3.114 \(\int \frac {1}{x (a+b (c+d x)^4)} \, dx\)

Optimal. Leaf size=393 \[ -\frac {\sqrt [4]{b} c \left (\sqrt {a}-\sqrt {b} c^2\right ) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} (c+d x)+\sqrt {a}+\sqrt {b} (c+d x)^2\right )}{4 \sqrt {2} a^{3/4} \left (a+b c^4\right )}+\frac {\sqrt [4]{b} c \left (\sqrt {a}-\sqrt {b} c^2\right ) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} (c+d x)+\sqrt {a}+\sqrt {b} (c+d x)^2\right )}{4 \sqrt {2} a^{3/4} \left (a+b c^4\right )}+\frac {\sqrt [4]{b} c \left (\sqrt {a}+\sqrt {b} c^2\right ) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} (c+d x)}{\sqrt [4]{a}}\right )}{2 \sqrt {2} a^{3/4} \left (a+b c^4\right )}-\frac {\sqrt [4]{b} c \left (\sqrt {a}+\sqrt {b} c^2\right ) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} (c+d x)}{\sqrt [4]{a}}+1\right )}{2 \sqrt {2} a^{3/4} \left (a+b c^4\right )}-\frac {\log \left (a+b (c+d x)^4\right )}{4 \left (a+b c^4\right )}+\frac {\log (x)}{a+b c^4}-\frac {\sqrt {b} c^2 \tan ^{-1}\left (\frac {\sqrt {b} (c+d x)^2}{\sqrt {a}}\right )}{2 \sqrt {a} \left (a+b c^4\right )} \]

[Out]

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

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Rubi [A]  time = 0.47, antiderivative size = 393, normalized size of antiderivative = 1.00, number of steps used = 18, number of rules used = 13, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.765, Rules used = {371, 6725, 1876, 1248, 635, 205, 260, 1168, 1162, 617, 204, 1165, 628} \[ -\frac {\sqrt [4]{b} c \left (\sqrt {a}-\sqrt {b} c^2\right ) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} (c+d x)+\sqrt {a}+\sqrt {b} (c+d x)^2\right )}{4 \sqrt {2} a^{3/4} \left (a+b c^4\right )}+\frac {\sqrt [4]{b} c \left (\sqrt {a}-\sqrt {b} c^2\right ) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} (c+d x)+\sqrt {a}+\sqrt {b} (c+d x)^2\right )}{4 \sqrt {2} a^{3/4} \left (a+b c^4\right )}+\frac {\sqrt [4]{b} c \left (\sqrt {a}+\sqrt {b} c^2\right ) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} (c+d x)}{\sqrt [4]{a}}\right )}{2 \sqrt {2} a^{3/4} \left (a+b c^4\right )}-\frac {\sqrt [4]{b} c \left (\sqrt {a}+\sqrt {b} c^2\right ) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} (c+d x)}{\sqrt [4]{a}}+1\right )}{2 \sqrt {2} a^{3/4} \left (a+b c^4\right )}-\frac {\log \left (a+b (c+d x)^4\right )}{4 \left (a+b c^4\right )}-\frac {\sqrt {b} c^2 \tan ^{-1}\left (\frac {\sqrt {b} (c+d x)^2}{\sqrt {a}}\right )}{2 \sqrt {a} \left (a+b c^4\right )}+\frac {\log (x)}{a+b c^4} \]

Antiderivative was successfully verified.

[In]

Int[1/(x*(a + b*(c + d*x)^4)),x]

[Out]

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

Rule 204

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

Rule 205

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

Rule 260

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 371

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 617

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 628

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 635

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 1162

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 1165

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 1168

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 1248

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 1876

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

Rule 6725

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rubi steps

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

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Mathematica [C]  time = 0.07, size = 163, normalized size = 0.41 \[ -\frac {\text {RootSum}\left [\text {$\#$1}^4 b d^4+4 \text {$\#$1}^3 b c d^3+6 \text {$\#$1}^2 b c^2 d^2+4 \text {$\#$1} b c^3 d+a+b c^4\& ,\frac {\text {$\#$1}^3 d^3 \log (x-\text {$\#$1})+4 \text {$\#$1}^2 c d^2 \log (x-\text {$\#$1})+4 c^3 \log (x-\text {$\#$1})+6 \text {$\#$1} c^2 d \log (x-\text {$\#$1})}{\text {$\#$1}^3 d^3+3 \text {$\#$1}^2 c d^2+3 \text {$\#$1} c^2 d+c^3}\& \right ]-4 \log (x)}{4 \left (a+b c^4\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x*(a + b*(c + d*x)^4)),x]

[Out]

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

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(a+b*(d*x+c)^4),x, algorithm="fricas")

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left ({\left (d x + c\right )}^{4} b + a\right )} x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(a+b*(d*x+c)^4),x, algorithm="giac")

[Out]

integrate(1/(((d*x + c)^4*b + a)*x), x)

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maple [C]  time = 0.01, size = 139, normalized size = 0.35 \[ \frac {\ln \relax (x )}{b \,c^{4}+a}-\frac {\left (d^{3} \RootOf \left (b \,d^{4} \textit {\_Z}^{4}+4 b \,d^{3} c \,\textit {\_Z}^{3}+6 b \,d^{2} c^{2} \textit {\_Z}^{2}+4 b \,c^{3} d \textit {\_Z} +b \,c^{4}+a \right )^{3}+4 \RootOf \left (b \,d^{4} \textit {\_Z}^{4}+4 b \,d^{3} c \,\textit {\_Z}^{3}+6 b \,d^{2} c^{2} \textit {\_Z}^{2}+4 b \,c^{3} d \textit {\_Z} +b \,c^{4}+a \right )^{2} c \,d^{2}+6 \RootOf \left (b \,d^{4} \textit {\_Z}^{4}+4 b \,d^{3} c \,\textit {\_Z}^{3}+6 b \,d^{2} c^{2} \textit {\_Z}^{2}+4 b \,c^{3} d \textit {\_Z} +b \,c^{4}+a \right ) c^{2} d +4 c^{3}\right ) \ln \left (-\RootOf \left (b \,d^{4} \textit {\_Z}^{4}+4 b \,d^{3} c \,\textit {\_Z}^{3}+6 b \,d^{2} c^{2} \textit {\_Z}^{2}+4 b \,c^{3} d \textit {\_Z} +b \,c^{4}+a \right )+x \right )}{4 \left (b \,c^{4}+a \right ) \left (d^{3} \RootOf \left (b \,d^{4} \textit {\_Z}^{4}+4 b \,d^{3} c \,\textit {\_Z}^{3}+6 b \,d^{2} c^{2} \textit {\_Z}^{2}+4 b \,c^{3} d \textit {\_Z} +b \,c^{4}+a \right )^{3}+3 \RootOf \left (b \,d^{4} \textit {\_Z}^{4}+4 b \,d^{3} c \,\textit {\_Z}^{3}+6 b \,d^{2} c^{2} \textit {\_Z}^{2}+4 b \,c^{3} d \textit {\_Z} +b \,c^{4}+a \right )^{2} c \,d^{2}+3 \RootOf \left (b \,d^{4} \textit {\_Z}^{4}+4 b \,d^{3} c \,\textit {\_Z}^{3}+6 b \,d^{2} c^{2} \textit {\_Z}^{2}+4 b \,c^{3} d \textit {\_Z} +b \,c^{4}+a \right ) c^{2} d +c^{3}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x/(a+b*(d*x+c)^4),x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {b d \int \frac {d^{3} x^{3} + 4 \, c d^{2} x^{2} + 6 \, c^{2} d x + 4 \, c^{3}}{b d^{4} x^{4} + 4 \, b c d^{3} x^{3} + 6 \, b c^{2} d^{2} x^{2} + 4 \, b c^{3} d x + b c^{4} + a}\,{d x}}{b c^{4} + a} + \frac {\log \relax (x)}{b c^{4} + a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(a+b*(d*x+c)^4),x, algorithm="maxima")

[Out]

-b*d*integrate((d^3*x^3 + 4*c*d^2*x^2 + 6*c^2*d*x + 4*c^3)/(b*d^4*x^4 + 4*b*c*d^3*x^3 + 6*b*c^2*d^2*x^2 + 4*b*
c^3*d*x + b*c^4 + a), x)/(b*c^4 + a) + log(x)/(b*c^4 + a)

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mupad [B]  time = 2.18, size = 882, normalized size = 2.24 \[ \frac {\ln \relax (x)}{b\,c^4+a}+\left (\sum _{k=1}^4\ln \left (-{\mathrm {root}\left (256\,a^3\,b\,c^4\,z^4+256\,a^4\,z^4+256\,a^3\,z^3+96\,a^2\,z^2+16\,a\,z+1,z,k\right )}^2\,b^5\,c^5\,d^{15}\,4+\mathrm {root}\left (256\,a^3\,b\,c^4\,z^4+256\,a^4\,z^4+256\,a^3\,z^3+96\,a^2\,z^2+16\,a\,z+1,z,k\right )\,b^4\,c\,d^{15}\,4+\mathrm {root}\left (256\,a^3\,b\,c^4\,z^4+256\,a^4\,z^4+256\,a^3\,z^3+96\,a^2\,z^2+16\,a\,z+1,z,k\right )\,b^4\,d^{16}\,x\,5-{\mathrm {root}\left (256\,a^3\,b\,c^4\,z^4+256\,a^4\,z^4+256\,a^3\,z^3+96\,a^2\,z^2+16\,a\,z+1,z,k\right )}^4\,a^2\,b^5\,c^5\,d^{15}\,64+{\mathrm {root}\left (256\,a^3\,b\,c^4\,z^4+256\,a^4\,z^4+256\,a^3\,z^3+96\,a^2\,z^2+16\,a\,z+1,z,k\right )}^2\,a\,b^4\,c\,d^{15}\,28+{\mathrm {root}\left (256\,a^3\,b\,c^4\,z^4+256\,a^4\,z^4+256\,a^3\,z^3+96\,a^2\,z^2+16\,a\,z+1,z,k\right )}^2\,a\,b^4\,d^{16}\,x\,60+{\mathrm {root}\left (256\,a^3\,b\,c^4\,z^4+256\,a^4\,z^4+256\,a^3\,z^3+96\,a^2\,z^2+16\,a\,z+1,z,k\right )}^3\,a^2\,b^4\,c\,d^{15}\,32-{\mathrm {root}\left (256\,a^3\,b\,c^4\,z^4+256\,a^4\,z^4+256\,a^3\,z^3+96\,a^2\,z^2+16\,a\,z+1,z,k\right )}^4\,a^3\,b^4\,c\,d^{15}\,64-{\mathrm {root}\left (256\,a^3\,b\,c^4\,z^4+256\,a^4\,z^4+256\,a^3\,z^3+96\,a^2\,z^2+16\,a\,z+1,z,k\right )}^3\,a\,b^5\,c^5\,d^{15}\,32+{\mathrm {root}\left (256\,a^3\,b\,c^4\,z^4+256\,a^4\,z^4+256\,a^3\,z^3+96\,a^2\,z^2+16\,a\,z+1,z,k\right )}^3\,a^2\,b^4\,d^{16}\,x\,240+{\mathrm {root}\left (256\,a^3\,b\,c^4\,z^4+256\,a^4\,z^4+256\,a^3\,z^3+96\,a^2\,z^2+16\,a\,z+1,z,k\right )}^4\,a^3\,b^4\,d^{16}\,x\,320-{\mathrm {root}\left (256\,a^3\,b\,c^4\,z^4+256\,a^4\,z^4+256\,a^3\,z^3+96\,a^2\,z^2+16\,a\,z+1,z,k\right )}^2\,b^5\,c^4\,d^{16}\,x\,4-{\mathrm {root}\left (256\,a^3\,b\,c^4\,z^4+256\,a^4\,z^4+256\,a^3\,z^3+96\,a^2\,z^2+16\,a\,z+1,z,k\right )}^3\,a\,b^5\,c^4\,d^{16}\,x\,48-{\mathrm {root}\left (256\,a^3\,b\,c^4\,z^4+256\,a^4\,z^4+256\,a^3\,z^3+96\,a^2\,z^2+16\,a\,z+1,z,k\right )}^4\,a^2\,b^5\,c^4\,d^{16}\,x\,192\right )\,\mathrm {root}\left (256\,a^3\,b\,c^4\,z^4+256\,a^4\,z^4+256\,a^3\,z^3+96\,a^2\,z^2+16\,a\,z+1,z,k\right )\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x*(a + b*(c + d*x)^4)),x)

[Out]

log(x)/(a + b*c^4) + symsum(log(4*root(256*a^3*b*c^4*z^4 + 256*a^4*z^4 + 256*a^3*z^3 + 96*a^2*z^2 + 16*a*z + 1
, z, k)*b^4*c*d^15 - 4*root(256*a^3*b*c^4*z^4 + 256*a^4*z^4 + 256*a^3*z^3 + 96*a^2*z^2 + 16*a*z + 1, z, k)^2*b
^5*c^5*d^15 + 5*root(256*a^3*b*c^4*z^4 + 256*a^4*z^4 + 256*a^3*z^3 + 96*a^2*z^2 + 16*a*z + 1, z, k)*b^4*d^16*x
- 64*root(256*a^3*b*c^4*z^4 + 256*a^4*z^4 + 256*a^3*z^3 + 96*a^2*z^2 + 16*a*z + 1, z, k)^4*a^2*b^5*c^5*d^15 +
28*root(256*a^3*b*c^4*z^4 + 256*a^4*z^4 + 256*a^3*z^3 + 96*a^2*z^2 + 16*a*z + 1, z, k)^2*a*b^4*c*d^15 + 60*ro
ot(256*a^3*b*c^4*z^4 + 256*a^4*z^4 + 256*a^3*z^3 + 96*a^2*z^2 + 16*a*z + 1, z, k)^2*a*b^4*d^16*x + 32*root(256
*a^3*b*c^4*z^4 + 256*a^4*z^4 + 256*a^3*z^3 + 96*a^2*z^2 + 16*a*z + 1, z, k)^3*a^2*b^4*c*d^15 - 64*root(256*a^3
*b*c^4*z^4 + 256*a^4*z^4 + 256*a^3*z^3 + 96*a^2*z^2 + 16*a*z + 1, z, k)^4*a^3*b^4*c*d^15 - 32*root(256*a^3*b*c
^4*z^4 + 256*a^4*z^4 + 256*a^3*z^3 + 96*a^2*z^2 + 16*a*z + 1, z, k)^3*a*b^5*c^5*d^15 + 240*root(256*a^3*b*c^4*
z^4 + 256*a^4*z^4 + 256*a^3*z^3 + 96*a^2*z^2 + 16*a*z + 1, z, k)^3*a^2*b^4*d^16*x + 320*root(256*a^3*b*c^4*z^4
+ 256*a^4*z^4 + 256*a^3*z^3 + 96*a^2*z^2 + 16*a*z + 1, z, k)^4*a^3*b^4*d^16*x - 4*root(256*a^3*b*c^4*z^4 + 25
6*a^4*z^4 + 256*a^3*z^3 + 96*a^2*z^2 + 16*a*z + 1, z, k)^2*b^5*c^4*d^16*x - 48*root(256*a^3*b*c^4*z^4 + 256*a^
4*z^4 + 256*a^3*z^3 + 96*a^2*z^2 + 16*a*z + 1, z, k)^3*a*b^5*c^4*d^16*x - 192*root(256*a^3*b*c^4*z^4 + 256*a^4
*z^4 + 256*a^3*z^3 + 96*a^2*z^2 + 16*a*z + 1, z, k)^4*a^2*b^5*c^4*d^16*x)*root(256*a^3*b*c^4*z^4 + 256*a^4*z^4
+ 256*a^3*z^3 + 96*a^2*z^2 + 16*a*z + 1, z, k), k, 1, 4)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(a+b*(d*x+c)**4),x)

[Out]

Timed out

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