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

Optimal. Leaf size=153 \[ -\frac {a+b \cot ^{-1}(c+d x)}{f (e+f x)}+\frac {b d \log \left (c^2+2 c d x+d^2 x^2+1\right )}{2 \left (\left (c^2+1\right ) f^2-2 c d e f+d^2 e^2\right )}-\frac {b d \log (e+f x)}{\left (c^2+1\right ) f^2-2 c d e f+d^2 e^2}-\frac {b d (d e-c f) \tan ^{-1}(c+d x)}{f \left (\left (c^2+1\right ) f^2-2 c d e f+d^2 e^2\right )} \]

[Out]

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

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Rubi [A]  time = 0.11, antiderivative size = 153, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {5046, 1982, 705, 31, 634, 618, 204, 628} \[ -\frac {a+b \cot ^{-1}(c+d x)}{f (e+f x)}+\frac {b d \log \left (c^2+2 c d x+d^2 x^2+1\right )}{2 \left (\left (c^2+1\right ) f^2-2 c d e f+d^2 e^2\right )}-\frac {b d \log (e+f x)}{\left (c^2+1\right ) f^2-2 c d e f+d^2 e^2}-\frac {b d (d e-c f) \tan ^{-1}(c+d x)}{f \left (\left (c^2+1\right ) f^2-2 c d e f+d^2 e^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 31

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

Rule 204

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

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 705

Int[1/(((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)), x_Symbol] :> Dist[e^2/(c*d^2 - b*d*e + a*e^2
), Int[1/(d + e*x), x], x] + Dist[1/(c*d^2 - b*d*e + a*e^2), Int[(c*d - b*e - c*e*x)/(a + b*x + c*x^2), x], x]
/; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0]

Rule 1982

Int[(u_)^(m_.)*(v_)^(p_.), x_Symbol] :> Int[ExpandToSum[u, x]^m*ExpandToSum[v, x]^p, x] /; FreeQ[{m, p}, x] &&
LinearQ[u, x] && QuadraticQ[v, x] &&  !(LinearMatchQ[u, x] && QuadraticMatchQ[v, x])

Rule 5046

Int[((a_.) + ArcCot[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^(m_), x_Symbol] :> Simp[((e + f*x)^(m
+ 1)*(a + b*ArcCot[c + d*x])^p)/(f*(m + 1)), x] + Dist[(b*d*p)/(f*(m + 1)), Int[((e + f*x)^(m + 1)*(a + b*Arc
Cot[c + d*x])^(p - 1))/(1 + (c + d*x)^2), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[p, 0] && ILtQ[m, -1]

Rubi steps

\begin {align*} \int \frac {a+b \cot ^{-1}(c+d x)}{(e+f x)^2} \, dx &=-\frac {a+b \cot ^{-1}(c+d x)}{f (e+f x)}-\frac {(b d) \int \frac {1}{(e+f x) \left (1+(c+d x)^2\right )} \, dx}{f}\\ &=-\frac {a+b \cot ^{-1}(c+d x)}{f (e+f x)}-\frac {(b d) \int \frac {1}{(e+f x) \left (1+c^2+2 c d x+d^2 x^2\right )} \, dx}{f}\\ &=-\frac {a+b \cot ^{-1}(c+d x)}{f (e+f x)}-\frac {(b d) \int \frac {d^2 e-2 c d f-d^2 f x}{1+c^2+2 c d x+d^2 x^2} \, dx}{f \left (d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2\right )}-\frac {(b d f) \int \frac {1}{e+f x} \, dx}{d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2}\\ &=-\frac {a+b \cot ^{-1}(c+d x)}{f (e+f x)}-\frac {b d \log (e+f x)}{d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2}+\frac {(b d) \int \frac {2 c d+2 d^2 x}{1+c^2+2 c d x+d^2 x^2} \, dx}{2 \left (d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2\right )}-\frac {\left (b d^2 (d e-c f)\right ) \int \frac {1}{1+c^2+2 c d x+d^2 x^2} \, dx}{f \left (d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2\right )}\\ &=-\frac {a+b \cot ^{-1}(c+d x)}{f (e+f x)}-\frac {b d \log (e+f x)}{d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2}+\frac {b d \log \left (1+c^2+2 c d x+d^2 x^2\right )}{2 \left (d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2\right )}+\frac {\left (2 b d^2 (d e-c f)\right ) \operatorname {Subst}\left (\int \frac {1}{-4 d^2-x^2} \, dx,x,2 c d+2 d^2 x\right )}{f \left (d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2\right )}\\ &=-\frac {a+b \cot ^{-1}(c+d x)}{f (e+f x)}-\frac {b d (d e-c f) \tan ^{-1}(c+d x)}{f \left (d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2\right )}-\frac {b d \log (e+f x)}{d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2}+\frac {b d \log \left (1+c^2+2 c d x+d^2 x^2\right )}{2 \left (d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2\right )}\\ \end {align*}

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Mathematica [C]  time = 0.20, size = 118, normalized size = 0.77 \[ \frac {-\frac {a+b \cot ^{-1}(c+d x)}{e+f x}+\frac {b d ((-i c f+i d e+f) \log (-c-d x+i)+(i c f-i d e+f) \log (c+d x+i)-2 f \log (d (e+f x)))}{2 \left (\left (c^2+1\right ) f^2-2 c d e f+d^2 e^2\right )}}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-((a + b*ArcCot[c + d*x])/(e + f*x)) + (b*d*((I*d*e + f - I*c*f)*Log[I - c - d*x] + ((-I)*d*e + f + I*c*f)*Lo
g[I + c + d*x] - 2*f*Log[d*(e + f*x)]))/(2*(d^2*e^2 - 2*c*d*e*f + (1 + c^2)*f^2)))/f

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fricas [A]  time = 1.12, size = 223, normalized size = 1.46 \[ -\frac {2 \, a d^{2} e^{2} - 4 \, a c d e f + 2 \, {\left (a c^{2} + a\right )} f^{2} + 2 \, {\left (b d^{2} e^{2} - 2 \, b c d e f + {\left (b c^{2} + b\right )} f^{2}\right )} \operatorname {arccot}\left (d x + c\right ) + 2 \, {\left (b d^{2} e^{2} - b c d e f + {\left (b d^{2} e f - b c d f^{2}\right )} x\right )} \arctan \left (d x + c\right ) - {\left (b d f^{2} x + b d e f\right )} \log \left (d^{2} x^{2} + 2 \, c d x + c^{2} + 1\right ) + 2 \, {\left (b d f^{2} x + b d e f\right )} \log \left (f x + e\right )}{2 \, {\left (d^{2} e^{3} f - 2 \, c d e^{2} f^{2} + {\left (c^{2} + 1\right )} e f^{3} + {\left (d^{2} e^{2} f^{2} - 2 \, c d e f^{3} + {\left (c^{2} + 1\right )} f^{4}\right )} x\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.87, size = 1278, normalized size = 8.35 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*b*c*f*arctan(1/(d*x + c))*tan(1/2*arctan(1/(d*x + c)))^2 - 2*b*d*arctan(1/(d*x + c))*e*tan(1/2*arctan(
1/(d*x + c)))^2 + 2*b*c*f*log(4*(4*c^2*f^2*tan(1/2*arctan(1/(d*x + c)))^2 - 8*c*d*f*e*tan(1/2*arctan(1/(d*x +
c)))^2 + 4*c*f^2*tan(1/2*arctan(1/(d*x + c)))^3 - 4*d*f*e*tan(1/2*arctan(1/(d*x + c)))^3 + f^2*tan(1/2*arctan(
1/(d*x + c)))^4 + 4*d^2*e^2*tan(1/2*arctan(1/(d*x + c)))^2 - 4*c*f^2*tan(1/2*arctan(1/(d*x + c))) + 4*d*f*e*ta
n(1/2*arctan(1/(d*x + c))) - 2*f^2*tan(1/2*arctan(1/(d*x + c)))^2 + f^2)/(tan(1/2*arctan(1/(d*x + c)))^4 + 2*t
an(1/2*arctan(1/(d*x + c)))^2 + 1))*tan(1/2*arctan(1/(d*x + c))) - 2*b*d*e*log(4*(4*c^2*f^2*tan(1/2*arctan(1/(
d*x + c)))^2 - 8*c*d*f*e*tan(1/2*arctan(1/(d*x + c)))^2 + 4*c*f^2*tan(1/2*arctan(1/(d*x + c)))^3 - 4*d*f*e*tan
(1/2*arctan(1/(d*x + c)))^3 + f^2*tan(1/2*arctan(1/(d*x + c)))^4 + 4*d^2*e^2*tan(1/2*arctan(1/(d*x + c)))^2 -
4*c*f^2*tan(1/2*arctan(1/(d*x + c))) + 4*d*f*e*tan(1/2*arctan(1/(d*x + c))) - 2*f^2*tan(1/2*arctan(1/(d*x + c)
))^2 + f^2)/(tan(1/2*arctan(1/(d*x + c)))^4 + 2*tan(1/2*arctan(1/(d*x + c)))^2 + 1))*tan(1/2*arctan(1/(d*x + c
))) + 2*a*c*f*tan(1/2*arctan(1/(d*x + c)))^2 - 2*a*d*e*tan(1/2*arctan(1/(d*x + c)))^2 + b*f*log(4*(4*c^2*f^2*t
an(1/2*arctan(1/(d*x + c)))^2 - 8*c*d*f*e*tan(1/2*arctan(1/(d*x + c)))^2 + 4*c*f^2*tan(1/2*arctan(1/(d*x + c))
)^3 - 4*d*f*e*tan(1/2*arctan(1/(d*x + c)))^3 + f^2*tan(1/2*arctan(1/(d*x + c)))^4 + 4*d^2*e^2*tan(1/2*arctan(1
/(d*x + c)))^2 - 4*c*f^2*tan(1/2*arctan(1/(d*x + c))) + 4*d*f*e*tan(1/2*arctan(1/(d*x + c))) - 2*f^2*tan(1/2*a
rctan(1/(d*x + c)))^2 + f^2)/(tan(1/2*arctan(1/(d*x + c)))^4 + 2*tan(1/2*arctan(1/(d*x + c)))^2 + 1))*tan(1/2*
arctan(1/(d*x + c)))^2 - 2*b*c*f*arctan(1/(d*x + c)) + 2*b*d*arctan(1/(d*x + c))*e - 4*b*f*arctan(1/(d*x + c))
*tan(1/2*arctan(1/(d*x + c))) - 2*a*c*f + 2*a*d*e - b*f*log(4*(4*c^2*f^2*tan(1/2*arctan(1/(d*x + c)))^2 - 8*c*
d*f*e*tan(1/2*arctan(1/(d*x + c)))^2 + 4*c*f^2*tan(1/2*arctan(1/(d*x + c)))^3 - 4*d*f*e*tan(1/2*arctan(1/(d*x
+ c)))^3 + f^2*tan(1/2*arctan(1/(d*x + c)))^4 + 4*d^2*e^2*tan(1/2*arctan(1/(d*x + c)))^2 - 4*c*f^2*tan(1/2*arc
tan(1/(d*x + c))) + 4*d*f*e*tan(1/2*arctan(1/(d*x + c))) - 2*f^2*tan(1/2*arctan(1/(d*x + c)))^2 + f^2)/(tan(1/
2*arctan(1/(d*x + c)))^4 + 2*tan(1/2*arctan(1/(d*x + c)))^2 + 1)) - 4*a*f*tan(1/2*arctan(1/(d*x + c))))*d/(2*c
^3*f^3*tan(1/2*arctan(1/(d*x + c))) - 6*c^2*d*f^2*e*tan(1/2*arctan(1/(d*x + c))) + c^2*f^3*tan(1/2*arctan(1/(d
*x + c)))^2 - 2*c*d*f^2*e*tan(1/2*arctan(1/(d*x + c)))^2 + 6*c*d^2*f*e^2*tan(1/2*arctan(1/(d*x + c))) + d^2*f*
e^2*tan(1/2*arctan(1/(d*x + c)))^2 - c^2*f^3 + 2*c*d*f^2*e + 2*c*f^3*tan(1/2*arctan(1/(d*x + c))) - 2*d^3*e^3*
tan(1/2*arctan(1/(d*x + c))) - 2*d*f^2*e*tan(1/2*arctan(1/(d*x + c))) + f^3*tan(1/2*arctan(1/(d*x + c)))^2 - d
^2*f*e^2 - f^3)

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maple [A]  time = 0.05, size = 206, normalized size = 1.35 \[ -\frac {d a}{\left (d f x +d e \right ) f}-\frac {d b \,\mathrm {arccot}\left (d x +c \right )}{\left (d f x +d e \right ) f}-\frac {d b \ln \left (f \left (d x +c \right )-c f +d e \right )}{c^{2} f^{2}-2 c d e f +d^{2} e^{2}+f^{2}}+\frac {d b \ln \left (1+\left (d x +c \right )^{2}\right )}{2 c^{2} f^{2}-4 c d e f +2 d^{2} e^{2}+2 f^{2}}+\frac {d b \arctan \left (d x +c \right ) c}{c^{2} f^{2}-2 c d e f +d^{2} e^{2}+f^{2}}-\frac {d^{2} b \arctan \left (d x +c \right ) e}{f \left (c^{2} f^{2}-2 c d e f +d^{2} e^{2}+f^{2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.44, size = 177, normalized size = 1.16 \[ -\frac {1}{2} \, {\left (d {\left (\frac {2 \, {\left (d^{2} e - c d f\right )} \arctan \left (\frac {d^{2} x + c d}{d}\right )}{{\left (d^{2} e^{2} f - 2 \, c d e f^{2} + {\left (c^{2} + 1\right )} f^{3}\right )} d} - \frac {\log \left (d^{2} x^{2} + 2 \, c d x + c^{2} + 1\right )}{d^{2} e^{2} - 2 \, c d e f + {\left (c^{2} + 1\right )} f^{2}} + \frac {2 \, \log \left (f x + e\right )}{d^{2} e^{2} - 2 \, c d e f + {\left (c^{2} + 1\right )} f^{2}}\right )} + \frac {2 \, \operatorname {arccot}\left (d x + c\right )}{f^{2} x + e f}\right )} b - \frac {a}{f^{2} x + e f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 2.08, size = 128, normalized size = 0.84 \[ -\frac {a}{x\,f^2+e\,f}-\frac {b\,\mathrm {acot}\left (c+d\,x\right )}{f\,\left (e+f\,x\right )}-\frac {b\,d\,\ln \left (e+f\,x\right )}{d^2\,e^2-2\,c\,d\,e\,f+\left (c^2+1\right )\,f^2}+\frac {b\,d\,\ln \left (c+d\,x-\mathrm {i}\right )\,1{}\mathrm {i}}{2\,f\,\left (d\,e-c\,f+f\,1{}\mathrm {i}\right )}+\frac {b\,d\,\ln \left (c+d\,x+1{}\mathrm {i}\right )}{2\,f\,\left (f-c\,f\,1{}\mathrm {i}+d\,e\,1{}\mathrm {i}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b*d*log(c + d*x - 1i)*1i)/(2*f*(f*1i - c*f + d*e)) - (b*acot(c + d*x))/(f*(e + f*x)) - (b*d*log(e + f*x))/(f^
2*(c^2 + 1) + d^2*e^2 - 2*c*d*e*f) - a/(e*f + f^2*x) + (b*d*log(c + d*x + 1i))/(2*f*(f - c*f*1i + d*e*1i))

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

[Out]

Timed out

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