3.3.32 \(\int \frac {(g x)^m (d^2-e^2 x^2)^{5/2}}{(d+e x)^3} \, dx\) [232]

3.3.32.1 Optimal result

Integrand size = 29, antiderivative size = 250 \[ \int \frac {(g x)^m \left (d^2-e^2 x^2\right )^{5/2}}{(d+e x)^3} \, dx=-\frac {3 d (g x)^{1+m} \sqrt {d^2-e^2 x^2}}{g (2+m)}+\frac {e (g x)^{2+m} \sqrt {d^2-e^2 x^2}}{g^2 (3+m)}+\frac {d^3 (5+4 m) (g x)^{1+m} \sqrt {1-\frac {e^2 x^2}{d^2}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},\frac {e^2 x^2}{d^2}\right )}{g (1+m) (2+m) \sqrt {d^2-e^2 x^2}}-\frac {d^2 e (11+4 m) (g x)^{2+m} \sqrt {1-\frac {e^2 x^2}{d^2}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+m}{2},\frac {4+m}{2},\frac {e^2 x^2}{d^2}\right )}{g^2 (2+m) (3+m) \sqrt {d^2-e^2 x^2}} \]

output

-3*d*(g*x)^(1+m)*(-e^2*x^2+d^2)^(1/2)/g/(2+m)+e*(g*x)^(2+m)*(-e^2*x^2+d^2) 
^(1/2)/g^2/(3+m)+d^3*(5+4*m)*(g*x)^(1+m)*hypergeom([1/2, 1/2+1/2*m],[3/2+1 
/2*m],e^2*x^2/d^2)*(1-e^2*x^2/d^2)^(1/2)/g/(1+m)/(2+m)/(-e^2*x^2+d^2)^(1/2 
)-d^2*e*(11+4*m)*(g*x)^(2+m)*hypergeom([1/2, 1+1/2*m],[2+1/2*m],e^2*x^2/d^ 
2)*(1-e^2*x^2/d^2)^(1/2)/g^2/(2+m)/(3+m)/(-e^2*x^2+d^2)^(1/2)
 

3.3.32.2 Mathematica [A] (verified)

Time = 0.74 (sec) , antiderivative size = 245, normalized size of antiderivative = 0.98 \[ \int \frac {(g x)^m \left (d^2-e^2 x^2\right )^{5/2}}{(d+e x)^3} \, dx=\frac {x (g x)^m \sqrt {d^2-e^2 x^2} \sqrt {1-\frac {e^2 x^2}{d^2}} \left (d^3 \left (24+26 m+9 m^2+m^3\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},\frac {e^2 x^2}{d^2}\right )-e (1+m) x \left (3 d^2 \left (12+7 m+m^2\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+m}{2},\frac {4+m}{2},\frac {e^2 x^2}{d^2}\right )+e (2+m) x \left (-3 d (4+m) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3+m}{2},\frac {5+m}{2},\frac {e^2 x^2}{d^2}\right )+e (3+m) x \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {4+m}{2},\frac {6+m}{2},\frac {e^2 x^2}{d^2}\right )\right )\right )\right )}{(1+m) (2+m) (3+m) (4+m) (d-e x) (d+e x)} \]

input

Integrate[((g*x)^m*(d^2 - e^2*x^2)^(5/2))/(d + e*x)^3,x]
 

output

(x*(g*x)^m*Sqrt[d^2 - e^2*x^2]*Sqrt[1 - (e^2*x^2)/d^2]*(d^3*(24 + 26*m + 9 
*m^2 + m^3)*Hypergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, (e^2*x^2)/d^2] - 
e*(1 + m)*x*(3*d^2*(12 + 7*m + m^2)*Hypergeometric2F1[1/2, (2 + m)/2, (4 + 
 m)/2, (e^2*x^2)/d^2] + e*(2 + m)*x*(-3*d*(4 + m)*Hypergeometric2F1[1/2, ( 
3 + m)/2, (5 + m)/2, (e^2*x^2)/d^2] + e*(3 + m)*x*Hypergeometric2F1[1/2, ( 
4 + m)/2, (6 + m)/2, (e^2*x^2)/d^2]))))/((1 + m)*(2 + m)*(3 + m)*(4 + m)*( 
d - e*x)*(d + e*x))
 

3.3.32.3 Rubi [A] (verified)

Time = 0.55 (sec) , antiderivative size = 262, normalized size of antiderivative = 1.05, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.310, Rules used = {570, 559, 25, 2340, 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.

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

\(\Big \downarrow \) 570

\(\displaystyle \int \frac {(d-e x)^3 (g x)^m}{\sqrt {d^2-e^2 x^2}}dx\)

\(\Big \downarrow \) 559

\(\displaystyle \frac {e \sqrt {d^2-e^2 x^2} (g x)^{m+2}}{g^2 (m+3)}-\frac {\int -\frac {(g x)^m \left (3 d (m+3) x^2 e^4-d^2 (4 m+11) x e^3+d^3 (m+3) e^2\right )}{\sqrt {d^2-e^2 x^2}}dx}{e^2 (m+3)}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\int \frac {(g x)^m \left (3 d (m+3) x^2 e^4-d^2 (4 m+11) x e^3+d^3 (m+3) e^2\right )}{\sqrt {d^2-e^2 x^2}}dx}{e^2 (m+3)}+\frac {e \sqrt {d^2-e^2 x^2} (g x)^{m+2}}{g^2 (m+3)}\)

\(\Big \downarrow \) 2340

\(\displaystyle \frac {-\frac {\int -\frac {d^2 e^4 (g x)^m (d (m+3) (4 m+5)-e (m+2) (4 m+11) x)}{\sqrt {d^2-e^2 x^2}}dx}{e^2 (m+2)}-\frac {3 d e^2 (m+3) \sqrt {d^2-e^2 x^2} (g x)^{m+1}}{g (m+2)}}{e^2 (m+3)}+\frac {e \sqrt {d^2-e^2 x^2} (g x)^{m+2}}{g^2 (m+3)}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\int \frac {d^2 e^4 (g x)^m (d (m+3) (4 m+5)-e (m+2) (4 m+11) x)}{\sqrt {d^2-e^2 x^2}}dx}{e^2 (m+2)}-\frac {3 d e^2 (m+3) \sqrt {d^2-e^2 x^2} (g x)^{m+1}}{g (m+2)}}{e^2 (m+3)}+\frac {e \sqrt {d^2-e^2 x^2} (g x)^{m+2}}{g^2 (m+3)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {d^2 e^2 \int \frac {(g x)^m (d (m+3) (4 m+5)-e (m+2) (4 m+11) x)}{\sqrt {d^2-e^2 x^2}}dx}{m+2}-\frac {3 d e^2 (m+3) \sqrt {d^2-e^2 x^2} (g x)^{m+1}}{g (m+2)}}{e^2 (m+3)}+\frac {e \sqrt {d^2-e^2 x^2} (g x)^{m+2}}{g^2 (m+3)}\)

\(\Big \downarrow \) 557

\(\displaystyle \frac {\frac {d^2 e^2 \left (d (m+3) (4 m+5) \int \frac {(g x)^m}{\sqrt {d^2-e^2 x^2}}dx-\frac {e (m+2) (4 m+11) \int \frac {(g x)^{m+1}}{\sqrt {d^2-e^2 x^2}}dx}{g}\right )}{m+2}-\frac {3 d e^2 (m+3) \sqrt {d^2-e^2 x^2} (g x)^{m+1}}{g (m+2)}}{e^2 (m+3)}+\frac {e \sqrt {d^2-e^2 x^2} (g x)^{m+2}}{g^2 (m+3)}\)

\(\Big \downarrow \) 279

\(\displaystyle \frac {\frac {d^2 e^2 \left (\frac {d (m+3) (4 m+5) \sqrt {1-\frac {e^2 x^2}{d^2}} \int \frac {(g x)^m}{\sqrt {1-\frac {e^2 x^2}{d^2}}}dx}{\sqrt {d^2-e^2 x^2}}-\frac {e (m+2) (4 m+11) \sqrt {1-\frac {e^2 x^2}{d^2}} \int \frac {(g x)^{m+1}}{\sqrt {1-\frac {e^2 x^2}{d^2}}}dx}{g \sqrt {d^2-e^2 x^2}}\right )}{m+2}-\frac {3 d e^2 (m+3) \sqrt {d^2-e^2 x^2} (g x)^{m+1}}{g (m+2)}}{e^2 (m+3)}+\frac {e \sqrt {d^2-e^2 x^2} (g x)^{m+2}}{g^2 (m+3)}\)

\(\Big \downarrow \) 278

\(\displaystyle \frac {\frac {d^2 e^2 \left (\frac {d (m+3) (4 m+5) \sqrt {1-\frac {e^2 x^2}{d^2}} (g x)^{m+1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+1}{2},\frac {m+3}{2},\frac {e^2 x^2}{d^2}\right )}{g (m+1) \sqrt {d^2-e^2 x^2}}-\frac {e (4 m+11) \sqrt {1-\frac {e^2 x^2}{d^2}} (g x)^{m+2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+2}{2},\frac {m+4}{2},\frac {e^2 x^2}{d^2}\right )}{g^2 \sqrt {d^2-e^2 x^2}}\right )}{m+2}-\frac {3 d e^2 (m+3) \sqrt {d^2-e^2 x^2} (g x)^{m+1}}{g (m+2)}}{e^2 (m+3)}+\frac {e \sqrt {d^2-e^2 x^2} (g x)^{m+2}}{g^2 (m+3)}\)

input

Int[((g*x)^m*(d^2 - e^2*x^2)^(5/2))/(d + e*x)^3,x]
 

output

(e*(g*x)^(2 + m)*Sqrt[d^2 - e^2*x^2])/(g^2*(3 + m)) + ((-3*d*e^2*(3 + m)*( 
g*x)^(1 + m)*Sqrt[d^2 - e^2*x^2])/(g*(2 + m)) + (d^2*e^2*((d*(3 + m)*(5 + 
4*m)*(g*x)^(1 + m)*Sqrt[1 - (e^2*x^2)/d^2]*Hypergeometric2F1[1/2, (1 + m)/ 
2, (3 + m)/2, (e^2*x^2)/d^2])/(g*(1 + m)*Sqrt[d^2 - e^2*x^2]) - (e*(11 + 4 
*m)*(g*x)^(2 + m)*Sqrt[1 - (e^2*x^2)/d^2]*Hypergeometric2F1[1/2, (2 + m)/2 
, (4 + m)/2, (e^2*x^2)/d^2])/(g^2*Sqrt[d^2 - e^2*x^2])))/(2 + m))/(e^2*(3 
+ m))
 

3.3.32.3.1 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 559

Int[((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), 
x_Symbol] :> Simp[d^n*(e*x)^(m + n - 1)*((a + b*x^2)^(p + 1)/(b*e^(n - 1)*( 
m + n + 2*p + 1))), x] + Simp[1/(b*(m + n + 2*p + 1))   Int[(e*x)^m*(a + b* 
x^2)^p*ExpandToSum[b*(m + n + 2*p + 1)*(c + d*x)^n - b*d^n*(m + n + 2*p + 1 
)*x^n - a*d^n*(m + n - 1)*x^(n - 2), x], x], x] /; FreeQ[{a, b, c, d, e, m, 
 p}, x] && IGtQ[n, 1] &&  !IntegerQ[m] && NeQ[m + n + 2*p + 1, 0]
 

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])
 

rule 2340

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[ 
{q = Expon[Pq, x], f = Coeff[Pq, x, Expon[Pq, x]]}, Simp[f*(c*x)^(m + q - 1 
)*((a + b*x^2)^(p + 1)/(b*c^(q - 1)*(m + q + 2*p + 1))), x] + Simp[1/(b*(m 
+ q + 2*p + 1))   Int[(c*x)^m*(a + b*x^2)^p*ExpandToSum[b*(m + q + 2*p + 1) 
*Pq - b*f*(m + q + 2*p + 1)*x^q - a*f*(m + q - 1)*x^(q - 2), x], x], x] /; 
GtQ[q, 1] && NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, b, c, m, p}, x] && PolyQ 
[Pq, x] && ( !IGtQ[m, 0] || IGtQ[p + 1/2, -1])
 

3.3.32.4 Maple [F]

\[\int \frac {\left (g x \right )^{m} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}{\left (e x +d \right )^{3}}d x\]

input

int((g*x)^m*(-e^2*x^2+d^2)^(5/2)/(e*x+d)^3,x)
 

output

int((g*x)^m*(-e^2*x^2+d^2)^(5/2)/(e*x+d)^3,x)
 

3.3.32.5 Fricas [F]

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

input

integrate((g*x)^m*(-e^2*x^2+d^2)^(5/2)/(e*x+d)^3,x, algorithm="fricas")
 

output

integral((e^2*x^2 - 2*d*e*x + d^2)*sqrt(-e^2*x^2 + d^2)*(g*x)^m/(e*x + d), 
 x)
 

3.3.32.6 Sympy [F(-2)]

Exception generated. \[ \int \frac {(g x)^m \left (d^2-e^2 x^2\right )^{5/2}}{(d+e x)^3} \, dx=\text {Exception raised: HeuristicGCDFailed} \]

input

integrate((g*x)**m*(-e**2*x**2+d**2)**(5/2)/(e*x+d)**3,x)
 

output

Exception raised: HeuristicGCDFailed >> no luck
 

3.3.32.7 Maxima [F]

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

input

integrate((g*x)^m*(-e^2*x^2+d^2)^(5/2)/(e*x+d)^3,x, algorithm="maxima")
 

output

integrate((-e^2*x^2 + d^2)^(5/2)*(g*x)^m/(e*x + d)^3, x)
 

3.3.32.8 Giac [F]

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

input

integrate((g*x)^m*(-e^2*x^2+d^2)^(5/2)/(e*x+d)^3,x, algorithm="giac")
 

output

integrate((-e^2*x^2 + d^2)^(5/2)*(g*x)^m/(e*x + d)^3, x)
 

3.3.32.9 Mupad [F(-1)]

Timed out. \[ \int \frac {(g x)^m \left (d^2-e^2 x^2\right )^{5/2}}{(d+e x)^3} \, dx=\int \frac {{\left (d^2-e^2\,x^2\right )}^{5/2}\,{\left (g\,x\right )}^m}{{\left (d+e\,x\right )}^3} \,d x \]

input

int(((d^2 - e^2*x^2)^(5/2)*(g*x)^m)/(d + e*x)^3,x)
 

output

int(((d^2 - e^2*x^2)^(5/2)*(g*x)^m)/(d + e*x)^3, x)