\(\int x^3 (A+B x^2) (b x^2+c x^4)^{3/2} \, dx\) [172]

Optimal result

Integrand size = 26, antiderivative size = 167 \[ \int x^3 \left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2} \, dx=-\frac {b^3 (7 b B-12 A c) \left (b+2 c x^2\right ) \sqrt {b x^2+c x^4}}{1024 c^4}+\frac {b (7 b B-12 A c) \left (b+2 c x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{384 c^3}-\frac {\left (7 b B-12 A c-10 B c x^2\right ) \left (b x^2+c x^4\right )^{5/2}}{120 c^2}+\frac {b^5 (7 b B-12 A c) \text {arctanh}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{1024 c^{9/2}} \] Output:

-1/1024*b^3*(-12*A*c+7*B*b)*(2*c*x^2+b)*(c*x^4+b*x^2)^(1/2)/c^4+1/384*b*(- 
12*A*c+7*B*b)*(2*c*x^2+b)*(c*x^4+b*x^2)^(3/2)/c^3-1/120*(-10*B*c*x^2-12*A* 
c+7*B*b)*(c*x^4+b*x^2)^(5/2)/c^2+1/1024*b^5*(-12*A*c+7*B*b)*arctanh(c^(1/2 
)*x^2/(c*x^4+b*x^2)^(1/2))/c^(9/2)
 

Mathematica [A] (verified)

Time = 0.99 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.35 \[ \int x^3 \left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2} \, dx=\frac {\left (x^2 \left (b+c x^2\right )\right )^{3/2} \left (-105 b^5 B x+180 A b^4 c x+70 b^4 B c x^3-120 A b^3 c^2 x^3-56 b^3 B c^2 x^5+96 A b^2 c^3 x^5+48 b^2 B c^3 x^7+2112 A b c^4 x^7+1664 b B c^4 x^9+1536 A c^5 x^9+1280 B c^5 x^{11}\right )}{15360 c^4 x^3 \left (b+c x^2\right )}+\frac {b^5 (7 b B-12 A c) \left (x^2 \left (b+c x^2\right )\right )^{3/2} \text {arctanh}\left (\frac {\sqrt {c} x}{-\sqrt {b}+\sqrt {b+c x^2}}\right )}{512 c^{9/2} x^3 \left (b+c x^2\right )^{3/2}} \] Input:

Integrate[x^3*(A + B*x^2)*(b*x^2 + c*x^4)^(3/2),x]
 

Output:

((x^2*(b + c*x^2))^(3/2)*(-105*b^5*B*x + 180*A*b^4*c*x + 70*b^4*B*c*x^3 - 
120*A*b^3*c^2*x^3 - 56*b^3*B*c^2*x^5 + 96*A*b^2*c^3*x^5 + 48*b^2*B*c^3*x^7 
 + 2112*A*b*c^4*x^7 + 1664*b*B*c^4*x^9 + 1536*A*c^5*x^9 + 1280*B*c^5*x^11) 
)/(15360*c^4*x^3*(b + c*x^2)) + (b^5*(7*b*B - 12*A*c)*(x^2*(b + c*x^2))^(3 
/2)*ArcTanh[(Sqrt[c]*x)/(-Sqrt[b] + Sqrt[b + c*x^2])])/(512*c^(9/2)*x^3*(b 
 + c*x^2)^(3/2))
 

Rubi [A] (verified)

Time = 0.57 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.01, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {1940, 1225, 1087, 1087, 1091, 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.

\(\displaystyle \int x^3 \left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2} \, dx\)

\(\Big \downarrow \) 1940

\(\displaystyle \frac {1}{2} \int x^2 \left (B x^2+A\right ) \left (c x^4+b x^2\right )^{3/2}dx^2\)

\(\Big \downarrow \) 1225

\(\displaystyle \frac {1}{2} \left (\frac {b (7 b B-12 A c) \int \left (c x^4+b x^2\right )^{3/2}dx^2}{24 c^2}-\frac {\left (b x^2+c x^4\right )^{5/2} \left (-12 A c+7 b B-10 B c x^2\right )}{60 c^2}\right )\)

\(\Big \downarrow \) 1087

\(\displaystyle \frac {1}{2} \left (\frac {b (7 b B-12 A c) \left (\frac {\left (b+2 c x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{8 c}-\frac {3 b^2 \int \sqrt {c x^4+b x^2}dx^2}{16 c}\right )}{24 c^2}-\frac {\left (b x^2+c x^4\right )^{5/2} \left (-12 A c+7 b B-10 B c x^2\right )}{60 c^2}\right )\)

\(\Big \downarrow \) 1087

\(\displaystyle \frac {1}{2} \left (\frac {b (7 b B-12 A c) \left (\frac {\left (b+2 c x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{8 c}-\frac {3 b^2 \left (\frac {\left (b+2 c x^2\right ) \sqrt {b x^2+c x^4}}{4 c}-\frac {b^2 \int \frac {1}{\sqrt {c x^4+b x^2}}dx^2}{8 c}\right )}{16 c}\right )}{24 c^2}-\frac {\left (b x^2+c x^4\right )^{5/2} \left (-12 A c+7 b B-10 B c x^2\right )}{60 c^2}\right )\)

\(\Big \downarrow \) 1091

\(\displaystyle \frac {1}{2} \left (\frac {b (7 b B-12 A c) \left (\frac {\left (b+2 c x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{8 c}-\frac {3 b^2 \left (\frac {\left (b+2 c x^2\right ) \sqrt {b x^2+c x^4}}{4 c}-\frac {b^2 \int \frac {1}{1-c x^4}d\frac {x^2}{\sqrt {c x^4+b x^2}}}{4 c}\right )}{16 c}\right )}{24 c^2}-\frac {\left (b x^2+c x^4\right )^{5/2} \left (-12 A c+7 b B-10 B c x^2\right )}{60 c^2}\right )\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{2} \left (\frac {b (7 b B-12 A c) \left (\frac {\left (b+2 c x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{8 c}-\frac {3 b^2 \left (\frac {\left (b+2 c x^2\right ) \sqrt {b x^2+c x^4}}{4 c}-\frac {b^2 \text {arctanh}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{4 c^{3/2}}\right )}{16 c}\right )}{24 c^2}-\frac {\left (b x^2+c x^4\right )^{5/2} \left (-12 A c+7 b B-10 B c x^2\right )}{60 c^2}\right )\)

Input:

Int[x^3*(A + B*x^2)*(b*x^2 + c*x^4)^(3/2),x]
 

Output:

(-1/60*((7*b*B - 12*A*c - 10*B*c*x^2)*(b*x^2 + c*x^4)^(5/2))/c^2 + (b*(7*b 
*B - 12*A*c)*(((b + 2*c*x^2)*(b*x^2 + c*x^4)^(3/2))/(8*c) - (3*b^2*(((b + 
2*c*x^2)*Sqrt[b*x^2 + c*x^4])/(4*c) - (b^2*ArcTanh[(Sqrt[c]*x^2)/Sqrt[b*x^ 
2 + c*x^4]])/(4*c^(3/2))))/(16*c)))/(24*c^2))/2
 

Defintions of rubi rules used

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 1087

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(b + 2*c*x) 
*((a + b*x + c*x^2)^p/(2*c*(2*p + 1))), x] - Simp[p*((b^2 - 4*a*c)/(2*c*(2* 
p + 1)))   Int[(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && 
GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[3*p])
 

rule 1091

Int[1/Sqrt[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Simp[2   Subst[Int[1/(1 
 - c*x^2), x], x, x/Sqrt[b*x + c*x^2]], x] /; FreeQ[{b, c}, x]
 

rule 1225

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*( 
x_)^2)^(p_), x_Symbol] :> Simp[(-(b*e*g*(p + 2) - c*(e*f + d*g)*(2*p + 3) - 
 2*c*e*g*(p + 1)*x))*((a + b*x + c*x^2)^(p + 1)/(2*c^2*(p + 1)*(2*p + 3))), 
 x] + Simp[(b^2*e*g*(p + 2) - 2*a*c*e*g + c*(2*c*d*f - b*(e*f + d*g))*(2*p 
+ 3))/(2*c^2*(2*p + 3))   Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c 
, d, e, f, g, p}, x] &&  !LeQ[p, -1]
 

rule 1940

Int[(x_)^(m_.)*((b_.)*(x_)^(k_.) + (a_.)*(x_)^(j_))^(p_)*((c_) + (d_.)*(x_) 
^(n_))^(q_.), x_Symbol] :> Simp[1/n   Subst[Int[x^(Simplify[(m + 1)/n] - 1) 
*(a*x^Simplify[j/n] + b*x^Simplify[k/n])^p*(c + d*x)^q, x], x, x^n], x] /; 
FreeQ[{a, b, c, d, j, k, m, n, p, q}, x] &&  !IntegerQ[p] && NeQ[k, j] && I 
ntegerQ[Simplify[j/n]] && IntegerQ[Simplify[k/n]] && IntegerQ[Simplify[(m + 
 1)/n]] && NeQ[n^2, 1]
 

Maple [A] (verified)

Time = 0.48 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.07

method	result	size
pseudoelliptic	\(\frac {\frac {3 \left (-\frac {1}{2} A \,b^{5} c +\frac {7}{24} B \,b^{6}\right ) \ln \left (\frac {2 c \,x^{2}+2 \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \sqrt {c}+b}{\sqrt {c}}\right )}{256}+\frac {3 \left (\frac {128 x^{8} \left (\frac {5 B \,x^{2}}{6}+A \right ) c^{\frac {11}{2}}}{15}+b \left (b^{3} \left (\frac {7 B \,x^{2}}{18}+A \right ) c^{\frac {3}{2}}-\frac {2 x^{2} \left (\frac {7 B \,x^{2}}{15}+A \right ) b^{2} c^{\frac {5}{2}}}{3}+\frac {8 \left (\frac {B \,x^{2}}{2}+A \right ) x^{4} b \,c^{\frac {7}{2}}}{15}+\frac {176 x^{6} \left (\frac {26 B \,x^{2}}{33}+A \right ) c^{\frac {9}{2}}}{15}-\frac {7 B \sqrt {c}\, b^{4}}{12}\right )\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{256}+\frac {3 \ln \left (2\right ) \left (A c -\frac {7 B b}{12}\right ) b^{5}}{512}}{c^{\frac {9}{2}}}\)	\(178\)
risch	\(\frac {\left (1280 B \,c^{5} x^{10}+1536 A \,c^{5} x^{8}+1664 B b \,c^{4} x^{8}+2112 A b \,c^{4} x^{6}+48 B \,b^{2} c^{3} x^{6}+96 A \,b^{2} c^{3} x^{4}-56 B \,b^{3} c^{2} x^{4}-120 A \,b^{3} c^{2} x^{2}+70 B \,b^{4} c \,x^{2}+180 A \,b^{4} c -105 b^{5} B \right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{15360 c^{4}}-\frac {b^{5} \left (12 A c -7 B b \right ) \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{1024 c^{\frac {9}{2}} x \sqrt {c \,x^{2}+b}}\)	\(188\)
default	\(\frac {\left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} \left (1280 B \left (c \,x^{2}+b \right )^{\frac {5}{2}} c^{\frac {7}{2}} x^{7}+1536 A \left (c \,x^{2}+b \right )^{\frac {5}{2}} c^{\frac {7}{2}} x^{5}-896 B \left (c \,x^{2}+b \right )^{\frac {5}{2}} c^{\frac {5}{2}} b \,x^{5}-960 A \left (c \,x^{2}+b \right )^{\frac {5}{2}} c^{\frac {5}{2}} b \,x^{3}+560 B \left (c \,x^{2}+b \right )^{\frac {5}{2}} c^{\frac {3}{2}} b^{2} x^{3}+480 A \left (c \,x^{2}+b \right )^{\frac {5}{2}} c^{\frac {3}{2}} b^{2} x -280 B \left (c \,x^{2}+b \right )^{\frac {5}{2}} \sqrt {c}\, b^{3} x -120 A \,c^{\frac {3}{2}} \left (c \,x^{2}+b \right )^{\frac {3}{2}} b^{3} x +70 B \sqrt {c}\, \left (c \,x^{2}+b \right )^{\frac {3}{2}} b^{4} x -180 A \,c^{\frac {3}{2}} \sqrt {c \,x^{2}+b}\, b^{4} x +105 B \sqrt {c}\, \sqrt {c \,x^{2}+b}\, b^{5} x -180 A \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right ) b^{5} c +105 B \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right ) b^{6}\right )}{15360 x^{3} \left (c \,x^{2}+b \right )^{\frac {3}{2}} c^{\frac {9}{2}}}\)	\(286\)

Input:

int(x^3*(B*x^2+A)*(c*x^4+b*x^2)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

3/256*((-1/2*A*b^5*c+7/24*B*b^6)*ln((2*c*x^2+2*(x^2*(c*x^2+b))^(1/2)*c^(1/ 
2)+b)/c^(1/2))+(128/15*x^8*(5/6*B*x^2+A)*c^(11/2)+b*(b^3*(7/18*B*x^2+A)*c^ 
(3/2)-2/3*x^2*(7/15*B*x^2+A)*b^2*c^(5/2)+8/15*(1/2*B*x^2+A)*x^4*b*c^(7/2)+ 
176/15*x^6*(26/33*B*x^2+A)*c^(9/2)-7/12*B*c^(1/2)*b^4))*(x^2*(c*x^2+b))^(1 
/2)+1/2*ln(2)*(A*c-7/12*B*b)*b^5)/c^(9/2)
 

Fricas [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 369, normalized size of antiderivative = 2.21 \[ \int x^3 \left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2} \, dx=\left [-\frac {15 \, {\left (7 \, B b^{6} - 12 \, A b^{5} c\right )} \sqrt {c} \log \left (-2 \, c x^{2} - b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right ) - 2 \, {\left (1280 \, B c^{6} x^{10} + 128 \, {\left (13 \, B b c^{5} + 12 \, A c^{6}\right )} x^{8} - 105 \, B b^{5} c + 180 \, A b^{4} c^{2} + 48 \, {\left (B b^{2} c^{4} + 44 \, A b c^{5}\right )} x^{6} - 8 \, {\left (7 \, B b^{3} c^{3} - 12 \, A b^{2} c^{4}\right )} x^{4} + 10 \, {\left (7 \, B b^{4} c^{2} - 12 \, A b^{3} c^{3}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{30720 \, c^{5}}, -\frac {15 \, {\left (7 \, B b^{6} - 12 \, A b^{5} c\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-c}}{c x^{2} + b}\right ) - {\left (1280 \, B c^{6} x^{10} + 128 \, {\left (13 \, B b c^{5} + 12 \, A c^{6}\right )} x^{8} - 105 \, B b^{5} c + 180 \, A b^{4} c^{2} + 48 \, {\left (B b^{2} c^{4} + 44 \, A b c^{5}\right )} x^{6} - 8 \, {\left (7 \, B b^{3} c^{3} - 12 \, A b^{2} c^{4}\right )} x^{4} + 10 \, {\left (7 \, B b^{4} c^{2} - 12 \, A b^{3} c^{3}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{15360 \, c^{5}}\right ] \] Input:

integrate(x^3*(B*x^2+A)*(c*x^4+b*x^2)^(3/2),x, algorithm="fricas")
 

Output:

[-1/30720*(15*(7*B*b^6 - 12*A*b^5*c)*sqrt(c)*log(-2*c*x^2 - b + 2*sqrt(c*x 
^4 + b*x^2)*sqrt(c)) - 2*(1280*B*c^6*x^10 + 128*(13*B*b*c^5 + 12*A*c^6)*x^ 
8 - 105*B*b^5*c + 180*A*b^4*c^2 + 48*(B*b^2*c^4 + 44*A*b*c^5)*x^6 - 8*(7*B 
*b^3*c^3 - 12*A*b^2*c^4)*x^4 + 10*(7*B*b^4*c^2 - 12*A*b^3*c^3)*x^2)*sqrt(c 
*x^4 + b*x^2))/c^5, -1/15360*(15*(7*B*b^6 - 12*A*b^5*c)*sqrt(-c)*arctan(sq 
rt(c*x^4 + b*x^2)*sqrt(-c)/(c*x^2 + b)) - (1280*B*c^6*x^10 + 128*(13*B*b*c 
^5 + 12*A*c^6)*x^8 - 105*B*b^5*c + 180*A*b^4*c^2 + 48*(B*b^2*c^4 + 44*A*b* 
c^5)*x^6 - 8*(7*B*b^3*c^3 - 12*A*b^2*c^4)*x^4 + 10*(7*B*b^4*c^2 - 12*A*b^3 
*c^3)*x^2)*sqrt(c*x^4 + b*x^2))/c^5]
                                                                                    
                                                                                    
 

Sympy [A] (verification not implemented)

Time = 1.09 (sec) , antiderivative size = 672, normalized size of antiderivative = 4.02 \[ \int x^3 \left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2} \, dx =\text {Too large to display} \] Input:

integrate(x**3*(B*x**2+A)*(c*x**4+b*x**2)**(3/2),x)
 

Output:

A*b*Piecewise((-5*b**4*Piecewise((log(b + 2*sqrt(c)*sqrt(b*x**2 + c*x**4) 
+ 2*c*x**2)/sqrt(c), Ne(b**2/c, 0)), ((b/(2*c) + x**2)*log(b/(2*c) + x**2) 
/sqrt(c*(b/(2*c) + x**2)**2), True))/(128*c**3) + sqrt(b*x**2 + c*x**4)*(5 
*b**3/(64*c**3) - 5*b**2*x**2/(96*c**2) + b*x**4/(24*c) + x**6/4), Ne(c, 0 
)), (2*(b*x**2)**(7/2)/(7*b**3), Ne(b, 0)), (0, True))/2 + A*c*Piecewise(( 
7*b**5*Piecewise((log(b + 2*sqrt(c)*sqrt(b*x**2 + c*x**4) + 2*c*x**2)/sqrt 
(c), Ne(b**2/c, 0)), ((b/(2*c) + x**2)*log(b/(2*c) + x**2)/sqrt(c*(b/(2*c) 
 + x**2)**2), True))/(256*c**4) + sqrt(b*x**2 + c*x**4)*(-7*b**4/(128*c**4 
) + 7*b**3*x**2/(192*c**3) - 7*b**2*x**4/(240*c**2) + b*x**6/(40*c) + x**8 
/5), Ne(c, 0)), (2*(b*x**2)**(9/2)/(9*b**4), Ne(b, 0)), (0, True))/2 + B*b 
*Piecewise((7*b**5*Piecewise((log(b + 2*sqrt(c)*sqrt(b*x**2 + c*x**4) + 2* 
c*x**2)/sqrt(c), Ne(b**2/c, 0)), ((b/(2*c) + x**2)*log(b/(2*c) + x**2)/sqr 
t(c*(b/(2*c) + x**2)**2), True))/(256*c**4) + sqrt(b*x**2 + c*x**4)*(-7*b* 
*4/(128*c**4) + 7*b**3*x**2/(192*c**3) - 7*b**2*x**4/(240*c**2) + b*x**6/( 
40*c) + x**8/5), Ne(c, 0)), (2*(b*x**2)**(9/2)/(9*b**4), Ne(b, 0)), (0, Tr 
ue))/2 + B*c*Piecewise((-21*b**6*Piecewise((log(b + 2*sqrt(c)*sqrt(b*x**2 
+ c*x**4) + 2*c*x**2)/sqrt(c), Ne(b**2/c, 0)), ((b/(2*c) + x**2)*log(b/(2* 
c) + x**2)/sqrt(c*(b/(2*c) + x**2)**2), True))/(1024*c**5) + sqrt(b*x**2 + 
 c*x**4)*(21*b**5/(512*c**5) - 7*b**4*x**2/(256*c**4) + 7*b**3*x**4/(320*c 
**3) - 3*b**2*x**6/(160*c**2) + b*x**8/(60*c) + x**10/6), Ne(c, 0)), (2...
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 315 vs. \(2 (147) = 294\).

Time = 0.05 (sec) , antiderivative size = 315, normalized size of antiderivative = 1.89 \[ \int x^3 \left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2} \, dx=\frac {1}{2560} \, {\left (\frac {60 \, \sqrt {c x^{4} + b x^{2}} b^{3} x^{2}}{c^{2}} - \frac {160 \, {\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} b x^{2}}{c} - \frac {15 \, b^{5} \log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right )}{c^{\frac {7}{2}}} + \frac {30 \, \sqrt {c x^{4} + b x^{2}} b^{4}}{c^{3}} - \frac {80 \, {\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} b^{2}}{c^{2}} + \frac {256 \, {\left (c x^{4} + b x^{2}\right )}^{\frac {5}{2}}}{c}\right )} A - \frac {1}{30720} \, {\left (\frac {420 \, \sqrt {c x^{4} + b x^{2}} b^{4} x^{2}}{c^{3}} - \frac {1120 \, {\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} b^{2} x^{2}}{c^{2}} - \frac {2560 \, {\left (c x^{4} + b x^{2}\right )}^{\frac {5}{2}} x^{2}}{c} - \frac {105 \, b^{6} \log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right )}{c^{\frac {9}{2}}} + \frac {210 \, \sqrt {c x^{4} + b x^{2}} b^{5}}{c^{4}} - \frac {560 \, {\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} b^{3}}{c^{3}} + \frac {1792 \, {\left (c x^{4} + b x^{2}\right )}^{\frac {5}{2}} b}{c^{2}}\right )} B \] Input:

integrate(x^3*(B*x^2+A)*(c*x^4+b*x^2)^(3/2),x, algorithm="maxima")
 

Output:

1/2560*(60*sqrt(c*x^4 + b*x^2)*b^3*x^2/c^2 - 160*(c*x^4 + b*x^2)^(3/2)*b*x 
^2/c - 15*b^5*log(2*c*x^2 + b + 2*sqrt(c*x^4 + b*x^2)*sqrt(c))/c^(7/2) + 3 
0*sqrt(c*x^4 + b*x^2)*b^4/c^3 - 80*(c*x^4 + b*x^2)^(3/2)*b^2/c^2 + 256*(c* 
x^4 + b*x^2)^(5/2)/c)*A - 1/30720*(420*sqrt(c*x^4 + b*x^2)*b^4*x^2/c^3 - 1 
120*(c*x^4 + b*x^2)^(3/2)*b^2*x^2/c^2 - 2560*(c*x^4 + b*x^2)^(5/2)*x^2/c - 
 105*b^6*log(2*c*x^2 + b + 2*sqrt(c*x^4 + b*x^2)*sqrt(c))/c^(9/2) + 210*sq 
rt(c*x^4 + b*x^2)*b^5/c^4 - 560*(c*x^4 + b*x^2)^(3/2)*b^3/c^3 + 1792*(c*x^ 
4 + b*x^2)^(5/2)*b/c^2)*B
 

Giac [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 246, normalized size of antiderivative = 1.47 \[ \int x^3 \left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2} \, dx=\frac {1}{15360} \, {\left (2 \, {\left (4 \, {\left (2 \, {\left (8 \, {\left (10 \, B c x^{2} \mathrm {sgn}\left (x\right ) + \frac {13 \, B b c^{10} \mathrm {sgn}\left (x\right ) + 12 \, A c^{11} \mathrm {sgn}\left (x\right )}{c^{10}}\right )} x^{2} + \frac {3 \, {\left (B b^{2} c^{9} \mathrm {sgn}\left (x\right ) + 44 \, A b c^{10} \mathrm {sgn}\left (x\right )\right )}}{c^{10}}\right )} x^{2} - \frac {7 \, B b^{3} c^{8} \mathrm {sgn}\left (x\right ) - 12 \, A b^{2} c^{9} \mathrm {sgn}\left (x\right )}{c^{10}}\right )} x^{2} + \frac {5 \, {\left (7 \, B b^{4} c^{7} \mathrm {sgn}\left (x\right ) - 12 \, A b^{3} c^{8} \mathrm {sgn}\left (x\right )\right )}}{c^{10}}\right )} x^{2} - \frac {15 \, {\left (7 \, B b^{5} c^{6} \mathrm {sgn}\left (x\right ) - 12 \, A b^{4} c^{7} \mathrm {sgn}\left (x\right )\right )}}{c^{10}}\right )} \sqrt {c x^{2} + b} x - \frac {{\left (7 \, B b^{6} \mathrm {sgn}\left (x\right ) - 12 \, A b^{5} c \mathrm {sgn}\left (x\right )\right )} \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + b} \right |}\right )}{1024 \, c^{\frac {9}{2}}} + \frac {{\left (7 \, B b^{6} \log \left ({\left | b \right |}\right ) - 12 \, A b^{5} c \log \left ({\left | b \right |}\right )\right )} \mathrm {sgn}\left (x\right )}{2048 \, c^{\frac {9}{2}}} \] Input:

integrate(x^3*(B*x^2+A)*(c*x^4+b*x^2)^(3/2),x, algorithm="giac")
 

Output:

1/15360*(2*(4*(2*(8*(10*B*c*x^2*sgn(x) + (13*B*b*c^10*sgn(x) + 12*A*c^11*s 
gn(x))/c^10)*x^2 + 3*(B*b^2*c^9*sgn(x) + 44*A*b*c^10*sgn(x))/c^10)*x^2 - ( 
7*B*b^3*c^8*sgn(x) - 12*A*b^2*c^9*sgn(x))/c^10)*x^2 + 5*(7*B*b^4*c^7*sgn(x 
) - 12*A*b^3*c^8*sgn(x))/c^10)*x^2 - 15*(7*B*b^5*c^6*sgn(x) - 12*A*b^4*c^7 
*sgn(x))/c^10)*sqrt(c*x^2 + b)*x - 1/1024*(7*B*b^6*sgn(x) - 12*A*b^5*c*sgn 
(x))*log(abs(-sqrt(c)*x + sqrt(c*x^2 + b)))/c^(9/2) + 1/2048*(7*B*b^6*log( 
abs(b)) - 12*A*b^5*c*log(abs(b)))*sgn(x)/c^(9/2)
 

Mupad [F(-1)]

Timed out. \[ \int x^3 \left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2} \, dx=\int x^3\,\left (B\,x^2+A\right )\,{\left (c\,x^4+b\,x^2\right )}^{3/2} \,d x \] Input:

int(x^3*(A + B*x^2)*(b*x^2 + c*x^4)^(3/2),x)
 

Output:

int(x^3*(A + B*x^2)*(b*x^2 + c*x^4)^(3/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 261, normalized size of antiderivative = 1.56 \[ \int x^3 \left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2} \, dx=\frac {180 \sqrt {c \,x^{2}+b}\, a \,b^{4} c^{2} x -120 \sqrt {c \,x^{2}+b}\, a \,b^{3} c^{3} x^{3}+96 \sqrt {c \,x^{2}+b}\, a \,b^{2} c^{4} x^{5}+2112 \sqrt {c \,x^{2}+b}\, a b \,c^{5} x^{7}+1536 \sqrt {c \,x^{2}+b}\, a \,c^{6} x^{9}-105 \sqrt {c \,x^{2}+b}\, b^{6} c x +70 \sqrt {c \,x^{2}+b}\, b^{5} c^{2} x^{3}-56 \sqrt {c \,x^{2}+b}\, b^{4} c^{3} x^{5}+48 \sqrt {c \,x^{2}+b}\, b^{3} c^{4} x^{7}+1664 \sqrt {c \,x^{2}+b}\, b^{2} c^{5} x^{9}+1280 \sqrt {c \,x^{2}+b}\, b \,c^{6} x^{11}-180 \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}+\sqrt {c}\, x}{\sqrt {b}}\right ) a \,b^{5} c +105 \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}+\sqrt {c}\, x}{\sqrt {b}}\right ) b^{7}}{15360 c^{5}} \] Input:

int(x^3*(B*x^2+A)*(c*x^4+b*x^2)^(3/2),x)
 

Output:

(180*sqrt(b + c*x**2)*a*b**4*c**2*x - 120*sqrt(b + c*x**2)*a*b**3*c**3*x** 
3 + 96*sqrt(b + c*x**2)*a*b**2*c**4*x**5 + 2112*sqrt(b + c*x**2)*a*b*c**5* 
x**7 + 1536*sqrt(b + c*x**2)*a*c**6*x**9 - 105*sqrt(b + c*x**2)*b**6*c*x + 
 70*sqrt(b + c*x**2)*b**5*c**2*x**3 - 56*sqrt(b + c*x**2)*b**4*c**3*x**5 + 
 48*sqrt(b + c*x**2)*b**3*c**4*x**7 + 1664*sqrt(b + c*x**2)*b**2*c**5*x**9 
 + 1280*sqrt(b + c*x**2)*b*c**6*x**11 - 180*sqrt(c)*log((sqrt(b + c*x**2) 
+ sqrt(c)*x)/sqrt(b))*a*b**5*c + 105*sqrt(c)*log((sqrt(b + c*x**2) + sqrt( 
c)*x)/sqrt(b))*b**7)/(15360*c**5)