3.342 \(\int \frac {x^2}{(c+d x) (a+b x^4)} \, dx\)

Optimal. Leaf size=417 \[ \frac {\sqrt {a} d^3 \tan ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )}{2 \sqrt {b} \left (a d^4+b c^4\right )}-\frac {c^2 d \log \left (a+b x^4\right )}{4 \left (a d^4+b c^4\right )}+\frac {c^2 d \log (c+d x)}{a d^4+b c^4}+\frac {c \left (\sqrt {a} d^2+\sqrt {b} c^2\right ) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {a}+\sqrt {b} x^2\right )}{4 \sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \left (a d^4+b c^4\right )}-\frac {c \left (\sqrt {a} d^2+\sqrt {b} c^2\right ) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {a}+\sqrt {b} x^2\right )}{4 \sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \left (a d^4+b c^4\right )}-\frac {c \left (\sqrt {b} c^2-\sqrt {a} d^2\right ) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{2 \sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \left (a d^4+b c^4\right )}+\frac {c \left (\sqrt {b} c^2-\sqrt {a} d^2\right ) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}+1\right )}{2 \sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \left (a d^4+b c^4\right )} \]

[Out]

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

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Rubi [A]  time = 0.55, antiderivative size = 417, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 12, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {6725, 1461, 1168, 1162, 617, 204, 1165, 628, 1248, 635, 205, 260} \[ -\frac {c^2 d \log \left (a+b x^4\right )}{4 \left (a d^4+b c^4\right )}+\frac {c \left (\sqrt {a} d^2+\sqrt {b} c^2\right ) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {a}+\sqrt {b} x^2\right )}{4 \sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \left (a d^4+b c^4\right )}-\frac {c \left (\sqrt {a} d^2+\sqrt {b} c^2\right ) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {a}+\sqrt {b} x^2\right )}{4 \sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \left (a d^4+b c^4\right )}+\frac {\sqrt {a} d^3 \tan ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )}{2 \sqrt {b} \left (a d^4+b c^4\right )}+\frac {c^2 d \log (c+d x)}{a d^4+b c^4}-\frac {c \left (\sqrt {b} c^2-\sqrt {a} d^2\right ) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{2 \sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \left (a d^4+b c^4\right )}+\frac {c \left (\sqrt {b} c^2-\sqrt {a} d^2\right ) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}+1\right )}{2 \sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \left (a d^4+b c^4\right )} \]

Antiderivative was successfully verified.

[In]

Int[x^2/((c + d*x)*(a + b*x^4)),x]

[Out]

(Sqrt[a]*d^3*ArcTan[(Sqrt[b]*x^2)/Sqrt[a]])/(2*Sqrt[b]*(b*c^4 + a*d^4)) - (c*(Sqrt[b]*c^2 - Sqrt[a]*d^2)*ArcTa
n[1 - (Sqrt[2]*b^(1/4)*x)/a^(1/4)])/(2*Sqrt[2]*a^(1/4)*b^(1/4)*(b*c^4 + a*d^4)) + (c*(Sqrt[b]*c^2 - Sqrt[a]*d^
2)*ArcTan[1 + (Sqrt[2]*b^(1/4)*x)/a^(1/4)])/(2*Sqrt[2]*a^(1/4)*b^(1/4)*(b*c^4 + a*d^4)) + (c^2*d*Log[c + d*x])
/(b*c^4 + a*d^4) + (c*(Sqrt[b]*c^2 + Sqrt[a]*d^2)*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*x + Sqrt[b]*x^2])/(4*S
qrt[2]*a^(1/4)*b^(1/4)*(b*c^4 + a*d^4)) - (c*(Sqrt[b]*c^2 + Sqrt[a]*d^2)*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)
*x + Sqrt[b]*x^2])/(4*Sqrt[2]*a^(1/4)*b^(1/4)*(b*c^4 + a*d^4)) - (c^2*d*Log[a + b*x^4])/(4*(b*c^4 + a*d^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 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 1461

Int[((A_) + (B_.)*(x_)^(m_.))*((d_) + (e_.)*(x_)^(n_))^(q_.)*((a_) + (c_.)*(x_)^(n2_))^(p_.), x_Symbol] :> Dis
t[A, Int[(d + e*x^n)^q*(a + c*x^(2*n))^p, x], x] + Dist[B, Int[x^m*(d + e*x^n)^q*(a + c*x^(2*n))^p, x], x] /;
FreeQ[{a, c, d, e, A, B, m, n, p, q}, x] && EqQ[n2, 2*n] && EqQ[m - n + 1, 0]

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

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Mathematica [A]  time = 0.23, size = 370, normalized size = 0.89 \[ \frac {-2 \left (2 a^{3/4} d^3-\sqrt {2} \sqrt {a} \sqrt [4]{b} c d^2+\sqrt {2} b^{3/4} c^3\right ) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )+2 \left (-2 a^{3/4} d^3-\sqrt {2} \sqrt {a} \sqrt [4]{b} c d^2+\sqrt {2} b^{3/4} c^3\right ) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}+1\right )+\sqrt [4]{b} c \left (\sqrt {2} \left (\sqrt {a} d^2+\sqrt {b} c^2\right ) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {a}+\sqrt {b} x^2\right )-\sqrt {2} \sqrt {b} c^2 \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {a}+\sqrt {b} x^2\right )-2 \sqrt [4]{a} \sqrt [4]{b} c d \log \left (a+b x^4\right )+8 \sqrt [4]{a} \sqrt [4]{b} c d \log (c+d x)-\sqrt {2} \sqrt {a} d^2 \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {a}+\sqrt {b} x^2\right )\right )}{8 \sqrt [4]{a} \sqrt {b} \left (a d^4+b c^4\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(Sqrt[2]*b^(3/4)*c^3 - Sqrt[2]*Sqrt[a]*b^(1/4)*c*d^2 + 2*a^(3/4)*d^3)*ArcTan[1 - (Sqrt[2]*b^(1/4)*x)/a^(1/
4)] + 2*(Sqrt[2]*b^(3/4)*c^3 - Sqrt[2]*Sqrt[a]*b^(1/4)*c*d^2 - 2*a^(3/4)*d^3)*ArcTan[1 + (Sqrt[2]*b^(1/4)*x)/a
^(1/4)] + b^(1/4)*c*(8*a^(1/4)*b^(1/4)*c*d*Log[c + d*x] + Sqrt[2]*(Sqrt[b]*c^2 + Sqrt[a]*d^2)*Log[Sqrt[a] - Sq
rt[2]*a^(1/4)*b^(1/4)*x + Sqrt[b]*x^2] - Sqrt[2]*Sqrt[b]*c^2*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*x + Sqrt[b]
*x^2] - Sqrt[2]*Sqrt[a]*d^2*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*x + Sqrt[b]*x^2] - 2*a^(1/4)*b^(1/4)*c*d*Log
[a + b*x^4]))/(8*a^(1/4)*Sqrt[b]*(b*c^4 + a*d^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(x^2/(d*x+c)/(b*x^4+a),x, algorithm="fricas")

[Out]

Timed out

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giac [A]  time = 0.57, size = 401, normalized size = 0.96 \[ \frac {c^{2} d^{2} \log \left ({\left | d x + c \right |}\right )}{b c^{4} d + a d^{5}} - \frac {c^{2} d \log \left ({\left | b x^{4} + a \right |}\right )}{4 \, {\left (b c^{4} + a d^{4}\right )}} + \frac {{\left (\sqrt {2} \sqrt {a b} b d + \left (a b^{3}\right )^{\frac {1}{4}} b c\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, x + \sqrt {2} \left (\frac {a}{b}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{2 \, {\left (\sqrt {2} a b^{2} d^{2} + \sqrt {2} \sqrt {a b} b^{2} c^{2} - 2 \, \left (a b^{3}\right )^{\frac {3}{4}} c d\right )}} + \frac {{\left (\sqrt {2} \sqrt {a b} b d + \left (a b^{3}\right )^{\frac {1}{4}} b c\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, x - \sqrt {2} \left (\frac {a}{b}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{2 \, {\left (\sqrt {2} a b^{2} d^{2} + \sqrt {2} \sqrt {a b} b^{2} c^{2} + 2 \, \left (a b^{3}\right )^{\frac {3}{4}} c d\right )}} - \frac {{\left (\left (a b^{3}\right )^{\frac {1}{4}} a b c d^{2} + \left (a b^{3}\right )^{\frac {3}{4}} c^{3}\right )} \log \left (x^{2} + \sqrt {2} x \left (\frac {a}{b}\right )^{\frac {1}{4}} + \sqrt {\frac {a}{b}}\right )}{4 \, {\left (\sqrt {2} a b^{3} c^{4} + \sqrt {2} a^{2} b^{2} d^{4}\right )}} + \frac {{\left (\left (a b^{3}\right )^{\frac {1}{4}} a b c d^{2} + \left (a b^{3}\right )^{\frac {3}{4}} c^{3}\right )} \log \left (x^{2} - \sqrt {2} x \left (\frac {a}{b}\right )^{\frac {1}{4}} + \sqrt {\frac {a}{b}}\right )}{4 \, {\left (\sqrt {2} a b^{3} c^{4} + \sqrt {2} a^{2} b^{2} d^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c^2*d^2*log(abs(d*x + c))/(b*c^4*d + a*d^5) - 1/4*c^2*d*log(abs(b*x^4 + a))/(b*c^4 + a*d^4) + 1/2*(sqrt(2)*sqr
t(a*b)*b*d + (a*b^3)^(1/4)*b*c)*arctan(1/2*sqrt(2)*(2*x + sqrt(2)*(a/b)^(1/4))/(a/b)^(1/4))/(sqrt(2)*a*b^2*d^2
+ sqrt(2)*sqrt(a*b)*b^2*c^2 - 2*(a*b^3)^(3/4)*c*d) + 1/2*(sqrt(2)*sqrt(a*b)*b*d + (a*b^3)^(1/4)*b*c)*arctan(1
/2*sqrt(2)*(2*x - sqrt(2)*(a/b)^(1/4))/(a/b)^(1/4))/(sqrt(2)*a*b^2*d^2 + sqrt(2)*sqrt(a*b)*b^2*c^2 + 2*(a*b^3)
^(3/4)*c*d) - 1/4*((a*b^3)^(1/4)*a*b*c*d^2 + (a*b^3)^(3/4)*c^3)*log(x^2 + sqrt(2)*x*(a/b)^(1/4) + sqrt(a/b))/(
sqrt(2)*a*b^3*c^4 + sqrt(2)*a^2*b^2*d^4) + 1/4*((a*b^3)^(1/4)*a*b*c*d^2 + (a*b^3)^(3/4)*c^3)*log(x^2 - sqrt(2)
*x*(a/b)^(1/4) + sqrt(a/b))/(sqrt(2)*a*b^3*c^4 + sqrt(2)*a^2*b^2*d^4)

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maple [A]  time = 0.01, size = 422, normalized size = 1.01 \[ \frac {a \,d^{3} \arctan \left (\sqrt {\frac {b}{a}}\, x^{2}\right )}{2 \left (a \,d^{4}+b \,c^{4}\right ) \sqrt {a b}}+\frac {\sqrt {2}\, c^{3} \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )}{4 \left (a \,d^{4}+b \,c^{4}\right ) \left (\frac {a}{b}\right )^{\frac {1}{4}}}+\frac {\sqrt {2}\, c^{3} \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )}{4 \left (a \,d^{4}+b \,c^{4}\right ) \left (\frac {a}{b}\right )^{\frac {1}{4}}}+\frac {\sqrt {2}\, c^{3} \ln \left (\frac {x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {\frac {a}{b}}}{x^{2}+\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {\frac {a}{b}}}\right )}{8 \left (a \,d^{4}+b \,c^{4}\right ) \left (\frac {a}{b}\right )^{\frac {1}{4}}}-\frac {c^{2} d \ln \left (b \,x^{4}+a \right )}{4 \left (a \,d^{4}+b \,c^{4}\right )}+\frac {c^{2} d \ln \left (d x +c \right )}{a \,d^{4}+b \,c^{4}}-\frac {\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, c \,d^{2} \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )}{4 \left (a \,d^{4}+b \,c^{4}\right )}-\frac {\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, c \,d^{2} \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )}{4 \left (a \,d^{4}+b \,c^{4}\right )}-\frac {\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, c \,d^{2} \ln \left (\frac {x^{2}+\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {\frac {a}{b}}}{x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {\frac {a}{b}}}\right )}{8 \left (a \,d^{4}+b \,c^{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(d*x+c)/(b*x^4+a),x)

[Out]

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

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maxima [A]  time = 1.48, size = 349, normalized size = 0.84 \[ \frac {c^{2} d \log \left (d x + c\right )}{b c^{4} + a d^{4}} - \frac {\frac {\sqrt {2} {\left (\sqrt {2} a^{\frac {3}{4}} b^{\frac {5}{4}} c^{2} d + \sqrt {a} b^{\frac {3}{2}} c^{3} + a b c d^{2}\right )} \log \left (\sqrt {b} x^{2} + \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} x + \sqrt {a}\right )}{a^{\frac {3}{4}} b^{\frac {5}{4}}} + \frac {\sqrt {2} {\left (\sqrt {2} a^{\frac {3}{4}} b^{\frac {5}{4}} c^{2} d - \sqrt {a} b^{\frac {3}{2}} c^{3} - a b c d^{2}\right )} \log \left (\sqrt {b} x^{2} - \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} x + \sqrt {a}\right )}{a^{\frac {3}{4}} b^{\frac {5}{4}}} - \frac {2 \, {\left (\sqrt {2} a^{\frac {3}{4}} b^{\frac {7}{4}} c^{3} - \sqrt {2} a^{\frac {5}{4}} b^{\frac {5}{4}} c d^{2} - 2 \, a^{\frac {3}{2}} b d^{3}\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {b} x + \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{a^{\frac {3}{4}} \sqrt {\sqrt {a} \sqrt {b}} b^{\frac {5}{4}}} - \frac {2 \, {\left (\sqrt {2} a^{\frac {3}{4}} b^{\frac {7}{4}} c^{3} - \sqrt {2} a^{\frac {5}{4}} b^{\frac {5}{4}} c d^{2} + 2 \, a^{\frac {3}{2}} b d^{3}\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {b} x - \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{a^{\frac {3}{4}} \sqrt {\sqrt {a} \sqrt {b}} b^{\frac {5}{4}}}}{8 \, {\left (b c^{4} + a d^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c^2*d*log(d*x + c)/(b*c^4 + a*d^4) - 1/8*(sqrt(2)*(sqrt(2)*a^(3/4)*b^(5/4)*c^2*d + sqrt(a)*b^(3/2)*c^3 + a*b*c
*d^2)*log(sqrt(b)*x^2 + sqrt(2)*a^(1/4)*b^(1/4)*x + sqrt(a))/(a^(3/4)*b^(5/4)) + sqrt(2)*(sqrt(2)*a^(3/4)*b^(5
/4)*c^2*d - sqrt(a)*b^(3/2)*c^3 - a*b*c*d^2)*log(sqrt(b)*x^2 - sqrt(2)*a^(1/4)*b^(1/4)*x + sqrt(a))/(a^(3/4)*b
^(5/4)) - 2*(sqrt(2)*a^(3/4)*b^(7/4)*c^3 - sqrt(2)*a^(5/4)*b^(5/4)*c*d^2 - 2*a^(3/2)*b*d^3)*arctan(1/2*sqrt(2)
*(2*sqrt(b)*x + sqrt(2)*a^(1/4)*b^(1/4))/sqrt(sqrt(a)*sqrt(b)))/(a^(3/4)*sqrt(sqrt(a)*sqrt(b))*b^(5/4)) - 2*(s
qrt(2)*a^(3/4)*b^(7/4)*c^3 - sqrt(2)*a^(5/4)*b^(5/4)*c*d^2 + 2*a^(3/2)*b*d^3)*arctan(1/2*sqrt(2)*(2*sqrt(b)*x
- sqrt(2)*a^(1/4)*b^(1/4))/sqrt(sqrt(a)*sqrt(b)))/(a^(3/4)*sqrt(sqrt(a)*sqrt(b))*b^(5/4)))/(b*c^4 + a*d^4)

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mupad [B]  time = 2.45, size = 823, normalized size = 1.97 \[ \left (\sum _{k=1}^4\ln \left (a\,b^2\,d\,\left (c\,d+d^2\,x-\mathrm {root}\left (256\,a^2\,b^2\,d^4\,z^4+256\,a\,b^3\,c^4\,z^4+256\,a\,b^2\,c^2\,d\,z^3+32\,a\,b\,d^2\,z^2+1,z,k\right )\,b\,c^3+{\mathrm {root}\left (256\,a^2\,b^2\,d^4\,z^4+256\,a\,b^3\,c^4\,z^4+256\,a\,b^2\,c^2\,d\,z^3+32\,a\,b\,d^2\,z^2+1,z,k\right )}^2\,b^2\,c^4\,x\,4+{\mathrm {root}\left (256\,a^2\,b^2\,d^4\,z^4+256\,a\,b^3\,c^4\,z^4+256\,a\,b^2\,c^2\,d\,z^3+32\,a\,b\,d^2\,z^2+1,z,k\right )}^2\,a\,b\,d^4\,x\,36-{\mathrm {root}\left (256\,a^2\,b^2\,d^4\,z^4+256\,a\,b^3\,c^4\,z^4+256\,a\,b^2\,c^2\,d\,z^3+32\,a\,b\,d^2\,z^2+1,z,k\right )}^4\,a\,b^3\,c^5\,d\,128-\mathrm {root}\left (256\,a^2\,b^2\,d^4\,z^4+256\,a\,b^3\,c^4\,z^4+256\,a\,b^2\,c^2\,d\,z^3+32\,a\,b\,d^2\,z^2+1,z,k\right )\,b\,c^2\,d\,x\,5+{\mathrm {root}\left (256\,a^2\,b^2\,d^4\,z^4+256\,a\,b^3\,c^4\,z^4+256\,a\,b^2\,c^2\,d\,z^3+32\,a\,b\,d^2\,z^2+1,z,k\right )}^3\,a\,b^2\,c^3\,d^2\,96+{\mathrm {root}\left (256\,a^2\,b^2\,d^4\,z^4+256\,a\,b^3\,c^4\,z^4+256\,a\,b^2\,c^2\,d\,z^3+32\,a\,b\,d^2\,z^2+1,z,k\right )}^4\,a^2\,b^2\,c\,d^5\,384+{\mathrm {root}\left (256\,a^2\,b^2\,d^4\,z^4+256\,a\,b^3\,c^4\,z^4+256\,a\,b^2\,c^2\,d\,z^3+32\,a\,b\,d^2\,z^2+1,z,k\right )}^4\,a^2\,b^2\,d^6\,x\,320+{\mathrm {root}\left (256\,a^2\,b^2\,d^4\,z^4+256\,a\,b^3\,c^4\,z^4+256\,a\,b^2\,c^2\,d\,z^3+32\,a\,b\,d^2\,z^2+1,z,k\right )}^2\,a\,b\,c\,d^3\,32+{\mathrm {root}\left (256\,a^2\,b^2\,d^4\,z^4+256\,a\,b^3\,c^4\,z^4+256\,a\,b^2\,c^2\,d\,z^3+32\,a\,b\,d^2\,z^2+1,z,k\right )}^3\,a\,b^2\,c^2\,d^3\,x\,160-{\mathrm {root}\left (256\,a^2\,b^2\,d^4\,z^4+256\,a\,b^3\,c^4\,z^4+256\,a\,b^2\,c^2\,d\,z^3+32\,a\,b\,d^2\,z^2+1,z,k\right )}^4\,a\,b^3\,c^4\,d^2\,x\,192\right )\right )\,\mathrm {root}\left (256\,a^2\,b^2\,d^4\,z^4+256\,a\,b^3\,c^4\,z^4+256\,a\,b^2\,c^2\,d\,z^3+32\,a\,b\,d^2\,z^2+1,z,k\right )\right )+\frac {c^2\,d\,\ln \left (c+d\,x\right )}{b\,c^4+a\,d^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/((a + b*x^4)*(c + d*x)),x)

[Out]

symsum(log(a*b^2*d*(c*d + d^2*x - root(256*a^2*b^2*d^4*z^4 + 256*a*b^3*c^4*z^4 + 256*a*b^2*c^2*d*z^3 + 32*a*b*
d^2*z^2 + 1, z, k)*b*c^3 + 4*root(256*a^2*b^2*d^4*z^4 + 256*a*b^3*c^4*z^4 + 256*a*b^2*c^2*d*z^3 + 32*a*b*d^2*z
^2 + 1, z, k)^2*b^2*c^4*x + 36*root(256*a^2*b^2*d^4*z^4 + 256*a*b^3*c^4*z^4 + 256*a*b^2*c^2*d*z^3 + 32*a*b*d^2
*z^2 + 1, z, k)^2*a*b*d^4*x - 128*root(256*a^2*b^2*d^4*z^4 + 256*a*b^3*c^4*z^4 + 256*a*b^2*c^2*d*z^3 + 32*a*b*
d^2*z^2 + 1, z, k)^4*a*b^3*c^5*d - 5*root(256*a^2*b^2*d^4*z^4 + 256*a*b^3*c^4*z^4 + 256*a*b^2*c^2*d*z^3 + 32*a
*b*d^2*z^2 + 1, z, k)*b*c^2*d*x + 96*root(256*a^2*b^2*d^4*z^4 + 256*a*b^3*c^4*z^4 + 256*a*b^2*c^2*d*z^3 + 32*a
*b*d^2*z^2 + 1, z, k)^3*a*b^2*c^3*d^2 + 384*root(256*a^2*b^2*d^4*z^4 + 256*a*b^3*c^4*z^4 + 256*a*b^2*c^2*d*z^3
+ 32*a*b*d^2*z^2 + 1, z, k)^4*a^2*b^2*c*d^5 + 320*root(256*a^2*b^2*d^4*z^4 + 256*a*b^3*c^4*z^4 + 256*a*b^2*c^
2*d*z^3 + 32*a*b*d^2*z^2 + 1, z, k)^4*a^2*b^2*d^6*x + 32*root(256*a^2*b^2*d^4*z^4 + 256*a*b^3*c^4*z^4 + 256*a*
b^2*c^2*d*z^3 + 32*a*b*d^2*z^2 + 1, z, k)^2*a*b*c*d^3 + 160*root(256*a^2*b^2*d^4*z^4 + 256*a*b^3*c^4*z^4 + 256
*a*b^2*c^2*d*z^3 + 32*a*b*d^2*z^2 + 1, z, k)^3*a*b^2*c^2*d^3*x - 192*root(256*a^2*b^2*d^4*z^4 + 256*a*b^3*c^4*
z^4 + 256*a*b^2*c^2*d*z^3 + 32*a*b*d^2*z^2 + 1, z, k)^4*a*b^3*c^4*d^2*x))*root(256*a^2*b^2*d^4*z^4 + 256*a*b^3
*c^4*z^4 + 256*a*b^2*c^2*d*z^3 + 32*a*b*d^2*z^2 + 1, z, k), k, 1, 4) + (c^2*d*log(c + d*x))/(a*d^4 + b*c^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(x**2/(d*x+c)/(b*x**4+a),x)

[Out]

Timed out

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