Integrand size = 14, antiderivative size = 207 \[ \int \frac {e^{n \arctan (a+b x)}}{x^3} \, dx=-\frac {(1-i a-i b x)^{1+\frac {i n}{2}} (1+i a+i b x)^{1-\frac {i n}{2}}}{2 \left (1+a^2\right ) x^2}-\frac {2 b^2 (2 a-n) (1-i a-i b x)^{1+\frac {i n}{2}} (1+i a+i b x)^{-1-\frac {i n}{2}} \operatorname {Hypergeometric2F1}\left (2,1+\frac {i n}{2},2+\frac {i n}{2},\frac {(i-a) (1-i a-i b x)}{(i+a) (1+i a+i b x)}\right )}{(i-a) (i+a)^3 (2 i-n)} \]
-1/2*(1-I*a-I*b*x)^(1+1/2*I*n)*(1+I*a+I*b*x)^(1-1/2*I*n)/(a^2+1)/x^2-2*b^2 *(2*a-n)*(1-I*a-I*b*x)^(1+1/2*I*n)*(1+I*a+I*b*x)^(-1-1/2*I*n)*hypergeom([2 , 1+1/2*I*n],[2+1/2*I*n],(I-a)*(1-I*a-I*b*x)/(I+a)/(1+I*a+I*b*x))/(I-a)/(I +a)^3/(2*I-n)
Time = 0.05 (sec) , antiderivative size = 173, normalized size of antiderivative = 0.84 \[ \int \frac {e^{n \arctan (a+b x)}}{x^3} \, dx=-\frac {i (1+i a+i b x)^{-\frac {i n}{2}} (-i (i+a+b x))^{1+\frac {i n}{2}} \left ((i+a)^2 (-2 i+n) (-i+a+b x)^2+4 b^2 (-2 a+n) x^2 \operatorname {Hypergeometric2F1}\left (2,1+\frac {i n}{2},2+\frac {i n}{2},\frac {1+a^2-i b x+a b x}{1+a^2+i b x+a b x}\right )\right )}{2 (-i+a) (i+a)^3 (-2 i+n) x^2 (-i+a+b x)} \]
((-1/2*I)*((-I)*(I + a + b*x))^(1 + (I/2)*n)*((I + a)^2*(-2*I + n)*(-I + a + b*x)^2 + 4*b^2*(-2*a + n)*x^2*Hypergeometric2F1[2, 1 + (I/2)*n, 2 + (I/ 2)*n, (1 + a^2 - I*b*x + a*b*x)/(1 + a^2 + I*b*x + a*b*x)]))/((-I + a)*(I + a)^3*(-2*I + n)*x^2*(1 + I*a + I*b*x)^((I/2)*n)*(-I + a + b*x))
Time = 0.34 (sec) , antiderivative size = 205, normalized size of antiderivative = 0.99, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {5618, 107, 141}
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.
\(\displaystyle \int \frac {e^{n \arctan (a+b x)}}{x^3} \, dx\) |
\(\Big \downarrow \) 5618 |
\(\displaystyle \int \frac {(-i a-i b x+1)^{\frac {i n}{2}} (i a+i b x+1)^{-\frac {i n}{2}}}{x^3}dx\) |
\(\Big \downarrow \) 107 |
\(\displaystyle -\frac {b (2 a-n) \int \frac {(-i a-i b x+1)^{\frac {i n}{2}} (i a+i b x+1)^{-\frac {i n}{2}}}{x^2}dx}{2 \left (a^2+1\right )}-\frac {(-i a-i b x+1)^{1+\frac {i n}{2}} (i a+i b x+1)^{1-\frac {i n}{2}}}{2 \left (a^2+1\right ) x^2}\) |
\(\Big \downarrow \) 141 |
\(\displaystyle \frac {2 b^2 (2 a-n) (-i a-i b x+1)^{1+\frac {i n}{2}} (i a+i b x+1)^{-1-\frac {i n}{2}} \operatorname {Hypergeometric2F1}\left (2,\frac {i n}{2}+1,\frac {i n}{2}+2,\frac {(i-a) (-i a-i b x+1)}{(a+i) (i a+i b x+1)}\right )}{(a+i)^2 \left (a^2+1\right ) (-n+2 i)}-\frac {(-i a-i b x+1)^{1+\frac {i n}{2}} (i a+i b x+1)^{1-\frac {i n}{2}}}{2 \left (a^2+1\right ) x^2}\) |
-1/2*((1 - I*a - I*b*x)^(1 + (I/2)*n)*(1 + I*a + I*b*x)^(1 - (I/2)*n))/((1 + a^2)*x^2) + (2*b^2*(2*a - n)*(1 - I*a - I*b*x)^(1 + (I/2)*n)*(1 + I*a + I*b*x)^(-1 - (I/2)*n)*Hypergeometric2F1[2, 1 + (I/2)*n, 2 + (I/2)*n, ((I - a)*(1 - I*a - I*b*x))/((I + a)*(1 + I*a + I*b*x))])/((I + a)^2*(1 + a^2) *(2*I - n))
3.3.43.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Simp[b*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1 )/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + Simp[(a*d*f*(m + 1) + b*c*f*(n + 1) + b*d*e*(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)) Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x ] && EqQ[Simplify[m + n + p + 3], 0] && (LtQ[m, -1] || SumSimplerQ[m, 1])
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Simp[(b*c - a*d)^n*((a + b*x)^(m + 1)/((m + 1)*(b*e - a*f)^( n + 1)*(e + f*x)^(m + 1)))*Hypergeometric2F1[m + 1, -n, m + 2, (-(d*e - c*f ))*((a + b*x)/((b*c - a*d)*(e + f*x)))], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[m + n + p + 2, 0] && ILtQ[n, 0] && (SumSimplerQ[m, 1] || !Su mSimplerQ[p, 1]) && !ILtQ[m, 0]
Int[E^(ArcTan[(c_.)*((a_) + (b_.)*(x_))]*(n_.))*((d_.) + (e_.)*(x_))^(m_.), x_Symbol] :> Int[(d + e*x)^m*((1 - I*a*c - I*b*c*x)^(I*(n/2))/(1 + I*a*c + I*b*c*x)^(I*(n/2))), x] /; FreeQ[{a, b, c, d, e, m, n}, x]
\[\int \frac {{\mathrm e}^{n \arctan \left (b x +a \right )}}{x^{3}}d x\]
\[ \int \frac {e^{n \arctan (a+b x)}}{x^3} \, dx=\int { \frac {e^{\left (n \arctan \left (b x + a\right )\right )}}{x^{3}} \,d x } \]
\[ \int \frac {e^{n \arctan (a+b x)}}{x^3} \, dx=\int \frac {e^{n \operatorname {atan}{\left (a + b x \right )}}}{x^{3}}\, dx \]
\[ \int \frac {e^{n \arctan (a+b x)}}{x^3} \, dx=\int { \frac {e^{\left (n \arctan \left (b x + a\right )\right )}}{x^{3}} \,d x } \]
Timed out. \[ \int \frac {e^{n \arctan (a+b x)}}{x^3} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {e^{n \arctan (a+b x)}}{x^3} \, dx=\int \frac {{\mathrm {e}}^{n\,\mathrm {atan}\left (a+b\,x\right )}}{x^3} \,d x \]