\(\int \frac {c+d x^4}{x^5 (a+b x^4)^{3/4}} \, dx\) [93]

Optimal result

Integrand size = 22, antiderivative size = 95 \[ \int \frac {c+d x^4}{x^5 \left (a+b x^4\right )^{3/4}} \, dx=-\frac {c \sqrt [4]{a+b x^4}}{4 a x^4}+\frac {(3 b c-4 a d) \arctan \left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{8 a^{7/4}}+\frac {(3 b c-4 a d) \text {arctanh}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{8 a^{7/4}} \] Output:

-1/4*c*(b*x^4+a)^(1/4)/a/x^4+1/8*(-4*a*d+3*b*c)*arctan((b*x^4+a)^(1/4)/a^( 
1/4))/a^(7/4)+1/8*(-4*a*d+3*b*c)*arctanh((b*x^4+a)^(1/4)/a^(1/4))/a^(7/4)
 

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.00 \[ \int \frac {c+d x^4}{x^5 \left (a+b x^4\right )^{3/4}} \, dx=-\frac {c \sqrt [4]{a+b x^4}}{4 a x^4}+\frac {(3 b c-4 a d) \arctan \left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{8 a^{7/4}}+\frac {(3 b c-4 a d) \text {arctanh}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{8 a^{7/4}} \] Input:

Integrate[(c + d*x^4)/(x^5*(a + b*x^4)^(3/4)),x]
 

Output:

-1/4*(c*(a + b*x^4)^(1/4))/(a*x^4) + ((3*b*c - 4*a*d)*ArcTan[(a + b*x^4)^( 
1/4)/a^(1/4)])/(8*a^(7/4)) + ((3*b*c - 4*a*d)*ArcTanh[(a + b*x^4)^(1/4)/a^ 
(1/4)])/(8*a^(7/4))
 

Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.04, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {948, 87, 73, 756, 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.

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

\(\Big \downarrow \) 948

\(\displaystyle \frac {1}{4} \int \frac {d x^4+c}{x^8 \left (b x^4+a\right )^{3/4}}dx^4\)

\(\Big \downarrow \) 87

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

\(\Big \downarrow \) 73

\(\displaystyle \frac {1}{4} \left (-\frac {(3 b c-4 a d) \int \frac {1}{\frac {x^{16}}{b}-\frac {a}{b}}d\sqrt [4]{b x^4+a}}{a b}-\frac {c \sqrt [4]{a+b x^4}}{a x^4}\right )\)

\(\Big \downarrow \) 756

\(\displaystyle \frac {1}{4} \left (-\frac {(3 b c-4 a d) \left (-\frac {b \int \frac {1}{\sqrt {a}-x^8}d\sqrt [4]{b x^4+a}}{2 \sqrt {a}}-\frac {b \int \frac {1}{x^8+\sqrt {a}}d\sqrt [4]{b x^4+a}}{2 \sqrt {a}}\right )}{a b}-\frac {c \sqrt [4]{a+b x^4}}{a x^4}\right )\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {1}{4} \left (-\frac {(3 b c-4 a d) \left (-\frac {b \int \frac {1}{\sqrt {a}-x^8}d\sqrt [4]{b x^4+a}}{2 \sqrt {a}}-\frac {b \arctan \left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{2 a^{3/4}}\right )}{a b}-\frac {c \sqrt [4]{a+b x^4}}{a x^4}\right )\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{4} \left (-\frac {(3 b c-4 a d) \left (-\frac {b \arctan \left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{2 a^{3/4}}-\frac {b \text {arctanh}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{2 a^{3/4}}\right )}{a b}-\frac {c \sqrt [4]{a+b x^4}}{a x^4}\right )\)

Input:

Int[(c + d*x^4)/(x^5*(a + b*x^4)^(3/4)),x]
 

Output:

(-((c*(a + b*x^4)^(1/4))/(a*x^4)) - ((3*b*c - 4*a*d)*(-1/2*(b*ArcTan[(a + 
b*x^4)^(1/4)/a^(1/4)])/a^(3/4) - (b*ArcTanh[(a + b*x^4)^(1/4)/a^(1/4)])/(2 
*a^(3/4))))/(a*b))/4
 

Defintions of rubi rules used

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 756

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2 
]], s = Denominator[Rt[-a/b, 2]]}, Simp[r/(2*a)   Int[1/(r - s*x^2), x], x] 
 + Simp[r/(2*a)   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 = 0.23 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.09

method	result	size
pseudoelliptic	\(\frac {x^{4} \left (-4 a^{2} d +3 a b c \right ) \ln \left (\frac {\left (b \,x^{4}+a \right )^{\frac {1}{4}}+a^{\frac {1}{4}}}{\left (b \,x^{4}+a \right )^{\frac {1}{4}}-a^{\frac {1}{4}}}\right )+2 x^{4} \left (-4 a^{2} d +3 a b c \right ) \arctan \left (\frac {\left (b \,x^{4}+a \right )^{\frac {1}{4}}}{a^{\frac {1}{4}}}\right )-4 c \left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{\frac {7}{4}}}{16 a^{\frac {11}{4}} x^{4}}\)	\(104\)

Input:

int((d*x^4+c)/x^5/(b*x^4+a)^(3/4),x,method=_RETURNVERBOSE)
 

Output:

1/16*(x^4*(-4*a^2*d+3*a*b*c)*ln(((b*x^4+a)^(1/4)+a^(1/4))/((b*x^4+a)^(1/4) 
-a^(1/4)))+2*x^4*(-4*a^2*d+3*a*b*c)*arctan((b*x^4+a)^(1/4)/a^(1/4))-4*c*(b 
*x^4+a)^(1/4)*a^(7/4))/a^(11/4)/x^4
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.13 (sec) , antiderivative size = 607, normalized size of antiderivative = 6.39 \[ \int \frac {c+d x^4}{x^5 \left (a+b x^4\right )^{3/4}} \, dx=-\frac {a x^{4} \left (\frac {81 \, b^{4} c^{4} - 432 \, a b^{3} c^{3} d + 864 \, a^{2} b^{2} c^{2} d^{2} - 768 \, a^{3} b c d^{3} + 256 \, a^{4} d^{4}}{a^{7}}\right )^{\frac {1}{4}} \log \left (a^{2} \left (\frac {81 \, b^{4} c^{4} - 432 \, a b^{3} c^{3} d + 864 \, a^{2} b^{2} c^{2} d^{2} - 768 \, a^{3} b c d^{3} + 256 \, a^{4} d^{4}}{a^{7}}\right )^{\frac {1}{4}} - {\left (b x^{4} + a\right )}^{\frac {1}{4}} {\left (3 \, b c - 4 \, a d\right )}\right ) + i \, a x^{4} \left (\frac {81 \, b^{4} c^{4} - 432 \, a b^{3} c^{3} d + 864 \, a^{2} b^{2} c^{2} d^{2} - 768 \, a^{3} b c d^{3} + 256 \, a^{4} d^{4}}{a^{7}}\right )^{\frac {1}{4}} \log \left (i \, a^{2} \left (\frac {81 \, b^{4} c^{4} - 432 \, a b^{3} c^{3} d + 864 \, a^{2} b^{2} c^{2} d^{2} - 768 \, a^{3} b c d^{3} + 256 \, a^{4} d^{4}}{a^{7}}\right )^{\frac {1}{4}} - {\left (b x^{4} + a\right )}^{\frac {1}{4}} {\left (3 \, b c - 4 \, a d\right )}\right ) - i \, a x^{4} \left (\frac {81 \, b^{4} c^{4} - 432 \, a b^{3} c^{3} d + 864 \, a^{2} b^{2} c^{2} d^{2} - 768 \, a^{3} b c d^{3} + 256 \, a^{4} d^{4}}{a^{7}}\right )^{\frac {1}{4}} \log \left (-i \, a^{2} \left (\frac {81 \, b^{4} c^{4} - 432 \, a b^{3} c^{3} d + 864 \, a^{2} b^{2} c^{2} d^{2} - 768 \, a^{3} b c d^{3} + 256 \, a^{4} d^{4}}{a^{7}}\right )^{\frac {1}{4}} - {\left (b x^{4} + a\right )}^{\frac {1}{4}} {\left (3 \, b c - 4 \, a d\right )}\right ) - a x^{4} \left (\frac {81 \, b^{4} c^{4} - 432 \, a b^{3} c^{3} d + 864 \, a^{2} b^{2} c^{2} d^{2} - 768 \, a^{3} b c d^{3} + 256 \, a^{4} d^{4}}{a^{7}}\right )^{\frac {1}{4}} \log \left (-a^{2} \left (\frac {81 \, b^{4} c^{4} - 432 \, a b^{3} c^{3} d + 864 \, a^{2} b^{2} c^{2} d^{2} - 768 \, a^{3} b c d^{3} + 256 \, a^{4} d^{4}}{a^{7}}\right )^{\frac {1}{4}} - {\left (b x^{4} + a\right )}^{\frac {1}{4}} {\left (3 \, b c - 4 \, a d\right )}\right ) + 4 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}} c}{16 \, a x^{4}} \] Input:

integrate((d*x^4+c)/x^5/(b*x^4+a)^(3/4),x, algorithm="fricas")
 

Output:

-1/16*(a*x^4*((81*b^4*c^4 - 432*a*b^3*c^3*d + 864*a^2*b^2*c^2*d^2 - 768*a^ 
3*b*c*d^3 + 256*a^4*d^4)/a^7)^(1/4)*log(a^2*((81*b^4*c^4 - 432*a*b^3*c^3*d 
 + 864*a^2*b^2*c^2*d^2 - 768*a^3*b*c*d^3 + 256*a^4*d^4)/a^7)^(1/4) - (b*x^ 
4 + a)^(1/4)*(3*b*c - 4*a*d)) + I*a*x^4*((81*b^4*c^4 - 432*a*b^3*c^3*d + 8 
64*a^2*b^2*c^2*d^2 - 768*a^3*b*c*d^3 + 256*a^4*d^4)/a^7)^(1/4)*log(I*a^2*( 
(81*b^4*c^4 - 432*a*b^3*c^3*d + 864*a^2*b^2*c^2*d^2 - 768*a^3*b*c*d^3 + 25 
6*a^4*d^4)/a^7)^(1/4) - (b*x^4 + a)^(1/4)*(3*b*c - 4*a*d)) - I*a*x^4*((81* 
b^4*c^4 - 432*a*b^3*c^3*d + 864*a^2*b^2*c^2*d^2 - 768*a^3*b*c*d^3 + 256*a^ 
4*d^4)/a^7)^(1/4)*log(-I*a^2*((81*b^4*c^4 - 432*a*b^3*c^3*d + 864*a^2*b^2* 
c^2*d^2 - 768*a^3*b*c*d^3 + 256*a^4*d^4)/a^7)^(1/4) - (b*x^4 + a)^(1/4)*(3 
*b*c - 4*a*d)) - a*x^4*((81*b^4*c^4 - 432*a*b^3*c^3*d + 864*a^2*b^2*c^2*d^ 
2 - 768*a^3*b*c*d^3 + 256*a^4*d^4)/a^7)^(1/4)*log(-a^2*((81*b^4*c^4 - 432* 
a*b^3*c^3*d + 864*a^2*b^2*c^2*d^2 - 768*a^3*b*c*d^3 + 256*a^4*d^4)/a^7)^(1 
/4) - (b*x^4 + a)^(1/4)*(3*b*c - 4*a*d)) + 4*(b*x^4 + a)^(1/4)*c)/(a*x^4)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 38.21 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.86 \[ \int \frac {c+d x^4}{x^5 \left (a+b x^4\right )^{3/4}} \, dx=- \frac {c \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {a e^{i \pi }}{b x^{4}}} \right )}}{4 b^{\frac {3}{4}} x^{7} \Gamma \left (\frac {11}{4}\right )} - \frac {d \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {a e^{i \pi }}{b x^{4}}} \right )}}{4 b^{\frac {3}{4}} x^{3} \Gamma \left (\frac {7}{4}\right )} \] Input:

integrate((d*x**4+c)/x**5/(b*x**4+a)**(3/4),x)
 

Output:

-c*gamma(7/4)*hyper((3/4, 7/4), (11/4,), a*exp_polar(I*pi)/(b*x**4))/(4*b* 
*(3/4)*x**7*gamma(11/4)) - d*gamma(3/4)*hyper((3/4, 3/4), (7/4,), a*exp_po 
lar(I*pi)/(b*x**4))/(4*b**(3/4)*x**3*gamma(7/4))
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 158 vs. \(2 (75) = 150\).

Time = 0.11 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.66 \[ \int \frac {c+d x^4}{x^5 \left (a+b x^4\right )^{3/4}} \, dx=-\frac {1}{16} \, c {\left (\frac {4 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}} b}{{\left (b x^{4} + a\right )} a - a^{2}} - \frac {3 \, {\left (\frac {2 \, b \arctan \left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{a^{\frac {1}{4}}}\right )}{a^{\frac {3}{4}}} - \frac {b \log \left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}} - a^{\frac {1}{4}}}{{\left (b x^{4} + a\right )}^{\frac {1}{4}} + a^{\frac {1}{4}}}\right )}{a^{\frac {3}{4}}}\right )}}{a}\right )} - \frac {1}{4} \, d {\left (\frac {2 \, \arctan \left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{a^{\frac {1}{4}}}\right )}{a^{\frac {3}{4}}} - \frac {\log \left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}} - a^{\frac {1}{4}}}{{\left (b x^{4} + a\right )}^{\frac {1}{4}} + a^{\frac {1}{4}}}\right )}{a^{\frac {3}{4}}}\right )} \] Input:

integrate((d*x^4+c)/x^5/(b*x^4+a)^(3/4),x, algorithm="maxima")
 

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 257 vs. \(2 (75) = 150\).

Time = 0.13 (sec) , antiderivative size = 257, normalized size of antiderivative = 2.71 \[ \int \frac {c+d x^4}{x^5 \left (a+b x^4\right )^{3/4}} \, dx=-\frac {1}{32} \, b {\left (\frac {2 \, \sqrt {2} {\left (3 \, b c - 4 \, a d\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} + 2 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{\left (-a\right )^{\frac {3}{4}} a b} + \frac {2 \, \sqrt {2} {\left (3 \, b c - 4 \, a d\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} - 2 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{\left (-a\right )^{\frac {3}{4}} a b} + \frac {\sqrt {2} {\left (3 \, b c - 4 \, a d\right )} \log \left (\sqrt {2} {\left (b x^{4} + a\right )}^{\frac {1}{4}} \left (-a\right )^{\frac {1}{4}} + \sqrt {b x^{4} + a} + \sqrt {-a}\right )}{\left (-a\right )^{\frac {3}{4}} a b} - \frac {\sqrt {2} {\left (3 \, b c - 4 \, a d\right )} \log \left (-\sqrt {2} {\left (b x^{4} + a\right )}^{\frac {1}{4}} \left (-a\right )^{\frac {1}{4}} + \sqrt {b x^{4} + a} + \sqrt {-a}\right )}{\left (-a\right )^{\frac {3}{4}} a b} + \frac {8 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}} c}{a b x^{4}}\right )} \] Input:

integrate((d*x^4+c)/x^5/(b*x^4+a)^(3/4),x, algorithm="giac")
 

Output:

-1/32*b*(2*sqrt(2)*(3*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)^(3/4)*a*b) + 2*sqrt(2)*(3*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)^(3/4)*a*b) + sqrt(2)*(3*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)^(3/4)*a*b) - sqrt(2)* 
(3*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)^(3/4)*a*b) + 8*(b*x^4 + a)^(1/4)*c/(a*b*x^4))
 

Mupad [B] (verification not implemented)

Time = 3.84 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.06 \[ \int \frac {c+d x^4}{x^5 \left (a+b x^4\right )^{3/4}} \, dx=\frac {3\,b\,c\,\mathrm {atan}\left (\frac {{\left (b\,x^4+a\right )}^{1/4}}{a^{1/4}}\right )}{8\,a^{7/4}}-\frac {d\,\mathrm {atanh}\left (\frac {{\left (b\,x^4+a\right )}^{1/4}}{a^{1/4}}\right )}{2\,a^{3/4}}-\frac {c\,{\left (b\,x^4+a\right )}^{1/4}}{4\,a\,x^4}-\frac {d\,\mathrm {atan}\left (\frac {{\left (b\,x^4+a\right )}^{1/4}}{a^{1/4}}\right )}{2\,a^{3/4}}+\frac {3\,b\,c\,\mathrm {atanh}\left (\frac {{\left (b\,x^4+a\right )}^{1/4}}{a^{1/4}}\right )}{8\,a^{7/4}} \] Input:

int((c + d*x^4)/(x^5*(a + b*x^4)^(3/4)),x)
 

Output:

(3*b*c*atan((a + b*x^4)^(1/4)/a^(1/4)))/(8*a^(7/4)) - (d*atanh((a + b*x^4) 
^(1/4)/a^(1/4)))/(2*a^(3/4)) - (c*(a + b*x^4)^(1/4))/(4*a*x^4) - (d*atan(( 
a + b*x^4)^(1/4)/a^(1/4)))/(2*a^(3/4)) + (3*b*c*atanh((a + b*x^4)^(1/4)/a^ 
(1/4)))/(8*a^(7/4))
 

Reduce [F]

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

int((d*x^4+c)/x^5/(b*x^4+a)^(3/4),x)
 

Output:

int(1/((a + b*x**4)**(3/4)*x**5),x)*c + int(1/((a + b*x**4)**(3/4)*x),x)*d