\(\int (e x)^m (c+d x)^2 (a+b x^2)^{3/2} \, dx\) [162]

Optimal result

Integrand size = 24, antiderivative size = 202 \[ \int (e x)^m (c+d x)^2 \left (a+b x^2\right )^{3/2} \, dx=\frac {d^2 (e x)^{1+m} \left (a+b x^2\right )^{5/2}}{b e (6+m)}-\frac {a \left (a d^2 (1+m)-b c^2 (6+m)\right ) (e x)^{1+m} \sqrt {a+b x^2} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )}{b e (1+m) (6+m) \sqrt {1+\frac {b x^2}{a}}}+\frac {2 a c d (e x)^{2+m} \sqrt {a+b x^2} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {2+m}{2},\frac {4+m}{2},-\frac {b x^2}{a}\right )}{e^2 (2+m) \sqrt {1+\frac {b x^2}{a}}} \] Output:

d^2*(e*x)^(1+m)*(b*x^2+a)^(5/2)/b/e/(6+m)-a*(a*d^2*(1+m)-b*c^2*(6+m))*(e*x 
)^(1+m)*(b*x^2+a)^(1/2)*hypergeom([-3/2, 1/2+1/2*m],[3/2+1/2*m],-b*x^2/a)/ 
b/e/(1+m)/(6+m)/(1+b*x^2/a)^(1/2)+2*a*c*d*(e*x)^(2+m)*(b*x^2+a)^(1/2)*hype 
rgeom([-3/2, 1+1/2*m],[2+1/2*m],-b*x^2/a)/e^2/(2+m)/(1+b*x^2/a)^(1/2)
 

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 161, normalized size of antiderivative = 0.80 \[ \int (e x)^m (c+d x)^2 \left (a+b x^2\right )^{3/2} \, dx=\frac {a x (e x)^m \sqrt {a+b x^2} \left (c^2 \left (6+5 m+m^2\right ) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )+d (1+m) x \left (2 c (3+m) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {2+m}{2},\frac {4+m}{2},-\frac {b x^2}{a}\right )+d (2+m) x \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {3+m}{2},\frac {5+m}{2},-\frac {b x^2}{a}\right )\right )\right )}{(1+m) (2+m) (3+m) \sqrt {1+\frac {b x^2}{a}}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.03, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {559, 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.

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

\(\Big \downarrow \) 559

\(\displaystyle \frac {\int -(e x)^m \left (-b (m+6) c^2-2 b d (m+6) x c+a d^2 (m+1)\right ) \left (b x^2+a\right )^{3/2}dx}{b (m+6)}+\frac {d^2 \left (a+b x^2\right )^{5/2} (e x)^{m+1}}{b e (m+6)}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {d^2 \left (a+b x^2\right )^{5/2} (e x)^{m+1}}{b e (m+6)}-\frac {\int (e x)^m \left (-b (m+6) c^2-2 b d (m+6) x c+a d^2 (m+1)\right ) \left (b x^2+a\right )^{3/2}dx}{b (m+6)}\)

\(\Big \downarrow \) 557

\(\displaystyle \frac {d^2 \left (a+b x^2\right )^{5/2} (e x)^{m+1}}{b e (m+6)}-\frac {\left (a d^2 (m+1)-b c^2 (m+6)\right ) \int (e x)^m \left (b x^2+a\right )^{3/2}dx-\frac {2 b c d (m+6) \int (e x)^{m+1} \left (b x^2+a\right )^{3/2}dx}{e}}{b (m+6)}\)

\(\Big \downarrow \) 279

\(\displaystyle \frac {d^2 \left (a+b x^2\right )^{5/2} (e x)^{m+1}}{b e (m+6)}-\frac {\frac {a \sqrt {a+b x^2} \left (a d^2 (m+1)-b c^2 (m+6)\right ) \int (e x)^m \left (\frac {b x^2}{a}+1\right )^{3/2}dx}{\sqrt {\frac {b x^2}{a}+1}}-\frac {2 a b c d (m+6) \sqrt {a+b x^2} \int (e x)^{m+1} \left (\frac {b x^2}{a}+1\right )^{3/2}dx}{e \sqrt {\frac {b x^2}{a}+1}}}{b (m+6)}\)

\(\Big \downarrow \) 278

\(\displaystyle \frac {d^2 \left (a+b x^2\right )^{5/2} (e x)^{m+1}}{b e (m+6)}-\frac {\frac {a \sqrt {a+b x^2} (e x)^{m+1} \left (a d^2 (m+1)-b c^2 (m+6)\right ) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {m+1}{2},\frac {m+3}{2},-\frac {b x^2}{a}\right )}{e (m+1) \sqrt {\frac {b x^2}{a}+1}}-\frac {2 a b c d (m+6) \sqrt {a+b x^2} (e x)^{m+2} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {m+2}{2},\frac {m+4}{2},-\frac {b x^2}{a}\right )}{e^2 (m+2) \sqrt {\frac {b x^2}{a}+1}}}{b (m+6)}\)

Input:

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

Output:

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

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 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]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 5.54 (sec) , antiderivative size = 364, normalized size of antiderivative = 1.80 \[ \int (e x)^m (c+d x)^2 \left (a+b x^2\right )^{3/2} \, dx=\frac {a^{\frac {3}{2}} c^{2} e^{m} x^{m + 1} \Gamma \left (\frac {m}{2} + \frac {1}{2}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {m}{2} + \frac {1}{2} \\ \frac {m}{2} + \frac {3}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \Gamma \left (\frac {m}{2} + \frac {3}{2}\right )} + \frac {a^{\frac {3}{2}} c d e^{m} x^{m + 2} \Gamma \left (\frac {m}{2} + 1\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {m}{2} + 1 \\ \frac {m}{2} + 2 \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{\Gamma \left (\frac {m}{2} + 2\right )} + \frac {a^{\frac {3}{2}} d^{2} e^{m} x^{m + 3} \Gamma \left (\frac {m}{2} + \frac {3}{2}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {m}{2} + \frac {3}{2} \\ \frac {m}{2} + \frac {5}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \Gamma \left (\frac {m}{2} + \frac {5}{2}\right )} + \frac {\sqrt {a} b c^{2} e^{m} x^{m + 3} \Gamma \left (\frac {m}{2} + \frac {3}{2}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {m}{2} + \frac {3}{2} \\ \frac {m}{2} + \frac {5}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \Gamma \left (\frac {m}{2} + \frac {5}{2}\right )} + \frac {\sqrt {a} b c d e^{m} x^{m + 4} \Gamma \left (\frac {m}{2} + 2\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {m}{2} + 2 \\ \frac {m}{2} + 3 \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{\Gamma \left (\frac {m}{2} + 3\right )} + \frac {\sqrt {a} b d^{2} e^{m} x^{m + 5} \Gamma \left (\frac {m}{2} + \frac {5}{2}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {m}{2} + \frac {5}{2} \\ \frac {m}{2} + \frac {7}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \Gamma \left (\frac {m}{2} + \frac {7}{2}\right )} \] Input:

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

Output:

a**(3/2)*c**2*e**m*x**(m + 1)*gamma(m/2 + 1/2)*hyper((-1/2, m/2 + 1/2), (m 
/2 + 3/2,), b*x**2*exp_polar(I*pi)/a)/(2*gamma(m/2 + 3/2)) + a**(3/2)*c*d* 
e**m*x**(m + 2)*gamma(m/2 + 1)*hyper((-1/2, m/2 + 1), (m/2 + 2,), b*x**2*e 
xp_polar(I*pi)/a)/gamma(m/2 + 2) + a**(3/2)*d**2*e**m*x**(m + 3)*gamma(m/2 
 + 3/2)*hyper((-1/2, m/2 + 3/2), (m/2 + 5/2,), b*x**2*exp_polar(I*pi)/a)/( 
2*gamma(m/2 + 5/2)) + sqrt(a)*b*c**2*e**m*x**(m + 3)*gamma(m/2 + 3/2)*hype 
r((-1/2, m/2 + 3/2), (m/2 + 5/2,), b*x**2*exp_polar(I*pi)/a)/(2*gamma(m/2 
+ 5/2)) + sqrt(a)*b*c*d*e**m*x**(m + 4)*gamma(m/2 + 2)*hyper((-1/2, m/2 + 
2), (m/2 + 3,), b*x**2*exp_polar(I*pi)/a)/gamma(m/2 + 3) + sqrt(a)*b*d**2* 
e**m*x**(m + 5)*gamma(m/2 + 5/2)*hyper((-1/2, m/2 + 5/2), (m/2 + 7/2,), b* 
x**2*exp_polar(I*pi)/a)/(2*gamma(m/2 + 7/2))
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

e**m*(int(x**m*sqrt(a + b*x**2)*x**4,x)*b*d**2 + 2*int(x**m*sqrt(a + b*x** 
2)*x**3,x)*b*c*d + int(x**m*sqrt(a + b*x**2)*x**2,x)*a*d**2 + int(x**m*sqr 
t(a + b*x**2)*x**2,x)*b*c**2 + 2*int(x**m*sqrt(a + b*x**2)*x,x)*a*c*d + in 
t(x**m*sqrt(a + b*x**2),x)*a*c**2)