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

Optimal. Leaf size=123 \[ -\frac {c d e \log \left (a+c x^2\right )}{\left (a e^2+c d^2\right )^2}-\frac {e}{(d+e x) \left (a e^2+c d^2\right )}+\frac {2 c d e \log (d+e x)}{\left (a e^2+c d^2\right )^2}+\frac {\sqrt {c} \left (c d^2-a e^2\right ) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{\sqrt {a} \left (a e^2+c d^2\right )^2} \]

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Rubi [A]  time = 0.10, antiderivative size = 123, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {710, 801, 635, 205, 260} \begin {gather*} -\frac {c d e \log \left (a+c x^2\right )}{\left (a e^2+c d^2\right )^2}-\frac {e}{(d+e x) \left (a e^2+c d^2\right )}+\frac {2 c d e \log (d+e x)}{\left (a e^2+c d^2\right )^2}+\frac {\sqrt {c} \left (c d^2-a e^2\right ) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{\sqrt {a} \left (a e^2+c d^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((d + e*x)^2*(a + c*x^2)),x]

[Out]

-(e/((c*d^2 + a*e^2)*(d + e*x))) + (Sqrt[c]*(c*d^2 - a*e^2)*ArcTan[(Sqrt[c]*x)/Sqrt[a]])/(Sqrt[a]*(c*d^2 + a*e
^2)^2) + (2*c*d*e*Log[d + e*x])/(c*d^2 + a*e^2)^2 - (c*d*e*Log[a + c*x^2])/(c*d^2 + a*e^2)^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 710

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

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]

Rubi steps

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

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Mathematica [A]  time = 0.08, size = 113, normalized size = 0.92 \begin {gather*} \frac {\sqrt {c} (d+e x) \left (c d^2-a e^2\right ) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right )-\sqrt {a} e \left (c d (d+e x) \log \left (a+c x^2\right )+a e^2+c d^2-2 c d (d+e x) \log (d+e x)\right )}{\sqrt {a} (d+e x) \left (a e^2+c d^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{(d+e x)^2 \left (a+c x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.54, size = 350, normalized size = 2.85 \begin {gather*} \left [-\frac {2 \, c d^{2} e + 2 \, a e^{3} + {\left (c d^{3} - a d e^{2} + {\left (c d^{2} e - a e^{3}\right )} x\right )} \sqrt {-\frac {c}{a}} \log \left (\frac {c x^{2} - 2 \, a x \sqrt {-\frac {c}{a}} - a}{c x^{2} + a}\right ) + 2 \, {\left (c d e^{2} x + c d^{2} e\right )} \log \left (c x^{2} + a\right ) - 4 \, {\left (c d e^{2} x + c d^{2} e\right )} \log \left (e x + d\right )}{2 \, {\left (c^{2} d^{5} + 2 \, a c d^{3} e^{2} + a^{2} d e^{4} + {\left (c^{2} d^{4} e + 2 \, a c d^{2} e^{3} + a^{2} e^{5}\right )} x\right )}}, -\frac {c d^{2} e + a e^{3} - {\left (c d^{3} - a d e^{2} + {\left (c d^{2} e - a e^{3}\right )} x\right )} \sqrt {\frac {c}{a}} \arctan \left (x \sqrt {\frac {c}{a}}\right ) + {\left (c d e^{2} x + c d^{2} e\right )} \log \left (c x^{2} + a\right ) - 2 \, {\left (c d e^{2} x + c d^{2} e\right )} \log \left (e x + d\right )}{c^{2} d^{5} + 2 \, a c d^{3} e^{2} + a^{2} d e^{4} + {\left (c^{2} d^{4} e + 2 \, a c d^{2} e^{3} + a^{2} e^{5}\right )} x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*(2*c*d^2*e + 2*a*e^3 + (c*d^3 - a*d*e^2 + (c*d^2*e - a*e^3)*x)*sqrt(-c/a)*log((c*x^2 - 2*a*x*sqrt(-c/a)
- a)/(c*x^2 + a)) + 2*(c*d*e^2*x + c*d^2*e)*log(c*x^2 + a) - 4*(c*d*e^2*x + c*d^2*e)*log(e*x + d))/(c^2*d^5 +
2*a*c*d^3*e^2 + a^2*d*e^4 + (c^2*d^4*e + 2*a*c*d^2*e^3 + a^2*e^5)*x), -(c*d^2*e + a*e^3 - (c*d^3 - a*d*e^2 + (
c*d^2*e - a*e^3)*x)*sqrt(c/a)*arctan(x*sqrt(c/a)) + (c*d*e^2*x + c*d^2*e)*log(c*x^2 + a) - 2*(c*d*e^2*x + c*d^
2*e)*log(e*x + d))/(c^2*d^5 + 2*a*c*d^3*e^2 + a^2*d*e^4 + (c^2*d^4*e + 2*a*c*d^2*e^3 + a^2*e^5)*x)]

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giac [A]  time = 0.16, size = 187, normalized size = 1.52 \begin {gather*} -\frac {c d e \log \left (c - \frac {2 \, c d}{x e + d} + \frac {c d^{2}}{{\left (x e + d\right )}^{2}} + \frac {a e^{2}}{{\left (x e + d\right )}^{2}}\right )}{c^{2} d^{4} + 2 \, a c d^{2} e^{2} + a^{2} e^{4}} + \frac {{\left (c^{2} d^{2} e^{2} - a c e^{4}\right )} \arctan \left (\frac {{\left (c d - \frac {c d^{2}}{x e + d} - \frac {a e^{2}}{x e + d}\right )} e^{\left (-1\right )}}{\sqrt {a c}}\right ) e^{\left (-2\right )}}{{\left (c^{2} d^{4} + 2 \, a c d^{2} e^{2} + a^{2} e^{4}\right )} \sqrt {a c}} - \frac {e^{3}}{{\left (c d^{2} e^{2} + a e^{4}\right )} {\left (x e + d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 143, normalized size = 1.16 \begin {gather*} -\frac {a c \,e^{2} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{\left (a \,e^{2}+c \,d^{2}\right )^{2} \sqrt {a c}}+\frac {c^{2} d^{2} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{\left (a \,e^{2}+c \,d^{2}\right )^{2} \sqrt {a c}}-\frac {c d e \ln \left (c \,x^{2}+a \right )}{\left (a \,e^{2}+c \,d^{2}\right )^{2}}+\frac {2 c d e \ln \left (e x +d \right )}{\left (a \,e^{2}+c \,d^{2}\right )^{2}}-\frac {e}{\left (a \,e^{2}+c \,d^{2}\right ) \left (e x +d \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 3.03, size = 167, normalized size = 1.36 \begin {gather*} -\frac {c d e \log \left (c x^{2} + a\right )}{c^{2} d^{4} + 2 \, a c d^{2} e^{2} + a^{2} e^{4}} + \frac {2 \, c d e \log \left (e x + d\right )}{c^{2} d^{4} + 2 \, a c d^{2} e^{2} + a^{2} e^{4}} + \frac {{\left (c^{2} d^{2} - a c e^{2}\right )} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{{\left (c^{2} d^{4} + 2 \, a c d^{2} e^{2} + a^{2} e^{4}\right )} \sqrt {a c}} - \frac {e}{c d^{3} + a d e^{2} + {\left (c d^{2} e + a e^{3}\right )} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.91, size = 452, normalized size = 3.67 \begin {gather*} \frac {\ln \left (a^5\,e^8\,\sqrt {-a\,c}-c^3\,d^8\,{\left (-a\,c\right )}^{3/2}-36\,a^3\,d^2\,e^6\,{\left (-a\,c\right )}^{3/2}+a\,c^5\,d^8\,x+a^5\,c\,e^8\,x+70\,a\,d^4\,e^4\,{\left (-a\,c\right )}^{5/2}+36\,c\,d^6\,e^2\,{\left (-a\,c\right )}^{5/2}+36\,a^2\,c^4\,d^6\,e^2\,x+70\,a^3\,c^3\,d^4\,e^4\,x+36\,a^4\,c^2\,d^2\,e^6\,x\right )\,\left (c\,\left (\frac {d^2\,\sqrt {-a\,c}}{2}-a\,d\,e\right )-\frac {a\,e^2\,\sqrt {-a\,c}}{2}\right )}{a^3\,e^4+2\,a^2\,c\,d^2\,e^2+a\,c^2\,d^4}+\frac {\ln \left (c^3\,d^8\,{\left (-a\,c\right )}^{3/2}-a^5\,e^8\,\sqrt {-a\,c}+36\,a^3\,d^2\,e^6\,{\left (-a\,c\right )}^{3/2}+a\,c^5\,d^8\,x+a^5\,c\,e^8\,x-70\,a\,d^4\,e^4\,{\left (-a\,c\right )}^{5/2}-36\,c\,d^6\,e^2\,{\left (-a\,c\right )}^{5/2}+36\,a^2\,c^4\,d^6\,e^2\,x+70\,a^3\,c^3\,d^4\,e^4\,x+36\,a^4\,c^2\,d^2\,e^6\,x\right )\,\left (a\,\left (\frac {e^2\,\sqrt {-a\,c}}{2}-c\,d\,e\right )-\frac {c\,d^2\,\sqrt {-a\,c}}{2}\right )}{a^3\,e^4+2\,a^2\,c\,d^2\,e^2+a\,c^2\,d^4}-\frac {e}{\left (c\,d^2+a\,e^2\right )\,\left (d+e\,x\right )}+\frac {2\,c\,d\,e\,\ln \left (d+e\,x\right )}{{\left (c\,d^2+a\,e^2\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a + c*x^2)*(d + e*x)^2),x)

[Out]

(log(a^5*e^8*(-a*c)^(1/2) - c^3*d^8*(-a*c)^(3/2) - 36*a^3*d^2*e^6*(-a*c)^(3/2) + a*c^5*d^8*x + a^5*c*e^8*x + 7
0*a*d^4*e^4*(-a*c)^(5/2) + 36*c*d^6*e^2*(-a*c)^(5/2) + 36*a^2*c^4*d^6*e^2*x + 70*a^3*c^3*d^4*e^4*x + 36*a^4*c^
2*d^2*e^6*x)*(c*((d^2*(-a*c)^(1/2))/2 - a*d*e) - (a*e^2*(-a*c)^(1/2))/2))/(a^3*e^4 + a*c^2*d^4 + 2*a^2*c*d^2*e
^2) + (log(c^3*d^8*(-a*c)^(3/2) - a^5*e^8*(-a*c)^(1/2) + 36*a^3*d^2*e^6*(-a*c)^(3/2) + a*c^5*d^8*x + a^5*c*e^8
*x - 70*a*d^4*e^4*(-a*c)^(5/2) - 36*c*d^6*e^2*(-a*c)^(5/2) + 36*a^2*c^4*d^6*e^2*x + 70*a^3*c^3*d^4*e^4*x + 36*
a^4*c^2*d^2*e^6*x)*(a*((e^2*(-a*c)^(1/2))/2 - c*d*e) - (c*d^2*(-a*c)^(1/2))/2))/(a^3*e^4 + a*c^2*d^4 + 2*a^2*c
*d^2*e^2) - e/((a*e^2 + c*d^2)*(d + e*x)) + (2*c*d*e*log(d + e*x))/(a*e^2 + c*d^2)^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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