\(\int \frac {(c+d x) (a+b x^2)^{3/2} (A+B x+C x^2)}{x^3} \, dx\) [53]

Optimal result

Integrand size = 30, antiderivative size = 267 \[ \int \frac {(c+d x) \left (a+b x^2\right )^{3/2} \left (A+B x+C x^2\right )}{x^3} \, dx=\frac {1}{2} (3 A b c+2 a (c C+B d)) \sqrt {a+b x^2}+\frac {3}{8} (a C d+4 b (B c+A d)) x \sqrt {a+b x^2}+\frac {(3 A b c+2 a (c C+B d)) \left (a+b x^2\right )^{3/2}}{6 a}+\frac {(a C d+4 b (B c+A d)) x \left (a+b x^2\right )^{3/2}}{4 a}-\frac {A c \left (a+b x^2\right )^{5/2}}{2 a x^2}-\frac {(B c+A d) \left (a+b x^2\right )^{5/2}}{a x}+\frac {3 a (a C d+4 b (B c+A d)) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{8 \sqrt {b}}-\frac {1}{2} \sqrt {a} (3 A b c+2 a (c C+B d)) \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right ) \] Output:

1/2*(3*A*b*c+2*a*(B*d+C*c))*(b*x^2+a)^(1/2)+3/8*(a*C*d+4*b*(A*d+B*c))*x*(b 
*x^2+a)^(1/2)+1/6*(3*A*b*c+2*a*(B*d+C*c))*(b*x^2+a)^(3/2)/a+1/4*(a*C*d+4*b 
*(A*d+B*c))*x*(b*x^2+a)^(3/2)/a-1/2*A*c*(b*x^2+a)^(5/2)/a/x^2-(A*d+B*c)*(b 
*x^2+a)^(5/2)/a/x+3/8*a*(a*C*d+4*b*(A*d+B*c))*arctanh(b^(1/2)*x/(b*x^2+a)^ 
(1/2))/b^(1/2)-1/2*a^(1/2)*(3*A*b*c+2*a*(B*d+C*c))*arctanh((b*x^2+a)^(1/2) 
/a^(1/2))
 

Mathematica [A] (verified)

Time = 1.35 (sec) , antiderivative size = 222, normalized size of antiderivative = 0.83 \[ \int \frac {(c+d x) \left (a+b x^2\right )^{3/2} \left (A+B x+C x^2\right )}{x^3} \, dx=\frac {\sqrt {a+b x^2} \left (2 b x^2 \left (6 A (2 c+d x)+x \left (6 B c+4 c C x+4 B d x+3 C d x^2\right )\right )+a \left (-12 A (c+2 d x)+x \left (-24 B c+32 c C x+32 B d x+15 C d x^2\right )\right )\right )}{24 x^2}+3 \sqrt {a} A b c \text {arctanh}\left (\frac {\sqrt {b} x-\sqrt {a+b x^2}}{\sqrt {a}}\right )-2 a^{3/2} (c C+B d) \text {arctanh}\left (\frac {-\sqrt {b} x+\sqrt {a+b x^2}}{\sqrt {a}}\right )-\frac {3 a (a C d+4 b (B c+A d)) \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{8 \sqrt {b}} \] Input:

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

Output:

(Sqrt[a + b*x^2]*(2*b*x^2*(6*A*(2*c + d*x) + x*(6*B*c + 4*c*C*x + 4*B*d*x 
+ 3*C*d*x^2)) + a*(-12*A*(c + 2*d*x) + x*(-24*B*c + 32*c*C*x + 32*B*d*x + 
15*C*d*x^2))))/(24*x^2) + 3*Sqrt[a]*A*b*c*ArcTanh[(Sqrt[b]*x - Sqrt[a + b* 
x^2])/Sqrt[a]] - 2*a^(3/2)*(c*C + B*d)*ArcTanh[(-(Sqrt[b]*x) + Sqrt[a + b* 
x^2])/Sqrt[a]] - (3*a*(a*C*d + 4*b*(B*c + A*d))*Log[-(Sqrt[b]*x) + Sqrt[a 
+ b*x^2]])/(8*Sqrt[b])
 

Rubi [A] (verified)

Time = 1.20 (sec) , antiderivative size = 253, normalized size of antiderivative = 0.95, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.467, Rules used = {2338, 25, 2338, 25, 27, 535, 27, 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.

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

\(\Big \downarrow \) 2338

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 2338

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 535

\(\displaystyle \frac {\frac {1}{4} a \int \frac {2 (2 (3 A b c+2 a (c C+B d))+3 (a C d+4 b (B c+A d)) x) \sqrt {b x^2+a}}{x}dx+\frac {1}{6} \left (a+b x^2\right )^{3/2} (3 x (a C d+4 b (A d+B c))+2 (2 a (B d+c C)+3 A b c))-\frac {2 \left (a+b x^2\right )^{5/2} (A d+B c)}{x}}{2 a}-\frac {A c \left (a+b x^2\right )^{5/2}}{2 a x^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{2} a \int \frac {(2 (3 A b c+2 a (c C+B d))+3 (a C d+4 b (B c+A d)) x) \sqrt {b x^2+a}}{x}dx+\frac {1}{6} \left (a+b x^2\right )^{3/2} (3 x (a C d+4 b (A d+B c))+2 (2 a (B d+c C)+3 A b c))-\frac {2 \left (a+b x^2\right )^{5/2} (A d+B c)}{x}}{2 a}-\frac {A c \left (a+b x^2\right )^{5/2}}{2 a x^2}\)

\(\Big \downarrow \) 535

\(\displaystyle \frac {\frac {1}{2} a \left (\frac {1}{2} a \int \frac {4 (3 A b c+2 a (c C+B d))+3 (a C d+4 b (B c+A d)) x}{x \sqrt {b x^2+a}}dx+\frac {1}{2} \sqrt {a+b x^2} (3 x (a C d+4 b (A d+B c))+4 (2 a (B d+c C)+3 A b c))\right )+\frac {1}{6} \left (a+b x^2\right )^{3/2} (3 x (a C d+4 b (A d+B c))+2 (2 a (B d+c C)+3 A b c))-\frac {2 \left (a+b x^2\right )^{5/2} (A d+B c)}{x}}{2 a}-\frac {A c \left (a+b x^2\right )^{5/2}}{2 a x^2}\)

\(\Big \downarrow \) 538

\(\displaystyle \frac {\frac {1}{2} a \left (\frac {1}{2} a \left (3 (a C d+4 b (A d+B c)) \int \frac {1}{\sqrt {b x^2+a}}dx+4 (2 a (B d+c C)+3 A b c) \int \frac {1}{x \sqrt {b x^2+a}}dx\right )+\frac {1}{2} \sqrt {a+b x^2} (3 x (a C d+4 b (A d+B c))+4 (2 a (B d+c C)+3 A b c))\right )+\frac {1}{6} \left (a+b x^2\right )^{3/2} (3 x (a C d+4 b (A d+B c))+2 (2 a (B d+c C)+3 A b c))-\frac {2 \left (a+b x^2\right )^{5/2} (A d+B c)}{x}}{2 a}-\frac {A c \left (a+b x^2\right )^{5/2}}{2 a x^2}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {\frac {1}{2} a \left (\frac {1}{2} a \left (4 (2 a (B d+c C)+3 A b c) \int \frac {1}{x \sqrt {b x^2+a}}dx+3 (a C d+4 b (A d+B c)) \int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}\right )+\frac {1}{2} \sqrt {a+b x^2} (3 x (a C d+4 b (A d+B c))+4 (2 a (B d+c C)+3 A b c))\right )+\frac {1}{6} \left (a+b x^2\right )^{3/2} (3 x (a C d+4 b (A d+B c))+2 (2 a (B d+c C)+3 A b c))-\frac {2 \left (a+b x^2\right )^{5/2} (A d+B c)}{x}}{2 a}-\frac {A c \left (a+b x^2\right )^{5/2}}{2 a x^2}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {1}{2} a \left (\frac {1}{2} a \left (4 (2 a (B d+c C)+3 A b c) \int \frac {1}{x \sqrt {b x^2+a}}dx+\frac {3 \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right ) (a C d+4 b (A d+B c))}{\sqrt {b}}\right )+\frac {1}{2} \sqrt {a+b x^2} (3 x (a C d+4 b (A d+B c))+4 (2 a (B d+c C)+3 A b c))\right )+\frac {1}{6} \left (a+b x^2\right )^{3/2} (3 x (a C d+4 b (A d+B c))+2 (2 a (B d+c C)+3 A b c))-\frac {2 \left (a+b x^2\right )^{5/2} (A d+B c)}{x}}{2 a}-\frac {A c \left (a+b x^2\right )^{5/2}}{2 a x^2}\)

\(\Big \downarrow \) 243

\(\displaystyle \frac {\frac {1}{2} a \left (\frac {1}{2} a \left (2 (2 a (B d+c C)+3 A b c) \int \frac {1}{x^2 \sqrt {b x^2+a}}dx^2+\frac {3 \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right ) (a C d+4 b (A d+B c))}{\sqrt {b}}\right )+\frac {1}{2} \sqrt {a+b x^2} (3 x (a C d+4 b (A d+B c))+4 (2 a (B d+c C)+3 A b c))\right )+\frac {1}{6} \left (a+b x^2\right )^{3/2} (3 x (a C d+4 b (A d+B c))+2 (2 a (B d+c C)+3 A b c))-\frac {2 \left (a+b x^2\right )^{5/2} (A d+B c)}{x}}{2 a}-\frac {A c \left (a+b x^2\right )^{5/2}}{2 a x^2}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {\frac {1}{2} a \left (\frac {1}{2} a \left (\frac {4 (2 a (B d+c C)+3 A b c) \int \frac {1}{\frac {x^4}{b}-\frac {a}{b}}d\sqrt {b x^2+a}}{b}+\frac {3 \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right ) (a C d+4 b (A d+B c))}{\sqrt {b}}\right )+\frac {1}{2} \sqrt {a+b x^2} (3 x (a C d+4 b (A d+B c))+4 (2 a (B d+c C)+3 A b c))\right )+\frac {1}{6} \left (a+b x^2\right )^{3/2} (3 x (a C d+4 b (A d+B c))+2 (2 a (B d+c C)+3 A b c))-\frac {2 \left (a+b x^2\right )^{5/2} (A d+B c)}{x}}{2 a}-\frac {A c \left (a+b x^2\right )^{5/2}}{2 a x^2}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {\frac {1}{2} a \left (\frac {1}{2} a \left (\frac {3 \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right ) (a C d+4 b (A d+B c))}{\sqrt {b}}-\frac {4 \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right ) (2 a (B d+c C)+3 A b c)}{\sqrt {a}}\right )+\frac {1}{2} \sqrt {a+b x^2} (3 x (a C d+4 b (A d+B c))+4 (2 a (B d+c C)+3 A b c))\right )+\frac {1}{6} \left (a+b x^2\right )^{3/2} (3 x (a C d+4 b (A d+B c))+2 (2 a (B d+c C)+3 A b c))-\frac {2 \left (a+b x^2\right )^{5/2} (A d+B c)}{x}}{2 a}-\frac {A c \left (a+b x^2\right )^{5/2}}{2 a x^2}\)

Input:

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

Output:

-1/2*(A*c*(a + b*x^2)^(5/2))/(a*x^2) + (((2*(3*A*b*c + 2*a*(c*C + B*d)) + 
3*(a*C*d + 4*b*(B*c + A*d))*x)*(a + b*x^2)^(3/2))/6 - (2*(B*c + A*d)*(a + 
b*x^2)^(5/2))/x + (a*(((4*(3*A*b*c + 2*a*(c*C + B*d)) + 3*(a*C*d + 4*b*(B* 
c + A*d))*x)*Sqrt[a + b*x^2])/2 + (a*((3*(a*C*d + 4*b*(B*c + A*d))*ArcTanh 
[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/Sqrt[b] - (4*(3*A*b*c + 2*a*(c*C + B*d))*Ar 
cTanh[Sqrt[a + b*x^2]/Sqrt[a]])/Sqrt[a]))/2))/2)/(2*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])
 

Maple [A] (verified)

Time = 0.21 (sec) , antiderivative size = 278, normalized size of antiderivative = 1.04

method	result	size
default	\(d C \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4}\right )+\left (A d +B c \right ) \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}}}{a x}+\frac {4 b \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4}\right )}{a}\right )+\left (B d +C c \right ) \left (\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}}}{3}+a \left (\sqrt {b \,x^{2}+a}-\sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )\right )\right )+A c \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}}}{2 a \,x^{2}}+\frac {3 b \left (\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}}}{3}+a \left (\sqrt {b \,x^{2}+a}-\sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )\right )\right )}{2 a}\right )\)	\(278\)
risch	\(-\frac {a \sqrt {b \,x^{2}+a}\, \left (2 A d x +2 B c x +A c \right )}{2 x^{2}}+\frac {b \,x^{2} \sqrt {b \,x^{2}+a}\, B d}{3}+\frac {b \,x^{2} \sqrt {b \,x^{2}+a}\, C c}{3}+\frac {4 a \sqrt {b \,x^{2}+a}\, B d}{3}+\frac {4 a \sqrt {b \,x^{2}+a}\, C c}{3}+\frac {b x \sqrt {b \,x^{2}+a}\, A d}{2}+\frac {b x \sqrt {b \,x^{2}+a}\, B c}{2}+\frac {5 x \sqrt {b \,x^{2}+a}\, a C d}{8}+\frac {3 A a \sqrt {b}\, d \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2}+\frac {3 B a \sqrt {b}\, c \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2}+\frac {3 C \,a^{2} d \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{8 \sqrt {b}}+A \sqrt {b \,x^{2}+a}\, b c -\frac {3 A \sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right ) b c}{2}-B \,a^{\frac {3}{2}} \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right ) d -C \,a^{\frac {3}{2}} \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right ) c +\frac {C b d \,x^{3} \sqrt {b \,x^{2}+a}}{4}\)	\(333\)

Input:

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

Output:

d*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*d+B*c)*(-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*d+C*c)*(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)))+A*c*(-1/2/a/x^2*(b*x^2+a)^(5/2) 
+3/2*b/a*(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))))
 

Fricas [A] (verification not implemented)

Time = 0.54 (sec) , antiderivative size = 907, normalized size of antiderivative = 3.40 \[ \int \frac {(c+d x) \left (a+b x^2\right )^{3/2} \left (A+B x+C x^2\right )}{x^3} \, dx =\text {Too large to display} \] Input:

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

Output:

[1/48*(9*(4*B*a*b*c + (C*a^2 + 4*A*a*b)*d)*sqrt(b)*x^2*log(-2*b*x^2 - 2*sq 
rt(b*x^2 + a)*sqrt(b)*x - a) + 12*(2*B*a*b*d + (2*C*a*b + 3*A*b^2)*c)*sqrt 
(a)*x^2*log(-(b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(a) + 2*a)/x^2) + 2*(6*C*b^2*d 
*x^5 + 8*(C*b^2*c + B*b^2*d)*x^4 - 12*A*a*b*c + 3*(4*B*b^2*c + (5*C*a*b + 
4*A*b^2)*d)*x^3 + 8*(4*B*a*b*d + (4*C*a*b + 3*A*b^2)*c)*x^2 - 24*(B*a*b*c 
+ A*a*b*d)*x)*sqrt(b*x^2 + a))/(b*x^2), -1/24*(9*(4*B*a*b*c + (C*a^2 + 4*A 
*a*b)*d)*sqrt(-b)*x^2*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) - 6*(2*B*a*b*d + 
(2*C*a*b + 3*A*b^2)*c)*sqrt(a)*x^2*log(-(b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(a) 
 + 2*a)/x^2) - (6*C*b^2*d*x^5 + 8*(C*b^2*c + B*b^2*d)*x^4 - 12*A*a*b*c + 3 
*(4*B*b^2*c + (5*C*a*b + 4*A*b^2)*d)*x^3 + 8*(4*B*a*b*d + (4*C*a*b + 3*A*b 
^2)*c)*x^2 - 24*(B*a*b*c + A*a*b*d)*x)*sqrt(b*x^2 + a))/(b*x^2), 1/48*(24* 
(2*B*a*b*d + (2*C*a*b + 3*A*b^2)*c)*sqrt(-a)*x^2*arctan(sqrt(b*x^2 + a)*sq 
rt(-a)/a) + 9*(4*B*a*b*c + (C*a^2 + 4*A*a*b)*d)*sqrt(b)*x^2*log(-2*b*x^2 - 
 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) + 2*(6*C*b^2*d*x^5 + 8*(C*b^2*c + B*b^2* 
d)*x^4 - 12*A*a*b*c + 3*(4*B*b^2*c + (5*C*a*b + 4*A*b^2)*d)*x^3 + 8*(4*B*a 
*b*d + (4*C*a*b + 3*A*b^2)*c)*x^2 - 24*(B*a*b*c + A*a*b*d)*x)*sqrt(b*x^2 + 
 a))/(b*x^2), -1/24*(9*(4*B*a*b*c + (C*a^2 + 4*A*a*b)*d)*sqrt(-b)*x^2*arct 
an(sqrt(-b)*x/sqrt(b*x^2 + a)) - 12*(2*B*a*b*d + (2*C*a*b + 3*A*b^2)*c)*sq 
rt(-a)*x^2*arctan(sqrt(b*x^2 + a)*sqrt(-a)/a) - (6*C*b^2*d*x^5 + 8*(C*b^2* 
c + B*b^2*d)*x^4 - 12*A*a*b*c + 3*(4*B*b^2*c + (5*C*a*b + 4*A*b^2)*d)*x...
 

Sympy [A] (verification not implemented)

Time = 5.67 (sec) , antiderivative size = 782, normalized size of antiderivative = 2.93 \[ \int \frac {(c+d x) \left (a+b x^2\right )^{3/2} \left (A+B x+C x^2\right )}{x^3} \, dx =\text {Too large to display} \] Input:

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

Output:

-A*a**(3/2)*d/(x*sqrt(1 + b*x**2/a)) - 3*A*sqrt(a)*b*c*asinh(sqrt(a)/(sqrt 
(b)*x))/2 - A*sqrt(a)*b*d*x/sqrt(1 + b*x**2/a) - A*a*sqrt(b)*c*sqrt(a/(b*x 
**2) + 1)/(2*x) + A*a*sqrt(b)*c/(x*sqrt(a/(b*x**2) + 1)) + A*a*sqrt(b)*d*a 
sinh(sqrt(b)*x/sqrt(a)) + A*b**(3/2)*c*x/sqrt(a/(b*x**2) + 1) + A*b*d*Piec 
ewise((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)) - B*a**(3/2)*c/(x*sqrt(1 + b*x**2/a)) - B*a**(3/2)*d*as 
inh(sqrt(a)/(sqrt(b)*x)) - B*sqrt(a)*b*c*x/sqrt(1 + b*x**2/a) + B*a**2*d/( 
sqrt(b)*x*sqrt(a/(b*x**2) + 1)) + B*a*sqrt(b)*c*asinh(sqrt(b)*x/sqrt(a)) + 
 B*a*sqrt(b)*d*x/sqrt(a/(b*x**2) + 1) + B*b*c*Piecewise((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)) + B*b 
*d*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**(3/2)*c*asinh(sqrt(a)/(sqrt(b)*x)) + C*a 
**2*c/(sqrt(b)*x*sqrt(a/(b*x**2) + 1)) + C*a*sqrt(b)*c*x/sqrt(a/(b*x**2) + 
 1) + C*a*d*Piecewise((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*c*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*b*d 
*Piecewise((-a**2*Piecewise((log(2*sqrt(b)*sqrt(a + b*x**2) + 2*b*x)/sq...
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 242, normalized size of antiderivative = 0.91 \[ \int \frac {(c+d x) \left (a+b x^2\right )^{3/2} \left (A+B x+C x^2\right )}{x^3} \, dx=\frac {1}{4} \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} C d x + \frac {3}{8} \, \sqrt {b x^{2} + a} C a d x + \frac {3 \, C a^{2} d \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {b}} - \frac {3}{2} \, A \sqrt {a} b c \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right ) + \frac {3}{2} \, \sqrt {b x^{2} + a} A b c + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} A b c}{2 \, a} + \frac {3}{2} \, \sqrt {b x^{2} + a} {\left (B c + A d\right )} b x + \frac {3}{2} \, {\left (B c + A d\right )} a \sqrt {b} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right ) - {\left (C c + B d\right )} a^{\frac {3}{2}} \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right ) + \frac {1}{3} \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} {\left (C c + B d\right )} + \sqrt {b x^{2} + a} {\left (C c + B d\right )} a - \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} A c}{2 \, a x^{2}} - \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} {\left (B c + A d\right )}}{x} \] Input:

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

Output:

1/4*(b*x^2 + a)^(3/2)*C*d*x + 3/8*sqrt(b*x^2 + a)*C*a*d*x + 3/8*C*a^2*d*ar 
csinh(b*x/sqrt(a*b))/sqrt(b) - 3/2*A*sqrt(a)*b*c*arcsinh(a/(sqrt(a*b)*abs( 
x))) + 3/2*sqrt(b*x^2 + a)*A*b*c + 1/2*(b*x^2 + a)^(3/2)*A*b*c/a + 3/2*sqr 
t(b*x^2 + a)*(B*c + A*d)*b*x + 3/2*(B*c + A*d)*a*sqrt(b)*arcsinh(b*x/sqrt( 
a*b)) - (C*c + B*d)*a^(3/2)*arcsinh(a/(sqrt(a*b)*abs(x))) + 1/3*(b*x^2 + a 
)^(3/2)*(C*c + B*d) + sqrt(b*x^2 + a)*(C*c + B*d)*a - 1/2*(b*x^2 + a)^(5/2 
)*A*c/(a*x^2) - (b*x^2 + a)^(3/2)*(B*c + A*d)/x
 

Giac [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 348, normalized size of antiderivative = 1.30 \[ \int \frac {(c+d x) \left (a+b x^2\right )^{3/2} \left (A+B x+C x^2\right )}{x^3} \, dx=\frac {1}{24} \, \sqrt {b x^{2} + a} {\left ({\left (2 \, {\left (3 \, C b d x + \frac {4 \, {\left (C b^{3} c + B b^{3} d\right )}}{b^{2}}\right )} x + \frac {3 \, {\left (4 \, B b^{3} c + 5 \, C a b^{2} d + 4 \, A b^{3} d\right )}}{b^{2}}\right )} x + \frac {8 \, {\left (4 \, C a b^{2} c + 3 \, A b^{3} c + 4 \, B a b^{2} d\right )}}{b^{2}}\right )} + \frac {{\left (2 \, C a^{2} c + 3 \, A a b c + 2 \, B a^{2} d\right )} \arctan \left (-\frac {\sqrt {b} x - \sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} - \frac {3 \, {\left (4 \, B a b c + C a^{2} d + 4 \, A a b d\right )} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{8 \, \sqrt {b}} + \frac {{\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{3} A a b c + 2 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} B a^{2} \sqrt {b} c + 2 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} A a^{2} \sqrt {b} d + {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )} A a^{2} b c - 2 \, B a^{3} \sqrt {b} c - 2 \, A a^{3} \sqrt {b} d}{{\left ({\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} - a\right )}^{2}} \] Input:

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

Output:

1/24*sqrt(b*x^2 + a)*((2*(3*C*b*d*x + 4*(C*b^3*c + B*b^3*d)/b^2)*x + 3*(4* 
B*b^3*c + 5*C*a*b^2*d + 4*A*b^3*d)/b^2)*x + 8*(4*C*a*b^2*c + 3*A*b^3*c + 4 
*B*a*b^2*d)/b^2) + (2*C*a^2*c + 3*A*a*b*c + 2*B*a^2*d)*arctan(-(sqrt(b)*x 
- sqrt(b*x^2 + a))/sqrt(-a))/sqrt(-a) - 3/8*(4*B*a*b*c + C*a^2*d + 4*A*a*b 
*d)*log(abs(-sqrt(b)*x + sqrt(b*x^2 + a)))/sqrt(b) + ((sqrt(b)*x - sqrt(b* 
x^2 + a))^3*A*a*b*c + 2*(sqrt(b)*x - sqrt(b*x^2 + a))^2*B*a^2*sqrt(b)*c + 
2*(sqrt(b)*x - sqrt(b*x^2 + a))^2*A*a^2*sqrt(b)*d + (sqrt(b)*x - sqrt(b*x^ 
2 + a))*A*a^2*b*c - 2*B*a^3*sqrt(b)*c - 2*A*a^3*sqrt(b)*d)/((sqrt(b)*x - s 
qrt(b*x^2 + a))^2 - a)^2
                                                                                    
                                                                                    
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 513, normalized size of antiderivative = 1.92 \[ \int \frac {(c+d x) \left (a+b x^2\right )^{3/2} \left (A+B x+C x^2\right )}{x^3} \, dx=\frac {-12 \sqrt {b \,x^{2}+a}\, a^{2} b c -24 \sqrt {b \,x^{2}+a}\, a^{2} b d x +24 \sqrt {b \,x^{2}+a}\, a \,b^{2} c \,x^{2}-24 \sqrt {b \,x^{2}+a}\, a \,b^{2} c x +12 \sqrt {b \,x^{2}+a}\, a \,b^{2} d \,x^{3}+32 \sqrt {b \,x^{2}+a}\, a \,b^{2} d \,x^{2}+32 \sqrt {b \,x^{2}+a}\, a b \,c^{2} x^{2}+15 \sqrt {b \,x^{2}+a}\, a b c d \,x^{3}+12 \sqrt {b \,x^{2}+a}\, b^{3} c \,x^{3}+8 \sqrt {b \,x^{2}+a}\, b^{3} d \,x^{4}+8 \sqrt {b \,x^{2}+a}\, b^{2} c^{2} x^{4}+6 \sqrt {b \,x^{2}+a}\, b^{2} c d \,x^{5}+36 \sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}-\sqrt {a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a \,b^{2} c \,x^{2}+24 \sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}-\sqrt {a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a \,b^{2} d \,x^{2}+24 \sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}-\sqrt {a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a b \,c^{2} x^{2}-36 \sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a \,b^{2} c \,x^{2}-24 \sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a \,b^{2} d \,x^{2}-24 \sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a b \,c^{2} x^{2}+36 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{2} b d \,x^{2}+9 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{2} c d \,x^{2}+36 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a \,b^{2} c \,x^{2}}{24 b \,x^{2}} \] Input:

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

Output:

( - 12*sqrt(a + b*x**2)*a**2*b*c - 24*sqrt(a + b*x**2)*a**2*b*d*x + 24*sqr 
t(a + b*x**2)*a*b**2*c*x**2 - 24*sqrt(a + b*x**2)*a*b**2*c*x + 12*sqrt(a + 
 b*x**2)*a*b**2*d*x**3 + 32*sqrt(a + b*x**2)*a*b**2*d*x**2 + 32*sqrt(a + b 
*x**2)*a*b*c**2*x**2 + 15*sqrt(a + b*x**2)*a*b*c*d*x**3 + 12*sqrt(a + b*x* 
*2)*b**3*c*x**3 + 8*sqrt(a + b*x**2)*b**3*d*x**4 + 8*sqrt(a + b*x**2)*b**2 
*c**2*x**4 + 6*sqrt(a + b*x**2)*b**2*c*d*x**5 + 36*sqrt(a)*log((sqrt(a + b 
*x**2) - sqrt(a) + sqrt(b)*x)/sqrt(a))*a*b**2*c*x**2 + 24*sqrt(a)*log((sqr 
t(a + b*x**2) - sqrt(a) + sqrt(b)*x)/sqrt(a))*a*b**2*d*x**2 + 24*sqrt(a)*l 
og((sqrt(a + b*x**2) - sqrt(a) + sqrt(b)*x)/sqrt(a))*a*b*c**2*x**2 - 36*sq 
rt(a)*log((sqrt(a + b*x**2) + sqrt(a) + sqrt(b)*x)/sqrt(a))*a*b**2*c*x**2 
- 24*sqrt(a)*log((sqrt(a + b*x**2) + sqrt(a) + sqrt(b)*x)/sqrt(a))*a*b**2* 
d*x**2 - 24*sqrt(a)*log((sqrt(a + b*x**2) + sqrt(a) + sqrt(b)*x)/sqrt(a))* 
a*b*c**2*x**2 + 36*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a** 
2*b*d*x**2 + 9*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a**2*c* 
d*x**2 + 36*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a*b**2*c*x 
**2)/(24*b*x**2)