\(\int \frac {c+d x^4}{x (a+b x^4)^{9/4}} \, dx\) [148]

Optimal result

Integrand size = 22, antiderivative size = 102 \[ \int \frac {c+d x^4}{x \left (a+b x^4\right )^{9/4}} \, dx=\frac {b c-a d}{5 a b \left (a+b x^4\right )^{5/4}}+\frac {c}{a^2 \sqrt [4]{a+b x^4}}+\frac {c \arctan \left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{2 a^{9/4}}-\frac {c \text {arctanh}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{2 a^{9/4}} \] Output:

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

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.92 \[ \int \frac {c+d x^4}{x \left (a+b x^4\right )^{9/4}} \, dx=\frac {\frac {2 \sqrt [4]{a} \left (6 a b c-a^2 d+5 b^2 c x^4\right )}{b \left (a+b x^4\right )^{5/4}}+5 c \arctan \left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )-5 c \text {arctanh}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{10 a^{9/4}} \] Input:

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

Output:

((2*a^(1/4)*(6*a*b*c - a^2*d + 5*b^2*c*x^4))/(b*(a + b*x^4)^(5/4)) + 5*c*A 
rcTan[(a + b*x^4)^(1/4)/a^(1/4)] - 5*c*ArcTanh[(a + b*x^4)^(1/4)/a^(1/4)]) 
/(10*a^(9/4))
 

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.14, 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.

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

\(\Big \downarrow \) 948

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

\(\Big \downarrow \) 87

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

\(\Big \downarrow \) 61

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 827

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

\(\Big \downarrow \) 216

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

((4*(b*c - a*d))/(5*a*b*(a + b*x^4)^(5/4)) + (c*(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))/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 = 0.16 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.02

method	result	size
pseudoelliptic	\(-\frac {-\frac {24 c b \,a^{\frac {5}{4}}}{5}+\frac {4 d \,a^{\frac {9}{4}}}{5}+b \left (-4 b \,x^{4} a^{\frac {1}{4}}+\left (b \,x^{4}+a \right )^{\frac {5}{4}} \left (\ln \left (\frac {\left (b \,x^{4}+a \right )^{\frac {1}{4}}+a^{\frac {1}{4}}}{\left (b \,x^{4}+a \right )^{\frac {1}{4}}-a^{\frac {1}{4}}}\right )-2 \arctan \left (\frac {\left (b \,x^{4}+a \right )^{\frac {1}{4}}}{a^{\frac {1}{4}}}\right )\right )\right ) c}{4 a^{\frac {9}{4}} \left (b \,x^{4}+a \right )^{\frac {5}{4}} b}\)	\(104\)

Input:

int((d*x^4+c)/x/(b*x^4+a)^(9/4),x,method=_RETURNVERBOSE)
 

Output:

-1/4/a^(9/4)*(-24/5*c*b*a^(5/4)+4/5*d*a^(9/4)+b*(-4*b*x^4*a^(1/4)+(b*x^4+a 
)^(5/4)*(ln(((b*x^4+a)^(1/4)+a^(1/4))/((b*x^4+a)^(1/4)-a^(1/4)))-2*arctan( 
(b*x^4+a)^(1/4)/a^(1/4))))*c)/(b*x^4+a)^(5/4)/b
                                                                                    
                                                                                    
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.13 (sec) , antiderivative size = 335, normalized size of antiderivative = 3.28 \[ \int \frac {c+d x^4}{x \left (a+b x^4\right )^{9/4}} \, dx=-\frac {5 \, {\left (a^{2} b^{3} x^{8} + 2 \, a^{3} b^{2} x^{4} + a^{4} b\right )} \left (\frac {c^{4}}{a^{9}}\right )^{\frac {1}{4}} \log \left (a^{7} \left (\frac {c^{4}}{a^{9}}\right )^{\frac {3}{4}} + {\left (b x^{4} + a\right )}^{\frac {1}{4}} c^{3}\right ) + 5 \, {\left (-i \, a^{2} b^{3} x^{8} - 2 i \, a^{3} b^{2} x^{4} - i \, a^{4} b\right )} \left (\frac {c^{4}}{a^{9}}\right )^{\frac {1}{4}} \log \left (i \, a^{7} \left (\frac {c^{4}}{a^{9}}\right )^{\frac {3}{4}} + {\left (b x^{4} + a\right )}^{\frac {1}{4}} c^{3}\right ) + 5 \, {\left (i \, a^{2} b^{3} x^{8} + 2 i \, a^{3} b^{2} x^{4} + i \, a^{4} b\right )} \left (\frac {c^{4}}{a^{9}}\right )^{\frac {1}{4}} \log \left (-i \, a^{7} \left (\frac {c^{4}}{a^{9}}\right )^{\frac {3}{4}} + {\left (b x^{4} + a\right )}^{\frac {1}{4}} c^{3}\right ) - 5 \, {\left (a^{2} b^{3} x^{8} + 2 \, a^{3} b^{2} x^{4} + a^{4} b\right )} \left (\frac {c^{4}}{a^{9}}\right )^{\frac {1}{4}} \log \left (-a^{7} \left (\frac {c^{4}}{a^{9}}\right )^{\frac {3}{4}} + {\left (b x^{4} + a\right )}^{\frac {1}{4}} c^{3}\right ) - 4 \, {\left (5 \, b^{2} c x^{4} + 6 \, a b c - a^{2} d\right )} {\left (b x^{4} + a\right )}^{\frac {3}{4}}}{20 \, {\left (a^{2} b^{3} x^{8} + 2 \, a^{3} b^{2} x^{4} + a^{4} b\right )}} \] Input:

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

Output:

-1/20*(5*(a^2*b^3*x^8 + 2*a^3*b^2*x^4 + a^4*b)*(c^4/a^9)^(1/4)*log(a^7*(c^ 
4/a^9)^(3/4) + (b*x^4 + a)^(1/4)*c^3) + 5*(-I*a^2*b^3*x^8 - 2*I*a^3*b^2*x^ 
4 - I*a^4*b)*(c^4/a^9)^(1/4)*log(I*a^7*(c^4/a^9)^(3/4) + (b*x^4 + a)^(1/4) 
*c^3) + 5*(I*a^2*b^3*x^8 + 2*I*a^3*b^2*x^4 + I*a^4*b)*(c^4/a^9)^(1/4)*log( 
-I*a^7*(c^4/a^9)^(3/4) + (b*x^4 + a)^(1/4)*c^3) - 5*(a^2*b^3*x^8 + 2*a^3*b 
^2*x^4 + a^4*b)*(c^4/a^9)^(1/4)*log(-a^7*(c^4/a^9)^(3/4) + (b*x^4 + a)^(1/ 
4)*c^3) - 4*(5*b^2*c*x^4 + 6*a*b*c - a^2*d)*(b*x^4 + a)^(3/4))/(a^2*b^3*x^ 
8 + 2*a^3*b^2*x^4 + a^4*b)
 

Sympy [A] (verification not implemented)

Time = 43.03 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.86 \[ \int \frac {c+d x^4}{x \left (a+b x^4\right )^{9/4}} \, dx=d \left (\begin {cases} - \frac {1}{5 a b \sqrt [4]{a + b x^{4}} + 5 b^{2} x^{4} \sqrt [4]{a + b x^{4}}} & \text {for}\: b \neq 0 \\\frac {x^{4}}{4 a^{\frac {9}{4}}} & \text {otherwise} \end {cases}\right ) - \frac {c \Gamma \left (\frac {9}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {9}{4}, \frac {9}{4} \\ \frac {13}{4} \end {matrix}\middle | {\frac {a e^{i \pi }}{b x^{4}}} \right )}}{4 b^{\frac {9}{4}} x^{9} \Gamma \left (\frac {13}{4}\right )} \] Input:

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

Output:

d*Piecewise((-1/(5*a*b*(a + b*x**4)**(1/4) + 5*b**2*x**4*(a + b*x**4)**(1/ 
4)), Ne(b, 0)), (x**4/(4*a**(9/4)), True)) - c*gamma(9/4)*hyper((9/4, 9/4) 
, (13/4,), a*exp_polar(I*pi)/(b*x**4))/(4*b**(9/4)*x**9*gamma(13/4))
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.03 \[ \int \frac {c+d x^4}{x \left (a+b x^4\right )^{9/4}} \, dx=\frac {1}{20} \, c {\left (\frac {5 \, {\left (\frac {2 \, \arctan \left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{a^{\frac {1}{4}}}\right )}{a^{\frac {1}{4}}} + \frac {\log \left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}} - a^{\frac {1}{4}}}{{\left (b x^{4} + a\right )}^{\frac {1}{4}} + a^{\frac {1}{4}}}\right )}{a^{\frac {1}{4}}}\right )}}{a^{2}} + \frac {4 \, {\left (5 \, b x^{4} + 6 \, a\right )}}{{\left (b x^{4} + a\right )}^{\frac {5}{4}} a^{2}}\right )} - \frac {d}{5 \, {\left (b x^{4} + a\right )}^{\frac {5}{4}} b} \] Input:

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

Output:

1/20*c*(5*(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 + 4*(5*b*x^4 
+ 6*a)/((b*x^4 + a)^(5/4)*a^2)) - 1/5*d/((b*x^4 + a)^(5/4)*b)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 229 vs. \(2 (80) = 160\).

Time = 0.13 (sec) , antiderivative size = 229, normalized size of antiderivative = 2.25 \[ \int \frac {c+d x^4}{x \left (a+b x^4\right )^{9/4}} \, dx=\frac {\sqrt {2} c \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} + 2 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{4 \, \left (-a\right )^{\frac {1}{4}} a^{2}} + \frac {\sqrt {2} c \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} - 2 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{4 \, \left (-a\right )^{\frac {1}{4}} a^{2}} + \frac {\sqrt {2} \left (-a\right )^{\frac {3}{4}} c \log \left (\sqrt {2} {\left (b x^{4} + a\right )}^{\frac {1}{4}} \left (-a\right )^{\frac {1}{4}} + \sqrt {b x^{4} + a} + \sqrt {-a}\right )}{8 \, a^{3}} + \frac {\sqrt {2} c \log \left (-\sqrt {2} {\left (b x^{4} + a\right )}^{\frac {1}{4}} \left (-a\right )^{\frac {1}{4}} + \sqrt {b x^{4} + a} + \sqrt {-a}\right )}{8 \, \left (-a\right )^{\frac {1}{4}} a^{2}} + \frac {5 \, {\left (b x^{4} + a\right )} b c + a b c - a^{2} d}{5 \, {\left (b x^{4} + a\right )}^{\frac {5}{4}} a^{2} b} \] Input:

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

Output:

1/4*sqrt(2)*c*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) + 1/4*sqrt(2)*c*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) + 1/8*sqr 
t(2)*(-a)^(3/4)*c*log(sqrt(2)*(b*x^4 + a)^(1/4)*(-a)^(1/4) + sqrt(b*x^4 + 
a) + sqrt(-a))/a^3 + 1/8*sqrt(2)*c*log(-sqrt(2)*(b*x^4 + a)^(1/4)*(-a)^(1/ 
4) + sqrt(b*x^4 + a) + sqrt(-a))/((-a)^(1/4)*a^2) + 1/5*(5*(b*x^4 + a)*b*c 
 + a*b*c - a^2*d)/((b*x^4 + a)^(5/4)*a^2*b)
 

Mupad [B] (verification not implemented)

Time = 3.74 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.83 \[ \int \frac {c+d x^4}{x \left (a+b x^4\right )^{9/4}} \, dx=\frac {\frac {c}{5\,a}+\frac {c\,\left (b\,x^4+a\right )}{a^2}}{{\left (b\,x^4+a\right )}^{5/4}}-\frac {d}{5\,b\,{\left (b\,x^4+a\right )}^{5/4}}+\frac {c\,\mathrm {atan}\left (\frac {{\left (b\,x^4+a\right )}^{1/4}}{a^{1/4}}\right )}{2\,a^{9/4}}-\frac {c\,\mathrm {atanh}\left (\frac {{\left (b\,x^4+a\right )}^{1/4}}{a^{1/4}}\right )}{2\,a^{9/4}} \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {c+d x^4}{x \left (a+b x^4\right )^{9/4}} \, dx=\left (\int \frac {x^{3}}{\left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{2}+2 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a b \,x^{4}+\left (b \,x^{4}+a \right )^{\frac {1}{4}} b^{2} x^{8}}d x \right ) d +\left (\int \frac {1}{\left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{2} x +2 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a b \,x^{5}+\left (b \,x^{4}+a \right )^{\frac {1}{4}} b^{2} x^{9}}d x \right ) c \] Input:

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

Output:

int(x**3/((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 + 2*(a 
 + b*x**4)**(1/4)*a*b*x**5 + (a + b*x**4)**(1/4)*b**2*x**9),x)*c