Integrand size = 18, antiderivative size = 177 \[ \int (e+f x)^m (a+b \arctan (c+d x)) \, dx=\frac {(e+f x)^{1+m} (a+b \arctan (c+d x))}{f (1+m)}-\frac {i b d (e+f x)^{2+m} \operatorname {Hypergeometric2F1}\left (1,2+m,3+m,\frac {d (e+f x)}{d e+i f-c f}\right )}{2 f (d e+(i-c) f) (1+m) (2+m)}+\frac {i b d (e+f x)^{2+m} \operatorname {Hypergeometric2F1}\left (1,2+m,3+m,\frac {d (e+f x)}{d e-(i+c) f}\right )}{2 f (d e-(i+c) f) (1+m) (2+m)} \]
(f*x+e)^(1+m)*(a+b*arctan(d*x+c))/f/(1+m)-1/2*I*b*d*(f*x+e)^(2+m)*hypergeo m([1, 2+m],[3+m],d*(f*x+e)/(d*e+I*f-c*f))/f/(d*e+(I-c)*f)/(1+m)/(2+m)+1/2* I*b*d*(f*x+e)^(2+m)*hypergeom([1, 2+m],[3+m],d*(f*x+e)/(d*e-(I+c)*f))/f/(d *e-(I+c)*f)/(1+m)/(2+m)
Time = 0.33 (sec) , antiderivative size = 162, normalized size of antiderivative = 0.92 \[ \int (e+f x)^m (a+b \arctan (c+d x)) \, dx=\frac {(e+f x)^{1+m} \left (2 (a+b \arctan (c+d x))+\frac {b d (e+f x) \left ((d e-(i+c) f) \operatorname {Hypergeometric2F1}\left (1,2+m,3+m,\frac {d (e+f x)}{d e-(-i+c) f}\right )+(-d e+(-i+c) f) \operatorname {Hypergeometric2F1}\left (1,2+m,3+m,\frac {d (e+f x)}{d e-(i+c) f}\right )\right )}{(i d e+f-i c f) (d e-(-i+c) f) (2+m)}\right )}{2 f (1+m)} \]
((e + f*x)^(1 + m)*(2*(a + b*ArcTan[c + d*x]) + (b*d*(e + f*x)*((d*e - (I + c)*f)*Hypergeometric2F1[1, 2 + m, 3 + m, (d*(e + f*x))/(d*e - (-I + c)*f )] + (-(d*e) + (-I + c)*f)*Hypergeometric2F1[1, 2 + m, 3 + m, (d*(e + f*x) )/(d*e - (I + c)*f)]))/((I*d*e + f - I*c*f)*(d*e - (-I + c)*f)*(2 + m))))/ (2*f*(1 + m))
Time = 0.45 (sec) , antiderivative size = 237, normalized size of antiderivative = 1.34, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {5570, 5387, 485, 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.
\(\displaystyle \int (e+f x)^m (a+b \arctan (c+d x)) \, dx\) |
\(\Big \downarrow \) 5570 |
\(\displaystyle \frac {\int \left (e-\frac {c f}{d}+\frac {f (c+d x)}{d}\right )^m (a+b \arctan (c+d x))d(c+d x)}{d}\) |
\(\Big \downarrow \) 5387 |
\(\displaystyle \frac {\frac {d (a+b \arctan (c+d x)) \left (\frac {f (c+d x)}{d}-\frac {c f}{d}+e\right )^{m+1}}{f (m+1)}-\frac {b d \int \frac {\left (e-\frac {c f}{d}+\frac {f (c+d x)}{d}\right )^{m+1}}{(c+d x)^2+1}d(c+d x)}{f (m+1)}}{d}\) |
\(\Big \downarrow \) 485 |
\(\displaystyle \frac {\frac {d (a+b \arctan (c+d x)) \left (\frac {f (c+d x)}{d}-\frac {c f}{d}+e\right )^{m+1}}{f (m+1)}-\frac {b d \int \left (\frac {i \left (e-\frac {c f}{d}+\frac {f (c+d x)}{d}\right )^{m+1}}{2 (-c-d x+i)}+\frac {i \left (e-\frac {c f}{d}+\frac {f (c+d x)}{d}\right )^{m+1}}{2 (c+d x+i)}\right )d(c+d x)}{f (m+1)}}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {d (a+b \arctan (c+d x)) \left (\frac {f (c+d x)}{d}-\frac {c f}{d}+e\right )^{m+1}}{f (m+1)}-\frac {b d \left (\frac {i d \left (\frac {f (c+d x)}{d}-\frac {c f}{d}+e\right )^{m+2} \operatorname {Hypergeometric2F1}\left (1,m+2,m+3,\frac {d e-c f+f (c+d x)}{d e-c f+i f}\right )}{2 (m+2) (d e+(-c+i) f)}-\frac {i d \left (\frac {f (c+d x)}{d}-\frac {c f}{d}+e\right )^{m+2} \operatorname {Hypergeometric2F1}\left (1,m+2,m+3,\frac {d e-c f+f (c+d x)}{d e-(c+i) f}\right )}{2 (m+2) (d e-(c+i) f)}\right )}{f (m+1)}}{d}\) |
((d*(e - (c*f)/d + (f*(c + d*x))/d)^(1 + m)*(a + b*ArcTan[c + d*x]))/(f*(1 + m)) - (b*d*(((I/2)*d*(e - (c*f)/d + (f*(c + d*x))/d)^(2 + m)*Hypergeome tric2F1[1, 2 + m, 3 + m, (d*e - c*f + f*(c + d*x))/(d*e + I*f - c*f)])/((d *e + (I - c)*f)*(2 + m)) - ((I/2)*d*(e - (c*f)/d + (f*(c + d*x))/d)^(2 + m )*Hypergeometric2F1[1, 2 + m, 3 + m, (d*e - c*f + f*(c + d*x))/(d*e - (I + c)*f)])/((d*e - (I + c)*f)*(2 + m))))/(f*(1 + m)))/d
3.1.41.3.1 Defintions of rubi rules used
Int[((c_) + (d_.)*(x_))^(n_)/((a_) + (b_.)*(x_)^2), x_Symbol] :> Int[Expand Integrand[(c + d*x)^n, 1/(a + b*x^2), x], x] /; FreeQ[{a, b, c, d, n}, x] & & !IntegerQ[2*n]
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]
\[\int \left (f x +e \right )^{m} \left (a +b \arctan \left (d x +c \right )\right )d x\]
\[ \int (e+f x)^m (a+b \arctan (c+d x)) \, dx=\int { {\left (b \arctan \left (d x + c\right ) + a\right )} {\left (f x + e\right )}^{m} \,d x } \]
Timed out. \[ \int (e+f x)^m (a+b \arctan (c+d x)) \, dx=\text {Timed out} \]
\[ \int (e+f x)^m (a+b \arctan (c+d x)) \, dx=\int { {\left (b \arctan \left (d x + c\right ) + a\right )} {\left (f x + e\right )}^{m} \,d x } \]
1/2*((3*e*m^2 + 2*e*m + (3*f*m^2 + 2*f*m + f)*x + e)*(f*x + e)^m*arctan(d* x + c) + (e*m + (f*m + f)*x + e)*(f*x + e)^m*log(d^2*x^2 + 2*c*d*x + c^2 + 1) + 2*(f*m^3 + f*m^2 + f*m + f)*integrate(-1/2*(2*((c^2 + 1)*f*m^3 + 2*( c^2 + 1)*f*m^2 + (c^2 + 1)*f*m + (d^2*f*m^3 + 2*d^2*f*m^2 + d^2*f*m)*x^2 + 2*(c*d*f*m^3 + 2*c*d*f*m^2 + c*d*f*m)*x)*(f*x + e)^m*arctan(d*x + c) - (( c^2 + 1)*f*m^3 - (c^2 + 1)*f*m + (d^2*f*m^3 - d^2*f*m)*x^2 + 2*(c*d*f*m^3 - c*d*f*m)*x)*(f*x + e)^m*log(d^2*x^2 + 2*c*d*x + c^2 + 1) - 2*((c - 1)*d* e*m^2 - (c + 1)*d*e*m + (d^2*f*m^2 - d^2*f*m)*x^2 + ((d^2*e + (c - 1)*d*f) *m^2 - (d^2*e + (c + 1)*d*f)*m)*x)*(f*x + e)^m)/((c^2 + 1)*f*m^3 + (c^2 + 1)*f*m^2 + (c^2 + 1)*f*m + (d^2*f*m^3 + d^2*f*m^2 + d^2*f*m + d^2*f)*x^2 + (c^2 + 1)*f + 2*(c*d*f*m^3 + c*d*f*m^2 + c*d*f*m + c*d*f)*x), x))*b/(f*m^ 3 + f*m^2 + f*m + f) + (f*x + e)^(m + 1)*a/(f*(m + 1))
\[ \int (e+f x)^m (a+b \arctan (c+d x)) \, dx=\int { {\left (b \arctan \left (d x + c\right ) + a\right )} {\left (f x + e\right )}^{m} \,d x } \]
Timed out. \[ \int (e+f x)^m (a+b \arctan (c+d x)) \, dx=\int {\left (e+f\,x\right )}^m\,\left (a+b\,\mathrm {atan}\left (c+d\,x\right )\right ) \,d x \]