\(\int (d+e x)^4 (a+b \arctan (c x)) \, dx\) [1]

Optimal result

Integrand size = 16, antiderivative size = 184 \[ \int (d+e x)^4 (a+b \arctan (c x)) \, dx=-\frac {b d e \left (2 c^2 d^2-e^2\right ) x}{c^3}-\frac {b e^2 \left (10 c^2 d^2-e^2\right ) x^2}{10 c^3}-\frac {b d e^3 x^3}{3 c}-\frac {b e^4 x^4}{20 c}-\frac {b d \left (c^4 d^4-10 c^2 d^2 e^2+5 e^4\right ) \arctan (c x)}{5 c^4 e}+\frac {(d+e x)^5 (a+b \arctan (c x))}{5 e}-\frac {b \left (5 c^4 d^4-10 c^2 d^2 e^2+e^4\right ) \log \left (1+c^2 x^2\right )}{10 c^5} \] Output:

-b*d*e*(2*c^2*d^2-e^2)*x/c^3-1/10*b*e^2*(10*c^2*d^2-e^2)*x^2/c^3-1/3*b*d*e 
^3*x^3/c-1/20*b*e^4*x^4/c-1/5*b*d*(c^4*d^4-10*c^2*d^2*e^2+5*e^4)*arctan(c* 
x)/c^4/e+1/5*(e*x+d)^5*(a+b*arctan(c*x))/e-1/10*b*(5*c^4*d^4-10*c^2*d^2*e^ 
2+e^4)*ln(c^2*x^2+1)/c^5
 

Mathematica [A] (verified)

Time = 0.31 (sec) , antiderivative size = 255, normalized size of antiderivative = 1.39 \[ \int (d+e x)^4 (a+b \arctan (c x)) \, dx=\frac {(d+e x)^5 (a+b \arctan (c x))-\frac {b \left (c^2 e^2 x \left (-6 e^2 (10 d+e x)+c^2 \left (120 d^3+60 d^2 e x+20 d e^2 x^2+3 e^3 x^3\right )\right )+6 \left (-10 c^2 d^2 e^2 \left (\sqrt {-c^2} d+e\right )+e^4 \left (5 \sqrt {-c^2} d+e\right )+c^4 d^4 \left (\sqrt {-c^2} d+5 e\right )\right ) \log \left (1-\sqrt {-c^2} x\right )-6 \left (c^4 d^4 \left (\sqrt {-c^2} d-5 e\right )-10 c^2 d^2 \left (\sqrt {-c^2} d-e\right ) e^2+\left (5 \sqrt {-c^2} d-e\right ) e^4\right ) \log \left (1+\sqrt {-c^2} x\right )\right )}{12 c^5}}{5 e} \] Input:

Integrate[(d + e*x)^4*(a + b*ArcTan[c*x]),x]
 

Output:

((d + e*x)^5*(a + b*ArcTan[c*x]) - (b*(c^2*e^2*x*(-6*e^2*(10*d + e*x) + c^ 
2*(120*d^3 + 60*d^2*e*x + 20*d*e^2*x^2 + 3*e^3*x^3)) + 6*(-10*c^2*d^2*e^2* 
(Sqrt[-c^2]*d + e) + e^4*(5*Sqrt[-c^2]*d + e) + c^4*d^4*(Sqrt[-c^2]*d + 5* 
e))*Log[1 - Sqrt[-c^2]*x] - 6*(c^4*d^4*(Sqrt[-c^2]*d - 5*e) - 10*c^2*d^2*( 
Sqrt[-c^2]*d - e)*e^2 + (5*Sqrt[-c^2]*d - e)*e^4)*Log[1 + Sqrt[-c^2]*x]))/ 
(12*c^5))/(5*e)
 

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.01, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {5387, 478, 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.

\(\displaystyle \int (d+e x)^4 (a+b \arctan (c x)) \, dx\)

\(\Big \downarrow \) 5387

\(\displaystyle \frac {(d+e x)^5 (a+b \arctan (c x))}{5 e}-\frac {b c \int \frac {(d+e x)^5}{c^2 x^2+1}dx}{5 e}\)

\(\Big \downarrow \) 478

\(\displaystyle \frac {(d+e x)^5 (a+b \arctan (c x))}{5 e}-\frac {b c \int \left (\frac {x^3 e^5}{c^2}+\frac {5 d x^2 e^4}{c^2}+\frac {\left (10 c^2 d^2-e^2\right ) x e^3}{c^4}+\frac {5 d \left (2 c^2 d^2-e^2\right ) e^2}{c^4}+\frac {c^4 d^5-10 c^2 e^2 d^3+5 e^4 d+e \left (5 c^4 d^4-10 c^2 e^2 d^2+e^4\right ) x}{c^4 \left (c^2 x^2+1\right )}\right )dx}{5 e}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {(d+e x)^5 (a+b \arctan (c x))}{5 e}-\frac {b c \left (\frac {d \arctan (c x) \left (c^4 d^4-10 c^2 d^2 e^2+5 e^4\right )}{c^5}+\frac {5 d e^4 x^3}{3 c^2}+\frac {e^5 x^4}{4 c^2}+\frac {5 d e^2 x \left (2 c^2 d^2-e^2\right )}{c^4}+\frac {e^3 x^2 \left (10 c^2 d^2-e^2\right )}{2 c^4}+\frac {e \left (5 c^4 d^4-10 c^2 d^2 e^2+e^4\right ) \log \left (c^2 x^2+1\right )}{2 c^6}\right )}{5 e}\)

Input:

Int[(d + e*x)^4*(a + b*ArcTan[c*x]),x]
 

Output:

((d + e*x)^5*(a + b*ArcTan[c*x]))/(5*e) - (b*c*((5*d*e^2*(2*c^2*d^2 - e^2) 
*x)/c^4 + (e^3*(10*c^2*d^2 - e^2)*x^2)/(2*c^4) + (5*d*e^4*x^3)/(3*c^2) + ( 
e^5*x^4)/(4*c^2) + (d*(c^4*d^4 - 10*c^2*d^2*e^2 + 5*e^4)*ArcTan[c*x])/c^5 
+ (e*(5*c^4*d^4 - 10*c^2*d^2*e^2 + e^4)*Log[1 + c^2*x^2])/(2*c^6)))/(5*e)
 

Defintions of rubi rules used

rule 478

Int[((c_) + (d_.)*(x_))^(n_)/((a_) + (b_.)*(x_)^2), x_Symbol] :> Int[Expand 
Integrand[(c + d*x)^n/(a + b*x^2), x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ 
[n, 1]
 

rule 2009

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

rule 5387

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] - Simp[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]
 

Maple [A] (verified)

Time = 0.53 (sec) , antiderivative size = 246, normalized size of antiderivative = 1.34

method	result	size
parts	\(\frac {a \left (e x +d \right )^{5}}{5 e}+\frac {b \left (\frac {c \,e^{4} \arctan \left (c x \right ) x^{5}}{5}+c \,e^{3} \arctan \left (c x \right ) x^{4} d +2 c \,e^{2} \arctan \left (c x \right ) x^{3} d^{2}+2 c e \arctan \left (c x \right ) x^{2} d^{3}+\arctan \left (c x \right ) c x \,d^{4}+\frac {c \arctan \left (c x \right ) d^{5}}{5 e}-\frac {10 c^{4} d^{3} e^{2} x +5 c^{4} d^{2} e^{3} x^{2}+\frac {5 c^{4} d \,e^{4} x^{3}}{3}+\frac {e^{5} c^{4} x^{4}}{4}-5 c^{2} d \,e^{4} x -\frac {e^{5} c^{2} x^{2}}{2}+\frac {\left (5 c^{4} d^{4} e -10 c^{2} d^{2} e^{3}+e^{5}\right ) \ln \left (c^{2} x^{2}+1\right )}{2}+\left (c^{5} d^{5}-10 c^{3} d^{3} e^{2}+5 c d \,e^{4}\right ) \arctan \left (c x \right )}{5 c^{4} e}\right )}{c}\)	\(246\)
derivativedivides	\(\frac {\frac {a \left (c e x +c d \right )^{5}}{5 c^{4} e}+\frac {b \left (\frac {\arctan \left (c x \right ) c^{5} d^{5}}{5 e}+\arctan \left (c x \right ) c^{5} d^{4} x +2 e \arctan \left (c x \right ) c^{5} d^{3} x^{2}+2 e^{2} \arctan \left (c x \right ) c^{5} d^{2} x^{3}+e^{3} \arctan \left (c x \right ) c^{5} d \,x^{4}+\frac {e^{4} \arctan \left (c x \right ) c^{5} x^{5}}{5}-\frac {10 c^{4} d^{3} e^{2} x +5 c^{4} d^{2} e^{3} x^{2}+\frac {5 c^{4} d \,e^{4} x^{3}}{3}+\frac {e^{5} c^{4} x^{4}}{4}-5 c^{2} d \,e^{4} x -\frac {e^{5} c^{2} x^{2}}{2}+\frac {\left (5 c^{4} d^{4} e -10 c^{2} d^{2} e^{3}+e^{5}\right ) \ln \left (c^{2} x^{2}+1\right )}{2}+\left (c^{5} d^{5}-10 c^{3} d^{3} e^{2}+5 c d \,e^{4}\right ) \arctan \left (c x \right )}{5 e}\right )}{c^{4}}}{c}\)	\(265\)
default	\(\frac {\frac {a \left (c e x +c d \right )^{5}}{5 c^{4} e}+\frac {b \left (\frac {\arctan \left (c x \right ) c^{5} d^{5}}{5 e}+\arctan \left (c x \right ) c^{5} d^{4} x +2 e \arctan \left (c x \right ) c^{5} d^{3} x^{2}+2 e^{2} \arctan \left (c x \right ) c^{5} d^{2} x^{3}+e^{3} \arctan \left (c x \right ) c^{5} d \,x^{4}+\frac {e^{4} \arctan \left (c x \right ) c^{5} x^{5}}{5}-\frac {10 c^{4} d^{3} e^{2} x +5 c^{4} d^{2} e^{3} x^{2}+\frac {5 c^{4} d \,e^{4} x^{3}}{3}+\frac {e^{5} c^{4} x^{4}}{4}-5 c^{2} d \,e^{4} x -\frac {e^{5} c^{2} x^{2}}{2}+\frac {\left (5 c^{4} d^{4} e -10 c^{2} d^{2} e^{3}+e^{5}\right ) \ln \left (c^{2} x^{2}+1\right )}{2}+\left (c^{5} d^{5}-10 c^{3} d^{3} e^{2}+5 c d \,e^{4}\right ) \arctan \left (c x \right )}{5 e}\right )}{c^{4}}}{c}\)	\(265\)
parallelrisch	\(-\frac {-12 x^{5} \arctan \left (c x \right ) b \,c^{5} e^{4}-12 x^{5} a \,c^{5} e^{4}-60 x^{4} \arctan \left (c x \right ) b \,c^{5} d \,e^{3}-60 x^{4} a \,c^{5} d \,e^{3}-120 x^{3} \arctan \left (c x \right ) b \,c^{5} d^{2} e^{2}+3 x^{4} b \,c^{4} e^{4}-120 x^{3} a \,c^{5} d^{2} e^{2}-120 x^{2} \arctan \left (c x \right ) b \,c^{5} d^{3} e +20 x^{3} b \,c^{4} d \,e^{3}-120 x^{2} a \,c^{5} d^{3} e -60 x \arctan \left (c x \right ) b \,c^{5} d^{4}+60 x^{2} b \,c^{4} d^{2} e^{2}-60 x a \,c^{5} d^{4}+30 \ln \left (c^{2} x^{2}+1\right ) b \,c^{4} d^{4}+120 x b \,c^{4} d^{3} e -6 x^{2} b \,c^{2} e^{4}-120 \arctan \left (c x \right ) b \,c^{3} d^{3} e -60 \ln \left (c^{2} x^{2}+1\right ) b \,c^{2} d^{2} e^{2}-60 x b \,c^{2} d \,e^{3}+60 \arctan \left (c x \right ) b c d \,e^{3}+6 \ln \left (c^{2} x^{2}+1\right ) b \,e^{4}}{60 c^{5}}\)	\(310\)
risch	\(\frac {i e^{4} b \,x^{5} \ln \left (-i c x +1\right )}{10}+\frac {i e^{3} b d \,x^{4} \ln \left (-i c x +1\right )}{2}+i e^{2} b \,d^{2} x^{3} \ln \left (-i c x +1\right )+\frac {i b \,d^{4} x \ln \left (-i c x +1\right )}{2}+\frac {x^{5} e^{4} a}{5}+\frac {i b \,d^{5} \ln \left (c^{2} x^{2}+1\right )}{20 e}+x^{4} e^{3} d a +i e b \,d^{3} x^{2} \ln \left (-i c x +1\right )+2 x^{3} e^{2} d^{2} a -\frac {b \,e^{4} x^{4}}{20 c}-\frac {i \left (e x +d \right )^{5} b \ln \left (i c x +1\right )}{10 e}+2 x^{2} e \,d^{3} a -\frac {b d \,e^{3} x^{3}}{3 c}-\frac {b \,d^{5} \arctan \left (c x \right )}{10 e}+x a \,d^{4}-\frac {e^{2} b \,d^{2} x^{2}}{c}-\frac {b \,d^{4} \ln \left (c^{2} x^{2}+1\right )}{2 c}-\frac {2 e b \,d^{3} x}{c}+\frac {2 e b \,d^{3} \arctan \left (c x \right )}{c^{2}}+\frac {e^{4} b \,x^{2}}{10 c^{3}}+\frac {e^{2} b \,d^{2} \ln \left (c^{2} x^{2}+1\right )}{c^{3}}+\frac {e^{3} b d x}{c^{3}}-\frac {e^{3} b d \arctan \left (c x \right )}{c^{4}}-\frac {e^{4} b \ln \left (c^{2} x^{2}+1\right )}{10 c^{5}}\)	\(356\)

Input:

int((e*x+d)^4*(a+b*arctan(c*x)),x,method=_RETURNVERBOSE)
 

Output:

1/5*a*(e*x+d)^5/e+b/c*(1/5*c*e^4*arctan(c*x)*x^5+c*e^3*arctan(c*x)*x^4*d+2 
*c*e^2*arctan(c*x)*x^3*d^2+2*c*e*arctan(c*x)*x^2*d^3+arctan(c*x)*c*x*d^4+1 
/5*c/e*arctan(c*x)*d^5-1/5/c^4/e*(10*c^4*d^3*e^2*x+5*c^4*d^2*e^3*x^2+5/3*c 
^4*d*e^4*x^3+1/4*e^5*c^4*x^4-5*c^2*d*e^4*x-1/2*e^5*c^2*x^2+1/2*(5*c^4*d^4* 
e-10*c^2*d^2*e^3+e^5)*ln(c^2*x^2+1)+(c^5*d^5-10*c^3*d^3*e^2+5*c*d*e^4)*arc 
tan(c*x)))
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 264, normalized size of antiderivative = 1.43 \[ \int (d+e x)^4 (a+b \arctan (c x)) \, dx=\frac {12 \, a c^{5} e^{4} x^{5} + 3 \, {\left (20 \, a c^{5} d e^{3} - b c^{4} e^{4}\right )} x^{4} + 20 \, {\left (6 \, a c^{5} d^{2} e^{2} - b c^{4} d e^{3}\right )} x^{3} + 6 \, {\left (20 \, a c^{5} d^{3} e - 10 \, b c^{4} d^{2} e^{2} + b c^{2} e^{4}\right )} x^{2} + 60 \, {\left (a c^{5} d^{4} - 2 \, b c^{4} d^{3} e + b c^{2} d e^{3}\right )} x + 12 \, {\left (b c^{5} e^{4} x^{5} + 5 \, b c^{5} d e^{3} x^{4} + 10 \, b c^{5} d^{2} e^{2} x^{3} + 10 \, b c^{5} d^{3} e x^{2} + 5 \, b c^{5} d^{4} x + 10 \, b c^{3} d^{3} e - 5 \, b c d e^{3}\right )} \arctan \left (c x\right ) - 6 \, {\left (5 \, b c^{4} d^{4} - 10 \, b c^{2} d^{2} e^{2} + b e^{4}\right )} \log \left (c^{2} x^{2} + 1\right )}{60 \, c^{5}} \] Input:

integrate((e*x+d)^4*(a+b*arctan(c*x)),x, algorithm="fricas")
 

Output:

1/60*(12*a*c^5*e^4*x^5 + 3*(20*a*c^5*d*e^3 - b*c^4*e^4)*x^4 + 20*(6*a*c^5* 
d^2*e^2 - b*c^4*d*e^3)*x^3 + 6*(20*a*c^5*d^3*e - 10*b*c^4*d^2*e^2 + b*c^2* 
e^4)*x^2 + 60*(a*c^5*d^4 - 2*b*c^4*d^3*e + b*c^2*d*e^3)*x + 12*(b*c^5*e^4* 
x^5 + 5*b*c^5*d*e^3*x^4 + 10*b*c^5*d^2*e^2*x^3 + 10*b*c^5*d^3*e*x^2 + 5*b* 
c^5*d^4*x + 10*b*c^3*d^3*e - 5*b*c*d*e^3)*arctan(c*x) - 6*(5*b*c^4*d^4 - 1 
0*b*c^2*d^2*e^2 + b*e^4)*log(c^2*x^2 + 1))/c^5
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 345 vs. \(2 (170) = 340\).

Time = 0.52 (sec) , antiderivative size = 345, normalized size of antiderivative = 1.88 \[ \int (d+e x)^4 (a+b \arctan (c x)) \, dx=\begin {cases} a d^{4} x + 2 a d^{3} e x^{2} + 2 a d^{2} e^{2} x^{3} + a d e^{3} x^{4} + \frac {a e^{4} x^{5}}{5} + b d^{4} x \operatorname {atan}{\left (c x \right )} + 2 b d^{3} e x^{2} \operatorname {atan}{\left (c x \right )} + 2 b d^{2} e^{2} x^{3} \operatorname {atan}{\left (c x \right )} + b d e^{3} x^{4} \operatorname {atan}{\left (c x \right )} + \frac {b e^{4} x^{5} \operatorname {atan}{\left (c x \right )}}{5} - \frac {b d^{4} \log {\left (x^{2} + \frac {1}{c^{2}} \right )}}{2 c} - \frac {2 b d^{3} e x}{c} - \frac {b d^{2} e^{2} x^{2}}{c} - \frac {b d e^{3} x^{3}}{3 c} - \frac {b e^{4} x^{4}}{20 c} + \frac {2 b d^{3} e \operatorname {atan}{\left (c x \right )}}{c^{2}} + \frac {b d^{2} e^{2} \log {\left (x^{2} + \frac {1}{c^{2}} \right )}}{c^{3}} + \frac {b d e^{3} x}{c^{3}} + \frac {b e^{4} x^{2}}{10 c^{3}} - \frac {b d e^{3} \operatorname {atan}{\left (c x \right )}}{c^{4}} - \frac {b e^{4} \log {\left (x^{2} + \frac {1}{c^{2}} \right )}}{10 c^{5}} & \text {for}\: c \neq 0 \\a \left (d^{4} x + 2 d^{3} e x^{2} + 2 d^{2} e^{2} x^{3} + d e^{3} x^{4} + \frac {e^{4} x^{5}}{5}\right ) & \text {otherwise} \end {cases} \] Input:

integrate((e*x+d)**4*(a+b*atan(c*x)),x)
 

Output:

Piecewise((a*d**4*x + 2*a*d**3*e*x**2 + 2*a*d**2*e**2*x**3 + a*d*e**3*x**4 
 + a*e**4*x**5/5 + b*d**4*x*atan(c*x) + 2*b*d**3*e*x**2*atan(c*x) + 2*b*d* 
*2*e**2*x**3*atan(c*x) + b*d*e**3*x**4*atan(c*x) + b*e**4*x**5*atan(c*x)/5 
 - b*d**4*log(x**2 + c**(-2))/(2*c) - 2*b*d**3*e*x/c - b*d**2*e**2*x**2/c 
- b*d*e**3*x**3/(3*c) - b*e**4*x**4/(20*c) + 2*b*d**3*e*atan(c*x)/c**2 + b 
*d**2*e**2*log(x**2 + c**(-2))/c**3 + b*d*e**3*x/c**3 + b*e**4*x**2/(10*c* 
*3) - b*d*e**3*atan(c*x)/c**4 - b*e**4*log(x**2 + c**(-2))/(10*c**5), Ne(c 
, 0)), (a*(d**4*x + 2*d**3*e*x**2 + 2*d**2*e**2*x**3 + d*e**3*x**4 + e**4* 
x**5/5), True))
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 252, normalized size of antiderivative = 1.37 \[ \int (d+e x)^4 (a+b \arctan (c x)) \, dx=\frac {1}{5} \, a e^{4} x^{5} + a d e^{3} x^{4} + 2 \, a d^{2} e^{2} x^{3} + 2 \, a d^{3} e x^{2} + 2 \, {\left (x^{2} \arctan \left (c x\right ) - c {\left (\frac {x}{c^{2}} - \frac {\arctan \left (c x\right )}{c^{3}}\right )}\right )} b d^{3} e + {\left (2 \, x^{3} \arctan \left (c x\right ) - c {\left (\frac {x^{2}}{c^{2}} - \frac {\log \left (c^{2} x^{2} + 1\right )}{c^{4}}\right )}\right )} b d^{2} e^{2} + \frac {1}{3} \, {\left (3 \, x^{4} \arctan \left (c x\right ) - c {\left (\frac {c^{2} x^{3} - 3 \, x}{c^{4}} + \frac {3 \, \arctan \left (c x\right )}{c^{5}}\right )}\right )} b d e^{3} + \frac {1}{20} \, {\left (4 \, x^{5} \arctan \left (c x\right ) - c {\left (\frac {c^{2} x^{4} - 2 \, x^{2}}{c^{4}} + \frac {2 \, \log \left (c^{2} x^{2} + 1\right )}{c^{6}}\right )}\right )} b e^{4} + a d^{4} x + \frac {{\left (2 \, c x \arctan \left (c x\right ) - \log \left (c^{2} x^{2} + 1\right )\right )} b d^{4}}{2 \, c} \] Input:

integrate((e*x+d)^4*(a+b*arctan(c*x)),x, algorithm="maxima")
 

Output:

1/5*a*e^4*x^5 + a*d*e^3*x^4 + 2*a*d^2*e^2*x^3 + 2*a*d^3*e*x^2 + 2*(x^2*arc 
tan(c*x) - c*(x/c^2 - arctan(c*x)/c^3))*b*d^3*e + (2*x^3*arctan(c*x) - c*( 
x^2/c^2 - log(c^2*x^2 + 1)/c^4))*b*d^2*e^2 + 1/3*(3*x^4*arctan(c*x) - c*(( 
c^2*x^3 - 3*x)/c^4 + 3*arctan(c*x)/c^5))*b*d*e^3 + 1/20*(4*x^5*arctan(c*x) 
 - c*((c^2*x^4 - 2*x^2)/c^4 + 2*log(c^2*x^2 + 1)/c^6))*b*e^4 + a*d^4*x + 1 
/2*(2*c*x*arctan(c*x) - log(c^2*x^2 + 1))*b*d^4/c
 

Giac [A] (verification not implemented)

Time = 0.53 (sec) , antiderivative size = 309, normalized size of antiderivative = 1.68 \[ \int (d+e x)^4 (a+b \arctan (c x)) \, dx=\frac {12 \, b c^{5} e^{4} x^{5} \arctan \left (c x\right ) + 12 \, a c^{5} e^{4} x^{5} + 60 \, b c^{5} d e^{3} x^{4} \arctan \left (c x\right ) + 60 \, a c^{5} d e^{3} x^{4} + 120 \, b c^{5} d^{2} e^{2} x^{3} \arctan \left (c x\right ) + 120 \, a c^{5} d^{2} e^{2} x^{3} - 3 \, b c^{4} e^{4} x^{4} + 120 \, b c^{5} d^{3} e x^{2} \arctan \left (c x\right ) + 120 \, a c^{5} d^{3} e x^{2} - 20 \, b c^{4} d e^{3} x^{3} + 60 \, b c^{5} d^{4} x \arctan \left (c x\right ) + 60 \, a c^{5} d^{4} x - 60 \, b c^{4} d^{2} e^{2} x^{2} - 120 \, b c^{4} d^{3} e x - 30 \, b c^{4} d^{4} \log \left (c^{2} x^{2} + 1\right ) + 6 \, b c^{2} e^{4} x^{2} + 120 \, b c^{3} d^{3} e \arctan \left (c x\right ) + 60 \, b c^{2} d e^{3} x + 60 \, b c^{2} d^{2} e^{2} \log \left (c^{2} x^{2} + 1\right ) - 60 \, b c d e^{3} \arctan \left (c x\right ) - 6 \, b e^{4} \log \left (c^{2} x^{2} + 1\right )}{60 \, c^{5}} \] Input:

integrate((e*x+d)^4*(a+b*arctan(c*x)),x, algorithm="giac")
 

Output:

1/60*(12*b*c^5*e^4*x^5*arctan(c*x) + 12*a*c^5*e^4*x^5 + 60*b*c^5*d*e^3*x^4 
*arctan(c*x) + 60*a*c^5*d*e^3*x^4 + 120*b*c^5*d^2*e^2*x^3*arctan(c*x) + 12 
0*a*c^5*d^2*e^2*x^3 - 3*b*c^4*e^4*x^4 + 120*b*c^5*d^3*e*x^2*arctan(c*x) + 
120*a*c^5*d^3*e*x^2 - 20*b*c^4*d*e^3*x^3 + 60*b*c^5*d^4*x*arctan(c*x) + 60 
*a*c^5*d^4*x - 60*b*c^4*d^2*e^2*x^2 - 120*b*c^4*d^3*e*x - 30*b*c^4*d^4*log 
(c^2*x^2 + 1) + 6*b*c^2*e^4*x^2 + 120*b*c^3*d^3*e*arctan(c*x) + 60*b*c^2*d 
*e^3*x + 60*b*c^2*d^2*e^2*log(c^2*x^2 + 1) - 60*b*c*d*e^3*arctan(c*x) - 6* 
b*e^4*log(c^2*x^2 + 1))/c^5
 

Mupad [B] (verification not implemented)

Time = 1.42 (sec) , antiderivative size = 273, normalized size of antiderivative = 1.48 \[ \int (d+e x)^4 (a+b \arctan (c x)) \, dx=\frac {a\,e^4\,x^5}{5}+a\,d^4\,x-\frac {b\,d^4\,\ln \left (c^2\,x^2+1\right )}{2\,c}-\frac {b\,e^4\,\ln \left (c^2\,x^2+1\right )}{10\,c^5}+2\,a\,d^2\,e^2\,x^3-\frac {b\,e^4\,x^4}{20\,c}+\frac {b\,e^4\,x^2}{10\,c^3}+b\,d^4\,x\,\mathrm {atan}\left (c\,x\right )+2\,a\,d^3\,e\,x^2+a\,d\,e^3\,x^4+\frac {b\,e^4\,x^5\,\mathrm {atan}\left (c\,x\right )}{5}-\frac {2\,b\,d^3\,e\,x}{c}+\frac {b\,d\,e^3\,x}{c^3}+\frac {2\,b\,d^3\,e\,\mathrm {atan}\left (c\,x\right )}{c^2}-\frac {b\,d\,e^3\,\mathrm {atan}\left (c\,x\right )}{c^4}+2\,b\,d^3\,e\,x^2\,\mathrm {atan}\left (c\,x\right )+b\,d\,e^3\,x^4\,\mathrm {atan}\left (c\,x\right )-\frac {b\,d\,e^3\,x^3}{3\,c}+2\,b\,d^2\,e^2\,x^3\,\mathrm {atan}\left (c\,x\right )+\frac {b\,d^2\,e^2\,\ln \left (c^2\,x^2+1\right )}{c^3}-\frac {b\,d^2\,e^2\,x^2}{c} \] Input:

int((a + b*atan(c*x))*(d + e*x)^4,x)
 

Output:

(a*e^4*x^5)/5 + a*d^4*x - (b*d^4*log(c^2*x^2 + 1))/(2*c) - (b*e^4*log(c^2* 
x^2 + 1))/(10*c^5) + 2*a*d^2*e^2*x^3 - (b*e^4*x^4)/(20*c) + (b*e^4*x^2)/(1 
0*c^3) + b*d^4*x*atan(c*x) + 2*a*d^3*e*x^2 + a*d*e^3*x^4 + (b*e^4*x^5*atan 
(c*x))/5 - (2*b*d^3*e*x)/c + (b*d*e^3*x)/c^3 + (2*b*d^3*e*atan(c*x))/c^2 - 
 (b*d*e^3*atan(c*x))/c^4 + 2*b*d^3*e*x^2*atan(c*x) + b*d*e^3*x^4*atan(c*x) 
 - (b*d*e^3*x^3)/(3*c) + 2*b*d^2*e^2*x^3*atan(c*x) + (b*d^2*e^2*log(c^2*x^ 
2 + 1))/c^3 - (b*d^2*e^2*x^2)/c
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 309, normalized size of antiderivative = 1.68 \[ \int (d+e x)^4 (a+b \arctan (c x)) \, dx=\frac {60 \mathit {atan} \left (c x \right ) b \,c^{5} d^{4} x +120 \mathit {atan} \left (c x \right ) b \,c^{5} d^{3} e \,x^{2}+120 \mathit {atan} \left (c x \right ) b \,c^{5} d^{2} e^{2} x^{3}+60 \mathit {atan} \left (c x \right ) b \,c^{5} d \,e^{3} x^{4}+12 \mathit {atan} \left (c x \right ) b \,c^{5} e^{4} x^{5}+120 \mathit {atan} \left (c x \right ) b \,c^{3} d^{3} e -60 \mathit {atan} \left (c x \right ) b c d \,e^{3}-30 \,\mathrm {log}\left (c^{2} x^{2}+1\right ) b \,c^{4} d^{4}+60 \,\mathrm {log}\left (c^{2} x^{2}+1\right ) b \,c^{2} d^{2} e^{2}-6 \,\mathrm {log}\left (c^{2} x^{2}+1\right ) b \,e^{4}+60 a \,c^{5} d^{4} x +120 a \,c^{5} d^{3} e \,x^{2}+120 a \,c^{5} d^{2} e^{2} x^{3}+60 a \,c^{5} d \,e^{3} x^{4}+12 a \,c^{5} e^{4} x^{5}-120 b \,c^{4} d^{3} e x -60 b \,c^{4} d^{2} e^{2} x^{2}-20 b \,c^{4} d \,e^{3} x^{3}-3 b \,c^{4} e^{4} x^{4}+60 b \,c^{2} d \,e^{3} x +6 b \,c^{2} e^{4} x^{2}}{60 c^{5}} \] Input:

int((e*x+d)^4*(a+b*atan(c*x)),x)
 

Output:

(60*atan(c*x)*b*c**5*d**4*x + 120*atan(c*x)*b*c**5*d**3*e*x**2 + 120*atan( 
c*x)*b*c**5*d**2*e**2*x**3 + 60*atan(c*x)*b*c**5*d*e**3*x**4 + 12*atan(c*x 
)*b*c**5*e**4*x**5 + 120*atan(c*x)*b*c**3*d**3*e - 60*atan(c*x)*b*c*d*e**3 
 - 30*log(c**2*x**2 + 1)*b*c**4*d**4 + 60*log(c**2*x**2 + 1)*b*c**2*d**2*e 
**2 - 6*log(c**2*x**2 + 1)*b*e**4 + 60*a*c**5*d**4*x + 120*a*c**5*d**3*e*x 
**2 + 120*a*c**5*d**2*e**2*x**3 + 60*a*c**5*d*e**3*x**4 + 12*a*c**5*e**4*x 
**5 - 120*b*c**4*d**3*e*x - 60*b*c**4*d**2*e**2*x**2 - 20*b*c**4*d*e**3*x* 
*3 - 3*b*c**4*e**4*x**4 + 60*b*c**2*d*e**3*x + 6*b*c**2*e**4*x**2)/(60*c** 
5)