Integrand size = 24, antiderivative size = 526 \[ \int \frac {(e x)^m (c+d x)^{7/2}}{\left (a+b x^2\right )^2} \, dx=-\frac {c d^2 (e x)^{1+m} \sqrt {c+d x}}{a b e}-\frac {d^2 (e x)^{1+m} (c+d x)^{3/2}}{2 a b e}+\frac {(e x)^{1+m} (c+d x)^{7/2}}{2 a e \left (a+b x^2\right )}+\frac {\left (a b^{3/2} c^3 d (1-8 m)+2 \sqrt {-a} b^2 c^4 (1-m)-(-a)^{5/2} d^4 (5+2 m)-3 (-a)^{3/2} b c^2 d^2 (3+4 m)+a^2 \sqrt {b} c d^3 (13+8 m)\right ) (e x)^{1+m} \sqrt {1+\frac {d x}{c}} \operatorname {AppellF1}\left (1+m,\frac {1}{2},1,2+m,-\frac {d x}{c},-\frac {\sqrt {b} x}{\sqrt {-a}}\right )}{8 (-a)^{5/2} b^2 e (1+m) \sqrt {c+d x}}-\frac {\left (a b^{3/2} c^3 d (1-8 m)-2 \sqrt {-a} b^2 c^4 (1-m)+(-a)^{5/2} d^4 (5+2 m)+3 (-a)^{3/2} b c^2 d^2 (3+4 m)+a^2 \sqrt {b} c d^3 (13+8 m)\right ) (e x)^{1+m} \sqrt {1+\frac {d x}{c}} \operatorname {AppellF1}\left (1+m,\frac {1}{2},1,2+m,-\frac {d x}{c},\frac {\sqrt {b} x}{\sqrt {-a}}\right )}{8 (-a)^{5/2} b^2 e (1+m) \sqrt {c+d x}}-\frac {d \left (3 b c^2 (1+2 m)-a d^2 (5+2 m)\right ) \left (-\frac {d x}{c}\right )^{-m} (e x)^m \sqrt {c+d x} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-m,\frac {3}{2},1+\frac {d x}{c}\right )}{2 a b^2} \] Output:
-c*d^2*(e*x)^(1+m)*(d*x+c)^(1/2)/a/b/e-1/2*d^2*(e*x)^(1+m)*(d*x+c)^(3/2)/a /b/e+1/2*(e*x)^(1+m)*(d*x+c)^(7/2)/a/e/(b*x^2+a)+1/8*(a*b^(3/2)*c^3*d*(1-8 *m)+2*(-a)^(1/2)*b^2*c^4*(1-m)-(-a)^(5/2)*d^4*(5+2*m)-3*(-a)^(3/2)*b*c^2*d ^2*(3+4*m)+a^2*b^(1/2)*c*d^3*(13+8*m))*(e*x)^(1+m)*(1+d*x/c)^(1/2)*AppellF 1(1+m,1,1/2,2+m,-b^(1/2)*x/(-a)^(1/2),-d*x/c)/(-a)^(5/2)/b^2/e/(1+m)/(d*x+ c)^(1/2)-1/8*(a*b^(3/2)*c^3*d*(1-8*m)-2*(-a)^(1/2)*b^2*c^4*(1-m)+(-a)^(5/2 )*d^4*(5+2*m)+3*(-a)^(3/2)*b*c^2*d^2*(3+4*m)+a^2*b^(1/2)*c*d^3*(13+8*m))*( e*x)^(1+m)*(1+d*x/c)^(1/2)*AppellF1(1+m,1,1/2,2+m,b^(1/2)*x/(-a)^(1/2),-d* x/c)/(-a)^(5/2)/b^2/e/(1+m)/(d*x+c)^(1/2)-1/2*d*(3*b*c^2*(1+2*m)-a*d^2*(5+ 2*m))*(e*x)^m*(d*x+c)^(1/2)*hypergeom([1/2, -m],[3/2],1+d*x/c)/a/b^2/((-d* x/c)^m)
\[ \int \frac {(e x)^m (c+d x)^{7/2}}{\left (a+b x^2\right )^2} \, dx=\int \frac {(e x)^m (c+d x)^{7/2}}{\left (a+b x^2\right )^2} \, dx \] Input:
Integrate[((e*x)^m*(c + d*x)^(7/2))/(a + b*x^2)^2,x]
Output:
Integrate[((e*x)^m*(c + d*x)^(7/2))/(a + b*x^2)^2, x]
Time = 0.59 (sec) , antiderivative size = 315, normalized size of antiderivative = 0.60, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {615, 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.
\(\displaystyle \int \frac {(c+d x)^{7/2} (e x)^m}{\left (a+b x^2\right )^2} \, dx\) |
\(\Big \downarrow \) 615 |
\(\displaystyle \int \left (-\frac {b (c+d x)^{7/2} (e x)^m}{2 a \left (-a b-b^2 x^2\right )}-\frac {b (c+d x)^{7/2} (e x)^m}{4 a \left (\sqrt {-a} \sqrt {b}-b x\right )^2}-\frac {b (c+d x)^{7/2} (e x)^m}{4 a \left (\sqrt {-a} \sqrt {b}+b x\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {c^3 \sqrt {c+d x} (e x)^{m+1} \operatorname {AppellF1}\left (m+1,-\frac {7}{2},1,m+2,-\frac {d x}{c},-\frac {\sqrt {b} x}{\sqrt {-a}}\right )}{4 a^2 e (m+1) \sqrt {\frac {d x}{c}+1}}+\frac {c^3 \sqrt {c+d x} (e x)^{m+1} \operatorname {AppellF1}\left (m+1,-\frac {7}{2},1,m+2,-\frac {d x}{c},\frac {\sqrt {b} x}{\sqrt {-a}}\right )}{4 a^2 e (m+1) \sqrt {\frac {d x}{c}+1}}+\frac {c^3 \sqrt {c+d x} (e x)^{m+1} \operatorname {AppellF1}\left (m+1,-\frac {7}{2},2,m+2,-\frac {d x}{c},-\frac {\sqrt {b} x}{\sqrt {-a}}\right )}{4 a^2 e (m+1) \sqrt {\frac {d x}{c}+1}}+\frac {c^3 \sqrt {c+d x} (e x)^{m+1} \operatorname {AppellF1}\left (m+1,-\frac {7}{2},2,m+2,-\frac {d x}{c},\frac {\sqrt {b} x}{\sqrt {-a}}\right )}{4 a^2 e (m+1) \sqrt {\frac {d x}{c}+1}}\) |
Input:
Int[((e*x)^m*(c + d*x)^(7/2))/(a + b*x^2)^2,x]
Output:
(c^3*(e*x)^(1 + m)*Sqrt[c + d*x]*AppellF1[1 + m, -7/2, 1, 2 + m, -((d*x)/c ), -((Sqrt[b]*x)/Sqrt[-a])])/(4*a^2*e*(1 + m)*Sqrt[1 + (d*x)/c]) + (c^3*(e *x)^(1 + m)*Sqrt[c + d*x]*AppellF1[1 + m, -7/2, 1, 2 + m, -((d*x)/c), (Sqr t[b]*x)/Sqrt[-a]])/(4*a^2*e*(1 + m)*Sqrt[1 + (d*x)/c]) + (c^3*(e*x)^(1 + m )*Sqrt[c + d*x]*AppellF1[1 + m, -7/2, 2, 2 + m, -((d*x)/c), -((Sqrt[b]*x)/ Sqrt[-a])])/(4*a^2*e*(1 + m)*Sqrt[1 + (d*x)/c]) + (c^3*(e*x)^(1 + m)*Sqrt[ c + d*x]*AppellF1[1 + m, -7/2, 2, 2 + m, -((d*x)/c), (Sqrt[b]*x)/Sqrt[-a]] )/(4*a^2*e*(1 + m)*Sqrt[1 + (d*x)/c])
Int[((e_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*(c + d*x)^n*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && ILtQ[p, 0]
\[\int \frac {\left (e x \right )^{m} \left (d x +c \right )^{\frac {7}{2}}}{\left (b \,x^{2}+a \right )^{2}}d x\]
Input:
int((e*x)^m*(d*x+c)^(7/2)/(b*x^2+a)^2,x)
Output:
int((e*x)^m*(d*x+c)^(7/2)/(b*x^2+a)^2,x)
\[ \int \frac {(e x)^m (c+d x)^{7/2}}{\left (a+b x^2\right )^2} \, dx=\int { \frac {{\left (d x + c\right )}^{\frac {7}{2}} \left (e x\right )^{m}}{{\left (b x^{2} + a\right )}^{2}} \,d x } \] Input:
integrate((e*x)^m*(d*x+c)^(7/2)/(b*x^2+a)^2,x, algorithm="fricas")
Output:
integral((d^3*x^3 + 3*c*d^2*x^2 + 3*c^2*d*x + c^3)*sqrt(d*x + c)*(e*x)^m/( b^2*x^4 + 2*a*b*x^2 + a^2), x)
Timed out. \[ \int \frac {(e x)^m (c+d x)^{7/2}}{\left (a+b x^2\right )^2} \, dx=\text {Timed out} \] Input:
integrate((e*x)**m*(d*x+c)**(7/2)/(b*x**2+a)**2,x)
Output:
Timed out
\[ \int \frac {(e x)^m (c+d x)^{7/2}}{\left (a+b x^2\right )^2} \, dx=\int { \frac {{\left (d x + c\right )}^{\frac {7}{2}} \left (e x\right )^{m}}{{\left (b x^{2} + a\right )}^{2}} \,d x } \] Input:
integrate((e*x)^m*(d*x+c)^(7/2)/(b*x^2+a)^2,x, algorithm="maxima")
Output:
integrate((d*x + c)^(7/2)*(e*x)^m/(b*x^2 + a)^2, x)
\[ \int \frac {(e x)^m (c+d x)^{7/2}}{\left (a+b x^2\right )^2} \, dx=\int { \frac {{\left (d x + c\right )}^{\frac {7}{2}} \left (e x\right )^{m}}{{\left (b x^{2} + a\right )}^{2}} \,d x } \] Input:
integrate((e*x)^m*(d*x+c)^(7/2)/(b*x^2+a)^2,x, algorithm="giac")
Output:
integrate((d*x + c)^(7/2)*(e*x)^m/(b*x^2 + a)^2, x)
Timed out. \[ \int \frac {(e x)^m (c+d x)^{7/2}}{\left (a+b x^2\right )^2} \, dx=\int \frac {{\left (e\,x\right )}^m\,{\left (c+d\,x\right )}^{7/2}}{{\left (b\,x^2+a\right )}^2} \,d x \] Input:
int(((e*x)^m*(c + d*x)^(7/2))/(a + b*x^2)^2,x)
Output:
int(((e*x)^m*(c + d*x)^(7/2))/(a + b*x^2)^2, x)
\[ \int \frac {(e x)^m (c+d x)^{7/2}}{\left (a+b x^2\right )^2} \, dx=\text {too large to display} \] Input:
int((e*x)^m*(d*x+c)^(7/2)/(b*x^2+a)^2,x)
Output:
(e**m*( - 16*x**m*sqrt(c + d*x)*a*d**3*m**2 - 32*x**m*sqrt(c + d*x)*a*d**3 *m - 8*x**m*sqrt(c + d*x)*a*d**3 + 16*x**m*sqrt(c + d*x)*b*c**2*d*m**2 - 4 *x**m*sqrt(c + d*x)*b*c**2*d + 12*x**m*sqrt(c + d*x)*b*c*d**2*m**2*x - 16* x**m*sqrt(c + d*x)*b*c*d**2*m*x - 16*x**m*sqrt(c + d*x)*b*c*d**2*x + 4*x** m*sqrt(c + d*x)*b*d**3*m**2*x**2 - 10*x**m*sqrt(c + d*x)*b*d**3*m*x**2 + 4 *x**m*sqrt(c + d*x)*b*d**3*x**2 + 48*int((x**m*sqrt(c + d*x)*x**2)/(4*a**2 *c*m**3 - 8*a**2*c*m**2 - a**2*c*m + 2*a**2*c + 4*a**2*d*m**3*x - 8*a**2*d *m**2*x - a**2*d*m*x + 2*a**2*d*x + 8*a*b*c*m**3*x**2 - 16*a*b*c*m**2*x**2 - 2*a*b*c*m*x**2 + 4*a*b*c*x**2 + 8*a*b*d*m**3*x**3 - 16*a*b*d*m**2*x**3 - 2*a*b*d*m*x**3 + 4*a*b*d*x**3 + 4*b**2*c*m**3*x**4 - 8*b**2*c*m**2*x**4 - b**2*c*m*x**4 + 2*b**2*c*x**4 + 4*b**2*d*m**3*x**5 - 8*b**2*d*m**2*x**5 - b**2*d*m*x**5 + 2*b**2*d*x**5),x)*a**2*b*d**4*m**6 - 64*int((x**m*sqrt(c + d*x)*x**2)/(4*a**2*c*m**3 - 8*a**2*c*m**2 - a**2*c*m + 2*a**2*c + 4*a** 2*d*m**3*x - 8*a**2*d*m**2*x - a**2*d*m*x + 2*a**2*d*x + 8*a*b*c*m**3*x**2 - 16*a*b*c*m**2*x**2 - 2*a*b*c*m*x**2 + 4*a*b*c*x**2 + 8*a*b*d*m**3*x**3 - 16*a*b*d*m**2*x**3 - 2*a*b*d*m*x**3 + 4*a*b*d*x**3 + 4*b**2*c*m**3*x**4 - 8*b**2*c*m**2*x**4 - b**2*c*m*x**4 + 2*b**2*c*x**4 + 4*b**2*d*m**3*x**5 - 8*b**2*d*m**2*x**5 - b**2*d*m*x**5 + 2*b**2*d*x**5),x)*a**2*b*d**4*m**5 - 152*int((x**m*sqrt(c + d*x)*x**2)/(4*a**2*c*m**3 - 8*a**2*c*m**2 - a**2* c*m + 2*a**2*c + 4*a**2*d*m**3*x - 8*a**2*d*m**2*x - a**2*d*m*x + 2*a**...