\(\int x^3 \sqrt {a+b x^2} (A+B x+C x^2+D x^3) \, dx\) [61]

Optimal result

Integrand size = 30, antiderivative size = 227 \[ \int x^3 \sqrt {a+b x^2} \left (A+B x+C x^2+D x^3\right ) \, dx=-\frac {a^2 (8 b B-5 a D) x \sqrt {a+b x^2}}{128 b^3}+\frac {a (8 b B-5 a D) x^3 \sqrt {a+b x^2}}{192 b^2}+\frac {(8 b B-5 a D) x^5 \sqrt {a+b x^2}}{48 b}-\frac {a (A b-a C) \left (a+b x^2\right )^{3/2}}{3 b^3}+\frac {D x^5 \left (a+b x^2\right )^{3/2}}{8 b}+\frac {(A b-2 a C) \left (a+b x^2\right )^{5/2}}{5 b^3}+\frac {C \left (a+b x^2\right )^{7/2}}{7 b^3}+\frac {a^3 (8 b B-5 a D) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{128 b^{7/2}} \] Output:

-1/128*a^2*(8*B*b-5*D*a)*x*(b*x^2+a)^(1/2)/b^3+1/192*a*(8*B*b-5*D*a)*x^3*( 
b*x^2+a)^(1/2)/b^2+1/48*(8*B*b-5*D*a)*x^5*(b*x^2+a)^(1/2)/b-1/3*a*(A*b-C*a 
)*(b*x^2+a)^(3/2)/b^3+1/8*D*x^5*(b*x^2+a)^(3/2)/b+1/5*(A*b-2*C*a)*(b*x^2+a 
)^(5/2)/b^3+1/7*C*(b*x^2+a)^(7/2)/b^3+1/128*a^3*(8*B*b-5*D*a)*arctanh(b^(1 
/2)*x/(b*x^2+a)^(1/2))/b^(7/2)
 

Mathematica [A] (verified)

Time = 0.79 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.70 \[ \int x^3 \sqrt {a+b x^2} \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {\sqrt {b} \sqrt {a+b x^2} \left (a^3 (1024 C+525 D x)+8 a b^2 x^2 \left (112 A+x \left (70 B+48 C x+35 D x^2\right )\right )-2 a^2 b \left (896 A+x \left (420 B+256 C x+175 D x^2\right )\right )+16 b^3 x^4 (168 A+5 x (28 B+3 x (8 C+7 D x)))\right )+105 a^3 (-8 b B+5 a D) \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{13440 b^{7/2}} \] Input:

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

Output:

(Sqrt[b]*Sqrt[a + b*x^2]*(a^3*(1024*C + 525*D*x) + 8*a*b^2*x^2*(112*A + x* 
(70*B + 48*C*x + 35*D*x^2)) - 2*a^2*b*(896*A + x*(420*B + 256*C*x + 175*D* 
x^2)) + 16*b^3*x^4*(168*A + 5*x*(28*B + 3*x*(8*C + 7*D*x)))) + 105*a^3*(-8 
*b*B + 5*a*D)*Log[-(Sqrt[b]*x) + Sqrt[a + b*x^2]])/(13440*b^(7/2))
 

Rubi [A] (verified)

Time = 0.64 (sec) , antiderivative size = 244, normalized size of antiderivative = 1.07, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.433, Rules used = {2340, 2340, 27, 533, 27, 533, 25, 27, 533, 455, 211, 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.

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

\(\Big \downarrow \) 2340

\(\displaystyle \frac {\int x^3 \sqrt {b x^2+a} \left (8 b C x^2+(8 b B-5 a D) x+8 A b\right )dx}{8 b}+\frac {D x^5 \left (a+b x^2\right )^{3/2}}{8 b}\)

\(\Big \downarrow \) 2340

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 533

\(\displaystyle \frac {\frac {1}{7} \left (\frac {7 x^3 \left (a+b x^2\right )^{3/2} (8 b B-5 a D)}{6 b}-\frac {\int 3 x^2 (7 a (8 b B-5 a D)-16 b (7 A b-4 a C) x) \sqrt {b x^2+a}dx}{6 b}\right )+\frac {8}{7} C x^4 \left (a+b x^2\right )^{3/2}}{8 b}+\frac {D x^5 \left (a+b x^2\right )^{3/2}}{8 b}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{7} \left (\frac {7 x^3 \left (a+b x^2\right )^{3/2} (8 b B-5 a D)}{6 b}-\frac {\int x^2 (7 a (8 b B-5 a D)-16 b (7 A b-4 a C) x) \sqrt {b x^2+a}dx}{2 b}\right )+\frac {8}{7} C x^4 \left (a+b x^2\right )^{3/2}}{8 b}+\frac {D x^5 \left (a+b x^2\right )^{3/2}}{8 b}\)

\(\Big \downarrow \) 533

\(\displaystyle \frac {\frac {1}{7} \left (\frac {7 x^3 \left (a+b x^2\right )^{3/2} (8 b B-5 a D)}{6 b}-\frac {-\frac {\int -a b x (32 (7 A b-4 a C)+35 (8 b B-5 a D) x) \sqrt {b x^2+a}dx}{5 b}-\frac {16}{5} x^2 \left (a+b x^2\right )^{3/2} (7 A b-4 a C)}{2 b}\right )+\frac {8}{7} C x^4 \left (a+b x^2\right )^{3/2}}{8 b}+\frac {D x^5 \left (a+b x^2\right )^{3/2}}{8 b}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {1}{7} \left (\frac {7 x^3 \left (a+b x^2\right )^{3/2} (8 b B-5 a D)}{6 b}-\frac {\frac {\int a b x (32 (7 A b-4 a C)+35 (8 b B-5 a D) x) \sqrt {b x^2+a}dx}{5 b}-\frac {16}{5} x^2 \left (a+b x^2\right )^{3/2} (7 A b-4 a C)}{2 b}\right )+\frac {8}{7} C x^4 \left (a+b x^2\right )^{3/2}}{8 b}+\frac {D x^5 \left (a+b x^2\right )^{3/2}}{8 b}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{7} \left (\frac {7 x^3 \left (a+b x^2\right )^{3/2} (8 b B-5 a D)}{6 b}-\frac {\frac {1}{5} a \int x (32 (7 A b-4 a C)+35 (8 b B-5 a D) x) \sqrt {b x^2+a}dx-\frac {16}{5} x^2 \left (a+b x^2\right )^{3/2} (7 A b-4 a C)}{2 b}\right )+\frac {8}{7} C x^4 \left (a+b x^2\right )^{3/2}}{8 b}+\frac {D x^5 \left (a+b x^2\right )^{3/2}}{8 b}\)

\(\Big \downarrow \) 533

\(\displaystyle \frac {\frac {1}{7} \left (\frac {7 x^3 \left (a+b x^2\right )^{3/2} (8 b B-5 a D)}{6 b}-\frac {\frac {1}{5} a \left (\frac {35 x \left (a+b x^2\right )^{3/2} (8 b B-5 a D)}{4 b}-\frac {\int (35 a (8 b B-5 a D)-128 b (7 A b-4 a C) x) \sqrt {b x^2+a}dx}{4 b}\right )-\frac {16}{5} x^2 \left (a+b x^2\right )^{3/2} (7 A b-4 a C)}{2 b}\right )+\frac {8}{7} C x^4 \left (a+b x^2\right )^{3/2}}{8 b}+\frac {D x^5 \left (a+b x^2\right )^{3/2}}{8 b}\)

\(\Big \downarrow \) 455

\(\displaystyle \frac {\frac {1}{7} \left (\frac {7 x^3 \left (a+b x^2\right )^{3/2} (8 b B-5 a D)}{6 b}-\frac {\frac {1}{5} a \left (\frac {35 x \left (a+b x^2\right )^{3/2} (8 b B-5 a D)}{4 b}-\frac {35 a (8 b B-5 a D) \int \sqrt {b x^2+a}dx-\frac {128}{3} \left (a+b x^2\right )^{3/2} (7 A b-4 a C)}{4 b}\right )-\frac {16}{5} x^2 \left (a+b x^2\right )^{3/2} (7 A b-4 a C)}{2 b}\right )+\frac {8}{7} C x^4 \left (a+b x^2\right )^{3/2}}{8 b}+\frac {D x^5 \left (a+b x^2\right )^{3/2}}{8 b}\)

\(\Big \downarrow \) 211

\(\displaystyle \frac {\frac {1}{7} \left (\frac {7 x^3 \left (a+b x^2\right )^{3/2} (8 b B-5 a D)}{6 b}-\frac {\frac {1}{5} a \left (\frac {35 x \left (a+b x^2\right )^{3/2} (8 b B-5 a D)}{4 b}-\frac {35 a (8 b B-5 a D) \left (\frac {1}{2} a \int \frac {1}{\sqrt {b x^2+a}}dx+\frac {1}{2} x \sqrt {a+b x^2}\right )-\frac {128}{3} \left (a+b x^2\right )^{3/2} (7 A b-4 a C)}{4 b}\right )-\frac {16}{5} x^2 \left (a+b x^2\right )^{3/2} (7 A b-4 a C)}{2 b}\right )+\frac {8}{7} C x^4 \left (a+b x^2\right )^{3/2}}{8 b}+\frac {D x^5 \left (a+b x^2\right )^{3/2}}{8 b}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {\frac {1}{7} \left (\frac {7 x^3 \left (a+b x^2\right )^{3/2} (8 b B-5 a D)}{6 b}-\frac {\frac {1}{5} a \left (\frac {35 x \left (a+b x^2\right )^{3/2} (8 b B-5 a D)}{4 b}-\frac {35 a (8 b B-5 a D) \left (\frac {1}{2} a \int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}+\frac {1}{2} x \sqrt {a+b x^2}\right )-\frac {128}{3} \left (a+b x^2\right )^{3/2} (7 A b-4 a C)}{4 b}\right )-\frac {16}{5} x^2 \left (a+b x^2\right )^{3/2} (7 A b-4 a C)}{2 b}\right )+\frac {8}{7} C x^4 \left (a+b x^2\right )^{3/2}}{8 b}+\frac {D x^5 \left (a+b x^2\right )^{3/2}}{8 b}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {1}{7} \left (\frac {7 x^3 \left (a+b x^2\right )^{3/2} (8 b B-5 a D)}{6 b}-\frac {\frac {1}{5} a \left (\frac {35 x \left (a+b x^2\right )^{3/2} (8 b B-5 a D)}{4 b}-\frac {35 a \left (\frac {a \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{2 \sqrt {b}}+\frac {1}{2} x \sqrt {a+b x^2}\right ) (8 b B-5 a D)-\frac {128}{3} \left (a+b x^2\right )^{3/2} (7 A b-4 a C)}{4 b}\right )-\frac {16}{5} x^2 \left (a+b x^2\right )^{3/2} (7 A b-4 a C)}{2 b}\right )+\frac {8}{7} C x^4 \left (a+b x^2\right )^{3/2}}{8 b}+\frac {D x^5 \left (a+b x^2\right )^{3/2}}{8 b}\)

Input:

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

Output:

(D*x^5*(a + b*x^2)^(3/2))/(8*b) + ((8*C*x^4*(a + b*x^2)^(3/2))/7 + ((7*(8* 
b*B - 5*a*D)*x^3*(a + b*x^2)^(3/2))/(6*b) - ((-16*(7*A*b - 4*a*C)*x^2*(a + 
 b*x^2)^(3/2))/5 + (a*((35*(8*b*B - 5*a*D)*x*(a + b*x^2)^(3/2))/(4*b) - (( 
-128*(7*A*b - 4*a*C)*(a + b*x^2)^(3/2))/3 + 35*a*(8*b*B - 5*a*D)*((x*Sqrt[ 
a + b*x^2])/2 + (a*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(2*Sqrt[b])))/(4* 
b)))/5)/(2*b))/7)/(8*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 211

Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x*((a + b*x^2)^p/(2*p + 1 
)), x] + Simp[2*a*(p/(2*p + 1))   Int[(a + b*x^2)^(p - 1), x], x] /; FreeQ[ 
{a, b}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[6*p])
 

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 533

Int[(x_)^(m_.)*((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> 
 Simp[d*x^m*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 2))), x] - Simp[1/(b*(m + 2* 
p + 2))   Int[x^(m - 1)*(a + b*x^2)^p*Simp[a*d*m - b*c*(m + 2*p + 2)*x, x], 
 x], x] /; FreeQ[{a, b, c, d, p}, x] && IGtQ[m, 0] && GtQ[p, -1] && Integer 
Q[2*p]
 

rule 2340

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[ 
{q = Expon[Pq, x], f = Coeff[Pq, x, Expon[Pq, x]]}, Simp[f*(c*x)^(m + q - 1 
)*((a + b*x^2)^(p + 1)/(b*c^(q - 1)*(m + q + 2*p + 1))), x] + Simp[1/(b*(m 
+ q + 2*p + 1))   Int[(c*x)^m*(a + b*x^2)^p*ExpandToSum[b*(m + q + 2*p + 1) 
*Pq - b*f*(m + q + 2*p + 1)*x^q - a*f*(m + q - 1)*x^(q - 2), x], x], x] /; 
GtQ[q, 1] && NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, b, c, m, p}, x] && PolyQ 
[Pq, x] && ( !IGtQ[m, 0] || IGtQ[p + 1/2, -1])
 

Maple [A] (verified)

Time = 0.59 (sec) , antiderivative size = 286, normalized size of antiderivative = 1.26

method	result	size
default	\(A \left (\frac {x^{2} \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{5 b}-\frac {2 a \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{15 b^{2}}\right )+B \left (\frac {x^{3} \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{6 b}-\frac {a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4 b}-\frac {a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4 b}\right )}{2 b}\right )+C \left (\frac {x^{4} \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{7 b}-\frac {4 a \left (\frac {x^{2} \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{5 b}-\frac {2 a \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{15 b^{2}}\right )}{7 b}\right )+D \left (\frac {x^{5} \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{8 b}-\frac {5 a \left (\frac {x^{3} \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{6 b}-\frac {a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4 b}-\frac {a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4 b}\right )}{2 b}\right )}{8 b}\right )\)	\(286\)

Input:

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

Output:

A*(1/5*x^2*(b*x^2+a)^(3/2)/b-2/15*a/b^2*(b*x^2+a)^(3/2))+B*(1/6*x^3*(b*x^2 
+a)^(3/2)/b-1/2*a/b*(1/4*x*(b*x^2+a)^(3/2)/b-1/4*a/b*(1/2*x*(b*x^2+a)^(1/2 
)+1/2*a/b^(1/2)*ln(b^(1/2)*x+(b*x^2+a)^(1/2)))))+C*(1/7*x^4*(b*x^2+a)^(3/2 
)/b-4/7*a/b*(1/5*x^2*(b*x^2+a)^(3/2)/b-2/15*a/b^2*(b*x^2+a)^(3/2)))+D*(1/8 
*x^5*(b*x^2+a)^(3/2)/b-5/8*a/b*(1/6*x^3*(b*x^2+a)^(3/2)/b-1/2*a/b*(1/4*x*( 
b*x^2+a)^(3/2)/b-1/4*a/b*(1/2*x*(b*x^2+a)^(1/2)+1/2*a/b^(1/2)*ln(b^(1/2)*x 
+(b*x^2+a)^(1/2))))))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 387, normalized size of antiderivative = 1.70 \[ \int x^3 \sqrt {a+b x^2} \left (A+B x+C x^2+D x^3\right ) \, dx=\left [-\frac {105 \, {\left (5 \, D a^{4} - 8 \, B a^{3} b\right )} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) - 2 \, {\left (1680 \, D b^{4} x^{7} + 1920 \, C b^{4} x^{6} + 280 \, {\left (D a b^{3} + 8 \, B b^{4}\right )} x^{5} + 1024 \, C a^{3} b - 1792 \, A a^{2} b^{2} + 384 \, {\left (C a b^{3} + 7 \, A b^{4}\right )} x^{4} - 70 \, {\left (5 \, D a^{2} b^{2} - 8 \, B a b^{3}\right )} x^{3} - 128 \, {\left (4 \, C a^{2} b^{2} - 7 \, A a b^{3}\right )} x^{2} + 105 \, {\left (5 \, D a^{3} b - 8 \, B a^{2} b^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{26880 \, b^{4}}, \frac {105 \, {\left (5 \, D a^{4} - 8 \, B a^{3} b\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) + {\left (1680 \, D b^{4} x^{7} + 1920 \, C b^{4} x^{6} + 280 \, {\left (D a b^{3} + 8 \, B b^{4}\right )} x^{5} + 1024 \, C a^{3} b - 1792 \, A a^{2} b^{2} + 384 \, {\left (C a b^{3} + 7 \, A b^{4}\right )} x^{4} - 70 \, {\left (5 \, D a^{2} b^{2} - 8 \, B a b^{3}\right )} x^{3} - 128 \, {\left (4 \, C a^{2} b^{2} - 7 \, A a b^{3}\right )} x^{2} + 105 \, {\left (5 \, D a^{3} b - 8 \, B a^{2} b^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{13440 \, b^{4}}\right ] \] Input:

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

Output:

[-1/26880*(105*(5*D*a^4 - 8*B*a^3*b)*sqrt(b)*log(-2*b*x^2 - 2*sqrt(b*x^2 + 
 a)*sqrt(b)*x - a) - 2*(1680*D*b^4*x^7 + 1920*C*b^4*x^6 + 280*(D*a*b^3 + 8 
*B*b^4)*x^5 + 1024*C*a^3*b - 1792*A*a^2*b^2 + 384*(C*a*b^3 + 7*A*b^4)*x^4 
- 70*(5*D*a^2*b^2 - 8*B*a*b^3)*x^3 - 128*(4*C*a^2*b^2 - 7*A*a*b^3)*x^2 + 1 
05*(5*D*a^3*b - 8*B*a^2*b^2)*x)*sqrt(b*x^2 + a))/b^4, 1/13440*(105*(5*D*a^ 
4 - 8*B*a^3*b)*sqrt(-b)*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) + (1680*D*b^4*x 
^7 + 1920*C*b^4*x^6 + 280*(D*a*b^3 + 8*B*b^4)*x^5 + 1024*C*a^3*b - 1792*A* 
a^2*b^2 + 384*(C*a*b^3 + 7*A*b^4)*x^4 - 70*(5*D*a^2*b^2 - 8*B*a*b^3)*x^3 - 
 128*(4*C*a^2*b^2 - 7*A*a*b^3)*x^2 + 105*(5*D*a^3*b - 8*B*a^2*b^2)*x)*sqrt 
(b*x^2 + a))/b^4]
 

Sympy [A] (verification not implemented)

Time = 0.80 (sec) , antiderivative size = 264, normalized size of antiderivative = 1.16 \[ \int x^3 \sqrt {a+b x^2} \left (A+B x+C x^2+D x^3\right ) \, dx=\begin {cases} \frac {3 a^{2} \left (B a - \frac {5 a \left (B b + \frac {D a}{8}\right )}{6 b}\right ) \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {b x^{2}}} & \text {otherwise} \end {cases}\right )}{8 b^{2}} + \sqrt {a + b x^{2}} \left (\frac {C x^{6}}{7} + \frac {D x^{7}}{8} - \frac {3 a x \left (B a - \frac {5 a \left (B b + \frac {D a}{8}\right )}{6 b}\right )}{8 b^{2}} - \frac {2 a \left (A a - \frac {4 a \left (A b + \frac {C a}{7}\right )}{5 b}\right )}{3 b^{2}} + \frac {x^{5} \left (B b + \frac {D a}{8}\right )}{6 b} + \frac {x^{4} \left (A b + \frac {C a}{7}\right )}{5 b} + \frac {x^{3} \left (B a - \frac {5 a \left (B b + \frac {D a}{8}\right )}{6 b}\right )}{4 b} + \frac {x^{2} \left (A a - \frac {4 a \left (A b + \frac {C a}{7}\right )}{5 b}\right )}{3 b}\right ) & \text {for}\: b \neq 0 \\\sqrt {a} \left (\frac {A x^{4}}{4} + \frac {B x^{5}}{5} + \frac {C x^{6}}{6} + \frac {D x^{7}}{7}\right ) & \text {otherwise} \end {cases} \] Input:

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

Output:

Piecewise((3*a**2*(B*a - 5*a*(B*b + D*a/8)/(6*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), Tru 
e))/(8*b**2) + sqrt(a + b*x**2)*(C*x**6/7 + D*x**7/8 - 3*a*x*(B*a - 5*a*(B 
*b + D*a/8)/(6*b))/(8*b**2) - 2*a*(A*a - 4*a*(A*b + C*a/7)/(5*b))/(3*b**2) 
 + x**5*(B*b + D*a/8)/(6*b) + x**4*(A*b + C*a/7)/(5*b) + x**3*(B*a - 5*a*( 
B*b + D*a/8)/(6*b))/(4*b) + x**2*(A*a - 4*a*(A*b + C*a/7)/(5*b))/(3*b)), N 
e(b, 0)), (sqrt(a)*(A*x**4/4 + B*x**5/5 + C*x**6/6 + D*x**7/7), True))
 

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 255, normalized size of antiderivative = 1.12 \[ \int x^3 \sqrt {a+b x^2} \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} D x^{5}}{8 \, b} + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} C x^{4}}{7 \, b} - \frac {5 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} D a x^{3}}{48 \, b^{2}} + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} B x^{3}}{6 \, b} - \frac {4 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} C a x^{2}}{35 \, b^{2}} + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} A x^{2}}{5 \, b} + \frac {5 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} D a^{2} x}{64 \, b^{3}} - \frac {5 \, \sqrt {b x^{2} + a} D a^{3} x}{128 \, b^{3}} - \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} B a x}{8 \, b^{2}} + \frac {\sqrt {b x^{2} + a} B a^{2} x}{16 \, b^{2}} - \frac {5 \, D a^{4} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{128 \, b^{\frac {7}{2}}} + \frac {B a^{3} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{16 \, b^{\frac {5}{2}}} + \frac {8 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} C a^{2}}{105 \, b^{3}} - \frac {2 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} A a}{15 \, b^{2}} \] Input:

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

Output:

1/8*(b*x^2 + a)^(3/2)*D*x^5/b + 1/7*(b*x^2 + a)^(3/2)*C*x^4/b - 5/48*(b*x^ 
2 + a)^(3/2)*D*a*x^3/b^2 + 1/6*(b*x^2 + a)^(3/2)*B*x^3/b - 4/35*(b*x^2 + a 
)^(3/2)*C*a*x^2/b^2 + 1/5*(b*x^2 + a)^(3/2)*A*x^2/b + 5/64*(b*x^2 + a)^(3/ 
2)*D*a^2*x/b^3 - 5/128*sqrt(b*x^2 + a)*D*a^3*x/b^3 - 1/8*(b*x^2 + a)^(3/2) 
*B*a*x/b^2 + 1/16*sqrt(b*x^2 + a)*B*a^2*x/b^2 - 5/128*D*a^4*arcsinh(b*x/sq 
rt(a*b))/b^(7/2) + 1/16*B*a^3*arcsinh(b*x/sqrt(a*b))/b^(5/2) + 8/105*(b*x^ 
2 + a)^(3/2)*C*a^2/b^3 - 2/15*(b*x^2 + a)^(3/2)*A*a/b^2
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 207, normalized size of antiderivative = 0.91 \[ \int x^3 \sqrt {a+b x^2} \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {1}{13440} \, \sqrt {b x^{2} + a} {\left ({\left (2 \, {\left ({\left (4 \, {\left (5 \, {\left (6 \, {\left (7 \, D x + 8 \, C\right )} x + \frac {7 \, {\left (D a b^{5} + 8 \, B b^{6}\right )}}{b^{6}}\right )} x + \frac {48 \, {\left (C a b^{5} + 7 \, A b^{6}\right )}}{b^{6}}\right )} x - \frac {35 \, {\left (5 \, D a^{2} b^{4} - 8 \, B a b^{5}\right )}}{b^{6}}\right )} x - \frac {64 \, {\left (4 \, C a^{2} b^{4} - 7 \, A a b^{5}\right )}}{b^{6}}\right )} x + \frac {105 \, {\left (5 \, D a^{3} b^{3} - 8 \, B a^{2} b^{4}\right )}}{b^{6}}\right )} x + \frac {256 \, {\left (4 \, C a^{3} b^{3} - 7 \, A a^{2} b^{4}\right )}}{b^{6}}\right )} + \frac {{\left (5 \, D a^{4} - 8 \, B a^{3} b\right )} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{128 \, b^{\frac {7}{2}}} \] Input:

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

Output:

1/13440*sqrt(b*x^2 + a)*((2*((4*(5*(6*(7*D*x + 8*C)*x + 7*(D*a*b^5 + 8*B*b 
^6)/b^6)*x + 48*(C*a*b^5 + 7*A*b^6)/b^6)*x - 35*(5*D*a^2*b^4 - 8*B*a*b^5)/ 
b^6)*x - 64*(4*C*a^2*b^4 - 7*A*a*b^5)/b^6)*x + 105*(5*D*a^3*b^3 - 8*B*a^2* 
b^4)/b^6)*x + 256*(4*C*a^3*b^3 - 7*A*a^2*b^4)/b^6) + 1/128*(5*D*a^4 - 8*B* 
a^3*b)*log(abs(-sqrt(b)*x + sqrt(b*x^2 + a)))/b^(7/2)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 305, normalized size of antiderivative = 1.34 \[ \int x^3 \sqrt {a+b x^2} \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {-1792 \sqrt {b \,x^{2}+a}\, a^{3} b^{2}+1024 \sqrt {b \,x^{2}+a}\, a^{3} b c +525 \sqrt {b \,x^{2}+a}\, a^{3} b d x +896 \sqrt {b \,x^{2}+a}\, a^{2} b^{3} x^{2}-840 \sqrt {b \,x^{2}+a}\, a^{2} b^{3} x -512 \sqrt {b \,x^{2}+a}\, a^{2} b^{2} c \,x^{2}-350 \sqrt {b \,x^{2}+a}\, a^{2} b^{2} d \,x^{3}+2688 \sqrt {b \,x^{2}+a}\, a \,b^{4} x^{4}+560 \sqrt {b \,x^{2}+a}\, a \,b^{4} x^{3}+384 \sqrt {b \,x^{2}+a}\, a \,b^{3} c \,x^{4}+280 \sqrt {b \,x^{2}+a}\, a \,b^{3} d \,x^{5}+2240 \sqrt {b \,x^{2}+a}\, b^{5} x^{5}+1920 \sqrt {b \,x^{2}+a}\, b^{4} c \,x^{6}+1680 \sqrt {b \,x^{2}+a}\, b^{4} d \,x^{7}-525 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{4} d +840 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{3} b^{2}}{13440 b^{4}} \] Input:

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

Output:

( - 1792*sqrt(a + b*x**2)*a**3*b**2 + 1024*sqrt(a + b*x**2)*a**3*b*c + 525 
*sqrt(a + b*x**2)*a**3*b*d*x + 896*sqrt(a + b*x**2)*a**2*b**3*x**2 - 840*s 
qrt(a + b*x**2)*a**2*b**3*x - 512*sqrt(a + b*x**2)*a**2*b**2*c*x**2 - 350* 
sqrt(a + b*x**2)*a**2*b**2*d*x**3 + 2688*sqrt(a + b*x**2)*a*b**4*x**4 + 56 
0*sqrt(a + b*x**2)*a*b**4*x**3 + 384*sqrt(a + b*x**2)*a*b**3*c*x**4 + 280* 
sqrt(a + b*x**2)*a*b**3*d*x**5 + 2240*sqrt(a + b*x**2)*b**5*x**5 + 1920*sq 
rt(a + b*x**2)*b**4*c*x**6 + 1680*sqrt(a + b*x**2)*b**4*d*x**7 - 525*sqrt( 
b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a**4*d + 840*sqrt(b)*log((s 
qrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a**3*b**2)/(13440*b**4)