\(\int \frac {x^3 (A+B x+C x^2+D x^3)}{(a+b x^2)^{5/2}} \, dx\) [110]

Optimal result

Integrand size = 30, antiderivative size = 151 \[ \int \frac {x^3 \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^{5/2}} \, dx=\frac {a (A b-a C+(b B-a D) x)}{3 b^3 \left (a+b x^2\right )^{3/2}}-\frac {3 (A b-2 a C)+(4 b B-7 a D) x}{3 b^3 \sqrt {a+b x^2}}+\frac {C \sqrt {a+b x^2}}{b^3}+\frac {D x \sqrt {a+b x^2}}{2 b^3}+\frac {(2 b B-5 a D) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{2 b^{7/2}} \] Output:

1/3*a*(A*b-C*a+(B*b-D*a)*x)/b^3/(b*x^2+a)^(3/2)-1/3*(3*A*b-6*C*a+(4*B*b-7* 
D*a)*x)/b^3/(b*x^2+a)^(1/2)+C*(b*x^2+a)^(1/2)/b^3+1/2*D*x*(b*x^2+a)^(1/2)/ 
b^3+1/2*(2*B*b-5*D*a)*arctanh(b^(1/2)*x/(b*x^2+a)^(1/2))/b^(7/2)
 

Mathematica [A] (verified)

Time = 0.89 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.81 \[ \int \frac {x^3 \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^{5/2}} \, dx=\frac {a^2 (16 C+15 D x)+b^2 x^2 (-6 A+x (-8 B+3 x (2 C+D x)))+2 a b (-2 A+x (-3 B+2 x (6 C+5 D x)))}{6 b^3 \left (a+b x^2\right )^{3/2}}+\frac {(-2 b B+5 a D) \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{2 b^{7/2}} \] Input:

Integrate[(x^3*(A + B*x + C*x^2 + D*x^3))/(a + b*x^2)^(5/2),x]
 

Output:

(a^2*(16*C + 15*D*x) + b^2*x^2*(-6*A + x*(-8*B + 3*x*(2*C + D*x))) + 2*a*b 
*(-2*A + x*(-3*B + 2*x*(6*C + 5*D*x))))/(6*b^3*(a + b*x^2)^(3/2)) + ((-2*b 
*B + 5*a*D)*Log[-(Sqrt[b]*x) + Sqrt[a + b*x^2]])/(2*b^(7/2))
 

Rubi [A] (verified)

Time = 0.59 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.29, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {2335, 25, 2335, 27, 533, 455, 224, 219}

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.

\(\displaystyle \int \frac {x^3 \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^{5/2}} \, dx\)

\(\Big \downarrow \) 2335

\(\displaystyle -\frac {\int -\frac {x^2 \left (3 a D x^2-(A b-4 a C) x+3 a \left (B-\frac {a D}{b}\right )\right )}{\left (b x^2+a\right )^{3/2}}dx}{3 a b}-\frac {x^3 \left (a \left (B-\frac {a D}{b}\right )-x (A b-a C)\right )}{3 a b \left (a+b x^2\right )^{3/2}}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\int \frac {x^2 \left (3 a D x^2-(A b-4 a C) x+3 a \left (B-\frac {a D}{b}\right )\right )}{\left (b x^2+a\right )^{3/2}}dx}{3 a b}-\frac {x^3 \left (a \left (B-\frac {a D}{b}\right )-x (A b-a C)\right )}{3 a b \left (a+b x^2\right )^{3/2}}\)

\(\Big \downarrow \) 2335

\(\displaystyle \frac {\frac {x^2 (3 x (b B-2 a D)-4 a C+A b)}{b \sqrt {a+b x^2}}-\frac {\int \frac {a x (2 (A b-4 a C)+3 (2 b B-5 a D) x)}{\sqrt {b x^2+a}}dx}{a b}}{3 a b}-\frac {x^3 \left (a \left (B-\frac {a D}{b}\right )-x (A b-a C)\right )}{3 a b \left (a+b x^2\right )^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {x^2 (3 x (b B-2 a D)-4 a C+A b)}{b \sqrt {a+b x^2}}-\frac {\int \frac {x (2 (A b-4 a C)+3 (2 b B-5 a D) x)}{\sqrt {b x^2+a}}dx}{b}}{3 a b}-\frac {x^3 \left (a \left (B-\frac {a D}{b}\right )-x (A b-a C)\right )}{3 a b \left (a+b x^2\right )^{3/2}}\)

\(\Big \downarrow \) 533

\(\displaystyle \frac {\frac {x^2 (3 x (b B-2 a D)-4 a C+A b)}{b \sqrt {a+b x^2}}-\frac {\frac {3 x \sqrt {a+b x^2} (2 b B-5 a D)}{2 b}-\frac {\int \frac {3 a (2 b B-5 a D)-4 b (A b-4 a C) x}{\sqrt {b x^2+a}}dx}{2 b}}{b}}{3 a b}-\frac {x^3 \left (a \left (B-\frac {a D}{b}\right )-x (A b-a C)\right )}{3 a b \left (a+b x^2\right )^{3/2}}\)

\(\Big \downarrow \) 455

\(\displaystyle \frac {\frac {x^2 (3 x (b B-2 a D)-4 a C+A b)}{b \sqrt {a+b x^2}}-\frac {\frac {3 x \sqrt {a+b x^2} (2 b B-5 a D)}{2 b}-\frac {3 a (2 b B-5 a D) \int \frac {1}{\sqrt {b x^2+a}}dx-4 \sqrt {a+b x^2} (A b-4 a C)}{2 b}}{b}}{3 a b}-\frac {x^3 \left (a \left (B-\frac {a D}{b}\right )-x (A b-a C)\right )}{3 a b \left (a+b x^2\right )^{3/2}}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {\frac {x^2 (3 x (b B-2 a D)-4 a C+A b)}{b \sqrt {a+b x^2}}-\frac {\frac {3 x \sqrt {a+b x^2} (2 b B-5 a D)}{2 b}-\frac {3 a (2 b B-5 a D) \int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}-4 \sqrt {a+b x^2} (A b-4 a C)}{2 b}}{b}}{3 a b}-\frac {x^3 \left (a \left (B-\frac {a D}{b}\right )-x (A b-a C)\right )}{3 a b \left (a+b x^2\right )^{3/2}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {x^2 (3 x (b B-2 a D)-4 a C+A b)}{b \sqrt {a+b x^2}}-\frac {\frac {3 x \sqrt {a+b x^2} (2 b B-5 a D)}{2 b}-\frac {\frac {3 a \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right ) (2 b B-5 a D)}{\sqrt {b}}-4 \sqrt {a+b x^2} (A b-4 a C)}{2 b}}{b}}{3 a b}-\frac {x^3 \left (a \left (B-\frac {a D}{b}\right )-x (A b-a C)\right )}{3 a b \left (a+b x^2\right )^{3/2}}\)

Input:

Int[(x^3*(A + B*x + C*x^2 + D*x^3))/(a + b*x^2)^(5/2),x]
 

Output:

-1/3*(x^3*(a*(B - (a*D)/b) - (A*b - a*C)*x))/(a*b*(a + b*x^2)^(3/2)) + ((x 
^2*(A*b - 4*a*C + 3*(b*B - 2*a*D)*x))/(b*Sqrt[a + b*x^2]) - ((3*(2*b*B - 5 
*a*D)*x*Sqrt[a + b*x^2])/(2*b) - (-4*(A*b - 4*a*C)*Sqrt[a + b*x^2] + (3*a* 
(2*b*B - 5*a*D)*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/Sqrt[b])/(2*b))/b)/( 
3*a*b)
 

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 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 224

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], 
x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] &&  !GtQ[a, 0]
 

rule 455

Int[((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[d*(( 
a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] + Simp[c   Int[(a + b*x^2)^p, x], x] 
/; FreeQ[{a, b, c, d, p}, x] &&  !LeQ[p, -1]
 

rule 533

Int[(x_)^(m_.)*((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> 
 Simp[d*x^m*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 2))), x] - Simp[1/(b*(m + 2* 
p + 2))   Int[x^(m - 1)*(a + b*x^2)^p*Simp[a*d*m - b*c*(m + 2*p + 2)*x, x], 
 x], x] /; FreeQ[{a, b, c, d, p}, x] && IGtQ[m, 0] && GtQ[p, -1] && Integer 
Q[2*p]
 

rule 2335

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[ 
{Q = PolynomialQuotient[Pq, a + b*x^2, x], f = Coeff[PolynomialRemainder[Pq 
, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 
 1]}, Simp[(c*x)^m*(a + b*x^2)^(p + 1)*((a*g - b*f*x)/(2*a*b*(p + 1))), x] 
+ Simp[c/(2*a*b*(p + 1))   Int[(c*x)^(m - 1)*(a + b*x^2)^(p + 1)*ExpandToSu 
m[2*a*b*(p + 1)*x*Q - a*g*m + b*f*(m + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, 
 b, c}, x] && PolyQ[Pq, x] && LtQ[p, -1] && GtQ[m, 0]
 

Maple [A] (verified)

Time = 0.60 (sec) , antiderivative size = 239, normalized size of antiderivative = 1.58

method	result	size
default	\(A \left (-\frac {x^{2}}{b \left (b \,x^{2}+a \right )^{\frac {3}{2}}}-\frac {2 a}{3 b^{2} \left (b \,x^{2}+a \right )^{\frac {3}{2}}}\right )+B \left (-\frac {x^{3}}{3 b \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {-\frac {x}{b \sqrt {b \,x^{2}+a}}+\frac {\ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{b^{\frac {3}{2}}}}{b}\right )+C \left (\frac {x^{4}}{b \left (b \,x^{2}+a \right )^{\frac {3}{2}}}-\frac {4 a \left (-\frac {x^{2}}{b \left (b \,x^{2}+a \right )^{\frac {3}{2}}}-\frac {2 a}{3 b^{2} \left (b \,x^{2}+a \right )^{\frac {3}{2}}}\right )}{b}\right )+D \left (\frac {x^{5}}{2 b \left (b \,x^{2}+a \right )^{\frac {3}{2}}}-\frac {5 a \left (-\frac {x^{3}}{3 b \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {-\frac {x}{b \sqrt {b \,x^{2}+a}}+\frac {\ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{b^{\frac {3}{2}}}}{b}\right )}{2 b}\right )\)	\(239\)

Input:

int(x^3*(D*x^3+C*x^2+B*x+A)/(b*x^2+a)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

A*(-x^2/b/(b*x^2+a)^(3/2)-2/3*a/b^2/(b*x^2+a)^(3/2))+B*(-1/3*x^3/b/(b*x^2+ 
a)^(3/2)+1/b*(-x/b/(b*x^2+a)^(1/2)+1/b^(3/2)*ln(b^(1/2)*x+(b*x^2+a)^(1/2)) 
))+C*(x^4/b/(b*x^2+a)^(3/2)-4*a/b*(-x^2/b/(b*x^2+a)^(3/2)-2/3*a/b^2/(b*x^2 
+a)^(3/2)))+D*(1/2*x^5/b/(b*x^2+a)^(3/2)-5/2*a/b*(-1/3*x^3/b/(b*x^2+a)^(3/ 
2)+1/b*(-x/b/(b*x^2+a)^(1/2)+1/b^(3/2)*ln(b^(1/2)*x+(b*x^2+a)^(1/2)))))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 417, normalized size of antiderivative = 2.76 \[ \int \frac {x^3 \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^{5/2}} \, dx=\left [-\frac {3 \, {\left ({\left (5 \, D a b^{2} - 2 \, B b^{3}\right )} x^{4} + 5 \, D a^{3} - 2 \, B a^{2} b + 2 \, {\left (5 \, D a^{2} b - 2 \, B a b^{2}\right )} x^{2}\right )} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) - 2 \, {\left (3 \, D b^{3} x^{5} + 6 \, C b^{3} x^{4} + 16 \, C a^{2} b - 4 \, A a b^{2} + 4 \, {\left (5 \, D a b^{2} - 2 \, B b^{3}\right )} x^{3} + 6 \, {\left (4 \, C a b^{2} - A b^{3}\right )} x^{2} + 3 \, {\left (5 \, D a^{2} b - 2 \, B a b^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{12 \, {\left (b^{6} x^{4} + 2 \, a b^{5} x^{2} + a^{2} b^{4}\right )}}, \frac {3 \, {\left ({\left (5 \, D a b^{2} - 2 \, B b^{3}\right )} x^{4} + 5 \, D a^{3} - 2 \, B a^{2} b + 2 \, {\left (5 \, D a^{2} b - 2 \, B a b^{2}\right )} x^{2}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) + {\left (3 \, D b^{3} x^{5} + 6 \, C b^{3} x^{4} + 16 \, C a^{2} b - 4 \, A a b^{2} + 4 \, {\left (5 \, D a b^{2} - 2 \, B b^{3}\right )} x^{3} + 6 \, {\left (4 \, C a b^{2} - A b^{3}\right )} x^{2} + 3 \, {\left (5 \, D a^{2} b - 2 \, B a b^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{6 \, {\left (b^{6} x^{4} + 2 \, a b^{5} x^{2} + a^{2} b^{4}\right )}}\right ] \] Input:

integrate(x^3*(D*x^3+C*x^2+B*x+A)/(b*x^2+a)^(5/2),x, algorithm="fricas")
 

Output:

[-1/12*(3*((5*D*a*b^2 - 2*B*b^3)*x^4 + 5*D*a^3 - 2*B*a^2*b + 2*(5*D*a^2*b 
- 2*B*a*b^2)*x^2)*sqrt(b)*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) 
- 2*(3*D*b^3*x^5 + 6*C*b^3*x^4 + 16*C*a^2*b - 4*A*a*b^2 + 4*(5*D*a*b^2 - 2 
*B*b^3)*x^3 + 6*(4*C*a*b^2 - A*b^3)*x^2 + 3*(5*D*a^2*b - 2*B*a*b^2)*x)*sqr 
t(b*x^2 + a))/(b^6*x^4 + 2*a*b^5*x^2 + a^2*b^4), 1/6*(3*((5*D*a*b^2 - 2*B* 
b^3)*x^4 + 5*D*a^3 - 2*B*a^2*b + 2*(5*D*a^2*b - 2*B*a*b^2)*x^2)*sqrt(-b)*a 
rctan(sqrt(-b)*x/sqrt(b*x^2 + a)) + (3*D*b^3*x^5 + 6*C*b^3*x^4 + 16*C*a^2* 
b - 4*A*a*b^2 + 4*(5*D*a*b^2 - 2*B*b^3)*x^3 + 6*(4*C*a*b^2 - A*b^3)*x^2 + 
3*(5*D*a^2*b - 2*B*a*b^2)*x)*sqrt(b*x^2 + a))/(b^6*x^4 + 2*a*b^5*x^2 + a^2 
*b^4)]
 

Sympy [A] (verification not implemented)

Time = 14.29 (sec) , antiderivative size = 911, normalized size of antiderivative = 6.03 \[ \int \frac {x^3 \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^{5/2}} \, dx =\text {Too large to display} \] Input:

integrate(x**3*(D*x**3+C*x**2+B*x+A)/(b*x**2+a)**(5/2),x)
 

Output:

A*Piecewise((-2*a/(3*a*b**2*sqrt(a + b*x**2) + 3*b**3*x**2*sqrt(a + b*x**2 
)) - 3*b*x**2/(3*a*b**2*sqrt(a + b*x**2) + 3*b**3*x**2*sqrt(a + b*x**2)), 
Ne(b, 0)), (x**4/(4*a**(5/2)), True)) + B*(3*a**(39/2)*b**11*sqrt(1 + b*x* 
*2/a)*asinh(sqrt(b)*x/sqrt(a))/(3*a**(39/2)*b**(27/2)*sqrt(1 + b*x**2/a) + 
 3*a**(37/2)*b**(29/2)*x**2*sqrt(1 + b*x**2/a)) + 3*a**(37/2)*b**12*x**2*s 
qrt(1 + b*x**2/a)*asinh(sqrt(b)*x/sqrt(a))/(3*a**(39/2)*b**(27/2)*sqrt(1 + 
 b*x**2/a) + 3*a**(37/2)*b**(29/2)*x**2*sqrt(1 + b*x**2/a)) - 3*a**19*b**( 
23/2)*x/(3*a**(39/2)*b**(27/2)*sqrt(1 + b*x**2/a) + 3*a**(37/2)*b**(29/2)* 
x**2*sqrt(1 + b*x**2/a)) - 4*a**18*b**(25/2)*x**3/(3*a**(39/2)*b**(27/2)*s 
qrt(1 + b*x**2/a) + 3*a**(37/2)*b**(29/2)*x**2*sqrt(1 + b*x**2/a))) + C*Pi 
ecewise((8*a**2/(3*a*b**3*sqrt(a + b*x**2) + 3*b**4*x**2*sqrt(a + b*x**2)) 
 + 12*a*b*x**2/(3*a*b**3*sqrt(a + b*x**2) + 3*b**4*x**2*sqrt(a + b*x**2)) 
+ 3*b**2*x**4/(3*a*b**3*sqrt(a + b*x**2) + 3*b**4*x**2*sqrt(a + b*x**2)), 
Ne(b, 0)), (x**6/(6*a**(5/2)), True)) + D*(-15*a**(81/2)*b**22*sqrt(1 + b* 
x**2/a)*asinh(sqrt(b)*x/sqrt(a))/(6*a**(79/2)*b**(51/2)*sqrt(1 + b*x**2/a) 
 + 6*a**(77/2)*b**(53/2)*x**2*sqrt(1 + b*x**2/a)) - 15*a**(79/2)*b**23*x** 
2*sqrt(1 + b*x**2/a)*asinh(sqrt(b)*x/sqrt(a))/(6*a**(79/2)*b**(51/2)*sqrt( 
1 + b*x**2/a) + 6*a**(77/2)*b**(53/2)*x**2*sqrt(1 + b*x**2/a)) + 15*a**40* 
b**(45/2)*x/(6*a**(79/2)*b**(51/2)*sqrt(1 + b*x**2/a) + 6*a**(77/2)*b**(53 
/2)*x**2*sqrt(1 + b*x**2/a)) + 20*a**39*b**(47/2)*x**3/(6*a**(79/2)*b**...
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.64 \[ \int \frac {x^3 \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^{5/2}} \, dx=\frac {D x^{5}}{2 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} b} - \frac {1}{3} \, B x {\left (\frac {3 \, x^{2}}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} b} + \frac {2 \, a}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{2}}\right )} + \frac {5 \, D a x {\left (\frac {3 \, x^{2}}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} b} + \frac {2 \, a}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{2}}\right )}}{6 \, b} + \frac {C x^{4}}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} b} + \frac {4 \, C a x^{2}}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{2}} - \frac {A x^{2}}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} b} + \frac {5 \, D a x}{6 \, \sqrt {b x^{2} + a} b^{3}} - \frac {B x}{3 \, \sqrt {b x^{2} + a} b^{2}} - \frac {5 \, D a \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{2 \, b^{\frac {7}{2}}} + \frac {B \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{b^{\frac {5}{2}}} + \frac {8 \, C a^{2}}{3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{3}} - \frac {2 \, A a}{3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{2}} \] Input:

integrate(x^3*(D*x^3+C*x^2+B*x+A)/(b*x^2+a)^(5/2),x, algorithm="maxima")
 

Output:

1/2*D*x^5/((b*x^2 + a)^(3/2)*b) - 1/3*B*x*(3*x^2/((b*x^2 + a)^(3/2)*b) + 2 
*a/((b*x^2 + a)^(3/2)*b^2)) + 5/6*D*a*x*(3*x^2/((b*x^2 + a)^(3/2)*b) + 2*a 
/((b*x^2 + a)^(3/2)*b^2))/b + C*x^4/((b*x^2 + a)^(3/2)*b) + 4*C*a*x^2/((b* 
x^2 + a)^(3/2)*b^2) - A*x^2/((b*x^2 + a)^(3/2)*b) + 5/6*D*a*x/(sqrt(b*x^2 
+ a)*b^3) - 1/3*B*x/(sqrt(b*x^2 + a)*b^2) - 5/2*D*a*arcsinh(b*x/sqrt(a*b)) 
/b^(7/2) + B*arcsinh(b*x/sqrt(a*b))/b^(5/2) + 8/3*C*a^2/((b*x^2 + a)^(3/2) 
*b^3) - 2/3*A*a/((b*x^2 + a)^(3/2)*b^2)
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.15 \[ \int \frac {x^3 \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^{5/2}} \, dx=\frac {{\left ({\left ({\left (3 \, {\left (\frac {D x}{b} + \frac {2 \, C}{b}\right )} x + \frac {4 \, {\left (5 \, D a^{2} b^{4} - 2 \, B a b^{5}\right )}}{a b^{6}}\right )} x + \frac {6 \, {\left (4 \, C a^{2} b^{4} - A a b^{5}\right )}}{a b^{6}}\right )} x + \frac {3 \, {\left (5 \, D a^{3} b^{3} - 2 \, B a^{2} b^{4}\right )}}{a b^{6}}\right )} x + \frac {4 \, {\left (4 \, C a^{3} b^{3} - A a^{2} b^{4}\right )}}{a b^{6}}}{6 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}}} + \frac {{\left (5 \, D a - 2 \, B b\right )} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{2 \, b^{\frac {7}{2}}} \] Input:

integrate(x^3*(D*x^3+C*x^2+B*x+A)/(b*x^2+a)^(5/2),x, algorithm="giac")
 

Output:

1/6*((((3*(D*x/b + 2*C/b)*x + 4*(5*D*a^2*b^4 - 2*B*a*b^5)/(a*b^6))*x + 6*( 
4*C*a^2*b^4 - A*a*b^5)/(a*b^6))*x + 3*(5*D*a^3*b^3 - 2*B*a^2*b^4)/(a*b^6)) 
*x + 4*(4*C*a^3*b^3 - A*a^2*b^4)/(a*b^6))/(b*x^2 + a)^(3/2) + 1/2*(5*D*a - 
 2*B*b)*log(abs(-sqrt(b)*x + sqrt(b*x^2 + a)))/b^(7/2)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^3 \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^{5/2}} \, dx=\int \frac {x^3\,\left (A+B\,x+C\,x^2+x^3\,D\right )}{{\left (b\,x^2+a\right )}^{5/2}} \,d x \] Input:

int((x^3*(A + B*x + C*x^2 + x^3*D))/(a + b*x^2)^(5/2),x)
 

Output:

int((x^3*(A + B*x + C*x^2 + x^3*D))/(a + b*x^2)^(5/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 400, normalized size of antiderivative = 2.65 \[ \int \frac {x^3 \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^{5/2}} \, dx=\frac {-8 \sqrt {b \,x^{2}+a}\, a^{2} b^{2}+32 \sqrt {b \,x^{2}+a}\, a^{2} b c +30 \sqrt {b \,x^{2}+a}\, a^{2} b d x -12 \sqrt {b \,x^{2}+a}\, a \,b^{3} x^{2}-12 \sqrt {b \,x^{2}+a}\, a \,b^{3} x +48 \sqrt {b \,x^{2}+a}\, a \,b^{2} c \,x^{2}+40 \sqrt {b \,x^{2}+a}\, a \,b^{2} d \,x^{3}-16 \sqrt {b \,x^{2}+a}\, b^{4} x^{3}+12 \sqrt {b \,x^{2}+a}\, b^{3} c \,x^{4}+6 \sqrt {b \,x^{2}+a}\, b^{3} d \,x^{5}-30 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{3} d +12 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{2} b^{2}-60 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{2} b d \,x^{2}+24 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a \,b^{3} x^{2}-30 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a \,b^{2} d \,x^{4}+12 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) b^{4} x^{4}-5 \sqrt {b}\, a^{3} d -10 \sqrt {b}\, a^{2} b d \,x^{2}-5 \sqrt {b}\, a \,b^{2} d \,x^{4}}{12 b^{4} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )} \] Input:

int(x^3*(D*x^3+C*x^2+B*x+A)/(b*x^2+a)^(5/2),x)
 

Output:

( - 8*sqrt(a + b*x**2)*a**2*b**2 + 32*sqrt(a + b*x**2)*a**2*b*c + 30*sqrt( 
a + b*x**2)*a**2*b*d*x - 12*sqrt(a + b*x**2)*a*b**3*x**2 - 12*sqrt(a + b*x 
**2)*a*b**3*x + 48*sqrt(a + b*x**2)*a*b**2*c*x**2 + 40*sqrt(a + b*x**2)*a* 
b**2*d*x**3 - 16*sqrt(a + b*x**2)*b**4*x**3 + 12*sqrt(a + b*x**2)*b**3*c*x 
**4 + 6*sqrt(a + b*x**2)*b**3*d*x**5 - 30*sqrt(b)*log((sqrt(a + b*x**2) + 
sqrt(b)*x)/sqrt(a))*a**3*d + 12*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x) 
/sqrt(a))*a**2*b**2 - 60*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a 
))*a**2*b*d*x**2 + 24*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))* 
a*b**3*x**2 - 30*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a*b** 
2*d*x**4 + 12*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*b**4*x** 
4 - 5*sqrt(b)*a**3*d - 10*sqrt(b)*a**2*b*d*x**2 - 5*sqrt(b)*a*b**2*d*x**4) 
/(12*b**4*(a**2 + 2*a*b*x**2 + b**2*x**4))