\(\int \frac {(c+d x)^2 (A+B x+C x^2+D x^3)}{(a+b x^2)^{3/2}} \, dx\) [109]

Optimal result

Integrand size = 34, antiderivative size = 274 \[ \int \frac {(c+d x)^2 \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^{3/2}} \, dx=-\frac {b^2 c (B c+2 A d)+a^2 d^2 D-a b \left (2 c C d+B d^2+c^2 D\right )}{b^3 \sqrt {a+b x^2}}+\frac {\left (A b \left (b c^2-a d^2\right )-a (b c (c C+2 B d)-a d (C d+2 c D))\right ) x}{a b^2 \sqrt {a+b x^2}}-\frac {\left (2 a d^2 D-b \left (2 c C d+B d^2+c^2 D\right )\right ) \sqrt {a+b x^2}}{b^3}+\frac {d (C d+2 c D) x \sqrt {a+b x^2}}{2 b^2}+\frac {d^2 D \left (a+b x^2\right )^{3/2}}{3 b^3}+\frac {\left (2 b \left (c^2 C+2 B c d+A d^2\right )-3 a d (C d+2 c D)\right ) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{2 b^{5/2}} \] Output:

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

Mathematica [A] (verified)

Time = 1.33 (sec) , antiderivative size = 233, normalized size of antiderivative = 0.85 \[ \int \frac {(c+d x)^2 \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {-16 a^3 d^2 D+6 A b^3 c^2 x+a^2 b \left (12 c^2 D+6 c d (4 C+3 D x)+d^2 (12 B+x (9 C-8 D x))\right )+a b^2 \left (-6 A d (2 c+d x)-6 B \left (c^2+2 c d x-d^2 x^2\right )+x \left (-6 c^2 (C-D x)+6 c d x (2 C+D x)+d^2 x^2 (3 C+2 D x)\right )\right )+3 a \sqrt {b} \left (-2 b \left (c^2 C+2 B c d+A d^2\right )+3 a d (C d+2 c D)\right ) \sqrt {a+b x^2} \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{6 a b^3 \sqrt {a+b x^2}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.66 (sec) , antiderivative size = 242, normalized size of antiderivative = 0.88, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.206, Rules used = {2176, 25, 2185, 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.

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

\(\Big \downarrow \) 2176

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 2185

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 676

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

\(\Big \downarrow \) 224

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

-(((a*(B - (a*D)/b) - (A*b - a*C)*x)*(c + d*x)^2)/(a*b*Sqrt[a + b*x^2])) + 
 ((a*D*(c + d*x)^2*Sqrt[a + b*x^2])/(3*b) + ((-2*(3*A*b^2*c*d + a*(4*a*d^2 
*D - b*(6*c*C*d + 3*B*d^2 + c^2*D)))*Sqrt[a + b*x^2])/b - (d*(6*A*b*d - 9* 
a*C*d - 2*a*c*D)*x*Sqrt[a + b*x^2])/2 + (3*a*(2*b*(c^2*C + 2*B*c*d + A*d^2 
) - 3*a*d*(C*d + 2*c*D))*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(2*Sqrt[b]) 
)/(3*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 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 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 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 = 1.23 (sec) , antiderivative size = 276, normalized size of antiderivative = 1.01

method	result	size
default	\(\frac {A \,c^{2} x}{\sqrt {b \,x^{2}+a}\, a}-\frac {c \left (2 A d +B c \right )}{b \sqrt {b \,x^{2}+a}}+d \left (C d +2 D c \right ) \left (\frac {x^{3}}{2 b \sqrt {b \,x^{2}+a}}-\frac {3 a \left (-\frac {x}{b \sqrt {b \,x^{2}+a}}+\frac {\ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{b^{\frac {3}{2}}}\right )}{2 b}\right )+\left (A \,d^{2}+2 B c d +C \,c^{2}\right ) \left (-\frac {x}{b \sqrt {b \,x^{2}+a}}+\frac {\ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{b^{\frac {3}{2}}}\right )+\left (B \,d^{2}+2 C c d +D c^{2}\right ) \left (\frac {x^{2}}{b \sqrt {b \,x^{2}+a}}+\frac {2 a}{b^{2} \sqrt {b \,x^{2}+a}}\right )+D d^{2} \left (\frac {x^{4}}{3 b \sqrt {b \,x^{2}+a}}-\frac {4 a \left (\frac {x^{2}}{b \sqrt {b \,x^{2}+a}}+\frac {2 a}{b^{2} \sqrt {b \,x^{2}+a}}\right )}{3 b}\right )\)	\(276\)

Input:

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

Output:

A*c^2/(b*x^2+a)^(1/2)/a*x-c*(2*A*d+B*c)/b/(b*x^2+a)^(1/2)+d*(C*d+2*D*c)*(1 
/2*x^3/b/(b*x^2+a)^(1/2)-3/2*a/b*(-x/b/(b*x^2+a)^(1/2)+1/b^(3/2)*ln(b^(1/2 
)*x+(b*x^2+a)^(1/2))))+(A*d^2+2*B*c*d+C*c^2)*(-x/b/(b*x^2+a)^(1/2)+1/b^(3/ 
2)*ln(b^(1/2)*x+(b*x^2+a)^(1/2)))+(B*d^2+2*C*c*d+D*c^2)*(x^2/b/(b*x^2+a)^( 
1/2)+2*a/b^2/(b*x^2+a)^(1/2))+D*d^2*(1/3*x^4/b/(b*x^2+a)^(1/2)-4/3*a/b*(x^ 
2/b/(b*x^2+a)^(1/2)+2*a/b^2/(b*x^2+a)^(1/2)))
 

Fricas [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 728, normalized size of antiderivative = 2.66 \[ \int \frac {(c+d x)^2 \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^{3/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

[1/12*(3*(2*C*a^2*b*c^2 - 2*(3*D*a^3 - 2*B*a^2*b)*c*d - (3*C*a^3 - 2*A*a^2 
*b)*d^2 + (2*C*a*b^2*c^2 - 2*(3*D*a^2*b - 2*B*a*b^2)*c*d - (3*C*a^2*b - 2* 
A*a*b^2)*d^2)*x^2)*sqrt(b)*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) 
 + 2*(2*D*a*b^2*d^2*x^4 + 3*(2*D*a*b^2*c*d + C*a*b^2*d^2)*x^3 + 6*(2*D*a^2 
*b - B*a*b^2)*c^2 + 12*(2*C*a^2*b - A*a*b^2)*c*d - 4*(4*D*a^3 - 3*B*a^2*b) 
*d^2 + 2*(3*D*a*b^2*c^2 + 6*C*a*b^2*c*d - (4*D*a^2*b - 3*B*a*b^2)*d^2)*x^2 
 - 3*(2*(C*a*b^2 - A*b^3)*c^2 - 2*(3*D*a^2*b - 2*B*a*b^2)*c*d - (3*C*a^2*b 
 - 2*A*a*b^2)*d^2)*x)*sqrt(b*x^2 + a))/(a*b^4*x^2 + a^2*b^3), -1/6*(3*(2*C 
*a^2*b*c^2 - 2*(3*D*a^3 - 2*B*a^2*b)*c*d - (3*C*a^3 - 2*A*a^2*b)*d^2 + (2* 
C*a*b^2*c^2 - 2*(3*D*a^2*b - 2*B*a*b^2)*c*d - (3*C*a^2*b - 2*A*a*b^2)*d^2) 
*x^2)*sqrt(-b)*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) - (2*D*a*b^2*d^2*x^4 + 3 
*(2*D*a*b^2*c*d + C*a*b^2*d^2)*x^3 + 6*(2*D*a^2*b - B*a*b^2)*c^2 + 12*(2*C 
*a^2*b - A*a*b^2)*c*d - 4*(4*D*a^3 - 3*B*a^2*b)*d^2 + 2*(3*D*a*b^2*c^2 + 6 
*C*a*b^2*c*d - (4*D*a^2*b - 3*B*a*b^2)*d^2)*x^2 - 3*(2*(C*a*b^2 - A*b^3)*c 
^2 - 2*(3*D*a^2*b - 2*B*a*b^2)*c*d - (3*C*a^2*b - 2*A*a*b^2)*d^2)*x)*sqrt( 
b*x^2 + a))/(a*b^4*x^2 + a^2*b^3)]
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 322, normalized size of antiderivative = 1.18 \[ \int \frac {(c+d x)^2 \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {D d^{2} x^{4}}{3 \, \sqrt {b x^{2} + a} b} - \frac {4 \, D a d^{2} x^{2}}{3 \, \sqrt {b x^{2} + a} b^{2}} + \frac {A c^{2} x}{\sqrt {b x^{2} + a} a} + \frac {{\left (2 \, D c d + C d^{2}\right )} x^{3}}{2 \, \sqrt {b x^{2} + a} b} - \frac {B c^{2}}{\sqrt {b x^{2} + a} b} - \frac {2 \, A c d}{\sqrt {b x^{2} + a} b} - \frac {8 \, D a^{2} d^{2}}{3 \, \sqrt {b x^{2} + a} b^{3}} + \frac {{\left (D c^{2} + 2 \, C c d + B d^{2}\right )} x^{2}}{\sqrt {b x^{2} + a} b} + \frac {3 \, {\left (2 \, D c d + C d^{2}\right )} a x}{2 \, \sqrt {b x^{2} + a} b^{2}} - \frac {{\left (C c^{2} + 2 \, B c d + A d^{2}\right )} x}{\sqrt {b x^{2} + a} b} - \frac {3 \, {\left (2 \, D c d + C d^{2}\right )} a \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{2 \, b^{\frac {5}{2}}} + \frac {{\left (C c^{2} + 2 \, B c d + A d^{2}\right )} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{b^{\frac {3}{2}}} + \frac {2 \, {\left (D c^{2} + 2 \, C c d + B d^{2}\right )} a}{\sqrt {b x^{2} + a} b^{2}} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 311, normalized size of antiderivative = 1.14 \[ \int \frac {(c+d x)^2 \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {{\left ({\left ({\left (\frac {2 \, D d^{2} x}{b} + \frac {3 \, {\left (2 \, D a b^{4} c d + C a b^{4} d^{2}\right )}}{a b^{5}}\right )} x + \frac {2 \, {\left (3 \, D a b^{4} c^{2} + 6 \, C a b^{4} c d - 4 \, D a^{2} b^{3} d^{2} + 3 \, B a b^{4} d^{2}\right )}}{a b^{5}}\right )} x - \frac {3 \, {\left (2 \, C a b^{4} c^{2} - 2 \, A b^{5} c^{2} - 6 \, D a^{2} b^{3} c d + 4 \, B a b^{4} c d - 3 \, C a^{2} b^{3} d^{2} + 2 \, A a b^{4} d^{2}\right )}}{a b^{5}}\right )} x + \frac {2 \, {\left (6 \, D a^{2} b^{3} c^{2} - 3 \, B a b^{4} c^{2} + 12 \, C a^{2} b^{3} c d - 6 \, A a b^{4} c d - 8 \, D a^{3} b^{2} d^{2} + 6 \, B a^{2} b^{3} d^{2}\right )}}{a b^{5}}}{6 \, \sqrt {b x^{2} + a}} - \frac {{\left (2 \, C b c^{2} - 6 \, D a c d + 4 \, B b c d - 3 \, C a d^{2} + 2 \, A b d^{2}\right )} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{2 \, b^{\frac {5}{2}}} \] Input:

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

Output:

1/6*((((2*D*d^2*x/b + 3*(2*D*a*b^4*c*d + C*a*b^4*d^2)/(a*b^5))*x + 2*(3*D* 
a*b^4*c^2 + 6*C*a*b^4*c*d - 4*D*a^2*b^3*d^2 + 3*B*a*b^4*d^2)/(a*b^5))*x - 
3*(2*C*a*b^4*c^2 - 2*A*b^5*c^2 - 6*D*a^2*b^3*c*d + 4*B*a*b^4*c*d - 3*C*a^2 
*b^3*d^2 + 2*A*a*b^4*d^2)/(a*b^5))*x + 2*(6*D*a^2*b^3*c^2 - 3*B*a*b^4*c^2 
+ 12*C*a^2*b^3*c*d - 6*A*a*b^4*c*d - 8*D*a^3*b^2*d^2 + 6*B*a^2*b^3*d^2)/(a 
*b^5))/sqrt(b*x^2 + a) - 1/2*(2*C*b*c^2 - 6*D*a*c*d + 4*B*b*c*d - 3*C*a*d^ 
2 + 2*A*b*d^2)*log(abs(-sqrt(b)*x + sqrt(b*x^2 + a)))/b^(5/2)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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