Integrand size = 21, antiderivative size = 104 \[ \int (e x)^m \left (a+\frac {b}{c+d x^2}\right )^p \, dx=\frac {(e x)^{1+m} \left (1+\frac {d x^2}{c}\right )^p \left (1+\frac {a d x^2}{b+a c}\right )^{-p} \left (a+\frac {b}{c+d x^2}\right )^p \operatorname {AppellF1}\left (\frac {1+m}{2},p,-p,\frac {3+m}{2},-\frac {d x^2}{c},-\frac {a d x^2}{b+a c}\right )}{e (1+m)} \] Output:
(e*x)^(1+m)*(1+d*x^2/c)^p*(a+b/(d*x^2+c))^p*AppellF1(1/2+1/2*m,p,-p,3/2+1/ 2*m,-d*x^2/c,-a*d*x^2/(a*c+b))/e/(1+m)/((1+a*d*x^2/(a*c+b))^p)
\[ \int (e x)^m \left (a+\frac {b}{c+d x^2}\right )^p \, dx=\int (e x)^m \left (a+\frac {b}{c+d x^2}\right )^p \, dx \] Input:
Integrate[(e*x)^m*(a + b/(c + d*x^2))^p,x]
Output:
Integrate[(e*x)^m*(a + b/(c + d*x^2))^p, x]
Time = 0.49 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.08, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {2057, 2058, 395, 395, 394}
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.
\(\displaystyle \int (e x)^m \left (a+\frac {b}{c+d x^2}\right )^p \, dx\) |
\(\Big \downarrow \) 2057 |
\(\displaystyle \int (e x)^m \left (\frac {a c+a d x^2+b}{c+d x^2}\right )^pdx\) |
\(\Big \downarrow \) 2058 |
\(\displaystyle \left (c+d x^2\right )^p \left (a c+a d x^2+b\right )^{-p} \left (\frac {a c+a d x^2+b}{c+d x^2}\right )^p \int (e x)^m \left (d x^2+c\right )^{-p} \left (a d x^2+b+a c\right )^pdx\) |
\(\Big \downarrow \) 395 |
\(\displaystyle \left (\frac {d x^2}{c}+1\right )^p \left (a c+a d x^2+b\right )^{-p} \left (\frac {a c+a d x^2+b}{c+d x^2}\right )^p \int (e x)^m \left (a d x^2+b+a c\right )^p \left (\frac {d x^2}{c}+1\right )^{-p}dx\) |
\(\Big \downarrow \) 395 |
\(\displaystyle \left (\frac {d x^2}{c}+1\right )^p \left (\frac {a c+a d x^2+b}{c+d x^2}\right )^p \left (\frac {a d x^2}{a c+b}+1\right )^{-p} \int (e x)^m \left (\frac {d x^2}{c}+1\right )^{-p} \left (\frac {a d x^2}{b+a c}+1\right )^pdx\) |
\(\Big \downarrow \) 394 |
\(\displaystyle \frac {(e x)^{m+1} \left (\frac {d x^2}{c}+1\right )^p \left (\frac {a c+a d x^2+b}{c+d x^2}\right )^p \left (\frac {a d x^2}{a c+b}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {m+1}{2},p,-p,\frac {m+3}{2},-\frac {d x^2}{c},-\frac {a d x^2}{b+a c}\right )}{e (m+1)}\) |
Input:
Int[(e*x)^m*(a + b/(c + d*x^2))^p,x]
Output:
((e*x)^(1 + m)*((b + a*c + a*d*x^2)/(c + d*x^2))^p*(1 + (d*x^2)/c)^p*Appel lF1[(1 + m)/2, p, -p, (3 + m)/2, -((d*x^2)/c), -((a*d*x^2)/(b + a*c))])/(e *(1 + m)*(1 + (a*d*x^2)/(b + a*c))^p)
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ ), x_Symbol] :> Simp[a^p*c^q*((e*x)^(m + 1)/(e*(m + 1)))*AppellF1[(m + 1)/2 , -p, -q, 1 + (m + 1)/2, (-b)*(x^2/a), (-d)*(x^2/c)], x] /; FreeQ[{a, b, c, d, e, m, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, 1] && (Int egerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ ), x_Symbol] :> Simp[a^IntPart[p]*((a + b*x^2)^FracPart[p]/(1 + b*(x^2/a))^ FracPart[p]) Int[(e*x)^m*(1 + b*(x^2/a))^p*(c + d*x^2)^q, x], x] /; FreeQ [{a, b, c, d, e, m, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, 1] && !(IntegerQ[p] || GtQ[a, 0])
Int[(u_.)*((a_) + (b_.)/((c_) + (d_.)*(x_)^(n_)))^(p_), x_Symbol] :> Int[u* ((b + a*c + a*d*x^n)/(c + d*x^n))^p, x] /; FreeQ[{a, b, c, d, n, p}, x]
Int[(u_.)*((e_.)*((a_.) + (b_.)*(x_)^(n_.))^(q_.)*((c_) + (d_.)*(x_)^(n_))^ (r_.))^(p_), x_Symbol] :> Simp[Simp[(e*(a + b*x^n)^q*(c + d*x^n)^r)^p/((a + b*x^n)^(p*q)*(c + d*x^n)^(p*r))] Int[u*(a + b*x^n)^(p*q)*(c + d*x^n)^(p* r), x], x] /; FreeQ[{a, b, c, d, e, n, p, q, r}, x]
\[\int \left (e x \right )^{m} \left (a +\frac {b}{d \,x^{2}+c}\right )^{p}d x\]
Input:
int((e*x)^m*(a+b/(d*x^2+c))^p,x)
Output:
int((e*x)^m*(a+b/(d*x^2+c))^p,x)
\[ \int (e x)^m \left (a+\frac {b}{c+d x^2}\right )^p \, dx=\int { \left (e x\right )^{m} {\left (a + \frac {b}{d x^{2} + c}\right )}^{p} \,d x } \] Input:
integrate((e*x)^m*(a+b/(d*x^2+c))^p,x, algorithm="fricas")
Output:
integral((e*x)^m*((a*d*x^2 + a*c + b)/(d*x^2 + c))^p, x)
Timed out. \[ \int (e x)^m \left (a+\frac {b}{c+d x^2}\right )^p \, dx=\text {Timed out} \] Input:
integrate((e*x)**m*(a+b/(d*x**2+c))**p,x)
Output:
Timed out
\[ \int (e x)^m \left (a+\frac {b}{c+d x^2}\right )^p \, dx=\int { \left (e x\right )^{m} {\left (a + \frac {b}{d x^{2} + c}\right )}^{p} \,d x } \] Input:
integrate((e*x)^m*(a+b/(d*x^2+c))^p,x, algorithm="maxima")
Output:
integrate((e*x)^m*(a + b/(d*x^2 + c))^p, x)
\[ \int (e x)^m \left (a+\frac {b}{c+d x^2}\right )^p \, dx=\int { \left (e x\right )^{m} {\left (a + \frac {b}{d x^{2} + c}\right )}^{p} \,d x } \] Input:
integrate((e*x)^m*(a+b/(d*x^2+c))^p,x, algorithm="giac")
Output:
integrate((e*x)^m*(a + b/(d*x^2 + c))^p, x)
Timed out. \[ \int (e x)^m \left (a+\frac {b}{c+d x^2}\right )^p \, dx=\int {\left (e\,x\right )}^m\,{\left (a+\frac {b}{d\,x^2+c}\right )}^p \,d x \] Input:
int((e*x)^m*(a + b/(c + d*x^2))^p,x)
Output:
int((e*x)^m*(a + b/(c + d*x^2))^p, x)
\[ \int (e x)^m \left (a+\frac {b}{c+d x^2}\right )^p \, dx=e^{m} \left (\int \frac {x^{m} \left (a d \,x^{2}+a c +b \right )^{p}}{\left (d \,x^{2}+c \right )^{p}}d x \right ) \] Input:
int((e*x)^m*(a+b/(d*x^2+c))^p,x)
Output:
e**m*int((x**m*(a*c + a*d*x**2 + b)**p)/(c + d*x**2)**p,x)