Integrand size = 26, antiderivative size = 291 \[ \int \frac {e^{n \arctan (a x)} x^2}{\sqrt {c+a^2 c x^2}} \, dx=-\frac {(1+i n) (1-i a x)^{\frac {1}{2} (1+i n)} (1+i a x)^{\frac {1}{2} (1-i n)} \sqrt {1+a^2 x^2}}{2 a^3 (i+n) \sqrt {c+a^2 c x^2}}+\frac {x (1-i a x)^{\frac {1}{2} (1+i n)} (1+i a x)^{\frac {1}{2} (1-i n)} \sqrt {1+a^2 x^2}}{2 a^2 \sqrt {c+a^2 c x^2}}-\frac {i 2^{\frac {1}{2}-\frac {i n}{2}} \left (1-n^2\right ) (1-i a x)^{\frac {1}{2} (1+i n)} \sqrt {1+a^2 x^2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (-1+i n),\frac {1}{2} (1+i n),\frac {1}{2} (3+i n),\frac {1}{2} (1-i a x)\right )}{a^3 \left (1+n^2\right ) \sqrt {c+a^2 c x^2}} \]
-1/2*(1+I*n)*(1-I*a*x)^(1/2+1/2*I*n)*(1+I*a*x)^(1/2-1/2*I*n)*(a^2*x^2+1)^( 1/2)/a^3/(I+n)/(a^2*c*x^2+c)^(1/2)+1/2*x*(1-I*a*x)^(1/2+1/2*I*n)*(1+I*a*x) ^(1/2-1/2*I*n)*(a^2*x^2+1)^(1/2)/a^2/(a^2*c*x^2+c)^(1/2)-I*2^(1/2-1/2*I*n) *(-n^2+1)*(1-I*a*x)^(1/2+1/2*I*n)*hypergeom([-1/2+1/2*I*n, 1/2+1/2*I*n],[3 /2+1/2*I*n],1/2-1/2*I*a*x)*(a^2*x^2+1)^(1/2)/a^3/(n^2+1)/(a^2*c*x^2+c)^(1/ 2)
Time = 0.14 (sec) , antiderivative size = 206, normalized size of antiderivative = 0.71 \[ \int \frac {e^{n \arctan (a x)} x^2}{\sqrt {c+a^2 c x^2}} \, dx=\frac {2^{-1-\frac {i n}{2}} (1-i a x)^{\frac {1}{2}+\frac {i n}{2}} (1+i a x)^{-\frac {i n}{2}} \sqrt {1+a^2 x^2} \left (2^{\frac {i n}{2}} (-i+n) \sqrt {1+i a x} (-1+i a x+n (-i+a x))+2 i \sqrt {2} \left (-1+n^2\right ) (1+i a x)^{\frac {i n}{2}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (1+i n),\frac {1}{2} i (i+n),\frac {1}{2} (3+i n),\frac {1}{2} (1-i a x)\right )\right )}{a^3 \left (1+n^2\right ) \sqrt {c+a^2 c x^2}} \]
(2^(-1 - (I/2)*n)*(1 - I*a*x)^(1/2 + (I/2)*n)*Sqrt[1 + a^2*x^2]*(2^((I/2)* n)*(-I + n)*Sqrt[1 + I*a*x]*(-1 + I*a*x + n*(-I + a*x)) + (2*I)*Sqrt[2]*(- 1 + n^2)*(1 + I*a*x)^((I/2)*n)*Hypergeometric2F1[(1 + I*n)/2, (I/2)*(I + n ), (3 + I*n)/2, (1 - I*a*x)/2]))/(a^3*(1 + n^2)*(1 + I*a*x)^((I/2)*n)*Sqrt [c + a^2*c*x^2])
Time = 0.60 (sec) , antiderivative size = 252, normalized size of antiderivative = 0.87, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {5608, 5605, 101, 25, 88, 79}
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.
\(\displaystyle \int \frac {x^2 e^{n \arctan (a x)}}{\sqrt {a^2 c x^2+c}} \, dx\) |
\(\Big \downarrow \) 5608 |
\(\displaystyle \frac {\sqrt {a^2 x^2+1} \int \frac {e^{n \arctan (a x)} x^2}{\sqrt {a^2 x^2+1}}dx}{\sqrt {a^2 c x^2+c}}\) |
\(\Big \downarrow \) 5605 |
\(\displaystyle \frac {\sqrt {a^2 x^2+1} \int x^2 (1-i a x)^{\frac {1}{2} (i n-1)} (i a x+1)^{\frac {1}{2} (-i n-1)}dx}{\sqrt {a^2 c x^2+c}}\) |
\(\Big \downarrow \) 101 |
\(\displaystyle \frac {\sqrt {a^2 x^2+1} \left (\frac {\int -(1-i a x)^{\frac {1}{2} (i n-1)} (i a x+1)^{\frac {1}{2} (-i n-1)} (a n x+1)dx}{2 a^2}+\frac {x (1-i a x)^{\frac {1}{2} (1+i n)} (1+i a x)^{\frac {1}{2} (1-i n)}}{2 a^2}\right )}{\sqrt {a^2 c x^2+c}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\sqrt {a^2 x^2+1} \left (\frac {x (1-i a x)^{\frac {1}{2} (1+i n)} (1+i a x)^{\frac {1}{2} (1-i n)}}{2 a^2}-\frac {\int (1-i a x)^{\frac {1}{2} (i n-1)} (i a x+1)^{\frac {1}{2} (-i n-1)} (a n x+1)dx}{2 a^2}\right )}{\sqrt {a^2 c x^2+c}}\) |
\(\Big \downarrow \) 88 |
\(\displaystyle \frac {\sqrt {a^2 x^2+1} \left (\frac {x (1-i a x)^{\frac {1}{2} (1+i n)} (1+i a x)^{\frac {1}{2} (1-i n)}}{2 a^2}-\frac {\frac {\left (1-n^2\right ) \int (1-i a x)^{\frac {1}{2} (i n-1)} (i a x+1)^{\frac {1}{2} (1-i n)}dx}{1-i n}+\frac {(1+i n) (1-i a x)^{\frac {1}{2} (1+i n)} (1+i a x)^{\frac {1}{2} (1-i n)}}{a (n+i)}}{2 a^2}\right )}{\sqrt {a^2 c x^2+c}}\) |
\(\Big \downarrow \) 79 |
\(\displaystyle \frac {\sqrt {a^2 x^2+1} \left (\frac {x (1-i a x)^{\frac {1}{2} (1+i n)} (1+i a x)^{\frac {1}{2} (1-i n)}}{2 a^2}-\frac {\frac {(1+i n) (1-i a x)^{\frac {1}{2} (1+i n)} (1+i a x)^{\frac {1}{2} (1-i n)}}{a (n+i)}-\frac {2^{\frac {3}{2}-\frac {i n}{2}} \left (1-n^2\right ) (1-i a x)^{\frac {1}{2} (1+i n)} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (i n-1),\frac {1}{2} (i n+1),\frac {1}{2} (i n+3),\frac {1}{2} (1-i a x)\right )}{a (-n+i) (1-i n)}}{2 a^2}\right )}{\sqrt {a^2 c x^2+c}}\) |
(Sqrt[1 + a^2*x^2]*((x*(1 - I*a*x)^((1 + I*n)/2)*(1 + I*a*x)^((1 - I*n)/2) )/(2*a^2) - (((1 + I*n)*(1 - I*a*x)^((1 + I*n)/2)*(1 + I*a*x)^((1 - I*n)/2 ))/(a*(I + n)) - (2^(3/2 - (I/2)*n)*(1 - n^2)*(1 - I*a*x)^((1 + I*n)/2)*Hy pergeometric2F1[(-1 + I*n)/2, (1 + I*n)/2, (3 + I*n)/2, (1 - I*a*x)/2])/(a *(I - n)*(1 - I*n)))/(2*a^2)))/Sqrt[c + a^2*c*x^2]
3.4.54.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(( a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c - a*d))^n))*Hypergeometric2F1[-n, m + 1 , m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m, n}, x] && !IntegerQ[m] && !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] || !(RationalQ[n] && GtQ[-d/(b*c - a*d), 0]))
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)) Int[(c + d*x)^n*(e + f*x)^Simplify[p + 1], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && !RationalQ[p] && SumSimpl erQ[p, 1]
Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^( p_), x_] :> Simp[b*(a + b*x)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 3))), x] + Simp[1/(d*f*(n + p + 3)) Int[(c + d*x)^n*(e + f*x)^p*Simp [a^2*d*f*(n + p + 3) - b*(b*c*e + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f *(n + p + 4) - b*(d*e*(n + 2) + c*f*(p + 2)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 3, 0]
Int[E^(ArcTan[(a_.)*(x_)]*(n_.))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_.), x_ Symbol] :> Simp[c^p Int[x^m*(1 - I*a*x)^(p + I*(n/2))*(1 + I*a*x)^(p - I* (n/2)), x], x] /; FreeQ[{a, c, d, m, n, p}, x] && EqQ[d, a^2*c] && (Integer Q[p] || GtQ[c, 0])
Int[E^(ArcTan[(a_.)*(x_)]*(n_.))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_), x_S ymbol] :> Simp[c^IntPart[p]*((c + d*x^2)^FracPart[p]/(1 + a^2*x^2)^FracPart [p]) Int[x^m*(1 + a^2*x^2)^p*E^(n*ArcTan[a*x]), x], x] /; FreeQ[{a, c, d, m, n, p}, x] && EqQ[d, a^2*c] && !(IntegerQ[p] || GtQ[c, 0])
\[\int \frac {{\mathrm e}^{n \arctan \left (a x \right )} x^{2}}{\sqrt {a^{2} c \,x^{2}+c}}d x\]
\[ \int \frac {e^{n \arctan (a x)} x^2}{\sqrt {c+a^2 c x^2}} \, dx=\int { \frac {x^{2} e^{\left (n \arctan \left (a x\right )\right )}}{\sqrt {a^{2} c x^{2} + c}} \,d x } \]
\[ \int \frac {e^{n \arctan (a x)} x^2}{\sqrt {c+a^2 c x^2}} \, dx=\int \frac {x^{2} e^{n \operatorname {atan}{\left (a x \right )}}}{\sqrt {c \left (a^{2} x^{2} + 1\right )}}\, dx \]
\[ \int \frac {e^{n \arctan (a x)} x^2}{\sqrt {c+a^2 c x^2}} \, dx=\int { \frac {x^{2} e^{\left (n \arctan \left (a x\right )\right )}}{\sqrt {a^{2} c x^{2} + c}} \,d x } \]
\[ \int \frac {e^{n \arctan (a x)} x^2}{\sqrt {c+a^2 c x^2}} \, dx=\int { \frac {x^{2} e^{\left (n \arctan \left (a x\right )\right )}}{\sqrt {a^{2} c x^{2} + c}} \,d x } \]
Timed out. \[ \int \frac {e^{n \arctan (a x)} x^2}{\sqrt {c+a^2 c x^2}} \, dx=\int \frac {x^2\,{\mathrm {e}}^{n\,\mathrm {atan}\left (a\,x\right )}}{\sqrt {c\,a^2\,x^2+c}} \,d x \]