\(\int \frac {(a+b x^2)^{3/2} (A+B x+C x^2+D x^3)}{x^2} \, dx\) [79]

Optimal result

Integrand size = 30, antiderivative size = 168 \[ \int \frac {\left (a+b x^2\right )^{3/2} \left (A+B x+C x^2+D x^3\right )}{x^2} \, dx=\frac {1}{8} (8 a B+3 (4 A b+a C) x) \sqrt {a+b x^2}+\frac {(4 a B+3 (4 A b+a C) x) \left (a+b x^2\right )^{3/2}}{12 a}+\frac {D \left (a+b x^2\right )^{5/2}}{5 b}-\frac {A \left (a+b x^2\right )^{5/2}}{a x}+\frac {3 a (4 A b+a C) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{8 \sqrt {b}}-a^{3/2} B \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right ) \] Output:

1/8*(8*B*a+3*(4*A*b+C*a)*x)*(b*x^2+a)^(1/2)+1/12*(4*B*a+3*(4*A*b+C*a)*x)*( 
b*x^2+a)^(3/2)/a+1/5*D*(b*x^2+a)^(5/2)/b-A*(b*x^2+a)^(5/2)/a/x+3/8*a*(4*A* 
b+C*a)*arctanh(b^(1/2)*x/(b*x^2+a)^(1/2))/b^(1/2)-a^(3/2)*B*arctanh((b*x^2 
+a)^(1/2)/a^(1/2))
 

Mathematica [A] (verified)

Time = 0.84 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.93 \[ \int \frac {\left (a+b x^2\right )^{3/2} \left (A+B x+C x^2+D x^3\right )}{x^2} \, dx=\frac {\sqrt {a+b x^2} \left (24 a^2 D x+a b \left (-120 A+x \left (160 B+75 C x+48 D x^2\right )\right )+2 b^2 x^2 (30 A+x (20 B+3 x (5 C+4 D x)))\right )}{120 b x}+2 a^{3/2} B \text {arctanh}\left (\frac {\sqrt {b} x-\sqrt {a+b x^2}}{\sqrt {a}}\right )-\frac {3 a (4 A b+a C) \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{8 \sqrt {b}} \] Input:

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

Output:

(Sqrt[a + b*x^2]*(24*a^2*D*x + a*b*(-120*A + x*(160*B + 75*C*x + 48*D*x^2) 
) + 2*b^2*x^2*(30*A + x*(20*B + 3*x*(5*C + 4*D*x)))))/(120*b*x) + 2*a^(3/2 
)*B*ArcTanh[(Sqrt[b]*x - Sqrt[a + b*x^2])/Sqrt[a]] - (3*a*(4*A*b + a*C)*Lo 
g[-(Sqrt[b]*x) + Sqrt[a + b*x^2]])/(8*Sqrt[b])
 

Rubi [A] (verified)

Time = 0.53 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.07, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {2338, 25, 2340, 27, 535, 535, 538, 224, 219, 243, 73, 221}

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.

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

\(\Big \downarrow \) 2338

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 2340

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 535

\(\displaystyle \frac {\frac {1}{4} a \int \frac {(4 a B+3 (4 A b+a C) x) \sqrt {b x^2+a}}{x}dx+\frac {1}{12} \left (a+b x^2\right )^{3/2} (3 x (a C+4 A b)+4 a B)+\frac {a D \left (a+b x^2\right )^{5/2}}{5 b}}{a}-\frac {A \left (a+b x^2\right )^{5/2}}{a x}\)

\(\Big \downarrow \) 535

\(\displaystyle \frac {\frac {1}{4} a \left (\frac {1}{2} a \int \frac {8 a B+3 (4 A b+a C) x}{x \sqrt {b x^2+a}}dx+\frac {1}{2} \sqrt {a+b x^2} (3 x (a C+4 A b)+8 a B)\right )+\frac {1}{12} \left (a+b x^2\right )^{3/2} (3 x (a C+4 A b)+4 a B)+\frac {a D \left (a+b x^2\right )^{5/2}}{5 b}}{a}-\frac {A \left (a+b x^2\right )^{5/2}}{a x}\)

\(\Big \downarrow \) 538

\(\displaystyle \frac {\frac {1}{4} a \left (\frac {1}{2} a \left (3 (a C+4 A b) \int \frac {1}{\sqrt {b x^2+a}}dx+8 a B \int \frac {1}{x \sqrt {b x^2+a}}dx\right )+\frac {1}{2} \sqrt {a+b x^2} (3 x (a C+4 A b)+8 a B)\right )+\frac {1}{12} \left (a+b x^2\right )^{3/2} (3 x (a C+4 A b)+4 a B)+\frac {a D \left (a+b x^2\right )^{5/2}}{5 b}}{a}-\frac {A \left (a+b x^2\right )^{5/2}}{a x}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {\frac {1}{4} a \left (\frac {1}{2} a \left (3 (a C+4 A b) \int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}+8 a B \int \frac {1}{x \sqrt {b x^2+a}}dx\right )+\frac {1}{2} \sqrt {a+b x^2} (3 x (a C+4 A b)+8 a B)\right )+\frac {1}{12} \left (a+b x^2\right )^{3/2} (3 x (a C+4 A b)+4 a B)+\frac {a D \left (a+b x^2\right )^{5/2}}{5 b}}{a}-\frac {A \left (a+b x^2\right )^{5/2}}{a x}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {1}{4} a \left (\frac {1}{2} a \left (8 a B \int \frac {1}{x \sqrt {b x^2+a}}dx+\frac {3 (a C+4 A b) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{\sqrt {b}}\right )+\frac {1}{2} \sqrt {a+b x^2} (3 x (a C+4 A b)+8 a B)\right )+\frac {1}{12} \left (a+b x^2\right )^{3/2} (3 x (a C+4 A b)+4 a B)+\frac {a D \left (a+b x^2\right )^{5/2}}{5 b}}{a}-\frac {A \left (a+b x^2\right )^{5/2}}{a x}\)

\(\Big \downarrow \) 243

\(\displaystyle \frac {\frac {1}{4} a \left (\frac {1}{2} a \left (4 a B \int \frac {1}{x^2 \sqrt {b x^2+a}}dx^2+\frac {3 (a C+4 A b) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{\sqrt {b}}\right )+\frac {1}{2} \sqrt {a+b x^2} (3 x (a C+4 A b)+8 a B)\right )+\frac {1}{12} \left (a+b x^2\right )^{3/2} (3 x (a C+4 A b)+4 a B)+\frac {a D \left (a+b x^2\right )^{5/2}}{5 b}}{a}-\frac {A \left (a+b x^2\right )^{5/2}}{a x}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {\frac {1}{4} a \left (\frac {1}{2} a \left (\frac {8 a B \int \frac {1}{\frac {x^4}{b}-\frac {a}{b}}d\sqrt {b x^2+a}}{b}+\frac {3 (a C+4 A b) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{\sqrt {b}}\right )+\frac {1}{2} \sqrt {a+b x^2} (3 x (a C+4 A b)+8 a B)\right )+\frac {1}{12} \left (a+b x^2\right )^{3/2} (3 x (a C+4 A b)+4 a B)+\frac {a D \left (a+b x^2\right )^{5/2}}{5 b}}{a}-\frac {A \left (a+b x^2\right )^{5/2}}{a x}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {\frac {1}{4} a \left (\frac {1}{2} a \left (\frac {3 (a C+4 A b) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{\sqrt {b}}-8 \sqrt {a} B \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )\right )+\frac {1}{2} \sqrt {a+b x^2} (3 x (a C+4 A b)+8 a B)\right )+\frac {1}{12} \left (a+b x^2\right )^{3/2} (3 x (a C+4 A b)+4 a B)+\frac {a D \left (a+b x^2\right )^{5/2}}{5 b}}{a}-\frac {A \left (a+b x^2\right )^{5/2}}{a x}\)

Input:

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

Output:

-((A*(a + b*x^2)^(5/2))/(a*x)) + (((4*a*B + 3*(4*A*b + a*C)*x)*(a + b*x^2) 
^(3/2))/12 + (a*D*(a + b*x^2)^(5/2))/(5*b) + (a*(((8*a*B + 3*(4*A*b + a*C) 
*x)*Sqrt[a + b*x^2])/2 + (a*((3*(4*A*b + a*C)*ArcTanh[(Sqrt[b]*x)/Sqrt[a + 
 b*x^2]])/Sqrt[b] - 8*Sqrt[a]*B*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]]))/2))/4)/ 
a
 

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 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, 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 221

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

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 243

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2   Subst[In 
t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I 
ntegerQ[(m - 1)/2]
 

rule 535

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

rule 538

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

rule 2338

Int[(Pq_)*((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{ 
Q = PolynomialQuotient[Pq, c*x, x], R = PolynomialRemainder[Pq, c*x, x]}, S 
imp[R*(c*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] + Simp[1/(a*c*( 
m + 1))   Int[(c*x)^(m + 1)*(a + b*x^2)^p*ExpandToSum[a*c*(m + 1)*Q - b*R*( 
m + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x] && Lt 
Q[m, -1] && (IntegerQ[2*p] || NeQ[Expon[Pq, x], 1])
 

rule 2340

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[ 
{q = Expon[Pq, x], f = Coeff[Pq, x, Expon[Pq, x]]}, Simp[f*(c*x)^(m + q - 1 
)*((a + b*x^2)^(p + 1)/(b*c^(q - 1)*(m + q + 2*p + 1))), x] + Simp[1/(b*(m 
+ q + 2*p + 1))   Int[(c*x)^m*(a + b*x^2)^p*ExpandToSum[b*(m + q + 2*p + 1) 
*Pq - b*f*(m + q + 2*p + 1)*x^q - a*f*(m + q - 1)*x^(q - 2), x], x], x] /; 
GtQ[q, 1] && NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, b, c, m, p}, x] && PolyQ 
[Pq, x] && ( !IGtQ[m, 0] || IGtQ[p + 1/2, -1])
 

Maple [A] (verified)

Time = 0.52 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.20

method	result	size
default	\(C \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4}\right )+A \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}}}{a x}+\frac {4 b \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4}\right )}{a}\right )+B \left (\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}}}{3}+a \left (\sqrt {b \,x^{2}+a}-\sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )\right )\right )+\frac {D \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{5 b}\)	\(201\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 663, normalized size of antiderivative = 3.95 \[ \int \frac {\left (a+b x^2\right )^{3/2} \left (A+B x+C x^2+D x^3\right )}{x^2} \, dx=\left [\frac {120 \, B a^{\frac {3}{2}} b x \log \left (-\frac {b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) + 45 \, {\left (C a^{2} + 4 \, A a b\right )} \sqrt {b} x \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 2 \, {\left (24 \, D b^{2} x^{5} + 30 \, C b^{2} x^{4} + 8 \, {\left (6 \, D a b + 5 \, B b^{2}\right )} x^{3} - 120 \, A a b + 15 \, {\left (5 \, C a b + 4 \, A b^{2}\right )} x^{2} + 8 \, {\left (3 \, D a^{2} + 20 \, B a b\right )} x\right )} \sqrt {b x^{2} + a}}{240 \, b x}, \frac {60 \, B a^{\frac {3}{2}} b x \log \left (-\frac {b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) - 45 \, {\left (C a^{2} + 4 \, A a b\right )} \sqrt {-b} x \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) + {\left (24 \, D b^{2} x^{5} + 30 \, C b^{2} x^{4} + 8 \, {\left (6 \, D a b + 5 \, B b^{2}\right )} x^{3} - 120 \, A a b + 15 \, {\left (5 \, C a b + 4 \, A b^{2}\right )} x^{2} + 8 \, {\left (3 \, D a^{2} + 20 \, B a b\right )} x\right )} \sqrt {b x^{2} + a}}{120 \, b x}, \frac {240 \, B \sqrt {-a} a b x \arctan \left (\frac {\sqrt {b x^{2} + a} \sqrt {-a}}{a}\right ) + 45 \, {\left (C a^{2} + 4 \, A a b\right )} \sqrt {b} x \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 2 \, {\left (24 \, D b^{2} x^{5} + 30 \, C b^{2} x^{4} + 8 \, {\left (6 \, D a b + 5 \, B b^{2}\right )} x^{3} - 120 \, A a b + 15 \, {\left (5 \, C a b + 4 \, A b^{2}\right )} x^{2} + 8 \, {\left (3 \, D a^{2} + 20 \, B a b\right )} x\right )} \sqrt {b x^{2} + a}}{240 \, b x}, \frac {120 \, B \sqrt {-a} a b x \arctan \left (\frac {\sqrt {b x^{2} + a} \sqrt {-a}}{a}\right ) - 45 \, {\left (C a^{2} + 4 \, A a b\right )} \sqrt {-b} x \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) + {\left (24 \, D b^{2} x^{5} + 30 \, C b^{2} x^{4} + 8 \, {\left (6 \, D a b + 5 \, B b^{2}\right )} x^{3} - 120 \, A a b + 15 \, {\left (5 \, C a b + 4 \, A b^{2}\right )} x^{2} + 8 \, {\left (3 \, D a^{2} + 20 \, B a b\right )} x\right )} \sqrt {b x^{2} + a}}{120 \, b x}\right ] \] Input:

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

Output:

[1/240*(120*B*a^(3/2)*b*x*log(-(b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(a) + 2*a)/x 
^2) + 45*(C*a^2 + 4*A*a*b)*sqrt(b)*x*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt 
(b)*x - a) + 2*(24*D*b^2*x^5 + 30*C*b^2*x^4 + 8*(6*D*a*b + 5*B*b^2)*x^3 - 
120*A*a*b + 15*(5*C*a*b + 4*A*b^2)*x^2 + 8*(3*D*a^2 + 20*B*a*b)*x)*sqrt(b* 
x^2 + a))/(b*x), 1/120*(60*B*a^(3/2)*b*x*log(-(b*x^2 - 2*sqrt(b*x^2 + a)*s 
qrt(a) + 2*a)/x^2) - 45*(C*a^2 + 4*A*a*b)*sqrt(-b)*x*arctan(sqrt(-b)*x/sqr 
t(b*x^2 + a)) + (24*D*b^2*x^5 + 30*C*b^2*x^4 + 8*(6*D*a*b + 5*B*b^2)*x^3 - 
 120*A*a*b + 15*(5*C*a*b + 4*A*b^2)*x^2 + 8*(3*D*a^2 + 20*B*a*b)*x)*sqrt(b 
*x^2 + a))/(b*x), 1/240*(240*B*sqrt(-a)*a*b*x*arctan(sqrt(b*x^2 + a)*sqrt( 
-a)/a) + 45*(C*a^2 + 4*A*a*b)*sqrt(b)*x*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*s 
qrt(b)*x - a) + 2*(24*D*b^2*x^5 + 30*C*b^2*x^4 + 8*(6*D*a*b + 5*B*b^2)*x^3 
 - 120*A*a*b + 15*(5*C*a*b + 4*A*b^2)*x^2 + 8*(3*D*a^2 + 20*B*a*b)*x)*sqrt 
(b*x^2 + a))/(b*x), 1/120*(120*B*sqrt(-a)*a*b*x*arctan(sqrt(b*x^2 + a)*sqr 
t(-a)/a) - 45*(C*a^2 + 4*A*a*b)*sqrt(-b)*x*arctan(sqrt(-b)*x/sqrt(b*x^2 + 
a)) + (24*D*b^2*x^5 + 30*C*b^2*x^4 + 8*(6*D*a*b + 5*B*b^2)*x^3 - 120*A*a*b 
 + 15*(5*C*a*b + 4*A*b^2)*x^2 + 8*(3*D*a^2 + 20*B*a*b)*x)*sqrt(b*x^2 + a)) 
/(b*x)]
 

Sympy [A] (verification not implemented)

Time = 3.95 (sec) , antiderivative size = 520, normalized size of antiderivative = 3.10 \[ \int \frac {\left (a+b x^2\right )^{3/2} \left (A+B x+C x^2+D x^3\right )}{x^2} \, dx =\text {Too large to display} \] Input:

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

Output:

-A*a**(3/2)/(x*sqrt(1 + b*x**2/a)) - A*sqrt(a)*b*x/sqrt(1 + b*x**2/a) + A* 
a*sqrt(b)*asinh(sqrt(b)*x/sqrt(a)) + A*b*Piecewise((a*Piecewise((log(2*sqr 
t(b)*sqrt(a + b*x**2) + 2*b*x)/sqrt(b), Ne(a, 0)), (x*log(x)/sqrt(b*x**2), 
 True))/2 + x*sqrt(a + b*x**2)/2, Ne(b, 0)), (sqrt(a)*x, True)) - B*a**(3/ 
2)*asinh(sqrt(a)/(sqrt(b)*x)) + B*a**2/(sqrt(b)*x*sqrt(a/(b*x**2) + 1)) + 
B*a*sqrt(b)*x/sqrt(a/(b*x**2) + 1) + B*b*Piecewise((a*sqrt(a + b*x**2)/(3* 
b) + x**2*sqrt(a + b*x**2)/3, Ne(b, 0)), (sqrt(a)*x**2/2, True)) + C*a*Pie 
cewise((a*Piecewise((log(2*sqrt(b)*sqrt(a + b*x**2) + 2*b*x)/sqrt(b), Ne(a 
, 0)), (x*log(x)/sqrt(b*x**2), True))/2 + x*sqrt(a + b*x**2)/2, Ne(b, 0)), 
 (sqrt(a)*x, True)) + C*b*Piecewise((-a**2*Piecewise((log(2*sqrt(b)*sqrt(a 
 + b*x**2) + 2*b*x)/sqrt(b), Ne(a, 0)), (x*log(x)/sqrt(b*x**2), True))/(8* 
b) + a*x*sqrt(a + b*x**2)/(8*b) + x**3*sqrt(a + b*x**2)/4, Ne(b, 0)), (sqr 
t(a)*x**3/3, True)) + D*a*Piecewise((a*sqrt(a + b*x**2)/(3*b) + x**2*sqrt( 
a + b*x**2)/3, Ne(b, 0)), (sqrt(a)*x**2/2, True)) + D*b*Piecewise((-2*a**2 
*sqrt(a + b*x**2)/(15*b**2) + a*x**2*sqrt(a + b*x**2)/(15*b) + x**4*sqrt(a 
 + b*x**2)/5, Ne(b, 0)), (sqrt(a)*x**4/4, True))
 

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.88 \[ \int \frac {\left (a+b x^2\right )^{3/2} \left (A+B x+C x^2+D x^3\right )}{x^2} \, dx=\frac {1}{4} \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} C x + \frac {3}{8} \, \sqrt {b x^{2} + a} C a x + \frac {3}{2} \, \sqrt {b x^{2} + a} A b x + \frac {3 \, C a^{2} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {b}} + \frac {3}{2} \, A a \sqrt {b} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right ) - B a^{\frac {3}{2}} \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right ) + \frac {1}{3} \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} B + \sqrt {b x^{2} + a} B a + \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} D}{5 \, b} - \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} A}{x} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.17 \[ \int \frac {\left (a+b x^2\right )^{3/2} \left (A+B x+C x^2+D x^3\right )}{x^2} \, dx=\frac {2 \, B a^{2} \arctan \left (-\frac {\sqrt {b} x - \sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} + \frac {2 \, A a^{2} \sqrt {b}}{{\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} - a} + \frac {1}{120} \, \sqrt {b x^{2} + a} {\left ({\left (2 \, {\left (3 \, {\left (4 \, D b x + 5 \, C b\right )} x + \frac {4 \, {\left (6 \, D a b^{3} + 5 \, B b^{4}\right )}}{b^{3}}\right )} x + \frac {15 \, {\left (5 \, C a b^{3} + 4 \, A b^{4}\right )}}{b^{3}}\right )} x + \frac {8 \, {\left (3 \, D a^{2} b^{2} + 20 \, B a b^{3}\right )}}{b^{3}}\right )} - \frac {3 \, {\left (C a^{2} + 4 \, A a b\right )} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{8 \, \sqrt {b}} \] Input:

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

Output:

2*B*a^2*arctan(-(sqrt(b)*x - sqrt(b*x^2 + a))/sqrt(-a))/sqrt(-a) + 2*A*a^2 
*sqrt(b)/((sqrt(b)*x - sqrt(b*x^2 + a))^2 - a) + 1/120*sqrt(b*x^2 + a)*((2 
*(3*(4*D*b*x + 5*C*b)*x + 4*(6*D*a*b^3 + 5*B*b^4)/b^3)*x + 15*(5*C*a*b^3 + 
 4*A*b^4)/b^3)*x + 8*(3*D*a^2*b^2 + 20*B*a*b^3)/b^3) - 3/8*(C*a^2 + 4*A*a* 
b)*log(abs(-sqrt(b)*x + sqrt(b*x^2 + a)))/sqrt(b)
 

Mupad [B] (verification not implemented)

Time = 3.04 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.98 \[ \int \frac {\left (a+b x^2\right )^{3/2} \left (A+B x+C x^2+D x^3\right )}{x^2} \, dx=\frac {B\,{\left (b\,x^2+a\right )}^{3/2}}{3}-B\,a^{3/2}\,\mathrm {atanh}\left (\frac {\sqrt {b\,x^2+a}}{\sqrt {a}}\right )+B\,a\,\sqrt {b\,x^2+a}-\frac {A\,{\left (b\,x^2+a\right )}^{3/2}\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{2},-\frac {1}{2};\ \frac {1}{2};\ -\frac {b\,x^2}{a}\right )}{x\,{\left (\frac {b\,x^2}{a}+1\right )}^{3/2}}+\frac {x^2\,{\left (b\,x^2+a\right )}^{3/2}\,D\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{2},1;\ 2;\ -\frac {b\,x^2}{a}\right )}{2\,{\left (\frac {b\,x^2}{a}+1\right )}^{3/2}}+\frac {C\,x\,{\left (b\,x^2+a\right )}^{3/2}\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{2},\frac {1}{2};\ \frac {3}{2};\ -\frac {b\,x^2}{a}\right )}{{\left (\frac {b\,x^2}{a}+1\right )}^{3/2}} \] Input:

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

Output:

(B*(a + b*x^2)^(3/2))/3 - B*a^(3/2)*atanh((a + b*x^2)^(1/2)/a^(1/2)) + B*a 
*(a + b*x^2)^(1/2) - (A*(a + b*x^2)^(3/2)*hypergeom([-3/2, -1/2], 1/2, -(b 
*x^2)/a))/(x*((b*x^2)/a + 1)^(3/2)) + (x^2*(a + b*x^2)^(3/2)*D*hypergeom([ 
-3/2, 1], 2, -(b*x^2)/a))/(2*((b*x^2)/a + 1)^(3/2)) + (C*x*(a + b*x^2)^(3/ 
2)*hypergeom([-3/2, 1/2], 3/2, -(b*x^2)/a))/((b*x^2)/a + 1)^(3/2)
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 288, normalized size of antiderivative = 1.71 \[ \int \frac {\left (a+b x^2\right )^{3/2} \left (A+B x+C x^2+D x^3\right )}{x^2} \, dx=\frac {-120 \sqrt {b \,x^{2}+a}\, a^{2} b +24 \sqrt {b \,x^{2}+a}\, a^{2} d x +60 \sqrt {b \,x^{2}+a}\, a \,b^{2} x^{2}+160 \sqrt {b \,x^{2}+a}\, a \,b^{2} x +75 \sqrt {b \,x^{2}+a}\, a b c \,x^{2}+48 \sqrt {b \,x^{2}+a}\, a b d \,x^{3}+40 \sqrt {b \,x^{2}+a}\, b^{3} x^{3}+30 \sqrt {b \,x^{2}+a}\, b^{2} c \,x^{4}+24 \sqrt {b \,x^{2}+a}\, b^{2} d \,x^{5}+120 \sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}-\sqrt {a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a \,b^{2} x -120 \sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a \,b^{2} x +180 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{2} b x +45 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{2} c x -135 \sqrt {b}\, a^{2} b x -15 \sqrt {b}\, a^{2} c x}{120 b x} \] Input:

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

Output:

( - 120*sqrt(a + b*x**2)*a**2*b + 24*sqrt(a + b*x**2)*a**2*d*x + 60*sqrt(a 
 + b*x**2)*a*b**2*x**2 + 160*sqrt(a + b*x**2)*a*b**2*x + 75*sqrt(a + b*x** 
2)*a*b*c*x**2 + 48*sqrt(a + b*x**2)*a*b*d*x**3 + 40*sqrt(a + b*x**2)*b**3* 
x**3 + 30*sqrt(a + b*x**2)*b**2*c*x**4 + 24*sqrt(a + b*x**2)*b**2*d*x**5 + 
 120*sqrt(a)*log((sqrt(a + b*x**2) - sqrt(a) + sqrt(b)*x)/sqrt(a))*a*b**2* 
x - 120*sqrt(a)*log((sqrt(a + b*x**2) + sqrt(a) + sqrt(b)*x)/sqrt(a))*a*b* 
*2*x + 180*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a**2*b*x + 
45*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a**2*c*x - 135*sqrt 
(b)*a**2*b*x - 15*sqrt(b)*a**2*c*x)/(120*b*x)