\(\int \frac {x (c+d x) (A+B x+C x^2)}{\sqrt {a+b x^2}} \, dx\) [96]

Optimal result

Integrand size = 28, antiderivative size = 159 \[ \int \frac {x (c+d x) \left (A+B x+C x^2\right )}{\sqrt {a+b x^2}} \, dx=\frac {(A b c-a (c C+B d)) \sqrt {a+b x^2}}{b^2}-\frac {(3 a C d-4 b (B c+A d)) x \sqrt {a+b x^2}}{8 b^2}+\frac {C d x^3 \sqrt {a+b x^2}}{4 b}+\frac {(c C+B d) \left (a+b x^2\right )^{3/2}}{3 b^2}+\frac {a (3 a C d-4 b (B c+A d)) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{8 b^{5/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.58 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.78 \[ \int \frac {x (c+d x) \left (A+B x+C x^2\right )}{\sqrt {a+b x^2}} \, dx=\frac {\sqrt {a+b x^2} \left (24 A b c-16 a c C-16 a B d+12 b B c x+12 A b d x-9 a C d x+8 b c C x^2+8 b B d x^2+6 b C d x^3\right )}{24 b^2}-\frac {a (-4 b B c-4 A b d+3 a C d) \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{8 b^{5/2}} \] Input:

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

Output:

(Sqrt[a + b*x^2]*(24*A*b*c - 16*a*c*C - 16*a*B*d + 12*b*B*c*x + 12*A*b*d*x 
 - 9*a*C*d*x + 8*b*c*C*x^2 + 8*b*B*d*x^2 + 6*b*C*d*x^3))/(24*b^2) - (a*(-4 
*b*B*c - 4*A*b*d + 3*a*C*d)*Log[-(Sqrt[b]*x) + Sqrt[a + b*x^2]])/(8*b^(5/2 
))
 

Rubi [A] (verified)

Time = 1.09 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.42, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {2185, 25, 2185, 25, 27, 676, 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.

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

\(\Big \downarrow \) 2185

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 2185

\(\displaystyle \frac {C \sqrt {a+b x^2} (c+d x)^3}{4 b d^2}-\frac {\frac {\int -\frac {b d^3 (c+d x) \left (a d (c C-8 B d)-\left (9 a C d^2-2 b \left (C c^2-2 B d c+6 A d^2\right )\right ) x\right )}{\sqrt {b x^2+a}}dx}{3 b d^2}+\frac {1}{3} d \sqrt {a+b x^2} (c+d x)^2 (5 c C-4 B d)}{4 b d^3}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {C \sqrt {a+b x^2} (c+d x)^3}{4 b d^2}-\frac {\frac {1}{3} d \sqrt {a+b x^2} (c+d x)^2 (5 c C-4 B d)-\frac {\int \frac {b d^3 (c+d x) \left (a d (c C-8 B d)-\left (9 a C d^2-2 b \left (C c^2-2 B d c+6 A d^2\right )\right ) x\right )}{\sqrt {b x^2+a}}dx}{3 b d^2}}{4 b d^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {C \sqrt {a+b x^2} (c+d x)^3}{4 b d^2}-\frac {\frac {1}{3} d \sqrt {a+b x^2} (c+d x)^2 (5 c C-4 B d)-\frac {1}{3} d \int \frac {(c+d x) \left (a d (c C-8 B d)-\left (9 a C d^2-2 b \left (C c^2-2 B d c+6 A d^2\right )\right ) x\right )}{\sqrt {b x^2+a}}dx}{4 b d^3}\)

\(\Big \downarrow \) 676

\(\displaystyle \frac {C \sqrt {a+b x^2} (c+d x)^3}{4 b d^2}-\frac {\frac {1}{3} d \sqrt {a+b x^2} (c+d x)^2 (5 c C-4 B d)-\frac {1}{3} d \left (\frac {3 a d^2 (3 a C d-4 b (A d+B c)) \int \frac {1}{\sqrt {b x^2+a}}dx}{2 b}-\frac {d x \sqrt {a+b x^2} \left (9 a C d^2-2 b \left (6 A d^2-2 B c d+c^2 C\right )\right )}{2 b}-\frac {2 \sqrt {a+b x^2} \left (4 a d^2 (B d+c C)-b c \left (6 A d^2-2 B c d+c^2 C\right )\right )}{b}\right )}{4 b d^3}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {C \sqrt {a+b x^2} (c+d x)^3}{4 b d^2}-\frac {\frac {1}{3} d \sqrt {a+b x^2} (c+d x)^2 (5 c C-4 B d)-\frac {1}{3} d \left (\frac {3 a d^2 (3 a C d-4 b (A d+B c)) \int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}}{2 b}-\frac {d x \sqrt {a+b x^2} \left (9 a C d^2-2 b \left (6 A d^2-2 B c d+c^2 C\right )\right )}{2 b}-\frac {2 \sqrt {a+b x^2} \left (4 a d^2 (B d+c C)-b c \left (6 A d^2-2 B c d+c^2 C\right )\right )}{b}\right )}{4 b d^3}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {C \sqrt {a+b x^2} (c+d x)^3}{4 b d^2}-\frac {\frac {1}{3} d \sqrt {a+b x^2} (c+d x)^2 (5 c C-4 B d)-\frac {1}{3} d \left (\frac {3 a d^2 \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right ) (3 a C d-4 b (A d+B c))}{2 b^{3/2}}-\frac {d x \sqrt {a+b x^2} \left (9 a C d^2-2 b \left (6 A d^2-2 B c d+c^2 C\right )\right )}{2 b}-\frac {2 \sqrt {a+b x^2} \left (4 a d^2 (B d+c C)-b c \left (6 A d^2-2 B c d+c^2 C\right )\right )}{b}\right )}{4 b d^3}\)

Input:

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

Output:

(C*(c + d*x)^3*Sqrt[a + b*x^2])/(4*b*d^2) - ((d*(5*c*C - 4*B*d)*(c + d*x)^ 
2*Sqrt[a + b*x^2])/3 - (d*((-2*(4*a*d^2*(c*C + B*d) - b*c*(c^2*C - 2*B*c*d 
 + 6*A*d^2))*Sqrt[a + b*x^2])/b - (d*(9*a*C*d^2 - 2*b*(c^2*C - 2*B*c*d + 6 
*A*d^2))*x*Sqrt[a + b*x^2])/(2*b) + (3*a*d^2*(3*a*C*d - 4*b*(B*c + A*d))*A 
rcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(2*b^(3/2))))/3)/(4*b*d^3)
 

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 676

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x 
_Symbol] :> Simp[(e*f + d*g)*((a + c*x^2)^(p + 1)/(2*c*(p + 1))), x] + (Sim 
p[e*g*x*((a + c*x^2)^(p + 1)/(c*(2*p + 3))), x] - Simp[(a*e*g - c*d*f*(2*p 
+ 3))/(c*(2*p + 3))   Int[(a + c*x^2)^p, x], x]) /; FreeQ[{a, c, d, e, f, g 
, p}, x] &&  !LeQ[p, -1]
 

rule 2185

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] : 
> With[{q = Expon[Pq, x], f = Coeff[Pq, x, Expon[Pq, x]]}, Simp[f*(d + e*x) 
^(m + q - 1)*((a + b*x^2)^(p + 1)/(b*e^(q - 1)*(m + q + 2*p + 1))), x] + Si 
mp[1/(b*e^q*(m + q + 2*p + 1))   Int[(d + e*x)^m*(a + b*x^2)^p*ExpandToSum[ 
b*e^q*(m + q + 2*p + 1)*Pq - b*f*(m + q + 2*p + 1)*(d + e*x)^q - f*(d + e*x 
)^(q - 2)*(a*e^2*(m + q - 1) - b*d^2*(m + q + 2*p + 1) - 2*b*d*e*(m + q + p 
)*x), x], x], x] /; GtQ[q, 1] && NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, b, d 
, e, m, p}, x] && PolyQ[Pq, x] && NeQ[b*d^2 + a*e^2, 0] &&  !(EqQ[d, 0] && 
True) &&  !(IGtQ[m, 0] && RationalQ[a, b, d, e] && (IntegerQ[p] || ILtQ[p + 
 1/2, 0]))
 

Maple [A] (verified)

Time = 0.19 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.70

method	result	size
risch	\(\frac {\left (6 C b d \,x^{3}+8 B b d \,x^{2}+8 C c b \,x^{2}+12 x A b d +12 B b c x -9 C a d x +24 A b c -16 B a d -16 C a c \right ) \sqrt {b \,x^{2}+a}}{24 b^{2}}-\frac {a \left (4 A b d +4 B b c -3 a C d \right ) \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{8 b^{\frac {5}{2}}}\)	\(112\)
default	\(\left (A d +B c \right ) \left (\frac {x \sqrt {b \,x^{2}+a}}{2 b}-\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 b^{\frac {3}{2}}}\right )+\left (B d +C c \right ) \left (\frac {x^{2} \sqrt {b \,x^{2}+a}}{3 b}-\frac {2 a \sqrt {b \,x^{2}+a}}{3 b^{2}}\right )+\frac {A c \sqrt {b \,x^{2}+a}}{b}+d C \left (\frac {x^{3} \sqrt {b \,x^{2}+a}}{4 b}-\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2 b}-\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 b^{\frac {3}{2}}}\right )}{4 b}\right )\)	\(169\)

Input:

int(x*(d*x+c)*(C*x^2+B*x+A)/(b*x^2+a)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

1/24*(6*C*b*d*x^3+8*B*b*d*x^2+8*C*b*c*x^2+12*A*b*d*x+12*B*b*c*x-9*C*a*d*x+ 
24*A*b*c-16*B*a*d-16*C*a*c)*(b*x^2+a)^(1/2)/b^2-1/8*a*(4*A*b*d+4*B*b*c-3*C 
*a*d)/b^(5/2)*ln(b^(1/2)*x+(b*x^2+a)^(1/2))
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 285, normalized size of antiderivative = 1.79 \[ \int \frac {x (c+d x) \left (A+B x+C x^2\right )}{\sqrt {a+b x^2}} \, dx=\left [\frac {3 \, {\left (4 \, B a b c - {\left (3 \, C a^{2} - 4 \, A a b\right )} d\right )} \sqrt {b} \log \left (-2 \, b x^{2} + 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 2 \, {\left (6 \, C b^{2} d x^{3} - 16 \, B a b d + 8 \, {\left (C b^{2} c + B b^{2} d\right )} x^{2} - 8 \, {\left (2 \, C a b - 3 \, A b^{2}\right )} c + 3 \, {\left (4 \, B b^{2} c - {\left (3 \, C a b - 4 \, A b^{2}\right )} d\right )} x\right )} \sqrt {b x^{2} + a}}{48 \, b^{3}}, \frac {3 \, {\left (4 \, B a b c - {\left (3 \, C a^{2} - 4 \, A a b\right )} d\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) + {\left (6 \, C b^{2} d x^{3} - 16 \, B a b d + 8 \, {\left (C b^{2} c + B b^{2} d\right )} x^{2} - 8 \, {\left (2 \, C a b - 3 \, A b^{2}\right )} c + 3 \, {\left (4 \, B b^{2} c - {\left (3 \, C a b - 4 \, A b^{2}\right )} d\right )} x\right )} \sqrt {b x^{2} + a}}{24 \, b^{3}}\right ] \] Input:

integrate(x*(d*x+c)*(C*x^2+B*x+A)/(b*x^2+a)^(1/2),x, algorithm="fricas")
 

Output:

[1/48*(3*(4*B*a*b*c - (3*C*a^2 - 4*A*a*b)*d)*sqrt(b)*log(-2*b*x^2 + 2*sqrt 
(b*x^2 + a)*sqrt(b)*x - a) + 2*(6*C*b^2*d*x^3 - 16*B*a*b*d + 8*(C*b^2*c + 
B*b^2*d)*x^2 - 8*(2*C*a*b - 3*A*b^2)*c + 3*(4*B*b^2*c - (3*C*a*b - 4*A*b^2 
)*d)*x)*sqrt(b*x^2 + a))/b^3, 1/24*(3*(4*B*a*b*c - (3*C*a^2 - 4*A*a*b)*d)* 
sqrt(-b)*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) + (6*C*b^2*d*x^3 - 16*B*a*b*d 
+ 8*(C*b^2*c + B*b^2*d)*x^2 - 8*(2*C*a*b - 3*A*b^2)*c + 3*(4*B*b^2*c - (3* 
C*a*b - 4*A*b^2)*d)*x)*sqrt(b*x^2 + a))/b^3]
 

Sympy [A] (verification not implemented)

Time = 0.55 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.16 \[ \int \frac {x (c+d x) \left (A+B x+C x^2\right )}{\sqrt {a+b x^2}} \, dx=\begin {cases} - \frac {a \left (A d + B c - \frac {3 C a d}{4 b}\right ) \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {b x^{2}}} & \text {otherwise} \end {cases}\right )}{2 b} + \sqrt {a + b x^{2}} \left (\frac {C d x^{3}}{4 b} + \frac {x^{2} \left (B d + C c\right )}{3 b} + \frac {x \left (A d + B c - \frac {3 C a d}{4 b}\right )}{2 b} + \frac {A c - \frac {2 a \left (B d + C c\right )}{3 b}}{b}\right ) & \text {for}\: b \neq 0 \\\frac {\frac {A c x^{2}}{2} + \frac {C d x^{5}}{5} + \frac {x^{4} \left (B d + C c\right )}{4} + \frac {x^{3} \left (A d + B c\right )}{3}}{\sqrt {a}} & \text {otherwise} \end {cases} \] Input:

integrate(x*(d*x+c)*(C*x**2+B*x+A)/(b*x**2+a)**(1/2),x)
 

Output:

Piecewise((-a*(A*d + B*c - 3*C*a*d/(4*b))*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*b 
) + sqrt(a + b*x**2)*(C*d*x**3/(4*b) + x**2*(B*d + C*c)/(3*b) + x*(A*d + B 
*c - 3*C*a*d/(4*b))/(2*b) + (A*c - 2*a*(B*d + C*c)/(3*b))/b), Ne(b, 0)), ( 
(A*c*x**2/2 + C*d*x**5/5 + x**4*(B*d + C*c)/4 + x**3*(A*d + B*c)/3)/sqrt(a 
), True))
 

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.02 \[ \int \frac {x (c+d x) \left (A+B x+C x^2\right )}{\sqrt {a+b x^2}} \, dx=\frac {\sqrt {b x^{2} + a} C d x^{3}}{4 \, b} - \frac {3 \, \sqrt {b x^{2} + a} C a d x}{8 \, b^{2}} + \frac {\sqrt {b x^{2} + a} {\left (C c + B d\right )} x^{2}}{3 \, b} + \frac {3 \, C a^{2} d \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{8 \, b^{\frac {5}{2}}} + \frac {\sqrt {b x^{2} + a} A c}{b} + \frac {\sqrt {b x^{2} + a} {\left (B c + A d\right )} x}{2 \, b} - \frac {{\left (B c + A d\right )} a \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{2 \, b^{\frac {3}{2}}} - \frac {2 \, \sqrt {b x^{2} + a} {\left (C c + B d\right )} a}{3 \, b^{2}} \] Input:

integrate(x*(d*x+c)*(C*x^2+B*x+A)/(b*x^2+a)^(1/2),x, algorithm="maxima")
 

Output:

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

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.92 \[ \int \frac {x (c+d x) \left (A+B x+C x^2\right )}{\sqrt {a+b x^2}} \, dx=\frac {1}{24} \, \sqrt {b x^{2} + a} {\left ({\left (2 \, {\left (\frac {3 \, C d x}{b} + \frac {4 \, {\left (C b^{3} c + B b^{3} d\right )}}{b^{4}}\right )} x + \frac {3 \, {\left (4 \, B b^{3} c - 3 \, C a b^{2} d + 4 \, A b^{3} d\right )}}{b^{4}}\right )} x - \frac {8 \, {\left (2 \, C a b^{2} c - 3 \, A b^{3} c + 2 \, B a b^{2} d\right )}}{b^{4}}\right )} + \frac {{\left (4 \, B a b c - 3 \, C a^{2} d + 4 \, A a b d\right )} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{8 \, b^{\frac {5}{2}}} \] Input:

integrate(x*(d*x+c)*(C*x^2+B*x+A)/(b*x^2+a)^(1/2),x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

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

int((x*(c + d*x)*(A + B*x + C*x^2))/(a + b*x^2)^(1/2),x)
 

Output:

int((x*(c + d*x)*(A + B*x + C*x^2))/(a + b*x^2)^(1/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 235, normalized size of antiderivative = 1.48 \[ \int \frac {x (c+d x) \left (A+B x+C x^2\right )}{\sqrt {a+b x^2}} \, dx=\frac {24 \sqrt {b \,x^{2}+a}\, a \,b^{2} c +12 \sqrt {b \,x^{2}+a}\, a \,b^{2} d x -16 \sqrt {b \,x^{2}+a}\, a \,b^{2} d -16 \sqrt {b \,x^{2}+a}\, a b \,c^{2}-9 \sqrt {b \,x^{2}+a}\, a b c d x +12 \sqrt {b \,x^{2}+a}\, b^{3} c x +8 \sqrt {b \,x^{2}+a}\, b^{3} d \,x^{2}+8 \sqrt {b \,x^{2}+a}\, b^{2} c^{2} x^{2}+6 \sqrt {b \,x^{2}+a}\, b^{2} c d \,x^{3}-12 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{2} b d +9 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{2} c d -12 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a \,b^{2} c}{24 b^{3}} \] Input:

int(x*(d*x+c)*(C*x^2+B*x+A)/(b*x^2+a)^(1/2),x)
 

Output:

(24*sqrt(a + b*x**2)*a*b**2*c + 12*sqrt(a + b*x**2)*a*b**2*d*x - 16*sqrt(a 
 + b*x**2)*a*b**2*d - 16*sqrt(a + b*x**2)*a*b*c**2 - 9*sqrt(a + b*x**2)*a* 
b*c*d*x + 12*sqrt(a + b*x**2)*b**3*c*x + 8*sqrt(a + b*x**2)*b**3*d*x**2 + 
8*sqrt(a + b*x**2)*b**2*c**2*x**2 + 6*sqrt(a + b*x**2)*b**2*c*d*x**3 - 12* 
sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a**2*b*d + 9*sqrt(b)*l 
og((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a**2*c*d - 12*sqrt(b)*log((sqrt 
(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a*b**2*c)/(24*b**3)