\(\int \frac {x^9 (A+B x^2)}{(b x^2+c x^4)^{3/2}} \, dx\) [210]

Optimal result

Integrand size = 26, antiderivative size = 172 \[ \int \frac {x^9 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {b^2 (b B-A c) x^2}{c^4 \sqrt {b x^2+c x^4}}+\frac {b (19 b B-14 A c) \sqrt {b x^2+c x^4}}{16 c^4}-\frac {(11 b B-6 A c) x^2 \sqrt {b x^2+c x^4}}{24 c^3}+\frac {B x^4 \sqrt {b x^2+c x^4}}{6 c^2}-\frac {5 b^2 (7 b B-6 A c) \text {arctanh}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{16 c^{9/2}} \] Output:

b^2*(-A*c+B*b)*x^2/c^4/(c*x^4+b*x^2)^(1/2)+1/16*b*(-14*A*c+19*B*b)*(c*x^4+ 
b*x^2)^(1/2)/c^4-1/24*(-6*A*c+11*B*b)*x^2*(c*x^4+b*x^2)^(1/2)/c^3+1/6*B*x^ 
4*(c*x^4+b*x^2)^(1/2)/c^2-5/16*b^2*(-6*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 = 0.53 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.91 \[ \int \frac {x^9 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {x^3 \left (\sqrt {c} x \left (b+c x^2\right ) \left (105 b^3 B+4 c^3 x^4 \left (3 A+2 B x^2\right )-2 b c^2 x^2 \left (15 A+7 B x^2\right )+b^2 \left (-90 A c+35 B c x^2\right )\right )-30 b^2 (7 b B-6 A c) \left (b+c x^2\right )^{3/2} \text {arctanh}\left (\frac {\sqrt {c} x}{-\sqrt {b}+\sqrt {b+c x^2}}\right )\right )}{48 c^{9/2} \left (x^2 \left (b+c x^2\right )\right )^{3/2}} \] Input:

Integrate[(x^9*(A + B*x^2))/(b*x^2 + c*x^4)^(3/2),x]
 

Output:

(x^3*(Sqrt[c]*x*(b + c*x^2)*(105*b^3*B + 4*c^3*x^4*(3*A + 2*B*x^2) - 2*b*c 
^2*x^2*(15*A + 7*B*x^2) + b^2*(-90*A*c + 35*B*c*x^2)) - 30*b^2*(7*b*B - 6* 
A*c)*(b + c*x^2)^(3/2)*ArcTanh[(Sqrt[c]*x)/(-Sqrt[b] + Sqrt[b + c*x^2])])) 
/(48*c^(9/2)*(x^2*(b + c*x^2))^(3/2))
 

Rubi [A] (verified)

Time = 1.01 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.09, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {1940, 1211, 25, 2192, 27, 2192, 27, 1160, 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.

\(\displaystyle \int \frac {x^9 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^{3/2}} \, dx\)

\(\Big \downarrow \) 1940

\(\displaystyle \frac {1}{2} \int \frac {x^8 \left (B x^2+A\right )}{\left (c x^4+b x^2\right )^{3/2}}dx^2\)

\(\Big \downarrow \) 1211

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 2192

\(\displaystyle \frac {1}{2} \left (\frac {2 b^2 x^2 (b B-A c)}{c^4 \sqrt {b x^2+c x^4}}-\frac {\frac {\int \frac {c^3 (11 b B-6 A c) x^4-6 b c^2 (b B-A c) x^2+6 b^2 c (b B-A c)}{2 \sqrt {c x^4+b x^2}}dx^2}{3 c}-\frac {1}{3} B c^2 x^4 \sqrt {b x^2+c x^4}}{c^4}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{2} \left (\frac {2 b^2 x^2 (b B-A c)}{c^4 \sqrt {b x^2+c x^4}}-\frac {\frac {\int \frac {c^3 (11 b B-6 A c) x^4-6 b c^2 (b B-A c) x^2+6 b^2 c (b B-A c)}{\sqrt {c x^4+b x^2}}dx^2}{6 c}-\frac {1}{3} B c^2 x^4 \sqrt {b x^2+c x^4}}{c^4}\right )\)

\(\Big \downarrow \) 2192

\(\displaystyle \frac {1}{2} \left (\frac {2 b^2 x^2 (b B-A c)}{c^4 \sqrt {b x^2+c x^4}}-\frac {\frac {\frac {\int \frac {3 b c^2 \left (8 b (b B-A c)-c (19 b B-14 A c) x^2\right )}{2 \sqrt {c x^4+b x^2}}dx^2}{2 c}+\frac {1}{2} c^2 x^2 \sqrt {b x^2+c x^4} (11 b B-6 A c)}{6 c}-\frac {1}{3} B c^2 x^4 \sqrt {b x^2+c x^4}}{c^4}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{2} \left (\frac {2 b^2 x^2 (b B-A c)}{c^4 \sqrt {b x^2+c x^4}}-\frac {\frac {\frac {3}{4} b c \int \frac {8 b (b B-A c)-c (19 b B-14 A c) x^2}{\sqrt {c x^4+b x^2}}dx^2+\frac {1}{2} c^2 x^2 \sqrt {b x^2+c x^4} (11 b B-6 A c)}{6 c}-\frac {1}{3} B c^2 x^4 \sqrt {b x^2+c x^4}}{c^4}\right )\)

\(\Big \downarrow \) 1160

\(\displaystyle \frac {1}{2} \left (\frac {2 b^2 x^2 (b B-A c)}{c^4 \sqrt {b x^2+c x^4}}-\frac {\frac {\frac {3}{4} b c \left (\frac {5}{2} b (7 b B-6 A c) \int \frac {1}{\sqrt {c x^4+b x^2}}dx^2-\sqrt {b x^2+c x^4} (19 b B-14 A c)\right )+\frac {1}{2} c^2 x^2 \sqrt {b x^2+c x^4} (11 b B-6 A c)}{6 c}-\frac {1}{3} B c^2 x^4 \sqrt {b x^2+c x^4}}{c^4}\right )\)

\(\Big \downarrow \) 1091

\(\displaystyle \frac {1}{2} \left (\frac {2 b^2 x^2 (b B-A c)}{c^4 \sqrt {b x^2+c x^4}}-\frac {\frac {\frac {3}{4} b c \left (5 b (7 b B-6 A c) \int \frac {1}{1-c x^4}d\frac {x^2}{\sqrt {c x^4+b x^2}}-\sqrt {b x^2+c x^4} (19 b B-14 A c)\right )+\frac {1}{2} c^2 x^2 \sqrt {b x^2+c x^4} (11 b B-6 A c)}{6 c}-\frac {1}{3} B c^2 x^4 \sqrt {b x^2+c x^4}}{c^4}\right )\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{2} \left (\frac {2 b^2 x^2 (b B-A c)}{c^4 \sqrt {b x^2+c x^4}}-\frac {\frac {\frac {3}{4} b c \left (\frac {5 b (7 b B-6 A c) \text {arctanh}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{\sqrt {c}}-\sqrt {b x^2+c x^4} (19 b B-14 A c)\right )+\frac {1}{2} c^2 x^2 \sqrt {b x^2+c x^4} (11 b B-6 A c)}{6 c}-\frac {1}{3} B c^2 x^4 \sqrt {b x^2+c x^4}}{c^4}\right )\)

Input:

Int[(x^9*(A + B*x^2))/(b*x^2 + c*x^4)^(3/2),x]
 

Output:

((2*b^2*(b*B - A*c)*x^2)/(c^4*Sqrt[b*x^2 + c*x^4]) - (-1/3*(B*c^2*x^4*Sqrt 
[b*x^2 + c*x^4]) + ((c^2*(11*b*B - 6*A*c)*x^2*Sqrt[b*x^2 + c*x^4])/2 + (3* 
b*c*(-((19*b*B - 14*A*c)*Sqrt[b*x^2 + c*x^4]) + (5*b*(7*b*B - 6*A*c)*ArcTa 
nh[(Sqrt[c]*x^2)/Sqrt[b*x^2 + c*x^4]])/Sqrt[c]))/4)/(6*c))/c^4)/2
 

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 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 1160

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

rule 1211

Int[(((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.))/((a_.) + (b_.)* 
(x_) + (c_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[-2*(2*c*d - b*e)^(m - 2)*(c*( 
e*f + d*g) - b*e*g)^n*((d + e*x)/(c^(m + n - 1)*e^(n - 1)*Sqrt[a + b*x + c* 
x^2])), x] + Simp[1/(c^(m + n - 1)*e^(n - 2))   Int[ExpandToSum[((2*c*d - b 
*e)^(m - 1)*(c*(e*f + d*g) - b*e*g)^n - c^(m + n - 1)*e^n*(d + e*x)^(m - 1) 
*(f + g*x)^n)/(c*d - b*e - c*e*x), x]/Sqrt[a + b*x + c*x^2], x], x] /; Free 
Q[{a, b, c, d, e, f, g}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[m, 0] 
&& IGtQ[n, 0]
 

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]
 

rule 2192

Int[(Pq_)*((a_.) + (b_.)*(x_) + (c_.)*(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 + 
 c*x^2)^(p + 1)/(c*(q + 2*p + 1))), x] + Simp[1/(c*(q + 2*p + 1))   Int[(a 
+ b*x + c*x^2)^p*ExpandToSum[c*(q + 2*p + 1)*Pq - a*e*(q - 1)*x^(q - 2) - b 
*e*(q + p)*x^(q - 1) - c*e*(q + 2*p + 1)*x^q, x], x], x]] /; FreeQ[{a, b, c 
, p}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] &&  !LeQ[p, -1]
 

Maple [A] (verified)

Time = 0.49 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.84

method	result	size
pseudoelliptic	\(\frac {-\frac {15 x^{2} \left (-\frac {7 B \,x^{2}}{18}+A \right ) b^{2} c^{\frac {3}{2}}}{8}-\frac {5 x^{4} \left (\frac {7 B \,x^{2}}{15}+A \right ) b \,c^{\frac {5}{2}}}{8}+\frac {x^{6} \left (\frac {2 B \,x^{2}}{3}+A \right ) c^{\frac {7}{2}}}{4}+\frac {15 \left (\frac {7 B b \,x^{2} \sqrt {c}}{3}+\left (A c -\frac {7 B b}{6}\right ) \left (-\ln \left (2\right )+\ln \left (\frac {2 c \,x^{2}+2 \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \sqrt {c}+b}{\sqrt {c}}\right )\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}\right ) b^{2}}{16}}{c^{\frac {9}{2}} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}\)	\(145\)
default	\(\frac {x^{3} \left (c \,x^{2}+b \right ) \left (8 x^{7} B \,c^{\frac {9}{2}}+12 A \,c^{\frac {9}{2}} x^{5}-14 x^{5} B b \,c^{\frac {7}{2}}-30 A \,c^{\frac {7}{2}} b \,x^{3}+35 b^{2} x^{3} B \,c^{\frac {5}{2}}-90 A \,c^{\frac {5}{2}} b^{2} x +105 B x \,b^{3} c^{\frac {3}{2}}+90 A \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right ) \sqrt {c \,x^{2}+b}\, b^{2} c^{2}-105 B \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right ) \sqrt {c \,x^{2}+b}\, b^{3} c \right )}{48 \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} c^{\frac {11}{2}}}\)	\(166\)
risch	\(-\frac {x^{2} \left (-8 B \,c^{2} x^{4}-12 A \,c^{2} x^{2}+22 x^{2} B b c +42 A b c -57 B \,b^{2}\right ) \left (c \,x^{2}+b \right )}{48 c^{4} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}+\frac {b^{2} \left (5 c \left (6 A c -7 B b \right ) \left (-\frac {x}{c \sqrt {c \,x^{2}+b}}+\frac {\ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right )}{c^{\frac {3}{2}}}\right )-\frac {19 B b x}{\sqrt {c \,x^{2}+b}}+\frac {14 A c x}{\sqrt {c \,x^{2}+b}}\right ) x \sqrt {c \,x^{2}+b}}{16 c^{4} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}\)	\(176\)

Input:

int(x^9*(B*x^2+A)/(c*x^4+b*x^2)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

15/16*(-2*x^2*(-7/18*B*x^2+A)*b^2*c^(3/2)-2/3*x^4*(7/15*B*x^2+A)*b*c^(5/2) 
+4/15*x^6*(2/3*B*x^2+A)*c^(7/2)+(7/3*B*b*x^2*c^(1/2)+(A*c-7/6*B*b)*(-ln(2) 
+ln((2*c*x^2+2*(x^2*(c*x^2+b))^(1/2)*c^(1/2)+b)/c^(1/2)))*(x^2*(c*x^2+b))^ 
(1/2))*b^2)/c^(9/2)/(x^2*(c*x^2+b))^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 340, normalized size of antiderivative = 1.98 \[ \int \frac {x^9 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\left [-\frac {15 \, {\left (7 \, B b^{4} - 6 \, A b^{3} c + {\left (7 \, B b^{3} c - 6 \, A b^{2} c^{2}\right )} x^{2}\right )} \sqrt {c} \log \left (-2 \, c x^{2} - b - 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right ) - 2 \, {\left (8 \, B c^{4} x^{6} + 105 \, B b^{3} c - 90 \, A b^{2} c^{2} - 2 \, {\left (7 \, B b c^{3} - 6 \, A c^{4}\right )} x^{4} + 5 \, {\left (7 \, B b^{2} c^{2} - 6 \, A b c^{3}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{96 \, {\left (c^{6} x^{2} + b c^{5}\right )}}, \frac {15 \, {\left (7 \, B b^{4} - 6 \, A b^{3} c + {\left (7 \, B b^{3} c - 6 \, A b^{2} c^{2}\right )} x^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-c}}{c x^{2} + b}\right ) + {\left (8 \, B c^{4} x^{6} + 105 \, B b^{3} c - 90 \, A b^{2} c^{2} - 2 \, {\left (7 \, B b c^{3} - 6 \, A c^{4}\right )} x^{4} + 5 \, {\left (7 \, B b^{2} c^{2} - 6 \, A b c^{3}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{48 \, {\left (c^{6} x^{2} + b c^{5}\right )}}\right ] \] Input:

integrate(x^9*(B*x^2+A)/(c*x^4+b*x^2)^(3/2),x, algorithm="fricas")
 

Output:

[-1/96*(15*(7*B*b^4 - 6*A*b^3*c + (7*B*b^3*c - 6*A*b^2*c^2)*x^2)*sqrt(c)*l 
og(-2*c*x^2 - b - 2*sqrt(c*x^4 + b*x^2)*sqrt(c)) - 2*(8*B*c^4*x^6 + 105*B* 
b^3*c - 90*A*b^2*c^2 - 2*(7*B*b*c^3 - 6*A*c^4)*x^4 + 5*(7*B*b^2*c^2 - 6*A* 
b*c^3)*x^2)*sqrt(c*x^4 + b*x^2))/(c^6*x^2 + b*c^5), 1/48*(15*(7*B*b^4 - 6* 
A*b^3*c + (7*B*b^3*c - 6*A*b^2*c^2)*x^2)*sqrt(-c)*arctan(sqrt(c*x^4 + b*x^ 
2)*sqrt(-c)/(c*x^2 + b)) + (8*B*c^4*x^6 + 105*B*b^3*c - 90*A*b^2*c^2 - 2*( 
7*B*b*c^3 - 6*A*c^4)*x^4 + 5*(7*B*b^2*c^2 - 6*A*b*c^3)*x^2)*sqrt(c*x^4 + b 
*x^2))/(c^6*x^2 + b*c^5)]
 

Sympy [F]

\[ \int \frac {x^9 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\int \frac {x^{9} \left (A + B x^{2}\right )}{\left (x^{2} \left (b + c x^{2}\right )\right )^{\frac {3}{2}}}\, dx \] Input:

integrate(x**9*(B*x**2+A)/(c*x**4+b*x**2)**(3/2),x)
 

Output:

Integral(x**9*(A + B*x**2)/(x**2*(b + c*x**2))**(3/2), x)
 

Maxima [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 237, normalized size of antiderivative = 1.38 \[ \int \frac {x^9 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {1}{16} \, {\left (\frac {4 \, x^{6}}{\sqrt {c x^{4} + b x^{2}} c} - \frac {10 \, b x^{4}}{\sqrt {c x^{4} + b x^{2}} c^{2}} - \frac {30 \, b^{2} x^{2}}{\sqrt {c x^{4} + b x^{2}} c^{3}} + \frac {15 \, b^{2} \log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right )}{c^{\frac {7}{2}}}\right )} A + \frac {1}{96} \, {\left (\frac {16 \, x^{8}}{\sqrt {c x^{4} + b x^{2}} c} - \frac {28 \, b x^{6}}{\sqrt {c x^{4} + b x^{2}} c^{2}} + \frac {70 \, b^{2} x^{4}}{\sqrt {c x^{4} + b x^{2}} c^{3}} + \frac {210 \, b^{3} x^{2}}{\sqrt {c x^{4} + b x^{2}} c^{4}} - \frac {105 \, b^{3} \log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right )}{c^{\frac {9}{2}}}\right )} B \] Input:

integrate(x^9*(B*x^2+A)/(c*x^4+b*x^2)^(3/2),x, algorithm="maxima")
 

Output:

1/16*(4*x^6/(sqrt(c*x^4 + b*x^2)*c) - 10*b*x^4/(sqrt(c*x^4 + b*x^2)*c^2) - 
 30*b^2*x^2/(sqrt(c*x^4 + b*x^2)*c^3) + 15*b^2*log(2*c*x^2 + b + 2*sqrt(c* 
x^4 + b*x^2)*sqrt(c))/c^(7/2))*A + 1/96*(16*x^8/(sqrt(c*x^4 + b*x^2)*c) - 
28*b*x^6/(sqrt(c*x^4 + b*x^2)*c^2) + 70*b^2*x^4/(sqrt(c*x^4 + b*x^2)*c^3) 
+ 210*b^3*x^2/(sqrt(c*x^4 + b*x^2)*c^4) - 105*b^3*log(2*c*x^2 + b + 2*sqrt 
(c*x^4 + b*x^2)*sqrt(c))/c^(9/2))*B
 

Giac [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.06 \[ \int \frac {x^9 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {{\left ({\left (2 \, x^{2} {\left (\frac {4 \, B x^{2}}{c \mathrm {sgn}\left (x\right )} - \frac {7 \, B b c^{5} \mathrm {sgn}\left (x\right ) - 6 \, A c^{6} \mathrm {sgn}\left (x\right )}{c^{7}}\right )} + \frac {5 \, {\left (7 \, B b^{2} c^{4} \mathrm {sgn}\left (x\right ) - 6 \, A b c^{5} \mathrm {sgn}\left (x\right )\right )}}{c^{7}}\right )} x^{2} + \frac {15 \, {\left (7 \, B b^{3} c^{3} \mathrm {sgn}\left (x\right ) - 6 \, A b^{2} c^{4} \mathrm {sgn}\left (x\right )\right )}}{c^{7}}\right )} x}{48 \, \sqrt {c x^{2} + b}} - \frac {5 \, {\left (7 \, B b^{3} \log \left ({\left | b \right |}\right ) - 6 \, A b^{2} c \log \left ({\left | b \right |}\right )\right )} \mathrm {sgn}\left (x\right )}{32 \, c^{\frac {9}{2}}} + \frac {5 \, {\left (7 \, B b^{3} - 6 \, A b^{2} c\right )} \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + b} \right |}\right )}{16 \, c^{\frac {9}{2}} \mathrm {sgn}\left (x\right )} \] Input:

integrate(x^9*(B*x^2+A)/(c*x^4+b*x^2)^(3/2),x, algorithm="giac")
 

Output:

1/48*((2*x^2*(4*B*x^2/(c*sgn(x)) - (7*B*b*c^5*sgn(x) - 6*A*c^6*sgn(x))/c^7 
) + 5*(7*B*b^2*c^4*sgn(x) - 6*A*b*c^5*sgn(x))/c^7)*x^2 + 15*(7*B*b^3*c^3*s 
gn(x) - 6*A*b^2*c^4*sgn(x))/c^7)*x/sqrt(c*x^2 + b) - 5/32*(7*B*b^3*log(abs 
(b)) - 6*A*b^2*c*log(abs(b)))*sgn(x)/c^(9/2) + 5/16*(7*B*b^3 - 6*A*b^2*c)* 
log(abs(-sqrt(c)*x + sqrt(c*x^2 + b)))/(c^(9/2)*sgn(x))
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^9 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\int \frac {x^9\,\left (B\,x^2+A\right )}{{\left (c\,x^4+b\,x^2\right )}^{3/2}} \,d x \] Input:

int((x^9*(A + B*x^2))/(b*x^2 + c*x^4)^(3/2),x)
 

Output:

int((x^9*(A + B*x^2))/(b*x^2 + c*x^4)^(3/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 296, normalized size of antiderivative = 1.72 \[ \int \frac {x^9 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {-720 \sqrt {c \,x^{2}+b}\, a \,b^{2} c^{2} x -240 \sqrt {c \,x^{2}+b}\, a b \,c^{3} x^{3}+96 \sqrt {c \,x^{2}+b}\, a \,c^{4} x^{5}+840 \sqrt {c \,x^{2}+b}\, b^{4} c x +280 \sqrt {c \,x^{2}+b}\, b^{3} c^{2} x^{3}-112 \sqrt {c \,x^{2}+b}\, b^{2} c^{3} x^{5}+64 \sqrt {c \,x^{2}+b}\, b \,c^{4} x^{7}+720 \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}+\sqrt {c}\, x}{\sqrt {b}}\right ) a \,b^{3} c +720 \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}+\sqrt {c}\, x}{\sqrt {b}}\right ) a \,b^{2} c^{2} x^{2}-840 \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}+\sqrt {c}\, x}{\sqrt {b}}\right ) b^{5}-840 \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}+\sqrt {c}\, x}{\sqrt {b}}\right ) b^{4} c \,x^{2}-480 \sqrt {c}\, a \,b^{3} c -480 \sqrt {c}\, a \,b^{2} c^{2} x^{2}+525 \sqrt {c}\, b^{5}+525 \sqrt {c}\, b^{4} c \,x^{2}}{384 c^{5} \left (c \,x^{2}+b \right )} \] Input:

int(x^9*(B*x^2+A)/(c*x^4+b*x^2)^(3/2),x)
 

Output:

( - 720*sqrt(b + c*x**2)*a*b**2*c**2*x - 240*sqrt(b + c*x**2)*a*b*c**3*x** 
3 + 96*sqrt(b + c*x**2)*a*c**4*x**5 + 840*sqrt(b + c*x**2)*b**4*c*x + 280* 
sqrt(b + c*x**2)*b**3*c**2*x**3 - 112*sqrt(b + c*x**2)*b**2*c**3*x**5 + 64 
*sqrt(b + c*x**2)*b*c**4*x**7 + 720*sqrt(c)*log((sqrt(b + c*x**2) + sqrt(c 
)*x)/sqrt(b))*a*b**3*c + 720*sqrt(c)*log((sqrt(b + c*x**2) + sqrt(c)*x)/sq 
rt(b))*a*b**2*c**2*x**2 - 840*sqrt(c)*log((sqrt(b + c*x**2) + sqrt(c)*x)/s 
qrt(b))*b**5 - 840*sqrt(c)*log((sqrt(b + c*x**2) + sqrt(c)*x)/sqrt(b))*b** 
4*c*x**2 - 480*sqrt(c)*a*b**3*c - 480*sqrt(c)*a*b**2*c**2*x**2 + 525*sqrt( 
c)*b**5 + 525*sqrt(c)*b**4*c*x**2)/(384*c**5*(b + c*x**2))