3.47 \(\int \frac {A+B x+C x^2}{(d+e x) (a+c x^2)} \, dx\)

Optimal. Leaf size=133 \[ \frac {\log \left (a+c x^2\right ) (a C e-A c e+B c d)}{2 c \left (a e^2+c d^2\right )}+\frac {\log (d+e x) \left (A e^2-B d e+C d^2\right )}{e \left (a e^2+c d^2\right )}+\frac {\tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right ) (a B e-a C d+A c d)}{\sqrt {a} \sqrt {c} \left (a e^2+c d^2\right )} \]

[Out]

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

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Rubi [A]  time = 0.16, antiderivative size = 133, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {1629, 635, 205, 260} \[ \frac {\log \left (a+c x^2\right ) (a C e-A c e+B c d)}{2 c \left (a e^2+c d^2\right )}+\frac {\log (d+e x) \left (A e^2-B d e+C d^2\right )}{e \left (a e^2+c d^2\right )}+\frac {\tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right ) (a B e-a C d+A c d)}{\sqrt {a} \sqrt {c} \left (a e^2+c d^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x + C*x^2)/((d + e*x)*(a + c*x^2)),x]

[Out]

((A*c*d - a*C*d + a*B*e)*ArcTan[(Sqrt[c]*x)/Sqrt[a]])/(Sqrt[a]*Sqrt[c]*(c*d^2 + a*e^2)) + ((C*d^2 - B*d*e + A*
e^2)*Log[d + e*x])/(e*(c*d^2 + a*e^2)) + ((B*c*d - A*c*e + a*C*e)*Log[a + c*x^2])/(2*c*(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 1629

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*
Pq*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rubi steps

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

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Mathematica [A]  time = 0.10, size = 120, normalized size = 0.90 \[ \frac {\sqrt {a} \left (e \log \left (a+c x^2\right ) (a C e-A c e+B c d)+2 c \log (d+e x) \left (A e^2-B d e+C d^2\right )\right )+2 \sqrt {c} e \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right ) (a B e-a C d+A c d)}{2 \sqrt {a} c e \left (a e^2+c d^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 14.21, size = 262, normalized size = 1.97 \[ \left [-\frac {{\left (B a e^{2} - {\left (C a - A c\right )} d e\right )} \sqrt {-a c} \log \left (\frac {c x^{2} - 2 \, \sqrt {-a c} x - a}{c x^{2} + a}\right ) - {\left (B a c d e + {\left (C a^{2} - A a c\right )} e^{2}\right )} \log \left (c x^{2} + a\right ) - 2 \, {\left (C a c d^{2} - B a c d e + A a c e^{2}\right )} \log \left (e x + d\right )}{2 \, {\left (a c^{2} d^{2} e + a^{2} c e^{3}\right )}}, \frac {2 \, {\left (B a e^{2} - {\left (C a - A c\right )} d e\right )} \sqrt {a c} \arctan \left (\frac {\sqrt {a c} x}{a}\right ) + {\left (B a c d e + {\left (C a^{2} - A a c\right )} e^{2}\right )} \log \left (c x^{2} + a\right ) + 2 \, {\left (C a c d^{2} - B a c d e + A a c e^{2}\right )} \log \left (e x + d\right )}{2 \, {\left (a c^{2} d^{2} e + a^{2} c e^{3}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.16, size = 125, normalized size = 0.94 \[ \frac {{\left (B c d + C a e - A c e\right )} \log \left (c x^{2} + a\right )}{2 \, {\left (c^{2} d^{2} + a c e^{2}\right )}} + \frac {{\left (C d^{2} - B d e + A e^{2}\right )} \log \left ({\left | x e + d \right |}\right )}{c d^{2} e + a e^{3}} - \frac {{\left (C a d - A c d - B a e\right )} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{{\left (c d^{2} + a e^{2}\right )} \sqrt {a c}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 247, normalized size = 1.86 \[ \frac {A c d \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{\left (a \,e^{2}+c \,d^{2}\right ) \sqrt {a c}}+\frac {B a e \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{\left (a \,e^{2}+c \,d^{2}\right ) \sqrt {a c}}-\frac {C a d \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{\left (a \,e^{2}+c \,d^{2}\right ) \sqrt {a c}}-\frac {A e \ln \left (c \,x^{2}+a \right )}{2 \left (a \,e^{2}+c \,d^{2}\right )}+\frac {A e \ln \left (e x +d \right )}{a \,e^{2}+c \,d^{2}}+\frac {B d \ln \left (c \,x^{2}+a \right )}{2 a \,e^{2}+2 c \,d^{2}}-\frac {B d \ln \left (e x +d \right )}{a \,e^{2}+c \,d^{2}}+\frac {C a e \ln \left (c \,x^{2}+a \right )}{2 \left (a \,e^{2}+c \,d^{2}\right ) c}+\frac {C \,d^{2} \ln \left (e x +d \right )}{\left (a \,e^{2}+c \,d^{2}\right ) e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.97, size = 123, normalized size = 0.92 \[ \frac {{\left (B c d + {\left (C a - A c\right )} e\right )} \log \left (c x^{2} + a\right )}{2 \, {\left (c^{2} d^{2} + a c e^{2}\right )}} + \frac {{\left (C d^{2} - B d e + A e^{2}\right )} \log \left (e x + d\right )}{c d^{2} e + a e^{3}} + \frac {{\left (B a e - {\left (C a - A c\right )} d\right )} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{{\left (c d^{2} + a e^{2}\right )} \sqrt {a c}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 6.49, size = 840, normalized size = 6.32 \[ \frac {\ln \left (d+e\,x\right )\,\left (C\,d^2-B\,d\,e+A\,e^2\right )}{c\,d^2\,e+a\,e^3}-\frac {\ln \left (x\,\left (c\,e\,B^2-c\,d\,B\,C+a\,e\,C^2-A\,c\,e\,C\right )+C^2\,a\,d+\frac {\left (c^2\,\left (\frac {A\,a\,e}{2}-\frac {B\,a\,d}{2}\right )-c\,\left (\frac {C\,a^2\,e}{2}-\frac {A\,d\,\sqrt {-a\,c^3}}{2}\right )+\frac {B\,a\,e\,\sqrt {-a\,c^3}}{2}-\frac {C\,a\,d\,\sqrt {-a\,c^3}}{2}\right )\,\left (\frac {\left (x\,\left (6\,a\,c^2\,e^3-2\,c^3\,d^2\,e\right )+8\,a\,c^2\,d\,e^2\right )\,\left (c^2\,\left (\frac {A\,a\,e}{2}-\frac {B\,a\,d}{2}\right )-c\,\left (\frac {C\,a^2\,e}{2}-\frac {A\,d\,\sqrt {-a\,c^3}}{2}\right )+\frac {B\,a\,e\,\sqrt {-a\,c^3}}{2}-\frac {C\,a\,d\,\sqrt {-a\,c^3}}{2}\right )}{a^2\,c^2\,e^2+a\,c^3\,d^2}-x\,\left (2\,C\,c^2\,d^2-B\,c^2\,d\,e+3\,A\,c^2\,e^2-5\,C\,a\,c\,e^2\right )+B\,a\,c\,e^2-A\,c^2\,d\,e+5\,C\,a\,c\,d\,e\right )}{a^2\,c^2\,e^2+a\,c^3\,d^2}+A\,B\,c\,e-A\,C\,c\,d\right )\,\left (c^2\,\left (\frac {A\,a\,e}{2}-\frac {B\,a\,d}{2}\right )-c\,\left (\frac {C\,a^2\,e}{2}-\frac {A\,d\,\sqrt {-a\,c^3}}{2}\right )+\frac {B\,a\,e\,\sqrt {-a\,c^3}}{2}-\frac {C\,a\,d\,\sqrt {-a\,c^3}}{2}\right )}{a^2\,c^2\,e^2+a\,c^3\,d^2}-\frac {\ln \left (x\,\left (c\,e\,B^2-c\,d\,B\,C+a\,e\,C^2-A\,c\,e\,C\right )+C^2\,a\,d+\frac {\left (c^2\,\left (\frac {A\,a\,e}{2}-\frac {B\,a\,d}{2}\right )-c\,\left (\frac {C\,a^2\,e}{2}+\frac {A\,d\,\sqrt {-a\,c^3}}{2}\right )-\frac {B\,a\,e\,\sqrt {-a\,c^3}}{2}+\frac {C\,a\,d\,\sqrt {-a\,c^3}}{2}\right )\,\left (\frac {\left (x\,\left (6\,a\,c^2\,e^3-2\,c^3\,d^2\,e\right )+8\,a\,c^2\,d\,e^2\right )\,\left (c^2\,\left (\frac {A\,a\,e}{2}-\frac {B\,a\,d}{2}\right )-c\,\left (\frac {C\,a^2\,e}{2}+\frac {A\,d\,\sqrt {-a\,c^3}}{2}\right )-\frac {B\,a\,e\,\sqrt {-a\,c^3}}{2}+\frac {C\,a\,d\,\sqrt {-a\,c^3}}{2}\right )}{a^2\,c^2\,e^2+a\,c^3\,d^2}-x\,\left (2\,C\,c^2\,d^2-B\,c^2\,d\,e+3\,A\,c^2\,e^2-5\,C\,a\,c\,e^2\right )+B\,a\,c\,e^2-A\,c^2\,d\,e+5\,C\,a\,c\,d\,e\right )}{a^2\,c^2\,e^2+a\,c^3\,d^2}+A\,B\,c\,e-A\,C\,c\,d\right )\,\left (c^2\,\left (\frac {A\,a\,e}{2}-\frac {B\,a\,d}{2}\right )-c\,\left (\frac {C\,a^2\,e}{2}+\frac {A\,d\,\sqrt {-a\,c^3}}{2}\right )-\frac {B\,a\,e\,\sqrt {-a\,c^3}}{2}+\frac {C\,a\,d\,\sqrt {-a\,c^3}}{2}\right )}{a^2\,c^2\,e^2+a\,c^3\,d^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(log(d + e*x)*(A*e^2 + C*d^2 - B*d*e))/(a*e^3 + c*d^2*e) - (log(x*(C^2*a*e + B^2*c*e - A*C*c*e - B*C*c*d) + C^
2*a*d + ((c^2*((A*a*e)/2 - (B*a*d)/2) - c*((C*a^2*e)/2 - (A*d*(-a*c^3)^(1/2))/2) + (B*a*e*(-a*c^3)^(1/2))/2 -
(C*a*d*(-a*c^3)^(1/2))/2)*(((x*(6*a*c^2*e^3 - 2*c^3*d^2*e) + 8*a*c^2*d*e^2)*(c^2*((A*a*e)/2 - (B*a*d)/2) - c*(
(C*a^2*e)/2 - (A*d*(-a*c^3)^(1/2))/2) + (B*a*e*(-a*c^3)^(1/2))/2 - (C*a*d*(-a*c^3)^(1/2))/2))/(a*c^3*d^2 + a^2
*c^2*e^2) - x*(3*A*c^2*e^2 + 2*C*c^2*d^2 - 5*C*a*c*e^2 - B*c^2*d*e) + B*a*c*e^2 - A*c^2*d*e + 5*C*a*c*d*e))/(a
*c^3*d^2 + a^2*c^2*e^2) + A*B*c*e - A*C*c*d)*(c^2*((A*a*e)/2 - (B*a*d)/2) - c*((C*a^2*e)/2 - (A*d*(-a*c^3)^(1/
2))/2) + (B*a*e*(-a*c^3)^(1/2))/2 - (C*a*d*(-a*c^3)^(1/2))/2))/(a*c^3*d^2 + a^2*c^2*e^2) - (log(x*(C^2*a*e + B
^2*c*e - A*C*c*e - B*C*c*d) + C^2*a*d + ((c^2*((A*a*e)/2 - (B*a*d)/2) - c*((C*a^2*e)/2 + (A*d*(-a*c^3)^(1/2))/
2) - (B*a*e*(-a*c^3)^(1/2))/2 + (C*a*d*(-a*c^3)^(1/2))/2)*(((x*(6*a*c^2*e^3 - 2*c^3*d^2*e) + 8*a*c^2*d*e^2)*(c
^2*((A*a*e)/2 - (B*a*d)/2) - c*((C*a^2*e)/2 + (A*d*(-a*c^3)^(1/2))/2) - (B*a*e*(-a*c^3)^(1/2))/2 + (C*a*d*(-a*
c^3)^(1/2))/2))/(a*c^3*d^2 + a^2*c^2*e^2) - x*(3*A*c^2*e^2 + 2*C*c^2*d^2 - 5*C*a*c*e^2 - B*c^2*d*e) + B*a*c*e^
2 - A*c^2*d*e + 5*C*a*c*d*e))/(a*c^3*d^2 + a^2*c^2*e^2) + A*B*c*e - A*C*c*d)*(c^2*((A*a*e)/2 - (B*a*d)/2) - c*
((C*a^2*e)/2 + (A*d*(-a*c^3)^(1/2))/2) - (B*a*e*(-a*c^3)^(1/2))/2 + (C*a*d*(-a*c^3)^(1/2))/2))/(a*c^3*d^2 + a^
2*c^2*e^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((C*x**2+B*x+A)/(e*x+d)/(c*x**2+a),x)

[Out]

Timed out

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