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

Optimal. Leaf size=227 \[ -\frac {a+b \tan ^{-1}(c+d x)}{2 f (e+f x)^2}-\frac {b d^2 (d e-c f) \log \left (c^2+2 c d x+d^2 x^2+1\right )}{2 \left (\left (c^2+1\right ) f^2-2 c d e f+d^2 e^2\right )^2}-\frac {b d}{2 (e+f x) \left (\left (c^2+1\right ) f^2-2 c d e f+d^2 e^2\right )}+\frac {b d^2 (d e-c f) \log (e+f x)}{\left (\left (c^2+1\right ) f^2-2 c d e f+d^2 e^2\right )^2}+\frac {b d^2 (-c f+d e+f) (d e-(c+1) f) \tan ^{-1}(c+d x)}{2 f \left (\left (c^2+1\right ) f^2-2 c d e f+d^2 e^2\right )^2} \]

[Out]

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

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Rubi [A]  time = 0.30, antiderivative size = 227, normalized size of antiderivative = 1.00, 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 = {5045, 1982, 709, 800, 634, 618, 204, 628} \[ -\frac {a+b \tan ^{-1}(c+d x)}{2 f (e+f x)^2}-\frac {b d^2 (d e-c f) \log \left (c^2+2 c d x+d^2 x^2+1\right )}{2 \left (\left (c^2+1\right ) f^2-2 c d e f+d^2 e^2\right )^2}-\frac {b d}{2 (e+f x) \left (\left (c^2+1\right ) f^2-2 c d e f+d^2 e^2\right )}+\frac {b d^2 (d e-c f) \log (e+f x)}{\left (\left (c^2+1\right ) f^2-2 c d e f+d^2 e^2\right )^2}+\frac {b d^2 (-c f+d e+f) (d e-(c+1) f) \tan ^{-1}(c+d x)}{2 f \left (\left (c^2+1\right ) f^2-2 c d e f+d^2 e^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTan[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)) + (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) - (a + b*ArcTan[c + d*x])/(2*f*(e + f*x)^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 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 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 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 5045

Int[((a_.) + ArcTan[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^(m_), x_Symbol] :> Simp[((e + f*x)^(m
+ 1)*(a + b*ArcTan[c + d*x])^p)/(f*(m + 1)), x] - Dist[(b*d*p)/(f*(m + 1)), Int[((e + f*x)^(m + 1)*(a + b*Arc
Tan[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 \tan ^{-1}(c+d x)}{(e+f x)^3} \, dx &=-\frac {a+b \tan ^{-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 \tan ^{-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 \tan ^{-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 \tan ^{-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 \tan ^{-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 \tan ^{-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 \tan ^{-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 {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 {a+b \tan ^{-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}\\ \end {align*}

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Mathematica [C]  time = 0.80, size = 175, normalized size = 0.77 \[ \frac {-\frac {a+b \tan ^{-1}(c+d x)}{(e+f x)^2}+\frac {1}{2} b d^2 \left (-\frac {2 f}{d (e+f x) \left (\left (c^2+1\right ) f^2-2 c d e f+d^2 e^2\right )}-\frac {4 f (c f-d e) \log (d (e+f x))}{\left (\left (c^2+1\right ) f^2-2 c d e f+d^2 e^2\right )^2}-\frac {i \log (-c-d x+i)}{(d e-(c-i) f)^2}+\frac {i \log (c+d x+i)}{(d e-(c+i) f)^2}\right )}{2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-((a + b*ArcTan[c + d*x])/(e + f*x)^2) + (b*d^2*((-2*f)/(d*(d^2*e^2 - 2*c*d*e*f + (1 + c^2)*f^2)*(e + f*x)) -
(I*Log[I - c - d*x])/(d*e - (-I + c)*f)^2 + (I*Log[I + c + d*x])/(d*e - (I + c)*f)^2 - (4*f*(-(d*e) + c*f)*Lo
g[d*(e + f*x)])/(d^2*e^2 - 2*c*d*e*f + (1 + c^2)*f^2)^2))/2)/(2*f)

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fricas [B]  time = 2.13, size = 682, normalized size = 3.00 \[ -\frac {a d^{4} e^{4} - {\left (4 \, a c - b\right )} d^{3} e^{3} f + 2 \, {\left (3 \, a c^{2} - b c + a\right )} d^{2} e^{2} f^{2} - {\left (4 \, a c^{3} - b c^{2} + 4 \, a c - b\right )} d e f^{3} + {\left (a c^{4} + 2 \, a c^{2} + a\right )} f^{4} + {\left (b d^{3} e^{2} f^{2} - 2 \, b c d^{2} e f^{3} + {\left (b c^{2} + b\right )} d f^{4}\right )} x - {\left (2 \, b c d^{3} e^{3} f - {\left (5 \, b c^{2} + 3 \, b\right )} d^{2} e^{2} f^{2} + 4 \, {\left (b c^{3} + b c\right )} d e f^{3} - {\left (b c^{4} + 2 \, b c^{2} + b\right )} f^{4} + {\left (b d^{4} e^{2} f^{2} - 2 \, b c d^{3} e f^{3} + {\left (b c^{2} - b\right )} d^{2} f^{4}\right )} x^{2} + 2 \, {\left (b d^{4} e^{3} f - 2 \, b c d^{3} e^{2} f^{2} + {\left (b c^{2} - b\right )} d^{2} e f^{3}\right )} x\right )} \arctan \left (d x + c\right ) + {\left (b d^{3} e^{3} f - b c d^{2} e^{2} f^{2} + {\left (b d^{3} e f^{3} - b c d^{2} f^{4}\right )} x^{2} + 2 \, {\left (b d^{3} e^{2} f^{2} - b c d^{2} e f^{3}\right )} x\right )} \log \left (d^{2} x^{2} + 2 \, c d x + c^{2} + 1\right ) - 2 \, {\left (b d^{3} e^{3} f - b c d^{2} e^{2} f^{2} + {\left (b d^{3} e f^{3} - b c d^{2} f^{4}\right )} x^{2} + 2 \, {\left (b d^{3} e^{2} f^{2} - b c d^{2} e f^{3}\right )} x\right )} \log \left (f x + e\right )}{2 \, {\left (d^{4} e^{6} f - 4 \, c d^{3} e^{5} f^{2} + 2 \, {\left (3 \, c^{2} + 1\right )} d^{2} e^{4} f^{3} - 4 \, {\left (c^{3} + c\right )} d e^{3} f^{4} + {\left (c^{4} + 2 \, c^{2} + 1\right )} e^{2} f^{5} + {\left (d^{4} e^{4} f^{3} - 4 \, c d^{3} e^{3} f^{4} + 2 \, {\left (3 \, c^{2} + 1\right )} d^{2} e^{2} f^{5} - 4 \, {\left (c^{3} + c\right )} d e f^{6} + {\left (c^{4} + 2 \, c^{2} + 1\right )} f^{7}\right )} x^{2} + 2 \, {\left (d^{4} e^{5} f^{2} - 4 \, c d^{3} e^{4} f^{3} + 2 \, {\left (3 \, c^{2} + 1\right )} d^{2} e^{3} f^{4} - 4 \, {\left (c^{3} + c\right )} d e^{2} f^{5} + {\left (c^{4} + 2 \, c^{2} + 1\right )} e f^{6}\right )} x\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(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 - (2*b*c*d^3*e^3*f
- (5*b*c^2 + 3*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 + (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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giac [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*arctan(d*x+c))/(f*x+e)^3,x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.05, size = 438, normalized size = 1.93 \[ -\frac {d^{2} a}{2 \left (d f x +d e \right )^{2} f}-\frac {d^{2} b \arctan \left (d x +c \right )}{2 \left (d f x +d e \right )^{2} f}-\frac {d^{2} b}{2 \left (c^{2} f^{2}-2 c d e f +d^{2} e^{2}+f^{2}\right ) \left (d f x +d e \right )}-\frac {d^{2} b f \ln \left (f \left (d x +c \right )-c f +d e \right ) c}{\left (c^{2} f^{2}-2 c d e f +d^{2} e^{2}+f^{2}\right )^{2}}+\frac {d^{3} b \ln \left (f \left (d x +c \right )-c f +d e \right ) e}{\left (c^{2} f^{2}-2 c d e f +d^{2} e^{2}+f^{2}\right )^{2}}+\frac {d^{2} b f \arctan \left (d x +c \right ) c^{2}}{2 \left (c^{2} f^{2}-2 c d e f +d^{2} e^{2}+f^{2}\right )^{2}}-\frac {d^{3} b \arctan \left (d x +c \right ) c e}{\left (c^{2} f^{2}-2 c d e f +d^{2} e^{2}+f^{2}\right )^{2}}+\frac {d^{4} b \arctan \left (d x +c \right ) e^{2}}{2 f \left (c^{2} f^{2}-2 c d e f +d^{2} e^{2}+f^{2}\right )^{2}}+\frac {d^{2} b f \ln \left (1+\left (d x +c \right )^{2}\right ) c}{2 \left (c^{2} f^{2}-2 c d e f +d^{2} e^{2}+f^{2}\right )^{2}}-\frac {d^{3} b \ln \left (1+\left (d x +c \right )^{2}\right ) e}{2 \left (c^{2} f^{2}-2 c d e f +d^{2} e^{2}+f^{2}\right )^{2}}-\frac {d^{2} b f \arctan \left (d x +c \right )}{2 \left (c^{2} f^{2}-2 c d e f +d^{2} e^{2}+f^{2}\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctan(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*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+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)

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maxima [A]  time = 0.43, size = 409, normalized size = 1.80 \[ -\frac {1}{2} \, {\left (d {\left (\frac {{\left (d^{2} e - c d f\right )} \log \left (d^{2} x^{2} + 2 \, c d x + c^{2} + 1\right )}{d^{4} e^{4} - 4 \, c d^{3} e^{3} f + 2 \, {\left (3 \, c^{2} + 1\right )} d^{2} e^{2} f^{2} - 4 \, {\left (c^{3} + c\right )} d e f^{3} + {\left (c^{4} + 2 \, c^{2} + 1\right )} f^{4}} - \frac {2 \, {\left (d^{2} e - c d f\right )} \log \left (f x + e\right )}{d^{4} e^{4} - 4 \, c d^{3} e^{3} f + 2 \, {\left (3 \, c^{2} + 1\right )} d^{2} e^{2} f^{2} - 4 \, {\left (c^{3} + c\right )} d e f^{3} + {\left (c^{4} + 2 \, c^{2} + 1\right )} f^{4}} - \frac {{\left (d^{4} e^{2} - 2 \, c d^{3} e f + {\left (c^{2} - 1\right )} d^{2} f^{2}\right )} \arctan \left (\frac {d^{2} x + c d}{d}\right )}{{\left (d^{4} e^{4} f - 4 \, c d^{3} e^{3} f^{2} + 2 \, {\left (3 \, c^{2} + 1\right )} d^{2} e^{2} f^{3} - 4 \, {\left (c^{3} + c\right )} d e f^{4} + {\left (c^{4} + 2 \, c^{2} + 1\right )} f^{5}\right )} d} + \frac {1}{d^{2} e^{3} - 2 \, c d e^{2} f + {\left (c^{2} + 1\right )} e f^{2} + {\left (d^{2} e^{2} f - 2 \, c d e f^{2} + {\left (c^{2} + 1\right )} f^{3}\right )} x}\right )} + \frac {\arctan \left (d x + c\right )}{f^{3} x^{2} + 2 \, e f^{2} x + e^{2} f}\right )} b - \frac {a}{2 \, {\left (f^{3} x^{2} + 2 \, e f^{2} x + e^{2} f\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(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)) + arctan(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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mupad [B]  time = 7.53, size = 399, normalized size = 1.76 \[ \frac {b\,d^3\,e\,\ln \left (e+f\,x\right )}{{\left (c^2\,f^2-2\,c\,d\,e\,f+d^2\,e^2+f^2\right )}^2}-\frac {a\,f}{2\,{\left (e+f\,x\right )}^2\,\left (c^2\,f^2-2\,c\,d\,e\,f+d^2\,e^2+f^2\right )}-\frac {b\,d\,e}{2\,{\left (e+f\,x\right )}^2\,\left (c^2\,f^2-2\,c\,d\,e\,f+d^2\,e^2+f^2\right )}-\frac {a\,c^2\,f}{2\,{\left (e+f\,x\right )}^2\,\left (c^2\,f^2-2\,c\,d\,e\,f+d^2\,e^2+f^2\right )}-\frac {b\,\mathrm {atan}\left (c+d\,x\right )}{2\,f\,{\left (e+f\,x\right )}^2}-\frac {b\,c\,d^2\,f\,\ln \left (e+f\,x\right )}{{\left (c^2\,f^2-2\,c\,d\,e\,f+d^2\,e^2+f^2\right )}^2}+\frac {a\,c\,d\,e}{{\left (e+f\,x\right )}^2\,\left (c^2\,f^2-2\,c\,d\,e\,f+d^2\,e^2+f^2\right )}-\frac {b\,d\,f\,x}{2\,{\left (e+f\,x\right )}^2\,\left (c^2\,f^2-2\,c\,d\,e\,f+d^2\,e^2+f^2\right )}-\frac {a\,d^2\,e^2}{2\,f\,{\left (e+f\,x\right )}^2\,\left (c^2\,f^2-2\,c\,d\,e\,f+d^2\,e^2+f^2\right )}-\frac {b\,d^2\,\ln \left (c+d\,x-\mathrm {i}\right )\,1{}\mathrm {i}}{4\,f\,{\left (d\,e-c\,f+f\,1{}\mathrm {i}\right )}^2}+\frac {b\,d^2\,\ln \left (c+d\,x+1{}\mathrm {i}\right )\,1{}\mathrm {i}}{4\,f\,{\left (c\,f-d\,e+f\,1{}\mathrm {i}\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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*atan(d*x+c))/(f*x+e)**3,x)

[Out]

Timed out

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