\(\int \frac {(d+e x)^3}{(a+c x^4)^3} \, dx\) [185]

Optimal result

Integrand size = 17, antiderivative size = 328 \[ \int \frac {(d+e x)^3}{\left (a+c x^4\right )^3} \, dx=-\frac {e^3}{8 c \left (a+c x^4\right )^2}+\frac {x \left (d^3+3 d^2 e x+3 d e^2 x^2\right )}{8 a \left (a+c x^4\right )^2}+\frac {x \left (7 d^3+18 d^2 e x+15 d e^2 x^2\right )}{32 a^2 \left (a+c x^4\right )}+\frac {9 d^2 e \arctan \left (\frac {\sqrt {c} x^2}{\sqrt {a}}\right )}{16 a^{5/2} \sqrt {c}}-\frac {3 d \left (7 \sqrt {c} d^2+5 \sqrt {a} e^2\right ) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{64 \sqrt {2} a^{11/4} c^{3/4}}+\frac {3 d \left (7 \sqrt {c} d^2+5 \sqrt {a} e^2\right ) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{64 \sqrt {2} a^{11/4} c^{3/4}}+\frac {3 d \left (7 \sqrt {c} d^2-5 \sqrt {a} e^2\right ) \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x}{\sqrt {a}+\sqrt {c} x^2}\right )}{64 \sqrt {2} a^{11/4} c^{3/4}} \] Output:

-1/8*e^3/c/(c*x^4+a)^2+1/8*x*(3*d*e^2*x^2+3*d^2*e*x+d^3)/a/(c*x^4+a)^2+1/3 
2*x*(15*d*e^2*x^2+18*d^2*e*x+7*d^3)/a^2/(c*x^4+a)+9/16*d^2*e*arctan(c^(1/2 
)*x^2/a^(1/2))/a^(5/2)/c^(1/2)+3/128*d*(7*c^(1/2)*d^2+5*a^(1/2)*e^2)*arcta 
n(-1+2^(1/2)*c^(1/4)*x/a^(1/4))*2^(1/2)/a^(11/4)/c^(3/4)+3/128*d*(7*c^(1/2 
)*d^2+5*a^(1/2)*e^2)*arctan(1+2^(1/2)*c^(1/4)*x/a^(1/4))*2^(1/2)/a^(11/4)/ 
c^(3/4)+3/128*d*(7*c^(1/2)*d^2-5*a^(1/2)*e^2)*arctanh(2^(1/2)*a^(1/4)*c^(1 
/4)*x/(a^(1/2)+c^(1/2)*x^2))*2^(1/2)/a^(11/4)/c^(3/4)
 

Mathematica [A] (verified)

Time = 0.39 (sec) , antiderivative size = 388, normalized size of antiderivative = 1.18 \[ \int \frac {(d+e x)^3}{\left (a+c x^4\right )^3} \, dx=\frac {\frac {8 a d x \left (7 d^2+18 d e x+15 e^2 x^2\right )}{a+c x^4}-\frac {32 a^2 \left (a e^3-c d x \left (d^2+3 d e x+3 e^2 x^2\right )\right )}{c \left (a+c x^4\right )^2}-\frac {6 \sqrt [4]{a} d \left (7 \sqrt {2} \sqrt {c} d^2+24 \sqrt [4]{a} \sqrt [4]{c} d e+5 \sqrt {2} \sqrt {a} e^2\right ) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{c^{3/4}}+\frac {6 \sqrt [4]{a} d \left (7 \sqrt {2} \sqrt {c} d^2-24 \sqrt [4]{a} \sqrt [4]{c} d e+5 \sqrt {2} \sqrt {a} e^2\right ) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{c^{3/4}}+\frac {3 \sqrt {2} \left (-7 \sqrt [4]{a} \sqrt {c} d^3+5 a^{3/4} d e^2\right ) \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )}{c^{3/4}}+\frac {3 \sqrt {2} \left (7 \sqrt [4]{a} \sqrt {c} d^3-5 a^{3/4} d e^2\right ) \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )}{c^{3/4}}}{256 a^3} \] Input:

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

Output:

((8*a*d*x*(7*d^2 + 18*d*e*x + 15*e^2*x^2))/(a + c*x^4) - (32*a^2*(a*e^3 - 
c*d*x*(d^2 + 3*d*e*x + 3*e^2*x^2)))/(c*(a + c*x^4)^2) - (6*a^(1/4)*d*(7*Sq 
rt[2]*Sqrt[c]*d^2 + 24*a^(1/4)*c^(1/4)*d*e + 5*Sqrt[2]*Sqrt[a]*e^2)*ArcTan 
[1 - (Sqrt[2]*c^(1/4)*x)/a^(1/4)])/c^(3/4) + (6*a^(1/4)*d*(7*Sqrt[2]*Sqrt[ 
c]*d^2 - 24*a^(1/4)*c^(1/4)*d*e + 5*Sqrt[2]*Sqrt[a]*e^2)*ArcTan[1 + (Sqrt[ 
2]*c^(1/4)*x)/a^(1/4)])/c^(3/4) + (3*Sqrt[2]*(-7*a^(1/4)*Sqrt[c]*d^3 + 5*a 
^(3/4)*d*e^2)*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*c^(1/4)*x + Sqrt[c]*x^2])/c^(3 
/4) + (3*Sqrt[2]*(7*a^(1/4)*Sqrt[c]*d^3 - 5*a^(3/4)*d*e^2)*Log[Sqrt[a] + S 
qrt[2]*a^(1/4)*c^(1/4)*x + Sqrt[c]*x^2])/c^(3/4))/(256*a^3)
 

Rubi [A] (verified)

Time = 1.03 (sec) , antiderivative size = 408, normalized size of antiderivative = 1.24, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {2393, 25, 2394, 27, 2415, 2009}

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.

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

\(\Big \downarrow \) 2393

\(\displaystyle -\frac {\int -\frac {7 d^3+18 e x d^2+15 e^2 x^2 d}{\left (c x^4+a\right )^2}dx}{8 a}-\frac {a e^3-c x \left (d^3+3 d^2 e x+3 d e^2 x^2\right )}{8 a c \left (a+c x^4\right )^2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\int \frac {7 d^3+18 e x d^2+15 e^2 x^2 d}{\left (c x^4+a\right )^2}dx}{8 a}-\frac {a e^3-c x \left (d^3+3 d^2 e x+3 d e^2 x^2\right )}{8 a c \left (a+c x^4\right )^2}\)

\(\Big \downarrow \) 2394

\(\displaystyle \frac {\frac {x \left (7 d^3+18 d^2 e x+15 d e^2 x^2\right )}{4 a \left (a+c x^4\right )}-\frac {\int -\frac {3 \left (7 d^3+12 e x d^2+5 e^2 x^2 d\right )}{c x^4+a}dx}{4 a}}{8 a}-\frac {a e^3-c x \left (d^3+3 d^2 e x+3 d e^2 x^2\right )}{8 a c \left (a+c x^4\right )^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {3 \int \frac {7 d^3+12 e x d^2+5 e^2 x^2 d}{c x^4+a}dx}{4 a}+\frac {x \left (7 d^3+18 d^2 e x+15 d e^2 x^2\right )}{4 a \left (a+c x^4\right )}}{8 a}-\frac {a e^3-c x \left (d^3+3 d^2 e x+3 d e^2 x^2\right )}{8 a c \left (a+c x^4\right )^2}\)

\(\Big \downarrow \) 2415

\(\displaystyle \frac {\frac {3 \int \left (\frac {12 e x d^2}{c x^4+a}+\frac {7 d^3+5 e^2 x^2 d}{c x^4+a}\right )dx}{4 a}+\frac {x \left (7 d^3+18 d^2 e x+15 d e^2 x^2\right )}{4 a \left (a+c x^4\right )}}{8 a}-\frac {a e^3-c x \left (d^3+3 d^2 e x+3 d e^2 x^2\right )}{8 a c \left (a+c x^4\right )^2}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\frac {3 \left (-\frac {d \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right ) \left (5 \sqrt {a} e^2+7 \sqrt {c} d^2\right )}{2 \sqrt {2} a^{3/4} c^{3/4}}+\frac {d \arctan \left (\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}+1\right ) \left (5 \sqrt {a} e^2+7 \sqrt {c} d^2\right )}{2 \sqrt {2} a^{3/4} c^{3/4}}-\frac {d \left (7 \sqrt {c} d^2-5 \sqrt {a} e^2\right ) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {a}+\sqrt {c} x^2\right )}{4 \sqrt {2} a^{3/4} c^{3/4}}+\frac {d \left (7 \sqrt {c} d^2-5 \sqrt {a} e^2\right ) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {a}+\sqrt {c} x^2\right )}{4 \sqrt {2} a^{3/4} c^{3/4}}+\frac {6 d^2 e \arctan \left (\frac {\sqrt {c} x^2}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {c}}\right )}{4 a}+\frac {x \left (7 d^3+18 d^2 e x+15 d e^2 x^2\right )}{4 a \left (a+c x^4\right )}}{8 a}-\frac {a e^3-c x \left (d^3+3 d^2 e x+3 d e^2 x^2\right )}{8 a c \left (a+c x^4\right )^2}\)

Input:

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

Output:

-1/8*(a*e^3 - c*x*(d^3 + 3*d^2*e*x + 3*d*e^2*x^2))/(a*c*(a + c*x^4)^2) + ( 
(x*(7*d^3 + 18*d^2*e*x + 15*d*e^2*x^2))/(4*a*(a + c*x^4)) + (3*((6*d^2*e*A 
rcTan[(Sqrt[c]*x^2)/Sqrt[a]])/(Sqrt[a]*Sqrt[c]) - (d*(7*Sqrt[c]*d^2 + 5*Sq 
rt[a]*e^2)*ArcTan[1 - (Sqrt[2]*c^(1/4)*x)/a^(1/4)])/(2*Sqrt[2]*a^(3/4)*c^( 
3/4)) + (d*(7*Sqrt[c]*d^2 + 5*Sqrt[a]*e^2)*ArcTan[1 + (Sqrt[2]*c^(1/4)*x)/ 
a^(1/4)])/(2*Sqrt[2]*a^(3/4)*c^(3/4)) - (d*(7*Sqrt[c]*d^2 - 5*Sqrt[a]*e^2) 
*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*c^(1/4)*x + Sqrt[c]*x^2])/(4*Sqrt[2]*a^(3/4 
)*c^(3/4)) + (d*(7*Sqrt[c]*d^2 - 5*Sqrt[a]*e^2)*Log[Sqrt[a] + Sqrt[2]*a^(1 
/4)*c^(1/4)*x + Sqrt[c]*x^2])/(4*Sqrt[2]*a^(3/4)*c^(3/4))))/(4*a))/(8*a)
 

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 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2393

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Module[{q = Expon[Pq, 
 x], i}, Simp[(a*Coeff[Pq, x, q] - b*x*ExpandToSum[Pq - Coeff[Pq, x, q]*x^q 
, x])*((a + b*x^n)^(p + 1)/(a*b*n*(p + 1))), x] + Simp[1/(a*n*(p + 1))   In 
t[Sum[(n*(p + 1) + i + 1)*Coeff[Pq, x, i]*x^i, {i, 0, q - 1}]*(a + b*x^n)^( 
p + 1), x], x] /; q == n - 1] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && IGtQ[n 
, 0] && LtQ[p, -1]
 

rule 2394

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(-x)*Pq*((a + b 
*x^n)^(p + 1)/(a*n*(p + 1))), x] + Simp[1/(a*n*(p + 1))   Int[ExpandToSum[n 
*(p + 1)*Pq + D[x*Pq, x], x]*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x 
] && PolyQ[Pq, x] && IGtQ[n, 0] && LtQ[p, -1] && LtQ[Expon[Pq, x], n - 1]
 

rule 2415

Int[(Pq_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = Sum[x^ii*((Coeff 
[Pq, x, ii] + Coeff[Pq, x, n/2 + ii]*x^(n/2))/(a + b*x^n)), {ii, 0, n/2 - 1 
}]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && IGtQ[n/2, 
 0] && Expon[Pq, x] < n
 

Maple [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.18 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.43

method	result	size
risch	\(\frac {\frac {15 c d \,e^{2} x^{7}}{32 a^{2}}+\frac {9 d^{2} e c \,x^{6}}{16 a^{2}}+\frac {7 c \,d^{3} x^{5}}{32 a^{2}}+\frac {27 d \,e^{2} x^{3}}{32 a}+\frac {15 d^{2} e \,x^{2}}{16 a}+\frac {11 d^{3} x}{32 a}-\frac {e^{3}}{8 c}}{\left (c \,x^{4}+a \right )^{2}}+\frac {3 d \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (c \,\textit {\_Z}^{4}+a \right )}{\sum }\frac {\left (5 e^{2} \textit {\_R}^{2}+12 d e \textit {\_R} +7 d^{2}\right ) \ln \left (x -\textit {\_R} \right )}{\textit {\_R}^{3}}\right )}{128 a^{2} c}\)	\(141\)
default	\(d^{3} \left (\frac {x}{8 a \left (c \,x^{4}+a \right )^{2}}+\frac {\frac {7 x}{32 a \left (c \,x^{4}+a \right )}+\frac {21 \left (\frac {a}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x^{2}+\left (\frac {a}{c}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {a}{c}}}{x^{2}-\left (\frac {a}{c}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {a}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{256 a^{2}}}{a}\right )+3 d^{2} e \left (\frac {x^{2}}{8 a \left (c \,x^{4}+a \right )^{2}}+\frac {\frac {3 x^{2}}{16 a \left (c \,x^{4}+a \right )}+\frac {3 \arctan \left (\sqrt {\frac {c}{a}}\, x^{2}\right )}{16 a \sqrt {a c}}}{a}\right )+3 d \,e^{2} \left (\frac {x^{3}}{8 a \left (c \,x^{4}+a \right )^{2}}+\frac {\frac {5 x^{3}}{32 a \left (c \,x^{4}+a \right )}+\frac {5 \sqrt {2}\, \left (\ln \left (\frac {x^{2}-\left (\frac {a}{c}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {a}{c}}}{x^{2}+\left (\frac {a}{c}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {a}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{256 a c \left (\frac {a}{c}\right )^{\frac {1}{4}}}}{a}\right )+e^{3} \left (\frac {x^{4}}{8 a \left (c \,x^{4}+a \right )^{2}}+\frac {x^{4}}{8 a^{2} \left (c \,x^{4}+a \right )}\right )\)	\(403\)

Input:

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

Output:

(15/32*c*d*e^2/a^2*x^7+9/16*d^2*e*c/a^2*x^6+7/32*c*d^3/a^2*x^5+27/32*d*e^2 
/a*x^3+15/16*d^2*e/a*x^2+11/32*d^3/a*x-1/8*e^3/c)/(c*x^4+a)^2+3/128/a^2*d/ 
c*sum((5*_R^2*e^2+12*_R*d*e+7*d^2)/_R^3*ln(x-_R),_R=RootOf(_Z^4*c+a))
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 15.07 (sec) , antiderivative size = 95566, normalized size of antiderivative = 291.36 \[ \int \frac {(d+e x)^3}{\left (a+c x^4\right )^3} \, dx=\text {Too large to display} \] Input:

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

Output:

Too large to include
 

Sympy [A] (verification not implemented)

Time = 6.25 (sec) , antiderivative size = 413, normalized size of antiderivative = 1.26 \[ \int \frac {(d+e x)^3}{\left (a+c x^4\right )^3} \, dx=\operatorname {RootSum} {\left (268435456 t^{4} a^{11} c^{3} + 63111168 t^{2} a^{6} c^{2} d^{4} e^{2} + t \left (4147200 a^{4} c d^{4} e^{5} - 8128512 a^{3} c^{2} d^{8} e\right ) + 50625 a^{2} d^{4} e^{8} + 245106 a c d^{8} e^{4} + 194481 c^{2} d^{12}, \left ( t \mapsto t \log {\left (x + \frac {262144000 t^{3} a^{10} c^{2} e^{6} + 3714056192 t^{3} a^{9} c^{3} d^{4} e^{2} - 539688960 t^{2} a^{7} c^{2} d^{4} e^{5} + 202309632 t^{2} a^{6} c^{3} d^{8} e + 77328000 t a^{5} c d^{4} e^{8} + 660699648 t a^{4} c^{2} d^{8} e^{4} + 19361664 t a^{3} c^{3} d^{12} + 3037500 a^{3} d^{4} e^{11} - 26360640 a^{2} c d^{8} e^{7} - 60566940 a c^{2} d^{12} e^{3}}{421875 a^{3} d^{3} e^{12} - 29598075 a^{2} c d^{7} e^{8} - 58012227 a c^{2} d^{11} e^{4} + 3176523 c^{3} d^{15}} \right )} \right )\right )} + \frac {- 4 a^{2} e^{3} + 11 a c d^{3} x + 30 a c d^{2} e x^{2} + 27 a c d e^{2} x^{3} + 7 c^{2} d^{3} x^{5} + 18 c^{2} d^{2} e x^{6} + 15 c^{2} d e^{2} x^{7}}{32 a^{4} c + 64 a^{3} c^{2} x^{4} + 32 a^{2} c^{3} x^{8}} \] Input:

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

Output:

RootSum(268435456*_t**4*a**11*c**3 + 63111168*_t**2*a**6*c**2*d**4*e**2 + 
_t*(4147200*a**4*c*d**4*e**5 - 8128512*a**3*c**2*d**8*e) + 50625*a**2*d**4 
*e**8 + 245106*a*c*d**8*e**4 + 194481*c**2*d**12, Lambda(_t, _t*log(x + (2 
62144000*_t**3*a**10*c**2*e**6 + 3714056192*_t**3*a**9*c**3*d**4*e**2 - 53 
9688960*_t**2*a**7*c**2*d**4*e**5 + 202309632*_t**2*a**6*c**3*d**8*e + 773 
28000*_t*a**5*c*d**4*e**8 + 660699648*_t*a**4*c**2*d**8*e**4 + 19361664*_t 
*a**3*c**3*d**12 + 3037500*a**3*d**4*e**11 - 26360640*a**2*c*d**8*e**7 - 6 
0566940*a*c**2*d**12*e**3)/(421875*a**3*d**3*e**12 - 29598075*a**2*c*d**7* 
e**8 - 58012227*a*c**2*d**11*e**4 + 3176523*c**3*d**15)))) + (-4*a**2*e**3 
 + 11*a*c*d**3*x + 30*a*c*d**2*e*x**2 + 27*a*c*d*e**2*x**3 + 7*c**2*d**3*x 
**5 + 18*c**2*d**2*e*x**6 + 15*c**2*d*e**2*x**7)/(32*a**4*c + 64*a**3*c**2 
*x**4 + 32*a**2*c**3*x**8)
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 392, normalized size of antiderivative = 1.20 \[ \int \frac {(d+e x)^3}{\left (a+c x^4\right )^3} \, dx=\frac {15 \, c^{2} d e^{2} x^{7} + 18 \, c^{2} d^{2} e x^{6} + 7 \, c^{2} d^{3} x^{5} + 27 \, a c d e^{2} x^{3} + 30 \, a c d^{2} e x^{2} + 11 \, a c d^{3} x - 4 \, a^{2} e^{3}}{32 \, {\left (a^{2} c^{3} x^{8} + 2 \, a^{3} c^{2} x^{4} + a^{4} c\right )}} + \frac {3 \, d {\left (\frac {\sqrt {2} {\left (7 \, \sqrt {c} d^{2} - 5 \, \sqrt {a} e^{2}\right )} \log \left (\sqrt {c} x^{2} + \sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}} x + \sqrt {a}\right )}{a^{\frac {3}{4}} c^{\frac {3}{4}}} - \frac {\sqrt {2} {\left (7 \, \sqrt {c} d^{2} - 5 \, \sqrt {a} e^{2}\right )} \log \left (\sqrt {c} x^{2} - \sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}} x + \sqrt {a}\right )}{a^{\frac {3}{4}} c^{\frac {3}{4}}} + \frac {2 \, {\left (7 \, \sqrt {2} a^{\frac {1}{4}} c^{\frac {3}{4}} d^{2} + 5 \, \sqrt {2} a^{\frac {3}{4}} c^{\frac {1}{4}} e^{2} - 24 \, \sqrt {a} \sqrt {c} d e\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {c} x + \sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {c}}}\right )}{a^{\frac {3}{4}} \sqrt {\sqrt {a} \sqrt {c}} c^{\frac {3}{4}}} + \frac {2 \, {\left (7 \, \sqrt {2} a^{\frac {1}{4}} c^{\frac {3}{4}} d^{2} + 5 \, \sqrt {2} a^{\frac {3}{4}} c^{\frac {1}{4}} e^{2} + 24 \, \sqrt {a} \sqrt {c} d e\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {c} x - \sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {c}}}\right )}{a^{\frac {3}{4}} \sqrt {\sqrt {a} \sqrt {c}} c^{\frac {3}{4}}}\right )}}{256 \, a^{2}} \] Input:

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

Output:

1/32*(15*c^2*d*e^2*x^7 + 18*c^2*d^2*e*x^6 + 7*c^2*d^3*x^5 + 27*a*c*d*e^2*x 
^3 + 30*a*c*d^2*e*x^2 + 11*a*c*d^3*x - 4*a^2*e^3)/(a^2*c^3*x^8 + 2*a^3*c^2 
*x^4 + a^4*c) + 3/256*d*(sqrt(2)*(7*sqrt(c)*d^2 - 5*sqrt(a)*e^2)*log(sqrt( 
c)*x^2 + sqrt(2)*a^(1/4)*c^(1/4)*x + sqrt(a))/(a^(3/4)*c^(3/4)) - sqrt(2)* 
(7*sqrt(c)*d^2 - 5*sqrt(a)*e^2)*log(sqrt(c)*x^2 - sqrt(2)*a^(1/4)*c^(1/4)* 
x + sqrt(a))/(a^(3/4)*c^(3/4)) + 2*(7*sqrt(2)*a^(1/4)*c^(3/4)*d^2 + 5*sqrt 
(2)*a^(3/4)*c^(1/4)*e^2 - 24*sqrt(a)*sqrt(c)*d*e)*arctan(1/2*sqrt(2)*(2*sq 
rt(c)*x + sqrt(2)*a^(1/4)*c^(1/4))/sqrt(sqrt(a)*sqrt(c)))/(a^(3/4)*sqrt(sq 
rt(a)*sqrt(c))*c^(3/4)) + 2*(7*sqrt(2)*a^(1/4)*c^(3/4)*d^2 + 5*sqrt(2)*a^( 
3/4)*c^(1/4)*e^2 + 24*sqrt(a)*sqrt(c)*d*e)*arctan(1/2*sqrt(2)*(2*sqrt(c)*x 
 - sqrt(2)*a^(1/4)*c^(1/4))/sqrt(sqrt(a)*sqrt(c)))/(a^(3/4)*sqrt(sqrt(a)*s 
qrt(c))*c^(3/4)))/a^2
 

Giac [A] (verification not implemented)

Time = 0.32 (sec) , antiderivative size = 392, normalized size of antiderivative = 1.20 \[ \int \frac {(d+e x)^3}{\left (a+c x^4\right )^3} \, dx=\frac {3 \, \sqrt {2} {\left (12 \, \sqrt {2} \sqrt {a c} c^{2} d^{2} e + 7 \, \left (a c^{3}\right )^{\frac {1}{4}} c^{2} d^{3} + 5 \, \left (a c^{3}\right )^{\frac {3}{4}} d e^{2}\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, x + \sqrt {2} \left (\frac {a}{c}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {a}{c}\right )^{\frac {1}{4}}}\right )}{128 \, a^{3} c^{3}} + \frac {3 \, \sqrt {2} {\left (12 \, \sqrt {2} \sqrt {a c} c^{2} d^{2} e + 7 \, \left (a c^{3}\right )^{\frac {1}{4}} c^{2} d^{3} + 5 \, \left (a c^{3}\right )^{\frac {3}{4}} d e^{2}\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, x - \sqrt {2} \left (\frac {a}{c}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {a}{c}\right )^{\frac {1}{4}}}\right )}{128 \, a^{3} c^{3}} + \frac {3 \, \sqrt {2} {\left (7 \, \left (a c^{3}\right )^{\frac {1}{4}} c^{2} d^{3} - 5 \, \left (a c^{3}\right )^{\frac {3}{4}} d e^{2}\right )} \log \left (x^{2} + \sqrt {2} x \left (\frac {a}{c}\right )^{\frac {1}{4}} + \sqrt {\frac {a}{c}}\right )}{256 \, a^{3} c^{3}} - \frac {3 \, \sqrt {2} {\left (7 \, \left (a c^{3}\right )^{\frac {1}{4}} c^{2} d^{3} - 5 \, \left (a c^{3}\right )^{\frac {3}{4}} d e^{2}\right )} \log \left (x^{2} - \sqrt {2} x \left (\frac {a}{c}\right )^{\frac {1}{4}} + \sqrt {\frac {a}{c}}\right )}{256 \, a^{3} c^{3}} + \frac {15 \, c^{2} d e^{2} x^{7} + 18 \, c^{2} d^{2} e x^{6} + 7 \, c^{2} d^{3} x^{5} + 27 \, a c d e^{2} x^{3} + 30 \, a c d^{2} e x^{2} + 11 \, a c d^{3} x - 4 \, a^{2} e^{3}}{32 \, {\left (c x^{4} + a\right )}^{2} a^{2} c} \] Input:

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

Output:

3/128*sqrt(2)*(12*sqrt(2)*sqrt(a*c)*c^2*d^2*e + 7*(a*c^3)^(1/4)*c^2*d^3 + 
5*(a*c^3)^(3/4)*d*e^2)*arctan(1/2*sqrt(2)*(2*x + sqrt(2)*(a/c)^(1/4))/(a/c 
)^(1/4))/(a^3*c^3) + 3/128*sqrt(2)*(12*sqrt(2)*sqrt(a*c)*c^2*d^2*e + 7*(a* 
c^3)^(1/4)*c^2*d^3 + 5*(a*c^3)^(3/4)*d*e^2)*arctan(1/2*sqrt(2)*(2*x - sqrt 
(2)*(a/c)^(1/4))/(a/c)^(1/4))/(a^3*c^3) + 3/256*sqrt(2)*(7*(a*c^3)^(1/4)*c 
^2*d^3 - 5*(a*c^3)^(3/4)*d*e^2)*log(x^2 + sqrt(2)*x*(a/c)^(1/4) + sqrt(a/c 
))/(a^3*c^3) - 3/256*sqrt(2)*(7*(a*c^3)^(1/4)*c^2*d^3 - 5*(a*c^3)^(3/4)*d* 
e^2)*log(x^2 - sqrt(2)*x*(a/c)^(1/4) + sqrt(a/c))/(a^3*c^3) + 1/32*(15*c^2 
*d*e^2*x^7 + 18*c^2*d^2*e*x^6 + 7*c^2*d^3*x^5 + 27*a*c*d*e^2*x^3 + 30*a*c* 
d^2*e*x^2 + 11*a*c*d^3*x - 4*a^2*e^3)/((c*x^4 + a)^2*a^2*c)
 

Mupad [B] (verification not implemented)

Time = 0.49 (sec) , antiderivative size = 721, normalized size of antiderivative = 2.20 \[ \int \frac {(d+e x)^3}{\left (a+c x^4\right )^3} \, dx=\frac {\frac {11\,d^3\,x}{32\,a}-\frac {e^3}{8\,c}+\frac {7\,c\,d^3\,x^5}{32\,a^2}+\frac {15\,d^2\,e\,x^2}{16\,a}+\frac {27\,d\,e^2\,x^3}{32\,a}+\frac {9\,c\,d^2\,e\,x^6}{16\,a^2}+\frac {15\,c\,d\,e^2\,x^7}{32\,a^2}}{a^2+2\,a\,c\,x^4+c^2\,x^8}+\left (\sum _{k=1}^4\ln \left (\frac {c\,d^2\,\left (6867\,c\,d^5\,e^2-1125\,a\,d\,e^6+7992\,c\,d^4\,e^3\,x-{\mathrm {root}\left (268435456\,a^{11}\,c^3\,z^4+63111168\,a^6\,c^2\,d^4\,e^2\,z^2-8128512\,a^3\,c^2\,d^8\,e\,z+4147200\,a^4\,c\,d^4\,e^5\,z+245106\,a\,c\,d^8\,e^4+50625\,a^2\,d^4\,e^8+194481\,c^2\,d^{12},z,k\right )}^2\,a^5\,c^2\,d\,114688+\mathrm {root}\left (268435456\,a^{11}\,c^3\,z^4+63111168\,a^6\,c^2\,d^4\,e^2\,z^2-8128512\,a^3\,c^2\,d^8\,e\,z+4147200\,a^4\,c\,d^4\,e^5\,z+245106\,a\,c\,d^8\,e^4+50625\,a^2\,d^4\,e^8+194481\,c^2\,d^{12},z,k\right )\,a^3\,c\,e^4\,x\,9600-\mathrm {root}\left (268435456\,a^{11}\,c^3\,z^4+63111168\,a^6\,c^2\,d^4\,e^2\,z^2-8128512\,a^3\,c^2\,d^8\,e\,z+4147200\,a^4\,c\,d^4\,e^5\,z+245106\,a\,c\,d^8\,e^4+50625\,a^2\,d^4\,e^8+194481\,c^2\,d^{12},z,k\right )\,a^2\,c^2\,d^4\,x\,18816+{\mathrm {root}\left (268435456\,a^{11}\,c^3\,z^4+63111168\,a^6\,c^2\,d^4\,e^2\,z^2-8128512\,a^3\,c^2\,d^8\,e\,z+4147200\,a^4\,c\,d^4\,e^5\,z+245106\,a\,c\,d^8\,e^4+50625\,a^2\,d^4\,e^8+194481\,c^2\,d^{12},z,k\right )}^2\,a^5\,c^2\,e\,x\,196608-\mathrm {root}\left (268435456\,a^{11}\,c^3\,z^4+63111168\,a^6\,c^2\,d^4\,e^2\,z^2-8128512\,a^3\,c^2\,d^8\,e\,z+4147200\,a^4\,c\,d^4\,e^5\,z+245106\,a\,c\,d^8\,e^4+50625\,a^2\,d^4\,e^8+194481\,c^2\,d^{12},z,k\right )\,a^3\,c\,d\,e^3\,46080\right )\,3}{a^6\,32768}\right )\,\mathrm {root}\left (268435456\,a^{11}\,c^3\,z^4+63111168\,a^6\,c^2\,d^4\,e^2\,z^2-8128512\,a^3\,c^2\,d^8\,e\,z+4147200\,a^4\,c\,d^4\,e^5\,z+245106\,a\,c\,d^8\,e^4+50625\,a^2\,d^4\,e^8+194481\,c^2\,d^{12},z,k\right )\right ) \] Input:

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

Output:

((11*d^3*x)/(32*a) - e^3/(8*c) + (7*c*d^3*x^5)/(32*a^2) + (15*d^2*e*x^2)/( 
16*a) + (27*d*e^2*x^3)/(32*a) + (9*c*d^2*e*x^6)/(16*a^2) + (15*c*d*e^2*x^7 
)/(32*a^2))/(a^2 + c^2*x^8 + 2*a*c*x^4) + symsum(log((3*c*d^2*(6867*c*d^5* 
e^2 - 1125*a*d*e^6 + 7992*c*d^4*e^3*x - 114688*root(268435456*a^11*c^3*z^4 
 + 63111168*a^6*c^2*d^4*e^2*z^2 - 8128512*a^3*c^2*d^8*e*z + 4147200*a^4*c* 
d^4*e^5*z + 245106*a*c*d^8*e^4 + 50625*a^2*d^4*e^8 + 194481*c^2*d^12, z, k 
)^2*a^5*c^2*d + 9600*root(268435456*a^11*c^3*z^4 + 63111168*a^6*c^2*d^4*e^ 
2*z^2 - 8128512*a^3*c^2*d^8*e*z + 4147200*a^4*c*d^4*e^5*z + 245106*a*c*d^8 
*e^4 + 50625*a^2*d^4*e^8 + 194481*c^2*d^12, z, k)*a^3*c*e^4*x - 18816*root 
(268435456*a^11*c^3*z^4 + 63111168*a^6*c^2*d^4*e^2*z^2 - 8128512*a^3*c^2*d 
^8*e*z + 4147200*a^4*c*d^4*e^5*z + 245106*a*c*d^8*e^4 + 50625*a^2*d^4*e^8 
+ 194481*c^2*d^12, z, k)*a^2*c^2*d^4*x + 196608*root(268435456*a^11*c^3*z^ 
4 + 63111168*a^6*c^2*d^4*e^2*z^2 - 8128512*a^3*c^2*d^8*e*z + 4147200*a^4*c 
*d^4*e^5*z + 245106*a*c*d^8*e^4 + 50625*a^2*d^4*e^8 + 194481*c^2*d^12, z, 
k)^2*a^5*c^2*e*x - 46080*root(268435456*a^11*c^3*z^4 + 63111168*a^6*c^2*d^ 
4*e^2*z^2 - 8128512*a^3*c^2*d^8*e*z + 4147200*a^4*c*d^4*e^5*z + 245106*a*c 
*d^8*e^4 + 50625*a^2*d^4*e^8 + 194481*c^2*d^12, z, k)*a^3*c*d*e^3))/(32768 
*a^6))*root(268435456*a^11*c^3*z^4 + 63111168*a^6*c^2*d^4*e^2*z^2 - 812851 
2*a^3*c^2*d^8*e*z + 4147200*a^4*c*d^4*e^5*z + 245106*a*c*d^8*e^4 + 50625*a 
^2*d^4*e^8 + 194481*c^2*d^12, z, k), k, 1, 4)
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 1303, normalized size of antiderivative = 3.97 \[ \int \frac {(d+e x)^3}{\left (a+c x^4\right )^3} \, dx =\text {Too large to display} \] Input:

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

Output:

( - 30*c**(1/4)*a**(3/4)*sqrt(2)*atan((c**(1/4)*a**(1/4)*sqrt(2) - 2*sqrt( 
c)*x)/(c**(1/4)*a**(1/4)*sqrt(2)))*a**2*d*e**2 - 60*c**(1/4)*a**(3/4)*sqrt 
(2)*atan((c**(1/4)*a**(1/4)*sqrt(2) - 2*sqrt(c)*x)/(c**(1/4)*a**(1/4)*sqrt 
(2)))*a*c*d*e**2*x**4 - 30*c**(1/4)*a**(3/4)*sqrt(2)*atan((c**(1/4)*a**(1/ 
4)*sqrt(2) - 2*sqrt(c)*x)/(c**(1/4)*a**(1/4)*sqrt(2)))*c**2*d*e**2*x**8 - 
42*c**(3/4)*a**(1/4)*sqrt(2)*atan((c**(1/4)*a**(1/4)*sqrt(2) - 2*sqrt(c)*x 
)/(c**(1/4)*a**(1/4)*sqrt(2)))*a**2*d**3 - 84*c**(3/4)*a**(1/4)*sqrt(2)*at 
an((c**(1/4)*a**(1/4)*sqrt(2) - 2*sqrt(c)*x)/(c**(1/4)*a**(1/4)*sqrt(2)))* 
a*c*d**3*x**4 - 42*c**(3/4)*a**(1/4)*sqrt(2)*atan((c**(1/4)*a**(1/4)*sqrt( 
2) - 2*sqrt(c)*x)/(c**(1/4)*a**(1/4)*sqrt(2)))*c**2*d**3*x**8 - 144*sqrt(c 
)*sqrt(a)*atan((c**(1/4)*a**(1/4)*sqrt(2) - 2*sqrt(c)*x)/(c**(1/4)*a**(1/4 
)*sqrt(2)))*a**2*d**2*e - 288*sqrt(c)*sqrt(a)*atan((c**(1/4)*a**(1/4)*sqrt 
(2) - 2*sqrt(c)*x)/(c**(1/4)*a**(1/4)*sqrt(2)))*a*c*d**2*e*x**4 - 144*sqrt 
(c)*sqrt(a)*atan((c**(1/4)*a**(1/4)*sqrt(2) - 2*sqrt(c)*x)/(c**(1/4)*a**(1 
/4)*sqrt(2)))*c**2*d**2*e*x**8 + 30*c**(1/4)*a**(3/4)*sqrt(2)*atan((c**(1/ 
4)*a**(1/4)*sqrt(2) + 2*sqrt(c)*x)/(c**(1/4)*a**(1/4)*sqrt(2)))*a**2*d*e** 
2 + 60*c**(1/4)*a**(3/4)*sqrt(2)*atan((c**(1/4)*a**(1/4)*sqrt(2) + 2*sqrt( 
c)*x)/(c**(1/4)*a**(1/4)*sqrt(2)))*a*c*d*e**2*x**4 + 30*c**(1/4)*a**(3/4)* 
sqrt(2)*atan((c**(1/4)*a**(1/4)*sqrt(2) + 2*sqrt(c)*x)/(c**(1/4)*a**(1/4)* 
sqrt(2)))*c**2*d*e**2*x**8 + 42*c**(3/4)*a**(1/4)*sqrt(2)*atan((c**(1/4...