\(\int (d x)^m (a^2+2 a b x^2+b^2 x^4)^{3/2} \, dx\) [679]

Optimal result

Integrand size = 28, antiderivative size = 205 \[ \int (d x)^m \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2} \, dx=\frac {a^3 (d x)^{1+m} \sqrt {a^2+2 a b x^2+b^2 x^4}}{d (1+m) \left (a+b x^2\right )}+\frac {3 a^2 b (d x)^{3+m} \sqrt {a^2+2 a b x^2+b^2 x^4}}{d^3 (3+m) \left (a+b x^2\right )}+\frac {3 a b^2 (d x)^{5+m} \sqrt {a^2+2 a b x^2+b^2 x^4}}{d^5 (5+m) \left (a+b x^2\right )}+\frac {b^3 (d x)^{7+m} \sqrt {a^2+2 a b x^2+b^2 x^4}}{d^7 (7+m) \left (a+b x^2\right )} \] Output:

a^3*(d*x)^(1+m)*((b*x^2+a)^2)^(1/2)/d/(1+m)/(b*x^2+a)+3*a^2*b*(d*x)^(3+m)* 
((b*x^2+a)^2)^(1/2)/d^3/(3+m)/(b*x^2+a)+3*a*b^2*(d*x)^(5+m)*((b*x^2+a)^2)^ 
(1/2)/d^5/(5+m)/(b*x^2+a)+b^3*(d*x)^(7+m)*((b*x^2+a)^2)^(1/2)/d^7/(7+m)/(b 
*x^2+a)
 

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.64 \[ \int (d x)^m \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2} \, dx=\frac {x (d x)^m \sqrt {\left (a+b x^2\right )^2} \left (a^3 \left (105+71 m+15 m^2+m^3\right )+3 a^2 b \left (35+47 m+13 m^2+m^3\right ) x^2+3 a b^2 \left (21+31 m+11 m^2+m^3\right ) x^4+b^3 \left (15+23 m+9 m^2+m^3\right ) x^6\right )}{(1+m) (3+m) (5+m) (7+m) \left (a+b x^2\right )} \] Input:

Integrate[(d*x)^m*(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2),x]
 

Output:

(x*(d*x)^m*Sqrt[(a + b*x^2)^2]*(a^3*(105 + 71*m + 15*m^2 + m^3) + 3*a^2*b* 
(35 + 47*m + 13*m^2 + m^3)*x^2 + 3*a*b^2*(21 + 31*m + 11*m^2 + m^3)*x^4 + 
b^3*(15 + 23*m + 9*m^2 + m^3)*x^6))/((1 + m)*(3 + m)*(5 + m)*(7 + m)*(a + 
b*x^2))
 

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.55, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {1384, 27, 244, 2009}

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.

\(\displaystyle \int \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2} (d x)^m \, dx\)

\(\Big \downarrow \) 1384

\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int b^3 (d x)^m \left (b x^2+a\right )^3dx}{b^3 \left (a+b x^2\right )}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int (d x)^m \left (b x^2+a\right )^3dx}{a+b x^2}\)

\(\Big \downarrow \) 244

\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \left (a^3 (d x)^m+\frac {3 a^2 b (d x)^{m+2}}{d^2}+\frac {3 a b^2 (d x)^{m+4}}{d^4}+\frac {b^3 (d x)^{m+6}}{d^6}\right )dx}{a+b x^2}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \left (\frac {a^3 (d x)^{m+1}}{d (m+1)}+\frac {3 a^2 b (d x)^{m+3}}{d^3 (m+3)}+\frac {3 a b^2 (d x)^{m+5}}{d^5 (m+5)}+\frac {b^3 (d x)^{m+7}}{d^7 (m+7)}\right )}{a+b x^2}\)

Input:

Int[(d*x)^m*(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2),x]
 

Output:

(Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]*((a^3*(d*x)^(1 + m))/(d*(1 + m)) + (3*a^2 
*b*(d*x)^(3 + m))/(d^3*(3 + m)) + (3*a*b^2*(d*x)^(5 + m))/(d^5*(5 + m)) + 
(b^3*(d*x)^(7 + m))/(d^7*(7 + m))))/(a + b*x^2)
 

Defintions of rubi rules used

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 244

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[Expand 
Integrand[(c*x)^m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && IGtQ[p 
, 0]
 

rule 1384

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> S 
imp[(a + b*x^n + c*x^(2*n))^FracPart[p]/(c^IntPart[p]*(b/2 + c*x^n)^(2*Frac 
Part[p]))   Int[u*(b/2 + c*x^n)^(2*p), x], x] /; FreeQ[{a, b, c, n, p}, x] 
&& EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2] && NeQ[u, x^(n 
- 1)] && NeQ[u, x^(2*n - 1)] &&  !(EqQ[p, 1/2] && EqQ[u, x^(-2*n - 1)])
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 199, normalized size of antiderivative = 0.97

method	result	size
gosper	\(\frac {x \left (b^{3} m^{3} x^{6}+9 b^{3} m^{2} x^{6}+3 a \,b^{2} m^{3} x^{4}+23 m \,x^{6} b^{3}+33 a \,b^{2} m^{2} x^{4}+15 b^{3} x^{6}+3 a^{2} b \,m^{3} x^{2}+93 m \,x^{4} b^{2} a +39 a^{2} b \,m^{2} x^{2}+63 b^{2} x^{4} a +a^{3} m^{3}+141 m \,x^{2} a^{2} b +15 a^{3} m^{2}+105 a^{2} b \,x^{2}+71 m \,a^{3}+105 a^{3}\right ) \left (d x \right )^{m} {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {3}{2}}}{\left (7+m \right ) \left (5+m \right ) \left (3+m \right ) \left (1+m \right ) \left (b \,x^{2}+a \right )^{3}}\)	\(199\)
risch	\(\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (b^{3} m^{3} x^{6}+9 b^{3} m^{2} x^{6}+3 a \,b^{2} m^{3} x^{4}+23 m \,x^{6} b^{3}+33 a \,b^{2} m^{2} x^{4}+15 b^{3} x^{6}+3 a^{2} b \,m^{3} x^{2}+93 m \,x^{4} b^{2} a +39 a^{2} b \,m^{2} x^{2}+63 b^{2} x^{4} a +a^{3} m^{3}+141 m \,x^{2} a^{2} b +15 a^{3} m^{2}+105 a^{2} b \,x^{2}+71 m \,a^{3}+105 a^{3}\right ) x \left (d x \right )^{m}}{\left (b \,x^{2}+a \right ) \left (7+m \right ) \left (5+m \right ) \left (3+m \right ) \left (1+m \right )}\)	\(199\)
orering	\(\frac {\left (b^{3} m^{3} x^{6}+9 b^{3} m^{2} x^{6}+3 a \,b^{2} m^{3} x^{4}+23 m \,x^{6} b^{3}+33 a \,b^{2} m^{2} x^{4}+15 b^{3} x^{6}+3 a^{2} b \,m^{3} x^{2}+93 m \,x^{4} b^{2} a +39 a^{2} b \,m^{2} x^{2}+63 b^{2} x^{4} a +a^{3} m^{3}+141 m \,x^{2} a^{2} b +15 a^{3} m^{2}+105 a^{2} b \,x^{2}+71 m \,a^{3}+105 a^{3}\right ) x \left (d x \right )^{m} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{\frac {3}{2}}}{\left (7+m \right ) \left (5+m \right ) \left (3+m \right ) \left (1+m \right ) \left (b \,x^{2}+a \right )^{3}}\)	\(208\)

Input:

int((d*x)^m*(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

x*(b^3*m^3*x^6+9*b^3*m^2*x^6+3*a*b^2*m^3*x^4+23*b^3*m*x^6+33*a*b^2*m^2*x^4 
+15*b^3*x^6+3*a^2*b*m^3*x^2+93*a*b^2*m*x^4+39*a^2*b*m^2*x^2+63*a*b^2*x^4+a 
^3*m^3+141*a^2*b*m*x^2+15*a^3*m^2+105*a^2*b*x^2+71*a^3*m+105*a^3)*(d*x)^m* 
((b*x^2+a)^2)^(3/2)/(7+m)/(5+m)/(3+m)/(1+m)/(b*x^2+a)^3
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.78 \[ \int (d x)^m \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2} \, dx=\frac {{\left ({\left (b^{3} m^{3} + 9 \, b^{3} m^{2} + 23 \, b^{3} m + 15 \, b^{3}\right )} x^{7} + 3 \, {\left (a b^{2} m^{3} + 11 \, a b^{2} m^{2} + 31 \, a b^{2} m + 21 \, a b^{2}\right )} x^{5} + 3 \, {\left (a^{2} b m^{3} + 13 \, a^{2} b m^{2} + 47 \, a^{2} b m + 35 \, a^{2} b\right )} x^{3} + {\left (a^{3} m^{3} + 15 \, a^{3} m^{2} + 71 \, a^{3} m + 105 \, a^{3}\right )} x\right )} \left (d x\right )^{m}}{m^{4} + 16 \, m^{3} + 86 \, m^{2} + 176 \, m + 105} \] Input:

integrate((d*x)^m*(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="fricas")
 

Output:

((b^3*m^3 + 9*b^3*m^2 + 23*b^3*m + 15*b^3)*x^7 + 3*(a*b^2*m^3 + 11*a*b^2*m 
^2 + 31*a*b^2*m + 21*a*b^2)*x^5 + 3*(a^2*b*m^3 + 13*a^2*b*m^2 + 47*a^2*b*m 
 + 35*a^2*b)*x^3 + (a^3*m^3 + 15*a^3*m^2 + 71*a^3*m + 105*a^3)*x)*(d*x)^m/ 
(m^4 + 16*m^3 + 86*m^2 + 176*m + 105)
 

Sympy [F]

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

integrate((d*x)**m*(b**2*x**4+2*a*b*x**2+a**2)**(3/2),x)
 

Output:

Integral((d*x)**m*((a + b*x**2)**2)**(3/2), x)
 

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.58 \[ \int (d x)^m \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2} \, dx=\frac {{\left ({\left (m^{3} + 9 \, m^{2} + 23 \, m + 15\right )} b^{3} d^{m} x^{7} + 3 \, {\left (m^{3} + 11 \, m^{2} + 31 \, m + 21\right )} a b^{2} d^{m} x^{5} + 3 \, {\left (m^{3} + 13 \, m^{2} + 47 \, m + 35\right )} a^{2} b d^{m} x^{3} + {\left (m^{3} + 15 \, m^{2} + 71 \, m + 105\right )} a^{3} d^{m} x\right )} x^{m}}{m^{4} + 16 \, m^{3} + 86 \, m^{2} + 176 \, m + 105} \] Input:

integrate((d*x)^m*(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="maxima")
 

Output:

((m^3 + 9*m^2 + 23*m + 15)*b^3*d^m*x^7 + 3*(m^3 + 11*m^2 + 31*m + 21)*a*b^ 
2*d^m*x^5 + 3*(m^3 + 13*m^2 + 47*m + 35)*a^2*b*d^m*x^3 + (m^3 + 15*m^2 + 7 
1*m + 105)*a^3*d^m*x)*x^m/(m^4 + 16*m^3 + 86*m^2 + 176*m + 105)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 384 vs. \(2 (161) = 322\).

Time = 0.15 (sec) , antiderivative size = 384, normalized size of antiderivative = 1.87 \[ \int (d x)^m \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2} \, dx=\frac {\left (d x\right )^{m} b^{3} m^{3} x^{7} \mathrm {sgn}\left (b x^{2} + a\right ) + 9 \, \left (d x\right )^{m} b^{3} m^{2} x^{7} \mathrm {sgn}\left (b x^{2} + a\right ) + 3 \, \left (d x\right )^{m} a b^{2} m^{3} x^{5} \mathrm {sgn}\left (b x^{2} + a\right ) + 23 \, \left (d x\right )^{m} b^{3} m x^{7} \mathrm {sgn}\left (b x^{2} + a\right ) + 33 \, \left (d x\right )^{m} a b^{2} m^{2} x^{5} \mathrm {sgn}\left (b x^{2} + a\right ) + 15 \, \left (d x\right )^{m} b^{3} x^{7} \mathrm {sgn}\left (b x^{2} + a\right ) + 3 \, \left (d x\right )^{m} a^{2} b m^{3} x^{3} \mathrm {sgn}\left (b x^{2} + a\right ) + 93 \, \left (d x\right )^{m} a b^{2} m x^{5} \mathrm {sgn}\left (b x^{2} + a\right ) + 39 \, \left (d x\right )^{m} a^{2} b m^{2} x^{3} \mathrm {sgn}\left (b x^{2} + a\right ) + 63 \, \left (d x\right )^{m} a b^{2} x^{5} \mathrm {sgn}\left (b x^{2} + a\right ) + \left (d x\right )^{m} a^{3} m^{3} x \mathrm {sgn}\left (b x^{2} + a\right ) + 141 \, \left (d x\right )^{m} a^{2} b m x^{3} \mathrm {sgn}\left (b x^{2} + a\right ) + 15 \, \left (d x\right )^{m} a^{3} m^{2} x \mathrm {sgn}\left (b x^{2} + a\right ) + 105 \, \left (d x\right )^{m} a^{2} b x^{3} \mathrm {sgn}\left (b x^{2} + a\right ) + 71 \, \left (d x\right )^{m} a^{3} m x \mathrm {sgn}\left (b x^{2} + a\right ) + 105 \, \left (d x\right )^{m} a^{3} x \mathrm {sgn}\left (b x^{2} + a\right )}{m^{4} + 16 \, m^{3} + 86 \, m^{2} + 176 \, m + 105} \] Input:

integrate((d*x)^m*(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="giac")
 

Output:

((d*x)^m*b^3*m^3*x^7*sgn(b*x^2 + a) + 9*(d*x)^m*b^3*m^2*x^7*sgn(b*x^2 + a) 
 + 3*(d*x)^m*a*b^2*m^3*x^5*sgn(b*x^2 + a) + 23*(d*x)^m*b^3*m*x^7*sgn(b*x^2 
 + a) + 33*(d*x)^m*a*b^2*m^2*x^5*sgn(b*x^2 + a) + 15*(d*x)^m*b^3*x^7*sgn(b 
*x^2 + a) + 3*(d*x)^m*a^2*b*m^3*x^3*sgn(b*x^2 + a) + 93*(d*x)^m*a*b^2*m*x^ 
5*sgn(b*x^2 + a) + 39*(d*x)^m*a^2*b*m^2*x^3*sgn(b*x^2 + a) + 63*(d*x)^m*a* 
b^2*x^5*sgn(b*x^2 + a) + (d*x)^m*a^3*m^3*x*sgn(b*x^2 + a) + 141*(d*x)^m*a^ 
2*b*m*x^3*sgn(b*x^2 + a) + 15*(d*x)^m*a^3*m^2*x*sgn(b*x^2 + a) + 105*(d*x) 
^m*a^2*b*x^3*sgn(b*x^2 + a) + 71*(d*x)^m*a^3*m*x*sgn(b*x^2 + a) + 105*(d*x 
)^m*a^3*x*sgn(b*x^2 + a))/(m^4 + 16*m^3 + 86*m^2 + 176*m + 105)
 

Mupad [F(-1)]

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

int((d*x)^m*(a^2 + b^2*x^4 + 2*a*b*x^2)^(3/2),x)
 

Output:

int((d*x)^m*(a^2 + b^2*x^4 + 2*a*b*x^2)^(3/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 179, normalized size of antiderivative = 0.87 \[ \int (d x)^m \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2} \, dx=\frac {x^{m} d^{m} x \left (b^{3} m^{3} x^{6}+9 b^{3} m^{2} x^{6}+3 a \,b^{2} m^{3} x^{4}+23 b^{3} m \,x^{6}+33 a \,b^{2} m^{2} x^{4}+15 b^{3} x^{6}+3 a^{2} b \,m^{3} x^{2}+93 a \,b^{2} m \,x^{4}+39 a^{2} b \,m^{2} x^{2}+63 a \,b^{2} x^{4}+a^{3} m^{3}+141 a^{2} b m \,x^{2}+15 a^{3} m^{2}+105 a^{2} b \,x^{2}+71 a^{3} m +105 a^{3}\right )}{m^{4}+16 m^{3}+86 m^{2}+176 m +105} \] Input:

int((d*x)^m*(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x)
 

Output:

(x**m*d**m*x*(a**3*m**3 + 15*a**3*m**2 + 71*a**3*m + 105*a**3 + 3*a**2*b*m 
**3*x**2 + 39*a**2*b*m**2*x**2 + 141*a**2*b*m*x**2 + 105*a**2*b*x**2 + 3*a 
*b**2*m**3*x**4 + 33*a*b**2*m**2*x**4 + 93*a*b**2*m*x**4 + 63*a*b**2*x**4 
+ b**3*m**3*x**6 + 9*b**3*m**2*x**6 + 23*b**3*m*x**6 + 15*b**3*x**6))/(m** 
4 + 16*m**3 + 86*m**2 + 176*m + 105)