\(\int \frac {(e x)^m (c+d x)^2}{(a+b x^2)^{5/2}} \, dx\) [180]

Optimal result

Integrand size = 24, antiderivative size = 211 \[ \int \frac {(e x)^m (c+d x)^2}{\left (a+b x^2\right )^{5/2}} \, dx=-\frac {d^2 (e x)^{1+m}}{b e (2-m) \left (a+b x^2\right )^{3/2}}+\frac {\left (b c^2 (2-m)+a d^2 (1+m)\right ) (e x)^{1+m} \sqrt {1+\frac {b x^2}{a}} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )}{a^2 b e (2-m) (1+m) \sqrt {a+b x^2}}+\frac {2 c d (e x)^{2+m} \sqrt {1+\frac {b x^2}{a}} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},\frac {2+m}{2},\frac {4+m}{2},-\frac {b x^2}{a}\right )}{a^2 e^2 (2+m) \sqrt {a+b x^2}} \] Output:

-d^2*(e*x)^(1+m)/b/e/(2-m)/(b*x^2+a)^(3/2)+(b*c^2*(2-m)+a*d^2*(1+m))*(e*x) 
^(1+m)*(1+b*x^2/a)^(1/2)*hypergeom([5/2, 1/2+1/2*m],[3/2+1/2*m],-b*x^2/a)/ 
a^2/b/e/(2-m)/(1+m)/(b*x^2+a)^(1/2)+2*c*d*(e*x)^(2+m)*(1+b*x^2/a)^(1/2)*hy 
pergeom([5/2, 1+1/2*m],[2+1/2*m],-b*x^2/a)/a^2/e^2/(2+m)/(b*x^2+a)^(1/2)
 

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.77 \[ \int \frac {(e x)^m (c+d x)^2}{\left (a+b x^2\right )^{5/2}} \, dx=\frac {x (e x)^m \sqrt {1+\frac {b x^2}{a}} \left (c^2 \left (6+5 m+m^2\right ) \operatorname {Hypergeometric2F1}\left (\frac {5}{2},\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )+d (1+m) x \left (2 c (3+m) \operatorname {Hypergeometric2F1}\left (\frac {5}{2},\frac {2+m}{2},\frac {4+m}{2},-\frac {b x^2}{a}\right )+d (2+m) x \operatorname {Hypergeometric2F1}\left (\frac {5}{2},\frac {3+m}{2},\frac {5+m}{2},-\frac {b x^2}{a}\right )\right )\right )}{a^2 (1+m) (2+m) (3+m) \sqrt {a+b x^2}} \] Input:

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

Output:

(x*(e*x)^m*Sqrt[1 + (b*x^2)/a]*(c^2*(6 + 5*m + m^2)*Hypergeometric2F1[5/2, 
 (1 + m)/2, (3 + m)/2, -((b*x^2)/a)] + d*(1 + m)*x*(2*c*(3 + m)*Hypergeome 
tric2F1[5/2, (2 + m)/2, (4 + m)/2, -((b*x^2)/a)] + d*(2 + m)*x*Hypergeomet 
ric2F1[5/2, (3 + m)/2, (5 + m)/2, -((b*x^2)/a)])))/(a^2*(1 + m)*(2 + m)*(3 
 + m)*Sqrt[a + b*x^2])
 

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.07, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {558, 25, 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.

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

\(\Big \downarrow \) 558

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 557

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

\(\Big \downarrow \) 279

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

\(\Big \downarrow \) 278

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

Input:

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

Output:

((e*x)^(1 + m)*(c^2 - (a*d^2)/b + 2*c*d*x))/(3*a*e*(a + b*x^2)^(3/2)) + (( 
(c^2*(2 - m) + (a*d^2*(1 + m))/b)*(e*x)^(1 + m)*Sqrt[1 + (b*x^2)/a]*Hyperg 
eometric2F1[3/2, (1 + m)/2, (3 + m)/2, -((b*x^2)/a)])/(a*e*(1 + m)*Sqrt[a 
+ b*x^2]) + (2*c*d*(1 - m)*(e*x)^(2 + m)*Sqrt[1 + (b*x^2)/a]*Hypergeometri 
c2F1[3/2, (2 + m)/2, (4 + m)/2, -((b*x^2)/a)])/(a*e^2*(2 + m)*Sqrt[a + b*x 
^2]))/(3*a)
 

Defintions of rubi rules used

rule 25

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

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

Integral((e*x)**m*(c + d*x)**2/(a + b*x**2)**(5/2), x)
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {(e x)^m (c+d x)^2}{\left (a+b x^2\right )^{5/2}} \, dx=e^{m} \left (\left (\int \frac {x^{m}}{\sqrt {b \,x^{2}+a}\, a^{2}+2 \sqrt {b \,x^{2}+a}\, a b \,x^{2}+\sqrt {b \,x^{2}+a}\, b^{2} x^{4}}d x \right ) c^{2}+\left (\int \frac {x^{m} x^{2}}{\sqrt {b \,x^{2}+a}\, a^{2}+2 \sqrt {b \,x^{2}+a}\, a b \,x^{2}+\sqrt {b \,x^{2}+a}\, b^{2} x^{4}}d x \right ) d^{2}+2 \left (\int \frac {x^{m} x}{\sqrt {b \,x^{2}+a}\, a^{2}+2 \sqrt {b \,x^{2}+a}\, a b \,x^{2}+\sqrt {b \,x^{2}+a}\, b^{2} x^{4}}d x \right ) c d \right ) \] Input:

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

Output:

e**m*(int(x**m/(sqrt(a + b*x**2)*a**2 + 2*sqrt(a + b*x**2)*a*b*x**2 + sqrt 
(a + b*x**2)*b**2*x**4),x)*c**2 + int((x**m*x**2)/(sqrt(a + b*x**2)*a**2 + 
 2*sqrt(a + b*x**2)*a*b*x**2 + sqrt(a + b*x**2)*b**2*x**4),x)*d**2 + 2*int 
((x**m*x)/(sqrt(a + b*x**2)*a**2 + 2*sqrt(a + b*x**2)*a*b*x**2 + sqrt(a + 
b*x**2)*b**2*x**4),x)*c*d)