\(\int \frac {(e x)^m}{(c+d x)^2 (c^2-d^2 x^2)^{3/2}} \, dx\) [1187]

Optimal result

Integrand size = 29, antiderivative size = 209 \[ \int \frac {(e x)^m}{(c+d x)^2 \left (c^2-d^2 x^2\right )^{3/2}} \, dx=\frac {(e x)^{1+m}}{e (4-m) \left (c^2-d^2 x^2\right )^{5/2}}+\frac {(3-2 m) (e x)^{1+m} \sqrt {1-\frac {d^2 x^2}{c^2}} \operatorname {Hypergeometric2F1}\left (\frac {7}{2},\frac {1+m}{2},\frac {3+m}{2},\frac {d^2 x^2}{c^2}\right )}{c^4 e (4-m) (1+m) \sqrt {c^2-d^2 x^2}}-\frac {2 d (e x)^{2+m} \sqrt {1-\frac {d^2 x^2}{c^2}} \operatorname {Hypergeometric2F1}\left (\frac {7}{2},\frac {2+m}{2},\frac {4+m}{2},\frac {d^2 x^2}{c^2}\right )}{c^5 e^2 (2+m) \sqrt {c^2-d^2 x^2}} \] Output:

(e*x)^(1+m)/e/(4-m)/(-d^2*x^2+c^2)^(5/2)+(3-2*m)*(e*x)^(1+m)*(1-d^2*x^2/c^ 
2)^(1/2)*hypergeom([7/2, 1/2+1/2*m],[3/2+1/2*m],d^2*x^2/c^2)/c^4/e/(4-m)/( 
1+m)/(-d^2*x^2+c^2)^(1/2)-2*d*(e*x)^(2+m)*(1-d^2*x^2/c^2)^(1/2)*hypergeom( 
[7/2, 1+1/2*m],[2+1/2*m],d^2*x^2/c^2)/c^5/e^2/(2+m)/(-d^2*x^2+c^2)^(1/2)
 

Mathematica [A] (verified)

Time = 0.64 (sec) , antiderivative size = 176, normalized size of antiderivative = 0.84 \[ \int \frac {(e x)^m}{(c+d x)^2 \left (c^2-d^2 x^2\right )^{3/2}} \, dx=\frac {x (e x)^m \sqrt {1-\frac {d^2 x^2}{c^2}} \left (c^2 \left (6+5 m+m^2\right ) \operatorname {Hypergeometric2F1}\left (\frac {7}{2},\frac {1+m}{2},\frac {3+m}{2},\frac {d^2 x^2}{c^2}\right )-d (1+m) x \left (2 c (3+m) \operatorname {Hypergeometric2F1}\left (\frac {7}{2},\frac {2+m}{2},\frac {4+m}{2},\frac {d^2 x^2}{c^2}\right )-d (2+m) x \operatorname {Hypergeometric2F1}\left (\frac {7}{2},\frac {3+m}{2},\frac {5+m}{2},\frac {d^2 x^2}{c^2}\right )\right )\right )}{c^6 (1+m) (2+m) (3+m) \sqrt {c^2-d^2 x^2}} \] Input:

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

Output:

(x*(e*x)^m*Sqrt[1 - (d^2*x^2)/c^2]*(c^2*(6 + 5*m + m^2)*Hypergeometric2F1[ 
7/2, (1 + m)/2, (3 + m)/2, (d^2*x^2)/c^2] - d*(1 + m)*x*(2*c*(3 + m)*Hyper 
geometric2F1[7/2, (2 + m)/2, (4 + m)/2, (d^2*x^2)/c^2] - d*(2 + m)*x*Hyper 
geometric2F1[7/2, (3 + m)/2, (5 + m)/2, (d^2*x^2)/c^2])))/(c^6*(1 + m)*(2 
+ m)*(3 + m)*Sqrt[c^2 - d^2*x^2])
 

Rubi [A] (verified)

Time = 0.59 (sec) , antiderivative size = 220, normalized size of antiderivative = 1.05, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {570, 558, 25, 27, 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.

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

\(\Big \downarrow \) 570

\(\displaystyle \int \frac {(c-d x)^2 (e x)^m}{\left (c^2-d^2 x^2\right )^{7/2}}dx\)

\(\Big \downarrow \) 558

\(\displaystyle \frac {2 (c-d x) (e x)^{m+1}}{5 c e \left (c^2-d^2 x^2\right )^{5/2}}-\frac {\int -\frac {c (e x)^m (c (3-2 m)-2 d (3-m) x)}{\left (c^2-d^2 x^2\right )^{5/2}}dx}{5 c^2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\int \frac {c (e x)^m (c (3-2 m)-2 d (3-m) x)}{\left (c^2-d^2 x^2\right )^{5/2}}dx}{5 c^2}+\frac {2 (c-d x) (e x)^{m+1}}{5 c e \left (c^2-d^2 x^2\right )^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {(e x)^m (c (3-2 m)-2 d (3-m) x)}{\left (c^2-d^2 x^2\right )^{5/2}}dx}{5 c}+\frac {2 (c-d x) (e x)^{m+1}}{5 c e \left (c^2-d^2 x^2\right )^{5/2}}\)

\(\Big \downarrow \) 557

\(\displaystyle \frac {c (3-2 m) \int \frac {(e x)^m}{\left (c^2-d^2 x^2\right )^{5/2}}dx-\frac {2 d (3-m) \int \frac {(e x)^{m+1}}{\left (c^2-d^2 x^2\right )^{5/2}}dx}{e}}{5 c}+\frac {2 (c-d x) (e x)^{m+1}}{5 c e \left (c^2-d^2 x^2\right )^{5/2}}\)

\(\Big \downarrow \) 279

\(\displaystyle \frac {\frac {(3-2 m) \sqrt {1-\frac {d^2 x^2}{c^2}} \int \frac {(e x)^m}{\left (1-\frac {d^2 x^2}{c^2}\right )^{5/2}}dx}{c^3 \sqrt {c^2-d^2 x^2}}-\frac {2 d (3-m) \sqrt {1-\frac {d^2 x^2}{c^2}} \int \frac {(e x)^{m+1}}{\left (1-\frac {d^2 x^2}{c^2}\right )^{5/2}}dx}{c^4 e \sqrt {c^2-d^2 x^2}}}{5 c}+\frac {2 (c-d x) (e x)^{m+1}}{5 c e \left (c^2-d^2 x^2\right )^{5/2}}\)

\(\Big \downarrow \) 278

\(\displaystyle \frac {2 (c-d x) (e x)^{m+1}}{5 c e \left (c^2-d^2 x^2\right )^{5/2}}+\frac {\frac {(3-2 m) \sqrt {1-\frac {d^2 x^2}{c^2}} (e x)^{m+1} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},\frac {m+1}{2},\frac {m+3}{2},\frac {d^2 x^2}{c^2}\right )}{c^3 e (m+1) \sqrt {c^2-d^2 x^2}}-\frac {2 d (3-m) \sqrt {1-\frac {d^2 x^2}{c^2}} (e x)^{m+2} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},\frac {m+2}{2},\frac {m+4}{2},\frac {d^2 x^2}{c^2}\right )}{c^4 e^2 (m+2) \sqrt {c^2-d^2 x^2}}}{5 c}\)

Input:

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

Output:

(2*(e*x)^(1 + m)*(c - d*x))/(5*c*e*(c^2 - d^2*x^2)^(5/2)) + (((3 - 2*m)*(e 
*x)^(1 + m)*Sqrt[1 - (d^2*x^2)/c^2]*Hypergeometric2F1[5/2, (1 + m)/2, (3 + 
 m)/2, (d^2*x^2)/c^2])/(c^3*e*(1 + m)*Sqrt[c^2 - d^2*x^2]) - (2*d*(3 - m)* 
(e*x)^(2 + m)*Sqrt[1 - (d^2*x^2)/c^2]*Hypergeometric2F1[5/2, (2 + m)/2, (4 
 + m)/2, (d^2*x^2)/c^2])/(c^4*e^2*(2 + m)*Sqrt[c^2 - d^2*x^2]))/(5*c)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

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 558

Int[((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_), 
 x_Symbol] :> With[{Qx = PolynomialQuotient[(c + d*x)^n, a + b*x^2, x], f = 
 Coeff[PolynomialRemainder[(c + d*x)^n, a + b*x^2, x], x, 0], g = Coeff[Pol 
ynomialRemainder[(c + d*x)^n, a + b*x^2, x], x, 1]}, Simp[(-(e*x)^(m + 1))* 
(f + g*x)*((a + b*x^2)^(p + 1)/(2*a*e*(p + 1))), x] + Simp[1/(2*a*(p + 1)) 
  Int[(e*x)^m*(a + b*x^2)^(p + 1)*ExpandToSum[2*a*(p + 1)*Qx + f*(m + 2*p + 
 3) + g*(m + 2*p + 4)*x, x], x], x]] /; FreeQ[{a, b, c, d, e, m}, x] && IGt 
Q[n, 1] &&  !IntegerQ[m] && LtQ[p, -1]
 

rule 570

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, -1] &&  !(IGtQ[m, 0] && ILtQ[m + n, 0] &&  !GtQ[p, 1])
 

Maple [F]

\[\int \frac {\left (e x \right )^{m}}{\left (d x +c \right )^{2} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}d x\]

Input:

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

Output:

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

Fricas [F]

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

integrate((e*x)^m/(d*x+c)^2/(-d^2*x^2+c^2)^(3/2),x, algorithm="fricas")
 

Output:

integral(sqrt(-d^2*x^2 + c^2)*(e*x)^m/(d^6*x^6 + 2*c*d^5*x^5 - c^2*d^4*x^4 
 - 4*c^3*d^3*x^3 - c^4*d^2*x^2 + 2*c^5*d*x + c^6), x)
 

Sympy [F]

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

integrate((e*x)**m/(d*x+c)**2/(-d**2*x**2+c**2)**(3/2),x)
 

Output:

Integral((e*x)**m/((-(-c + d*x)*(c + d*x))**(3/2)*(c + d*x)**2), x)
 

Maxima [F]

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

integrate((e*x)^m/(d*x+c)^2/(-d^2*x^2+c^2)^(3/2),x, algorithm="maxima")
 

Output:

integrate((e*x)^m/((-d^2*x^2 + c^2)^(3/2)*(d*x + c)^2), x)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {(e x)^m}{(c+d x)^2 \left (c^2-d^2 x^2\right )^{3/2}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((e*x)^m/(d*x+c)^2/(-d^2*x^2+c^2)^(3/2),x, algorithm="giac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Unable to divide, perhaps due to ro 
unding error%%%{1,[0,1,4,0,0,0]%%%}+%%%{3,[0,1,2,0,0,0]%%%}+%%%{3,[0,1,0,0 
,0,0]%%%}
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {(e x)^m}{(c+d x)^2 \left (c^2-d^2 x^2\right )^{3/2}} \, dx=e^{m} \left (\int \frac {x^{m}}{\sqrt {-d^{2} x^{2}+c^{2}}\, c^{4}+2 \sqrt {-d^{2} x^{2}+c^{2}}\, c^{3} d x -2 \sqrt {-d^{2} x^{2}+c^{2}}\, c \,d^{3} x^{3}-\sqrt {-d^{2} x^{2}+c^{2}}\, d^{4} x^{4}}d x \right ) \] Input:

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

Output:

e**m*int(x**m/(sqrt(c**2 - d**2*x**2)*c**4 + 2*sqrt(c**2 - d**2*x**2)*c**3 
*d*x - 2*sqrt(c**2 - d**2*x**2)*c*d**3*x**3 - sqrt(c**2 - d**2*x**2)*d**4* 
x**4),x)