3.135 \(\int \frac{a+b \cot ^{-1}(c+d x)}{(e+f x)^3} \, dx\)

Optimal. Leaf size=228 \[ -\frac{a+b \cot ^{-1}(c+d x)}{2 f (e+f x)^2}+\frac{b d^2 (d e-c f) \log \left (c^2+2 c d x+d^2 x^2+1\right )}{2 \left (\left (c^2+1\right ) f^2-2 c d e f+d^2 e^2\right )^2}+\frac{b d}{2 (e+f x) \left (\left (c^2+1\right ) f^2-2 c d e f+d^2 e^2\right )}-\frac{b d^2 (d e-c f) \log (e+f x)}{\left (\left (c^2+1\right ) f^2-2 c d e f+d^2 e^2\right )^2}-\frac{b d^2 (-c f+d e+f) (d e-(c+1) f) \tan ^{-1}(c+d x)}{2 f \left (\left (c^2+1\right ) f^2-2 c d e f+d^2 e^2\right )^2} \]

[Out]

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

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Rubi [A]  time = 0.278451, antiderivative size = 228, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.444, Rules used = {5046, 1982, 709, 800, 634, 618, 204, 628} \[ -\frac{a+b \cot ^{-1}(c+d x)}{2 f (e+f x)^2}+\frac{b d^2 (d e-c f) \log \left (c^2+2 c d x+d^2 x^2+1\right )}{2 \left (\left (c^2+1\right ) f^2-2 c d e f+d^2 e^2\right )^2}+\frac{b d}{2 (e+f x) \left (\left (c^2+1\right ) f^2-2 c d e f+d^2 e^2\right )}-\frac{b d^2 (d e-c f) \log (e+f x)}{\left (\left (c^2+1\right ) f^2-2 c d e f+d^2 e^2\right )^2}-\frac{b d^2 (-c f+d e+f) (d e-(c+1) f) \tan ^{-1}(c+d x)}{2 f \left (\left (c^2+1\right ) f^2-2 c d e f+d^2 e^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcCot[c + d*x])/(e + f*x)^3,x]

[Out]

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

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]

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 709

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

Rule 800

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Int[Exp
andIntegrand[((d + e*x)^m*(f + g*x))/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 -
4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[m]

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 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 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 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]

Rubi steps

\begin{align*} \int \frac{a+b \cot ^{-1}(c+d x)}{(e+f x)^3} \, dx &=-\frac{a+b \cot ^{-1}(c+d x)}{2 f (e+f x)^2}-\frac{(b d) \int \frac{1}{(e+f x)^2 \left (1+(c+d x)^2\right )} \, dx}{2 f}\\ &=-\frac{a+b \cot ^{-1}(c+d x)}{2 f (e+f x)^2}-\frac{(b d) \int \frac{1}{(e+f x)^2 \left (1+c^2+2 c d x+d^2 x^2\right )} \, dx}{2 f}\\ &=\frac{b d}{2 \left (d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2\right ) (e+f x)}-\frac{a+b \cot ^{-1}(c+d x)}{2 f (e+f x)^2}-\frac{(b d) \int \frac{d (d e-2 c f)-d^2 f x}{(e+f x) \left (1+c^2+2 c d x+d^2 x^2\right )} \, dx}{2 f \left (d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2\right )}\\ &=\frac{b d}{2 \left (d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2\right ) (e+f x)}-\frac{a+b \cot ^{-1}(c+d x)}{2 f (e+f x)^2}-\frac{(b d) \int \left (\frac{2 d f^2 (d e-c f)}{\left (d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2\right ) (e+f x)}+\frac{d^2 \left (d^2 e^2-4 c d e f-\left (1-3 c^2\right ) f^2-2 d f (d e-c f) x\right )}{\left (d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2\right ) \left (1+c^2+2 c d x+d^2 x^2\right )}\right ) \, dx}{2 f \left (d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2\right )}\\ &=\frac{b d}{2 \left (d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2\right ) (e+f x)}-\frac{a+b \cot ^{-1}(c+d x)}{2 f (e+f x)^2}-\frac{b d^2 (d e-c f) \log (e+f x)}{\left (d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2\right )^2}-\frac{\left (b d^3\right ) \int \frac{d^2 e^2-4 c d e f-\left (1-3 c^2\right ) f^2-2 d f (d e-c f) x}{1+c^2+2 c d x+d^2 x^2} \, dx}{2 f \left (d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2\right )^2}\\ &=\frac{b d}{2 \left (d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2\right ) (e+f x)}-\frac{a+b \cot ^{-1}(c+d x)}{2 f (e+f x)^2}-\frac{b d^2 (d e-c f) \log (e+f x)}{\left (d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2\right )^2}+\frac{\left (b d^2 (d e-c f)\right ) \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 )^2}-\frac{\left (b d \left (4 c d^2 f (d e-c f)+2 d^2 \left (d^2 e^2-4 c d e f-\left (1-3 c^2\right ) f^2\right )\right )\right ) \int \frac{1}{1+c^2+2 c d x+d^2 x^2} \, dx}{4 f \left (d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2\right )^2}\\ &=\frac{b d}{2 \left (d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2\right ) (e+f x)}-\frac{a+b \cot ^{-1}(c+d x)}{2 f (e+f x)^2}-\frac{b d^2 (d e-c f) \log (e+f x)}{\left (d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2\right )^2}+\frac{b d^2 (d e-c f) \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 )^2}+\frac{\left (b d \left (4 c d^2 f (d e-c f)+2 d^2 \left (d^2 e^2-4 c d e f-\left (1-3 c^2\right ) f^2\right )\right )\right ) \operatorname{Subst}\left (\int \frac{1}{-4 d^2-x^2} \, dx,x,2 c d+2 d^2 x\right )}{2 f \left (d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2\right )^2}\\ &=\frac{b d}{2 \left (d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2\right ) (e+f x)}-\frac{a+b \cot ^{-1}(c+d x)}{2 f (e+f x)^2}-\frac{b d^2 (d e-f-c f) (d e+f-c f) \tan ^{-1}(c+d x)}{2 f \left (d^2 e^2-2 c d e f+f^2+c^2 f^2\right )^2}-\frac{b d^2 (d e-c f) \log (e+f x)}{\left (d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2\right )^2}+\frac{b d^2 (d e-c f) \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 )^2}\\ \end{align*}

Mathematica [C]  time = 0.502029, size = 180, normalized size = 0.79 \[ \frac{-\frac{a+b \cot ^{-1}(c+d x)}{(e+f x)^2}+\frac{b d f}{(e+f x) \left (\left (c^2+1\right ) f^2-2 c d e f+d^2 e^2\right )}-\frac{2 b d^2 f (d e-c f) \log (d (e+f x))}{\left (\left (c^2+1\right ) f^2-2 c d e f+d^2 e^2\right )^2}+\frac{i b d^2 \log (-c-d x+i)}{2 (d e-(c-i) f)^2}-\frac{i b d^2 \log (c+d x+i)}{2 (d e-(c+i) f)^2}}{2 f} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcCot[c + d*x])/(e + f*x)^3,x]

[Out]

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

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Maple [A]  time = 0.059, size = 437, normalized size = 1.9 \begin{align*} -{\frac{{d}^{2}a}{2\, \left ( dfx+de \right ) ^{2}f}}-{\frac{b{d}^{2}{\rm arccot} \left (dx+c\right )}{2\, \left ( dfx+de \right ) ^{2}f}}-{\frac{b{d}^{2}f\arctan \left ( dx+c \right ){c}^{2}}{2\, \left ({c}^{2}{f}^{2}-2\,cdef+{d}^{2}{e}^{2}+{f}^{2} \right ) ^{2}}}+{\frac{b{d}^{3}\arctan \left ( dx+c \right ) ce}{ \left ({c}^{2}{f}^{2}-2\,cdef+{d}^{2}{e}^{2}+{f}^{2} \right ) ^{2}}}-{\frac{{d}^{4}b\arctan \left ( dx+c \right ){e}^{2}}{2\, \left ({c}^{2}{f}^{2}-2\,cdef+{d}^{2}{e}^{2}+{f}^{2} \right ) ^{2}f}}-{\frac{b{d}^{2}f\ln \left ( 1+ \left ( dx+c \right ) ^{2} \right ) c}{2\, \left ({c}^{2}{f}^{2}-2\,cdef+{d}^{2}{e}^{2}+{f}^{2} \right ) ^{2}}}+{\frac{b{d}^{3}\ln \left ( 1+ \left ( dx+c \right ) ^{2} \right ) e}{2\, \left ({c}^{2}{f}^{2}-2\,cdef+{d}^{2}{e}^{2}+{f}^{2} \right ) ^{2}}}+{\frac{b{d}^{2}f\arctan \left ( dx+c \right ) }{2\, \left ({c}^{2}{f}^{2}-2\,cdef+{d}^{2}{e}^{2}+{f}^{2} \right ) ^{2}}}+{\frac{b{d}^{2}}{ \left ( 2\,{c}^{2}{f}^{2}-4\,cdef+2\,{d}^{2}{e}^{2}+2\,{f}^{2} \right ) \left ( dfx+de \right ) }}+{\frac{b{d}^{2}f\ln \left ( f \left ( dx+c \right ) -cf+de \right ) c}{ \left ({c}^{2}{f}^{2}-2\,cdef+{d}^{2}{e}^{2}+{f}^{2} \right ) ^{2}}}-{\frac{b{d}^{3}\ln \left ( f \left ( dx+c \right ) -cf+de \right ) e}{ \left ({c}^{2}{f}^{2}-2\,cdef+{d}^{2}{e}^{2}+{f}^{2} \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arccot(d*x+c))/(f*x+e)^3,x)

[Out]

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

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Maxima [A]  time = 1.52127, size = 554, normalized size = 2.43 \begin{align*} \frac{1}{2} \,{\left (d{\left (\frac{{\left (d^{2} e - c d f\right )} \log \left (d^{2} x^{2} + 2 \, c d x + c^{2} + 1\right )}{d^{4} e^{4} - 4 \, c d^{3} e^{3} f + 2 \,{\left (3 \, c^{2} + 1\right )} d^{2} e^{2} f^{2} - 4 \,{\left (c^{3} + c\right )} d e f^{3} +{\left (c^{4} + 2 \, c^{2} + 1\right )} f^{4}} - \frac{2 \,{\left (d^{2} e - c d f\right )} \log \left (f x + e\right )}{d^{4} e^{4} - 4 \, c d^{3} e^{3} f + 2 \,{\left (3 \, c^{2} + 1\right )} d^{2} e^{2} f^{2} - 4 \,{\left (c^{3} + c\right )} d e f^{3} +{\left (c^{4} + 2 \, c^{2} + 1\right )} f^{4}} - \frac{{\left (d^{4} e^{2} - 2 \, c d^{3} e f +{\left (c^{2} - 1\right )} d^{2} f^{2}\right )} \arctan \left (\frac{d^{2} x + c d}{d}\right )}{{\left (d^{4} e^{4} f - 4 \, c d^{3} e^{3} f^{2} + 2 \,{\left (3 \, c^{2} + 1\right )} d^{2} e^{2} f^{3} - 4 \,{\left (c^{3} + c\right )} d e f^{4} +{\left (c^{4} + 2 \, c^{2} + 1\right )} f^{5}\right )} d} + \frac{1}{d^{2} e^{3} - 2 \, c d e^{2} f +{\left (c^{2} + 1\right )} e f^{2} +{\left (d^{2} e^{2} f - 2 \, c d e f^{2} +{\left (c^{2} + 1\right )} f^{3}\right )} x}\right )} - \frac{\operatorname{arccot}\left (d x + c\right )}{f^{3} x^{2} + 2 \, e f^{2} x + e^{2} f}\right )} b - \frac{a}{2 \,{\left (f^{3} x^{2} + 2 \, e f^{2} x + e^{2} f\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arccot(d*x+c))/(f*x+e)^3,x, algorithm="maxima")

[Out]

1/2*(d*((d^2*e - c*d*f)*log(d^2*x^2 + 2*c*d*x + c^2 + 1)/(d^4*e^4 - 4*c*d^3*e^3*f + 2*(3*c^2 + 1)*d^2*e^2*f^2
- 4*(c^3 + c)*d*e*f^3 + (c^4 + 2*c^2 + 1)*f^4) - 2*(d^2*e - c*d*f)*log(f*x + e)/(d^4*e^4 - 4*c*d^3*e^3*f + 2*(
3*c^2 + 1)*d^2*e^2*f^2 - 4*(c^3 + c)*d*e*f^3 + (c^4 + 2*c^2 + 1)*f^4) - (d^4*e^2 - 2*c*d^3*e*f + (c^2 - 1)*d^2
*f^2)*arctan((d^2*x + c*d)/d)/((d^4*e^4*f - 4*c*d^3*e^3*f^2 + 2*(3*c^2 + 1)*d^2*e^2*f^3 - 4*(c^3 + c)*d*e*f^4
+ (c^4 + 2*c^2 + 1)*f^5)*d) + 1/(d^2*e^3 - 2*c*d*e^2*f + (c^2 + 1)*e*f^2 + (d^2*e^2*f - 2*c*d*e*f^2 + (c^2 + 1
)*f^3)*x)) - arccot(d*x + c)/(f^3*x^2 + 2*e*f^2*x + e^2*f))*b - 1/2*a/(f^3*x^2 + 2*e*f^2*x + e^2*f)

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Fricas [B]  time = 20.4374, size = 1531, normalized size = 6.71 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arccot(d*x+c))/(f*x+e)^3,x, algorithm="fricas")

[Out]

-1/2*(a*d^4*e^4 - (4*a*c + b)*d^3*e^3*f + 2*(3*a*c^2 + b*c + a)*d^2*e^2*f^2 - (4*a*c^3 + b*c^2 + 4*a*c + b)*d*
e*f^3 + (a*c^4 + 2*a*c^2 + a)*f^4 - (b*d^3*e^2*f^2 - 2*b*c*d^2*e*f^3 + (b*c^2 + b)*d*f^4)*x + (b*d^4*e^4 - 4*b
*c*d^3*e^3*f + 2*(3*b*c^2 + b)*d^2*e^2*f^2 - 4*(b*c^3 + b*c)*d*e*f^3 + (b*c^4 + 2*b*c^2 + b)*f^4)*arccot(d*x +
c) + (b*d^4*e^4 - 2*b*c*d^3*e^3*f + (b*c^2 - b)*d^2*e^2*f^2 + (b*d^4*e^2*f^2 - 2*b*c*d^3*e*f^3 + (b*c^2 - b)*
d^2*f^4)*x^2 + 2*(b*d^4*e^3*f - 2*b*c*d^3*e^2*f^2 + (b*c^2 - b)*d^2*e*f^3)*x)*arctan(d*x + c) - (b*d^3*e^3*f -
b*c*d^2*e^2*f^2 + (b*d^3*e*f^3 - b*c*d^2*f^4)*x^2 + 2*(b*d^3*e^2*f^2 - b*c*d^2*e*f^3)*x)*log(d^2*x^2 + 2*c*d*
x + c^2 + 1) + 2*(b*d^3*e^3*f - b*c*d^2*e^2*f^2 + (b*d^3*e*f^3 - b*c*d^2*f^4)*x^2 + 2*(b*d^3*e^2*f^2 - b*c*d^2
*e*f^3)*x)*log(f*x + e))/(d^4*e^6*f - 4*c*d^3*e^5*f^2 + 2*(3*c^2 + 1)*d^2*e^4*f^3 - 4*(c^3 + c)*d*e^3*f^4 + (c
^4 + 2*c^2 + 1)*e^2*f^5 + (d^4*e^4*f^3 - 4*c*d^3*e^3*f^4 + 2*(3*c^2 + 1)*d^2*e^2*f^5 - 4*(c^3 + c)*d*e*f^6 + (
c^4 + 2*c^2 + 1)*f^7)*x^2 + 2*(d^4*e^5*f^2 - 4*c*d^3*e^4*f^3 + 2*(3*c^2 + 1)*d^2*e^3*f^4 - 4*(c^3 + c)*d*e^2*f
^5 + (c^4 + 2*c^2 + 1)*e*f^6)*x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*acot(d*x+c))/(f*x+e)**3,x)

[Out]

Timed out

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Giac [B]  time = 6.61449, size = 1550, normalized size = 6.8 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arccot(d*x+c))/(f*x+e)^3,x, algorithm="giac")

[Out]

-1/2*(b*c^2*d^2*f^4*x^2*arctan(d*x + c) - 2*b*c*d^3*f^3*x^2*arctan(d*x + c)*e + b*d^4*f^2*x^2*arctan(d*x + c)*
e^2 + 2*b*c^2*d^2*f^3*x*arctan(d*x + c)*e + b*c*d^2*f^4*x^2*log(d^2*x^2 + 2*c*d*x + c^2 + 1) - b*d^3*f^3*x^2*e
*log(d^2*x^2 + 2*c*d*x + c^2 + 1) - 2*b*c*d^2*f^4*x^2*log(abs(f*x + e)) + 2*b*d^3*f^3*x^2*e*log(abs(f*x + e))
- b*d^2*f^4*x^2*arctan(d*x + c) + b*c^4*f^4*arctan(1/(d*x + c)) - 4*b*c*d^3*f^2*x*arctan(d*x + c)*e^2 - 4*b*c^
3*d*f^3*arctan(1/(d*x + c))*e + 2*b*c*d^2*f^3*x*e*log(d^2*x^2 + 2*c*d*x + c^2 + 1) - 4*b*c*d^2*f^3*x*e*log(abs
(f*x + e)) + a*c^4*f^4 - b*c^2*d*f^4*x + 2*b*d^4*f*x*arctan(d*x + c)*e^3 + b*c^2*d^2*f^2*arctan(d*x + c)*e^2 +
6*b*c^2*d^2*f^2*arctan(1/(d*x + c))*e^2 - 4*a*c^3*d*f^3*e + 2*b*c*d^2*f^3*x*e - 2*b*d^2*f^3*x*arctan(d*x + c)
*e - 2*b*d^3*f^2*x*e^2*log(d^2*x^2 + 2*c*d*x + c^2 + 1) + 4*b*d^3*f^2*x*e^2*log(abs(f*x + e)) + 2*b*c^2*f^4*ar
ctan(1/(d*x + c)) - 2*b*c*d^3*f*arctan(d*x + c)*e^3 - 4*b*c*d^3*f*arctan(1/(d*x + c))*e^3 + 6*a*c^2*d^2*f^2*e^
2 - b*d^3*f^2*x*e^2 - b*c^2*d*f^3*e - 4*b*c*d*f^3*arctan(1/(d*x + c))*e + b*c*d^2*f^2*e^2*log(d^2*x^2 + 2*c*d*
x + c^2 + 1) - 2*b*c*d^2*f^2*e^2*log(abs(f*x + e)) + 2*a*c^2*f^4 - b*d*f^4*x + b*d^4*arctan(d*x + c)*e^4 + b*d
^4*arctan(1/(d*x + c))*e^4 - 4*a*c*d^3*f*e^3 + 2*b*c*d^2*f^2*e^2 - b*d^2*f^2*arctan(d*x + c)*e^2 + 2*b*d^2*f^2
*arctan(1/(d*x + c))*e^2 - 4*a*c*d*f^3*e - b*d^3*f*e^3*log(d^2*x^2 + 2*c*d*x + c^2 + 1) + 2*b*d^3*f*e^3*log(ab
s(f*x + e)) + b*f^4*arctan(1/(d*x + c)) + a*d^4*e^4 - b*d^3*f*e^3 + 2*a*d^2*f^2*e^2 - b*d*f^3*e + a*f^4)/(c^4*
f^7*x^2 - 4*c^3*d*f^6*x^2*e + 6*c^2*d^2*f^5*x^2*e^2 + 2*c^4*f^6*x*e + 2*c^2*f^7*x^2 - 4*c*d^3*f^4*x^2*e^3 - 8*
c^3*d*f^5*x*e^2 - 4*c*d*f^6*x^2*e + d^4*f^3*x^2*e^4 + 12*c^2*d^2*f^4*x*e^3 + c^4*f^5*e^2 + 2*d^2*f^5*x^2*e^2 +
4*c^2*f^6*x*e + f^7*x^2 - 8*c*d^3*f^3*x*e^4 - 4*c^3*d*f^4*e^3 - 8*c*d*f^5*x*e^2 + 2*d^4*f^2*x*e^5 + 6*c^2*d^2
*f^3*e^4 + 4*d^2*f^4*x*e^3 + 2*c^2*f^5*e^2 + 2*f^6*x*e - 4*c*d^3*f^2*e^5 - 4*c*d*f^4*e^3 + d^4*f*e^6 + 2*d^2*f
^3*e^4 + f^5*e^2)