3.6 \(\int \frac {a+b \tan ^{-1}(c x)}{(d+e x)^2} \, dx\)

Optimal. Leaf size=98 \[ -\frac {a+b \tan ^{-1}(c x)}{e (d+e x)}-\frac {b c \log \left (c^2 x^2+1\right )}{2 \left (c^2 d^2+e^2\right )}+\frac {b c \log (d+e x)}{c^2 d^2+e^2}+\frac {b c^2 d \tan ^{-1}(c x)}{e \left (c^2 d^2+e^2\right )} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {4862, 706, 31, 635, 203, 260} \[ -\frac {a+b \tan ^{-1}(c x)}{e (d+e x)}-\frac {b c \log \left (c^2 x^2+1\right )}{2 \left (c^2 d^2+e^2\right )}+\frac {b c \log (d+e x)}{c^2 d^2+e^2}+\frac {b c^2 d \tan ^{-1}(c x)}{e \left (c^2 d^2+e^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*c^2*d*ArcTan[c*x])/(e*(c^2*d^2 + e^2)) - (a + b*ArcTan[c*x])/(e*(d + e*x)) + (b*c*Log[d + e*x])/(c^2*d^2 +
e^2) - (b*c*Log[1 + c^2*x^2])/(2*(c^2*d^2 + 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 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

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]

Rule 4862

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

Rubi steps

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

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Mathematica [A]  time = 0.20, size = 111, normalized size = 1.13 \[ \frac {\frac {b c \left (\left (\sqrt {-c^2} d-e\right ) \log \left (1-\sqrt {-c^2} x\right )-\left (\sqrt {-c^2} d+e\right ) \log \left (\sqrt {-c^2} x+1\right )+2 e \log (d+e x)\right )}{2 \left (c^2 d^2+e^2\right )}-\frac {a+b \tan ^{-1}(c x)}{d+e x}}{e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-((a + b*ArcTan[c*x])/(d + e*x)) + (b*c*((Sqrt[-c^2]*d - e)*Log[1 - Sqrt[-c^2]*x] - (Sqrt[-c^2]*d + e)*Log[1
+ Sqrt[-c^2]*x] + 2*e*Log[d + e*x]))/(2*(c^2*d^2 + e^2)))/e

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fricas [A]  time = 0.53, size = 116, normalized size = 1.18 \[ -\frac {2 \, a c^{2} d^{2} + 2 \, a e^{2} - 2 \, {\left (b c^{2} d e x - b e^{2}\right )} \arctan \left (c x\right ) + {\left (b c e^{2} x + b c d e\right )} \log \left (c^{2} x^{2} + 1\right ) - 2 \, {\left (b c e^{2} x + b c d e\right )} \log \left (e x + d\right )}{2 \, {\left (c^{2} d^{3} e + d e^{3} + {\left (c^{2} d^{2} e^{2} + e^{4}\right )} x\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {sage}_{0} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x

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maple [A]  time = 0.04, size = 118, normalized size = 1.20 \[ -\frac {c a}{\left (c e x +d c \right ) e}-\frac {c b \arctan \left (c x \right )}{\left (c e x +d c \right ) e}+\frac {c b \ln \left (c e x +d c \right )}{c^{2} d^{2}+e^{2}}-\frac {b c \ln \left (c^{2} x^{2}+1\right )}{2 \left (c^{2} d^{2}+e^{2}\right )}+\frac {b \,c^{2} d \arctan \left (c x \right )}{e \left (c^{2} d^{2}+e^{2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.42, size = 107, normalized size = 1.09 \[ \frac {1}{2} \, {\left ({\left (\frac {2 \, c d \arctan \left (c x\right )}{c^{2} d^{2} e + e^{3}} - \frac {\log \left (c^{2} x^{2} + 1\right )}{c^{2} d^{2} + e^{2}} + \frac {2 \, \log \left (e x + d\right )}{c^{2} d^{2} + e^{2}}\right )} c - \frac {2 \, \arctan \left (c x\right )}{e^{2} x + d e}\right )} b - \frac {a}{e^{2} x + d e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 3.46, size = 112, normalized size = 1.14 \[ \frac {d^2\,\left (b\,c\,\ln \left (d+e\,x\right )-\frac {b\,c\,\ln \left (c^2\,x^2+1\right )}{2}+a\,c^2\,x+b\,c^2\,x\,\mathrm {atan}\left (c\,x\right )\right )-d\,e\,\left (b\,\mathrm {atan}\left (c\,x\right )-b\,c\,x\,\ln \left (d+e\,x\right )+\frac {b\,c\,x\,\ln \left (c^2\,x^2+1\right )}{2}\right )+a\,e^2\,x}{d\,\left (c^2\,d^2+e^2\right )\,\left (d+e\,x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atan(c*x))/(d + e*x)^2,x)

[Out]

(d^2*(b*c*log(d + e*x) - (b*c*log(c^2*x^2 + 1))/2 + a*c^2*x + b*c^2*x*atan(c*x)) - d*e*(b*atan(c*x) - b*c*x*lo
g(d + e*x) + (b*c*x*log(c^2*x^2 + 1))/2) + a*e^2*x)/(d*(e^2 + c^2*d^2)*(d + e*x))

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sympy [A]  time = 4.26, size = 695, normalized size = 7.09 \[ \begin {cases} - \frac {a}{d e + e^{2} x} & \text {for}\: c = 0 \\\frac {a x + b x \operatorname {atan}{\left (c x \right )} - \frac {b \log {\left (x^{2} + \frac {1}{c^{2}} \right )}}{2 c}}{d^{2}} & \text {for}\: e = 0 \\\frac {2 a d}{- 2 d^{2} e - 2 d e^{2} x} - \frac {i b d \operatorname {atanh}{\left (\frac {e x}{d} \right )}}{- 2 d^{2} e - 2 d e^{2} x} - \frac {i b d}{- 2 d^{2} e - 2 d e^{2} x} + \frac {i b e x \operatorname {atanh}{\left (\frac {e x}{d} \right )}}{- 2 d^{2} e - 2 d e^{2} x} & \text {for}\: c = - \frac {i e}{d} \\\frac {2 a d}{- 2 d^{2} e - 2 d e^{2} x} + \frac {i b d \operatorname {atanh}{\left (\frac {e x}{d} \right )}}{- 2 d^{2} e - 2 d e^{2} x} + \frac {i b d}{- 2 d^{2} e - 2 d e^{2} x} - \frac {i b e x \operatorname {atanh}{\left (\frac {e x}{d} \right )}}{- 2 d^{2} e - 2 d e^{2} x} & \text {for}\: c = \frac {i e}{d} \\\tilde {\infty } \left (a x + b x \operatorname {atan}{\left (c x \right )} - \frac {b \log {\left (x^{2} + \frac {1}{c^{2}} \right )}}{2 c}\right ) & \text {for}\: d = - e x \\- \frac {2 a c^{2} d^{2}}{2 c^{2} d^{3} e + 2 c^{2} d^{2} e^{2} x + 2 d e^{3} + 2 e^{4} x} - \frac {2 a e^{2}}{2 c^{2} d^{3} e + 2 c^{2} d^{2} e^{2} x + 2 d e^{3} + 2 e^{4} x} + \frac {2 b c^{2} d e x \operatorname {atan}{\left (c x \right )}}{2 c^{2} d^{3} e + 2 c^{2} d^{2} e^{2} x + 2 d e^{3} + 2 e^{4} x} - \frac {b c d e \log {\left (x^{2} + \frac {1}{c^{2}} \right )}}{2 c^{2} d^{3} e + 2 c^{2} d^{2} e^{2} x + 2 d e^{3} + 2 e^{4} x} + \frac {2 b c d e \log {\left (\frac {d}{e} + x \right )}}{2 c^{2} d^{3} e + 2 c^{2} d^{2} e^{2} x + 2 d e^{3} + 2 e^{4} x} - \frac {b c e^{2} x \log {\left (x^{2} + \frac {1}{c^{2}} \right )}}{2 c^{2} d^{3} e + 2 c^{2} d^{2} e^{2} x + 2 d e^{3} + 2 e^{4} x} + \frac {2 b c e^{2} x \log {\left (\frac {d}{e} + x \right )}}{2 c^{2} d^{3} e + 2 c^{2} d^{2} e^{2} x + 2 d e^{3} + 2 e^{4} x} - \frac {2 b e^{2} \operatorname {atan}{\left (c x \right )}}{2 c^{2} d^{3} e + 2 c^{2} d^{2} e^{2} x + 2 d e^{3} + 2 e^{4} x} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-a/(d*e + e**2*x), Eq(c, 0)), ((a*x + b*x*atan(c*x) - b*log(x**2 + c**(-2))/(2*c))/d**2, Eq(e, 0)),
(2*a*d/(-2*d**2*e - 2*d*e**2*x) - I*b*d*atanh(e*x/d)/(-2*d**2*e - 2*d*e**2*x) - I*b*d/(-2*d**2*e - 2*d*e**2*x
) + I*b*e*x*atanh(e*x/d)/(-2*d**2*e - 2*d*e**2*x), Eq(c, -I*e/d)), (2*a*d/(-2*d**2*e - 2*d*e**2*x) + I*b*d*ata
nh(e*x/d)/(-2*d**2*e - 2*d*e**2*x) + I*b*d/(-2*d**2*e - 2*d*e**2*x) - I*b*e*x*atanh(e*x/d)/(-2*d**2*e - 2*d*e*
*2*x), Eq(c, I*e/d)), (zoo*(a*x + b*x*atan(c*x) - b*log(x**2 + c**(-2))/(2*c)), Eq(d, -e*x)), (-2*a*c**2*d**2/
(2*c**2*d**3*e + 2*c**2*d**2*e**2*x + 2*d*e**3 + 2*e**4*x) - 2*a*e**2/(2*c**2*d**3*e + 2*c**2*d**2*e**2*x + 2*
d*e**3 + 2*e**4*x) + 2*b*c**2*d*e*x*atan(c*x)/(2*c**2*d**3*e + 2*c**2*d**2*e**2*x + 2*d*e**3 + 2*e**4*x) - b*c
*d*e*log(x**2 + c**(-2))/(2*c**2*d**3*e + 2*c**2*d**2*e**2*x + 2*d*e**3 + 2*e**4*x) + 2*b*c*d*e*log(d/e + x)/(
2*c**2*d**3*e + 2*c**2*d**2*e**2*x + 2*d*e**3 + 2*e**4*x) - b*c*e**2*x*log(x**2 + c**(-2))/(2*c**2*d**3*e + 2*
c**2*d**2*e**2*x + 2*d*e**3 + 2*e**4*x) + 2*b*c*e**2*x*log(d/e + x)/(2*c**2*d**3*e + 2*c**2*d**2*e**2*x + 2*d*
e**3 + 2*e**4*x) - 2*b*e**2*atan(c*x)/(2*c**2*d**3*e + 2*c**2*d**2*e**2*x + 2*d*e**3 + 2*e**4*x), True))

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