\(\int \frac {(d+e x)^3}{(a+c x^4)^2} \, dx\) [401]

Optimal result

Integrand size = 17, antiderivative size = 349 \[ \int \frac {(d+e x)^3}{\left (a+c x^4\right )^2} \, dx=-\frac {a e^3-c x \left (d^3+3 d^2 e x+3 d e^2 x^2\right )}{4 a c \left (a+c x^4\right )}+\frac {3 d^2 e \arctan \left (\frac {\sqrt {c} x^2}{\sqrt {a}}\right )}{4 a^{3/2} \sqrt {c}}-\frac {3 d \left (\sqrt {c} d^2+\sqrt {a} e^2\right ) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{8 \sqrt {2} a^{7/4} c^{3/4}}+\frac {3 d \left (\sqrt {c} d^2+\sqrt {a} e^2\right ) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{8 \sqrt {2} a^{7/4} c^{3/4}}-\frac {3 d \left (\sqrt {c} d^2-\sqrt {a} e^2\right ) \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )}{16 \sqrt {2} a^{7/4} c^{3/4}}+\frac {3 d \left (\sqrt {c} d^2-\sqrt {a} e^2\right ) \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )}{16 \sqrt {2} a^{7/4} c^{3/4}} \]

[Out]

1/4*(-a*e^3+c*x*(3*d*e^2*x^2+3*d^2*e*x+d^3))/a/c/(c*x^4+a)+3/4*d^2*e*arctan(x^2*c^(1/2)/a^(1/2))/a^(3/2)/c^(1/
2)-3/32*d*ln(-a^(1/4)*c^(1/4)*x*2^(1/2)+a^(1/2)+x^2*c^(1/2))*(-e^2*a^(1/2)+d^2*c^(1/2))/a^(7/4)/c^(3/4)*2^(1/2
)+3/32*d*ln(a^(1/4)*c^(1/4)*x*2^(1/2)+a^(1/2)+x^2*c^(1/2))*(-e^2*a^(1/2)+d^2*c^(1/2))/a^(7/4)/c^(3/4)*2^(1/2)+
3/16*d*arctan(-1+c^(1/4)*x*2^(1/2)/a^(1/4))*(e^2*a^(1/2)+d^2*c^(1/2))/a^(7/4)/c^(3/4)*2^(1/2)+3/16*d*arctan(1+
c^(1/4)*x*2^(1/2)/a^(1/4))*(e^2*a^(1/2)+d^2*c^(1/2))/a^(7/4)/c^(3/4)*2^(1/2)

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 349, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.706, Rules used = {1868, 27, 12, 1890, 281, 211, 1182, 1176, 631, 210, 1179, 642} \[ \int \frac {(d+e x)^3}{\left (a+c x^4\right )^2} \, dx=-\frac {3 d \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right ) \left (\sqrt {a} e^2+\sqrt {c} d^2\right )}{8 \sqrt {2} a^{7/4} c^{3/4}}+\frac {3 d \arctan \left (\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}+1\right ) \left (\sqrt {a} e^2+\sqrt {c} d^2\right )}{8 \sqrt {2} a^{7/4} c^{3/4}}+\frac {3 d^2 e \arctan \left (\frac {\sqrt {c} x^2}{\sqrt {a}}\right )}{4 a^{3/2} \sqrt {c}}-\frac {3 d \left (\sqrt {c} d^2-\sqrt {a} e^2\right ) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {a}+\sqrt {c} x^2\right )}{16 \sqrt {2} a^{7/4} c^{3/4}}+\frac {3 d \left (\sqrt {c} d^2-\sqrt {a} e^2\right ) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {a}+\sqrt {c} x^2\right )}{16 \sqrt {2} a^{7/4} c^{3/4}}-\frac {a e^3-c x \left (d^3+3 d^2 e x+3 d e^2 x^2\right )}{4 a c \left (a+c x^4\right )} \]

[In]

Int[(d + e*x)^3/(a + c*x^4)^2,x]

[Out]

-1/4*(a*e^3 - c*x*(d^3 + 3*d^2*e*x + 3*d*e^2*x^2))/(a*c*(a + c*x^4)) + (3*d^2*e*ArcTan[(Sqrt[c]*x^2)/Sqrt[a]])
/(4*a^(3/2)*Sqrt[c]) - (3*d*(Sqrt[c]*d^2 + Sqrt[a]*e^2)*ArcTan[1 - (Sqrt[2]*c^(1/4)*x)/a^(1/4)])/(8*Sqrt[2]*a^
(7/4)*c^(3/4)) + (3*d*(Sqrt[c]*d^2 + Sqrt[a]*e^2)*ArcTan[1 + (Sqrt[2]*c^(1/4)*x)/a^(1/4)])/(8*Sqrt[2]*a^(7/4)*
c^(3/4)) - (3*d*(Sqrt[c]*d^2 - Sqrt[a]*e^2)*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*c^(1/4)*x + Sqrt[c]*x^2])/(16*Sqrt[2
]*a^(7/4)*c^(3/4)) + (3*d*(Sqrt[c]*d^2 - Sqrt[a]*e^2)*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*c^(1/4)*x + Sqrt[c]*x^2])/
(16*Sqrt[2]*a^(7/4)*c^(3/4))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 281

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

Rule 631

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[a*(c/b^2)]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[d*(Log[RemoveContent[a + b*x +
c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1176

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[2*(d/e), 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1179

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[-2*(d/e), 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 1182

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[a*c, 2]}, Dist[(d*q + a*e)/(2*a*c),
Int[(q + c*x^2)/(a + c*x^4), x], x] + Dist[(d*q - a*e)/(2*a*c), Int[(q - c*x^2)/(a + c*x^4), x], x]] /; FreeQ
[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && NeQ[c*d^2 - a*e^2, 0] && NegQ[(-a)*c]

Rule 1868

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] + Dist[1/(a*n*(p + 1))
, Int[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 1890

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

Rubi steps \begin{align*} \text {integral}& = -\frac {a e^3-c x \left (d^3+3 d^2 e x+3 d e^2 x^2\right )}{4 a c \left (a+c x^4\right )}-\frac {\int \frac {-3 d^3-6 d^2 e x-3 d e^2 x^2}{a+c x^4} \, dx}{4 a} \\ & = -\frac {a e^3-c x \left (d^3+3 d^2 e x+3 d e^2 x^2\right )}{4 a c \left (a+c x^4\right )}-\frac {\int -\frac {3 d (d+e x)^2}{a+c x^4} \, dx}{4 a} \\ & = -\frac {a e^3-c x \left (d^3+3 d^2 e x+3 d e^2 x^2\right )}{4 a c \left (a+c x^4\right )}+\frac {(3 d) \int \frac {(d+e x)^2}{a+c x^4} \, dx}{4 a} \\ & = -\frac {a e^3-c x \left (d^3+3 d^2 e x+3 d e^2 x^2\right )}{4 a c \left (a+c x^4\right )}+\frac {(3 d) \int \left (\frac {2 d e x}{a+c x^4}+\frac {d^2+e^2 x^2}{a+c x^4}\right ) \, dx}{4 a} \\ & = -\frac {a e^3-c x \left (d^3+3 d^2 e x+3 d e^2 x^2\right )}{4 a c \left (a+c x^4\right )}+\frac {(3 d) \int \frac {d^2+e^2 x^2}{a+c x^4} \, dx}{4 a}+\frac {\left (3 d^2 e\right ) \int \frac {x}{a+c x^4} \, dx}{2 a} \\ & = -\frac {a e^3-c x \left (d^3+3 d^2 e x+3 d e^2 x^2\right )}{4 a c \left (a+c x^4\right )}+\frac {\left (3 d^2 e\right ) \text {Subst}\left (\int \frac {1}{a+c x^2} \, dx,x,x^2\right )}{4 a}+\frac {\left (3 d \left (\frac {\sqrt {c} d^2}{\sqrt {a}}-e^2\right )\right ) \int \frac {\sqrt {a} \sqrt {c}-c x^2}{a+c x^4} \, dx}{8 a c}+\frac {\left (3 d \left (\frac {\sqrt {c} d^2}{\sqrt {a}}+e^2\right )\right ) \int \frac {\sqrt {a} \sqrt {c}+c x^2}{a+c x^4} \, dx}{8 a c} \\ & = -\frac {a e^3-c x \left (d^3+3 d^2 e x+3 d e^2 x^2\right )}{4 a c \left (a+c x^4\right )}+\frac {3 d^2 e \tan ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {a}}\right )}{4 a^{3/2} \sqrt {c}}+\frac {\left (3 d \left (\frac {\sqrt {c} d^2}{\sqrt {a}}+e^2\right )\right ) \int \frac {1}{\frac {\sqrt {a}}{\sqrt {c}}-\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{c}}+x^2} \, dx}{16 a c}+\frac {\left (3 d \left (\frac {\sqrt {c} d^2}{\sqrt {a}}+e^2\right )\right ) \int \frac {1}{\frac {\sqrt {a}}{\sqrt {c}}+\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{c}}+x^2} \, dx}{16 a c}-\frac {\left (3 d \left (\sqrt {c} d^2-\sqrt {a} e^2\right )\right ) \int \frac {\frac {\sqrt {2} \sqrt [4]{a}}{\sqrt [4]{c}}+2 x}{-\frac {\sqrt {a}}{\sqrt {c}}-\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{c}}-x^2} \, dx}{16 \sqrt {2} a^{7/4} c^{3/4}}-\frac {\left (3 d \left (\sqrt {c} d^2-\sqrt {a} e^2\right )\right ) \int \frac {\frac {\sqrt {2} \sqrt [4]{a}}{\sqrt [4]{c}}-2 x}{-\frac {\sqrt {a}}{\sqrt {c}}+\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{c}}-x^2} \, dx}{16 \sqrt {2} a^{7/4} c^{3/4}} \\ & = -\frac {a e^3-c x \left (d^3+3 d^2 e x+3 d e^2 x^2\right )}{4 a c \left (a+c x^4\right )}+\frac {3 d^2 e \tan ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {a}}\right )}{4 a^{3/2} \sqrt {c}}-\frac {3 d \left (\sqrt {c} d^2-\sqrt {a} e^2\right ) \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )}{16 \sqrt {2} a^{7/4} c^{3/4}}+\frac {3 d \left (\sqrt {c} d^2-\sqrt {a} e^2\right ) \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )}{16 \sqrt {2} a^{7/4} c^{3/4}}+\frac {\left (3 d \left (\sqrt {c} d^2+\sqrt {a} e^2\right )\right ) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{8 \sqrt {2} a^{7/4} c^{3/4}}-\frac {\left (3 d \left (\sqrt {c} d^2+\sqrt {a} e^2\right )\right ) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{8 \sqrt {2} a^{7/4} c^{3/4}} \\ & = -\frac {a e^3-c x \left (d^3+3 d^2 e x+3 d e^2 x^2\right )}{4 a c \left (a+c x^4\right )}+\frac {3 d^2 e \tan ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {a}}\right )}{4 a^{3/2} \sqrt {c}}-\frac {3 d \left (\sqrt {c} d^2+\sqrt {a} e^2\right ) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{8 \sqrt {2} a^{7/4} c^{3/4}}+\frac {3 d \left (\sqrt {c} d^2+\sqrt {a} e^2\right ) \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{8 \sqrt {2} a^{7/4} c^{3/4}}-\frac {3 d \left (\sqrt {c} d^2-\sqrt {a} e^2\right ) \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )}{16 \sqrt {2} a^{7/4} c^{3/4}}+\frac {3 d \left (\sqrt {c} d^2-\sqrt {a} e^2\right ) \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )}{16 \sqrt {2} a^{7/4} c^{3/4}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 347, normalized size of antiderivative = 0.99 \[ \int \frac {(d+e x)^3}{\left (a+c x^4\right )^2} \, dx=\frac {-\frac {8 a \left (a e^3-c d x \left (d^2+3 d e x+3 e^2 x^2\right )\right )}{a+c x^4}-6 \sqrt [4]{a} \sqrt [4]{c} d \left (\sqrt {2} \sqrt {c} d^2+4 \sqrt [4]{a} \sqrt [4]{c} d e+\sqrt {2} \sqrt {a} e^2\right ) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )+6 \sqrt [4]{a} \sqrt [4]{c} d \left (\sqrt {2} \sqrt {c} d^2-4 \sqrt [4]{a} \sqrt [4]{c} d e+\sqrt {2} \sqrt {a} e^2\right ) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )+3 \sqrt {2} \sqrt [4]{c} \left (-\sqrt [4]{a} \sqrt {c} d^3+a^{3/4} d e^2\right ) \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )+3 \sqrt {2} \sqrt [4]{c} \left (\sqrt [4]{a} \sqrt {c} d^3-a^{3/4} d e^2\right ) \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )}{32 a^2 c} \]

[In]

Integrate[(d + e*x)^3/(a + c*x^4)^2,x]

[Out]

((-8*a*(a*e^3 - c*d*x*(d^2 + 3*d*e*x + 3*e^2*x^2)))/(a + c*x^4) - 6*a^(1/4)*c^(1/4)*d*(Sqrt[2]*Sqrt[c]*d^2 + 4
*a^(1/4)*c^(1/4)*d*e + Sqrt[2]*Sqrt[a]*e^2)*ArcTan[1 - (Sqrt[2]*c^(1/4)*x)/a^(1/4)] + 6*a^(1/4)*c^(1/4)*d*(Sqr
t[2]*Sqrt[c]*d^2 - 4*a^(1/4)*c^(1/4)*d*e + Sqrt[2]*Sqrt[a]*e^2)*ArcTan[1 + (Sqrt[2]*c^(1/4)*x)/a^(1/4)] + 3*Sq
rt[2]*c^(1/4)*(-(a^(1/4)*Sqrt[c]*d^3) + a^(3/4)*d*e^2)*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*c^(1/4)*x + Sqrt[c]*x^2]
+ 3*Sqrt[2]*c^(1/4)*(a^(1/4)*Sqrt[c]*d^3 - a^(3/4)*d*e^2)*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*c^(1/4)*x + Sqrt[c]*x^
2])/(32*a^2*c)

Maple [C] (verified)

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

Time = 0.85 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.29

method	result	size
risch	\(\frac {\frac {3 d \,e^{2} x^{3}}{4 a}+\frac {3 e \,x^{2} d^{2}}{4 a}+\frac {d^{3} x}{4 a}-\frac {e^{3}}{4 c}}{c \,x^{4}+a}+\frac {3 d \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (c \,\textit {\_Z}^{4}+a \right )}{\sum }\frac {\left (e^{2} \textit {\_R}^{2}+2 e d \textit {\_R} +d^{2}\right ) \ln \left (x -\textit {\_R} \right )}{\textit {\_R}^{3}}\right )}{16 a c}\)	\(100\)
default	\(d^{3} \left (\frac {x}{4 a \left (c \,x^{4}+a \right )}+\frac {3 \left (\frac {a}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x^{2}+\left (\frac {a}{c}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {a}{c}}}{x^{2}-\left (\frac {a}{c}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {a}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{32 a^{2}}\right )+3 d^{2} e \left (\frac {x^{2}}{4 a \left (c \,x^{4}+a \right )}+\frac {\arctan \left (x^{2} \sqrt {\frac {c}{a}}\right )}{4 a \sqrt {a c}}\right )+3 d \,e^{2} \left (\frac {x^{3}}{4 a \left (c \,x^{4}+a \right )}+\frac {\sqrt {2}\, \left (\ln \left (\frac {x^{2}-\left (\frac {a}{c}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {a}{c}}}{x^{2}+\left (\frac {a}{c}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {a}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{32 a c \left (\frac {a}{c}\right )^{\frac {1}{4}}}\right )+\frac {e^{3} x^{4}}{4 a \left (c \,x^{4}+a \right )}\)	\(317\)

[In]

int((e*x+d)^3/(c*x^4+a)^2,x,method=_RETURNVERBOSE)

[Out]

(3/4*d*e^2/a*x^3+3/4*e/a*x^2*d^2+1/4*d^3/a*x-1/4*e^3/c)/(c*x^4+a)+3/16*d/a/c*sum((_R^2*e^2+2*_R*d*e+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 = 12.49 (sec) , antiderivative size = 91191, normalized size of antiderivative = 261.29 \[ \int \frac {(d+e x)^3}{\left (a+c x^4\right )^2} \, dx=\text {Too large to display} \]

[In]

integrate((e*x+d)^3/(c*x^4+a)^2,x, algorithm="fricas")

[Out]

Too large to include

Sympy [F(-1)]

Timed out. \[ \int \frac {(d+e x)^3}{\left (a+c x^4\right )^2} \, dx=\text {Timed out} \]

[In]

integrate((e*x+d)**3/(c*x**4+a)**2,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 332, normalized size of antiderivative = 0.95 \[ \int \frac {(d+e x)^3}{\left (a+c x^4\right )^2} \, dx=\frac {3 \, d {\left (\frac {\sqrt {2} {\left (\sqrt {c} d^{2} - \sqrt {a} e^{2}\right )} \log \left (\sqrt {c} x^{2} + \sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}} x + \sqrt {a}\right )}{a^{\frac {3}{4}} c^{\frac {3}{4}}} - \frac {\sqrt {2} {\left (\sqrt {c} d^{2} - \sqrt {a} e^{2}\right )} \log \left (\sqrt {c} x^{2} - \sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}} x + \sqrt {a}\right )}{a^{\frac {3}{4}} c^{\frac {3}{4}}} + \frac {2 \, {\left (\sqrt {2} a^{\frac {1}{4}} c^{\frac {3}{4}} d^{2} + \sqrt {2} a^{\frac {3}{4}} c^{\frac {1}{4}} e^{2} - 4 \, \sqrt {a} \sqrt {c} d e\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {c} x + \sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {c}}}\right )}{a^{\frac {3}{4}} \sqrt {\sqrt {a} \sqrt {c}} c^{\frac {3}{4}}} + \frac {2 \, {\left (\sqrt {2} a^{\frac {1}{4}} c^{\frac {3}{4}} d^{2} + \sqrt {2} a^{\frac {3}{4}} c^{\frac {1}{4}} e^{2} + 4 \, \sqrt {a} \sqrt {c} d e\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {c} x - \sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {c}}}\right )}{a^{\frac {3}{4}} \sqrt {\sqrt {a} \sqrt {c}} c^{\frac {3}{4}}}\right )}}{32 \, a} + \frac {3 \, c d e^{2} x^{3} + 3 \, c d^{2} e x^{2} + c d^{3} x - a e^{3}}{4 \, {\left (a c^{2} x^{4} + a^{2} c\right )}} \]

[In]

integrate((e*x+d)^3/(c*x^4+a)^2,x, algorithm="maxima")

[Out]

3/32*d*(sqrt(2)*(sqrt(c)*d^2 - 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)*(sqrt(c)*d^2 - 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*(sqrt(2)*a^(1/4)*c^(3/4)*d^2 + sqrt(2)*a^(3/4)*c^(1/4)*e^2 - 4*sqrt(a)*sqrt(c)*d*e)*arctan(1/2*sqr
t(2)*(2*sqrt(c)*x + sqrt(2)*a^(1/4)*c^(1/4))/sqrt(sqrt(a)*sqrt(c)))/(a^(3/4)*sqrt(sqrt(a)*sqrt(c))*c^(3/4)) +
2*(sqrt(2)*a^(1/4)*c^(3/4)*d^2 + sqrt(2)*a^(3/4)*c^(1/4)*e^2 + 4*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(sqrt(a)*sqrt(c))*c^(3/4)))/a + 1/4*(3*
c*d*e^2*x^3 + 3*c*d^2*e*x^2 + c*d^3*x - a*e^3)/(a*c^2*x^4 + a^2*c)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 345, normalized size of antiderivative = 0.99 \[ \int \frac {(d+e x)^3}{\left (a+c x^4\right )^2} \, dx=\frac {3 \, c d e^{2} x^{3} + 3 \, c d^{2} e x^{2} + c d^{3} x - a e^{3}}{4 \, {\left (c x^{4} + a\right )} a c} + \frac {3 \, \sqrt {2} {\left (2 \, \sqrt {2} \sqrt {a c} c^{2} d^{2} e + \left (a c^{3}\right )^{\frac {1}{4}} c^{2} d^{3} + \left (a c^{3}\right )^{\frac {3}{4}} d e^{2}\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, x + \sqrt {2} \left (\frac {a}{c}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {a}{c}\right )^{\frac {1}{4}}}\right )}{16 \, a^{2} c^{3}} + \frac {3 \, \sqrt {2} {\left (2 \, \sqrt {2} \sqrt {a c} c^{2} d^{2} e + \left (a c^{3}\right )^{\frac {1}{4}} c^{2} d^{3} + \left (a c^{3}\right )^{\frac {3}{4}} d e^{2}\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, x - \sqrt {2} \left (\frac {a}{c}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {a}{c}\right )^{\frac {1}{4}}}\right )}{16 \, a^{2} c^{3}} + \frac {3 \, \sqrt {2} {\left (\left (a c^{3}\right )^{\frac {1}{4}} c^{2} d^{3} - \left (a c^{3}\right )^{\frac {3}{4}} d e^{2}\right )} \log \left (x^{2} + \sqrt {2} x \left (\frac {a}{c}\right )^{\frac {1}{4}} + \sqrt {\frac {a}{c}}\right )}{32 \, a^{2} c^{3}} - \frac {3 \, \sqrt {2} {\left (\left (a c^{3}\right )^{\frac {1}{4}} c^{2} d^{3} - \left (a c^{3}\right )^{\frac {3}{4}} d e^{2}\right )} \log \left (x^{2} - \sqrt {2} x \left (\frac {a}{c}\right )^{\frac {1}{4}} + \sqrt {\frac {a}{c}}\right )}{32 \, a^{2} c^{3}} \]

[In]

integrate((e*x+d)^3/(c*x^4+a)^2,x, algorithm="giac")

[Out]

1/4*(3*c*d*e^2*x^3 + 3*c*d^2*e*x^2 + c*d^3*x - a*e^3)/((c*x^4 + a)*a*c) + 3/16*sqrt(2)*(2*sqrt(2)*sqrt(a*c)*c^
2*d^2*e + (a*c^3)^(1/4)*c^2*d^3 + (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^2*c^3) + 3/16*sqrt(2)*(2*sqrt(2)*sqrt(a*c)*c^2*d^2*e + (a*c^3)^(1/4)*c^2*d^3 + (a*c^3)^(3/4)*d*e^2)*ar
ctan(1/2*sqrt(2)*(2*x - sqrt(2)*(a/c)^(1/4))/(a/c)^(1/4))/(a^2*c^3) + 3/32*sqrt(2)*((a*c^3)^(1/4)*c^2*d^3 - (a
*c^3)^(3/4)*d*e^2)*log(x^2 + sqrt(2)*x*(a/c)^(1/4) + sqrt(a/c))/(a^2*c^3) - 3/32*sqrt(2)*((a*c^3)^(1/4)*c^2*d^
3 - (a*c^3)^(3/4)*d*e^2)*log(x^2 - sqrt(2)*x*(a/c)^(1/4) + sqrt(a/c))/(a^2*c^3)

Mupad [B] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 670, normalized size of antiderivative = 1.92 \[ \int \frac {(d+e x)^3}{\left (a+c x^4\right )^2} \, dx=\left (\sum _{k=1}^4\ln \left (\frac {c\,d^2\,\left (27\,c\,d^5\,e^2-9\,a\,d\,e^6+36\,c\,d^4\,e^3\,x-{\mathrm {root}\left (65536\,a^7\,c^3\,z^4+27648\,a^4\,c^2\,d^4\,e^2\,z^2+3456\,a^3\,c\,d^4\,e^5\,z-3456\,a^2\,c^2\,d^8\,e\,z+162\,a\,c\,d^8\,e^4+81\,a^2\,d^4\,e^8+81\,c^2\,d^{12},z,k\right )}^2\,a^3\,c^2\,d\,256-\mathrm {root}\left (65536\,a^7\,c^3\,z^4+27648\,a^4\,c^2\,d^4\,e^2\,z^2+3456\,a^3\,c\,d^4\,e^5\,z-3456\,a^2\,c^2\,d^8\,e\,z+162\,a\,c\,d^8\,e^4+81\,a^2\,d^4\,e^8+81\,c^2\,d^{12},z,k\right )\,a\,c^2\,d^4\,x\,48+\mathrm {root}\left (65536\,a^7\,c^3\,z^4+27648\,a^4\,c^2\,d^4\,e^2\,z^2+3456\,a^3\,c\,d^4\,e^5\,z-3456\,a^2\,c^2\,d^8\,e\,z+162\,a\,c\,d^8\,e^4+81\,a^2\,d^4\,e^8+81\,c^2\,d^{12},z,k\right )\,a^2\,c\,e^4\,x\,48+{\mathrm {root}\left (65536\,a^7\,c^3\,z^4+27648\,a^4\,c^2\,d^4\,e^2\,z^2+3456\,a^3\,c\,d^4\,e^5\,z-3456\,a^2\,c^2\,d^8\,e\,z+162\,a\,c\,d^8\,e^4+81\,a^2\,d^4\,e^8+81\,c^2\,d^{12},z,k\right )}^2\,a^3\,c^2\,e\,x\,512-\mathrm {root}\left (65536\,a^7\,c^3\,z^4+27648\,a^4\,c^2\,d^4\,e^2\,z^2+3456\,a^3\,c\,d^4\,e^5\,z-3456\,a^2\,c^2\,d^8\,e\,z+162\,a\,c\,d^8\,e^4+81\,a^2\,d^4\,e^8+81\,c^2\,d^{12},z,k\right )\,a^2\,c\,d\,e^3\,192\right )\,3}{a^3\,64}\right )\,\mathrm {root}\left (65536\,a^7\,c^3\,z^4+27648\,a^4\,c^2\,d^4\,e^2\,z^2+3456\,a^3\,c\,d^4\,e^5\,z-3456\,a^2\,c^2\,d^8\,e\,z+162\,a\,c\,d^8\,e^4+81\,a^2\,d^4\,e^8+81\,c^2\,d^{12},z,k\right )\right )+\frac {\frac {d^3\,x}{4\,a}-\frac {e^3}{4\,c}+\frac {3\,d^2\,e\,x^2}{4\,a}+\frac {3\,d\,e^2\,x^3}{4\,a}}{c\,x^4+a} \]

[In]

int((d + e*x)^3/(a + c*x^4)^2,x)

[Out]

symsum(log((3*c*d^2*(27*c*d^5*e^2 - 9*a*d*e^6 + 36*c*d^4*e^3*x - 256*root(65536*a^7*c^3*z^4 + 27648*a^4*c^2*d^
4*e^2*z^2 + 3456*a^3*c*d^4*e^5*z - 3456*a^2*c^2*d^8*e*z + 162*a*c*d^8*e^4 + 81*a^2*d^4*e^8 + 81*c^2*d^12, z, k
)^2*a^3*c^2*d - 48*root(65536*a^7*c^3*z^4 + 27648*a^4*c^2*d^4*e^2*z^2 + 3456*a^3*c*d^4*e^5*z - 3456*a^2*c^2*d^
8*e*z + 162*a*c*d^8*e^4 + 81*a^2*d^4*e^8 + 81*c^2*d^12, z, k)*a*c^2*d^4*x + 48*root(65536*a^7*c^3*z^4 + 27648*
a^4*c^2*d^4*e^2*z^2 + 3456*a^3*c*d^4*e^5*z - 3456*a^2*c^2*d^8*e*z + 162*a*c*d^8*e^4 + 81*a^2*d^4*e^8 + 81*c^2*
d^12, z, k)*a^2*c*e^4*x + 512*root(65536*a^7*c^3*z^4 + 27648*a^4*c^2*d^4*e^2*z^2 + 3456*a^3*c*d^4*e^5*z - 3456
*a^2*c^2*d^8*e*z + 162*a*c*d^8*e^4 + 81*a^2*d^4*e^8 + 81*c^2*d^12, z, k)^2*a^3*c^2*e*x - 192*root(65536*a^7*c^
3*z^4 + 27648*a^4*c^2*d^4*e^2*z^2 + 3456*a^3*c*d^4*e^5*z - 3456*a^2*c^2*d^8*e*z + 162*a*c*d^8*e^4 + 81*a^2*d^4
*e^8 + 81*c^2*d^12, z, k)*a^2*c*d*e^3))/(64*a^3))*root(65536*a^7*c^3*z^4 + 27648*a^4*c^2*d^4*e^2*z^2 + 3456*a^
3*c*d^4*e^5*z - 3456*a^2*c^2*d^8*e*z + 162*a*c*d^8*e^4 + 81*a^2*d^4*e^8 + 81*c^2*d^12, z, k), k, 1, 4) + ((d^3
*x)/(4*a) - e^3/(4*c) + (3*d^2*e*x^2)/(4*a) + (3*d*e^2*x^3)/(4*a))/(a + c*x^4)