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

Optimal. Leaf size=109 \[ \frac {\log \left (a+c x^2\right ) (B d-A e)}{2 \left (a e^2+c d^2\right )}-\frac {(B d-A e) \log (d+e x)}{a e^2+c d^2}+\frac {\tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right ) (a B e+A c d)}{\sqrt {a} \sqrt {c} \left (a e^2+c d^2\right )} \]

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Rubi [A]  time = 0.11, antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {801, 635, 205, 260} \begin {gather*} \frac {\log \left (a+c x^2\right ) (B d-A e)}{2 \left (a e^2+c d^2\right )}-\frac {(B d-A e) \log (d+e x)}{a e^2+c d^2}+\frac {\tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right ) (a B e+A c d)}{\sqrt {a} \sqrt {c} \left (a e^2+c d^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((A*c*d + a*B*e)*ArcTan[(Sqrt[c]*x)/Sqrt[a]])/(Sqrt[a]*Sqrt[c]*(c*d^2 + a*e^2)) - ((B*d - A*e)*Log[d + e*x])/(
c*d^2 + a*e^2) + ((B*d - A*e)*Log[a + c*x^2])/(2*(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 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 {A+B x}{(d+e x) \left (a+c x^2\right )} \, dx &=\int \left (\frac {e (-B d+A e)}{\left (c d^2+a e^2\right ) (d+e x)}+\frac {A c d+a B e+c (B d-A e) x}{\left (c d^2+a e^2\right ) \left (a+c x^2\right )}\right ) \, dx\\ &=-\frac {(B d-A e) \log (d+e x)}{c d^2+a e^2}+\frac {\int \frac {A c d+a B e+c (B d-A e) x}{a+c x^2} \, dx}{c d^2+a e^2}\\ &=-\frac {(B d-A e) \log (d+e x)}{c d^2+a e^2}+\frac {(c (B d-A e)) \int \frac {x}{a+c x^2} \, dx}{c d^2+a e^2}+\frac {(A c d+a B e) \int \frac {1}{a+c x^2} \, dx}{c d^2+a e^2}\\ &=\frac {(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 {(B d-A e) \log (d+e x)}{c d^2+a e^2}+\frac {(B d-A 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 = 0.06, size = 91, normalized size = 0.83 \begin {gather*} \frac {2 \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right ) (a B e+A c d)-\sqrt {a} \sqrt {c} (B d-A e) \left (2 \log (d+e x)-\log \left (a+c x^2\right )\right )}{2 \sqrt {a} \sqrt {c} \left (a e^2+c d^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 1.35, size = 200, normalized size = 1.83 \begin {gather*} \left [-\frac {{\left (A c d + B a 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 - A a c e\right )} \log \left (c x^{2} + a\right ) + 2 \, {\left (B a c d - A a c e\right )} \log \left (e x + d\right )}{2 \, {\left (a c^{2} d^{2} + a^{2} c e^{2}\right )}}, \frac {2 \, {\left (A c d + B a e\right )} \sqrt {a c} \arctan \left (\frac {\sqrt {a c} x}{a}\right ) + {\left (B a c d - A a c e\right )} \log \left (c x^{2} + a\right ) - 2 \, {\left (B a c d - A a c e\right )} \log \left (e x + d\right )}{2 \, {\left (a c^{2} d^{2} + a^{2} c e^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.16, size = 104, normalized size = 0.95 \begin {gather*} \frac {{\left (B d - A e\right )} \log \left (c x^{2} + a\right )}{2 \, {\left (c d^{2} + a e^{2}\right )}} - \frac {{\left (B d e - A e^{2}\right )} \log \left ({\left | x e + d \right |}\right )}{c d^{2} e + a e^{3}} + \frac {{\left (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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(B*d - A*e)*log(c*x^2 + a)/(c*d^2 + a*e^2) - (B*d*e - A*e^2)*log(abs(x*e + d))/(c*d^2*e + a*e^3) + (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 = 0.06, size = 159, normalized size = 1.46 \begin {gather*} \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 {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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.31, size = 98, normalized size = 0.90 \begin {gather*} \frac {{\left (B d - A e\right )} \log \left (c x^{2} + a\right )}{2 \, {\left (c d^{2} + a e^{2}\right )}} - \frac {{\left (B d - A e\right )} \log \left (e x + d\right )}{c d^{2} + a e^{2}} + \frac {{\left (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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(B*d - A*e)*log(c*x^2 + a)/(c*d^2 + a*e^2) - (B*d - A*e)*log(e*x + d)/(c*d^2 + a*e^2) + (A*c*d + B*a*e)*ar
ctan(c*x/sqrt(a*c))/((c*d^2 + a*e^2)*sqrt(a*c))

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mupad [B]  time = 3.13, size = 535, normalized size = 4.91 \begin {gather*} \frac {\ln \left (d+e\,x\right )\,\left (A\,e-B\,d\right )}{c\,d^2+a\,e^2}-\frac {\ln \left (B^2\,c\,e\,x-\frac {\left (c\,\left (a\,\left (\frac {A\,e}{2}-\frac {B\,d}{2}\right )+\frac {A\,d\,\sqrt {-a\,c}}{2}\right )+\frac {B\,a\,e\,\sqrt {-a\,c}}{2}\right )\,\left (x\,\left (3\,A\,c^2\,e^2-B\,c^2\,d\,e\right )-\frac {\left (c\,\left (a\,\left (\frac {A\,e}{2}-\frac {B\,d}{2}\right )+\frac {A\,d\,\sqrt {-a\,c}}{2}\right )+\frac {B\,a\,e\,\sqrt {-a\,c}}{2}\right )\,\left (x\,\left (6\,a\,c^2\,e^3-2\,c^3\,d^2\,e\right )+8\,a\,c^2\,d\,e^2\right )}{a^2\,c\,e^2+a\,c^2\,d^2}-B\,a\,c\,e^2+A\,c^2\,d\,e\right )}{a^2\,c\,e^2+a\,c^2\,d^2}+A\,B\,c\,e\right )\,\left (c\,\left (a\,\left (\frac {A\,e}{2}-\frac {B\,d}{2}\right )+\frac {A\,d\,\sqrt {-a\,c}}{2}\right )+\frac {B\,a\,e\,\sqrt {-a\,c}}{2}\right )}{a^2\,c\,e^2+a\,c^2\,d^2}-\frac {\ln \left (B^2\,c\,e\,x-\frac {\left (c\,\left (a\,\left (\frac {A\,e}{2}-\frac {B\,d}{2}\right )-\frac {A\,d\,\sqrt {-a\,c}}{2}\right )-\frac {B\,a\,e\,\sqrt {-a\,c}}{2}\right )\,\left (x\,\left (3\,A\,c^2\,e^2-B\,c^2\,d\,e\right )-\frac {\left (c\,\left (a\,\left (\frac {A\,e}{2}-\frac {B\,d}{2}\right )-\frac {A\,d\,\sqrt {-a\,c}}{2}\right )-\frac {B\,a\,e\,\sqrt {-a\,c}}{2}\right )\,\left (x\,\left (6\,a\,c^2\,e^3-2\,c^3\,d^2\,e\right )+8\,a\,c^2\,d\,e^2\right )}{a^2\,c\,e^2+a\,c^2\,d^2}-B\,a\,c\,e^2+A\,c^2\,d\,e\right )}{a^2\,c\,e^2+a\,c^2\,d^2}+A\,B\,c\,e\right )\,\left (c\,\left (a\,\left (\frac {A\,e}{2}-\frac {B\,d}{2}\right )-\frac {A\,d\,\sqrt {-a\,c}}{2}\right )-\frac {B\,a\,e\,\sqrt {-a\,c}}{2}\right )}{a^2\,c\,e^2+a\,c^2\,d^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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