Integrand size = 24, antiderivative size = 252 \[ \int \frac {(e x)^m}{(c+d x)^2 \sqrt {a+b x^2}} \, dx=\frac {(e x)^{1+m} \sqrt {1+\frac {b x^2}{a}} \operatorname {AppellF1}\left (\frac {1+m}{2},\frac {1}{2},2,\frac {3+m}{2},-\frac {b x^2}{a},\frac {d^2 x^2}{c^2}\right )}{c^2 e (1+m) \sqrt {a+b x^2}}-\frac {2 d (e x)^{2+m} \sqrt {1+\frac {b x^2}{a}} \operatorname {AppellF1}\left (\frac {2+m}{2},\frac {1}{2},2,\frac {4+m}{2},-\frac {b x^2}{a},\frac {d^2 x^2}{c^2}\right )}{c^3 e^2 (2+m) \sqrt {a+b x^2}}+\frac {d^2 (e x)^{3+m} \sqrt {1+\frac {b x^2}{a}} \operatorname {AppellF1}\left (\frac {3+m}{2},\frac {1}{2},2,\frac {5+m}{2},-\frac {b x^2}{a},\frac {d^2 x^2}{c^2}\right )}{c^4 e^3 (3+m) \sqrt {a+b x^2}} \] Output:
(e*x)^(1+m)*(1+b*x^2/a)^(1/2)*AppellF1(1/2+1/2*m,2,1/2,3/2+1/2*m,d^2*x^2/c ^2,-b*x^2/a)/c^2/e/(1+m)/(b*x^2+a)^(1/2)-2*d*(e*x)^(2+m)*(1+b*x^2/a)^(1/2) *AppellF1(1+1/2*m,2,1/2,2+1/2*m,d^2*x^2/c^2,-b*x^2/a)/c^3/e^2/(2+m)/(b*x^2 +a)^(1/2)+d^2*(e*x)^(3+m)*(1+b*x^2/a)^(1/2)*AppellF1(3/2+1/2*m,2,1/2,5/2+1 /2*m,d^2*x^2/c^2,-b*x^2/a)/c^4/e^3/(3+m)/(b*x^2+a)^(1/2)
\[ \int \frac {(e x)^m}{(c+d x)^2 \sqrt {a+b x^2}} \, dx=\int \frac {(e x)^m}{(c+d x)^2 \sqrt {a+b x^2}} \, dx \] Input:
Integrate[(e*x)^m/((c + d*x)^2*Sqrt[a + b*x^2]),x]
Output:
Integrate[(e*x)^m/((c + d*x)^2*Sqrt[a + b*x^2]), x]
Time = 0.46 (sec) , antiderivative size = 248, normalized size of antiderivative = 0.98, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {623, 622, 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.
\(\displaystyle \int \frac {(e x)^m}{\sqrt {a+b x^2} (c+d x)^2} \, dx\) |
\(\Big \downarrow \) 623 |
\(\displaystyle x^{-m} (e x)^m \int \frac {x^m}{(c+d x)^2 \sqrt {b x^2+a}}dx\) |
\(\Big \downarrow \) 622 |
\(\displaystyle x^{-m} (e x)^m \int \left (\frac {c^2 x^m}{\sqrt {b x^2+a} \left (c^2-d^2 x^2\right )^2}-\frac {2 c d x^{m+1}}{\sqrt {b x^2+a} \left (c^2-d^2 x^2\right )^2}+\frac {d^2 x^{m+2}}{\sqrt {b x^2+a} \left (d^2 x^2-c^2\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle x^{-m} (e x)^m \left (\frac {x^{m+1} \sqrt {\frac {b x^2}{a}+1} \operatorname {AppellF1}\left (\frac {m+1}{2},\frac {1}{2},2,\frac {m+3}{2},-\frac {b x^2}{a},\frac {d^2 x^2}{c^2}\right )}{c^2 (m+1) \sqrt {a+b x^2}}+\frac {d^2 x^{m+3} \sqrt {\frac {b x^2}{a}+1} \operatorname {AppellF1}\left (\frac {m+3}{2},\frac {1}{2},2,\frac {m+5}{2},-\frac {b x^2}{a},\frac {d^2 x^2}{c^2}\right )}{c^4 (m+3) \sqrt {a+b x^2}}-\frac {2 d x^{m+2} \sqrt {\frac {b x^2}{a}+1} \operatorname {AppellF1}\left (\frac {m+2}{2},\frac {1}{2},2,\frac {m+4}{2},-\frac {b x^2}{a},\frac {d^2 x^2}{c^2}\right )}{c^3 (m+2) \sqrt {a+b x^2}}\right )\) |
Input:
Int[(e*x)^m/((c + d*x)^2*Sqrt[a + b*x^2]),x]
Output:
((e*x)^m*((x^(1 + m)*Sqrt[1 + (b*x^2)/a]*AppellF1[(1 + m)/2, 1/2, 2, (3 + m)/2, -((b*x^2)/a), (d^2*x^2)/c^2])/(c^2*(1 + m)*Sqrt[a + b*x^2]) - (2*d*x ^(2 + m)*Sqrt[1 + (b*x^2)/a]*AppellF1[(2 + m)/2, 1/2, 2, (4 + m)/2, -((b*x ^2)/a), (d^2*x^2)/c^2])/(c^3*(2 + m)*Sqrt[a + b*x^2]) + (d^2*x^(3 + m)*Sqr t[1 + (b*x^2)/a]*AppellF1[(3 + m)/2, 1/2, 2, (5 + m)/2, -((b*x^2)/a), (d^2 *x^2)/c^2])/(c^4*(3 + m)*Sqrt[a + b*x^2])))/x^m
Int[(x_)^(m_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo l] :> Int[ExpandIntegrand[x^m*(a + b*x^2)^p, (c/(c^2 - d^2*x^2) - d*(x/(c^2 - d^2*x^2)))^(-n), x], x] /; FreeQ[{a, b, c, d, m, p}, x] && ILtQ[n, -1]
Int[((e_)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*x)^m/x^m Int[x^m*(c + d*x)^n*(a + b*x^2)^p, x], x] / ; FreeQ[{a, b, c, d, e, m, p}, x] && ILtQ[n, 0]
\[\int \frac {\left (e x \right )^{m}}{\left (d x +c \right )^{2} \sqrt {b \,x^{2}+a}}d x\]
Input:
int((e*x)^m/(d*x+c)^2/(b*x^2+a)^(1/2),x)
Output:
int((e*x)^m/(d*x+c)^2/(b*x^2+a)^(1/2),x)
\[ \int \frac {(e x)^m}{(c+d x)^2 \sqrt {a+b x^2}} \, dx=\int { \frac {\left (e x\right )^{m}}{\sqrt {b x^{2} + a} {\left (d x + c\right )}^{2}} \,d x } \] Input:
integrate((e*x)^m/(d*x+c)^2/(b*x^2+a)^(1/2),x, algorithm="fricas")
Output:
integral(sqrt(b*x^2 + a)*(e*x)^m/(b*d^2*x^4 + 2*b*c*d*x^3 + 2*a*c*d*x + a* c^2 + (b*c^2 + a*d^2)*x^2), x)
\[ \int \frac {(e x)^m}{(c+d x)^2 \sqrt {a+b x^2}} \, dx=\int \frac {\left (e x\right )^{m}}{\sqrt {a + b x^{2}} \left (c + d x\right )^{2}}\, dx \] Input:
integrate((e*x)**m/(d*x+c)**2/(b*x**2+a)**(1/2),x)
Output:
Integral((e*x)**m/(sqrt(a + b*x**2)*(c + d*x)**2), x)
\[ \int \frac {(e x)^m}{(c+d x)^2 \sqrt {a+b x^2}} \, dx=\int { \frac {\left (e x\right )^{m}}{\sqrt {b x^{2} + a} {\left (d x + c\right )}^{2}} \,d x } \] Input:
integrate((e*x)^m/(d*x+c)^2/(b*x^2+a)^(1/2),x, algorithm="maxima")
Output:
integrate((e*x)^m/(sqrt(b*x^2 + a)*(d*x + c)^2), x)
\[ \int \frac {(e x)^m}{(c+d x)^2 \sqrt {a+b x^2}} \, dx=\int { \frac {\left (e x\right )^{m}}{\sqrt {b x^{2} + a} {\left (d x + c\right )}^{2}} \,d x } \] Input:
integrate((e*x)^m/(d*x+c)^2/(b*x^2+a)^(1/2),x, algorithm="giac")
Output:
integrate((e*x)^m/(sqrt(b*x^2 + a)*(d*x + c)^2), x)
Timed out. \[ \int \frac {(e x)^m}{(c+d x)^2 \sqrt {a+b x^2}} \, dx=\int \frac {{\left (e\,x\right )}^m}{\sqrt {b\,x^2+a}\,{\left (c+d\,x\right )}^2} \,d x \] Input:
int((e*x)^m/((a + b*x^2)^(1/2)*(c + d*x)^2),x)
Output:
int((e*x)^m/((a + b*x^2)^(1/2)*(c + d*x)^2), x)
\[ \int \frac {(e x)^m}{(c+d x)^2 \sqrt {a+b x^2}} \, dx=e^{m} \left (\int \frac {x^{m} \sqrt {b \,x^{2}+a}}{b \,d^{2} x^{4}+2 b c d \,x^{3}+a \,d^{2} x^{2}+b \,c^{2} x^{2}+2 a c d x +a \,c^{2}}d x \right ) \] Input:
int((e*x)^m/(d*x+c)^2/(b*x^2+a)^(1/2),x)
Output:
e**m*int((x**m*sqrt(a + b*x**2))/(a*c**2 + 2*a*c*d*x + a*d**2*x**2 + b*c** 2*x**2 + 2*b*c*d*x**3 + b*d**2*x**4),x)