\(\int \frac {x^5 (A+B x^2)}{(b x^2+c x^4)^{3/2}} \, dx\) [212]

Optimal result

Integrand size = 26, antiderivative size = 97 \[ \int \frac {x^5 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {(b B-A c) x^2}{c^2 \sqrt {b x^2+c x^4}}+\frac {B \sqrt {b x^2+c x^4}}{2 c^2}-\frac {(3 b B-2 A c) \text {arctanh}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{2 c^{5/2}} \] Output:

(-A*c+B*b)*x^2/c^2/(c*x^4+b*x^2)^(1/2)+1/2*B*(c*x^4+b*x^2)^(1/2)/c^2-1/2*( 
-2*A*c+3*B*b)*arctanh(c^(1/2)*x^2/(c*x^4+b*x^2)^(1/2))/c^(5/2)
 

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.11 \[ \int \frac {x^5 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {x^3 \left (\sqrt {c} x \left (b+c x^2\right ) \left (3 b B-2 A c+B c x^2\right )+2 (-3 b B+2 A c) \left (b+c x^2\right )^{3/2} \text {arctanh}\left (\frac {\sqrt {c} x}{-\sqrt {b}+\sqrt {b+c x^2}}\right )\right )}{2 c^{5/2} \left (x^2 \left (b+c x^2\right )\right )^{3/2}} \] Input:

Integrate[(x^5*(A + B*x^2))/(b*x^2 + c*x^4)^(3/2),x]
 

Output:

(x^3*(Sqrt[c]*x*(b + c*x^2)*(3*b*B - 2*A*c + B*c*x^2) + 2*(-3*b*B + 2*A*c) 
*(b + c*x^2)^(3/2)*ArcTanh[(Sqrt[c]*x)/(-Sqrt[b] + Sqrt[b + c*x^2])]))/(2* 
c^(5/2)*(x^2*(b + c*x^2))^(3/2))
 

Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.03, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {1940, 1211, 25, 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.

\(\displaystyle \int \frac {x^5 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^{3/2}} \, dx\)

\(\Big \downarrow \) 1940

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

\(\Big \downarrow \) 1211

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 1160

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

\(\Big \downarrow \) 1091

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

\(\Big \downarrow \) 219

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

Input:

Int[(x^5*(A + B*x^2))/(b*x^2 + c*x^4)^(3/2),x]
 

Output:

((2*(b*B - A*c)*x^2)/(c^2*Sqrt[b*x^2 + c*x^4]) - (-(B*Sqrt[b*x^2 + c*x^4]) 
 + ((3*b*B - 2*A*c)*ArcTanh[(Sqrt[c]*x^2)/Sqrt[b*x^2 + c*x^4]])/Sqrt[c])/c 
^2)/2
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], 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]
 

Maple [A] (verified)

Time = 0.46 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.18

method	result	size
default	\(\frac {x^{3} \left (c \,x^{2}+b \right ) \left (x^{3} B \,c^{\frac {5}{2}}-2 A \,c^{\frac {5}{2}} x +3 B x b \,c^{\frac {3}{2}}+2 A \sqrt {c \,x^{2}+b}\, \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right ) c^{2}-3 B \sqrt {c \,x^{2}+b}\, \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right ) b c \right )}{2 \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} c^{\frac {7}{2}}}\)	\(114\)
risch	\(\frac {B \,x^{2} \left (c \,x^{2}+b \right )}{2 c^{2} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}+\frac {\left (c \left (2 A c -3 B b \right ) \left (-\frac {x}{c \sqrt {c \,x^{2}+b}}+\frac {\ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right )}{c^{\frac {3}{2}}}\right )-\frac {B b x}{\sqrt {c \,x^{2}+b}}\right ) x \sqrt {c \,x^{2}+b}}{2 c^{2} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}\)	\(121\)
pseudoelliptic	\(\frac {2 B \,c^{\frac {3}{2}} x^{4}-4 A \,c^{\frac {3}{2}} x^{2}+6 B b \,x^{2} \sqrt {c}+2 A \ln \left (\frac {2 c \,x^{2}+2 \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \sqrt {c}+b}{\sqrt {c}}\right ) c \sqrt {x^{2} \left (c \,x^{2}+b \right )}-2 A \ln \left (2\right ) c \sqrt {x^{2} \left (c \,x^{2}+b \right )}-3 B \ln \left (\frac {2 c \,x^{2}+2 \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \sqrt {c}+b}{\sqrt {c}}\right ) b \sqrt {x^{2} \left (c \,x^{2}+b \right )}+3 B \ln \left (2\right ) b \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{4 c^{\frac {5}{2}} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}\)	\(182\)

Input:

int(x^5*(B*x^2+A)/(c*x^4+b*x^2)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

1/2*x^3*(c*x^2+b)*(x^3*B*c^(5/2)-2*A*c^(5/2)*x+3*B*x*b*c^(3/2)+2*A*(c*x^2+ 
b)^(1/2)*ln(c^(1/2)*x+(c*x^2+b)^(1/2))*c^2-3*B*(c*x^2+b)^(1/2)*ln(c^(1/2)* 
x+(c*x^2+b)^(1/2))*b*c)/(c*x^4+b*x^2)^(3/2)/c^(7/2)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 230, normalized size of antiderivative = 2.37 \[ \int \frac {x^5 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\left [-\frac {{\left (3 \, B b^{2} - 2 \, A b c + {\left (3 \, B b c - 2 \, A c^{2}\right )} x^{2}\right )} \sqrt {c} \log \left (-2 \, c x^{2} - b - 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right ) - 2 \, {\left (B c^{2} x^{2} + 3 \, B b c - 2 \, A c^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{4 \, {\left (c^{4} x^{2} + b c^{3}\right )}}, \frac {{\left (3 \, B b^{2} - 2 \, A b c + {\left (3 \, B b c - 2 \, A c^{2}\right )} x^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-c}}{c x^{2} + b}\right ) + {\left (B c^{2} x^{2} + 3 \, B b c - 2 \, A c^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{2 \, {\left (c^{4} x^{2} + b c^{3}\right )}}\right ] \] Input:

integrate(x^5*(B*x^2+A)/(c*x^4+b*x^2)^(3/2),x, algorithm="fricas")
 

Output:

[-1/4*((3*B*b^2 - 2*A*b*c + (3*B*b*c - 2*A*c^2)*x^2)*sqrt(c)*log(-2*c*x^2 
- b - 2*sqrt(c*x^4 + b*x^2)*sqrt(c)) - 2*(B*c^2*x^2 + 3*B*b*c - 2*A*c^2)*s 
qrt(c*x^4 + b*x^2))/(c^4*x^2 + b*c^3), 1/2*((3*B*b^2 - 2*A*b*c + (3*B*b*c 
- 2*A*c^2)*x^2)*sqrt(-c)*arctan(sqrt(c*x^4 + b*x^2)*sqrt(-c)/(c*x^2 + b)) 
+ (B*c^2*x^2 + 3*B*b*c - 2*A*c^2)*sqrt(c*x^4 + b*x^2))/(c^4*x^2 + b*c^3)]
 

Sympy [F]

\[ \int \frac {x^5 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\int \frac {x^{5} \left (A + B x^{2}\right )}{\left (x^{2} \left (b + c x^{2}\right )\right )^{\frac {3}{2}}}\, dx \] Input:

integrate(x**5*(B*x**2+A)/(c*x**4+b*x**2)**(3/2),x)
 

Output:

Integral(x**5*(A + B*x**2)/(x**2*(b + c*x**2))**(3/2), x)
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.42 \[ \int \frac {x^5 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {1}{4} \, {\left (\frac {2 \, x^{4}}{\sqrt {c x^{4} + b x^{2}} c} + \frac {6 \, b x^{2}}{\sqrt {c x^{4} + b x^{2}} c^{2}} - \frac {3 \, b \log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right )}{c^{\frac {5}{2}}}\right )} B - \frac {1}{2} \, A {\left (\frac {2 \, x^{2}}{\sqrt {c x^{4} + b x^{2}} c} - \frac {\log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right )}{c^{\frac {3}{2}}}\right )} \] Input:

integrate(x^5*(B*x^2+A)/(c*x^4+b*x^2)^(3/2),x, algorithm="maxima")
 

Output:

1/4*(2*x^4/(sqrt(c*x^4 + b*x^2)*c) + 6*b*x^2/(sqrt(c*x^4 + b*x^2)*c^2) - 3 
*b*log(2*c*x^2 + b + 2*sqrt(c*x^4 + b*x^2)*sqrt(c))/c^(5/2))*B - 1/2*A*(2* 
x^2/(sqrt(c*x^4 + b*x^2)*c) - log(2*c*x^2 + b + 2*sqrt(c*x^4 + b*x^2)*sqrt 
(c))/c^(3/2))
 

Giac [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.07 \[ \int \frac {x^5 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {x {\left (\frac {B x^{2}}{c \mathrm {sgn}\left (x\right )} + \frac {3 \, B b c \mathrm {sgn}\left (x\right ) - 2 \, A c^{2} \mathrm {sgn}\left (x\right )}{c^{3}}\right )}}{2 \, \sqrt {c x^{2} + b}} - \frac {{\left (3 \, B b \log \left ({\left | b \right |}\right ) - 2 \, A c \log \left ({\left | b \right |}\right )\right )} \mathrm {sgn}\left (x\right )}{4 \, c^{\frac {5}{2}}} + \frac {{\left (3 \, B b - 2 \, A c\right )} \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + b} \right |}\right )}{2 \, c^{\frac {5}{2}} \mathrm {sgn}\left (x\right )} \] Input:

integrate(x^5*(B*x^2+A)/(c*x^4+b*x^2)^(3/2),x, algorithm="giac")
 

Output:

1/2*x*(B*x^2/(c*sgn(x)) + (3*B*b*c*sgn(x) - 2*A*c^2*sgn(x))/c^3)/sqrt(c*x^ 
2 + b) - 1/4*(3*B*b*log(abs(b)) - 2*A*c*log(abs(b)))*sgn(x)/c^(5/2) + 1/2* 
(3*B*b - 2*A*c)*log(abs(-sqrt(c)*x + sqrt(c*x^2 + b)))/(c^(5/2)*sgn(x))
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^5 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\int \frac {x^5\,\left (B\,x^2+A\right )}{{\left (c\,x^4+b\,x^2\right )}^{3/2}} \,d x \] Input:

int((x^5*(A + B*x^2))/(b*x^2 + c*x^4)^(3/2),x)
 

Output:

int((x^5*(A + B*x^2))/(b*x^2 + c*x^4)^(3/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 210, normalized size of antiderivative = 2.16 \[ \int \frac {x^5 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {-8 \sqrt {c \,x^{2}+b}\, a \,c^{2} x +12 \sqrt {c \,x^{2}+b}\, b^{2} c x +4 \sqrt {c \,x^{2}+b}\, b \,c^{2} x^{3}+8 \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}+\sqrt {c}\, x}{\sqrt {b}}\right ) a b c +8 \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}+\sqrt {c}\, x}{\sqrt {b}}\right ) a \,c^{2} x^{2}-12 \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}+\sqrt {c}\, x}{\sqrt {b}}\right ) b^{3}-12 \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}+\sqrt {c}\, x}{\sqrt {b}}\right ) b^{2} c \,x^{2}-8 \sqrt {c}\, a b c -8 \sqrt {c}\, a \,c^{2} x^{2}+9 \sqrt {c}\, b^{3}+9 \sqrt {c}\, b^{2} c \,x^{2}}{8 c^{3} \left (c \,x^{2}+b \right )} \] Input:

int(x^5*(B*x^2+A)/(c*x^4+b*x^2)^(3/2),x)
 

Output:

( - 8*sqrt(b + c*x**2)*a*c**2*x + 12*sqrt(b + c*x**2)*b**2*c*x + 4*sqrt(b 
+ c*x**2)*b*c**2*x**3 + 8*sqrt(c)*log((sqrt(b + c*x**2) + sqrt(c)*x)/sqrt( 
b))*a*b*c + 8*sqrt(c)*log((sqrt(b + c*x**2) + sqrt(c)*x)/sqrt(b))*a*c**2*x 
**2 - 12*sqrt(c)*log((sqrt(b + c*x**2) + sqrt(c)*x)/sqrt(b))*b**3 - 12*sqr 
t(c)*log((sqrt(b + c*x**2) + sqrt(c)*x)/sqrt(b))*b**2*c*x**2 - 8*sqrt(c)*a 
*b*c - 8*sqrt(c)*a*c**2*x**2 + 9*sqrt(c)*b**3 + 9*sqrt(c)*b**2*c*x**2)/(8* 
c**3*(b + c*x**2))