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

Optimal. Leaf size=73 \[ \frac {e \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{2 a^{3/2} \sqrt {c}}-\frac {d \log \left (a+c x^2\right )}{2 a^2}+\frac {d \log (x)}{a^2}+\frac {d+e x}{2 a \left (a+c x^2\right )} \]

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Rubi [A]  time = 0.06, antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {823, 801, 635, 205, 260} \begin {gather*} -\frac {d \log \left (a+c x^2\right )}{2 a^2}+\frac {e \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{2 a^{3/2} \sqrt {c}}+\frac {d \log (x)}{a^2}+\frac {d+e x}{2 a \left (a+c x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d + e*x)/(x*(a + c*x^2)^2),x]

[Out]

(d + e*x)/(2*a*(a + c*x^2)) + (e*ArcTan[(Sqrt[c]*x)/Sqrt[a]])/(2*a^(3/2)*Sqrt[c]) + (d*Log[x])/a^2 - (d*Log[a
+ c*x^2])/(2*a^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 801

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
(d + e*x)^m*(f + g*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

Rule 823

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(
m + 1)*(f*a*c*e - a*g*c*d + c*(c*d*f + a*e*g)*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), x] + Di
st[1/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*Simp[f*(c^2*d^2*(2*p + 3) + a*c*e^2*
(m + 2*p + 3)) - a*c*d*e*g*m + c*e*(c*d*f + a*e*g)*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x]
&& NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rubi steps

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

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Mathematica [A]  time = 0.06, size = 65, normalized size = 0.89 \begin {gather*} \frac {\frac {a (d+e x)}{a+c x^2}-d \log \left (a+c x^2\right )+\frac {\sqrt {a} e \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{\sqrt {c}}+2 d \log (x)}{2 a^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {d+e x}{x \left (a+c x^2\right )^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[(d + e*x)/(x*(a + c*x^2)^2),x]

[Out]

IntegrateAlgebraic[(d + e*x)/(x*(a + c*x^2)^2), x]

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fricas [A]  time = 0.43, size = 217, normalized size = 2.97 \begin {gather*} \left [\frac {2 \, a c e x + 2 \, a c d - {\left (c e x^{2} + a e\right )} \sqrt {-a c} \log \left (\frac {c x^{2} - 2 \, \sqrt {-a c} x - a}{c x^{2} + a}\right ) - 2 \, {\left (c^{2} d x^{2} + a c d\right )} \log \left (c x^{2} + a\right ) + 4 \, {\left (c^{2} d x^{2} + a c d\right )} \log \relax (x)}{4 \, {\left (a^{2} c^{2} x^{2} + a^{3} c\right )}}, \frac {a c e x + a c d + {\left (c e x^{2} + a e\right )} \sqrt {a c} \arctan \left (\frac {\sqrt {a c} x}{a}\right ) - {\left (c^{2} d x^{2} + a c d\right )} \log \left (c x^{2} + a\right ) + 2 \, {\left (c^{2} d x^{2} + a c d\right )} \log \relax (x)}{2 \, {\left (a^{2} c^{2} x^{2} + a^{3} c\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.15, size = 67, normalized size = 0.92 \begin {gather*} \frac {\arctan \left (\frac {c x}{\sqrt {a c}}\right ) e}{2 \, \sqrt {a c} a} - \frac {d \log \left (c x^{2} + a\right )}{2 \, a^{2}} + \frac {d \log \left ({\left | x \right |}\right )}{a^{2}} + \frac {a x e + a d}{2 \, {\left (c x^{2} + a\right )} a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 74, normalized size = 1.01 \begin {gather*} \frac {e x}{2 \left (c \,x^{2}+a \right ) a}+\frac {e \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{2 \sqrt {a c}\, a}+\frac {d}{2 \left (c \,x^{2}+a \right ) a}+\frac {d \ln \relax (x )}{a^{2}}-\frac {d \ln \left (c \,x^{2}+a \right )}{2 a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.28, size = 61, normalized size = 0.84 \begin {gather*} \frac {e \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{2 \, \sqrt {a c} a} + \frac {e x + d}{2 \, {\left (a c x^{2} + a^{2}\right )}} - \frac {d \log \left (c x^{2} + a\right )}{2 \, a^{2}} + \frac {d \log \relax (x)}{a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.26, size = 165, normalized size = 2.26 \begin {gather*} \frac {\frac {d}{2\,a}+\frac {e\,x}{2\,a}}{c\,x^2+a}+\frac {d\,\ln \relax (x)}{a^2}+\frac {\ln \left (a\,e\,\sqrt {-a^5\,c}-6\,a^3\,c\,d+a^3\,c\,e\,x+6\,c\,d\,x\,\sqrt {-a^5\,c}\right )\,\left (e\,\sqrt {-a^5\,c}-2\,a^2\,c\,d\right )}{4\,a^4\,c}-\frac {\ln \left (a\,e\,\sqrt {-a^5\,c}+6\,a^3\,c\,d-a^3\,c\,e\,x+6\,c\,d\,x\,\sqrt {-a^5\,c}\right )\,\left (e\,\sqrt {-a^5\,c}+2\,a^2\,c\,d\right )}{4\,a^4\,c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)/(x*(a + c*x^2)^2),x)

[Out]

(d/(2*a) + (e*x)/(2*a))/(a + c*x^2) + (d*log(x))/a^2 + (log(a*e*(-a^5*c)^(1/2) - 6*a^3*c*d + a^3*c*e*x + 6*c*d
*x*(-a^5*c)^(1/2))*(e*(-a^5*c)^(1/2) - 2*a^2*c*d))/(4*a^4*c) - (log(a*e*(-a^5*c)^(1/2) + 6*a^3*c*d - a^3*c*e*x
+ 6*c*d*x*(-a^5*c)^(1/2))*(e*(-a^5*c)^(1/2) + 2*a^2*c*d))/(4*a^4*c)

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sympy [B]  time = 1.71, size = 359, normalized size = 4.92 \begin {gather*} \left (- \frac {d}{2 a^{2}} - \frac {e \sqrt {- a^{5} c}}{4 a^{4} c}\right ) \log {\left (x + \frac {- 96 a^{4} c d \left (- \frac {d}{2 a^{2}} - \frac {e \sqrt {- a^{5} c}}{4 a^{4} c}\right )^{2} + 4 a^{3} e^{2} \left (- \frac {d}{2 a^{2}} - \frac {e \sqrt {- a^{5} c}}{4 a^{4} c}\right ) + 48 a^{2} c d^{2} \left (- \frac {d}{2 a^{2}} - \frac {e \sqrt {- a^{5} c}}{4 a^{4} c}\right ) - 4 a d e^{2} + 48 c d^{3}}{a e^{3} + 36 c d^{2} e} \right )} + \left (- \frac {d}{2 a^{2}} + \frac {e \sqrt {- a^{5} c}}{4 a^{4} c}\right ) \log {\left (x + \frac {- 96 a^{4} c d \left (- \frac {d}{2 a^{2}} + \frac {e \sqrt {- a^{5} c}}{4 a^{4} c}\right )^{2} + 4 a^{3} e^{2} \left (- \frac {d}{2 a^{2}} + \frac {e \sqrt {- a^{5} c}}{4 a^{4} c}\right ) + 48 a^{2} c d^{2} \left (- \frac {d}{2 a^{2}} + \frac {e \sqrt {- a^{5} c}}{4 a^{4} c}\right ) - 4 a d e^{2} + 48 c d^{3}}{a e^{3} + 36 c d^{2} e} \right )} + \frac {d + e x}{2 a^{2} + 2 a c x^{2}} + \frac {d \log {\relax (x )}}{a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-d/(2*a**2) - e*sqrt(-a**5*c)/(4*a**4*c))*log(x + (-96*a**4*c*d*(-d/(2*a**2) - e*sqrt(-a**5*c)/(4*a**4*c))**2
+ 4*a**3*e**2*(-d/(2*a**2) - e*sqrt(-a**5*c)/(4*a**4*c)) + 48*a**2*c*d**2*(-d/(2*a**2) - e*sqrt(-a**5*c)/(4*a
**4*c)) - 4*a*d*e**2 + 48*c*d**3)/(a*e**3 + 36*c*d**2*e)) + (-d/(2*a**2) + e*sqrt(-a**5*c)/(4*a**4*c))*log(x +
(-96*a**4*c*d*(-d/(2*a**2) + e*sqrt(-a**5*c)/(4*a**4*c))**2 + 4*a**3*e**2*(-d/(2*a**2) + e*sqrt(-a**5*c)/(4*a
**4*c)) + 48*a**2*c*d**2*(-d/(2*a**2) + e*sqrt(-a**5*c)/(4*a**4*c)) - 4*a*d*e**2 + 48*c*d**3)/(a*e**3 + 36*c*d
**2*e)) + (d + e*x)/(2*a**2 + 2*a*c*x**2) + d*log(x)/a**2

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