Integrand size = 28, antiderivative size = 283 \[ \int \frac {(e x)^m (a-b x)^{5/2}}{(a c+b c x)^{3/2}} \, dx=\frac {4 (e x)^{1+m} (a-b x)^{3/2}}{c e \sqrt {a c+b c x}}+\frac {(9+4 m) (e x)^{1+m} \sqrt {a-b x} \sqrt {a c+b c x}}{c^2 e (2+m)}-\frac {a^2 \left (15+24 m+8 m^2\right ) (e x)^{1+m} \sqrt {1-\frac {b^2 x^2}{a^2}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},\frac {b^2 x^2}{a^2}\right )}{c e (1+m) (2+m) \sqrt {a-b x} \sqrt {a c+b c x}}+\frac {4 a b (3+2 m) (e x)^{2+m} \sqrt {1-\frac {b^2 x^2}{a^2}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+m}{2},\frac {4+m}{2},\frac {b^2 x^2}{a^2}\right )}{c e^2 (2+m) \sqrt {a-b x} \sqrt {a c+b c x}} \] Output:
4*(e*x)^(1+m)*(-b*x+a)^(3/2)/c/e/(b*c*x+a*c)^(1/2)+(9+4*m)*(e*x)^(1+m)*(-b *x+a)^(1/2)*(b*c*x+a*c)^(1/2)/c^2/e/(2+m)-a^2*(8*m^2+24*m+15)*(e*x)^(1+m)* (1-b^2*x^2/a^2)^(1/2)*hypergeom([1/2, 1/2+1/2*m],[3/2+1/2*m],b^2*x^2/a^2)/ c/e/(1+m)/(2+m)/(-b*x+a)^(1/2)/(b*c*x+a*c)^(1/2)+4*a*b*(3+2*m)*(e*x)^(2+m) *(1-b^2*x^2/a^2)^(1/2)*hypergeom([1/2, 1+1/2*m],[2+1/2*m],b^2*x^2/a^2)/c/e ^2/(2+m)/(-b*x+a)^(1/2)/(b*c*x+a*c)^(1/2)
Time = 10.26 (sec) , antiderivative size = 244, normalized size of antiderivative = 0.86 \[ \int \frac {(e x)^m (a-b x)^{5/2}}{(a c+b c x)^{3/2}} \, dx=\frac {x (e x)^m \sqrt {1-\frac {b^2 x^2}{a^2}} \left (\frac {a^2 \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {1+m}{2},\frac {3+m}{2},\frac {b^2 x^2}{a^2}\right )}{1+m}+b x \left (-\frac {4 a \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {2+m}{2},\frac {4+m}{2},\frac {b^2 x^2}{a^2}\right )}{2+m}+b x \left (\frac {6 \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {3+m}{2},\frac {5+m}{2},\frac {b^2 x^2}{a^2}\right )}{3+m}+\frac {b x \left (-\frac {4 a \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {4+m}{2},\frac {6+m}{2},\frac {b^2 x^2}{a^2}\right )}{4+m}+\frac {b x \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {5+m}{2},\frac {7+m}{2},\frac {b^2 x^2}{a^2}\right )}{5+m}\right )}{a^2}\right )\right )\right )}{c \sqrt {a-b x} \sqrt {c (a+b x)}} \] Input:
Integrate[((e*x)^m*(a - b*x)^(5/2))/(a*c + b*c*x)^(3/2),x]
Output:
(x*(e*x)^m*Sqrt[1 - (b^2*x^2)/a^2]*((a^2*Hypergeometric2F1[3/2, (1 + m)/2, (3 + m)/2, (b^2*x^2)/a^2])/(1 + m) + b*x*((-4*a*Hypergeometric2F1[3/2, (2 + m)/2, (4 + m)/2, (b^2*x^2)/a^2])/(2 + m) + b*x*((6*Hypergeometric2F1[3/ 2, (3 + m)/2, (5 + m)/2, (b^2*x^2)/a^2])/(3 + m) + (b*x*((-4*a*Hypergeomet ric2F1[3/2, (4 + m)/2, (6 + m)/2, (b^2*x^2)/a^2])/(4 + m) + (b*x*Hypergeom etric2F1[3/2, (5 + m)/2, (7 + m)/2, (b^2*x^2)/a^2])/(5 + m)))/a^2))))/(c*S qrt[a - b*x]*Sqrt[c*(a + b*x)])
Time = 0.62 (sec) , antiderivative size = 454, normalized size of antiderivative = 1.60, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {147, 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.
| \(\displaystyle \int \frac {(a-b x)^{5/2} (e x)^m}{(a c+b c x)^{3/2}} \, dx\) |
| \(\Big \downarrow \) 147 |
| \(\displaystyle \int \left (\frac {a^4 (e x)^m}{(a-b x)^{3/2} (a c+b c x)^{3/2}}-\frac {4 a^3 b (e x)^{m+1}}{e (a-b x)^{3/2} (a c+b c x)^{3/2}}+\frac {6 a^2 b^2 (e x)^{m+2}}{e^2 (a-b x)^{3/2} (a c+b c x)^{3/2}}+\frac {b^4 (e x)^{m+4}}{e^4 (a-b x)^{3/2} (a c+b c x)^{3/2}}-\frac {4 a b^3 (e x)^{m+3}}{e^3 (a-b x)^{3/2} (a c+b c x)^{3/2}}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {6 b^2 \sqrt {1-\frac {b^2 x^2}{a^2}} (e x)^{m+3} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {m+3}{2},\frac {m+5}{2},\frac {b^2 x^2}{a^2}\right )}{c e^3 (m+3) \sqrt {a-b x} \sqrt {a c+b c x}}-\frac {4 a b \sqrt {1-\frac {b^2 x^2}{a^2}} (e x)^{m+2} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {m+2}{2},\frac {m+4}{2},\frac {b^2 x^2}{a^2}\right )}{c e^2 (m+2) \sqrt {a-b x} \sqrt {a c+b c x}}+\frac {a^2 \sqrt {1-\frac {b^2 x^2}{a^2}} (e x)^{m+1} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {m+1}{2},\frac {m+3}{2},\frac {b^2 x^2}{a^2}\right )}{c e (m+1) \sqrt {a-b x} \sqrt {a c+b c x}}+\frac {b^4 \sqrt {1-\frac {b^2 x^2}{a^2}} (e x)^{m+5} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {m+5}{2},\frac {m+7}{2},\frac {b^2 x^2}{a^2}\right )}{a^2 c e^5 (m+5) \sqrt {a-b x} \sqrt {a c+b c x}}-\frac {4 b^3 \sqrt {1-\frac {b^2 x^2}{a^2}} (e x)^{m+4} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {m+4}{2},\frac {m+6}{2},\frac {b^2 x^2}{a^2}\right )}{a c e^4 (m+4) \sqrt {a-b x} \sqrt {a c+b c x}}\) |
Input:
Int[((e*x)^m*(a - b*x)^(5/2))/(a*c + b*c*x)^(3/2),x]
Output:
(a^2*(e*x)^(1 + m)*Sqrt[1 - (b^2*x^2)/a^2]*Hypergeometric2F1[3/2, (1 + m)/ 2, (3 + m)/2, (b^2*x^2)/a^2])/(c*e*(1 + m)*Sqrt[a - b*x]*Sqrt[a*c + b*c*x] ) - (4*a*b*(e*x)^(2 + m)*Sqrt[1 - (b^2*x^2)/a^2]*Hypergeometric2F1[3/2, (2 + m)/2, (4 + m)/2, (b^2*x^2)/a^2])/(c*e^2*(2 + m)*Sqrt[a - b*x]*Sqrt[a*c + b*c*x]) + (6*b^2*(e*x)^(3 + m)*Sqrt[1 - (b^2*x^2)/a^2]*Hypergeometric2F1 [3/2, (3 + m)/2, (5 + m)/2, (b^2*x^2)/a^2])/(c*e^3*(3 + m)*Sqrt[a - b*x]*S qrt[a*c + b*c*x]) - (4*b^3*(e*x)^(4 + m)*Sqrt[1 - (b^2*x^2)/a^2]*Hypergeom etric2F1[3/2, (4 + m)/2, (6 + m)/2, (b^2*x^2)/a^2])/(a*c*e^4*(4 + m)*Sqrt[ a - b*x]*Sqrt[a*c + b*c*x]) + (b^4*(e*x)^(5 + m)*Sqrt[1 - (b^2*x^2)/a^2]*H ypergeometric2F1[3/2, (5 + m)/2, (7 + m)/2, (b^2*x^2)/a^2])/(a^2*c*e^5*(5 + m)*Sqrt[a - b*x]*Sqrt[a*c + b*c*x])
Int[((f_.)*(x_))^(p_)*((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_] :> Int[ExpandIntegrand[(a + b*x)^n*(c + d*x)^n*(f*x)^p, (a + b*x)^(m - n), x], x] /; FreeQ[{a, b, c, d, f, m, n, p}, x] && EqQ[b*c + a*d, 0] && IG tQ[m - n, 0] && NeQ[m + n + p + 2, 0]
\[\int \frac {\left (e x \right )^{m} \left (-b x +a \right )^{\frac {5}{2}}}{\left (b c x +a c \right )^{\frac {3}{2}}}d x\]
Input:
int((e*x)^m*(-b*x+a)^(5/2)/(b*c*x+a*c)^(3/2),x)
Output:
int((e*x)^m*(-b*x+a)^(5/2)/(b*c*x+a*c)^(3/2),x)
\[ \int \frac {(e x)^m (a-b x)^{5/2}}{(a c+b c x)^{3/2}} \, dx=\int { \frac {{\left (-b x + a\right )}^{\frac {5}{2}} \left (e x\right )^{m}}{{\left (b c x + a c\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate((e*x)^m*(-b*x+a)^(5/2)/(b*c*x+a*c)^(3/2),x, algorithm="fricas")
Output:
integral((b^2*x^2 - 2*a*b*x + a^2)*sqrt(b*c*x + a*c)*sqrt(-b*x + a)*(e*x)^ m/(b^2*c^2*x^2 + 2*a*b*c^2*x + a^2*c^2), x)
\[ \int \frac {(e x)^m (a-b x)^{5/2}}{(a c+b c x)^{3/2}} \, dx=\int \frac {\left (e x\right )^{m} \left (a - b x\right )^{\frac {5}{2}}}{\left (c \left (a + b x\right )\right )^{\frac {3}{2}}}\, dx \] Input:
integrate((e*x)**m*(-b*x+a)**(5/2)/(b*c*x+a*c)**(3/2),x)
Output:
Integral((e*x)**m*(a - b*x)**(5/2)/(c*(a + b*x))**(3/2), x)
\[ \int \frac {(e x)^m (a-b x)^{5/2}}{(a c+b c x)^{3/2}} \, dx=\int { \frac {{\left (-b x + a\right )}^{\frac {5}{2}} \left (e x\right )^{m}}{{\left (b c x + a c\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate((e*x)^m*(-b*x+a)^(5/2)/(b*c*x+a*c)^(3/2),x, algorithm="maxima")
Output:
integrate((-b*x + a)^(5/2)*(e*x)^m/(b*c*x + a*c)^(3/2), x)
\[ \int \frac {(e x)^m (a-b x)^{5/2}}{(a c+b c x)^{3/2}} \, dx=\int { \frac {{\left (-b x + a\right )}^{\frac {5}{2}} \left (e x\right )^{m}}{{\left (b c x + a c\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate((e*x)^m*(-b*x+a)^(5/2)/(b*c*x+a*c)^(3/2),x, algorithm="giac")
Output:
integrate((-b*x + a)^(5/2)*(e*x)^m/(b*c*x + a*c)^(3/2), x)
Timed out. \[ \int \frac {(e x)^m (a-b x)^{5/2}}{(a c+b c x)^{3/2}} \, dx=\int \frac {{\left (e\,x\right )}^m\,{\left (a-b\,x\right )}^{5/2}}{{\left (a\,c+b\,c\,x\right )}^{3/2}} \,d x \] Input:
int(((e*x)^m*(a - b*x)^(5/2))/(a*c + b*c*x)^(3/2),x)
Output:
int(((e*x)^m*(a - b*x)^(5/2))/(a*c + b*c*x)^(3/2), x)
\[ \int \frac {(e x)^m (a-b x)^{5/2}}{(a c+b c x)^{3/2}} \, dx=\frac {e^{m} \left (\left (\int \frac {x^{m} \sqrt {-b x +a}\, x^{2}}{\sqrt {b x +a}\, a +\sqrt {b x +a}\, b x}d x \right ) b^{2}-2 \left (\int \frac {x^{m} \sqrt {-b x +a}\, x}{\sqrt {b x +a}\, a +\sqrt {b x +a}\, b x}d x \right ) a b +\left (\int \frac {x^{m} \sqrt {-b x +a}}{\sqrt {b x +a}\, a +\sqrt {b x +a}\, b x}d x \right ) a^{2}\right )}{\sqrt {c}\, c} \] Input:
int((e*x)^m*(-b*x+a)^(5/2)/(b*c*x+a*c)^(3/2),x)
Output:
(e**m*(int((x**m*sqrt(a - b*x)*x**2)/(sqrt(a + b*x)*a + sqrt(a + b*x)*b*x) ,x)*b**2 - 2*int((x**m*sqrt(a - b*x)*x)/(sqrt(a + b*x)*a + sqrt(a + b*x)*b *x),x)*a*b + int((x**m*sqrt(a - b*x))/(sqrt(a + b*x)*a + sqrt(a + b*x)*b*x ),x)*a**2))/(sqrt(c)*c)