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

Optimal. Leaf size=86 \[ -\frac {e \log \left (a+c x^2\right )}{2 \left (a e^2+c d^2\right )}+\frac {e \log (d+e x)}{a e^2+c d^2}+\frac {\sqrt {c} d \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{\sqrt {a} \left (a e^2+c d^2\right )} \]

[Out]

e*ln(e*x+d)/(a*e^2+c*d^2)-1/2*e*ln(c*x^2+a)/(a*e^2+c*d^2)+d*arctan(x*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 = 0.04, antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {706, 31, 635, 205, 260} \[ -\frac {e \log \left (a+c x^2\right )}{2 \left (a e^2+c d^2\right )}+\frac {e \log (d+e x)}{a e^2+c d^2}+\frac {\sqrt {c} d \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{\sqrt {a} \left (a e^2+c d^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[1/((d + e*x)*(a + c*x^2)),x]

[Out]

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

Rule 31

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

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 706

Int[1/(((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)), x_Symbol] :> Dist[e^2/(c*d^2 + a*e^2), Int[1/(d + e*x), x],
x] + Dist[1/(c*d^2 + a*e^2), Int[(c*d - c*e*x)/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a
*e^2, 0]

Rubi steps

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

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Mathematica [A]  time = 0.04, size = 63, normalized size = 0.73 \[ \frac {\frac {2 \sqrt {c} d \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{\sqrt {a}}-e \log \left (a+c x^2\right )+2 e \log (d+e x)}{2 a e^2+2 c d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.25, size = 135, normalized size = 1.57 \[ \left [\frac {d \sqrt {-\frac {c}{a}} \log \left (\frac {c x^{2} + 2 \, a x \sqrt {-\frac {c}{a}} - a}{c x^{2} + a}\right ) - e \log \left (c x^{2} + a\right ) + 2 \, e \log \left (e x + d\right )}{2 \, {\left (c d^{2} + a e^{2}\right )}}, \frac {2 \, d \sqrt {\frac {c}{a}} \arctan \left (x \sqrt {\frac {c}{a}}\right ) - e \log \left (c x^{2} + a\right ) + 2 \, e \log \left (e x + d\right )}{2 \, {\left (c d^{2} + a e^{2}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.17, size = 79, normalized size = 0.92 \[ \frac {c d \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{{\left (c d^{2} + a e^{2}\right )} \sqrt {a c}} - \frac {e \log \left (c x^{2} + a\right )}{2 \, {\left (c d^{2} + a e^{2}\right )}} + \frac {e^{2} \log \left ({\left | x e + d \right |}\right )}{c d^{2} e + a e^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 77, normalized size = 0.90 \[ \frac {c d \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{\left (a \,e^{2}+c \,d^{2}\right ) \sqrt {a c}}-\frac {e \ln \left (c \,x^{2}+a \right )}{2 \left (a \,e^{2}+c \,d^{2}\right )}+\frac {e \ln \left (e x +d \right )}{a \,e^{2}+c \,d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.96, size = 76, normalized size = 0.88 \[ \frac {c d \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{{\left (c d^{2} + a e^{2}\right )} \sqrt {a c}} - \frac {e \log \left (c x^{2} + a\right )}{2 \, {\left (c d^{2} + a e^{2}\right )}} + \frac {e \log \left (e x + d\right )}{c d^{2} + a e^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.82, size = 230, normalized size = 2.67 \[ \frac {e\,\ln \left (d+e\,x\right )}{c\,d^2+a\,e^2}-\frac {\ln \left (3\,c^2\,e^2\,x+c^2\,d\,e-\frac {c^2\,e\,\left (a\,e-d\,\sqrt {-a\,c}\right )\,\left (-c\,x\,d^2+4\,a\,d\,e+3\,a\,x\,e^2\right )}{a^2\,e^2+c\,a\,d^2}\right )\,\left (a\,e-d\,\sqrt {-a\,c}\right )}{2\,\left (a^2\,e^2+c\,a\,d^2\right )}-\frac {\ln \left (3\,c^2\,e^2\,x+c^2\,d\,e-\frac {c^2\,e\,\left (a\,e+d\,\sqrt {-a\,c}\right )\,\left (-c\,x\,d^2+4\,a\,d\,e+3\,a\,x\,e^2\right )}{a^2\,e^2+c\,a\,d^2}\right )\,\left (a\,e+d\,\sqrt {-a\,c}\right )}{2\,\left (a^2\,e^2+c\,a\,d^2\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a + c*x^2)*(d + e*x)),x)

[Out]

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

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

[Out]

Timed out

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