\(\int \frac {(c+d x)^2 (A+B x+C x^2+D x^3)}{\sqrt {a+b x^2}} \, dx\) [102]

Optimal result

Integrand size = 34, antiderivative size = 280 \[ \int \frac {(c+d x)^2 \left (A+B x+C x^2+D x^3\right )}{\sqrt {a+b x^2}} \, dx=\frac {\left (b^2 c (B c+2 A d)+a^2 d^2 D-a b \left (2 c C d+B d^2+c^2 D\right )\right ) \sqrt {a+b x^2}}{b^3}+\frac {\left (4 b \left (c^2 C+2 B c d+A d^2\right )-3 a d (C d+2 c D)\right ) x \sqrt {a+b x^2}}{8 b^2}+\frac {d (C d+2 c D) x^3 \sqrt {a+b x^2}}{4 b}-\frac {\left (2 a d^2 D-b \left (2 c C d+B d^2+c^2 D\right )\right ) \left (a+b x^2\right )^{3/2}}{3 b^3}+\frac {d^2 D \left (a+b x^2\right )^{5/2}}{5 b^3}+\frac {\left (4 A b \left (2 b c^2-a d^2\right )-a (4 b c (c C+2 B d)-3 a d (C d+2 c D))\right ) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{8 b^{5/2}} \] 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*x^2+a)^(1/2)/b^ 
3+1/8*(4*b*(A*d^2+2*B*c*d+C*c^2)-3*a*d*(C*d+2*D*c))*x*(b*x^2+a)^(1/2)/b^2+ 
1/4*d*(C*d+2*D*c)*x^3*(b*x^2+a)^(1/2)/b-1/3*(2*a*d^2*D-b*(B*d^2+2*C*c*d+D* 
c^2))*(b*x^2+a)^(3/2)/b^3+1/5*d^2*D*(b*x^2+a)^(5/2)/b^3+1/8*(4*A*b*(-a*d^2 
+2*b*c^2)-a*(4*b*c*(2*B*d+C*c)-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 = 1.30 (sec) , antiderivative size = 226, normalized size of antiderivative = 0.81 \[ \int \frac {(c+d x)^2 \left (A+B x+C x^2+D x^3\right )}{\sqrt {a+b x^2}} \, dx=\frac {\sqrt {a+b x^2} \left (64 a^2 d^2 D+2 b^2 \left (30 A d (4 c+d x)+20 B \left (3 c^2+3 c d x+d^2 x^2\right )+x \left (10 c^2 (3 C+2 D x)+10 c d x (4 C+3 D x)+3 d^2 x^2 (5 C+4 D x)\right )\right )-a b \left (80 c^2 D+10 c d (16 C+9 D x)+d^2 (80 B+x (45 C+32 D x))\right )\right )+15 \sqrt {b} \left (4 A b \left (-2 b c^2+a d^2\right )+a (4 b c (c C+2 B d)-3 a d (C d+2 c D))\right ) \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{120 b^3} \] Input:

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

Output:

(Sqrt[a + b*x^2]*(64*a^2*d^2*D + 2*b^2*(30*A*d*(4*c + d*x) + 20*B*(3*c^2 + 
 3*c*d*x + d^2*x^2) + x*(10*c^2*(3*C + 2*D*x) + 10*c*d*x*(4*C + 3*D*x) + 3 
*d^2*x^2*(5*C + 4*D*x))) - a*b*(80*c^2*D + 10*c*d*(16*C + 9*D*x) + d^2*(80 
*B + x*(45*C + 32*D*x)))) + 15*Sqrt[b]*(4*A*b*(-2*b*c^2 + a*d^2) + a*(4*b* 
c*(c*C + 2*B*d) - 3*a*d*(C*d + 2*c*D)))*Log[-(Sqrt[b]*x) + Sqrt[a + b*x^2] 
])/(120*b^3)
 

Rubi [A] (verified)

Time = 0.90 (sec) , antiderivative size = 363, normalized size of antiderivative = 1.30, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.206, Rules used = {2185, 2185, 27, 687, 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.

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

\(\Big \downarrow \) 2185

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

\(\Big \downarrow \) 2185

\(\displaystyle \frac {\frac {\int \frac {b d^3 (c+d x)^2 \left (d (20 A b d-15 a C d+2 a c D)-\left (16 a D d^2+b \left (-2 D c^2+5 C d c-20 B d^2\right )\right ) x\right )}{\sqrt {b x^2+a}}dx}{4 b d^2}+\frac {1}{4} d \sqrt {a+b x^2} (c+d x)^3 (5 C d-6 c D)}{5 b d^3}+\frac {D \sqrt {a+b x^2} (c+d x)^4}{5 b d^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{4} d \int \frac {(c+d x)^2 \left (d (20 A b d-15 a C d+2 a c D)-\left (16 a D d^2+b \left (-2 D c^2+5 C d c-20 B d^2\right )\right ) x\right )}{\sqrt {b x^2+a}}dx+\frac {1}{4} d \sqrt {a+b x^2} (c+d x)^3 (5 C d-6 c D)}{5 b d^3}+\frac {D \sqrt {a+b x^2} (c+d x)^4}{5 b d^2}\)

\(\Big \downarrow \) 687

\(\displaystyle \frac {\frac {1}{4} d \left (\frac {\int \frac {(c+d x) \left (d \left (60 A c d b^2+a \left (32 a d^2 D-b \left (-2 D c^2+35 C d c+40 B d^2\right )\right )\right )-b \left (a (45 C d+26 c D) d^2+2 b \left (-2 D c^3+5 C d c^2-20 B d^2 c-30 A d^3\right )\right ) x\right )}{\sqrt {b x^2+a}}dx}{3 b}-\frac {\sqrt {a+b x^2} (c+d x)^2 \left (16 a d^2 D+b \left (-20 B d^2-2 c^2 D+5 c C d\right )\right )}{3 b}\right )+\frac {1}{4} d \sqrt {a+b x^2} (c+d x)^3 (5 C d-6 c D)}{5 b d^3}+\frac {D \sqrt {a+b x^2} (c+d x)^4}{5 b d^2}\)

\(\Big \downarrow \) 676

\(\displaystyle \frac {\frac {1}{4} d \left (\frac {\frac {15}{2} d^2 \left (4 A b \left (2 b c^2-a d^2\right )-a (4 b c (2 B d+c C)-3 a d (2 c D+C d))\right ) \int \frac {1}{\sqrt {b x^2+a}}dx+\frac {2 \sqrt {a+b x^2} \left (16 a^2 d^4 D-4 a b d^2 \left (5 B d^2+3 c^2 D+10 c C d\right )-b^2 c \left (-60 A d^3-20 B c d^2-2 c^3 D+5 c^2 C d\right )\right )}{b}-\frac {1}{2} d x \sqrt {a+b x^2} \left (a d^2 (26 c D+45 C d)+2 b \left (-30 A d^3-20 B c d^2-2 c^3 D+5 c^2 C d\right )\right )}{3 b}-\frac {\sqrt {a+b x^2} (c+d x)^2 \left (16 a d^2 D+b \left (-20 B d^2-2 c^2 D+5 c C d\right )\right )}{3 b}\right )+\frac {1}{4} d \sqrt {a+b x^2} (c+d x)^3 (5 C d-6 c D)}{5 b d^3}+\frac {D \sqrt {a+b x^2} (c+d x)^4}{5 b d^2}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {\frac {1}{4} d \left (\frac {\frac {15}{2} d^2 \left (4 A b \left (2 b c^2-a d^2\right )-a (4 b c (2 B d+c C)-3 a d (2 c D+C d))\right ) \int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}+\frac {2 \sqrt {a+b x^2} \left (16 a^2 d^4 D-4 a b d^2 \left (5 B d^2+3 c^2 D+10 c C d\right )-b^2 c \left (-60 A d^3-20 B c d^2-2 c^3 D+5 c^2 C d\right )\right )}{b}-\frac {1}{2} d x \sqrt {a+b x^2} \left (a d^2 (26 c D+45 C d)+2 b \left (-30 A d^3-20 B c d^2-2 c^3 D+5 c^2 C d\right )\right )}{3 b}-\frac {\sqrt {a+b x^2} (c+d x)^2 \left (16 a d^2 D+b \left (-20 B d^2-2 c^2 D+5 c C d\right )\right )}{3 b}\right )+\frac {1}{4} d \sqrt {a+b x^2} (c+d x)^3 (5 C d-6 c D)}{5 b d^3}+\frac {D \sqrt {a+b x^2} (c+d x)^4}{5 b d^2}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {1}{4} d \left (\frac {\frac {2 \sqrt {a+b x^2} \left (16 a^2 d^4 D-4 a b d^2 \left (5 B d^2+3 c^2 D+10 c C d\right )-b^2 c \left (-60 A d^3-20 B c d^2-2 c^3 D+5 c^2 C d\right )\right )}{b}+\frac {15 d^2 \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right ) \left (4 A b \left (2 b c^2-a d^2\right )-a (4 b c (2 B d+c C)-3 a d (2 c D+C d))\right )}{2 \sqrt {b}}-\frac {1}{2} d x \sqrt {a+b x^2} \left (a d^2 (26 c D+45 C d)+2 b \left (-30 A d^3-20 B c d^2-2 c^3 D+5 c^2 C d\right )\right )}{3 b}-\frac {\sqrt {a+b x^2} (c+d x)^2 \left (16 a d^2 D+b \left (-20 B d^2-2 c^2 D+5 c C d\right )\right )}{3 b}\right )+\frac {1}{4} d \sqrt {a+b x^2} (c+d x)^3 (5 C d-6 c D)}{5 b d^3}+\frac {D \sqrt {a+b x^2} (c+d x)^4}{5 b d^2}\)

Input:

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

Output:

(D*(c + d*x)^4*Sqrt[a + b*x^2])/(5*b*d^2) + ((d*(5*C*d - 6*c*D)*(c + d*x)^ 
3*Sqrt[a + b*x^2])/4 + (d*(-1/3*((16*a*d^2*D + b*(5*c*C*d - 20*B*d^2 - 2*c 
^2*D))*(c + d*x)^2*Sqrt[a + b*x^2])/b + ((2*(16*a^2*d^4*D - 4*a*b*d^2*(10* 
c*C*d + 5*B*d^2 + 3*c^2*D) - b^2*c*(5*c^2*C*d - 20*B*c*d^2 - 60*A*d^3 - 2* 
c^3*D))*Sqrt[a + b*x^2])/b - (d*(a*d^2*(45*C*d + 26*c*D) + 2*b*(5*c^2*C*d 
- 20*B*c*d^2 - 30*A*d^3 - 2*c^3*D))*x*Sqrt[a + b*x^2])/2 + (15*d^2*(4*A*b* 
(2*b*c^2 - a*d^2) - a*(4*b*c*(c*C + 2*B*d) - 3*a*d*(C*d + 2*c*D)))*ArcTanh 
[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(2*Sqrt[b]))/(3*b)))/4)/(5*b*d^3)
 

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 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 687

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p 
_.), x_Symbol] :> Simp[g*(d + e*x)^m*((a + c*x^2)^(p + 1)/(c*(m + 2*p + 2)) 
), x] + Simp[1/(c*(m + 2*p + 2))   Int[(d + e*x)^(m - 1)*(a + c*x^2)^p*Simp 
[c*d*f*(m + 2*p + 2) - a*e*g*m + c*(e*f*(m + 2*p + 2) + d*g*m)*x, x], x], x 
] /; FreeQ[{a, c, d, e, f, g, p}, x] && GtQ[m, 0] && NeQ[m + 2*p + 2, 0] && 
 (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p]) &&  !(IGtQ[m, 0] && Eq 
Q[f, 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 = 1.22 (sec) , antiderivative size = 287, normalized size of antiderivative = 1.02

method	result	size
default	\(\frac {A \,c^{2} \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{\sqrt {b}}+\frac {c \left (2 A d +B c \right ) \sqrt {b \,x^{2}+a}}{b}+d \left (C d +2 D c \right ) \left (\frac {x^{3} \sqrt {b \,x^{2}+a}}{4 b}-\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2 b}-\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 b^{\frac {3}{2}}}\right )}{4 b}\right )+\left (A \,d^{2}+2 B c d +C \,c^{2}\right ) \left (\frac {x \sqrt {b \,x^{2}+a}}{2 b}-\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 b^{\frac {3}{2}}}\right )+\left (B \,d^{2}+2 C c d +D c^{2}\right ) \left (\frac {x^{2} \sqrt {b \,x^{2}+a}}{3 b}-\frac {2 a \sqrt {b \,x^{2}+a}}{3 b^{2}}\right )+D d^{2} \left (\frac {x^{4} \sqrt {b \,x^{2}+a}}{5 b}-\frac {4 a \left (\frac {x^{2} \sqrt {b \,x^{2}+a}}{3 b}-\frac {2 a \sqrt {b \,x^{2}+a}}{3 b^{2}}\right )}{5 b}\right )\)	\(287\)

Input:

int((d*x+c)^2*(D*x^3+C*x^2+B*x+A)/(b*x^2+a)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 531, normalized size of antiderivative = 1.90 \[ \int \frac {(c+d x)^2 \left (A+B x+C x^2+D x^3\right )}{\sqrt {a+b x^2}} \, dx=\left [\frac {15 \, {\left (4 \, {\left (C a b - 2 \, A b^{2}\right )} c^{2} - 2 \, {\left (3 \, D a^{2} - 4 \, B a b\right )} c d - {\left (3 \, C a^{2} - 4 \, A a b\right )} d^{2}\right )} \sqrt {b} \log \left (-2 \, b x^{2} + 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 2 \, {\left (24 \, D b^{2} d^{2} x^{4} + 30 \, {\left (2 \, D b^{2} c d + C b^{2} d^{2}\right )} x^{3} - 40 \, {\left (2 \, D a b - 3 \, B b^{2}\right )} c^{2} - 80 \, {\left (2 \, C a b - 3 \, A b^{2}\right )} c d + 16 \, {\left (4 \, D a^{2} - 5 \, B a b\right )} d^{2} + 8 \, {\left (5 \, D b^{2} c^{2} + 10 \, C b^{2} c d - {\left (4 \, D a b - 5 \, B b^{2}\right )} d^{2}\right )} x^{2} + 15 \, {\left (4 \, C b^{2} c^{2} - 2 \, {\left (3 \, D a b - 4 \, B b^{2}\right )} c d - {\left (3 \, C a b - 4 \, A b^{2}\right )} d^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{240 \, b^{3}}, \frac {15 \, {\left (4 \, {\left (C a b - 2 \, A b^{2}\right )} c^{2} - 2 \, {\left (3 \, D a^{2} - 4 \, B a b\right )} c d - {\left (3 \, C a^{2} - 4 \, A a b\right )} d^{2}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) + {\left (24 \, D b^{2} d^{2} x^{4} + 30 \, {\left (2 \, D b^{2} c d + C b^{2} d^{2}\right )} x^{3} - 40 \, {\left (2 \, D a b - 3 \, B b^{2}\right )} c^{2} - 80 \, {\left (2 \, C a b - 3 \, A b^{2}\right )} c d + 16 \, {\left (4 \, D a^{2} - 5 \, B a b\right )} d^{2} + 8 \, {\left (5 \, D b^{2} c^{2} + 10 \, C b^{2} c d - {\left (4 \, D a b - 5 \, B b^{2}\right )} d^{2}\right )} x^{2} + 15 \, {\left (4 \, C b^{2} c^{2} - 2 \, {\left (3 \, D a b - 4 \, B b^{2}\right )} c d - {\left (3 \, C a b - 4 \, A b^{2}\right )} d^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{120 \, b^{3}}\right ] \] Input:

integrate((d*x+c)^2*(D*x^3+C*x^2+B*x+A)/(b*x^2+a)^(1/2),x, algorithm="fric 
as")
                                                                                    
                                                                                    
 

Output:

[1/240*(15*(4*(C*a*b - 2*A*b^2)*c^2 - 2*(3*D*a^2 - 4*B*a*b)*c*d - (3*C*a^2 
 - 4*A*a*b)*d^2)*sqrt(b)*log(-2*b*x^2 + 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) + 
 2*(24*D*b^2*d^2*x^4 + 30*(2*D*b^2*c*d + C*b^2*d^2)*x^3 - 40*(2*D*a*b - 3* 
B*b^2)*c^2 - 80*(2*C*a*b - 3*A*b^2)*c*d + 16*(4*D*a^2 - 5*B*a*b)*d^2 + 8*( 
5*D*b^2*c^2 + 10*C*b^2*c*d - (4*D*a*b - 5*B*b^2)*d^2)*x^2 + 15*(4*C*b^2*c^ 
2 - 2*(3*D*a*b - 4*B*b^2)*c*d - (3*C*a*b - 4*A*b^2)*d^2)*x)*sqrt(b*x^2 + a 
))/b^3, 1/120*(15*(4*(C*a*b - 2*A*b^2)*c^2 - 2*(3*D*a^2 - 4*B*a*b)*c*d - ( 
3*C*a^2 - 4*A*a*b)*d^2)*sqrt(-b)*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) + (24* 
D*b^2*d^2*x^4 + 30*(2*D*b^2*c*d + C*b^2*d^2)*x^3 - 40*(2*D*a*b - 3*B*b^2)* 
c^2 - 80*(2*C*a*b - 3*A*b^2)*c*d + 16*(4*D*a^2 - 5*B*a*b)*d^2 + 8*(5*D*b^2 
*c^2 + 10*C*b^2*c*d - (4*D*a*b - 5*B*b^2)*d^2)*x^2 + 15*(4*C*b^2*c^2 - 2*( 
3*D*a*b - 4*B*b^2)*c*d - (3*C*a*b - 4*A*b^2)*d^2)*x)*sqrt(b*x^2 + a))/b^3]
 

Sympy [A] (verification not implemented)

Time = 0.72 (sec) , antiderivative size = 354, normalized size of antiderivative = 1.26 \[ \int \frac {(c+d x)^2 \left (A+B x+C x^2+D x^3\right )}{\sqrt {a+b x^2}} \, dx=\begin {cases} \sqrt {a + b x^{2}} \left (\frac {D d^{2} x^{4}}{5 b} + \frac {x^{3} \left (C d^{2} + 2 D c d\right )}{4 b} + \frac {x^{2} \left (B d^{2} + 2 C c d - \frac {4 D a d^{2}}{5 b} + D c^{2}\right )}{3 b} + \frac {x \left (A d^{2} + 2 B c d + C c^{2} - \frac {3 a \left (C d^{2} + 2 D c d\right )}{4 b}\right )}{2 b} + \frac {2 A c d + B c^{2} - \frac {2 a \left (B d^{2} + 2 C c d - \frac {4 D a d^{2}}{5 b} + D c^{2}\right )}{3 b}}{b}\right ) + \left (A c^{2} - \frac {a \left (A d^{2} + 2 B c d + C c^{2} - \frac {3 a \left (C d^{2} + 2 D c d\right )}{4 b}\right )}{2 b}\right ) \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {b x^{2}}} & \text {otherwise} \end {cases}\right ) & \text {for}\: b \neq 0 \\\frac {A c^{2} x + \frac {D d^{2} x^{6}}{6} + \frac {x^{5} \left (C d^{2} + 2 D c d\right )}{5} + \frac {x^{4} \left (B d^{2} + 2 C c d + D c^{2}\right )}{4} + \frac {x^{3} \left (A d^{2} + 2 B c d + C c^{2}\right )}{3} + \frac {x^{2} \cdot \left (2 A c d + B c^{2}\right )}{2}}{\sqrt {a}} & \text {otherwise} \end {cases} \] Input:

integrate((d*x+c)**2*(D*x**3+C*x**2+B*x+A)/(b*x**2+a)**(1/2),x)
 

Output:

Piecewise((sqrt(a + b*x**2)*(D*d**2*x**4/(5*b) + x**3*(C*d**2 + 2*D*c*d)/( 
4*b) + x**2*(B*d**2 + 2*C*c*d - 4*D*a*d**2/(5*b) + D*c**2)/(3*b) + x*(A*d* 
*2 + 2*B*c*d + C*c**2 - 3*a*(C*d**2 + 2*D*c*d)/(4*b))/(2*b) + (2*A*c*d + B 
*c**2 - 2*a*(B*d**2 + 2*C*c*d - 4*D*a*d**2/(5*b) + D*c**2)/(3*b))/b) + (A* 
c**2 - a*(A*d**2 + 2*B*c*d + C*c**2 - 3*a*(C*d**2 + 2*D*c*d)/(4*b))/(2*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)), Ne(b, 0)), ((A*c**2*x + D*d**2*x**6/6 + x**5 
*(C*d**2 + 2*D*c*d)/5 + x**4*(B*d**2 + 2*C*c*d + D*c**2)/4 + x**3*(A*d**2 
+ 2*B*c*d + C*c**2)/3 + x**2*(2*A*c*d + B*c**2)/2)/sqrt(a), True))
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 325, normalized size of antiderivative = 1.16 \[ \int \frac {(c+d x)^2 \left (A+B x+C x^2+D x^3\right )}{\sqrt {a+b x^2}} \, dx=\frac {\sqrt {b x^{2} + a} D d^{2} x^{4}}{5 \, b} - \frac {4 \, \sqrt {b x^{2} + a} D a d^{2} x^{2}}{15 \, b^{2}} + \frac {{\left (2 \, D c d + C d^{2}\right )} \sqrt {b x^{2} + a} x^{3}}{4 \, b} + \frac {A c^{2} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {b}} + \frac {\sqrt {b x^{2} + a} B c^{2}}{b} + \frac {2 \, \sqrt {b x^{2} + a} A c d}{b} + \frac {8 \, \sqrt {b x^{2} + a} D a^{2} d^{2}}{15 \, b^{3}} + \frac {{\left (D c^{2} + 2 \, C c d + B d^{2}\right )} \sqrt {b x^{2} + a} x^{2}}{3 \, b} - \frac {3 \, {\left (2 \, D c d + C d^{2}\right )} \sqrt {b x^{2} + a} a x}{8 \, b^{2}} + \frac {{\left (C c^{2} + 2 \, B c d + A d^{2}\right )} \sqrt {b x^{2} + a} x}{2 \, b} + \frac {3 \, {\left (2 \, D c d + C d^{2}\right )} a^{2} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{8 \, b^{\frac {5}{2}}} - \frac {{\left (C c^{2} + 2 \, B c d + A d^{2}\right )} a \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{2 \, b^{\frac {3}{2}}} - \frac {2 \, {\left (D c^{2} + 2 \, C c d + B d^{2}\right )} \sqrt {b x^{2} + a} a}{3 \, b^{2}} \] Input:

integrate((d*x+c)^2*(D*x^3+C*x^2+B*x+A)/(b*x^2+a)^(1/2),x, algorithm="maxi 
ma")
 

Output:

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

Giac [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 286, normalized size of antiderivative = 1.02 \[ \int \frac {(c+d x)^2 \left (A+B x+C x^2+D x^3\right )}{\sqrt {a+b x^2}} \, dx=\frac {1}{120} \, \sqrt {b x^{2} + a} {\left ({\left (2 \, {\left (3 \, {\left (\frac {4 \, D d^{2} x}{b} + \frac {5 \, {\left (2 \, D b^{4} c d + C b^{4} d^{2}\right )}}{b^{5}}\right )} x + \frac {4 \, {\left (5 \, D b^{4} c^{2} + 10 \, C b^{4} c d - 4 \, D a b^{3} d^{2} + 5 \, B b^{4} d^{2}\right )}}{b^{5}}\right )} x + \frac {15 \, {\left (4 \, C b^{4} c^{2} - 6 \, D a b^{3} c d + 8 \, B b^{4} c d - 3 \, C a b^{3} d^{2} + 4 \, A b^{4} d^{2}\right )}}{b^{5}}\right )} x - \frac {8 \, {\left (10 \, D a b^{3} c^{2} - 15 \, B b^{4} c^{2} + 20 \, C a b^{3} c d - 30 \, A b^{4} c d - 8 \, D a^{2} b^{2} d^{2} + 10 \, B a b^{3} d^{2}\right )}}{b^{5}}\right )} + \frac {{\left (4 \, C a b c^{2} - 8 \, A b^{2} c^{2} - 6 \, D a^{2} c d + 8 \, B a b c d - 3 \, C a^{2} d^{2} + 4 \, A a b d^{2}\right )} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{8 \, b^{\frac {5}{2}}} \] Input:

integrate((d*x+c)^2*(D*x^3+C*x^2+B*x+A)/(b*x^2+a)^(1/2),x, algorithm="giac 
")
 

Output:

1/120*sqrt(b*x^2 + a)*((2*(3*(4*D*d^2*x/b + 5*(2*D*b^4*c*d + C*b^4*d^2)/b^ 
5)*x + 4*(5*D*b^4*c^2 + 10*C*b^4*c*d - 4*D*a*b^3*d^2 + 5*B*b^4*d^2)/b^5)*x 
 + 15*(4*C*b^4*c^2 - 6*D*a*b^3*c*d + 8*B*b^4*c*d - 3*C*a*b^3*d^2 + 4*A*b^4 
*d^2)/b^5)*x - 8*(10*D*a*b^3*c^2 - 15*B*b^4*c^2 + 20*C*a*b^3*c*d - 30*A*b^ 
4*c*d - 8*D*a^2*b^2*d^2 + 10*B*a*b^3*d^2)/b^5) + 1/8*(4*C*a*b*c^2 - 8*A*b^ 
2*c^2 - 6*D*a^2*c*d + 8*B*a*b*c*d - 3*C*a^2*d^2 + 4*A*a*b*d^2)*log(abs(-sq 
rt(b)*x + sqrt(b*x^2 + a)))/b^(5/2)
 

Mupad [F(-1)]

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

int(((c + d*x)^2*(A + B*x + C*x^2 + x^3*D))/(a + b*x^2)^(1/2),x)
 

Output:

int(((c + d*x)^2*(A + B*x + C*x^2 + x^3*D))/(a + b*x^2)^(1/2), x)
 

Reduce [F]

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

int((d*x+c)^2*(D*x^3+C*x^2+B*x+A)/(b*x^2+a)^(1/2),x)
 

Output:

int((d*x+c)^2*(D*x^3+C*x^2+B*x+A)/(b*x^2+a)^(1/2),x)