\(\int \frac {x^{12} (c+d x^4)}{(a+b x^4)^{13/4}} \, dx\) [165]

Optimal result

Integrand size = 22, antiderivative size = 182 \[ \int \frac {x^{12} \left (c+d x^4\right )}{\left (a+b x^4\right )^{13/4}} \, dx=-\frac {a^2 (b c-a d) x}{9 b^4 \left (a+b x^4\right )^{9/4}}+\frac {a (19 b c-28 a d) x}{45 b^4 \left (a+b x^4\right )^{5/4}}-\frac {(59 b c-158 a d) x}{45 b^4 \sqrt [4]{a+b x^4}}+\frac {d x \left (a+b x^4\right )^{3/4}}{4 b^4}+\frac {(4 b c-13 a d) \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{8 b^{17/4}}+\frac {(4 b c-13 a d) \text {arctanh}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{8 b^{17/4}} \] Output:

-1/9*a^2*(-a*d+b*c)*x/b^4/(b*x^4+a)^(9/4)+1/45*a*(-28*a*d+19*b*c)*x/b^4/(b 
*x^4+a)^(5/4)-1/45*(-158*a*d+59*b*c)*x/b^4/(b*x^4+a)^(1/4)+1/4*d*x*(b*x^4+ 
a)^(3/4)/b^4+1/8*(-13*a*d+4*b*c)*arctan(b^(1/4)*x/(b*x^4+a)^(1/4))/b^(17/4 
)+1/8*(-13*a*d+4*b*c)*arctanh(b^(1/4)*x/(b*x^4+a)^(1/4))/b^(17/4)
 

Mathematica [A] (verified)

Time = 6.11 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.81 \[ \int \frac {x^{12} \left (c+d x^4\right )}{\left (a+b x^4\right )^{13/4}} \, dx=\frac {\frac {2 \sqrt [4]{b} x \left (585 a^3 d-9 a^2 b \left (20 c-143 d x^4\right )+b^3 x^8 \left (-236 c+45 d x^4\right )+a b^2 x^4 \left (-396 c+767 d x^4\right )\right )}{\left (a+b x^4\right )^{9/4}}+45 (4 b c-13 a d) \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )+45 (4 b c-13 a d) \text {arctanh}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{360 b^{17/4}} \] Input:

Integrate[(x^12*(c + d*x^4))/(a + b*x^4)^(13/4),x]
 

Output:

((2*b^(1/4)*x*(585*a^3*d - 9*a^2*b*(20*c - 143*d*x^4) + b^3*x^8*(-236*c + 
45*d*x^4) + a*b^2*x^4*(-396*c + 767*d*x^4)))/(a + b*x^4)^(9/4) + 45*(4*b*c 
 - 13*a*d)*ArcTan[(b^(1/4)*x)/(a + b*x^4)^(1/4)] + 45*(4*b*c - 13*a*d)*Arc 
Tanh[(b^(1/4)*x)/(a + b*x^4)^(1/4)])/(360*b^(17/4))
 

Rubi [A] (verified)

Time = 0.55 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.05, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {957, 817, 817, 843, 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.

\(\displaystyle \int \frac {x^{12} \left (c+d x^4\right )}{\left (a+b x^4\right )^{13/4}} \, dx\)

\(\Big \downarrow \) 957

\(\displaystyle \frac {x^{13} (b c-a d)}{9 a b \left (a+b x^4\right )^{9/4}}-\frac {(4 b c-13 a d) \int \frac {x^{12}}{\left (b x^4+a\right )^{9/4}}dx}{9 a b}\)

\(\Big \downarrow \) 817

\(\displaystyle \frac {x^{13} (b c-a d)}{9 a b \left (a+b x^4\right )^{9/4}}-\frac {(4 b c-13 a d) \left (\frac {9 \int \frac {x^8}{\left (b x^4+a\right )^{5/4}}dx}{5 b}-\frac {x^9}{5 b \left (a+b x^4\right )^{5/4}}\right )}{9 a b}\)

\(\Big \downarrow \) 817

\(\displaystyle \frac {x^{13} (b c-a d)}{9 a b \left (a+b x^4\right )^{9/4}}-\frac {(4 b c-13 a d) \left (\frac {9 \left (\frac {5 \int \frac {x^4}{\sqrt [4]{b x^4+a}}dx}{b}-\frac {x^5}{b \sqrt [4]{a+b x^4}}\right )}{5 b}-\frac {x^9}{5 b \left (a+b x^4\right )^{5/4}}\right )}{9 a b}\)

\(\Big \downarrow \) 843

\(\displaystyle \frac {x^{13} (b c-a d)}{9 a b \left (a+b x^4\right )^{9/4}}-\frac {(4 b c-13 a d) \left (\frac {9 \left (\frac {5 \left (\frac {x \left (a+b x^4\right )^{3/4}}{4 b}-\frac {a \int \frac {1}{\sqrt [4]{b x^4+a}}dx}{4 b}\right )}{b}-\frac {x^5}{b \sqrt [4]{a+b x^4}}\right )}{5 b}-\frac {x^9}{5 b \left (a+b x^4\right )^{5/4}}\right )}{9 a b}\)

\(\Big \downarrow \) 770

\(\displaystyle \frac {x^{13} (b c-a d)}{9 a b \left (a+b x^4\right )^{9/4}}-\frac {(4 b c-13 a d) \left (\frac {9 \left (\frac {5 \left (\frac {x \left (a+b x^4\right )^{3/4}}{4 b}-\frac {a \int \frac {1}{1-\frac {b x^4}{b x^4+a}}d\frac {x}{\sqrt [4]{b x^4+a}}}{4 b}\right )}{b}-\frac {x^5}{b \sqrt [4]{a+b x^4}}\right )}{5 b}-\frac {x^9}{5 b \left (a+b x^4\right )^{5/4}}\right )}{9 a b}\)

\(\Big \downarrow \) 756

\(\displaystyle \frac {x^{13} (b c-a d)}{9 a b \left (a+b x^4\right )^{9/4}}-\frac {(4 b c-13 a d) \left (\frac {9 \left (\frac {5 \left (\frac {x \left (a+b x^4\right )^{3/4}}{4 b}-\frac {a \left (\frac {1}{2} \int \frac {1}{1-\frac {\sqrt {b} x^2}{\sqrt {b x^4+a}}}d\frac {x}{\sqrt [4]{b x^4+a}}+\frac {1}{2} \int \frac {1}{\frac {\sqrt {b} x^2}{\sqrt {b x^4+a}}+1}d\frac {x}{\sqrt [4]{b x^4+a}}\right )}{4 b}\right )}{b}-\frac {x^5}{b \sqrt [4]{a+b x^4}}\right )}{5 b}-\frac {x^9}{5 b \left (a+b x^4\right )^{5/4}}\right )}{9 a b}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {x^{13} (b c-a d)}{9 a b \left (a+b x^4\right )^{9/4}}-\frac {(4 b c-13 a d) \left (\frac {9 \left (\frac {5 \left (\frac {x \left (a+b x^4\right )^{3/4}}{4 b}-\frac {a \left (\frac {1}{2} \int \frac {1}{1-\frac {\sqrt {b} x^2}{\sqrt {b x^4+a}}}d\frac {x}{\sqrt [4]{b x^4+a}}+\frac {\arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 \sqrt [4]{b}}\right )}{4 b}\right )}{b}-\frac {x^5}{b \sqrt [4]{a+b x^4}}\right )}{5 b}-\frac {x^9}{5 b \left (a+b x^4\right )^{5/4}}\right )}{9 a b}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {x^{13} (b c-a d)}{9 a b \left (a+b x^4\right )^{9/4}}-\frac {(4 b c-13 a d) \left (\frac {9 \left (\frac {5 \left (\frac {x \left (a+b x^4\right )^{3/4}}{4 b}-\frac {a \left (\frac {\arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 \sqrt [4]{b}}+\frac {\text {arctanh}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 \sqrt [4]{b}}\right )}{4 b}\right )}{b}-\frac {x^5}{b \sqrt [4]{a+b x^4}}\right )}{5 b}-\frac {x^9}{5 b \left (a+b x^4\right )^{5/4}}\right )}{9 a b}\)

Input:

Int[(x^12*(c + d*x^4))/(a + b*x^4)^(13/4),x]
 

Output:

((b*c - a*d)*x^13)/(9*a*b*(a + b*x^4)^(9/4)) - ((4*b*c - 13*a*d)*(-1/5*x^9 
/(b*(a + b*x^4)^(5/4)) + (9*(-(x^5/(b*(a + b*x^4)^(1/4))) + (5*((x*(a + b* 
x^4)^(3/4))/(4*b) - (a*(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))))/(4*b)))/b))/(5*b))) 
/(9*a*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 843

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*(m + n*p + 1))), x] - Simp[ 
a*c^n*((m - n + 1)/(b*(m + n*p + 1)))   Int[(c*x)^(m - n)*(a + b*x^n)^p, x] 
, x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n* 
p + 1, 0] && IntBinomialQ[a, b, c, n, m, p, x]
 

rule 957

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*n*(p + 1))), x] - Simp[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*b*n* 
(p + 1))   Int[(e*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, 
 m, n}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && (( !IntegerQ[p + 1/2] && N 
eQ[p, -5/4]) ||  !RationalQ[m] || (IGtQ[n, 0] && ILtQ[p + 1/2, 0] && LeQ[-1 
, m, (-n)*(p + 1)]))
 

Maple [A] (verified)

Time = 0.44 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.82

method	result	size
pseudoelliptic	\(-\frac {13 \left (\left (-\frac {4}{13} d \,x^{13}+\frac {944}{585} c \,x^{9}\right ) b^{\frac {13}{4}}+\frac {16 \left (-\frac {143 d \,x^{4}}{20}+c \right ) x \,a^{2} b^{\frac {5}{4}}}{13}+\frac {176 \left (-\frac {767 d \,x^{4}}{396}+c \right ) x^{5} a \,b^{\frac {9}{4}}}{65}-4 a^{3} d x \,b^{\frac {1}{4}}+\left (\ln \left (\frac {x \,b^{\frac {1}{4}}+\left (b \,x^{4}+a \right )^{\frac {1}{4}}}{-x \,b^{\frac {1}{4}}+\left (b \,x^{4}+a \right )^{\frac {1}{4}}}\right )-2 \arctan \left (\frac {\left (b \,x^{4}+a \right )^{\frac {1}{4}}}{x \,b^{\frac {1}{4}}}\right )\right ) \left (a d -\frac {4 c b}{13}\right ) \left (b \,x^{4}+a \right )^{\frac {9}{4}}\right )}{16 b^{\frac {17}{4}} \left (b \,x^{4}+a \right )^{\frac {9}{4}}}\)	\(150\)

Input:

int(x^12*(d*x^4+c)/(b*x^4+a)^(13/4),x,method=_RETURNVERBOSE)
 

Output:

-13/16/b^(17/4)*((-4/13*d*x^13+944/585*c*x^9)*b^(13/4)+16/13*(-143/20*d*x^ 
4+c)*x*a^2*b^(5/4)+176/65*(-767/396*d*x^4+c)*x^5*a*b^(9/4)-4*a^3*d*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)))-2*arctan( 
(b*x^4+a)^(1/4)/x/b^(1/4)))*(a*d-4/13*c*b)*(b*x^4+a)^(9/4))/(b*x^4+a)^(9/4 
)
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.12 (sec) , antiderivative size = 963, normalized size of antiderivative = 5.29 \[ \int \frac {x^{12} \left (c+d x^4\right )}{\left (a+b x^4\right )^{13/4}} \, dx =\text {Too large to display} \] Input:

integrate(x^12*(d*x^4+c)/(b*x^4+a)^(13/4),x, algorithm="fricas")
 

Output:

1/720*(45*(b^7*x^12 + 3*a*b^6*x^8 + 3*a^2*b^5*x^4 + a^3*b^4)*((256*b^4*c^4 
 - 3328*a*b^3*c^3*d + 16224*a^2*b^2*c^2*d^2 - 35152*a^3*b*c*d^3 + 28561*a^ 
4*d^4)/b^17)^(1/4)*log(-(b^13*x*((256*b^4*c^4 - 3328*a*b^3*c^3*d + 16224*a 
^2*b^2*c^2*d^2 - 35152*a^3*b*c*d^3 + 28561*a^4*d^4)/b^17)^(3/4) + (64*b^3* 
c^3 - 624*a*b^2*c^2*d + 2028*a^2*b*c*d^2 - 2197*a^3*d^3)*(b*x^4 + a)^(1/4) 
)/x) - 45*(b^7*x^12 + 3*a*b^6*x^8 + 3*a^2*b^5*x^4 + a^3*b^4)*((256*b^4*c^4 
 - 3328*a*b^3*c^3*d + 16224*a^2*b^2*c^2*d^2 - 35152*a^3*b*c*d^3 + 28561*a^ 
4*d^4)/b^17)^(1/4)*log((b^13*x*((256*b^4*c^4 - 3328*a*b^3*c^3*d + 16224*a^ 
2*b^2*c^2*d^2 - 35152*a^3*b*c*d^3 + 28561*a^4*d^4)/b^17)^(3/4) - (64*b^3*c 
^3 - 624*a*b^2*c^2*d + 2028*a^2*b*c*d^2 - 2197*a^3*d^3)*(b*x^4 + a)^(1/4)) 
/x) - 45*(-I*b^7*x^12 - 3*I*a*b^6*x^8 - 3*I*a^2*b^5*x^4 - I*a^3*b^4)*((256 
*b^4*c^4 - 3328*a*b^3*c^3*d + 16224*a^2*b^2*c^2*d^2 - 35152*a^3*b*c*d^3 + 
28561*a^4*d^4)/b^17)^(1/4)*log((I*b^13*x*((256*b^4*c^4 - 3328*a*b^3*c^3*d 
+ 16224*a^2*b^2*c^2*d^2 - 35152*a^3*b*c*d^3 + 28561*a^4*d^4)/b^17)^(3/4) - 
 (64*b^3*c^3 - 624*a*b^2*c^2*d + 2028*a^2*b*c*d^2 - 2197*a^3*d^3)*(b*x^4 + 
 a)^(1/4))/x) - 45*(I*b^7*x^12 + 3*I*a*b^6*x^8 + 3*I*a^2*b^5*x^4 + I*a^3*b 
^4)*((256*b^4*c^4 - 3328*a*b^3*c^3*d + 16224*a^2*b^2*c^2*d^2 - 35152*a^3*b 
*c*d^3 + 28561*a^4*d^4)/b^17)^(1/4)*log((-I*b^13*x*((256*b^4*c^4 - 3328*a* 
b^3*c^3*d + 16224*a^2*b^2*c^2*d^2 - 35152*a^3*b*c*d^3 + 28561*a^4*d^4)/b^1 
7)^(3/4) - (64*b^3*c^3 - 624*a*b^2*c^2*d + 2028*a^2*b*c*d^2 - 2197*a^3*...
 

Sympy [F(-1)]

Timed out. \[ \int \frac {x^{12} \left (c+d x^4\right )}{\left (a+b x^4\right )^{13/4}} \, dx=\text {Timed out} \] Input:

integrate(x**12*(d*x**4+c)/(b*x**4+a)**(13/4),x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 296, normalized size of antiderivative = 1.63 \[ \int \frac {x^{12} \left (c+d x^4\right )}{\left (a+b x^4\right )^{13/4}} \, dx=-\frac {1}{180} \, {\left (\frac {4 \, {\left (5 \, b^{2} + \frac {9 \, {\left (b x^{4} + a\right )} b}{x^{4}} + \frac {45 \, {\left (b x^{4} + a\right )}^{2}}{x^{8}}\right )} x^{9}}{{\left (b x^{4} + a\right )}^{\frac {9}{4}} b^{3}} + \frac {45 \, {\left (\frac {2 \, \arctan \left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{b^{\frac {1}{4}} x}\right )}{b^{\frac {1}{4}}} + \frac {\log \left (-\frac {b^{\frac {1}{4}} - \frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{x}}{b^{\frac {1}{4}} + \frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{x}}\right )}{b^{\frac {1}{4}}}\right )}}{b^{3}}\right )} c + \frac {1}{720} \, d {\left (\frac {4 \, {\left (20 \, a b^{3} + \frac {52 \, {\left (b x^{4} + a\right )} a b^{2}}{x^{4}} + \frac {468 \, {\left (b x^{4} + a\right )}^{2} a b}{x^{8}} - \frac {585 \, {\left (b x^{4} + a\right )}^{3} a}{x^{12}}\right )}}{\frac {{\left (b x^{4} + a\right )}^{\frac {9}{4}} b^{5}}{x^{9}} - \frac {{\left (b x^{4} + a\right )}^{\frac {13}{4}} b^{4}}{x^{13}}} + \frac {585 \, a {\left (\frac {2 \, \arctan \left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{b^{\frac {1}{4}} x}\right )}{b^{\frac {1}{4}}} + \frac {\log \left (-\frac {b^{\frac {1}{4}} - \frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{x}}{b^{\frac {1}{4}} + \frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{x}}\right )}{b^{\frac {1}{4}}}\right )}}{b^{4}}\right )} \] Input:

integrate(x^12*(d*x^4+c)/(b*x^4+a)^(13/4),x, algorithm="maxima")
 

Output:

-1/180*(4*(5*b^2 + 9*(b*x^4 + a)*b/x^4 + 45*(b*x^4 + a)^2/x^8)*x^9/((b*x^4 
 + a)^(9/4)*b^3) + 45*(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^3)*c + 1/720*d*(4*(20*a*b^3 + 52*(b*x^4 + a)*a*b^2/x^4 + 468*(b*x^4 
+ a)^2*a*b/x^8 - 585*(b*x^4 + a)^3*a/x^12)/((b*x^4 + a)^(9/4)*b^5/x^9 - (b 
*x^4 + a)^(13/4)*b^4/x^13) + 585*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^4)
 

Giac [F]

\[ \int \frac {x^{12} \left (c+d x^4\right )}{\left (a+b x^4\right )^{13/4}} \, dx=\int { \frac {{\left (d x^{4} + c\right )} x^{12}}{{\left (b x^{4} + a\right )}^{\frac {13}{4}}} \,d x } \] Input:

integrate(x^12*(d*x^4+c)/(b*x^4+a)^(13/4),x, algorithm="giac")
 

Output:

integrate((d*x^4 + c)*x^12/(b*x^4 + a)^(13/4), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^{12} \left (c+d x^4\right )}{\left (a+b x^4\right )^{13/4}} \, dx=\int \frac {x^{12}\,\left (d\,x^4+c\right )}{{\left (b\,x^4+a\right )}^{13/4}} \,d x \] Input:

int((x^12*(c + d*x^4))/(a + b*x^4)^(13/4),x)
 

Output:

int((x^12*(c + d*x^4))/(a + b*x^4)^(13/4), x)
 

Reduce [F]

\[ \int \frac {x^{12} \left (c+d x^4\right )}{\left (a+b x^4\right )^{13/4}} \, dx=\left (\int \frac {x^{16}}{\left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{3}+3 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{2} b \,x^{4}+3 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a \,b^{2} x^{8}+\left (b \,x^{4}+a \right )^{\frac {1}{4}} b^{3} x^{12}}d x \right ) d +\left (\int \frac {x^{12}}{\left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{3}+3 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{2} b \,x^{4}+3 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a \,b^{2} x^{8}+\left (b \,x^{4}+a \right )^{\frac {1}{4}} b^{3} x^{12}}d x \right ) c \] Input:

int(x^12*(d*x^4+c)/(b*x^4+a)^(13/4),x)
 

Output:

int(x**16/((a + b*x**4)**(1/4)*a**3 + 3*(a + b*x**4)**(1/4)*a**2*b*x**4 + 
3*(a + b*x**4)**(1/4)*a*b**2*x**8 + (a + b*x**4)**(1/4)*b**3*x**12),x)*d + 
 int(x**12/((a + b*x**4)**(1/4)*a**3 + 3*(a + b*x**4)**(1/4)*a**2*b*x**4 + 
 3*(a + b*x**4)**(1/4)*a*b**2*x**8 + (a + b*x**4)**(1/4)*b**3*x**12),x)*c