\(\int \frac {(e x)^m (c^2-d^2 x^2)^{5/2}}{c+d x} \, dx\) [1168]

Optimal result

Integrand size = 29, antiderivative size = 163 \[ \int \frac {(e x)^m \left (c^2-d^2 x^2\right )^{5/2}}{c+d x} \, dx=\frac {c^3 (e x)^{1+m} \sqrt {c^2-d^2 x^2} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {1+m}{2},\frac {3+m}{2},\frac {d^2 x^2}{c^2}\right )}{e (1+m) \sqrt {1-\frac {d^2 x^2}{c^2}}}-\frac {c^2 d (e x)^{2+m} \sqrt {c^2-d^2 x^2} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {2+m}{2},\frac {4+m}{2},\frac {d^2 x^2}{c^2}\right )}{e^2 (2+m) \sqrt {1-\frac {d^2 x^2}{c^2}}} \] Output:

c^3*(e*x)^(1+m)*(-d^2*x^2+c^2)^(1/2)*hypergeom([-3/2, 1/2+1/2*m],[3/2+1/2* 
m],d^2*x^2/c^2)/e/(1+m)/(1-d^2*x^2/c^2)^(1/2)-c^2*d*(e*x)^(2+m)*(-d^2*x^2+ 
c^2)^(1/2)*hypergeom([-3/2, 1+1/2*m],[2+1/2*m],d^2*x^2/c^2)/e^2/(2+m)/(1-d 
^2*x^2/c^2)^(1/2)
 

Mathematica [A] (verified)

Time = 0.71 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.75 \[ \int \frac {(e x)^m \left (c^2-d^2 x^2\right )^{5/2}}{c+d x} \, dx=\frac {c^2 x (e x)^m \sqrt {c^2-d^2 x^2} \left (-d (1+m) x \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1+\frac {m}{2},2+\frac {m}{2},\frac {d^2 x^2}{c^2}\right )+c (2+m) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {1+m}{2},\frac {3+m}{2},\frac {d^2 x^2}{c^2}\right )\right )}{(1+m) (2+m) \sqrt {1-\frac {d^2 x^2}{c^2}}} \] Input:

Integrate[((e*x)^m*(c^2 - d^2*x^2)^(5/2))/(c + d*x),x]
 

Output:

(c^2*x*(e*x)^m*Sqrt[c^2 - d^2*x^2]*(-(d*(1 + m)*x*Hypergeometric2F1[-3/2, 
1 + m/2, 2 + m/2, (d^2*x^2)/c^2]) + c*(2 + m)*Hypergeometric2F1[-3/2, (1 + 
 m)/2, (3 + m)/2, (d^2*x^2)/c^2]))/((1 + m)*(2 + m)*Sqrt[1 - (d^2*x^2)/c^2 
])
 

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {583, 557, 279, 278}

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.

\(\displaystyle \int \frac {\left (c^2-d^2 x^2\right )^{5/2} (e x)^m}{c+d x} \, dx\)

\(\Big \downarrow \) 583

\(\displaystyle \int (c-d x) \left (c^2-d^2 x^2\right )^{3/2} (e x)^mdx\)

\(\Big \downarrow \) 557

\(\displaystyle c \int (e x)^m \left (c^2-d^2 x^2\right )^{3/2}dx-\frac {d \int (e x)^{m+1} \left (c^2-d^2 x^2\right )^{3/2}dx}{e}\)

\(\Big \downarrow \) 279

\(\displaystyle \frac {c^3 \sqrt {c^2-d^2 x^2} \int (e x)^m \left (1-\frac {d^2 x^2}{c^2}\right )^{3/2}dx}{\sqrt {1-\frac {d^2 x^2}{c^2}}}-\frac {c^2 d \sqrt {c^2-d^2 x^2} \int (e x)^{m+1} \left (1-\frac {d^2 x^2}{c^2}\right )^{3/2}dx}{e \sqrt {1-\frac {d^2 x^2}{c^2}}}\)

\(\Big \downarrow \) 278

\(\displaystyle \frac {c^3 \sqrt {c^2-d^2 x^2} (e x)^{m+1} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {m+1}{2},\frac {m+3}{2},\frac {d^2 x^2}{c^2}\right )}{e (m+1) \sqrt {1-\frac {d^2 x^2}{c^2}}}-\frac {c^2 d \sqrt {c^2-d^2 x^2} (e x)^{m+2} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {m+2}{2},\frac {m+4}{2},\frac {d^2 x^2}{c^2}\right )}{e^2 (m+2) \sqrt {1-\frac {d^2 x^2}{c^2}}}\)

Input:

Int[((e*x)^m*(c^2 - d^2*x^2)^(5/2))/(c + d*x),x]
 

Output:

(c^3*(e*x)^(1 + m)*Sqrt[c^2 - d^2*x^2]*Hypergeometric2F1[-3/2, (1 + m)/2, 
(3 + m)/2, (d^2*x^2)/c^2])/(e*(1 + m)*Sqrt[1 - (d^2*x^2)/c^2]) - (c^2*d*(e 
*x)^(2 + m)*Sqrt[c^2 - d^2*x^2]*Hypergeometric2F1[-3/2, (2 + m)/2, (4 + m) 
/2, (d^2*x^2)/c^2])/(e^2*(2 + m)*Sqrt[1 - (d^2*x^2)/c^2])
 

Defintions of rubi rules used

rule 278

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^p*(( 
c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/2, (m + 1)/2 + 1, ( 
-b)*(x^2/a)], x] /; FreeQ[{a, b, c, m, p}, x] &&  !IGtQ[p, 0] && (ILtQ[p, 0 
] || GtQ[a, 0])
 

rule 279

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^IntP 
art[p]*((a + b*x^2)^FracPart[p]/(1 + b*(x^2/a))^FracPart[p])   Int[(c*x)^m* 
(1 + b*(x^2/a))^p, x], x] /; FreeQ[{a, b, c, m, p}, x] &&  !IGtQ[p, 0] && 
!(ILtQ[p, 0] || GtQ[a, 0])
 

rule 557

Int[((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_), x_Sym 
bol] :> Simp[c   Int[(e*x)^m*(a + b*x^2)^p, x], x] + Simp[d/e   Int[(e*x)^( 
m + 1)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x]
 

rule 583

Int[((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), 
x_Symbol] :> Simp[c^(2*n)/a^n   Int[(e*x)^m*((a + b*x^2)^(n + p)/(c - d*x)^ 
n), x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && EqQ[b*c^2 + a*d^2, 0] && I 
LtQ[n, 0]
 

Maple [F]

\[\int \frac {\left (e x \right )^{m} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {5}{2}}}{d x +c}d x\]

Input:

int((e*x)^m*(-d^2*x^2+c^2)^(5/2)/(d*x+c),x)
 

Output:

int((e*x)^m*(-d^2*x^2+c^2)^(5/2)/(d*x+c),x)
 

Fricas [F]

\[ \int \frac {(e x)^m \left (c^2-d^2 x^2\right )^{5/2}}{c+d x} \, dx=\int { \frac {{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {5}{2}} \left (e x\right )^{m}}{d x + c} \,d x } \] Input:

integrate((e*x)^m*(-d^2*x^2+c^2)^(5/2)/(d*x+c),x, algorithm="fricas")
 

Output:

integral((d^3*x^3 - c*d^2*x^2 - c^2*d*x + c^3)*sqrt(-d^2*x^2 + c^2)*(e*x)^ 
m, x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 11.12 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.49 \[ \int \frac {(e x)^m \left (c^2-d^2 x^2\right )^{5/2}}{c+d x} \, dx=\frac {c^{4} e^{m} x^{m + 1} \Gamma \left (\frac {m}{2} + \frac {1}{2}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {m}{2} + \frac {1}{2} \\ \frac {m}{2} + \frac {3}{2} \end {matrix}\middle | {\frac {d^{2} x^{2} e^{2 i \pi }}{c^{2}}} \right )}}{2 \Gamma \left (\frac {m}{2} + \frac {3}{2}\right )} - \frac {c^{3} d e^{m} x^{m + 2} \Gamma \left (\frac {m}{2} + 1\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {m}{2} + 1 \\ \frac {m}{2} + 2 \end {matrix}\middle | {\frac {d^{2} x^{2} e^{2 i \pi }}{c^{2}}} \right )}}{2 \Gamma \left (\frac {m}{2} + 2\right )} - \frac {c^{2} d^{2} e^{m} x^{m + 3} \Gamma \left (\frac {m}{2} + \frac {3}{2}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {m}{2} + \frac {3}{2} \\ \frac {m}{2} + \frac {5}{2} \end {matrix}\middle | {\frac {d^{2} x^{2} e^{2 i \pi }}{c^{2}}} \right )}}{2 \Gamma \left (\frac {m}{2} + \frac {5}{2}\right )} + \frac {c d^{3} e^{m} x^{m + 4} \Gamma \left (\frac {m}{2} + 2\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {m}{2} + 2 \\ \frac {m}{2} + 3 \end {matrix}\middle | {\frac {d^{2} x^{2} e^{2 i \pi }}{c^{2}}} \right )}}{2 \Gamma \left (\frac {m}{2} + 3\right )} \] Input:

integrate((e*x)**m*(-d**2*x**2+c**2)**(5/2)/(d*x+c),x)
 

Output:

c**4*e**m*x**(m + 1)*gamma(m/2 + 1/2)*hyper((-1/2, m/2 + 1/2), (m/2 + 3/2, 
), d**2*x**2*exp_polar(2*I*pi)/c**2)/(2*gamma(m/2 + 3/2)) - c**3*d*e**m*x* 
*(m + 2)*gamma(m/2 + 1)*hyper((-1/2, m/2 + 1), (m/2 + 2,), d**2*x**2*exp_p 
olar(2*I*pi)/c**2)/(2*gamma(m/2 + 2)) - c**2*d**2*e**m*x**(m + 3)*gamma(m/ 
2 + 3/2)*hyper((-1/2, m/2 + 3/2), (m/2 + 5/2,), d**2*x**2*exp_polar(2*I*pi 
)/c**2)/(2*gamma(m/2 + 5/2)) + c*d**3*e**m*x**(m + 4)*gamma(m/2 + 2)*hyper 
((-1/2, m/2 + 2), (m/2 + 3,), d**2*x**2*exp_polar(2*I*pi)/c**2)/(2*gamma(m 
/2 + 3))
 

Maxima [F]

\[ \int \frac {(e x)^m \left (c^2-d^2 x^2\right )^{5/2}}{c+d x} \, dx=\int { \frac {{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {5}{2}} \left (e x\right )^{m}}{d x + c} \,d x } \] Input:

integrate((e*x)^m*(-d^2*x^2+c^2)^(5/2)/(d*x+c),x, algorithm="maxima")
 

Output:

integrate((-d^2*x^2 + c^2)^(5/2)*(e*x)^m/(d*x + c), x)
 

Giac [F]

\[ \int \frac {(e x)^m \left (c^2-d^2 x^2\right )^{5/2}}{c+d x} \, dx=\int { \frac {{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {5}{2}} \left (e x\right )^{m}}{d x + c} \,d x } \] Input:

integrate((e*x)^m*(-d^2*x^2+c^2)^(5/2)/(d*x+c),x, algorithm="giac")
 

Output:

integrate((-d^2*x^2 + c^2)^(5/2)*(e*x)^m/(d*x + c), x)
                                                                                    
                                                                                    
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(e x)^m \left (c^2-d^2 x^2\right )^{5/2}}{c+d x} \, dx=\int \frac {{\left (c^2-d^2\,x^2\right )}^{5/2}\,{\left (e\,x\right )}^m}{c+d\,x} \,d x \] Input:

int(((c^2 - d^2*x^2)^(5/2)*(e*x)^m)/(c + d*x),x)
 

Output:

int(((c^2 - d^2*x^2)^(5/2)*(e*x)^m)/(c + d*x), x)
 

Reduce [F]

\[ \int \frac {(e x)^m \left (c^2-d^2 x^2\right )^{5/2}}{c+d x} \, dx=e^{m} \left (\left (\int x^{m} \sqrt {-d^{2} x^{2}+c^{2}}\, x^{3}d x \right ) d^{3}-\left (\int x^{m} \sqrt {-d^{2} x^{2}+c^{2}}\, x^{2}d x \right ) c \,d^{2}-\left (\int x^{m} \sqrt {-d^{2} x^{2}+c^{2}}\, x d x \right ) c^{2} d +\left (\int x^{m} \sqrt {-d^{2} x^{2}+c^{2}}d x \right ) c^{3}\right ) \] Input:

int((e*x)^m*(-d^2*x^2+c^2)^(5/2)/(d*x+c),x)
 

Output:

e**m*(int(x**m*sqrt(c**2 - d**2*x**2)*x**3,x)*d**3 - int(x**m*sqrt(c**2 - 
d**2*x**2)*x**2,x)*c*d**2 - int(x**m*sqrt(c**2 - d**2*x**2)*x,x)*c**2*d + 
int(x**m*sqrt(c**2 - d**2*x**2),x)*c**3)