\(\int \frac {x^4 (c+d x^4)}{(a+b x^4)^{9/4}} \, dx\) [151]

Optimal result

Integrand size = 22, antiderivative size = 109 \[ \int \frac {x^4 \left (c+d x^4\right )}{\left (a+b x^4\right )^{9/4}} \, dx=\frac {(b c-a d) x^5}{5 a b \left (a+b x^4\right )^{5/4}}-\frac {d x}{b^2 \sqrt [4]{a+b x^4}}+\frac {d \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 b^{9/4}}+\frac {d \text {arctanh}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 b^{9/4}} \] Output:

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

Mathematica [A] (verified)

Time = 1.34 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.91 \[ \int \frac {x^4 \left (c+d x^4\right )}{\left (a+b x^4\right )^{9/4}} \, dx=\frac {\frac {2 \sqrt [4]{b} x \left (-5 a^2 d+b^2 c x^4-6 a b d x^4\right )}{a \left (a+b x^4\right )^{5/4}}+5 d \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )+5 d \text {arctanh}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{10 b^{9/4}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.07, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {954, 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.

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

\(\Big \downarrow \) 954

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

\(\Big \downarrow \) 817

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

\(\Big \downarrow \) 770

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

\(\Big \downarrow \) 756

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

\(\Big \downarrow \) 216

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

((b*c - a*d)*x^5)/(5*a*b*(a + b*x^4)^(5/4)) + (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))/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 954

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

Maple [A] (verified)

Time = 0.22 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.07

method	result	size
pseudoelliptic	\(-\frac {\frac {12 a \,b^{\frac {5}{4}} d \,x^{5}}{5}-\frac {2 b^{\frac {9}{4}} c \,x^{5}}{5}+d \left (2 a x \,b^{\frac {1}{4}}+\frac {\left (b \,x^{4}+a \right )^{\frac {5}{4}} \left (2 \arctan \left (\frac {\left (b \,x^{4}+a \right )^{\frac {1}{4}}}{x \,b^{\frac {1}{4}}}\right )-\ln \left (\frac {x \,b^{\frac {1}{4}}+\left (b \,x^{4}+a \right )^{\frac {1}{4}}}{-x \,b^{\frac {1}{4}}+\left (b \,x^{4}+a \right )^{\frac {1}{4}}}\right )\right )}{2}\right ) a}{2 b^{\frac {9}{4}} \left (b \,x^{4}+a \right )^{\frac {5}{4}} a}\)	\(117\)

Input:

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

Output:

-1/2/b^(9/4)/(b*x^4+a)^(5/4)*(12/5*a*b^(5/4)*d*x^5-2/5*b^(9/4)*c*x^5+d*(2* 
a*x*b^(1/4)+1/2*(b*x^4+a)^(5/4)*(2*arctan((b*x^4+a)^(1/4)/x/b^(1/4))-ln((x 
*b^(1/4)+(b*x^4+a)^(1/4))/(-x*b^(1/4)+(b*x^4+a)^(1/4)))))*a)/a
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.11 (sec) , antiderivative size = 358, normalized size of antiderivative = 3.28 \[ \int \frac {x^4 \left (c+d x^4\right )}{\left (a+b x^4\right )^{9/4}} \, dx=\frac {5 \, {\left (a b^{4} x^{8} + 2 \, a^{2} b^{3} x^{4} + a^{3} b^{2}\right )} \left (\frac {d^{4}}{b^{9}}\right )^{\frac {1}{4}} \log \left (\frac {b^{7} x \left (\frac {d^{4}}{b^{9}}\right )^{\frac {3}{4}} + {\left (b x^{4} + a\right )}^{\frac {1}{4}} d^{3}}{x}\right ) - 5 \, {\left (a b^{4} x^{8} + 2 \, a^{2} b^{3} x^{4} + a^{3} b^{2}\right )} \left (\frac {d^{4}}{b^{9}}\right )^{\frac {1}{4}} \log \left (-\frac {b^{7} x \left (\frac {d^{4}}{b^{9}}\right )^{\frac {3}{4}} - {\left (b x^{4} + a\right )}^{\frac {1}{4}} d^{3}}{x}\right ) - 5 \, {\left (i \, a b^{4} x^{8} + 2 i \, a^{2} b^{3} x^{4} + i \, a^{3} b^{2}\right )} \left (\frac {d^{4}}{b^{9}}\right )^{\frac {1}{4}} \log \left (\frac {i \, b^{7} x \left (\frac {d^{4}}{b^{9}}\right )^{\frac {3}{4}} + {\left (b x^{4} + a\right )}^{\frac {1}{4}} d^{3}}{x}\right ) - 5 \, {\left (-i \, a b^{4} x^{8} - 2 i \, a^{2} b^{3} x^{4} - i \, a^{3} b^{2}\right )} \left (\frac {d^{4}}{b^{9}}\right )^{\frac {1}{4}} \log \left (\frac {-i \, b^{7} x \left (\frac {d^{4}}{b^{9}}\right )^{\frac {3}{4}} + {\left (b x^{4} + a\right )}^{\frac {1}{4}} d^{3}}{x}\right ) + 4 \, {\left ({\left (b^{2} c - 6 \, a b d\right )} x^{5} - 5 \, a^{2} d x\right )} {\left (b x^{4} + a\right )}^{\frac {3}{4}}}{20 \, {\left (a b^{4} x^{8} + 2 \, a^{2} b^{3} x^{4} + a^{3} b^{2}\right )}} \] Input:

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

Output:

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 27.47 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.94 \[ \int \frac {x^4 \left (c+d x^4\right )}{\left (a+b x^4\right )^{9/4}} \, dx=\frac {c x^{5} \Gamma \left (\frac {5}{4}\right )}{4 a^{\frac {9}{4}} \sqrt [4]{1 + \frac {b x^{4}}{a}} \Gamma \left (\frac {9}{4}\right ) + 4 a^{\frac {5}{4}} b x^{4} \sqrt [4]{1 + \frac {b x^{4}}{a}} \Gamma \left (\frac {9}{4}\right )} + \frac {d x^{9} \Gamma \left (\frac {9}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {9}{4}, \frac {9}{4} \\ \frac {13}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {9}{4}} \Gamma \left (\frac {13}{4}\right )} \] Input:

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

Output:

c*x**5*gamma(5/4)/(4*a**(9/4)*(1 + b*x**4/a)**(1/4)*gamma(9/4) + 4*a**(5/4 
)*b*x**4*(1 + b*x**4/a)**(1/4)*gamma(9/4)) + d*x**9*gamma(9/4)*hyper((9/4, 
 9/4), (13/4,), b*x**4*exp_polar(I*pi)/a)/(4*a**(9/4)*gamma(13/4))
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.16 \[ \int \frac {x^4 \left (c+d x^4\right )}{\left (a+b x^4\right )^{9/4}} \, dx=\frac {c x^{5}}{5 \, {\left (b x^{4} + a\right )}^{\frac {5}{4}} a} - \frac {1}{20} \, {\left (\frac {4 \, {\left (b + \frac {5 \, {\left (b x^{4} + a\right )}}{x^{4}}\right )} x^{5}}{{\left (b x^{4} + a\right )}^{\frac {5}{4}} b^{2}} + \frac {5 \, {\left (\frac {2 \, \arctan \left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{b^{\frac {1}{4}} x}\right )}{b^{\frac {1}{4}}} + \frac {\log \left (-\frac {b^{\frac {1}{4}} - \frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{x}}{b^{\frac {1}{4}} + \frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{x}}\right )}{b^{\frac {1}{4}}}\right )}}{b^{2}}\right )} d \] Input:

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

int(x**8/((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(x**4/((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)*c