\(\int \frac {c+d x^4}{x^5 (a+b x^4)^{5/4}} \, dx\) [111]

Optimal result

Integrand size = 22, antiderivative size = 119 \[ \int \frac {c+d x^4}{x^5 \left (a+b x^4\right )^{5/4}} \, dx=-\frac {b c-a d}{a^2 \sqrt [4]{a+b x^4}}-\frac {c \left (a+b x^4\right )^{3/4}}{4 a^2 x^4}-\frac {(5 b c-4 a d) \arctan \left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{8 a^{9/4}}+\frac {(5 b c-4 a d) \text {arctanh}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{8 a^{9/4}} \] Output:

-(-a*d+b*c)/a^2/(b*x^4+a)^(1/4)-1/4*c*(b*x^4+a)^(3/4)/a^2/x^4-1/8*(-4*a*d+ 
5*b*c)*arctan((b*x^4+a)^(1/4)/a^(1/4))/a^(9/4)+1/8*(-4*a*d+5*b*c)*arctanh( 
(b*x^4+a)^(1/4)/a^(1/4))/a^(9/4)
 

Mathematica [A] (verified)

Time = 0.29 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.88 \[ \int \frac {c+d x^4}{x^5 \left (a+b x^4\right )^{5/4}} \, dx=\frac {-\frac {2 \sqrt [4]{a} \left (a c+5 b c x^4-4 a d x^4\right )}{x^4 \sqrt [4]{a+b x^4}}+(-5 b c+4 a d) \arctan \left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )+(5 b c-4 a d) \text {arctanh}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{8 a^{9/4}} \] Input:

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

Output:

((-2*a^(1/4)*(a*c + 5*b*c*x^4 - 4*a*d*x^4))/(x^4*(a + b*x^4)^(1/4)) + (-5* 
b*c + 4*a*d)*ArcTan[(a + b*x^4)^(1/4)/a^(1/4)] + (5*b*c - 4*a*d)*ArcTanh[( 
a + b*x^4)^(1/4)/a^(1/4)])/(8*a^(9/4))
 

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.99, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.409, Rules used = {948, 87, 61, 73, 25, 27, 827, 216, 219}

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.

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

\(\Big \downarrow \) 948

\(\displaystyle \frac {1}{4} \int \frac {d x^4+c}{x^8 \left (b x^4+a\right )^{5/4}}dx^4\)

\(\Big \downarrow \) 87

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

\(\Big \downarrow \) 61

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

\(\Big \downarrow \) 73

\(\displaystyle \frac {1}{4} \left (-\frac {(5 b c-4 a d) \left (\frac {4 \int -\frac {b x^8}{a-x^{16}}d\sqrt [4]{b x^4+a}}{a b}+\frac {4}{a \sqrt [4]{a+b x^4}}\right )}{4 a}-\frac {c}{a x^4 \sqrt [4]{a+b x^4}}\right )\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{4} \left (-\frac {(5 b c-4 a d) \left (\frac {4}{a \sqrt [4]{a+b x^4}}-\frac {4 \int \frac {b x^8}{a-x^{16}}d\sqrt [4]{b x^4+a}}{a b}\right )}{4 a}-\frac {c}{a x^4 \sqrt [4]{a+b x^4}}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{4} \left (-\frac {(5 b c-4 a d) \left (\frac {4}{a \sqrt [4]{a+b x^4}}-\frac {4 \int \frac {x^8}{a-x^{16}}d\sqrt [4]{b x^4+a}}{a}\right )}{4 a}-\frac {c}{a x^4 \sqrt [4]{a+b x^4}}\right )\)

\(\Big \downarrow \) 827

\(\displaystyle \frac {1}{4} \left (-\frac {(5 b c-4 a d) \left (\frac {4}{a \sqrt [4]{a+b x^4}}-\frac {4 \left (\frac {1}{2} \int \frac {1}{\sqrt {a}-x^8}d\sqrt [4]{b x^4+a}-\frac {1}{2} \int \frac {1}{x^8+\sqrt {a}}d\sqrt [4]{b x^4+a}\right )}{a}\right )}{4 a}-\frac {c}{a x^4 \sqrt [4]{a+b x^4}}\right )\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {1}{4} \left (-\frac {(5 b c-4 a d) \left (\frac {4}{a \sqrt [4]{a+b x^4}}-\frac {4 \left (\frac {1}{2} \int \frac {1}{\sqrt {a}-x^8}d\sqrt [4]{b x^4+a}-\frac {\arctan \left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{2 \sqrt [4]{a}}\right )}{a}\right )}{4 a}-\frac {c}{a x^4 \sqrt [4]{a+b x^4}}\right )\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{4} \left (-\frac {(5 b c-4 a d) \left (\frac {4}{a \sqrt [4]{a+b x^4}}-\frac {4 \left (\frac {\text {arctanh}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{2 \sqrt [4]{a}}-\frac {\arctan \left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{2 \sqrt [4]{a}}\right )}{a}\right )}{4 a}-\frac {c}{a x^4 \sqrt [4]{a+b x^4}}\right )\)

Input:

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

Output:

(-(c/(a*x^4*(a + b*x^4)^(1/4))) - ((5*b*c - 4*a*d)*(4/(a*(a + b*x^4)^(1/4) 
) - (4*(-1/2*ArcTan[(a + b*x^4)^(1/4)/a^(1/4)]/a^(1/4) + ArcTanh[(a + b*x^ 
4)^(1/4)/a^(1/4)]/(2*a^(1/4))))/a))/(4*a))/4
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 61

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( 
m + n + 2)/((b*c - a*d)*(m + 1)))   Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], 
x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0 
] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d 
, m, n, x]
 

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 87

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p 
+ 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p 
+ 1)))/(f*(p + 1)*(c*f - d*e))   Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] 
/; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege 
rQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ[p, n]))))
 

rule 216

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 827

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 
 2]], s = Denominator[Rt[-a/b, 2]]}, Simp[s/(2*b)   Int[1/(r + s*x^2), x], 
x] - Simp[s/(2*b)   Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !GtQ 
[a/b, 0]
 

rule 948

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_. 
), x_Symbol] :> Simp[1/n   Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^ 
p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ 
[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]
 

Maple [A] (verified)

Time = 0.22 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.94

method	result	size
pseudoelliptic	\(\frac {\left (2 d \,x^{4}-\frac {c}{2}\right ) a^{\frac {5}{4}}+x^{4} \left (-\frac {5 b c \,a^{\frac {1}{4}}}{2}+\left (\arctan \left (\frac {\left (b \,x^{4}+a \right )^{\frac {1}{4}}}{a^{\frac {1}{4}}}\right )-\frac {\ln \left (\frac {\left (b \,x^{4}+a \right )^{\frac {1}{4}}+a^{\frac {1}{4}}}{\left (b \,x^{4}+a \right )^{\frac {1}{4}}-a^{\frac {1}{4}}}\right )}{2}\right ) \left (b \,x^{4}+a \right )^{\frac {1}{4}} \left (a d -\frac {5 c b}{4}\right )\right )}{2 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{\frac {9}{4}} x^{4}}\)	\(112\)

Input:

int((d*x^4+c)/x^5/(b*x^4+a)^(5/4),x,method=_RETURNVERBOSE)
 

Output:

1/2/(b*x^4+a)^(1/4)/a^(9/4)*((2*d*x^4-1/2*c)*a^(5/4)+x^4*(-5/2*b*c*a^(1/4) 
+(arctan((b*x^4+a)^(1/4)/a^(1/4))-1/2*ln(((b*x^4+a)^(1/4)+a^(1/4))/((b*x^4 
+a)^(1/4)-a^(1/4))))*(b*x^4+a)^(1/4)*(a*d-5/4*c*b)))/x^4
                                                                                    
                                                                                    
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.10 (sec) , antiderivative size = 797, normalized size of antiderivative = 6.70 \[ \int \frac {c+d x^4}{x^5 \left (a+b x^4\right )^{5/4}} \, dx =\text {Too large to display} \] Input:

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

Output:

-1/16*((a^2*b*x^8 + a^3*x^4)*((625*b^4*c^4 - 2000*a*b^3*c^3*d + 2400*a^2*b 
^2*c^2*d^2 - 1280*a^3*b*c*d^3 + 256*a^4*d^4)/a^9)^(1/4)*log(a^7*((625*b^4* 
c^4 - 2000*a*b^3*c^3*d + 2400*a^2*b^2*c^2*d^2 - 1280*a^3*b*c*d^3 + 256*a^4 
*d^4)/a^9)^(3/4) - (125*b^3*c^3 - 300*a*b^2*c^2*d + 240*a^2*b*c*d^2 - 64*a 
^3*d^3)*(b*x^4 + a)^(1/4)) + (-I*a^2*b*x^8 - I*a^3*x^4)*((625*b^4*c^4 - 20 
00*a*b^3*c^3*d + 2400*a^2*b^2*c^2*d^2 - 1280*a^3*b*c*d^3 + 256*a^4*d^4)/a^ 
9)^(1/4)*log(I*a^7*((625*b^4*c^4 - 2000*a*b^3*c^3*d + 2400*a^2*b^2*c^2*d^2 
 - 1280*a^3*b*c*d^3 + 256*a^4*d^4)/a^9)^(3/4) - (125*b^3*c^3 - 300*a*b^2*c 
^2*d + 240*a^2*b*c*d^2 - 64*a^3*d^3)*(b*x^4 + a)^(1/4)) + (I*a^2*b*x^8 + I 
*a^3*x^4)*((625*b^4*c^4 - 2000*a*b^3*c^3*d + 2400*a^2*b^2*c^2*d^2 - 1280*a 
^3*b*c*d^3 + 256*a^4*d^4)/a^9)^(1/4)*log(-I*a^7*((625*b^4*c^4 - 2000*a*b^3 
*c^3*d + 2400*a^2*b^2*c^2*d^2 - 1280*a^3*b*c*d^3 + 256*a^4*d^4)/a^9)^(3/4) 
 - (125*b^3*c^3 - 300*a*b^2*c^2*d + 240*a^2*b*c*d^2 - 64*a^3*d^3)*(b*x^4 + 
 a)^(1/4)) - (a^2*b*x^8 + a^3*x^4)*((625*b^4*c^4 - 2000*a*b^3*c^3*d + 2400 
*a^2*b^2*c^2*d^2 - 1280*a^3*b*c*d^3 + 256*a^4*d^4)/a^9)^(1/4)*log(-a^7*((6 
25*b^4*c^4 - 2000*a*b^3*c^3*d + 2400*a^2*b^2*c^2*d^2 - 1280*a^3*b*c*d^3 + 
256*a^4*d^4)/a^9)^(3/4) - (125*b^3*c^3 - 300*a*b^2*c^2*d + 240*a^2*b*c*d^2 
 - 64*a^3*d^3)*(b*x^4 + a)^(1/4)) + 4*((5*b*c - 4*a*d)*x^4 + a*c)*(b*x^4 + 
 a)^(3/4))/(a^2*b*x^8 + a^3*x^4)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 58.57 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.69 \[ \int \frac {c+d x^4}{x^5 \left (a+b x^4\right )^{5/4}} \, dx=- \frac {c \Gamma \left (\frac {9}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{4}, \frac {9}{4} \\ \frac {13}{4} \end {matrix}\middle | {\frac {a e^{i \pi }}{b x^{4}}} \right )}}{4 b^{\frac {5}{4}} x^{9} \Gamma \left (\frac {13}{4}\right )} - \frac {d \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{4}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {\frac {a e^{i \pi }}{b x^{4}}} \right )}}{4 b^{\frac {5}{4}} x^{5} \Gamma \left (\frac {9}{4}\right )} \] Input:

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

Output:

-c*gamma(9/4)*hyper((5/4, 9/4), (13/4,), a*exp_polar(I*pi)/(b*x**4))/(4*b* 
*(5/4)*x**9*gamma(13/4)) - d*gamma(5/4)*hyper((5/4, 5/4), (9/4,), a*exp_po 
lar(I*pi)/(b*x**4))/(4*b**(5/4)*x**5*gamma(9/4))
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.61 \[ \int \frac {c+d x^4}{x^5 \left (a+b x^4\right )^{5/4}} \, dx=-\frac {1}{16} \, c {\left (\frac {4 \, {\left (5 \, {\left (b x^{4} + a\right )} b - 4 \, a b\right )}}{{\left (b x^{4} + a\right )}^{\frac {5}{4}} a^{2} - {\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{3}} + \frac {5 \, b {\left (\frac {2 \, \arctan \left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{a^{\frac {1}{4}}}\right )}{a^{\frac {1}{4}}} + \frac {\log \left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}} - a^{\frac {1}{4}}}{{\left (b x^{4} + a\right )}^{\frac {1}{4}} + a^{\frac {1}{4}}}\right )}{a^{\frac {1}{4}}}\right )}}{a^{2}}\right )} + \frac {1}{4} \, d {\left (\frac {\frac {2 \, \arctan \left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{a^{\frac {1}{4}}}\right )}{a^{\frac {1}{4}}} + \frac {\log \left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}} - a^{\frac {1}{4}}}{{\left (b x^{4} + a\right )}^{\frac {1}{4}} + a^{\frac {1}{4}}}\right )}{a^{\frac {1}{4}}}}{a} + \frac {4}{{\left (b x^{4} + a\right )}^{\frac {1}{4}} a}\right )} \] Input:

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

Output:

-1/16*c*(4*(5*(b*x^4 + a)*b - 4*a*b)/((b*x^4 + a)^(5/4)*a^2 - (b*x^4 + a)^ 
(1/4)*a^3) + 5*b*(2*arctan((b*x^4 + a)^(1/4)/a^(1/4))/a^(1/4) + log(((b*x^ 
4 + a)^(1/4) - a^(1/4))/((b*x^4 + a)^(1/4) + a^(1/4)))/a^(1/4))/a^2) + 1/4 
*d*((2*arctan((b*x^4 + a)^(1/4)/a^(1/4))/a^(1/4) + log(((b*x^4 + a)^(1/4) 
- a^(1/4))/((b*x^4 + a)^(1/4) + a^(1/4)))/a^(1/4))/a + 4/((b*x^4 + a)^(1/4 
)*a))
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 285 vs. \(2 (97) = 194\).

Time = 0.13 (sec) , antiderivative size = 285, normalized size of antiderivative = 2.39 \[ \int \frac {c+d x^4}{x^5 \left (a+b x^4\right )^{5/4}} \, dx=-\frac {\sqrt {2} {\left (5 \, b c - 4 \, a d\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} + 2 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{16 \, \left (-a\right )^{\frac {1}{4}} a^{2}} - \frac {\sqrt {2} {\left (5 \, b c - 4 \, a d\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} - 2 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{16 \, \left (-a\right )^{\frac {1}{4}} a^{2}} + \frac {\sqrt {2} {\left (5 \, b c - 4 \, a d\right )} \log \left (\sqrt {2} {\left (b x^{4} + a\right )}^{\frac {1}{4}} \left (-a\right )^{\frac {1}{4}} + \sqrt {b x^{4} + a} + \sqrt {-a}\right )}{32 \, \left (-a\right )^{\frac {1}{4}} a^{2}} - \frac {\sqrt {2} {\left (5 \, b c - 4 \, a d\right )} \log \left (-\sqrt {2} {\left (b x^{4} + a\right )}^{\frac {1}{4}} \left (-a\right )^{\frac {1}{4}} + \sqrt {b x^{4} + a} + \sqrt {-a}\right )}{32 \, \left (-a\right )^{\frac {1}{4}} a^{2}} - \frac {5 \, {\left (b x^{4} + a\right )} b c - 4 \, a b c - 4 \, {\left (b x^{4} + a\right )} a d + 4 \, a^{2} d}{4 \, {\left ({\left (b x^{4} + a\right )}^{\frac {5}{4}} - {\left (b x^{4} + a\right )}^{\frac {1}{4}} a\right )} a^{2}} \] Input:

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

Output:

-1/16*sqrt(2)*(5*b*c - 4*a*d)*arctan(1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) + 2*( 
b*x^4 + a)^(1/4))/(-a)^(1/4))/((-a)^(1/4)*a^2) - 1/16*sqrt(2)*(5*b*c - 4*a 
*d)*arctan(-1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) - 2*(b*x^4 + a)^(1/4))/(-a)^(1 
/4))/((-a)^(1/4)*a^2) + 1/32*sqrt(2)*(5*b*c - 4*a*d)*log(sqrt(2)*(b*x^4 + 
a)^(1/4)*(-a)^(1/4) + sqrt(b*x^4 + a) + sqrt(-a))/((-a)^(1/4)*a^2) - 1/32* 
sqrt(2)*(5*b*c - 4*a*d)*log(-sqrt(2)*(b*x^4 + a)^(1/4)*(-a)^(1/4) + sqrt(b 
*x^4 + a) + sqrt(-a))/((-a)^(1/4)*a^2) - 1/4*(5*(b*x^4 + a)*b*c - 4*a*b*c 
- 4*(b*x^4 + a)*a*d + 4*a^2*d)/(((b*x^4 + a)^(5/4) - (b*x^4 + a)^(1/4)*a)* 
a^2)
                                                                                    
                                                                                    
 

Mupad [B] (verification not implemented)

Time = 4.28 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.22 \[ \int \frac {c+d x^4}{x^5 \left (a+b x^4\right )^{5/4}} \, dx=\frac {d}{a\,{\left (b\,x^4+a\right )}^{1/4}}-\frac {\frac {b\,c}{a}-\frac {5\,b\,c\,\left (b\,x^4+a\right )}{4\,a^2}}{a\,{\left (b\,x^4+a\right )}^{1/4}-{\left (b\,x^4+a\right )}^{5/4}}+\frac {d\,\mathrm {atan}\left (\frac {{\left (b\,x^4+a\right )}^{1/4}}{a^{1/4}}\right )}{2\,a^{5/4}}-\frac {d\,\mathrm {atanh}\left (\frac {{\left (b\,x^4+a\right )}^{1/4}}{a^{1/4}}\right )}{2\,a^{5/4}}-\frac {5\,b\,c\,\mathrm {atan}\left (\frac {{\left (b\,x^4+a\right )}^{1/4}}{a^{1/4}}\right )}{8\,a^{9/4}}+\frac {5\,b\,c\,\mathrm {atanh}\left (\frac {{\left (b\,x^4+a\right )}^{1/4}}{a^{1/4}}\right )}{8\,a^{9/4}} \] Input:

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

Output:

d/(a*(a + b*x^4)^(1/4)) - ((b*c)/a - (5*b*c*(a + b*x^4))/(4*a^2))/(a*(a + 
b*x^4)^(1/4) - (a + b*x^4)^(5/4)) + (d*atan((a + b*x^4)^(1/4)/a^(1/4)))/(2 
*a^(5/4)) - (d*atanh((a + b*x^4)^(1/4)/a^(1/4)))/(2*a^(5/4)) - (5*b*c*atan 
((a + b*x^4)^(1/4)/a^(1/4)))/(8*a^(9/4)) + (5*b*c*atanh((a + b*x^4)^(1/4)/ 
a^(1/4)))/(8*a^(9/4))
 

Reduce [F]

\[ \int \frac {c+d x^4}{x^5 \left (a+b x^4\right )^{5/4}} \, dx=\left (\int \frac {1}{\left (b \,x^{4}+a \right )^{\frac {1}{4}} a \,x^{5}+\left (b \,x^{4}+a \right )^{\frac {1}{4}} b \,x^{9}}d x \right ) c +\left (\int \frac {1}{\left (b \,x^{4}+a \right )^{\frac {1}{4}} a x +\left (b \,x^{4}+a \right )^{\frac {1}{4}} b \,x^{5}}d x \right ) d \] Input:

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

Output:

int(1/((a + b*x**4)**(1/4)*a*x**5 + (a + b*x**4)**(1/4)*b*x**9),x)*c + int 
(1/((a + b*x**4)**(1/4)*a*x + (a + b*x**4)**(1/4)*b*x**5),x)*d