\(\int \frac {(c+d x) (a+b x^2)^{3/2} (A+B x+C x^2)}{x} \, dx\) [51]

Optimal result

Integrand size = 30, antiderivative size = 221 \[ \int \frac {(c+d x) \left (a+b x^2\right )^{3/2} \left (A+B x+C x^2\right )}{x} \, dx=a A c \sqrt {a+b x^2}-\frac {a (a C d-6 b (B c+A d)) x \sqrt {a+b x^2}}{16 b}+\frac {1}{3} A c \left (a+b x^2\right )^{3/2}-\frac {(a C d-6 b (B c+A d)) x \left (a+b x^2\right )^{3/2}}{24 b}+\frac {(c C+B d) \left (a+b x^2\right )^{5/2}}{5 b}+\frac {C d x \left (a+b x^2\right )^{5/2}}{6 b}-\frac {a^2 (a C d-6 b (B c+A d)) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{16 b^{3/2}}-a^{3/2} A c \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right ) \] Output:

a*A*c*(b*x^2+a)^(1/2)-1/16*a*(a*C*d-6*b*(A*d+B*c))*x*(b*x^2+a)^(1/2)/b+1/3 
*A*c*(b*x^2+a)^(3/2)-1/24*(a*C*d-6*b*(A*d+B*c))*x*(b*x^2+a)^(3/2)/b+1/5*(B 
*d+C*c)*(b*x^2+a)^(5/2)/b+1/6*C*d*x*(b*x^2+a)^(5/2)/b-1/16*a^2*(a*C*d-6*b* 
(A*d+B*c))*arctanh(b^(1/2)*x/(b*x^2+a)^(1/2))/b^(3/2)-a^(3/2)*A*c*arctanh( 
(b*x^2+a)^(1/2)/a^(1/2))
 

Mathematica [A] (verified)

Time = 1.22 (sec) , antiderivative size = 210, normalized size of antiderivative = 0.95 \[ \int \frac {(c+d x) \left (a+b x^2\right )^{3/2} \left (A+B x+C x^2\right )}{x} \, dx=\frac {\sqrt {a+b x^2} \left (3 a^2 (16 c C+16 B d+5 C d x)+2 a b \left (5 A (32 c+15 d x)+x \left (75 B c+48 c C x+48 B d x+35 C d x^2\right )\right )+4 b^2 x^2 (5 A (4 c+3 d x)+x (3 B (5 c+4 d x)+2 C x (6 c+5 d x)))\right )}{240 b}+2 a^{3/2} A c \text {arctanh}\left (\frac {\sqrt {b} x-\sqrt {a+b x^2}}{\sqrt {a}}\right )+\frac {a^2 (a C d-6 b (B c+A d)) \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{16 b^{3/2}} \] Input:

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

Output:

(Sqrt[a + b*x^2]*(3*a^2*(16*c*C + 16*B*d + 5*C*d*x) + 2*a*b*(5*A*(32*c + 1 
5*d*x) + x*(75*B*c + 48*c*C*x + 48*B*d*x + 35*C*d*x^2)) + 4*b^2*x^2*(5*A*( 
4*c + 3*d*x) + x*(3*B*(5*c + 4*d*x) + 2*C*x*(6*c + 5*d*x)))))/(240*b) + 2* 
a^(3/2)*A*c*ArcTanh[(Sqrt[b]*x - Sqrt[a + b*x^2])/Sqrt[a]] + (a^2*(a*C*d - 
 6*b*(B*c + A*d))*Log[-(Sqrt[b]*x) + Sqrt[a + b*x^2]])/(16*b^(3/2))
 

Rubi [A] (verified)

Time = 1.12 (sec) , antiderivative size = 211, normalized size of antiderivative = 0.95, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {2340, 2340, 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.

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

\(\Big \downarrow \) 2340

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

\(\Big \downarrow \) 2340

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 535

\(\displaystyle \frac {\frac {1}{4} a \int \frac {3 (8 A b c-(a C d-6 b (B c+A d)) x) \sqrt {b x^2+a}}{x}dx+\frac {1}{4} \left (a+b x^2\right )^{3/2} (8 A b c-x (a C d-6 b (A d+B c)))+\frac {6}{5} \left (a+b x^2\right )^{5/2} (B d+c C)}{6 b}+\frac {C d x \left (a+b x^2\right )^{5/2}}{6 b}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {3}{4} a \int \frac {(8 A b c-(a C d-6 b (B c+A d)) x) \sqrt {b x^2+a}}{x}dx+\frac {1}{4} \left (a+b x^2\right )^{3/2} (8 A b c-x (a C d-6 b (A d+B c)))+\frac {6}{5} \left (a+b x^2\right )^{5/2} (B d+c C)}{6 b}+\frac {C d x \left (a+b x^2\right )^{5/2}}{6 b}\)

\(\Big \downarrow \) 535

\(\displaystyle \frac {\frac {3}{4} a \left (\frac {1}{2} a \int \frac {16 A b c-(a C d-6 b (B c+A d)) x}{x \sqrt {b x^2+a}}dx+\frac {1}{2} \sqrt {a+b x^2} (16 A b c-x (a C d-6 b (A d+B c)))\right )+\frac {1}{4} \left (a+b x^2\right )^{3/2} (8 A b c-x (a C d-6 b (A d+B c)))+\frac {6}{5} \left (a+b x^2\right )^{5/2} (B d+c C)}{6 b}+\frac {C d x \left (a+b x^2\right )^{5/2}}{6 b}\)

\(\Big \downarrow \) 538

\(\displaystyle \frac {\frac {3}{4} a \left (\frac {1}{2} a \left (16 A b c \int \frac {1}{x \sqrt {b x^2+a}}dx-(a C d-6 b (A d+B c)) \int \frac {1}{\sqrt {b x^2+a}}dx\right )+\frac {1}{2} \sqrt {a+b x^2} (16 A b c-x (a C d-6 b (A d+B c)))\right )+\frac {1}{4} \left (a+b x^2\right )^{3/2} (8 A b c-x (a C d-6 b (A d+B c)))+\frac {6}{5} \left (a+b x^2\right )^{5/2} (B d+c C)}{6 b}+\frac {C d x \left (a+b x^2\right )^{5/2}}{6 b}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {\frac {3}{4} a \left (\frac {1}{2} a \left (16 A b c \int \frac {1}{x \sqrt {b x^2+a}}dx-(a C d-6 b (A d+B c)) \int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}\right )+\frac {1}{2} \sqrt {a+b x^2} (16 A b c-x (a C d-6 b (A d+B c)))\right )+\frac {1}{4} \left (a+b x^2\right )^{3/2} (8 A b c-x (a C d-6 b (A d+B c)))+\frac {6}{5} \left (a+b x^2\right )^{5/2} (B d+c C)}{6 b}+\frac {C d x \left (a+b x^2\right )^{5/2}}{6 b}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {3}{4} a \left (\frac {1}{2} a \left (16 A b c \int \frac {1}{x \sqrt {b x^2+a}}dx-\frac {\text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right ) (a C d-6 b (A d+B c))}{\sqrt {b}}\right )+\frac {1}{2} \sqrt {a+b x^2} (16 A b c-x (a C d-6 b (A d+B c)))\right )+\frac {1}{4} \left (a+b x^2\right )^{3/2} (8 A b c-x (a C d-6 b (A d+B c)))+\frac {6}{5} \left (a+b x^2\right )^{5/2} (B d+c C)}{6 b}+\frac {C d x \left (a+b x^2\right )^{5/2}}{6 b}\)

\(\Big \downarrow \) 243

\(\displaystyle \frac {\frac {3}{4} a \left (\frac {1}{2} a \left (8 A b c \int \frac {1}{x^2 \sqrt {b x^2+a}}dx^2-\frac {\text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right ) (a C d-6 b (A d+B c))}{\sqrt {b}}\right )+\frac {1}{2} \sqrt {a+b x^2} (16 A b c-x (a C d-6 b (A d+B c)))\right )+\frac {1}{4} \left (a+b x^2\right )^{3/2} (8 A b c-x (a C d-6 b (A d+B c)))+\frac {6}{5} \left (a+b x^2\right )^{5/2} (B d+c C)}{6 b}+\frac {C d x \left (a+b x^2\right )^{5/2}}{6 b}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {\frac {3}{4} a \left (\frac {1}{2} a \left (16 A c \int \frac {1}{\frac {x^4}{b}-\frac {a}{b}}d\sqrt {b x^2+a}-\frac {\text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right ) (a C d-6 b (A d+B c))}{\sqrt {b}}\right )+\frac {1}{2} \sqrt {a+b x^2} (16 A b c-x (a C d-6 b (A d+B c)))\right )+\frac {1}{4} \left (a+b x^2\right )^{3/2} (8 A b c-x (a C d-6 b (A d+B c)))+\frac {6}{5} \left (a+b x^2\right )^{5/2} (B d+c C)}{6 b}+\frac {C d x \left (a+b x^2\right )^{5/2}}{6 b}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {\frac {3}{4} a \left (\frac {1}{2} a \left (-\frac {\text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right ) (a C d-6 b (A d+B c))}{\sqrt {b}}-\frac {16 A b c \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{\sqrt {a}}\right )+\frac {1}{2} \sqrt {a+b x^2} (16 A b c-x (a C d-6 b (A d+B c)))\right )+\frac {1}{4} \left (a+b x^2\right )^{3/2} (8 A b c-x (a C d-6 b (A d+B c)))+\frac {6}{5} \left (a+b x^2\right )^{5/2} (B d+c C)}{6 b}+\frac {C d x \left (a+b x^2\right )^{5/2}}{6 b}\)

Input:

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

Output:

(C*d*x*(a + b*x^2)^(5/2))/(6*b) + (((8*A*b*c - (a*C*d - 6*b*(B*c + A*d))*x 
)*(a + b*x^2)^(3/2))/4 + (6*(c*C + B*d)*(a + b*x^2)^(5/2))/5 + (3*a*(((16* 
A*b*c - (a*C*d - 6*b*(B*c + A*d))*x)*Sqrt[a + b*x^2])/2 + (a*(-(((a*C*d - 
6*b*(B*c + A*d))*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/Sqrt[b]) - (16*A*b* 
c*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]])/Sqrt[a]))/2))/4)/(6*b)
 

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 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 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 = 0.17 (sec) , antiderivative size = 273, normalized size of antiderivative = 1.24

method	result	size
default	\(A d \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4}\right )+B c \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4}\right )+A c \left (\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}}}{3}+a \left (\sqrt {b \,x^{2}+a}-\sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )\right )\right )+\frac {B d \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{5 b}+\frac {C c \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{5 b}+d C \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{6 b}-\frac {a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4}\right )}{6 b}\right )\)	\(273\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.63 (sec) , antiderivative size = 987, normalized size of antiderivative = 4.47 \[ \int \frac {(c+d x) \left (a+b x^2\right )^{3/2} \left (A+B x+C x^2\right )}{x} \, dx=\text {Too large to display} \] Input:

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

Output:

[1/480*(240*A*a^(3/2)*b^2*c*log(-(b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(a) + 2*a) 
/x^2) + 15*(6*B*a^2*b*c - (C*a^3 - 6*A*a^2*b)*d)*sqrt(b)*log(-2*b*x^2 - 2* 
sqrt(b*x^2 + a)*sqrt(b)*x - a) + 2*(40*C*b^3*d*x^5 + 48*B*a^2*b*d + 48*(C* 
b^3*c + B*b^3*d)*x^4 + 10*(6*B*b^3*c + (7*C*a*b^2 + 6*A*b^3)*d)*x^3 + 16*( 
6*B*a*b^2*d + (6*C*a*b^2 + 5*A*b^3)*c)*x^2 + 16*(3*C*a^2*b + 20*A*a*b^2)*c 
 + 15*(10*B*a*b^2*c + (C*a^2*b + 10*A*a*b^2)*d)*x)*sqrt(b*x^2 + a))/b^2, 1 
/240*(120*A*a^(3/2)*b^2*c*log(-(b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(a) + 2*a)/x 
^2) - 15*(6*B*a^2*b*c - (C*a^3 - 6*A*a^2*b)*d)*sqrt(-b)*arctan(sqrt(-b)*x/ 
sqrt(b*x^2 + a)) + (40*C*b^3*d*x^5 + 48*B*a^2*b*d + 48*(C*b^3*c + B*b^3*d) 
*x^4 + 10*(6*B*b^3*c + (7*C*a*b^2 + 6*A*b^3)*d)*x^3 + 16*(6*B*a*b^2*d + (6 
*C*a*b^2 + 5*A*b^3)*c)*x^2 + 16*(3*C*a^2*b + 20*A*a*b^2)*c + 15*(10*B*a*b^ 
2*c + (C*a^2*b + 10*A*a*b^2)*d)*x)*sqrt(b*x^2 + a))/b^2, 1/480*(480*A*sqrt 
(-a)*a*b^2*c*arctan(sqrt(b*x^2 + a)*sqrt(-a)/a) + 15*(6*B*a^2*b*c - (C*a^3 
 - 6*A*a^2*b)*d)*sqrt(b)*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) + 
 2*(40*C*b^3*d*x^5 + 48*B*a^2*b*d + 48*(C*b^3*c + B*b^3*d)*x^4 + 10*(6*B*b 
^3*c + (7*C*a*b^2 + 6*A*b^3)*d)*x^3 + 16*(6*B*a*b^2*d + (6*C*a*b^2 + 5*A*b 
^3)*c)*x^2 + 16*(3*C*a^2*b + 20*A*a*b^2)*c + 15*(10*B*a*b^2*c + (C*a^2*b + 
 10*A*a*b^2)*d)*x)*sqrt(b*x^2 + a))/b^2, 1/240*(240*A*sqrt(-a)*a*b^2*c*arc 
tan(sqrt(b*x^2 + a)*sqrt(-a)/a) - 15*(6*B*a^2*b*c - (C*a^3 - 6*A*a^2*b)*d) 
*sqrt(-b)*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) + (40*C*b^3*d*x^5 + 48*B*a...
 

Sympy [A] (verification not implemented)

Time = 11.75 (sec) , antiderivative size = 901, normalized size of antiderivative = 4.08 \[ \int \frac {(c+d x) \left (a+b x^2\right )^{3/2} \left (A+B x+C x^2\right )}{x} \, dx=\text {Too large to display} \] Input:

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

Output:

-A*a**(3/2)*c*asinh(sqrt(a)/(sqrt(b)*x)) + A*a**2*c/(sqrt(b)*x*sqrt(a/(b*x 
**2) + 1)) + A*a*sqrt(b)*c*x/sqrt(a/(b*x**2) + 1) + A*a*d*Piecewise((a*Pie 
cewise((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)) + A*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)) + A*b*d*Piecewise((-a**2*Piecewise( 
(log(2*sqrt(b)*sqrt(a + b*x**2) + 2*b*x)/sqrt(b), Ne(a, 0)), (x*log(x)/sqr 
t(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)), (sqrt(a)*x**3/3, True)) + B*a*c*Piecewise((a*Piecewise((l 
og(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*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)) + B*b*c*Piecewise((-a**2*Piecewise((log(2*sq 
rt(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)), (sqrt(a)*x**3/3, True)) + B*b*d*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)) + C*a*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*a*d*P 
iecewise((-a**2*Piecewise((log(2*sqrt(b)*sqrt(a + b*x**2) + 2*b*x)/sqrt...
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 215, normalized size of antiderivative = 0.97 \[ \int \frac {(c+d x) \left (a+b x^2\right )^{3/2} \left (A+B x+C x^2\right )}{x} \, dx=\frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} C d x}{6 \, b} - \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} C a d x}{24 \, b} - \frac {\sqrt {b x^{2} + a} C a^{2} d x}{16 \, b} - \frac {C a^{3} d \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{16 \, b^{\frac {3}{2}}} - A a^{\frac {3}{2}} c \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right ) + \frac {1}{3} \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} A c + \sqrt {b x^{2} + a} A a c + \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} C c}{5 \, b} + \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} B d}{5 \, b} + \frac {1}{4} \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} {\left (B c + A d\right )} x + \frac {3}{8} \, \sqrt {b x^{2} + a} {\left (B c + A d\right )} a x + \frac {3 \, {\left (B c + A d\right )} a^{2} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {b}} \] Input:

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

Output:

1/6*(b*x^2 + a)^(5/2)*C*d*x/b - 1/24*(b*x^2 + a)^(3/2)*C*a*d*x/b - 1/16*sq 
rt(b*x^2 + a)*C*a^2*d*x/b - 1/16*C*a^3*d*arcsinh(b*x/sqrt(a*b))/b^(3/2) - 
A*a^(3/2)*c*arcsinh(a/(sqrt(a*b)*abs(x))) + 1/3*(b*x^2 + a)^(3/2)*A*c + sq 
rt(b*x^2 + a)*A*a*c + 1/5*(b*x^2 + a)^(5/2)*C*c/b + 1/5*(b*x^2 + a)^(5/2)* 
B*d/b + 1/4*(b*x^2 + a)^(3/2)*(B*c + A*d)*x + 3/8*sqrt(b*x^2 + a)*(B*c + A 
*d)*a*x + 3/8*(B*c + A*d)*a^2*arcsinh(b*x/sqrt(a*b))/sqrt(b)
 

Giac [F(-2)]

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

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:index.cc index_m i_lex_is_greater E 
rror: Bad Argument Value
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 424, normalized size of antiderivative = 1.92 \[ \int \frac {(c+d x) \left (a+b x^2\right )^{3/2} \left (A+B x+C x^2\right )}{x} \, dx=\frac {320 \sqrt {b \,x^{2}+a}\, a^{2} b^{2} c +150 \sqrt {b \,x^{2}+a}\, a^{2} b^{2} d x +48 \sqrt {b \,x^{2}+a}\, a^{2} b^{2} d +48 \sqrt {b \,x^{2}+a}\, a^{2} b \,c^{2}+15 \sqrt {b \,x^{2}+a}\, a^{2} b c d x +80 \sqrt {b \,x^{2}+a}\, a \,b^{3} c \,x^{2}+150 \sqrt {b \,x^{2}+a}\, a \,b^{3} c x +60 \sqrt {b \,x^{2}+a}\, a \,b^{3} d \,x^{3}+96 \sqrt {b \,x^{2}+a}\, a \,b^{3} d \,x^{2}+96 \sqrt {b \,x^{2}+a}\, a \,b^{2} c^{2} x^{2}+70 \sqrt {b \,x^{2}+a}\, a \,b^{2} c d \,x^{3}+60 \sqrt {b \,x^{2}+a}\, b^{4} c \,x^{3}+48 \sqrt {b \,x^{2}+a}\, b^{4} d \,x^{4}+48 \sqrt {b \,x^{2}+a}\, b^{3} c^{2} x^{4}+40 \sqrt {b \,x^{2}+a}\, b^{3} c d \,x^{5}+240 \sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}-\sqrt {a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{2} b^{2} c -240 \sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{2} b^{2} c +90 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{3} b d -15 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{3} c d +90 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{2} b^{2} c}{240 b^{2}} \] Input:

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

Output:

(320*sqrt(a + b*x**2)*a**2*b**2*c + 150*sqrt(a + b*x**2)*a**2*b**2*d*x + 4 
8*sqrt(a + b*x**2)*a**2*b**2*d + 48*sqrt(a + b*x**2)*a**2*b*c**2 + 15*sqrt 
(a + b*x**2)*a**2*b*c*d*x + 80*sqrt(a + b*x**2)*a*b**3*c*x**2 + 150*sqrt(a 
 + b*x**2)*a*b**3*c*x + 60*sqrt(a + b*x**2)*a*b**3*d*x**3 + 96*sqrt(a + b* 
x**2)*a*b**3*d*x**2 + 96*sqrt(a + b*x**2)*a*b**2*c**2*x**2 + 70*sqrt(a + b 
*x**2)*a*b**2*c*d*x**3 + 60*sqrt(a + b*x**2)*b**4*c*x**3 + 48*sqrt(a + b*x 
**2)*b**4*d*x**4 + 48*sqrt(a + b*x**2)*b**3*c**2*x**4 + 40*sqrt(a + b*x**2 
)*b**3*c*d*x**5 + 240*sqrt(a)*log((sqrt(a + b*x**2) - sqrt(a) + sqrt(b)*x) 
/sqrt(a))*a**2*b**2*c - 240*sqrt(a)*log((sqrt(a + b*x**2) + sqrt(a) + sqrt 
(b)*x)/sqrt(a))*a**2*b**2*c + 90*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x 
)/sqrt(a))*a**3*b*d - 15*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a 
))*a**3*c*d + 90*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a**2* 
b**2*c)/(240*b**2)