\(\int \frac {x^4 (A+B x+C x^2+D x^3)}{(a+b x^2)^{5/2}} \, dx\) [109]

Optimal result

Integrand size = 30, antiderivative size = 186 \[ \int \frac {x^4 \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^{5/2}} \, dx=-\frac {a (a (b B-a D)-b (A b-a C) x)}{3 b^4 \left (a+b x^2\right )^{3/2}}+\frac {3 a (2 b B-3 a D)-b (4 A b-7 a C) x}{3 b^4 \sqrt {a+b x^2}}+\frac {(b B-3 a D) \sqrt {a+b x^2}}{b^4}+\frac {C x \sqrt {a+b x^2}}{2 b^3}+\frac {D \left (a+b x^2\right )^{3/2}}{3 b^4}+\frac {(2 A b-5 a C) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{2 b^{7/2}} \] Output:

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

Mathematica [A] (verified)

Time = 1.10 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.78 \[ \int \frac {x^4 \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^{5/2}} \, dx=\frac {-32 a^3 D+a^2 b (16 B+3 x (5 C-16 D x))+2 a b^2 x \left (-3 A+2 x \left (6 B+5 C x-3 D x^2\right )\right )+b^3 x^3 \left (-8 A+x \left (6 B+3 C x+2 D x^2\right )\right )}{6 b^4 \left (a+b x^2\right )^{3/2}}+\frac {(2 A b-5 a C) \text {arctanh}\left (\frac {\sqrt {b} x}{-\sqrt {a}+\sqrt {a+b x^2}}\right )}{b^{7/2}} \] Input:

Integrate[(x^4*(A + B*x + C*x^2 + D*x^3))/(a + b*x^2)^(5/2),x]
 

Output:

(-32*a^3*D + a^2*b*(16*B + 3*x*(5*C - 16*D*x)) + 2*a*b^2*x*(-3*A + 2*x*(6* 
B + 5*C*x - 3*D*x^2)) + b^3*x^3*(-8*A + x*(6*B + 3*C*x + 2*D*x^2)))/(6*b^4 
*(a + b*x^2)^(3/2)) + ((2*A*b - 5*a*C)*ArcTanh[(Sqrt[b]*x)/(-Sqrt[a] + Sqr 
t[a + b*x^2])])/b^(7/2)
 

Rubi [A] (verified)

Time = 0.68 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.24, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.367, Rules used = {2335, 25, 2335, 27, 533, 533, 25, 27, 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.

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

\(\Big \downarrow \) 2335

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 2335

\(\displaystyle \frac {\frac {x^3 (x (4 b B-7 a D)-5 a C+2 A b)}{b \sqrt {a+b x^2}}-\frac {\int \frac {3 a x^2 (2 A b-5 a C+4 (b B-2 a D) x)}{\sqrt {b x^2+a}}dx}{a b}}{3 a b}-\frac {x^4 \left (a \left (B-\frac {a D}{b}\right )-x (A b-a C)\right )}{3 a b \left (a+b x^2\right )^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {x^3 (x (4 b B-7 a D)-5 a C+2 A b)}{b \sqrt {a+b x^2}}-\frac {3 \int \frac {x^2 (2 A b-5 a C+4 (b B-2 a D) x)}{\sqrt {b x^2+a}}dx}{b}}{3 a b}-\frac {x^4 \left (a \left (B-\frac {a D}{b}\right )-x (A b-a C)\right )}{3 a b \left (a+b x^2\right )^{3/2}}\)

\(\Big \downarrow \) 533

\(\displaystyle \frac {\frac {x^3 (x (4 b B-7 a D)-5 a C+2 A b)}{b \sqrt {a+b x^2}}-\frac {3 \left (\frac {4 x^2 \sqrt {a+b x^2} (b B-2 a D)}{3 b}-\frac {\int \frac {x (8 a (b B-2 a D)-3 b (2 A b-5 a C) x)}{\sqrt {b x^2+a}}dx}{3 b}\right )}{b}}{3 a b}-\frac {x^4 \left (a \left (B-\frac {a D}{b}\right )-x (A b-a C)\right )}{3 a b \left (a+b x^2\right )^{3/2}}\)

\(\Big \downarrow \) 533

\(\displaystyle \frac {\frac {x^3 (x (4 b B-7 a D)-5 a C+2 A b)}{b \sqrt {a+b x^2}}-\frac {3 \left (\frac {4 x^2 \sqrt {a+b x^2} (b B-2 a D)}{3 b}-\frac {-\frac {\int -\frac {a b (3 (2 A b-5 a C)+16 (b B-2 a D) x)}{\sqrt {b x^2+a}}dx}{2 b}-\frac {3}{2} x \sqrt {a+b x^2} (2 A b-5 a C)}{3 b}\right )}{b}}{3 a b}-\frac {x^4 \left (a \left (B-\frac {a D}{b}\right )-x (A b-a C)\right )}{3 a b \left (a+b x^2\right )^{3/2}}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {x^3 (x (4 b B-7 a D)-5 a C+2 A b)}{b \sqrt {a+b x^2}}-\frac {3 \left (\frac {4 x^2 \sqrt {a+b x^2} (b B-2 a D)}{3 b}-\frac {\frac {\int \frac {a b (3 (2 A b-5 a C)+16 (b B-2 a D) x)}{\sqrt {b x^2+a}}dx}{2 b}-\frac {3}{2} x \sqrt {a+b x^2} (2 A b-5 a C)}{3 b}\right )}{b}}{3 a b}-\frac {x^4 \left (a \left (B-\frac {a D}{b}\right )-x (A b-a C)\right )}{3 a b \left (a+b x^2\right )^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {x^3 (x (4 b B-7 a D)-5 a C+2 A b)}{b \sqrt {a+b x^2}}-\frac {3 \left (\frac {4 x^2 \sqrt {a+b x^2} (b B-2 a D)}{3 b}-\frac {\frac {1}{2} a \int \frac {3 (2 A b-5 a C)+16 (b B-2 a D) x}{\sqrt {b x^2+a}}dx-\frac {3}{2} x \sqrt {a+b x^2} (2 A b-5 a C)}{3 b}\right )}{b}}{3 a b}-\frac {x^4 \left (a \left (B-\frac {a D}{b}\right )-x (A b-a C)\right )}{3 a b \left (a+b x^2\right )^{3/2}}\)

\(\Big \downarrow \) 455

\(\displaystyle \frac {\frac {x^3 (x (4 b B-7 a D)-5 a C+2 A b)}{b \sqrt {a+b x^2}}-\frac {3 \left (\frac {4 x^2 \sqrt {a+b x^2} (b B-2 a D)}{3 b}-\frac {\frac {1}{2} a \left (3 (2 A b-5 a C) \int \frac {1}{\sqrt {b x^2+a}}dx+\frac {16 \sqrt {a+b x^2} (b B-2 a D)}{b}\right )-\frac {3}{2} x \sqrt {a+b x^2} (2 A b-5 a C)}{3 b}\right )}{b}}{3 a b}-\frac {x^4 \left (a \left (B-\frac {a D}{b}\right )-x (A b-a C)\right )}{3 a b \left (a+b x^2\right )^{3/2}}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {\frac {x^3 (x (4 b B-7 a D)-5 a C+2 A b)}{b \sqrt {a+b x^2}}-\frac {3 \left (\frac {4 x^2 \sqrt {a+b x^2} (b B-2 a D)}{3 b}-\frac {\frac {1}{2} a \left (3 (2 A b-5 a C) \int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}+\frac {16 \sqrt {a+b x^2} (b B-2 a D)}{b}\right )-\frac {3}{2} x \sqrt {a+b x^2} (2 A b-5 a C)}{3 b}\right )}{b}}{3 a b}-\frac {x^4 \left (a \left (B-\frac {a D}{b}\right )-x (A b-a C)\right )}{3 a b \left (a+b x^2\right )^{3/2}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {x^3 (x (4 b B-7 a D)-5 a C+2 A b)}{b \sqrt {a+b x^2}}-\frac {3 \left (\frac {4 x^2 \sqrt {a+b x^2} (b B-2 a D)}{3 b}-\frac {\frac {1}{2} a \left (\frac {3 (2 A b-5 a C) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{\sqrt {b}}+\frac {16 \sqrt {a+b x^2} (b B-2 a D)}{b}\right )-\frac {3}{2} x \sqrt {a+b x^2} (2 A b-5 a C)}{3 b}\right )}{b}}{3 a b}-\frac {x^4 \left (a \left (B-\frac {a D}{b}\right )-x (A b-a C)\right )}{3 a b \left (a+b x^2\right )^{3/2}}\)

Input:

Int[(x^4*(A + B*x + C*x^2 + D*x^3))/(a + b*x^2)^(5/2),x]
 

Output:

-1/3*(x^4*(a*(B - (a*D)/b) - (A*b - a*C)*x))/(a*b*(a + b*x^2)^(3/2)) + ((x 
^3*(2*A*b - 5*a*C + (4*b*B - 7*a*D)*x))/(b*Sqrt[a + b*x^2]) - (3*((4*(b*B 
- 2*a*D)*x^2*Sqrt[a + b*x^2])/(3*b) - ((-3*(2*A*b - 5*a*C)*x*Sqrt[a + b*x^ 
2])/2 + (a*((16*(b*B - 2*a*D)*Sqrt[a + b*x^2])/b + (3*(2*A*b - 5*a*C)*ArcT 
anh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/Sqrt[b]))/2)/(3*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 = 0.62 (sec) , antiderivative size = 286, normalized size of antiderivative = 1.54

method	result	size
default	\(A \left (-\frac {x^{3}}{3 b \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {-\frac {x}{b \sqrt {b \,x^{2}+a}}+\frac {\ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{b^{\frac {3}{2}}}}{b}\right )+B \left (\frac {x^{4}}{b \left (b \,x^{2}+a \right )^{\frac {3}{2}}}-\frac {4 a \left (-\frac {x^{2}}{b \left (b \,x^{2}+a \right )^{\frac {3}{2}}}-\frac {2 a}{3 b^{2} \left (b \,x^{2}+a \right )^{\frac {3}{2}}}\right )}{b}\right )+C \left (\frac {x^{5}}{2 b \left (b \,x^{2}+a \right )^{\frac {3}{2}}}-\frac {5 a \left (-\frac {x^{3}}{3 b \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {-\frac {x}{b \sqrt {b \,x^{2}+a}}+\frac {\ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{b^{\frac {3}{2}}}}{b}\right )}{2 b}\right )+D \left (\frac {x^{6}}{3 b \left (b \,x^{2}+a \right )^{\frac {3}{2}}}-\frac {2 a \left (\frac {x^{4}}{b \left (b \,x^{2}+a \right )^{\frac {3}{2}}}-\frac {4 a \left (-\frac {x^{2}}{b \left (b \,x^{2}+a \right )^{\frac {3}{2}}}-\frac {2 a}{3 b^{2} \left (b \,x^{2}+a \right )^{\frac {3}{2}}}\right )}{b}\right )}{b}\right )\)	\(286\)

Input:

int(x^4*(D*x^3+C*x^2+B*x+A)/(b*x^2+a)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 455, normalized size of antiderivative = 2.45 \[ \int \frac {x^4 \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^{5/2}} \, dx=\left [-\frac {3 \, {\left ({\left (5 \, C a b^{2} - 2 \, A b^{3}\right )} x^{4} + 5 \, C a^{3} - 2 \, A a^{2} b + 2 \, {\left (5 \, C a^{2} b - 2 \, A a b^{2}\right )} x^{2}\right )} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) - 2 \, {\left (2 \, D b^{3} x^{6} + 3 \, C b^{3} x^{5} - 6 \, {\left (2 \, D a b^{2} - B b^{3}\right )} x^{4} - 32 \, D a^{3} + 16 \, B a^{2} b + 4 \, {\left (5 \, C a b^{2} - 2 \, A b^{3}\right )} x^{3} - 24 \, {\left (2 \, D a^{2} b - B a b^{2}\right )} x^{2} + 3 \, {\left (5 \, C a^{2} b - 2 \, A a b^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{12 \, {\left (b^{6} x^{4} + 2 \, a b^{5} x^{2} + a^{2} b^{4}\right )}}, \frac {3 \, {\left ({\left (5 \, C a b^{2} - 2 \, A b^{3}\right )} x^{4} + 5 \, C a^{3} - 2 \, A a^{2} b + 2 \, {\left (5 \, C a^{2} b - 2 \, A a b^{2}\right )} x^{2}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) + {\left (2 \, D b^{3} x^{6} + 3 \, C b^{3} x^{5} - 6 \, {\left (2 \, D a b^{2} - B b^{3}\right )} x^{4} - 32 \, D a^{3} + 16 \, B a^{2} b + 4 \, {\left (5 \, C a b^{2} - 2 \, A b^{3}\right )} x^{3} - 24 \, {\left (2 \, D a^{2} b - B a b^{2}\right )} x^{2} + 3 \, {\left (5 \, C a^{2} b - 2 \, A a b^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{6 \, {\left (b^{6} x^{4} + 2 \, a b^{5} x^{2} + a^{2} b^{4}\right )}}\right ] \] Input:

integrate(x^4*(D*x^3+C*x^2+B*x+A)/(b*x^2+a)^(5/2),x, algorithm="fricas")
 

Output:

[-1/12*(3*((5*C*a*b^2 - 2*A*b^3)*x^4 + 5*C*a^3 - 2*A*a^2*b + 2*(5*C*a^2*b 
- 2*A*a*b^2)*x^2)*sqrt(b)*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) 
- 2*(2*D*b^3*x^6 + 3*C*b^3*x^5 - 6*(2*D*a*b^2 - B*b^3)*x^4 - 32*D*a^3 + 16 
*B*a^2*b + 4*(5*C*a*b^2 - 2*A*b^3)*x^3 - 24*(2*D*a^2*b - B*a*b^2)*x^2 + 3* 
(5*C*a^2*b - 2*A*a*b^2)*x)*sqrt(b*x^2 + a))/(b^6*x^4 + 2*a*b^5*x^2 + a^2*b 
^4), 1/6*(3*((5*C*a*b^2 - 2*A*b^3)*x^4 + 5*C*a^3 - 2*A*a^2*b + 2*(5*C*a^2* 
b - 2*A*a*b^2)*x^2)*sqrt(-b)*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) + (2*D*b^3 
*x^6 + 3*C*b^3*x^5 - 6*(2*D*a*b^2 - B*b^3)*x^4 - 32*D*a^3 + 16*B*a^2*b + 4 
*(5*C*a*b^2 - 2*A*b^3)*x^3 - 24*(2*D*a^2*b - B*a*b^2)*x^2 + 3*(5*C*a^2*b - 
 2*A*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 = 14.92 (sec) , antiderivative size = 1003, normalized size of antiderivative = 5.39 \[ \int \frac {x^4 \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^{5/2}} \, dx =\text {Too large to display} \] Input:

integrate(x**4*(D*x**3+C*x**2+B*x+A)/(b*x**2+a)**(5/2),x)
 

Output:

A*(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*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**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**(2 
5/2)*x**3/(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))) + B*Piecewise((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)) + 
C*(-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/sqr 
t(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**(51/2)*sqrt(1 + b*x**2/a) + 6*a**(77/2)*b** 
(53/2)*x**2*sqrt(1 + b*x**2/a)) + 3*a**38*b**(49/2)*x**5/(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...
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 290, normalized size of antiderivative = 1.56 \[ \int \frac {x^4 \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^{5/2}} \, dx=\frac {D x^{6}}{3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} b} + \frac {C x^{5}}{2 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} b} - \frac {1}{3} \, A x {\left (\frac {3 \, x^{2}}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} b} + \frac {2 \, a}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{2}}\right )} + \frac {5 \, C a x {\left (\frac {3 \, x^{2}}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} b} + \frac {2 \, a}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{2}}\right )}}{6 \, b} - \frac {2 \, D a x^{4}}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{2}} + \frac {B x^{4}}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} b} - \frac {8 \, D a^{2} x^{2}}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{3}} + \frac {4 \, B a x^{2}}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{2}} + \frac {5 \, C a x}{6 \, \sqrt {b x^{2} + a} b^{3}} - \frac {A x}{3 \, \sqrt {b x^{2} + a} b^{2}} - \frac {5 \, C a \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{2 \, b^{\frac {7}{2}}} + \frac {A \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{b^{\frac {5}{2}}} - \frac {16 \, D a^{3}}{3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{4}} + \frac {8 \, B a^{2}}{3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{3}} \] Input:

integrate(x^4*(D*x^3+C*x^2+B*x+A)/(b*x^2+a)^(5/2),x, algorithm="maxima")
 

Output:

1/3*D*x^6/((b*x^2 + a)^(3/2)*b) + 1/2*C*x^5/((b*x^2 + a)^(3/2)*b) - 1/3*A* 
x*(3*x^2/((b*x^2 + a)^(3/2)*b) + 2*a/((b*x^2 + a)^(3/2)*b^2)) + 5/6*C*a*x* 
(3*x^2/((b*x^2 + a)^(3/2)*b) + 2*a/((b*x^2 + a)^(3/2)*b^2))/b - 2*D*a*x^4/ 
((b*x^2 + a)^(3/2)*b^2) + B*x^4/((b*x^2 + a)^(3/2)*b) - 8*D*a^2*x^2/((b*x^ 
2 + a)^(3/2)*b^3) + 4*B*a*x^2/((b*x^2 + a)^(3/2)*b^2) + 5/6*C*a*x/(sqrt(b* 
x^2 + a)*b^3) - 1/3*A*x/(sqrt(b*x^2 + a)*b^2) - 5/2*C*a*arcsinh(b*x/sqrt(a 
*b))/b^(7/2) + A*arcsinh(b*x/sqrt(a*b))/b^(5/2) - 16/3*D*a^3/((b*x^2 + a)^ 
(3/2)*b^4) + 8/3*B*a^2/((b*x^2 + a)^(3/2)*b^3)
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.10 \[ \int \frac {x^4 \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^{5/2}} \, dx=\frac {{\left ({\left ({\left ({\left ({\left (\frac {2 \, D x}{b} + \frac {3 \, C}{b}\right )} x - \frac {6 \, {\left (2 \, D a^{2} b^{5} - B a b^{6}\right )}}{a b^{7}}\right )} x + \frac {4 \, {\left (5 \, C a^{2} b^{5} - 2 \, A a b^{6}\right )}}{a b^{7}}\right )} x - \frac {24 \, {\left (2 \, D a^{3} b^{4} - B a^{2} b^{5}\right )}}{a b^{7}}\right )} x + \frac {3 \, {\left (5 \, C a^{3} b^{4} - 2 \, A a^{2} b^{5}\right )}}{a b^{7}}\right )} x - \frac {16 \, {\left (2 \, D a^{4} b^{3} - B a^{3} b^{4}\right )}}{a b^{7}}}{6 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}}} + \frac {{\left (5 \, C a - 2 \, A b\right )} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{2 \, b^{\frac {7}{2}}} \] Input:

integrate(x^4*(D*x^3+C*x^2+B*x+A)/(b*x^2+a)^(5/2),x, algorithm="giac")
 

Output:

1/6*((((((2*D*x/b + 3*C/b)*x - 6*(2*D*a^2*b^5 - B*a*b^6)/(a*b^7))*x + 4*(5 
*C*a^2*b^5 - 2*A*a*b^6)/(a*b^7))*x - 24*(2*D*a^3*b^4 - B*a^2*b^5)/(a*b^7)) 
*x + 3*(5*C*a^3*b^4 - 2*A*a^2*b^5)/(a*b^7))*x - 16*(2*D*a^4*b^3 - B*a^3*b^ 
4)/(a*b^7))/(b*x^2 + a)^(3/2) + 1/2*(5*C*a - 2*A*b)*log(abs(-sqrt(b)*x + s 
qrt(b*x^2 + a)))/b^(7/2)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^4 \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^{5/2}} \, dx=\int \frac {x^4\,\left (A+B\,x+C\,x^2+x^3\,D\right )}{{\left (b\,x^2+a\right )}^{5/2}} \,d x \] Input:

int((x^4*(A + B*x + C*x^2 + x^3*D))/(a + b*x^2)^(5/2),x)
 

Output:

int((x^4*(A + B*x + C*x^2 + x^3*D))/(a + b*x^2)^(5/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 437, normalized size of antiderivative = 2.35 \[ \int \frac {x^4 \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^{5/2}} \, dx=\frac {-64 \sqrt {b \,x^{2}+a}\, a^{3} d -12 \sqrt {b \,x^{2}+a}\, a^{2} b^{2} x +32 \sqrt {b \,x^{2}+a}\, a^{2} b^{2}+30 \sqrt {b \,x^{2}+a}\, a^{2} b c x -96 \sqrt {b \,x^{2}+a}\, a^{2} b d \,x^{2}-16 \sqrt {b \,x^{2}+a}\, a \,b^{3} x^{3}+48 \sqrt {b \,x^{2}+a}\, a \,b^{3} x^{2}+40 \sqrt {b \,x^{2}+a}\, a \,b^{2} c \,x^{3}-24 \sqrt {b \,x^{2}+a}\, a \,b^{2} d \,x^{4}+12 \sqrt {b \,x^{2}+a}\, b^{4} x^{4}+6 \sqrt {b \,x^{2}+a}\, b^{3} c \,x^{5}+4 \sqrt {b \,x^{2}+a}\, b^{3} d \,x^{6}+12 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{3} b -30 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{3} c +24 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{2} b^{2} x^{2}-60 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{2} b c \,x^{2}+12 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a \,b^{3} x^{4}-30 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a \,b^{2} c \,x^{4}-5 \sqrt {b}\, a^{3} c -10 \sqrt {b}\, a^{2} b c \,x^{2}-5 \sqrt {b}\, a \,b^{2} c \,x^{4}}{12 b^{4} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )} \] Input:

int(x^4*(D*x^3+C*x^2+B*x+A)/(b*x^2+a)^(5/2),x)
 

Output:

( - 64*sqrt(a + b*x**2)*a**3*d - 12*sqrt(a + b*x**2)*a**2*b**2*x + 32*sqrt 
(a + b*x**2)*a**2*b**2 + 30*sqrt(a + b*x**2)*a**2*b*c*x - 96*sqrt(a + b*x* 
*2)*a**2*b*d*x**2 - 16*sqrt(a + b*x**2)*a*b**3*x**3 + 48*sqrt(a + b*x**2)* 
a*b**3*x**2 + 40*sqrt(a + b*x**2)*a*b**2*c*x**3 - 24*sqrt(a + b*x**2)*a*b* 
*2*d*x**4 + 12*sqrt(a + b*x**2)*b**4*x**4 + 6*sqrt(a + b*x**2)*b**3*c*x**5 
 + 4*sqrt(a + b*x**2)*b**3*d*x**6 + 12*sqrt(b)*log((sqrt(a + b*x**2) + sqr 
t(b)*x)/sqrt(a))*a**3*b - 30*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sq 
rt(a))*a**3*c + 24*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a** 
2*b**2*x**2 - 60*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a**2* 
b*c*x**2 + 12*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a*b**3*x 
**4 - 30*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a*b**2*c*x**4 
 - 5*sqrt(b)*a**3*c - 10*sqrt(b)*a**2*b*c*x**2 - 5*sqrt(b)*a*b**2*c*x**4)/ 
(12*b**4*(a**2 + 2*a*b*x**2 + b**2*x**4))