\(\int \frac {a+b \arctan (c+d x)}{(e+f x)^3} \, dx\) [38]

Optimal result

Integrand size = 18, antiderivative size = 227 \[ \int \frac {a+b \arctan (c+d x)}{(e+f x)^3} \, dx=-\frac {b d}{2 \left (d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2\right ) (e+f x)}+\frac {b d^2 (d e+f-c f) (d e-(1+c) f) \arctan (c+d x)}{2 f \left (d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2\right )^2}-\frac {a+b \arctan (c+d x)}{2 f (e+f x)^2}+\frac {b d^2 (d e-c f) \log (e+f x)}{\left (d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2\right )^2}-\frac {b d^2 (d e-c f) \log \left (1+c^2+2 c d x+d^2 x^2\right )}{2 \left (d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2\right )^2} \] Output:

-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*arc 
tan(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
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.57 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.77 \[ \int \frac {a+b \arctan (c+d x)}{(e+f x)^3} \, dx=\frac {-\frac {a+b \arctan (c+d x)}{(e+f x)^2}+\frac {1}{2} b d^2 \left (-\frac {2 f}{d \left (d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2\right ) (e+f x)}-\frac {i \log (i-c-d x)}{(d e-(-i+c) f)^2}+\frac {i \log (i+c+d x)}{(d e-(i+c) f)^2}-\frac {4 f (-d e+c f) \log (d (e+f x))}{\left (d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2\right )^2}\right )}{2 f} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.59 (sec) , antiderivative size = 245, normalized size of antiderivative = 1.08, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {5568, 2081, 1145, 27, 1200, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {a+b \arctan (c+d x)}{(e+f x)^3} \, dx\)

\(\Big \downarrow \) 5568

\(\displaystyle \frac {b d \int \frac {1}{(e+f x)^2 \left ((c+d x)^2+1\right )}dx}{2 f}-\frac {a+b \arctan (c+d x)}{2 f (e+f x)^2}\)

\(\Big \downarrow \) 2081

\(\displaystyle \frac {b d \int \frac {1}{(e+f x)^2 \left (c^2+2 d x c+d^2 x^2+1\right )}dx}{2 f}-\frac {a+b \arctan (c+d x)}{2 f (e+f x)^2}\)

\(\Big \downarrow \) 1145

\(\displaystyle \frac {b d \left (\frac {\int \frac {d (d e-2 c f-d f x)}{(e+f x) \left (c^2+2 d x c+d^2 x^2+1\right )}dx}{\left (c^2+1\right ) f^2-2 c d e f+d^2 e^2}-\frac {f}{(e+f x) \left (\left (c^2+1\right ) f^2-2 c d e f+d^2 e^2\right )}\right )}{2 f}-\frac {a+b \arctan (c+d x)}{2 f (e+f x)^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {b d \left (\frac {d \int \frac {d e-2 c f-d f x}{(e+f x) \left (c^2+2 d x c+d^2 x^2+1\right )}dx}{\left (c^2+1\right ) f^2-2 c d e f+d^2 e^2}-\frac {f}{(e+f x) \left (\left (c^2+1\right ) f^2-2 c d e f+d^2 e^2\right )}\right )}{2 f}-\frac {a+b \arctan (c+d x)}{2 f (e+f x)^2}\)

\(\Big \downarrow \) 1200

\(\displaystyle \frac {b d \left (\frac {d \int \left (\frac {2 (d e-c f) f^2}{\left (d^2 e^2-2 c d f e+\left (c^2+1\right ) f^2\right ) (e+f x)}+\frac {d \left (d^2 e^2-4 c d f e-\left (1-3 c^2\right ) f^2-2 d f (d e-c f) x\right )}{\left (d^2 e^2-2 c d f e+\left (c^2+1\right ) f^2\right ) \left (c^2+2 d x c+d^2 x^2+1\right )}\right )dx}{\left (c^2+1\right ) f^2-2 c d e f+d^2 e^2}-\frac {f}{(e+f x) \left (\left (c^2+1\right ) f^2-2 c d e f+d^2 e^2\right )}\right )}{2 f}-\frac {a+b \arctan (c+d x)}{2 f (e+f x)^2}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {b d \left (\frac {d \left (\frac {\arctan (c+d x) (-c f+d e+f) (d e-(c+1) f)}{\left (c^2+1\right ) f^2-2 c d e f+d^2 e^2}-\frac {f (d e-c f) \log \left (c^2+2 c d x+d^2 x^2+1\right )}{\left (c^2+1\right ) f^2-2 c d e f+d^2 e^2}+\frac {2 f (d e-c f) \log (e+f x)}{\left (c^2+1\right ) f^2-2 c d e f+d^2 e^2}\right )}{\left (c^2+1\right ) f^2-2 c d e f+d^2 e^2}-\frac {f}{(e+f x) \left (\left (c^2+1\right ) f^2-2 c d e f+d^2 e^2\right )}\right )}{2 f}-\frac {a+b \arctan (c+d x)}{2 f (e+f x)^2}\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 1145

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] + Simp 
[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}, x] && ILtQ[m, -1]
 

rule 1200

Int[(((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.))/((a_.) + (b_.)* 
(x_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*((f + g* 
x)^n/(a + b*x + c*x^2)), x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && In 
tegersQ[n]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2081

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 5568

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] - Simp[b*d*(p/(f*(m + 1)))   Int[(e + f*x)^(m + 1)*((a + b*ArcTan[ 
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]
 

Maple [A] (verified)

Time = 0.60 (sec) , antiderivative size = 245, normalized size of antiderivative = 1.08

method	result	size
parts	\(-\frac {a}{2 \left (f x +e \right )^{2} f}+\frac {b \left (-\frac {d^{3} \arctan \left (d x +c \right )}{2 \left (f \left (d x +c \right )-c f +d e \right )^{2} f}+\frac {d^{3} \left (\frac {\frac {\left (2 c \,f^{2}-2 d e f \right ) \ln \left (1+\left (d x +c \right )^{2}\right )}{2}+\left (c^{2} f^{2}-2 c d e f +d^{2} e^{2}-f^{2}\right ) \arctan \left (d x +c \right )}{\left (c^{2} f^{2}-2 c d e f +d^{2} e^{2}+f^{2}\right )^{2}}-\frac {f}{\left (c^{2} f^{2}-2 c d e f +d^{2} e^{2}+f^{2}\right ) \left (f \left (d x +c \right )-c f +d e \right )}-\frac {2 \left (c f -d e \right ) f \ln \left (f \left (d x +c \right )-c f +d e \right )}{\left (c^{2} f^{2}-2 c d e f +d^{2} e^{2}+f^{2}\right )^{2}}\right )}{2 f}\right )}{d}\)	\(245\)
derivativedivides	\(\frac {-\frac {a \,d^{3}}{2 \left (c f -d e -f \left (d x +c \right )\right )^{2} f}-b \,d^{3} \left (\frac {\arctan \left (d x +c \right )}{2 \left (c f -d e -f \left (d x +c \right )\right )^{2} f}-\frac {\frac {f}{\left (c^{2} f^{2}-2 c d e f +d^{2} e^{2}+f^{2}\right ) \left (c f -d e -f \left (d x +c \right )\right )}-\frac {2 f \left (c f -d e \right ) \ln \left (c f -d e -f \left (d x +c \right )\right )}{\left (c^{2} f^{2}-2 c d e f +d^{2} e^{2}+f^{2}\right )^{2}}+\frac {\frac {\left (2 c \,f^{2}-2 d e f \right ) \ln \left (1+\left (d x +c \right )^{2}\right )}{2}+\left (c^{2} f^{2}-2 c d e f +d^{2} e^{2}-f^{2}\right ) \arctan \left (d x +c \right )}{\left (c^{2} f^{2}-2 c d e f +d^{2} e^{2}+f^{2}\right )^{2}}}{2 f}\right )}{d}\)	\(260\)
default	\(\frac {-\frac {a \,d^{3}}{2 \left (c f -d e -f \left (d x +c \right )\right )^{2} f}-b \,d^{3} \left (\frac {\arctan \left (d x +c \right )}{2 \left (c f -d e -f \left (d x +c \right )\right )^{2} f}-\frac {\frac {f}{\left (c^{2} f^{2}-2 c d e f +d^{2} e^{2}+f^{2}\right ) \left (c f -d e -f \left (d x +c \right )\right )}-\frac {2 f \left (c f -d e \right ) \ln \left (c f -d e -f \left (d x +c \right )\right )}{\left (c^{2} f^{2}-2 c d e f +d^{2} e^{2}+f^{2}\right )^{2}}+\frac {\frac {\left (2 c \,f^{2}-2 d e f \right ) \ln \left (1+\left (d x +c \right )^{2}\right )}{2}+\left (c^{2} f^{2}-2 c d e f +d^{2} e^{2}-f^{2}\right ) \arctan \left (d x +c \right )}{\left (c^{2} f^{2}-2 c d e f +d^{2} e^{2}+f^{2}\right )^{2}}}{2 f}\right )}{d}\)	\(260\)
parallelrisch	\(-\frac {4 \ln \left (f x +e \right ) x b c \,d^{4} e \,f^{4}-2 \ln \left (d^{2} x^{2}+2 c d x +c^{2}+1\right ) x b c \,d^{4} e \,f^{4}+2 x^{2} \arctan \left (d x +c \right ) b c \,d^{5} e \,f^{4}-2 x \arctan \left (d x +c \right ) b \,c^{2} d^{4} e \,f^{4}+4 x \arctan \left (d x +c \right ) b c \,d^{5} e^{2} f^{3}+2 \ln \left (f x +e \right ) x^{2} b c \,d^{4} f^{5}-2 \ln \left (f x +e \right ) x^{2} b \,d^{5} e \,f^{4}-\ln \left (d^{2} x^{2}+2 c d x +c^{2}+1\right ) x^{2} b c \,d^{4} f^{5}+\ln \left (d^{2} x^{2}+2 c d x +c^{2}+1\right ) x^{2} b \,d^{5} e \,f^{4}-4 \ln \left (f x +e \right ) x b \,d^{5} e^{2} f^{3}+2 \ln \left (d^{2} x^{2}+2 c d x +c^{2}+1\right ) x b \,d^{5} e^{2} f^{3}+2 \ln \left (f x +e \right ) b c \,d^{4} e^{2} f^{3}-\ln \left (d^{2} x^{2}+2 c d x +c^{2}+1\right ) b c \,d^{4} e^{2} f^{3}-2 x \arctan \left (d x +c \right ) b \,d^{6} e^{3} f^{2}-4 \arctan \left (d x +c \right ) b \,c^{3} d^{3} e \,f^{4}+5 \arctan \left (d x +c \right ) b \,c^{2} d^{4} e^{2} f^{3}-2 \arctan \left (d x +c \right ) b c \,d^{5} e^{3} f^{2}+2 x \arctan \left (d x +c \right ) b \,d^{4} e \,f^{4}-x^{2} \arctan \left (d x +c \right ) b \,c^{2} d^{4} f^{5}-x^{2} \arctan \left (d x +c \right ) b \,d^{6} e^{2} f^{3}+a \,c^{4} d^{2} f^{5}+2 a \,c^{2} d^{2} f^{5}+x b \,d^{3} f^{5}+\arctan \left (d x +c \right ) b \,d^{2} f^{5}+a \,d^{2} f^{5}-4 a \,c^{3} d^{3} e \,f^{4}+6 a \,c^{2} d^{4} e^{2} f^{3}+b \,c^{2} d^{3} e \,f^{4}-4 a c \,d^{3} e \,f^{4}-4 \arctan \left (d x +c \right ) b c \,d^{3} e \,f^{4}-2 x b c \,d^{4} e \,f^{4}+2 a \,d^{4} e^{2} f^{3}+b \,d^{3} e \,f^{4}-4 a c \,d^{5} e^{3} f^{2}-2 b c \,d^{4} e^{2} f^{3}-2 \ln \left (f x +e \right ) b \,d^{5} e^{3} f^{2}+\ln \left (d^{2} x^{2}+2 c d x +c^{2}+1\right ) b \,d^{5} e^{3} f^{2}+x^{2} \arctan \left (d x +c \right ) b \,d^{4} f^{5}+\arctan \left (d x +c \right ) b \,c^{4} d^{2} f^{5}+x b \,c^{2} d^{3} f^{5}+x b \,d^{5} e^{2} f^{3}+2 \arctan \left (d x +c \right ) b \,c^{2} d^{2} f^{5}+3 \arctan \left (d x +c \right ) b \,d^{4} e^{2} f^{3}+a \,e^{4} f \,d^{6}+b \,e^{3} f^{2} d^{5}}{2 \left (f x +e \right )^{2} \left (c^{4} f^{4}-4 c^{3} d e \,f^{3}+6 c^{2} d^{2} e^{2} f^{2}-4 c \,d^{3} e^{3} f +d^{4} e^{4}+2 c^{2} f^{4}-4 c d e \,f^{3}+2 e^{2} f^{2} d^{2}+f^{4}\right ) f^{2} d^{2}}\)	\(884\)
risch	\(\text {Expression too large to display}\)	\(13218\)

Input:

int((a+b*arctan(d*x+c))/(f*x+e)^3,x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 682 vs. \(2 (219) = 438\).

Time = 1.22 (sec) , antiderivative size = 682, normalized size of antiderivative = 3.00 \[ \int \frac {a+b \arctan (c+d x)}{(e+f x)^3} \, dx=-\frac {a d^{4} e^{4} - {\left (4 \, a c - b\right )} d^{3} e^{3} f + 2 \, {\left (3 \, a c^{2} - b c + a\right )} d^{2} e^{2} f^{2} - {\left (4 \, a c^{3} - b c^{2} + 4 \, a c - b\right )} d e f^{3} + {\left (a c^{4} + 2 \, a c^{2} + a\right )} f^{4} + {\left (b d^{3} e^{2} f^{2} - 2 \, b c d^{2} e f^{3} + {\left (b c^{2} + b\right )} d f^{4}\right )} x - {\left (2 \, b c d^{3} e^{3} f - {\left (5 \, b c^{2} + 3 \, b\right )} d^{2} e^{2} f^{2} + 4 \, {\left (b c^{3} + b c\right )} d e f^{3} - {\left (b c^{4} + 2 \, b c^{2} + b\right )} f^{4} + {\left (b d^{4} e^{2} f^{2} - 2 \, b c d^{3} e f^{3} + {\left (b c^{2} - b\right )} d^{2} f^{4}\right )} x^{2} + 2 \, {\left (b d^{4} e^{3} f - 2 \, b c d^{3} e^{2} f^{2} + {\left (b c^{2} - b\right )} d^{2} e f^{3}\right )} x\right )} \arctan \left (d x + c\right ) + {\left (b d^{3} e^{3} f - b c d^{2} e^{2} f^{2} + {\left (b d^{3} e f^{3} - b c d^{2} f^{4}\right )} x^{2} + 2 \, {\left (b d^{3} e^{2} f^{2} - b c d^{2} e f^{3}\right )} x\right )} \log \left (d^{2} x^{2} + 2 \, c d x + c^{2} + 1\right ) - 2 \, {\left (b d^{3} e^{3} f - b c d^{2} e^{2} f^{2} + {\left (b d^{3} e f^{3} - b c d^{2} f^{4}\right )} x^{2} + 2 \, {\left (b d^{3} e^{2} f^{2} - b c d^{2} e f^{3}\right )} x\right )} \log \left (f x + e\right )}{2 \, {\left (d^{4} e^{6} f - 4 \, c d^{3} e^{5} f^{2} + 2 \, {\left (3 \, c^{2} + 1\right )} d^{2} e^{4} f^{3} - 4 \, {\left (c^{3} + c\right )} d e^{3} f^{4} + {\left (c^{4} + 2 \, c^{2} + 1\right )} e^{2} f^{5} + {\left (d^{4} e^{4} f^{3} - 4 \, c d^{3} e^{3} f^{4} + 2 \, {\left (3 \, c^{2} + 1\right )} d^{2} e^{2} f^{5} - 4 \, {\left (c^{3} + c\right )} d e f^{6} + {\left (c^{4} + 2 \, c^{2} + 1\right )} f^{7}\right )} x^{2} + 2 \, {\left (d^{4} e^{5} f^{2} - 4 \, c d^{3} e^{4} f^{3} + 2 \, {\left (3 \, c^{2} + 1\right )} d^{2} e^{3} f^{4} - 4 \, {\left (c^{3} + c\right )} d e^{2} f^{5} + {\left (c^{4} + 2 \, c^{2} + 1\right )} e f^{6}\right )} x\right )}} \] Input:

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

Output:

-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)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {a+b \arctan (c+d x)}{(e+f x)^3} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 409, normalized size of antiderivative = 1.80 \[ \int \frac {a+b \arctan (c+d x)}{(e+f x)^3} \, dx=-\frac {1}{2} \, {\left (d {\left (\frac {{\left (d^{2} e - c d f\right )} \log \left (d^{2} x^{2} + 2 \, c d x + c^{2} + 1\right )}{d^{4} e^{4} - 4 \, c d^{3} e^{3} f + 2 \, {\left (3 \, c^{2} + 1\right )} d^{2} e^{2} f^{2} - 4 \, {\left (c^{3} + c\right )} d e f^{3} + {\left (c^{4} + 2 \, c^{2} + 1\right )} f^{4}} - \frac {2 \, {\left (d^{2} e - c d f\right )} \log \left (f x + e\right )}{d^{4} e^{4} - 4 \, c d^{3} e^{3} f + 2 \, {\left (3 \, c^{2} + 1\right )} d^{2} e^{2} f^{2} - 4 \, {\left (c^{3} + c\right )} d e f^{3} + {\left (c^{4} + 2 \, c^{2} + 1\right )} f^{4}} - \frac {{\left (d^{4} e^{2} - 2 \, c d^{3} e f + {\left (c^{2} - 1\right )} d^{2} f^{2}\right )} \arctan \left (\frac {d^{2} x + c d}{d}\right )}{{\left (d^{4} e^{4} f - 4 \, c d^{3} e^{3} f^{2} + 2 \, {\left (3 \, c^{2} + 1\right )} d^{2} e^{2} f^{3} - 4 \, {\left (c^{3} + c\right )} d e f^{4} + {\left (c^{4} + 2 \, c^{2} + 1\right )} f^{5}\right )} d} + \frac {1}{d^{2} e^{3} - 2 \, c d e^{2} f + {\left (c^{2} + 1\right )} e f^{2} + {\left (d^{2} e^{2} f - 2 \, c d e f^{2} + {\left (c^{2} + 1\right )} f^{3}\right )} x}\right )} + \frac {\arctan \left (d x + c\right )}{f^{3} x^{2} + 2 \, e f^{2} x + e^{2} f}\right )} b - \frac {a}{2 \, {\left (f^{3} x^{2} + 2 \, e f^{2} x + e^{2} f\right )}} \] Input:

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

Output:

-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)
 

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 3.59 (sec) , antiderivative size = 1561, normalized size of antiderivative = 6.88 \[ \int \frac {a+b \arctan (c+d x)}{(e+f x)^3} \, dx=\text {Too large to display} \] Input:

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

Output:

1/4*(I*b*d^4*e^2*f^2*x^2*log(I*d*x + I*c - 1) - 2*I*b*c*d^3*e*f^3*x^2*log( 
I*d*x + I*c - 1) + I*b*c^2*d^2*f^4*x^2*log(I*d*x + I*c - 1) - I*b*d^4*e^2* 
f^2*x^2*log(-I*d*x - I*c - 1) + 2*I*b*c*d^3*e*f^3*x^2*log(-I*d*x - I*c - 1 
) - I*b*c^2*d^2*f^4*x^2*log(-I*d*x - I*c - 1) + 2*I*b*d^4*e^3*f*x*log(I*d* 
x + I*c - 1) - 4*I*b*c*d^3*e^2*f^2*x*log(I*d*x + I*c - 1) + 2*I*b*c^2*d^2* 
e*f^3*x*log(I*d*x + I*c - 1) - 2*b*d^3*e*f^3*x^2*log(I*d*x + I*c - 1) + 2* 
b*c*d^2*f^4*x^2*log(I*d*x + I*c - 1) - 2*I*b*d^4*e^3*f*x*log(-I*d*x - I*c 
- 1) + 4*I*b*c*d^3*e^2*f^2*x*log(-I*d*x - I*c - 1) - 2*I*b*c^2*d^2*e*f^3*x 
*log(-I*d*x - I*c - 1) - 2*b*d^3*e*f^3*x^2*log(-I*d*x - I*c - 1) + 2*b*c*d 
^2*f^4*x^2*log(-I*d*x - I*c - 1) + 4*b*d^3*e*f^3*x^2*log(f*x + e) - 4*b*c* 
d^2*f^4*x^2*log(f*x + e) - 2*b*d^4*e^4*arctan(d*x + c) + 8*b*c*d^3*e^3*f*a 
rctan(d*x + c) - 12*b*c^2*d^2*e^2*f^2*arctan(d*x + c) + 8*b*c^3*d*e*f^3*ar 
ctan(d*x + c) - 2*b*c^4*f^4*arctan(d*x + c) + I*b*d^4*e^4*log(I*d*x + I*c 
- 1) - 2*I*b*c*d^3*e^3*f*log(I*d*x + I*c - 1) + I*b*c^2*d^2*e^2*f^2*log(I* 
d*x + I*c - 1) - 4*b*d^3*e^2*f^2*x*log(I*d*x + I*c - 1) + 4*b*c*d^2*e*f^3* 
x*log(I*d*x + I*c - 1) - I*b*d^2*f^4*x^2*log(I*d*x + I*c - 1) - I*b*d^4*e^ 
4*log(-I*d*x - I*c - 1) + 2*I*b*c*d^3*e^3*f*log(-I*d*x - I*c - 1) - I*b*c^ 
2*d^2*e^2*f^2*log(-I*d*x - I*c - 1) - 4*b*d^3*e^2*f^2*x*log(-I*d*x - I*c - 
 1) + 4*b*c*d^2*e*f^3*x*log(-I*d*x - I*c - 1) + I*b*d^2*f^4*x^2*log(-I*d*x 
 - I*c - 1) + 8*b*d^3*e^2*f^2*x*log(f*x + e) - 8*b*c*d^2*e*f^3*x*log(f*...
 

Mupad [B] (verification not implemented)

Time = 8.60 (sec) , antiderivative size = 399, normalized size of antiderivative = 1.76 \[ \int \frac {a+b \arctan (c+d x)}{(e+f x)^3} \, dx=\frac {b\,d^3\,e\,\ln \left (e+f\,x\right )}{{\left (c^2\,f^2-2\,c\,d\,e\,f+d^2\,e^2+f^2\right )}^2}-\frac {a\,f}{2\,{\left (e+f\,x\right )}^2\,\left (c^2\,f^2-2\,c\,d\,e\,f+d^2\,e^2+f^2\right )}-\frac {b\,d\,e}{2\,{\left (e+f\,x\right )}^2\,\left (c^2\,f^2-2\,c\,d\,e\,f+d^2\,e^2+f^2\right )}-\frac {a\,c^2\,f}{2\,{\left (e+f\,x\right )}^2\,\left (c^2\,f^2-2\,c\,d\,e\,f+d^2\,e^2+f^2\right )}-\frac {b\,\mathrm {atan}\left (c+d\,x\right )}{2\,f\,{\left (e+f\,x\right )}^2}-\frac {b\,c\,d^2\,f\,\ln \left (e+f\,x\right )}{{\left (c^2\,f^2-2\,c\,d\,e\,f+d^2\,e^2+f^2\right )}^2}+\frac {a\,c\,d\,e}{{\left (e+f\,x\right )}^2\,\left (c^2\,f^2-2\,c\,d\,e\,f+d^2\,e^2+f^2\right )}-\frac {b\,d\,f\,x}{2\,{\left (e+f\,x\right )}^2\,\left (c^2\,f^2-2\,c\,d\,e\,f+d^2\,e^2+f^2\right )}-\frac {a\,d^2\,e^2}{2\,f\,{\left (e+f\,x\right )}^2\,\left (c^2\,f^2-2\,c\,d\,e\,f+d^2\,e^2+f^2\right )}-\frac {b\,d^2\,\ln \left (c+d\,x-\mathrm {i}\right )\,1{}\mathrm {i}}{4\,f\,{\left (d\,e-c\,f+f\,1{}\mathrm {i}\right )}^2}+\frac {b\,d^2\,\ln \left (c+d\,x+1{}\mathrm {i}\right )\,1{}\mathrm {i}}{4\,f\,{\left (c\,f-d\,e+f\,1{}\mathrm {i}\right )}^2} \] Input:

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

Output:

(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*1 
i - 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))
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 1110, normalized size of antiderivative = 4.89 \[ \int \frac {a+b \arctan (c+d x)}{(e+f x)^3} \, dx =\text {Too large to display} \] Input:

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

Output:

( - 2*atan(c + d*x)*b*c**4*e*f**4 + 8*atan(c + d*x)*b*c**3*d*e**2*f**3 - 1 
0*atan(c + d*x)*b*c**2*d**2*e**3*f**2 + 4*atan(c + d*x)*b*c**2*d**2*e**2*f 
**3*x + 2*atan(c + d*x)*b*c**2*d**2*e*f**4*x**2 - 4*atan(c + d*x)*b*c**2*e 
*f**4 + 4*atan(c + d*x)*b*c*d**3*e**4*f - 8*atan(c + d*x)*b*c*d**3*e**3*f* 
*2*x - 4*atan(c + d*x)*b*c*d**3*e**2*f**3*x**2 + 8*atan(c + d*x)*b*c*d*e** 
2*f**3 + 4*atan(c + d*x)*b*d**4*e**4*f*x + 2*atan(c + d*x)*b*d**4*e**3*f** 
2*x**2 - 6*atan(c + d*x)*b*d**2*e**3*f**2 - 4*atan(c + d*x)*b*d**2*e**2*f* 
*3*x - 2*atan(c + d*x)*b*d**2*e*f**4*x**2 - 2*atan(c + d*x)*b*e*f**4 + 2*l 
og(c**2 + 2*c*d*x + d**2*x**2 + 1)*b*c*d**2*e**3*f**2 + 4*log(c**2 + 2*c*d 
*x + d**2*x**2 + 1)*b*c*d**2*e**2*f**3*x + 2*log(c**2 + 2*c*d*x + d**2*x** 
2 + 1)*b*c*d**2*e*f**4*x**2 - 2*log(c**2 + 2*c*d*x + d**2*x**2 + 1)*b*d**3 
*e**4*f - 4*log(c**2 + 2*c*d*x + d**2*x**2 + 1)*b*d**3*e**3*f**2*x - 2*log 
(c**2 + 2*c*d*x + d**2*x**2 + 1)*b*d**3*e**2*f**3*x**2 - 4*log(e + f*x)*b* 
c*d**2*e**3*f**2 - 8*log(e + f*x)*b*c*d**2*e**2*f**3*x - 4*log(e + f*x)*b* 
c*d**2*e*f**4*x**2 + 4*log(e + f*x)*b*d**3*e**4*f + 8*log(e + f*x)*b*d**3* 
e**3*f**2*x + 4*log(e + f*x)*b*d**3*e**2*f**3*x**2 - 2*a*c**4*e*f**4 + 8*a 
*c**3*d*e**2*f**3 - 12*a*c**2*d**2*e**3*f**2 - 4*a*c**2*e*f**4 + 8*a*c*d** 
3*e**4*f + 8*a*c*d*e**2*f**3 - 2*a*d**4*e**5 - 4*a*d**2*e**3*f**2 - 2*a*e* 
f**4 - b*c**2*d*e**2*f**3 + b*c**2*d*f**5*x**2 + 2*b*c*d**2*e**3*f**2 - 2* 
b*c*d**2*e*f**4*x**2 - b*d**3*e**4*f + b*d**3*e**2*f**3*x**2 - b*d*e**2...