\(\int \frac {x^3}{a+b (c+d x)^4} \, dx\) [110]

Optimal result

Integrand size = 17, antiderivative size = 356 \[ \int \frac {x^3}{a+b (c+d x)^4} \, dx=\frac {3 c^2 \arctan \left (\frac {\sqrt {b} (c+d x)^2}{\sqrt {a}}\right )}{2 \sqrt {a} \sqrt {b} d^4}+\frac {c \left (3 \sqrt {a}+\sqrt {b} c^2\right ) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} (c+d x)}{\sqrt [4]{a}}\right )}{2 \sqrt {2} a^{3/4} b^{3/4} d^4}-\frac {c \left (3 \sqrt {a}+\sqrt {b} c^2\right ) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{b} (c+d x)}{\sqrt [4]{a}}\right )}{2 \sqrt {2} a^{3/4} b^{3/4} d^4}-\frac {c \left (3 \sqrt {a}-\sqrt {b} c^2\right ) \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} (c+d x)+\sqrt {b} (c+d x)^2\right )}{4 \sqrt {2} a^{3/4} b^{3/4} d^4}+\frac {c \left (3 \sqrt {a}-\sqrt {b} c^2\right ) \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} (c+d x)+\sqrt {b} (c+d x)^2\right )}{4 \sqrt {2} a^{3/4} b^{3/4} d^4}+\frac {\log \left (a+b (c+d x)^4\right )}{4 b d^4} \]

[Out]

1/4*ln(a+b*(d*x+c)^4)/b/d^4+3/2*c^2*arctan((d*x+c)^2*b^(1/2)/a^(1/2))/d^4/a^(1/2)/b^(1/2)-1/8*c*ln(-a^(1/4)*b^
(1/4)*(d*x+c)*2^(1/2)+a^(1/2)+(d*x+c)^2*b^(1/2))*(3*a^(1/2)-b^(1/2)*c^2)/a^(3/4)/b^(3/4)/d^4*2^(1/2)+1/8*c*ln(
a^(1/4)*b^(1/4)*(d*x+c)*2^(1/2)+a^(1/2)+(d*x+c)^2*b^(1/2))*(3*a^(1/2)-b^(1/2)*c^2)/a^(3/4)/b^(3/4)/d^4*2^(1/2)
-1/4*c*arctan(-1+b^(1/4)*(d*x+c)*2^(1/2)/a^(1/4))*(3*a^(1/2)+b^(1/2)*c^2)/a^(3/4)/b^(3/4)/d^4*2^(1/2)-1/4*c*ar
ctan(1+b^(1/4)*(d*x+c)*2^(1/2)/a^(1/4))*(3*a^(1/2)+b^(1/2)*c^2)/a^(3/4)/b^(3/4)/d^4*2^(1/2)

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 356, 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 = {378, 1890, 1262, 649, 211, 266, 1182, 1176, 631, 210, 1179, 642} \[ \int \frac {x^3}{a+b (c+d x)^4} \, dx=\frac {c \left (3 \sqrt {a}+\sqrt {b} c^2\right ) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} (c+d x)}{\sqrt [4]{a}}\right )}{2 \sqrt {2} a^{3/4} b^{3/4} d^4}-\frac {c \left (3 \sqrt {a}+\sqrt {b} c^2\right ) \arctan \left (\frac {\sqrt {2} \sqrt [4]{b} (c+d x)}{\sqrt [4]{a}}+1\right )}{2 \sqrt {2} a^{3/4} b^{3/4} d^4}-\frac {c \left (3 \sqrt {a}-\sqrt {b} c^2\right ) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} (c+d x)+\sqrt {a}+\sqrt {b} (c+d x)^2\right )}{4 \sqrt {2} a^{3/4} b^{3/4} d^4}+\frac {c \left (3 \sqrt {a}-\sqrt {b} c^2\right ) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} (c+d x)+\sqrt {a}+\sqrt {b} (c+d x)^2\right )}{4 \sqrt {2} a^{3/4} b^{3/4} d^4}+\frac {3 c^2 \arctan \left (\frac {\sqrt {b} (c+d x)^2}{\sqrt {a}}\right )}{2 \sqrt {a} \sqrt {b} d^4}+\frac {\log \left (a+b (c+d x)^4\right )}{4 b d^4} \]

[In]

Int[x^3/(a + b*(c + d*x)^4),x]

[Out]

(3*c^2*ArcTan[(Sqrt[b]*(c + d*x)^2)/Sqrt[a]])/(2*Sqrt[a]*Sqrt[b]*d^4) + (c*(3*Sqrt[a] + Sqrt[b]*c^2)*ArcTan[1
- (Sqrt[2]*b^(1/4)*(c + d*x))/a^(1/4)])/(2*Sqrt[2]*a^(3/4)*b^(3/4)*d^4) - (c*(3*Sqrt[a] + Sqrt[b]*c^2)*ArcTan[
1 + (Sqrt[2]*b^(1/4)*(c + d*x))/a^(1/4)])/(2*Sqrt[2]*a^(3/4)*b^(3/4)*d^4) - (c*(3*Sqrt[a] - Sqrt[b]*c^2)*Log[S
qrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*(c + d*x) + Sqrt[b]*(c + d*x)^2])/(4*Sqrt[2]*a^(3/4)*b^(3/4)*d^4) + (c*(3*Sqr
t[a] - Sqrt[b]*c^2)*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*(c + d*x) + Sqrt[b]*(c + d*x)^2])/(4*Sqrt[2]*a^(3/4)
*b^(3/4)*d^4) + Log[a + b*(c + d*x)^4]/(4*b*d^4)

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 266

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 378

Int[((a_) + (b_.)*(v_)^(n_))^(p_.)*(x_)^(m_.), x_Symbol] :> With[{c = Coefficient[v, x, 0], d = Coefficient[v,
x, 1]}, Dist[1/d^(m + 1), Subst[Int[SimplifyIntegrand[(x - c)^m*(a + b*x^n)^p, x], x], x, v], x] /; NeQ[c, 0]
] /; FreeQ[{a, b, n, p}, x] && LinearQ[v, x] && 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 649

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[d, Int[1/(a + c*x^2), x], x] + Dist[e, Int[x/
(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] &&  !NiceSqrtQ[(-a)*c]

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 1262

Int[(x_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[(d + e*x)^q
*(a + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, c, d, e, p, q}, x]

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 {\text {Subst}\left (\int \frac {(-c+x)^3}{a+b x^4} \, dx,x,c+d x\right )}{d^4} \\ & = \frac {\text {Subst}\left (\int \left (\frac {x \left (3 c^2+x^2\right )}{a+b x^4}+\frac {-c^3-3 c x^2}{a+b x^4}\right ) \, dx,x,c+d x\right )}{d^4} \\ & = \frac {\text {Subst}\left (\int \frac {x \left (3 c^2+x^2\right )}{a+b x^4} \, dx,x,c+d x\right )}{d^4}+\frac {\text {Subst}\left (\int \frac {-c^3-3 c x^2}{a+b x^4} \, dx,x,c+d x\right )}{d^4} \\ & = \frac {\text {Subst}\left (\int \frac {3 c^2+x}{a+b x^2} \, dx,x,(c+d x)^2\right )}{2 d^4}+\frac {\left (c \left (3-\frac {\sqrt {b} c^2}{\sqrt {a}}\right )\right ) \text {Subst}\left (\int \frac {\sqrt {a} \sqrt {b}-b x^2}{a+b x^4} \, dx,x,c+d x\right )}{2 b d^4}-\frac {\left (c \left (3+\frac {\sqrt {b} c^2}{\sqrt {a}}\right )\right ) \text {Subst}\left (\int \frac {\sqrt {a} \sqrt {b}+b x^2}{a+b x^4} \, dx,x,c+d x\right )}{2 b d^4} \\ & = \frac {\text {Subst}\left (\int \frac {x}{a+b x^2} \, dx,x,(c+d x)^2\right )}{2 d^4}+\frac {\left (3 c^2\right ) \text {Subst}\left (\int \frac {1}{a+b x^2} \, dx,x,(c+d x)^2\right )}{2 d^4}-\frac {\left (c \left (3 \sqrt {a}-\sqrt {b} c^2\right )\right ) \text {Subst}\left (\int \frac {\frac {\sqrt {2} \sqrt [4]{a}}{\sqrt [4]{b}}+2 x}{-\frac {\sqrt {a}}{\sqrt {b}}-\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{b}}-x^2} \, dx,x,c+d x\right )}{4 \sqrt {2} a^{3/4} b^{3/4} d^4}-\frac {\left (c \left (3 \sqrt {a}-\sqrt {b} c^2\right )\right ) \text {Subst}\left (\int \frac {\frac {\sqrt {2} \sqrt [4]{a}}{\sqrt [4]{b}}-2 x}{-\frac {\sqrt {a}}{\sqrt {b}}+\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{b}}-x^2} \, dx,x,c+d x\right )}{4 \sqrt {2} a^{3/4} b^{3/4} d^4}-\frac {\left (c \left (3+\frac {\sqrt {b} c^2}{\sqrt {a}}\right )\right ) \text {Subst}\left (\int \frac {1}{\frac {\sqrt {a}}{\sqrt {b}}-\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{b}}+x^2} \, dx,x,c+d x\right )}{4 b d^4}-\frac {\left (c \left (3+\frac {\sqrt {b} c^2}{\sqrt {a}}\right )\right ) \text {Subst}\left (\int \frac {1}{\frac {\sqrt {a}}{\sqrt {b}}+\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{b}}+x^2} \, dx,x,c+d x\right )}{4 b d^4} \\ & = \frac {3 c^2 \tan ^{-1}\left (\frac {\sqrt {b} (c+d x)^2}{\sqrt {a}}\right )}{2 \sqrt {a} \sqrt {b} d^4}-\frac {c \left (3 \sqrt {a}-\sqrt {b} c^2\right ) \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} (c+d x)+\sqrt {b} (c+d x)^2\right )}{4 \sqrt {2} a^{3/4} b^{3/4} d^4}+\frac {c \left (3 \sqrt {a}-\sqrt {b} c^2\right ) \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} (c+d x)+\sqrt {b} (c+d x)^2\right )}{4 \sqrt {2} a^{3/4} b^{3/4} d^4}+\frac {\log \left (a+b (c+d x)^4\right )}{4 b d^4}-\frac {\left (c \left (3 \sqrt {a}+\sqrt {b} c^2\right )\right ) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt [4]{b} (c+d x)}{\sqrt [4]{a}}\right )}{2 \sqrt {2} a^{3/4} b^{3/4} d^4}+\frac {\left (c \left (3 \sqrt {a}+\sqrt {b} c^2\right )\right ) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt [4]{b} (c+d x)}{\sqrt [4]{a}}\right )}{2 \sqrt {2} a^{3/4} b^{3/4} d^4} \\ & = \frac {3 c^2 \tan ^{-1}\left (\frac {\sqrt {b} (c+d x)^2}{\sqrt {a}}\right )}{2 \sqrt {a} \sqrt {b} d^4}+\frac {c \left (3 \sqrt {a}+\sqrt {b} c^2\right ) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} (c+d x)}{\sqrt [4]{a}}\right )}{2 \sqrt {2} a^{3/4} b^{3/4} d^4}-\frac {c \left (3 \sqrt {a}+\sqrt {b} c^2\right ) \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{b} (c+d x)}{\sqrt [4]{a}}\right )}{2 \sqrt {2} a^{3/4} b^{3/4} d^4}-\frac {c \left (3 \sqrt {a}-\sqrt {b} c^2\right ) \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} (c+d x)+\sqrt {b} (c+d x)^2\right )}{4 \sqrt {2} a^{3/4} b^{3/4} d^4}+\frac {c \left (3 \sqrt {a}-\sqrt {b} c^2\right ) \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} (c+d x)+\sqrt {b} (c+d x)^2\right )}{4 \sqrt {2} a^{3/4} b^{3/4} d^4}+\frac {\log \left (a+b (c+d x)^4\right )}{4 b d^4} \\ \end{align*}

Mathematica [C] (verified)

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

Time = 0.03 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.30 \[ \int \frac {x^3}{a+b (c+d x)^4} \, dx=\frac {\text {RootSum}\left [a+b c^4+4 b c^3 d \text {$\#$1}+6 b c^2 d^2 \text {$\#$1}^2+4 b c d^3 \text {$\#$1}^3+b d^4 \text {$\#$1}^4\&,\frac {\log (x-\text {$\#$1}) \text {$\#$1}^3}{c^3+3 c^2 d \text {$\#$1}+3 c d^2 \text {$\#$1}^2+d^3 \text {$\#$1}^3}\&\right ]}{4 b d} \]

[In]

Integrate[x^3/(a + b*(c + d*x)^4),x]

[Out]

RootSum[a + b*c^4 + 4*b*c^3*d*#1 + 6*b*c^2*d^2*#1^2 + 4*b*c*d^3*#1^3 + b*d^4*#1^4 & , (Log[x - #1]*#1^3)/(c^3
+ 3*c^2*d*#1 + 3*c*d^2*#1^2 + d^3*#1^3) & ]/(4*b*d)

Maple [C] (verified)

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

Time = 0.10 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.27

method	result	size
default	\(\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (b \,d^{4} \textit {\_Z}^{4}+4 b c \,d^{3} \textit {\_Z}^{3}+6 b \,c^{2} d^{2} \textit {\_Z}^{2}+4 b \,c^{3} d \textit {\_Z} +b \,c^{4}+a \right )}{\sum }\frac {\textit {\_R}^{3} \ln \left (x -\textit {\_R} \right )}{d^{3} \textit {\_R}^{3}+3 c \,d^{2} \textit {\_R}^{2}+3 c^{2} d \textit {\_R} +c^{3}}}{4 b d}\)	\(97\)
risch	\(\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (b \,d^{4} \textit {\_Z}^{4}+4 b c \,d^{3} \textit {\_Z}^{3}+6 b \,c^{2} d^{2} \textit {\_Z}^{2}+4 b \,c^{3} d \textit {\_Z} +b \,c^{4}+a \right )}{\sum }\frac {\textit {\_R}^{3} \ln \left (x -\textit {\_R} \right )}{d^{3} \textit {\_R}^{3}+3 c \,d^{2} \textit {\_R}^{2}+3 c^{2} d \textit {\_R} +c^{3}}}{4 b d}\)	\(97\)

[In]

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

[Out]

1/4/b/d*sum(_R^3/(_R^3*d^3+3*_R^2*c*d^2+3*_R*c^2*d+c^3)*ln(x-_R),_R=RootOf(_Z^4*b*d^4+4*_Z^3*b*c*d^3+6*_Z^2*b*
c^2*d^2+4*_Z*b*c^3*d+b*c^4+a))

Fricas [F(-1)]

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

[In]

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

[Out]

Timed out

Sympy [A] (verification not implemented)

Time = 2.24 (sec) , antiderivative size = 374, normalized size of antiderivative = 1.05 \[ \int \frac {x^3}{a+b (c+d x)^4} \, dx=\operatorname {RootSum} {\left (256 t^{4} a^{3} b^{4} d^{16} - 256 t^{3} a^{3} b^{3} d^{12} + t^{2} \cdot \left (96 a^{3} b^{2} d^{8} + 480 a^{2} b^{3} c^{4} d^{8}\right ) + t \left (- 16 a^{3} b d^{4} + 192 a^{2} b^{2} c^{4} d^{4} - 48 a b^{3} c^{8} d^{4}\right ) + a^{3} + 3 a^{2} b c^{4} + 3 a b^{2} c^{8} + b^{3} c^{12}, \left ( t \mapsto t \log {\left (x + \frac {- 1728 t^{3} a^{4} b^{3} d^{12} - 960 t^{3} a^{3} b^{4} c^{4} d^{12} + 1296 t^{2} a^{4} b^{2} d^{8} + 2016 t^{2} a^{3} b^{3} c^{4} d^{8} - 48 t^{2} a^{2} b^{4} c^{8} d^{8} - 324 t a^{4} b d^{4} - 4716 t a^{3} b^{2} c^{4} d^{4} - 1452 t a^{2} b^{3} c^{8} d^{4} - 4 t a b^{4} c^{12} d^{4} + 27 a^{4} - 390 a^{3} b c^{4} - 444 a^{2} b^{2} c^{8} - 26 a b^{3} c^{12} + b^{4} c^{16}}{729 a^{3} b c^{3} d - 1053 a^{2} b^{2} c^{7} d - 117 a b^{3} c^{11} d + b^{4} c^{15} d} \right )} \right )\right )} \]

[In]

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

[Out]

RootSum(256*_t**4*a**3*b**4*d**16 - 256*_t**3*a**3*b**3*d**12 + _t**2*(96*a**3*b**2*d**8 + 480*a**2*b**3*c**4*
d**8) + _t*(-16*a**3*b*d**4 + 192*a**2*b**2*c**4*d**4 - 48*a*b**3*c**8*d**4) + a**3 + 3*a**2*b*c**4 + 3*a*b**2
*c**8 + b**3*c**12, Lambda(_t, _t*log(x + (-1728*_t**3*a**4*b**3*d**12 - 960*_t**3*a**3*b**4*c**4*d**12 + 1296
*_t**2*a**4*b**2*d**8 + 2016*_t**2*a**3*b**3*c**4*d**8 - 48*_t**2*a**2*b**4*c**8*d**8 - 324*_t*a**4*b*d**4 - 4
716*_t*a**3*b**2*c**4*d**4 - 1452*_t*a**2*b**3*c**8*d**4 - 4*_t*a*b**4*c**12*d**4 + 27*a**4 - 390*a**3*b*c**4
- 444*a**2*b**2*c**8 - 26*a*b**3*c**12 + b**4*c**16)/(729*a**3*b*c**3*d - 1053*a**2*b**2*c**7*d - 117*a*b**3*c
**11*d + b**4*c**15*d))))

Maxima [F]

\[ \int \frac {x^3}{a+b (c+d x)^4} \, dx=\int { \frac {x^{3}}{{\left (d x + c\right )}^{4} b + a} \,d x } \]

[In]

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

[Out]

integrate(x^3/((d*x + c)^4*b + a), x)

Giac [F]

\[ \int \frac {x^3}{a+b (c+d x)^4} \, dx=\int { \frac {x^{3}}{{\left (d x + c\right )}^{4} b + a} \,d x } \]

[In]

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

[Out]

integrate(x^3/((d*x + c)^4*b + a), x)

Mupad [B] (verification not implemented)

Time = 9.62 (sec) , antiderivative size = 1003, normalized size of antiderivative = 2.82 \[ \int \frac {x^3}{a+b (c+d x)^4} \, dx=\sum _{k=1}^4\ln \left (b\,c^2\,d\,\left (2\,a\,c+2\,b\,c^5-3\,a\,d\,x+5\,b\,c^4\,d\,x-\mathrm {root}\left (256\,a^3\,b^4\,d^{16}\,z^4-256\,a^3\,b^3\,d^{12}\,z^3+480\,a^2\,b^3\,c^4\,d^8\,z^2+96\,a^3\,b^2\,d^8\,z^2+192\,a^2\,b^2\,c^4\,d^4\,z-48\,a\,b^3\,c^8\,d^4\,z-16\,a^3\,b\,d^4\,z+3\,a\,b^2\,c^8+3\,a^2\,b\,c^4+b^3\,c^{12}+a^3,z,k\right )\,b^2\,c^5\,d^4\,2+{\mathrm {root}\left (256\,a^3\,b^4\,d^{16}\,z^4-256\,a^3\,b^3\,d^{12}\,z^3+480\,a^2\,b^3\,c^4\,d^8\,z^2+96\,a^3\,b^2\,d^8\,z^2+192\,a^2\,b^2\,c^4\,d^4\,z-48\,a\,b^3\,c^8\,d^4\,z-16\,a^3\,b\,d^4\,z+3\,a\,b^2\,c^8+3\,a^2\,b\,c^4+b^3\,c^{12}+a^3,z,k\right )}^2\,a\,b^2\,c\,d^8\,32+{\mathrm {root}\left (256\,a^3\,b^4\,d^{16}\,z^4-256\,a^3\,b^3\,d^{12}\,z^3+480\,a^2\,b^3\,c^4\,d^8\,z^2+96\,a^3\,b^2\,d^8\,z^2+192\,a^2\,b^2\,c^4\,d^4\,z-48\,a\,b^3\,c^8\,d^4\,z-16\,a^3\,b\,d^4\,z+3\,a\,b^2\,c^8+3\,a^2\,b\,c^4+b^3\,c^{12}+a^3,z,k\right )}^2\,a\,b^2\,d^9\,x\,24-\mathrm {root}\left (256\,a^3\,b^4\,d^{16}\,z^4-256\,a^3\,b^3\,d^{12}\,z^3+480\,a^2\,b^3\,c^4\,d^8\,z^2+96\,a^3\,b^2\,d^8\,z^2+192\,a^2\,b^2\,c^4\,d^4\,z-48\,a\,b^3\,c^8\,d^4\,z-16\,a^3\,b\,d^4\,z+3\,a\,b^2\,c^8+3\,a^2\,b\,c^4+b^3\,c^{12}+a^3,z,k\right )\,b^2\,c^4\,d^5\,x\,2+\mathrm {root}\left (256\,a^3\,b^4\,d^{16}\,z^4-256\,a^3\,b^3\,d^{12}\,z^3+480\,a^2\,b^3\,c^4\,d^8\,z^2+96\,a^3\,b^2\,d^8\,z^2+192\,a^2\,b^2\,c^4\,d^4\,z-48\,a\,b^3\,c^8\,d^4\,z-16\,a^3\,b\,d^4\,z+3\,a\,b^2\,c^8+3\,a^2\,b\,c^4+b^3\,c^{12}+a^3,z,k\right )\,a\,b\,c\,d^4\,38+\mathrm {root}\left (256\,a^3\,b^4\,d^{16}\,z^4-256\,a^3\,b^3\,d^{12}\,z^3+480\,a^2\,b^3\,c^4\,d^8\,z^2+96\,a^3\,b^2\,d^8\,z^2+192\,a^2\,b^2\,c^4\,d^4\,z-48\,a\,b^3\,c^8\,d^4\,z-16\,a^3\,b\,d^4\,z+3\,a\,b^2\,c^8+3\,a^2\,b\,c^4+b^3\,c^{12}+a^3,z,k\right )\,a\,b\,d^5\,x\,6\right )\,2\right )\,\mathrm {root}\left (256\,a^3\,b^4\,d^{16}\,z^4-256\,a^3\,b^3\,d^{12}\,z^3+480\,a^2\,b^3\,c^4\,d^8\,z^2+96\,a^3\,b^2\,d^8\,z^2+192\,a^2\,b^2\,c^4\,d^4\,z-48\,a\,b^3\,c^8\,d^4\,z-16\,a^3\,b\,d^4\,z+3\,a\,b^2\,c^8+3\,a^2\,b\,c^4+b^3\,c^{12}+a^3,z,k\right ) \]

[In]

int(x^3/(a + b*(c + d*x)^4),x)

[Out]

symsum(log(2*b*c^2*d*(2*a*c + 2*b*c^5 - 3*a*d*x + 5*b*c^4*d*x - 2*root(256*a^3*b^4*d^16*z^4 - 256*a^3*b^3*d^12
*z^3 + 480*a^2*b^3*c^4*d^8*z^2 + 96*a^3*b^2*d^8*z^2 + 192*a^2*b^2*c^4*d^4*z - 48*a*b^3*c^8*d^4*z - 16*a^3*b*d^
4*z + 3*a*b^2*c^8 + 3*a^2*b*c^4 + b^3*c^12 + a^3, z, k)*b^2*c^5*d^4 + 32*root(256*a^3*b^4*d^16*z^4 - 256*a^3*b
^3*d^12*z^3 + 480*a^2*b^3*c^4*d^8*z^2 + 96*a^3*b^2*d^8*z^2 + 192*a^2*b^2*c^4*d^4*z - 48*a*b^3*c^8*d^4*z - 16*a
^3*b*d^4*z + 3*a*b^2*c^8 + 3*a^2*b*c^4 + b^3*c^12 + a^3, z, k)^2*a*b^2*c*d^8 + 24*root(256*a^3*b^4*d^16*z^4 -
256*a^3*b^3*d^12*z^3 + 480*a^2*b^3*c^4*d^8*z^2 + 96*a^3*b^2*d^8*z^2 + 192*a^2*b^2*c^4*d^4*z - 48*a*b^3*c^8*d^4
*z - 16*a^3*b*d^4*z + 3*a*b^2*c^8 + 3*a^2*b*c^4 + b^3*c^12 + a^3, z, k)^2*a*b^2*d^9*x - 2*root(256*a^3*b^4*d^1
6*z^4 - 256*a^3*b^3*d^12*z^3 + 480*a^2*b^3*c^4*d^8*z^2 + 96*a^3*b^2*d^8*z^2 + 192*a^2*b^2*c^4*d^4*z - 48*a*b^3
*c^8*d^4*z - 16*a^3*b*d^4*z + 3*a*b^2*c^8 + 3*a^2*b*c^4 + b^3*c^12 + a^3, z, k)*b^2*c^4*d^5*x + 38*root(256*a^
3*b^4*d^16*z^4 - 256*a^3*b^3*d^12*z^3 + 480*a^2*b^3*c^4*d^8*z^2 + 96*a^3*b^2*d^8*z^2 + 192*a^2*b^2*c^4*d^4*z -
48*a*b^3*c^8*d^4*z - 16*a^3*b*d^4*z + 3*a*b^2*c^8 + 3*a^2*b*c^4 + b^3*c^12 + a^3, z, k)*a*b*c*d^4 + 6*root(25
6*a^3*b^4*d^16*z^4 - 256*a^3*b^3*d^12*z^3 + 480*a^2*b^3*c^4*d^8*z^2 + 96*a^3*b^2*d^8*z^2 + 192*a^2*b^2*c^4*d^4
*z - 48*a*b^3*c^8*d^4*z - 16*a^3*b*d^4*z + 3*a*b^2*c^8 + 3*a^2*b*c^4 + b^3*c^12 + a^3, z, k)*a*b*d^5*x))*root(
256*a^3*b^4*d^16*z^4 - 256*a^3*b^3*d^12*z^3 + 480*a^2*b^3*c^4*d^8*z^2 + 96*a^3*b^2*d^8*z^2 + 192*a^2*b^2*c^4*d
^4*z - 48*a*b^3*c^8*d^4*z - 16*a^3*b*d^4*z + 3*a*b^2*c^8 + 3*a^2*b*c^4 + b^3*c^12 + a^3, z, k), k, 1, 4)