\(\int \frac {x^3 (c+d x) (A+B x+C x^2)}{(a+b x^2)^{3/2}} \, dx\) [132]

Optimal result

Integrand size = 30, antiderivative size = 221 \[ \int \frac {x^3 (c+d x) \left (A+B x+C x^2\right )}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {a (A b c-a c C-a B d)}{b^3 \sqrt {a+b x^2}}+\frac {a (b B c+A b d-a C d) x}{b^3 \sqrt {a+b x^2}}+\frac {(A b c-2 a (c C+B d)) \sqrt {a+b x^2}}{b^3}-\frac {(7 a C d-4 b (B c+A d)) x \sqrt {a+b x^2}}{8 b^3}+\frac {C d x^3 \sqrt {a+b x^2}}{4 b^2}+\frac {(c C+B d) \left (a+b x^2\right )^{3/2}}{3 b^3}+\frac {3 a (5 a C d-4 b (B c+A d)) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{8 b^{7/2}} \] Output:

a*(A*b*c-B*a*d-C*a*c)/b^3/(b*x^2+a)^(1/2)+a*(A*b*d+B*b*c-C*a*d)*x/b^3/(b*x 
^2+a)^(1/2)+(A*b*c-2*a*(B*d+C*c))*(b*x^2+a)^(1/2)/b^3-1/8*(7*a*C*d-4*b*(A* 
d+B*c))*x*(b*x^2+a)^(1/2)/b^3+1/4*C*d*x^3*(b*x^2+a)^(1/2)/b^2+1/3*(B*d+C*c 
)*(b*x^2+a)^(3/2)/b^3+3/8*a*(5*a*C*d-4*b*(A*d+B*c))*arctanh(b^(1/2)*x/(b*x 
^2+a)^(1/2))/b^(7/2)
 

Mathematica [A] (verified)

Time = 0.96 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.77 \[ \int \frac {x^3 (c+d x) \left (A+B x+C x^2\right )}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {-a^2 (64 c C+64 B d+45 C d x)+a b \left (4 B x (9 c-8 d x)+12 A (4 c+3 d x)-C x^2 (32 c+15 d x)\right )+2 b^2 x^2 (6 A (2 c+d x)+x (C x (4 c+3 d x)+B (6 c+4 d x)))}{24 b^3 \sqrt {a+b x^2}}-\frac {3 a (5 a C d-4 b (B c+A d)) \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{8 b^{7/2}} \] Input:

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

Output:

(-(a^2*(64*c*C + 64*B*d + 45*C*d*x)) + a*b*(4*B*x*(9*c - 8*d*x) + 12*A*(4* 
c + 3*d*x) - C*x^2*(32*c + 15*d*x)) + 2*b^2*x^2*(6*A*(2*c + d*x) + x*(C*x* 
(4*c + 3*d*x) + B*(6*c + 4*d*x))))/(24*b^3*Sqrt[a + b*x^2]) - (3*a*(5*a*C* 
d - 4*b*(B*c + A*d))*Log[-(Sqrt[b]*x) + Sqrt[a + b*x^2]])/(8*b^(7/2))
 

Rubi [A] (verified)

Time = 1.63 (sec) , antiderivative size = 220, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {2176, 2346, 2346, 2346, 25, 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.

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

\(\Big \downarrow \) 2176

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

\(\Big \downarrow \) 2346

\(\displaystyle \frac {a (c+d x) (-a C+A b+b B x)}{b^3 \sqrt {a+b x^2}}-\frac {\frac {\int \frac {-4 a b (c C+B d) x^3+a (7 a C d-4 b (B c+A d)) x^2-4 a (A b c-a (c C+2 B d)) x+\frac {4 a^2 (b B c+A b d-a C d)}{b}}{\sqrt {b x^2+a}}dx}{4 b}-\frac {a C d x^3 \sqrt {a+b x^2}}{4 b}}{a b}\)

\(\Big \downarrow \) 2346

\(\displaystyle \frac {a (c+d x) (-a C+A b+b B x)}{b^3 \sqrt {a+b x^2}}-\frac {\frac {\frac {\int \frac {12 (b B c+A b d-a C d) a^2+3 b (7 a C d-4 b (B c+A d)) x^2 a-4 b (3 A b c-5 a C c-8 a B d) x a}{\sqrt {b x^2+a}}dx}{3 b}-\frac {4}{3} a x^2 \sqrt {a+b x^2} (B d+c C)}{4 b}-\frac {a C d x^3 \sqrt {a+b x^2}}{4 b}}{a b}\)

\(\Big \downarrow \) 2346

\(\displaystyle \frac {a (c+d x) (-a C+A b+b B x)}{b^3 \sqrt {a+b x^2}}-\frac {\frac {\frac {\frac {\int -\frac {a b (9 a (5 a C d-4 b (B c+A d))+8 b (3 A b c-5 a C c-8 a B d) x)}{\sqrt {b x^2+a}}dx}{2 b}+\frac {3}{2} a x \sqrt {a+b x^2} (7 a C d-4 b (A d+B c))}{3 b}-\frac {4}{3} a x^2 \sqrt {a+b x^2} (B d+c C)}{4 b}-\frac {a C d x^3 \sqrt {a+b x^2}}{4 b}}{a b}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {a (c+d x) (-a C+A b+b B x)}{b^3 \sqrt {a+b x^2}}-\frac {\frac {\frac {\frac {3}{2} a x \sqrt {a+b x^2} (7 a C d-4 b (A d+B c))-\frac {\int \frac {a b (9 a (5 a C d-4 b (B c+A d))+8 b (3 A b c-5 a C c-8 a B d) x)}{\sqrt {b x^2+a}}dx}{2 b}}{3 b}-\frac {4}{3} a x^2 \sqrt {a+b x^2} (B d+c C)}{4 b}-\frac {a C d x^3 \sqrt {a+b x^2}}{4 b}}{a b}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {a (c+d x) (-a C+A b+b B x)}{b^3 \sqrt {a+b x^2}}-\frac {\frac {\frac {\frac {3}{2} a x \sqrt {a+b x^2} (7 a C d-4 b (A d+B c))-\frac {1}{2} a \int \frac {9 a (5 a C d-4 b (B c+A d))+8 b (3 A b c-5 a C c-8 a B d) x}{\sqrt {b x^2+a}}dx}{3 b}-\frac {4}{3} a x^2 \sqrt {a+b x^2} (B d+c C)}{4 b}-\frac {a C d x^3 \sqrt {a+b x^2}}{4 b}}{a b}\)

\(\Big \downarrow \) 455

\(\displaystyle \frac {a (c+d x) (-a C+A b+b B x)}{b^3 \sqrt {a+b x^2}}-\frac {\frac {\frac {\frac {3}{2} a x \sqrt {a+b x^2} (7 a C d-4 b (A d+B c))-\frac {1}{2} a \left (9 a (5 a C d-4 b (A d+B c)) \int \frac {1}{\sqrt {b x^2+a}}dx+8 \sqrt {a+b x^2} (-8 a B d-5 a c C+3 A b c)\right )}{3 b}-\frac {4}{3} a x^2 \sqrt {a+b x^2} (B d+c C)}{4 b}-\frac {a C d x^3 \sqrt {a+b x^2}}{4 b}}{a b}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {a (c+d x) (-a C+A b+b B x)}{b^3 \sqrt {a+b x^2}}-\frac {\frac {\frac {\frac {3}{2} a x \sqrt {a+b x^2} (7 a C d-4 b (A d+B c))-\frac {1}{2} a \left (9 a (5 a C d-4 b (A d+B c)) \int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}+8 \sqrt {a+b x^2} (-8 a B d-5 a c C+3 A b c)\right )}{3 b}-\frac {4}{3} a x^2 \sqrt {a+b x^2} (B d+c C)}{4 b}-\frac {a C d x^3 \sqrt {a+b x^2}}{4 b}}{a b}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {a (c+d x) (-a C+A b+b B x)}{b^3 \sqrt {a+b x^2}}-\frac {\frac {\frac {\frac {3}{2} a x \sqrt {a+b x^2} (7 a C d-4 b (A d+B c))-\frac {1}{2} a \left (\frac {9 a \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right ) (5 a C d-4 b (A d+B c))}{\sqrt {b}}+8 \sqrt {a+b x^2} (-8 a B d-5 a c C+3 A b c)\right )}{3 b}-\frac {4}{3} a x^2 \sqrt {a+b x^2} (B d+c C)}{4 b}-\frac {a C d x^3 \sqrt {a+b x^2}}{4 b}}{a b}\)

Input:

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

Output:

(a*(A*b - a*C + b*B*x)*(c + d*x))/(b^3*Sqrt[a + b*x^2]) - (-1/4*(a*C*d*x^3 
*Sqrt[a + b*x^2])/b + ((-4*a*(c*C + B*d)*x^2*Sqrt[a + b*x^2])/3 + ((3*a*(7 
*a*C*d - 4*b*(B*c + A*d))*x*Sqrt[a + b*x^2])/2 - (a*(8*(3*A*b*c - 5*a*c*C 
- 8*a*B*d)*Sqrt[a + b*x^2] + (9*a*(5*a*C*d - 4*b*(B*c + A*d))*ArcTanh[(Sqr 
t[b]*x)/Sqrt[a + b*x^2]])/Sqrt[b]))/2)/(3*b))/(4*b))/(a*b)
 

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 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 = 0.30 (sec) , antiderivative size = 207, normalized size of antiderivative = 0.94

method	result	size
risch	\(\frac {\left (6 C b d \,x^{3}+8 B b d \,x^{2}+8 C c b \,x^{2}+12 x A b d +12 B b c x -21 C a d x +24 A b c -40 B a d -40 C a c \right ) \sqrt {b \,x^{2}+a}}{24 b^{3}}-\frac {a \left (3 b \left (4 A b d +4 B b c -5 a C d \right ) \left (-\frac {x}{b \sqrt {b \,x^{2}+a}}+\frac {\ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{b^{\frac {3}{2}}}\right )-\frac {8 \left (A b c -B a d -C a c \right )}{\sqrt {b \,x^{2}+a}}-\frac {7 C a d x}{\sqrt {b \,x^{2}+a}}+\frac {4 A b d x}{\sqrt {b \,x^{2}+a}}+\frac {4 B b c x}{\sqrt {b \,x^{2}+a}}\right )}{8 b^{3}}\)	\(207\)
default	\(\left (A d +B c \right ) \left (\frac {x^{3}}{2 b \sqrt {b \,x^{2}+a}}-\frac {3 a \left (-\frac {x}{b \sqrt {b \,x^{2}+a}}+\frac {\ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{b^{\frac {3}{2}}}\right )}{2 b}\right )+\left (B d +C c \right ) \left (\frac {x^{4}}{3 b \sqrt {b \,x^{2}+a}}-\frac {4 a \left (\frac {x^{2}}{b \sqrt {b \,x^{2}+a}}+\frac {2 a}{b^{2} \sqrt {b \,x^{2}+a}}\right )}{3 b}\right )+A c \left (\frac {x^{2}}{b \sqrt {b \,x^{2}+a}}+\frac {2 a}{b^{2} \sqrt {b \,x^{2}+a}}\right )+d C \left (\frac {x^{5}}{4 b \sqrt {b \,x^{2}+a}}-\frac {5 a \left (\frac {x^{3}}{2 b \sqrt {b \,x^{2}+a}}-\frac {3 a \left (-\frac {x}{b \sqrt {b \,x^{2}+a}}+\frac {\ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{b^{\frac {3}{2}}}\right )}{2 b}\right )}{4 b}\right )\)	\(256\)

Input:

int(x^3*(d*x+c)*(C*x^2+B*x+A)/(b*x^2+a)^(3/2),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-21*C*a*d*x 
+24*A*b*c-40*B*a*d-40*C*a*c)*(b*x^2+a)^(1/2)/b^3-1/8*a/b^3*(3*b*(4*A*b*d+4 
*B*b*c-5*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)))-8*(A*b*c-B*a*d-C*a*c)/(b*x^2+a)^(1/2)-7*C*a*d*x/(b*x^2+a)^(1/2)+4*A*b 
*d*x/(b*x^2+a)^(1/2)+4*B*b*c*x/(b*x^2+a)^(1/2))
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 517, normalized size of antiderivative = 2.34 \[ \int \frac {x^3 (c+d x) \left (A+B x+C x^2\right )}{\left (a+b x^2\right )^{3/2}} \, dx=\left [\frac {9 \, {\left (4 \, B a^{2} b c + {\left (4 \, B a b^{2} c - {\left (5 \, C a^{2} b - 4 \, A a b^{2}\right )} d\right )} x^{2} - {\left (5 \, C a^{3} - 4 \, A a^{2} b\right )} d\right )} \sqrt {b} \log \left (-2 \, b x^{2} + 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 2 \, {\left (6 \, C b^{3} d x^{5} - 64 \, B a^{2} b d + 8 \, {\left (C b^{3} c + B b^{3} d\right )} x^{4} + 3 \, {\left (4 \, B b^{3} c - {\left (5 \, C a b^{2} - 4 \, A b^{3}\right )} d\right )} x^{3} - 8 \, {\left (4 \, B a b^{2} d + {\left (4 \, C a b^{2} - 3 \, A b^{3}\right )} c\right )} x^{2} - 16 \, {\left (4 \, C a^{2} b - 3 \, A a b^{2}\right )} c + 9 \, {\left (4 \, B a b^{2} c - {\left (5 \, C a^{2} b - 4 \, A a b^{2}\right )} d\right )} x\right )} \sqrt {b x^{2} + a}}{48 \, {\left (b^{5} x^{2} + a b^{4}\right )}}, \frac {9 \, {\left (4 \, B a^{2} b c + {\left (4 \, B a b^{2} c - {\left (5 \, C a^{2} b - 4 \, A a b^{2}\right )} d\right )} x^{2} - {\left (5 \, C a^{3} - 4 \, A a^{2} b\right )} d\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) + {\left (6 \, C b^{3} d x^{5} - 64 \, B a^{2} b d + 8 \, {\left (C b^{3} c + B b^{3} d\right )} x^{4} + 3 \, {\left (4 \, B b^{3} c - {\left (5 \, C a b^{2} - 4 \, A b^{3}\right )} d\right )} x^{3} - 8 \, {\left (4 \, B a b^{2} d + {\left (4 \, C a b^{2} - 3 \, A b^{3}\right )} c\right )} x^{2} - 16 \, {\left (4 \, C a^{2} b - 3 \, A a b^{2}\right )} c + 9 \, {\left (4 \, B a b^{2} c - {\left (5 \, C a^{2} b - 4 \, A a b^{2}\right )} d\right )} x\right )} \sqrt {b x^{2} + a}}{24 \, {\left (b^{5} x^{2} + a b^{4}\right )}}\right ] \] Input:

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

Output:

[1/48*(9*(4*B*a^2*b*c + (4*B*a*b^2*c - (5*C*a^2*b - 4*A*a*b^2)*d)*x^2 - (5 
*C*a^3 - 4*A*a^2*b)*d)*sqrt(b)*log(-2*b*x^2 + 2*sqrt(b*x^2 + a)*sqrt(b)*x 
- a) + 2*(6*C*b^3*d*x^5 - 64*B*a^2*b*d + 8*(C*b^3*c + B*b^3*d)*x^4 + 3*(4* 
B*b^3*c - (5*C*a*b^2 - 4*A*b^3)*d)*x^3 - 8*(4*B*a*b^2*d + (4*C*a*b^2 - 3*A 
*b^3)*c)*x^2 - 16*(4*C*a^2*b - 3*A*a*b^2)*c + 9*(4*B*a*b^2*c - (5*C*a^2*b 
- 4*A*a*b^2)*d)*x)*sqrt(b*x^2 + a))/(b^5*x^2 + a*b^4), 1/24*(9*(4*B*a^2*b* 
c + (4*B*a*b^2*c - (5*C*a^2*b - 4*A*a*b^2)*d)*x^2 - (5*C*a^3 - 4*A*a^2*b)* 
d)*sqrt(-b)*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) + (6*C*b^3*d*x^5 - 64*B*a^2 
*b*d + 8*(C*b^3*c + B*b^3*d)*x^4 + 3*(4*B*b^3*c - (5*C*a*b^2 - 4*A*b^3)*d) 
*x^3 - 8*(4*B*a*b^2*d + (4*C*a*b^2 - 3*A*b^3)*c)*x^2 - 16*(4*C*a^2*b - 3*A 
*a*b^2)*c + 9*(4*B*a*b^2*c - (5*C*a^2*b - 4*A*a*b^2)*d)*x)*sqrt(b*x^2 + a) 
)/(b^5*x^2 + a*b^4)]
 

Sympy [A] (verification not implemented)

Time = 12.03 (sec) , antiderivative size = 449, normalized size of antiderivative = 2.03 \[ \int \frac {x^3 (c+d x) \left (A+B x+C x^2\right )}{\left (a+b x^2\right )^{3/2}} \, dx=A c \left (\begin {cases} \frac {2 a}{b^{2} \sqrt {a + b x^{2}}} + \frac {x^{2}}{b \sqrt {a + b x^{2}}} & \text {for}\: b \neq 0 \\\frac {x^{4}}{4 a^{\frac {3}{2}}} & \text {otherwise} \end {cases}\right ) + A d \left (\frac {3 \sqrt {a} x}{2 b^{2} \sqrt {1 + \frac {b x^{2}}{a}}} - \frac {3 a \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{2 b^{\frac {5}{2}}} + \frac {x^{3}}{2 \sqrt {a} b \sqrt {1 + \frac {b x^{2}}{a}}}\right ) + B c \left (\frac {3 \sqrt {a} x}{2 b^{2} \sqrt {1 + \frac {b x^{2}}{a}}} - \frac {3 a \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{2 b^{\frac {5}{2}}} + \frac {x^{3}}{2 \sqrt {a} b \sqrt {1 + \frac {b x^{2}}{a}}}\right ) + B d \left (\begin {cases} - \frac {8 a^{2}}{3 b^{3} \sqrt {a + b x^{2}}} - \frac {4 a x^{2}}{3 b^{2} \sqrt {a + b x^{2}}} + \frac {x^{4}}{3 b \sqrt {a + b x^{2}}} & \text {for}\: b \neq 0 \\\frac {x^{6}}{6 a^{\frac {3}{2}}} & \text {otherwise} \end {cases}\right ) + C c \left (\begin {cases} - \frac {8 a^{2}}{3 b^{3} \sqrt {a + b x^{2}}} - \frac {4 a x^{2}}{3 b^{2} \sqrt {a + b x^{2}}} + \frac {x^{4}}{3 b \sqrt {a + b x^{2}}} & \text {for}\: b \neq 0 \\\frac {x^{6}}{6 a^{\frac {3}{2}}} & \text {otherwise} \end {cases}\right ) + C d \left (- \frac {15 a^{\frac {3}{2}} x}{8 b^{3} \sqrt {1 + \frac {b x^{2}}{a}}} - \frac {5 \sqrt {a} x^{3}}{8 b^{2} \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {15 a^{2} \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{8 b^{\frac {7}{2}}} + \frac {x^{5}}{4 \sqrt {a} b \sqrt {1 + \frac {b x^{2}}{a}}}\right ) \] Input:

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

Output:

A*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)) + A*d*(3*sqrt(a)*x/(2*b**2*sqrt(1 + b* 
x**2/a)) - 3*a*asinh(sqrt(b)*x/sqrt(a))/(2*b**(5/2)) + x**3/(2*sqrt(a)*b*s 
qrt(1 + b*x**2/a))) + B*c*(3*sqrt(a)*x/(2*b**2*sqrt(1 + b*x**2/a)) - 3*a*a 
sinh(sqrt(b)*x/sqrt(a))/(2*b**(5/2)) + x**3/(2*sqrt(a)*b*sqrt(1 + b*x**2/a 
))) + B*d*Piecewise((-8*a**2/(3*b**3*sqrt(a + b*x**2)) - 4*a*x**2/(3*b**2* 
sqrt(a + b*x**2)) + x**4/(3*b*sqrt(a + b*x**2)), Ne(b, 0)), (x**6/(6*a**(3 
/2)), True)) + C*c*Piecewise((-8*a**2/(3*b**3*sqrt(a + b*x**2)) - 4*a*x**2 
/(3*b**2*sqrt(a + b*x**2)) + x**4/(3*b*sqrt(a + b*x**2)), Ne(b, 0)), (x**6 
/(6*a**(3/2)), True)) + C*d*(-15*a**(3/2)*x/(8*b**3*sqrt(1 + b*x**2/a)) - 
5*sqrt(a)*x**3/(8*b**2*sqrt(1 + b*x**2/a)) + 15*a**2*asinh(sqrt(b)*x/sqrt( 
a))/(8*b**(7/2)) + x**5/(4*sqrt(a)*b*sqrt(1 + b*x**2/a)))
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 256, normalized size of antiderivative = 1.16 \[ \int \frac {x^3 (c+d x) \left (A+B x+C x^2\right )}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {C d x^{5}}{4 \, \sqrt {b x^{2} + a} b} - \frac {5 \, C a d x^{3}}{8 \, \sqrt {b x^{2} + a} b^{2}} + \frac {{\left (C c + B d\right )} x^{4}}{3 \, \sqrt {b x^{2} + a} b} + \frac {A c x^{2}}{\sqrt {b x^{2} + a} b} + \frac {{\left (B c + A d\right )} x^{3}}{2 \, \sqrt {b x^{2} + a} b} - \frac {15 \, C a^{2} d x}{8 \, \sqrt {b x^{2} + a} b^{3}} - \frac {4 \, {\left (C c + B d\right )} a x^{2}}{3 \, \sqrt {b x^{2} + a} b^{2}} + \frac {15 \, C a^{2} d \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{8 \, b^{\frac {7}{2}}} + \frac {2 \, A a c}{\sqrt {b x^{2} + a} b^{2}} + \frac {3 \, {\left (B c + A d\right )} a x}{2 \, \sqrt {b x^{2} + a} b^{2}} - \frac {3 \, {\left (B c + A d\right )} a \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{2 \, b^{\frac {5}{2}}} - \frac {8 \, {\left (C c + B d\right )} a^{2}}{3 \, \sqrt {b x^{2} + a} b^{3}} \] Input:

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

Output:

1/4*C*d*x^5/(sqrt(b*x^2 + a)*b) - 5/8*C*a*d*x^3/(sqrt(b*x^2 + a)*b^2) + 1/ 
3*(C*c + B*d)*x^4/(sqrt(b*x^2 + a)*b) + A*c*x^2/(sqrt(b*x^2 + a)*b) + 1/2* 
(B*c + A*d)*x^3/(sqrt(b*x^2 + a)*b) - 15/8*C*a^2*d*x/(sqrt(b*x^2 + a)*b^3) 
 - 4/3*(C*c + B*d)*a*x^2/(sqrt(b*x^2 + a)*b^2) + 15/8*C*a^2*d*arcsinh(b*x/ 
sqrt(a*b))/b^(7/2) + 2*A*a*c/(sqrt(b*x^2 + a)*b^2) + 3/2*(B*c + A*d)*a*x/( 
sqrt(b*x^2 + a)*b^2) - 3/2*(B*c + A*d)*a*arcsinh(b*x/sqrt(a*b))/b^(5/2) - 
8/3*(C*c + B*d)*a^2/(sqrt(b*x^2 + a)*b^3)
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 218, normalized size of antiderivative = 0.99 \[ \int \frac {x^3 (c+d x) \left (A+B x+C x^2\right )}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {{\left ({\left ({\left (2 \, {\left (\frac {3 \, C d x}{b} + \frac {4 \, {\left (C b^{5} c + B b^{5} d\right )}}{b^{6}}\right )} x + \frac {3 \, {\left (4 \, B b^{5} c - 5 \, C a b^{4} d + 4 \, A b^{5} d\right )}}{b^{6}}\right )} x - \frac {8 \, {\left (4 \, C a b^{4} c - 3 \, A b^{5} c + 4 \, B a b^{4} d\right )}}{b^{6}}\right )} x + \frac {9 \, {\left (4 \, B a b^{4} c - 5 \, C a^{2} b^{3} d + 4 \, A a b^{4} d\right )}}{b^{6}}\right )} x - \frac {16 \, {\left (4 \, C a^{2} b^{3} c - 3 \, A a b^{4} c + 4 \, B a^{2} b^{3} d\right )}}{b^{6}}}{24 \, \sqrt {b x^{2} + a}} + \frac {3 \, {\left (4 \, B a b c - 5 \, C a^{2} d + 4 \, A a b d\right )} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{8 \, b^{\frac {7}{2}}} \] Input:

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

Output:

1/24*((((2*(3*C*d*x/b + 4*(C*b^5*c + B*b^5*d)/b^6)*x + 3*(4*B*b^5*c - 5*C* 
a*b^4*d + 4*A*b^5*d)/b^6)*x - 8*(4*C*a*b^4*c - 3*A*b^5*c + 4*B*a*b^4*d)/b^ 
6)*x + 9*(4*B*a*b^4*c - 5*C*a^2*b^3*d + 4*A*a*b^4*d)/b^6)*x - 16*(4*C*a^2* 
b^3*c - 3*A*a*b^4*c + 4*B*a^2*b^3*d)/b^6)/sqrt(b*x^2 + a) + 3/8*(4*B*a*b*c 
 - 5*C*a^2*d + 4*A*a*b*d)*log(abs(-sqrt(b)*x + sqrt(b*x^2 + a)))/b^(7/2)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 531, normalized size of antiderivative = 2.40 \[ \int \frac {x^3 (c+d x) \left (A+B x+C x^2\right )}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {27 \sqrt {b}\, a^{3} b d -30 \sqrt {b}\, a^{3} c d +27 \sqrt {b}\, a^{2} b^{2} c +27 \sqrt {b}\, a^{2} b^{2} d \,x^{2}+27 \sqrt {b}\, a \,b^{3} c \,x^{2}+45 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{2} b c d \,x^{2}-36 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{2} b^{2} d \,x^{2}-36 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a \,b^{3} c \,x^{2}-30 \sqrt {b}\, a^{2} b c d \,x^{2}+36 \sqrt {b \,x^{2}+a}\, a^{2} b^{2} d x +24 \sqrt {b \,x^{2}+a}\, a \,b^{3} c \,x^{2}+12 \sqrt {b \,x^{2}+a}\, a \,b^{3} d \,x^{3}-36 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{3} b d -64 \sqrt {b \,x^{2}+a}\, a^{2} b^{2} d -64 \sqrt {b \,x^{2}+a}\, a^{2} b \,c^{2}+12 \sqrt {b \,x^{2}+a}\, b^{4} c \,x^{3}+8 \sqrt {b \,x^{2}+a}\, b^{4} d \,x^{4}+8 \sqrt {b \,x^{2}+a}\, b^{3} c^{2} x^{4}+48 \sqrt {b \,x^{2}+a}\, a^{2} b^{2} c +36 \sqrt {b \,x^{2}+a}\, a \,b^{3} c x -32 \sqrt {b \,x^{2}+a}\, a \,b^{3} d \,x^{2}-32 \sqrt {b \,x^{2}+a}\, a \,b^{2} c^{2} x^{2}+6 \sqrt {b \,x^{2}+a}\, b^{3} c d \,x^{5}+45 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{3} c d -36 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{2} b^{2} c -45 \sqrt {b \,x^{2}+a}\, a^{2} b c d x -15 \sqrt {b \,x^{2}+a}\, a \,b^{2} c d \,x^{3}}{24 b^{4} \left (b \,x^{2}+a \right )} \] Input:

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

Output:

(48*sqrt(a + b*x**2)*a**2*b**2*c + 36*sqrt(a + b*x**2)*a**2*b**2*d*x - 64* 
sqrt(a + b*x**2)*a**2*b**2*d - 64*sqrt(a + b*x**2)*a**2*b*c**2 - 45*sqrt(a 
 + b*x**2)*a**2*b*c*d*x + 24*sqrt(a + b*x**2)*a*b**3*c*x**2 + 36*sqrt(a + 
b*x**2)*a*b**3*c*x + 12*sqrt(a + b*x**2)*a*b**3*d*x**3 - 32*sqrt(a + b*x** 
2)*a*b**3*d*x**2 - 32*sqrt(a + b*x**2)*a*b**2*c**2*x**2 - 15*sqrt(a + b*x* 
*2)*a*b**2*c*d*x**3 + 12*sqrt(a + b*x**2)*b**4*c*x**3 + 8*sqrt(a + b*x**2) 
*b**4*d*x**4 + 8*sqrt(a + b*x**2)*b**3*c**2*x**4 + 6*sqrt(a + b*x**2)*b**3 
*c*d*x**5 - 36*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a**3*b* 
d + 45*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a**3*c*d - 36*s 
qrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a**2*b**2*c - 36*sqrt(b 
)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a**2*b**2*d*x**2 + 45*sqrt(b 
)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a**2*b*c*d*x**2 - 36*sqrt(b) 
*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a*b**3*c*x**2 + 27*sqrt(b)*a* 
*3*b*d - 30*sqrt(b)*a**3*c*d + 27*sqrt(b)*a**2*b**2*c + 27*sqrt(b)*a**2*b* 
*2*d*x**2 - 30*sqrt(b)*a**2*b*c*d*x**2 + 27*sqrt(b)*a*b**3*c*x**2)/(24*b** 
4*(a + b*x**2))