Integrand size = 16, antiderivative size = 97 \[ \int (e+f x) (a+b \arctan (c+d x)) \, dx=-\frac {b f x}{2 d}-\frac {b (d e+f-c f) (d e-(1+c) f) \arctan (c+d x)}{2 d^2 f}+\frac {(e+f x)^2 (a+b \arctan (c+d x))}{2 f}-\frac {b (d e-c f) \log \left (1+(c+d x)^2\right )}{2 d^2} \]
-1/2*b*f*x/d-1/2*b*(-c*f+d*e+f)*(d*e-(1+c)*f)*arctan(d*x+c)/d^2/f+1/2*(f*x +e)^2*(a+b*arctan(d*x+c))/f-1/2*b*(-c*f+d*e)*ln(1+(d*x+c)^2)/d^2
Result contains complex when optimal does not.
Time = 0.05 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.68 \[ \int (e+f x) (a+b \arctan (c+d x)) \, dx=a e x+\frac {1}{2} a f x^2+b e x \arctan (c+d x)+\frac {b f \left (\frac {1}{2} d \left (-\frac {c}{d}+\frac {c+d x}{d}\right )^2 \arctan (c+d x)-\frac {1}{2} d \left (\frac {x}{d}-\frac {i (i-c)^2 \log (i-c-d x)}{2 d^2}+\frac {i (i+c)^2 \log (i+c+d x)}{2 d^2}\right )\right )}{d}-\frac {b e \left (-2 c \arctan (c+d x)+\log \left (1+c^2+2 c d x+d^2 x^2\right )\right )}{2 d} \]
a*e*x + (a*f*x^2)/2 + b*e*x*ArcTan[c + d*x] + (b*f*((d*(-(c/d) + (c + d*x) /d)^2*ArcTan[c + d*x])/2 - (d*(x/d - ((I/2)*(I - c)^2*Log[I - c - d*x])/d^ 2 + ((I/2)*(I + c)^2*Log[I + c + d*x])/d^2))/2))/d - (b*e*(-2*c*ArcTan[c + d*x] + Log[1 + c^2 + 2*c*d*x + d^2*x^2]))/(2*d)
Time = 0.32 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.06, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {5570, 27, 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.
\(\displaystyle \int (e+f x) (a+b \arctan (c+d x)) \, dx\) |
\(\Big \downarrow \) 5570 |
\(\displaystyle \frac {\int \frac {\left (d \left (e-\frac {c f}{d}\right )+f (c+d x)\right ) (a+b \arctan (c+d x))}{d}d(c+d x)}{d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int (d e-c f+f (c+d x)) (a+b \arctan (c+d x))d(c+d x)}{d^2}\) |
\(\Big \downarrow \) 5387 |
\(\displaystyle \frac {\frac {(f (c+d x)-c f+d e)^2 (a+b \arctan (c+d x))}{2 f}-\frac {b \int \frac {(d e-c f+f (c+d x))^2}{(c+d x)^2+1}d(c+d x)}{2 f}}{d^2}\) |
\(\Big \downarrow \) 478 |
\(\displaystyle \frac {\frac {(f (c+d x)-c f+d e)^2 (a+b \arctan (c+d x))}{2 f}-\frac {b \int \left (f^2+\frac {(d e-c f-f) (d e-c f+f)+2 f (d e-c f) (c+d x)}{(c+d x)^2+1}\right )d(c+d x)}{2 f}}{d^2}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {(f (c+d x)-c f+d e)^2 (a+b \arctan (c+d x))}{2 f}-\frac {b \left (\arctan (c+d x) (-c f+d e+f) (d e-(c+1) f)+f (d e-c f) \log \left ((c+d x)^2+1\right )+f^2 (c+d x)\right )}{2 f}}{d^2}\) |
(((d*e - c*f + f*(c + d*x))^2*(a + b*ArcTan[c + d*x]))/(2*f) - (b*(f^2*(c + d*x) + (d*e + f - c*f)*(d*e - (1 + c)*f)*ArcTan[c + d*x] + f*(d*e - c*f) *Log[1 + (c + d*x)^2]))/(2*f))/d^2
3.1.26.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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]
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]
Int[((a_.) + ArcTan[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^(m _.), x_Symbol] :> Simp[1/d Subst[Int[((d*e - c*f)/d + f*(x/d))^m*(a + b*A rcTan[x])^p, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && I GtQ[p, 0]
Time = 0.12 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.16
method | result | size |
parts | \(a \left (\frac {1}{2} f \,x^{2}+e x \right )+\frac {b \left (\frac {\arctan \left (d x +c \right ) \left (d x +c \right )^{2} f}{2 d}-\frac {\arctan \left (d x +c \right ) c f \left (d x +c \right )}{d}+\arctan \left (d x +c \right ) e \left (d x +c \right )-\frac {f \left (d x +c \right )+\frac {\left (-2 c f +2 d e \right ) \ln \left (1+\left (d x +c \right )^{2}\right )}{2}-f \arctan \left (d x +c \right )}{2 d}\right )}{d}\) | \(113\) |
derivativedivides | \(\frac {-\frac {a \left (f c \left (d x +c \right )-e d \left (d x +c \right )-\frac {f \left (d x +c \right )^{2}}{2}\right )}{d}-\frac {b \left (\arctan \left (d x +c \right ) f c \left (d x +c \right )-\arctan \left (d x +c \right ) e d \left (d x +c \right )-\frac {\arctan \left (d x +c \right ) f \left (d x +c \right )^{2}}{2}+\frac {f \left (d x +c \right )}{2}-\frac {\left (2 c f -2 d e \right ) \ln \left (1+\left (d x +c \right )^{2}\right )}{4}-\frac {f \arctan \left (d x +c \right )}{2}\right )}{d}}{d}\) | \(130\) |
default | \(\frac {-\frac {a \left (f c \left (d x +c \right )-e d \left (d x +c \right )-\frac {f \left (d x +c \right )^{2}}{2}\right )}{d}-\frac {b \left (\arctan \left (d x +c \right ) f c \left (d x +c \right )-\arctan \left (d x +c \right ) e d \left (d x +c \right )-\frac {\arctan \left (d x +c \right ) f \left (d x +c \right )^{2}}{2}+\frac {f \left (d x +c \right )}{2}-\frac {\left (2 c f -2 d e \right ) \ln \left (1+\left (d x +c \right )^{2}\right )}{4}-\frac {f \arctan \left (d x +c \right )}{2}\right )}{d}}{d}\) | \(130\) |
parallelrisch | \(\frac {\arctan \left (d x +c \right ) b \,d^{2} f \,x^{2}+a \,d^{2} f \,x^{2}+2 x \arctan \left (d x +c \right ) b \,d^{2} e +2 a \,d^{2} e x -\arctan \left (d x +c \right ) b \,c^{2} f +2 b c d e \arctan \left (d x +c \right )+b c f \ln \left (d^{2} x^{2}+2 c d x +c^{2}+1\right )-b d e \ln \left (d^{2} x^{2}+2 c d x +c^{2}+1\right )-b d f x -a \,c^{2} f -4 a c d e +\arctan \left (d x +c \right ) b f +2 b c f -a f}{2 d^{2}}\) | \(160\) |
risch | \(-\frac {i b \left (f \,x^{2}+2 e x \right ) \ln \left (1+i \left (d x +c \right )\right )}{4}+\frac {i b f \,x^{2} \ln \left (1-i \left (d x +c \right )\right )}{4}+\frac {i b e x \ln \left (1-i \left (d x +c \right )\right )}{2}+\frac {a f \,x^{2}}{2}-\frac {\arctan \left (d x +c \right ) b \,c^{2} f}{2 d^{2}}+\frac {b c e \arctan \left (d x +c \right )}{d}+a e x +\frac {b c f \ln \left (d^{2} x^{2}+2 c d x +c^{2}+1\right )}{2 d^{2}}-\frac {b e \ln \left (d^{2} x^{2}+2 c d x +c^{2}+1\right )}{2 d}-\frac {b f x}{2 d}+\frac {\arctan \left (d x +c \right ) b f}{2 d^{2}}\) | \(175\) |
a*(1/2*f*x^2+e*x)+b/d*(1/2/d*arctan(d*x+c)*(d*x+c)^2*f-1/d*arctan(d*x+c)*c *f*(d*x+c)+arctan(d*x+c)*e*(d*x+c)-1/2/d*(f*(d*x+c)+1/2*(-2*c*f+2*d*e)*ln( 1+(d*x+c)^2)-f*arctan(d*x+c)))
Time = 0.29 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.06 \[ \int (e+f x) (a+b \arctan (c+d x)) \, dx=\frac {a d^{2} f x^{2} + {\left (2 \, a d^{2} e - b d f\right )} x + {\left (b d^{2} f x^{2} + 2 \, b d^{2} e x + 2 \, b c d e - {\left (b c^{2} - b\right )} f\right )} \arctan \left (d x + c\right ) - {\left (b d e - b c f\right )} \log \left (d^{2} x^{2} + 2 \, c d x + c^{2} + 1\right )}{2 \, d^{2}} \]
1/2*(a*d^2*f*x^2 + (2*a*d^2*e - b*d*f)*x + (b*d^2*f*x^2 + 2*b*d^2*e*x + 2* b*c*d*e - (b*c^2 - b)*f)*arctan(d*x + c) - (b*d*e - b*c*f)*log(d^2*x^2 + 2 *c*d*x + c^2 + 1))/d^2
Result contains complex when optimal does not.
Time = 15.21 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.82 \[ \int (e+f x) (a+b \arctan (c+d x)) \, dx=\begin {cases} a e x + \frac {a f x^{2}}{2} - \frac {b c^{2} f \operatorname {atan}{\left (c + d x \right )}}{2 d^{2}} + \frac {b c e \operatorname {atan}{\left (c + d x \right )}}{d} + \frac {b c f \log {\left (\frac {c}{d} + x - \frac {i}{d} \right )}}{d^{2}} - \frac {i b c f \operatorname {atan}{\left (c + d x \right )}}{d^{2}} + b e x \operatorname {atan}{\left (c + d x \right )} + \frac {b f x^{2} \operatorname {atan}{\left (c + d x \right )}}{2} - \frac {b e \log {\left (\frac {c}{d} + x - \frac {i}{d} \right )}}{d} + \frac {i b e \operatorname {atan}{\left (c + d x \right )}}{d} - \frac {b f x}{2 d} + \frac {b f \operatorname {atan}{\left (c + d x \right )}}{2 d^{2}} & \text {for}\: d \neq 0 \\\left (a + b \operatorname {atan}{\left (c \right )}\right ) \left (e x + \frac {f x^{2}}{2}\right ) & \text {otherwise} \end {cases} \]
Piecewise((a*e*x + a*f*x**2/2 - b*c**2*f*atan(c + d*x)/(2*d**2) + b*c*e*at an(c + d*x)/d + b*c*f*log(c/d + x - I/d)/d**2 - I*b*c*f*atan(c + d*x)/d**2 + b*e*x*atan(c + d*x) + b*f*x**2*atan(c + d*x)/2 - b*e*log(c/d + x - I/d) /d + I*b*e*atan(c + d*x)/d - b*f*x/(2*d) + b*f*atan(c + d*x)/(2*d**2), Ne( d, 0)), ((a + b*atan(c))*(e*x + f*x**2/2), True))
Time = 0.29 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.20 \[ \int (e+f x) (a+b \arctan (c+d x)) \, dx=\frac {1}{2} \, a f x^{2} + \frac {1}{2} \, {\left (x^{2} \arctan \left (d x + c\right ) - d {\left (\frac {x}{d^{2}} + \frac {{\left (c^{2} - 1\right )} \arctan \left (\frac {d^{2} x + c d}{d}\right )}{d^{3}} - \frac {c \log \left (d^{2} x^{2} + 2 \, c d x + c^{2} + 1\right )}{d^{3}}\right )}\right )} b f + a e x + \frac {{\left (2 \, {\left (d x + c\right )} \arctan \left (d x + c\right ) - \log \left ({\left (d x + c\right )}^{2} + 1\right )\right )} b e}{2 \, d} \]
1/2*a*f*x^2 + 1/2*(x^2*arctan(d*x + c) - d*(x/d^2 + (c^2 - 1)*arctan((d^2* x + c*d)/d)/d^3 - c*log(d^2*x^2 + 2*c*d*x + c^2 + 1)/d^3))*b*f + a*e*x + 1 /2*(2*(d*x + c)*arctan(d*x + c) - log((d*x + c)^2 + 1))*b*e/d
\[ \int (e+f x) (a+b \arctan (c+d x)) \, dx=\int { {\left (f x + e\right )} {\left (b \arctan \left (d x + c\right ) + a\right )} \,d x } \]
Time = 1.93 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.40 \[ \int (e+f x) (a+b \arctan (c+d x)) \, dx=a\,e\,x+\frac {a\,f\,x^2}{2}-\frac {b\,e\,\ln \left (c^2+2\,c\,d\,x+d^2\,x^2+1\right )}{2\,d}+\frac {b\,f\,\mathrm {atan}\left (c+d\,x\right )}{2\,d^2}+\frac {b\,f\,x^2\,\mathrm {atan}\left (c+d\,x\right )}{2}-\frac {b\,f\,x}{2\,d}+b\,e\,x\,\mathrm {atan}\left (c+d\,x\right )-\frac {b\,c^2\,f\,\mathrm {atan}\left (c+d\,x\right )}{2\,d^2}+\frac {b\,c\,f\,\ln \left (c^2+2\,c\,d\,x+d^2\,x^2+1\right )}{2\,d^2}+\frac {b\,c\,e\,\mathrm {atan}\left (c+d\,x\right )}{d} \]