c-6 a d}{12 a^2 x^2 \sqrt [4]{a+b x^4}}+\frac {\sqrt {b} (7 b c-6 a d) \sqrt [4]{1+\frac {b x^4}{a}} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{4 a^{5/2} \sqrt [4]{a+b x^4}} \] Output:

-1/6*c/a/x^6/(b*x^4+a)^(1/4)+1/12*(-6*a*d+7*b*c)/a^2/x^2/(b*x^4+a)^(1/4)+1 
/4*b^(1/2)*(-6*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*x^4+a)^(1/4)
 

Mathematica [C] (verified)

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

Time = 10.03 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.59 \[ \int \frac {c+d x^4}{x^7 \left (a+b x^4\right )^{5/4}} \, dx=\frac {-2 a c+(7 b c-6 a d) x^4 \sqrt [4]{1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {5}{4},\frac {1}{2},-\frac {b x^4}{a}\right )}{12 a^2 x^6 \sqrt [4]{a+b x^4}} \] Input:

Integrate[(c + d*x^4)/(x^7*(a + b*x^4)^(5/4)),x]
 

Output:

(-2*a*c + (7*b*c - 6*a*d)*x^4*(1 + (b*x^4)/a)^(1/4)*Hypergeometric2F1[-1/2 
, 5/4, 1/2, -((b*x^4)/a)])/(12*a^2*x^6*(a + b*x^4)^(1/4))
 

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.96, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {955, 807, 251, 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.

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

\(\Big \downarrow \) 955

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

\(\Big \downarrow \) 807

\(\displaystyle -\frac {(7 b c-6 a d) \int \frac {1}{x^4 \left (b x^4+a\right )^{5/4}}dx^2}{12 a}-\frac {c}{6 a x^6 \sqrt [4]{a+b x^4}}\)

\(\Big \downarrow \) 251

\(\displaystyle -\frac {(7 b c-6 a d) \left (-\frac {3 b \int \frac {1}{\left (b x^4+a\right )^{5/4}}dx^2}{2 a}-\frac {1}{a x^2 \sqrt [4]{a+b x^4}}\right )}{12 a}-\frac {c}{6 a x^6 \sqrt [4]{a+b x^4}}\)

\(\Big \downarrow \) 213

\(\displaystyle -\frac {(7 b c-6 a d) \left (-\frac {3 b \sqrt [4]{\frac {b x^4}{a}+1} \int \frac {1}{\left (\frac {b x^4}{a}+1\right )^{5/4}}dx^2}{2 a^2 \sqrt [4]{a+b x^4}}-\frac {1}{a x^2 \sqrt [4]{a+b x^4}}\right )}{12 a}-\frac {c}{6 a x^6 \sqrt [4]{a+b x^4}}\)

\(\Big \downarrow \) 212

\(\displaystyle -\frac {(7 b c-6 a d) \left (-\frac {3 \sqrt {b} \sqrt [4]{\frac {b x^4}{a}+1} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{a^{3/2} \sqrt [4]{a+b x^4}}-\frac {1}{a x^2 \sqrt [4]{a+b x^4}}\right )}{12 a}-\frac {c}{6 a x^6 \sqrt [4]{a+b x^4}}\)

Input:

Int[(c + d*x^4)/(x^7*(a + b*x^4)^(5/4)),x]
 

Output:

-1/6*c/(a*x^6*(a + b*x^4)^(1/4)) - ((7*b*c - 6*a*d)*(-(1/(a*x^2*(a + b*x^4 
)^(1/4))) - (3*Sqrt[b]*(1 + (b*x^4)/a)^(1/4)*EllipticE[ArcTan[(Sqrt[b]*x^2 
)/Sqrt[a]]/2, 2])/(a^(3/2)*(a + b*x^4)^(1/4))))/(12*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 251

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

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]

\[\int \frac {d \,x^{4}+c}{x^{7} \left (b \,x^{4}+a \right )^{\frac {5}{4}}}d x\]

Input:

int((d*x^4+c)/x^7/(b*x^4+a)^(5/4),x)
 

Output:

int((d*x^4+c)/x^7/(b*x^4+a)^(5/4),x)
 

Fricas [F]

\[ \int \frac {c+d x^4}{x^7 \left (a+b x^4\right )^{5/4}} \, dx=\int { \frac {d x^{4} + c}{{\left (b x^{4} + a\right )}^{\frac {5}{4}} x^{7}} \,d x } \] Input:

integrate((d*x^4+c)/x^7/(b*x^4+a)^(5/4),x, algorithm="fricas")
 

Output:

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 27.52 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.54 \[ \int \frac {c+d x^4}{x^7 \left (a+b x^4\right )^{5/4}} \, dx=- \frac {c {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{2}, \frac {5}{4} \\ - \frac {1}{2} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{6 a^{\frac {5}{4}} x^{6}} - \frac {d {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {5}{4} \\ \frac {1}{2} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{2 a^{\frac {5}{4}} x^{2}} \] Input:

integrate((d*x**4+c)/x**7/(b*x**4+a)**(5/4),x)
 

Output:

-c*hyper((-3/2, 5/4), (-1/2,), b*x**4*exp_polar(I*pi)/a)/(6*a**(5/4)*x**6) 
 - d*hyper((-1/2, 5/4), (1/2,), b*x**4*exp_polar(I*pi)/a)/(2*a**(5/4)*x**2 
)
 

Maxima [F]

\[ \int \frac {c+d x^4}{x^7 \left (a+b x^4\right )^{5/4}} \, dx=\int { \frac {d x^{4} + c}{{\left (b x^{4} + a\right )}^{\frac {5}{4}} x^{7}} \,d x } \] Input:

integrate((d*x^4+c)/x^7/(b*x^4+a)^(5/4),x, algorithm="maxima")
 

Output:

integrate((d*x^4 + c)/((b*x^4 + a)^(5/4)*x^7), x)
 

Giac [F]

\[ \int \frac {c+d x^4}{x^7 \left (a+b x^4\right )^{5/4}} \, dx=\int { \frac {d x^{4} + c}{{\left (b x^{4} + a\right )}^{\frac {5}{4}} x^{7}} \,d x } \] Input:

integrate((d*x^4+c)/x^7/(b*x^4+a)^(5/4),x, algorithm="giac")
                                                                                    
                                                                                    
 

Output:

integrate((d*x^4 + c)/((b*x^4 + a)^(5/4)*x^7), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {c+d x^4}{x^7 \left (a+b x^4\right )^{5/4}} \, dx=\int \frac {d\,x^4+c}{x^7\,{\left (b\,x^4+a\right )}^{5/4}} \,d x \] Input:

int((c + d*x^4)/(x^7*(a + b*x^4)^(5/4)),x)
 

Output:

int((c + d*x^4)/(x^7*(a + b*x^4)^(5/4)), x)
 

Reduce [F]

\[ \int \frac {c+d x^4}{x^7 \left (a+b x^4\right )^{5/4}} \, dx=\left (\int \frac {1}{\left (b \,x^{4}+a \right )^{\frac {1}{4}} a \,x^{7}+\left (b \,x^{4}+a \right )^{\frac {1}{4}} b \,x^{11}}d x \right ) c +\left (\int \frac {1}{\left (b \,x^{4}+a \right )^{\frac {1}{4}} a \,x^{3}+\left (b \,x^{4}+a \right )^{\frac {1}{4}} b \,x^{7}}d x \right ) d \] Input:

int((d*x^4+c)/x^7/(b*x^4+a)^(5/4),x)
 

Output:

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