\(\int \frac {x^4 (c+d x^4)}{(a+b x^4)^{5/4}} \, dx\) [112]

Optimal result

Integrand size = 22, antiderivative size = 125 \[ \int \frac {x^4 \left (c+d x^4\right )}{\left (a+b x^4\right )^{5/4}} \, dx=-\frac {(4 b c-5 a d) x}{4 b^2 \sqrt [4]{a+b x^4}}+\frac {d x^5}{4 b \sqrt [4]{a+b x^4}}+\frac {(4 b c-5 a d) \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{8 b^{9/4}}+\frac {(4 b c-5 a d) \text {arctanh}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{8 b^{9/4}} \] Output:

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

Mathematica [A] (verified)

Time = 1.06 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.82 \[ \int \frac {x^4 \left (c+d x^4\right )}{\left (a+b x^4\right )^{5/4}} \, dx=\frac {\frac {2 \sqrt [4]{b} x \left (-4 b c+5 a d+b d x^4\right )}{\sqrt [4]{a+b x^4}}+(4 b c-5 a d) \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )+(4 b c-5 a d) \text {arctanh}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{8 b^{9/4}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.94, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {959, 817, 770, 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.

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

\(\Big \downarrow \) 959

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

\(\Big \downarrow \) 817

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

\(\Big \downarrow \) 770

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

\(\Big \downarrow \) 756

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

\(\Big \downarrow \) 216

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

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

Defintions of rubi rules used

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 770

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^(p + 1/n)   Subst[In 
t[1/(1 - b*x^n)^(p + 1/n + 1), x], x, x/(a + b*x^n)^(1/n)], x] /; FreeQ[{a, 
 b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[p, -2^(-1)] && IntegerQ[p + 1 
/n]
 

rule 817

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^( 
n - 1)*(c*x)^(m - n + 1)*((a + b*x^n)^(p + 1)/(b*n*(p + 1))), x] - Simp[c^n 
*((m - n + 1)/(b*n*(p + 1)))   Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x], x 
] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  ! 
ILtQ[(m + n*(p + 1) + 1)/n, 0] && IntBinomialQ[a, b, c, n, m, p, x]
 

rule 959

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

Maple [A] (verified)

Time = 0.26 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.57

method	result	size
pseudoelliptic	\(\frac {4 b^{\frac {5}{4}} d \,x^{5}-16 b^{\frac {5}{4}} c x +10 \arctan \left (\frac {\left (b \,x^{4}+a \right )^{\frac {1}{4}}}{x \,b^{\frac {1}{4}}}\right ) a d \left (b \,x^{4}+a \right )^{\frac {1}{4}}-8 \arctan \left (\frac {\left (b \,x^{4}+a \right )^{\frac {1}{4}}}{x \,b^{\frac {1}{4}}}\right ) c b \left (b \,x^{4}+a \right )^{\frac {1}{4}}-5 \ln \left (\frac {x \,b^{\frac {1}{4}}+\left (b \,x^{4}+a \right )^{\frac {1}{4}}}{-x \,b^{\frac {1}{4}}+\left (b \,x^{4}+a \right )^{\frac {1}{4}}}\right ) a d \left (b \,x^{4}+a \right )^{\frac {1}{4}}+4 \ln \left (\frac {x \,b^{\frac {1}{4}}+\left (b \,x^{4}+a \right )^{\frac {1}{4}}}{-x \,b^{\frac {1}{4}}+\left (b \,x^{4}+a \right )^{\frac {1}{4}}}\right ) c b \left (b \,x^{4}+a \right )^{\frac {1}{4}}+20 a d x \,b^{\frac {1}{4}}}{16 b^{\frac {9}{4}} \left (b \,x^{4}+a \right )^{\frac {1}{4}}}\)	\(196\)

Input:

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

Output:

1/16*(4*b^(5/4)*d*x^5-16*b^(5/4)*c*x+10*arctan((b*x^4+a)^(1/4)/x/b^(1/4))* 
a*d*(b*x^4+a)^(1/4)-8*arctan((b*x^4+a)^(1/4)/x/b^(1/4))*c*b*(b*x^4+a)^(1/4 
)-5*ln((x*b^(1/4)+(b*x^4+a)^(1/4))/(-x*b^(1/4)+(b*x^4+a)^(1/4)))*a*d*(b*x^ 
4+a)^(1/4)+4*ln((x*b^(1/4)+(b*x^4+a)^(1/4))/(-x*b^(1/4)+(b*x^4+a)^(1/4)))* 
c*b*(b*x^4+a)^(1/4)+20*a*d*x*b^(1/4))/b^(9/4)/(b*x^4+a)^(1/4)
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.12 (sec) , antiderivative size = 805, normalized size of antiderivative = 6.44 \[ \int \frac {x^4 \left (c+d x^4\right )}{\left (a+b x^4\right )^{5/4}} \, dx =\text {Too large to display} \] Input:

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

Output:

1/16*((b^3*x^4 + a*b^2)*((256*b^4*c^4 - 1280*a*b^3*c^3*d + 2400*a^2*b^2*c^ 
2*d^2 - 2000*a^3*b*c*d^3 + 625*a^4*d^4)/b^9)^(1/4)*log(-(b^7*x*((256*b^4*c 
^4 - 1280*a*b^3*c^3*d + 2400*a^2*b^2*c^2*d^2 - 2000*a^3*b*c*d^3 + 625*a^4* 
d^4)/b^9)^(3/4) + (64*b^3*c^3 - 240*a*b^2*c^2*d + 300*a^2*b*c*d^2 - 125*a^ 
3*d^3)*(b*x^4 + a)^(1/4))/x) - (b^3*x^4 + a*b^2)*((256*b^4*c^4 - 1280*a*b^ 
3*c^3*d + 2400*a^2*b^2*c^2*d^2 - 2000*a^3*b*c*d^3 + 625*a^4*d^4)/b^9)^(1/4 
)*log((b^7*x*((256*b^4*c^4 - 1280*a*b^3*c^3*d + 2400*a^2*b^2*c^2*d^2 - 200 
0*a^3*b*c*d^3 + 625*a^4*d^4)/b^9)^(3/4) - (64*b^3*c^3 - 240*a*b^2*c^2*d + 
300*a^2*b*c*d^2 - 125*a^3*d^3)*(b*x^4 + a)^(1/4))/x) - (-I*b^3*x^4 - I*a*b 
^2)*((256*b^4*c^4 - 1280*a*b^3*c^3*d + 2400*a^2*b^2*c^2*d^2 - 2000*a^3*b*c 
*d^3 + 625*a^4*d^4)/b^9)^(1/4)*log((I*b^7*x*((256*b^4*c^4 - 1280*a*b^3*c^3 
*d + 2400*a^2*b^2*c^2*d^2 - 2000*a^3*b*c*d^3 + 625*a^4*d^4)/b^9)^(3/4) - ( 
64*b^3*c^3 - 240*a*b^2*c^2*d + 300*a^2*b*c*d^2 - 125*a^3*d^3)*(b*x^4 + a)^ 
(1/4))/x) - (I*b^3*x^4 + I*a*b^2)*((256*b^4*c^4 - 1280*a*b^3*c^3*d + 2400* 
a^2*b^2*c^2*d^2 - 2000*a^3*b*c*d^3 + 625*a^4*d^4)/b^9)^(1/4)*log((-I*b^7*x 
*((256*b^4*c^4 - 1280*a*b^3*c^3*d + 2400*a^2*b^2*c^2*d^2 - 2000*a^3*b*c*d^ 
3 + 625*a^4*d^4)/b^9)^(3/4) - (64*b^3*c^3 - 240*a*b^2*c^2*d + 300*a^2*b*c* 
d^2 - 125*a^3*d^3)*(b*x^4 + a)^(1/4))/x) + 4*(b*d*x^5 - (4*b*c - 5*a*d)*x) 
*(b*x^4 + a)^(3/4))/(b^3*x^4 + a*b^2)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 11.84 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.64 \[ \int \frac {x^4 \left (c+d x^4\right )}{\left (a+b x^4\right )^{5/4}} \, dx=\frac {c x^{5} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{4}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {5}{4}} \Gamma \left (\frac {9}{4}\right )} + \frac {d x^{9} \Gamma \left (\frac {9}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{4}, \frac {9}{4} \\ \frac {13}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {5}{4}} \Gamma \left (\frac {13}{4}\right )} \] Input:

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

Output:

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 224 vs. \(2 (101) = 202\).

Time = 0.11 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.79 \[ \int \frac {x^4 \left (c+d x^4\right )}{\left (a+b x^4\right )^{5/4}} \, dx=\frac {1}{16} \, d {\left (\frac {4 \, {\left (4 \, a b - \frac {5 \, {\left (b x^{4} + a\right )} a}{x^{4}}\right )}}{\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}} b^{3}}{x} - \frac {{\left (b x^{4} + a\right )}^{\frac {5}{4}} b^{2}}{x^{5}}} + \frac {5 \, a {\left (\frac {2 \, \arctan \left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{b^{\frac {1}{4}} x}\right )}{b^{\frac {1}{4}}} + \frac {\log \left (-\frac {b^{\frac {1}{4}} - \frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{x}}{b^{\frac {1}{4}} + \frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{x}}\right )}{b^{\frac {1}{4}}}\right )}}{b^{2}}\right )} - \frac {1}{4} \, c {\left (\frac {\frac {2 \, \arctan \left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{b^{\frac {1}{4}} x}\right )}{b^{\frac {1}{4}}} + \frac {\log \left (-\frac {b^{\frac {1}{4}} - \frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{x}}{b^{\frac {1}{4}} + \frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{x}}\right )}{b^{\frac {1}{4}}}}{b} + \frac {4 \, x}{{\left (b x^{4} + a\right )}^{\frac {1}{4}} b}\right )} \] Input:

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

int(x**8/((a + b*x**4)**(1/4)*a + (a + b*x**4)**(1/4)*b*x**4),x)*d + int(x 
**4/((a + b*x**4)**(1/4)*a + (a + b*x**4)**(1/4)*b*x**4),x)*c