3.234 \(\int \frac {1}{x (d+e x^2) (a+c x^4)} \, dx\)

Optimal. Leaf size=114 \[ -\frac {c d \log \left (a+c x^4\right )}{4 a \left (a e^2+c d^2\right )}-\frac {e^2 \log \left (d+e x^2\right )}{2 d \left (a e^2+c d^2\right )}-\frac {\sqrt {c} e \tan ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {a}}\right )}{2 \sqrt {a} \left (a e^2+c d^2\right )}+\frac {\log (x)}{a d} \]

[Out]

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

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Rubi [A]  time = 0.12, antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {1252, 894, 635, 205, 260} \[ -\frac {e^2 \log \left (d+e x^2\right )}{2 d \left (a e^2+c d^2\right )}-\frac {c d \log \left (a+c x^4\right )}{4 a \left (a e^2+c d^2\right )}-\frac {\sqrt {c} e \tan ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {a}}\right )}{2 \sqrt {a} \left (a e^2+c d^2\right )}+\frac {\log (x)}{a d} \]

Antiderivative was successfully verified.

[In]

Int[1/(x*(d + e*x^2)*(a + c*x^4)),x]

[Out]

-(Sqrt[c]*e*ArcTan[(Sqrt[c]*x^2)/Sqrt[a]])/(2*Sqrt[a]*(c*d^2 + a*e^2)) + Log[x]/(a*d) - (e^2*Log[d + e*x^2])/(
2*d*(c*d^2 + a*e^2)) - (c*d*Log[a + c*x^4])/(4*a*(c*d^2 + a*e^2))

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 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 894

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIn
tegrand[(d + e*x)^m*(f + g*x)^n*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] &&
NeQ[c*d^2 + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && IntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

Rule 1252

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

Rubi steps

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

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Mathematica [A]  time = 0.07, size = 134, normalized size = 1.18 \[ \frac {-c d^2 \log \left (a+c x^4\right )+2 \sqrt {a} \sqrt {c} d e \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )+2 \sqrt {a} \sqrt {c} d e \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}+1\right )-2 a e^2 \log \left (d+e x^2\right )+4 a e^2 \log (x)+4 c d^2 \log (x)}{4 a^2 d e^2+4 a c d^3} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x*(d + e*x^2)*(a + c*x^4)),x]

[Out]

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

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fricas [A]  time = 11.29, size = 201, normalized size = 1.76 \[ \left [\frac {a d e \sqrt {-\frac {c}{a}} \log \left (\frac {c x^{4} - 2 \, a x^{2} \sqrt {-\frac {c}{a}} - a}{c x^{4} + a}\right ) - c d^{2} \log \left (c x^{4} + a\right ) - 2 \, a e^{2} \log \left (e x^{2} + d\right ) + 4 \, {\left (c d^{2} + a e^{2}\right )} \log \relax (x)}{4 \, {\left (a c d^{3} + a^{2} d e^{2}\right )}}, \frac {2 \, a d e \sqrt {\frac {c}{a}} \arctan \left (\frac {a \sqrt {\frac {c}{a}}}{c x^{2}}\right ) - c d^{2} \log \left (c x^{4} + a\right ) - 2 \, a e^{2} \log \left (e x^{2} + d\right ) + 4 \, {\left (c d^{2} + a e^{2}\right )} \log \relax (x)}{4 \, {\left (a c d^{3} + a^{2} d e^{2}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(e*x^2+d)/(c*x^4+a),x, algorithm="fricas")

[Out]

[1/4*(a*d*e*sqrt(-c/a)*log((c*x^4 - 2*a*x^2*sqrt(-c/a) - a)/(c*x^4 + a)) - c*d^2*log(c*x^4 + a) - 2*a*e^2*log(
e*x^2 + d) + 4*(c*d^2 + a*e^2)*log(x))/(a*c*d^3 + a^2*d*e^2), 1/4*(2*a*d*e*sqrt(c/a)*arctan(a*sqrt(c/a)/(c*x^2
)) - c*d^2*log(c*x^4 + a) - 2*a*e^2*log(e*x^2 + d) + 4*(c*d^2 + a*e^2)*log(x))/(a*c*d^3 + a^2*d*e^2)]

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giac [A]  time = 0.29, size = 102, normalized size = 0.89 \[ -\frac {c d \log \left (c x^{4} + a\right )}{4 \, {\left (a c d^{2} + a^{2} e^{2}\right )}} - \frac {c \arctan \left (\frac {c x^{2}}{\sqrt {a c}}\right ) e}{2 \, {\left (c d^{2} + a e^{2}\right )} \sqrt {a c}} - \frac {e^{3} \log \left ({\left | x^{2} e + d \right |}\right )}{2 \, {\left (c d^{3} e + a d e^{3}\right )}} + \frac {\log \left (x^{2}\right )}{2 \, a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(e*x^2+d)/(c*x^4+a),x, algorithm="giac")

[Out]

-1/4*c*d*log(c*x^4 + a)/(a*c*d^2 + a^2*e^2) - 1/2*c*arctan(c*x^2/sqrt(a*c))*e/((c*d^2 + a*e^2)*sqrt(a*c)) - 1/
2*e^3*log(abs(x^2*e + d))/(c*d^3*e + a*d*e^3) + 1/2*log(x^2)/(a*d)

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maple [A]  time = 0.01, size = 101, normalized size = 0.89 \[ -\frac {c e \arctan \left (\frac {c \,x^{2}}{\sqrt {a c}}\right )}{2 \left (a \,e^{2}+c \,d^{2}\right ) \sqrt {a c}}-\frac {c d \ln \left (c \,x^{4}+a \right )}{4 \left (a \,e^{2}+c \,d^{2}\right ) a}-\frac {e^{2} \ln \left (e \,x^{2}+d \right )}{2 \left (a \,e^{2}+c \,d^{2}\right ) d}+\frac {\ln \relax (x )}{a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x/(e*x^2+d)/(c*x^4+a),x)

[Out]

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

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maxima [A]  time = 1.97, size = 101, normalized size = 0.89 \[ -\frac {c d \log \left (c x^{4} + a\right )}{4 \, {\left (a c d^{2} + a^{2} e^{2}\right )}} - \frac {e^{2} \log \left (e x^{2} + d\right )}{2 \, {\left (c d^{3} + a d e^{2}\right )}} - \frac {c e \arctan \left (\frac {c x^{2}}{\sqrt {a c}}\right )}{2 \, {\left (c d^{2} + a e^{2}\right )} \sqrt {a c}} + \frac {\log \left (x^{2}\right )}{2 \, a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(e*x^2+d)/(c*x^4+a),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.96, size = 527, normalized size = 4.62 \[ \frac {\ln \left (64\,a^7\,c\,e^{10}\,x^2-64\,a^6\,e^{10}\,\sqrt {-a^3\,c}-25\,a\,c^5\,d^{10}\,\sqrt {-a^3\,c}+25\,a^2\,c^6\,d^{10}\,x^2+180\,a^2\,d^2\,e^8\,{\left (-a^3\,c\right )}^{3/2}-41\,c^2\,d^6\,e^4\,{\left (-a^3\,c\right )}^{3/2}-9\,a^3\,c^5\,d^8\,e^2\,x^2-41\,a^4\,c^4\,d^6\,e^4\,x^2+109\,a^5\,c^3\,d^4\,e^6\,x^2+180\,a^6\,c^2\,d^2\,e^8\,x^2+9\,a^2\,c^4\,d^8\,e^2\,\sqrt {-a^3\,c}+109\,a\,c\,d^4\,e^6\,{\left (-a^3\,c\right )}^{3/2}\right )\,\left (e\,\sqrt {-a^3\,c}-a\,c\,d\right )}{4\,a^3\,e^2+4\,c\,a^2\,d^2}-\frac {\ln \left (64\,a^6\,e^{10}\,\sqrt {-a^3\,c}+64\,a^7\,c\,e^{10}\,x^2+25\,a\,c^5\,d^{10}\,\sqrt {-a^3\,c}+25\,a^2\,c^6\,d^{10}\,x^2-180\,a^2\,d^2\,e^8\,{\left (-a^3\,c\right )}^{3/2}+41\,c^2\,d^6\,e^4\,{\left (-a^3\,c\right )}^{3/2}-9\,a^3\,c^5\,d^8\,e^2\,x^2-41\,a^4\,c^4\,d^6\,e^4\,x^2+109\,a^5\,c^3\,d^4\,e^6\,x^2+180\,a^6\,c^2\,d^2\,e^8\,x^2-9\,a^2\,c^4\,d^8\,e^2\,\sqrt {-a^3\,c}-109\,a\,c\,d^4\,e^6\,{\left (-a^3\,c\right )}^{3/2}\right )\,\left (e\,\sqrt {-a^3\,c}+a\,c\,d\right )}{4\,\left (a^3\,e^2+c\,a^2\,d^2\right )}-\frac {e^2\,\ln \left (e\,x^2+d\right )}{2\,c\,d^3+2\,a\,d\,e^2}+\frac {\ln \relax (x)}{a\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x*(a + c*x^4)*(d + e*x^2)),x)

[Out]

(log(64*a^7*c*e^10*x^2 - 64*a^6*e^10*(-a^3*c)^(1/2) - 25*a*c^5*d^10*(-a^3*c)^(1/2) + 25*a^2*c^6*d^10*x^2 + 180
*a^2*d^2*e^8*(-a^3*c)^(3/2) - 41*c^2*d^6*e^4*(-a^3*c)^(3/2) - 9*a^3*c^5*d^8*e^2*x^2 - 41*a^4*c^4*d^6*e^4*x^2 +
109*a^5*c^3*d^4*e^6*x^2 + 180*a^6*c^2*d^2*e^8*x^2 + 9*a^2*c^4*d^8*e^2*(-a^3*c)^(1/2) + 109*a*c*d^4*e^6*(-a^3*
c)^(3/2))*(e*(-a^3*c)^(1/2) - a*c*d))/(4*a^3*e^2 + 4*a^2*c*d^2) - (log(64*a^6*e^10*(-a^3*c)^(1/2) + 64*a^7*c*e
^10*x^2 + 25*a*c^5*d^10*(-a^3*c)^(1/2) + 25*a^2*c^6*d^10*x^2 - 180*a^2*d^2*e^8*(-a^3*c)^(3/2) + 41*c^2*d^6*e^4
*(-a^3*c)^(3/2) - 9*a^3*c^5*d^8*e^2*x^2 - 41*a^4*c^4*d^6*e^4*x^2 + 109*a^5*c^3*d^4*e^6*x^2 + 180*a^6*c^2*d^2*e
^8*x^2 - 9*a^2*c^4*d^8*e^2*(-a^3*c)^(1/2) - 109*a*c*d^4*e^6*(-a^3*c)^(3/2))*(e*(-a^3*c)^(1/2) + a*c*d))/(4*(a^
3*e^2 + a^2*c*d^2)) - (e^2*log(d + e*x^2))/(2*c*d^3 + 2*a*d*e^2) + log(x)/(a*d)

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

[Out]

Timed out

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