\(\int \frac {x (c+d x) (A+B x+C x^2)}{(a+b x^2)^{3/2}} \, dx\) [134]

Optimal result

Integrand size = 28, antiderivative size = 150 \[ \int \frac {x (c+d x) \left (A+B x+C x^2\right )}{\left (a+b x^2\right )^{3/2}} \, dx=-\frac {A b c-a (c C+B d)}{b^2 \sqrt {a+b x^2}}-\frac {(b B c+A b d-a C d) x}{b^2 \sqrt {a+b x^2}}+\frac {(c C+B d) \sqrt {a+b x^2}}{b^2}+\frac {C d x \sqrt {a+b x^2}}{2 b^2}-\frac {(3 a C d-2 b (B c+A d)) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{2 b^{5/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.70 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.77 \[ \int \frac {x (c+d x) \left (A+B x+C x^2\right )}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {\frac {\sqrt {b} \left (-2 A b (c+d x)+a (4 c C+4 B d+3 C d x)+b x \left (-2 B c+2 c C x+2 B d x+C d x^2\right )\right )}{\sqrt {a+b x^2}}+(3 a C d-2 b (B c+A d)) \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{2 b^{5/2}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.74 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.91, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {2176, 2346, 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.

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

\(\Big \downarrow \) 2176

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

\(\Big \downarrow \) 2346

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 455

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

\(\Big \downarrow \) 224

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

-(((A - (a*C)/b + B*x)*(c + d*x))/(b*Sqrt[a + b*x^2])) - (-1/2*(a*C*d*x*Sq 
rt[a + b*x^2])/b + (a*(-2*(c*C + 2*B*d)*Sqrt[a + b*x^2] + ((3*a*C*d - 2*b* 
(B*c + A*d))*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/Sqrt[b]))/(2*b))/(a*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 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 2176

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] : 
> With[{Qx = PolynomialQuotient[Pq, a + b*x^2, x], R = Coeff[PolynomialRema 
inder[Pq, a + b*x^2, x], x, 0], S = Coeff[PolynomialRemainder[Pq, a + b*x^2 
, x], x, 1]}, Simp[(d + e*x)^m*(a + b*x^2)^(p + 1)*((a*S - b*R*x)/(2*a*b*(p 
 + 1))), x] + Simp[1/(2*a*b*(p + 1))   Int[(d + e*x)^(m - 1)*(a + b*x^2)^(p 
 + 1)*ExpandToSum[2*a*b*(p + 1)*(d + e*x)*Qx - a*e*S*m + b*d*R*(2*p + 3) + 
b*e*R*(m + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, b, d, e}, x] && PolyQ[Pq, x 
] && NeQ[b*d^2 + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 0] &&  !(IGtQ[m, 0] && R 
ationalQ[a, b, d, e] && (IntegerQ[p] || ILtQ[p + 1/2, 0]))
 

rule 2346

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], 
e = Coeff[Pq, x, Expon[Pq, x]]}, Simp[e*x^(q - 1)*((a + b*x^2)^(p + 1)/(b*( 
q + 2*p + 1))), x] + Simp[1/(b*(q + 2*p + 1))   Int[(a + b*x^2)^p*ExpandToS 
um[b*(q + 2*p + 1)*Pq - a*e*(q - 1)*x^(q - 2) - b*e*(q + 2*p + 1)*x^q, x], 
x], x]] /; FreeQ[{a, b, p}, x] && PolyQ[Pq, x] &&  !LeQ[p, -1]
 

Maple [A] (verified)

Time = 0.21 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.87

method	result	size
risch	\(\frac {\left (C x d +2 B d +2 C c \right ) \sqrt {b \,x^{2}+a}}{2 b^{2}}+\frac {b \left (2 A b d +2 B b c -3 a C d \right ) \left (-\frac {x}{b \sqrt {b \,x^{2}+a}}+\frac {\ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{b^{\frac {3}{2}}}\right )-\frac {2 \left (A b c -B a d -C a c \right )}{\sqrt {b \,x^{2}+a}}-\frac {C a d x}{\sqrt {b \,x^{2}+a}}}{2 b^{2}}\)	\(130\)
default	\(\left (A d +B c \right ) \left (-\frac {x}{b \sqrt {b \,x^{2}+a}}+\frac {\ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{b^{\frac {3}{2}}}\right )+\left (B d +C c \right ) \left (\frac {x^{2}}{b \sqrt {b \,x^{2}+a}}+\frac {2 a}{b^{2} \sqrt {b \,x^{2}+a}}\right )-\frac {A c}{b \sqrt {b \,x^{2}+a}}+d C \left (\frac {x^{3}}{2 b \sqrt {b \,x^{2}+a}}-\frac {3 a \left (-\frac {x}{b \sqrt {b \,x^{2}+a}}+\frac {\ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{b^{\frac {3}{2}}}\right )}{2 b}\right )\)	\(165\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 360, normalized size of antiderivative = 2.40 \[ \int \frac {x (c+d x) \left (A+B x+C x^2\right )}{\left (a+b x^2\right )^{3/2}} \, dx=\left [\frac {{\left (2 \, B a b c + {\left (2 \, B b^{2} c - {\left (3 \, C a b - 2 \, A b^{2}\right )} d\right )} x^{2} - {\left (3 \, C a^{2} - 2 \, A a b\right )} d\right )} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 2 \, {\left (C b^{2} d x^{3} + 4 \, B a b d + 2 \, {\left (C b^{2} c + B b^{2} d\right )} x^{2} + 2 \, {\left (2 \, C a b - A b^{2}\right )} c - {\left (2 \, B b^{2} c - {\left (3 \, C a b - 2 \, A b^{2}\right )} d\right )} x\right )} \sqrt {b x^{2} + a}}{4 \, {\left (b^{4} x^{2} + a b^{3}\right )}}, -\frac {{\left (2 \, B a b c + {\left (2 \, B b^{2} c - {\left (3 \, C a b - 2 \, A b^{2}\right )} d\right )} x^{2} - {\left (3 \, C a^{2} - 2 \, A a b\right )} d\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - {\left (C b^{2} d x^{3} + 4 \, B a b d + 2 \, {\left (C b^{2} c + B b^{2} d\right )} x^{2} + 2 \, {\left (2 \, C a b - A b^{2}\right )} c - {\left (2 \, B b^{2} c - {\left (3 \, C a b - 2 \, A b^{2}\right )} d\right )} x\right )} \sqrt {b x^{2} + a}}{2 \, {\left (b^{4} x^{2} + a b^{3}\right )}}\right ] \] Input:

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

Output:

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

Sympy [A] (verification not implemented)

Time = 7.60 (sec) , antiderivative size = 280, normalized size of antiderivative = 1.87 \[ \int \frac {x (c+d x) \left (A+B x+C x^2\right )}{\left (a+b x^2\right )^{3/2}} \, dx=A c \left (\begin {cases} - \frac {1}{b \sqrt {a + b x^{2}}} & \text {for}\: b \neq 0 \\\frac {x^{2}}{2 a^{\frac {3}{2}}} & \text {otherwise} \end {cases}\right ) + A d \left (\frac {\operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{b^{\frac {3}{2}}} - \frac {x}{\sqrt {a} b \sqrt {1 + \frac {b x^{2}}{a}}}\right ) + B c \left (\frac {\operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{b^{\frac {3}{2}}} - \frac {x}{\sqrt {a} b \sqrt {1 + \frac {b x^{2}}{a}}}\right ) + B d \left (\begin {cases} \frac {2 a}{b^{2} \sqrt {a + b x^{2}}} + \frac {x^{2}}{b \sqrt {a + b x^{2}}} & \text {for}\: b \neq 0 \\\frac {x^{4}}{4 a^{\frac {3}{2}}} & \text {otherwise} \end {cases}\right ) + C c \left (\begin {cases} \frac {2 a}{b^{2} \sqrt {a + b x^{2}}} + \frac {x^{2}}{b \sqrt {a + b x^{2}}} & \text {for}\: b \neq 0 \\\frac {x^{4}}{4 a^{\frac {3}{2}}} & \text {otherwise} \end {cases}\right ) + C d \left (\frac {3 \sqrt {a} x}{2 b^{2} \sqrt {1 + \frac {b x^{2}}{a}}} - \frac {3 a \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{2 b^{\frac {5}{2}}} + \frac {x^{3}}{2 \sqrt {a} b \sqrt {1 + \frac {b x^{2}}{a}}}\right ) \] Input:

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

Output:

A*c*Piecewise((-1/(b*sqrt(a + b*x**2)), Ne(b, 0)), (x**2/(2*a**(3/2)), Tru 
e)) + A*d*(asinh(sqrt(b)*x/sqrt(a))/b**(3/2) - x/(sqrt(a)*b*sqrt(1 + b*x** 
2/a))) + B*c*(asinh(sqrt(b)*x/sqrt(a))/b**(3/2) - x/(sqrt(a)*b*sqrt(1 + b* 
x**2/a))) + B*d*Piecewise((2*a/(b**2*sqrt(a + b*x**2)) + x**2/(b*sqrt(a + 
b*x**2)), Ne(b, 0)), (x**4/(4*a**(3/2)), True)) + C*c*Piecewise((2*a/(b**2 
*sqrt(a + b*x**2)) + x**2/(b*sqrt(a + b*x**2)), Ne(b, 0)), (x**4/(4*a**(3/ 
2)), True)) + C*d*(3*sqrt(a)*x/(2*b**2*sqrt(1 + b*x**2/a)) - 3*a*asinh(sqr 
t(b)*x/sqrt(a))/(2*b**(5/2)) + x**3/(2*sqrt(a)*b*sqrt(1 + b*x**2/a)))
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.05 \[ \int \frac {x (c+d x) \left (A+B x+C x^2\right )}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {C d x^{3}}{2 \, \sqrt {b x^{2} + a} b} + \frac {3 \, C a d x}{2 \, \sqrt {b x^{2} + a} b^{2}} + \frac {{\left (C c + B d\right )} x^{2}}{\sqrt {b x^{2} + a} b} - \frac {3 \, C a d \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{2 \, b^{\frac {5}{2}}} - \frac {A c}{\sqrt {b x^{2} + a} b} - \frac {{\left (B c + A d\right )} x}{\sqrt {b x^{2} + a} b} + \frac {{\left (B c + A d\right )} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{b^{\frac {3}{2}}} + \frac {2 \, {\left (C c + B d\right )} a}{\sqrt {b x^{2} + a} b^{2}} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.93 \[ \int \frac {x (c+d x) \left (A+B x+C x^2\right )}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {{\left ({\left (\frac {C d x}{b} + \frac {2 \, {\left (C b^{3} c + B b^{3} d\right )}}{b^{4}}\right )} x - \frac {2 \, B b^{3} c - 3 \, C a b^{2} d + 2 \, A b^{3} d}{b^{4}}\right )} x + \frac {2 \, {\left (2 \, C a b^{2} c - A b^{3} c + 2 \, B a b^{2} d\right )}}{b^{4}}}{2 \, \sqrt {b x^{2} + a}} - \frac {{\left (2 \, B b c - 3 \, C a d + 2 \, A b d\right )} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{2 \, b^{\frac {5}{2}}} \] Input:

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 395, normalized size of antiderivative = 2.63 \[ \int \frac {x (c+d x) \left (A+B x+C x^2\right )}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {-2 \sqrt {b \,x^{2}+a}\, a \,b^{2} c -2 \sqrt {b \,x^{2}+a}\, a \,b^{2} d x +4 \sqrt {b \,x^{2}+a}\, a \,b^{2} d +4 \sqrt {b \,x^{2}+a}\, a b \,c^{2}+3 \sqrt {b \,x^{2}+a}\, a b c d x -2 \sqrt {b \,x^{2}+a}\, b^{3} c x +2 \sqrt {b \,x^{2}+a}\, b^{3} d \,x^{2}+2 \sqrt {b \,x^{2}+a}\, b^{2} c^{2} x^{2}+\sqrt {b \,x^{2}+a}\, b^{2} c d \,x^{3}+2 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{2} b d -3 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{2} c d +2 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a \,b^{2} c +2 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a \,b^{2} d \,x^{2}-3 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a b c d \,x^{2}+2 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) b^{3} c \,x^{2}-2 \sqrt {b}\, a^{2} b d +2 \sqrt {b}\, a^{2} c d -2 \sqrt {b}\, a \,b^{2} c -2 \sqrt {b}\, a \,b^{2} d \,x^{2}+2 \sqrt {b}\, a b c d \,x^{2}-2 \sqrt {b}\, b^{3} c \,x^{2}}{2 b^{3} \left (b \,x^{2}+a \right )} \] Input:

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

Output:

( - 2*sqrt(a + b*x**2)*a*b**2*c - 2*sqrt(a + b*x**2)*a*b**2*d*x + 4*sqrt(a 
 + b*x**2)*a*b**2*d + 4*sqrt(a + b*x**2)*a*b*c**2 + 3*sqrt(a + b*x**2)*a*b 
*c*d*x - 2*sqrt(a + b*x**2)*b**3*c*x + 2*sqrt(a + b*x**2)*b**3*d*x**2 + 2* 
sqrt(a + b*x**2)*b**2*c**2*x**2 + sqrt(a + b*x**2)*b**2*c*d*x**3 + 2*sqrt( 
b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a**2*b*d - 3*sqrt(b)*log((s 
qrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a**2*c*d + 2*sqrt(b)*log((sqrt(a + b 
*x**2) + sqrt(b)*x)/sqrt(a))*a*b**2*c + 2*sqrt(b)*log((sqrt(a + b*x**2) + 
sqrt(b)*x)/sqrt(a))*a*b**2*d*x**2 - 3*sqrt(b)*log((sqrt(a + b*x**2) + sqrt 
(b)*x)/sqrt(a))*a*b*c*d*x**2 + 2*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x 
)/sqrt(a))*b**3*c*x**2 - 2*sqrt(b)*a**2*b*d + 2*sqrt(b)*a**2*c*d - 2*sqrt( 
b)*a*b**2*c - 2*sqrt(b)*a*b**2*d*x**2 + 2*sqrt(b)*a*b*c*d*x**2 - 2*sqrt(b) 
*b**3*c*x**2)/(2*b**3*(a + b*x**2))