\(\int \frac {x^9}{(b x^2+c x^4)^{3/2}} \, dx\) [207]

Optimal result

Integrand size = 19, antiderivative size = 112 \[ \int \frac {x^9}{\left (b x^2+c x^4\right )^{3/2}} \, dx=-\frac {b^2 x^2}{c^3 \sqrt {b x^2+c x^4}}-\frac {7 b \sqrt {b x^2+c x^4}}{8 c^3}+\frac {x^2 \sqrt {b x^2+c x^4}}{4 c^2}+\frac {15 b^2 \text {arctanh}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{8 c^{7/2}} \] Output:

-b^2*x^2/c^3/(c*x^4+b*x^2)^(1/2)-7/8*b*(c*x^4+b*x^2)^(1/2)/c^3+1/4*x^2*(c* 
x^4+b*x^2)^(1/2)/c^2+15/8*b^2*arctanh(c^(1/2)*x^2/(c*x^4+b*x^2)^(1/2))/c^( 
7/2)
 

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.88 \[ \int \frac {x^9}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {x \left (\sqrt {c} x \left (-15 b^2-5 b c x^2+2 c^2 x^4\right )+30 b^2 \sqrt {b+c x^2} \text {arctanh}\left (\frac {\sqrt {c} x}{-\sqrt {b}+\sqrt {b+c x^2}}\right )\right )}{8 c^{7/2} \sqrt {x^2 \left (b+c x^2\right )}} \] Input:

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

Output:

(x*(Sqrt[c]*x*(-15*b^2 - 5*b*c*x^2 + 2*c^2*x^4) + 30*b^2*Sqrt[b + c*x^2]*A 
rcTanh[(Sqrt[c]*x)/(-Sqrt[b] + Sqrt[b + c*x^2])]))/(8*c^(7/2)*Sqrt[x^2*(b 
+ c*x^2)])
 

Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.03, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {1424, 1124, 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.

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

\(\Big \downarrow \) 1424

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

\(\Big \downarrow \) 1124

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

\(\Big \downarrow \) 2192

\(\displaystyle \frac {1}{2} \left (\frac {\frac {\int \frac {b c \left (4 b-7 c x^2\right )}{2 \sqrt {c x^4+b x^2}}dx^2}{2 c}+\frac {1}{2} c x^2 \sqrt {b x^2+c x^4}}{c^3}-\frac {2 b^2 x^2}{c^3 \sqrt {b x^2+c x^4}}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{2} \left (\frac {\frac {1}{4} b \int \frac {4 b-7 c x^2}{\sqrt {c x^4+b x^2}}dx^2+\frac {1}{2} c x^2 \sqrt {b x^2+c x^4}}{c^3}-\frac {2 b^2 x^2}{c^3 \sqrt {b x^2+c x^4}}\right )\)

\(\Big \downarrow \) 1160

\(\displaystyle \frac {1}{2} \left (\frac {\frac {1}{4} b \left (\frac {15}{2} b \int \frac {1}{\sqrt {c x^4+b x^2}}dx^2-7 \sqrt {b x^2+c x^4}\right )+\frac {1}{2} c x^2 \sqrt {b x^2+c x^4}}{c^3}-\frac {2 b^2 x^2}{c^3 \sqrt {b x^2+c x^4}}\right )\)

\(\Big \downarrow \) 1091

\(\displaystyle \frac {1}{2} \left (\frac {\frac {1}{4} b \left (15 b \int \frac {1}{1-c x^4}d\frac {x^2}{\sqrt {c x^4+b x^2}}-7 \sqrt {b x^2+c x^4}\right )+\frac {1}{2} c x^2 \sqrt {b x^2+c x^4}}{c^3}-\frac {2 b^2 x^2}{c^3 \sqrt {b x^2+c x^4}}\right )\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{2} \left (\frac {\frac {1}{4} b \left (\frac {15 b \text {arctanh}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{\sqrt {c}}-7 \sqrt {b x^2+c x^4}\right )+\frac {1}{2} c x^2 \sqrt {b x^2+c x^4}}{c^3}-\frac {2 b^2 x^2}{c^3 \sqrt {b x^2+c x^4}}\right )\)

Input:

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

Output:

((-2*b^2*x^2)/(c^3*Sqrt[b*x^2 + c*x^4]) + ((c*x^2*Sqrt[b*x^2 + c*x^4])/2 + 
 (b*(-7*Sqrt[b*x^2 + c*x^4] + (15*b*ArcTanh[(Sqrt[c]*x^2)/Sqrt[b*x^2 + c*x 
^4]])/Sqrt[c]))/4)/c^3)/2
 

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 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 1124

Int[((d_.) + (e_.)*(x_))^(m_.)/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(3/2), x 
_Symbol] :> Simp[-2*e*(2*c*d - b*e)^(m - 2)*((d + e*x)/(c^(m - 1)*Sqrt[a + 
b*x + c*x^2])), x] + Simp[e^2/c^(m - 1)   Int[(1/Sqrt[a + b*x + c*x^2])*Exp 
andToSum[((2*c*d - b*e)^(m - 1) - c^(m - 1)*(d + e*x)^(m - 1))/(c*d - b*e - 
 c*e*x), x], x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c*d^2 - b*d*e + a*e 
^2, 0] && IGtQ[m, 0]
 

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 1424

Int[(x_)^(m_.)*((b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[1/2 
Subst[Int[x^((m - 1)/2)*(b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{b, c, m, 
 p}, x] &&  !IntegerQ[p] && IntegerQ[(m - 1)/2]
 

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 = 0.20 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.78

method	result	size
default	\(\frac {x^{3} \left (c \,x^{2}+b \right ) \left (2 c^{\frac {7}{2}} x^{5}-5 c^{\frac {5}{2}} b \,x^{3}-15 c^{\frac {3}{2}} b^{2} x +15 \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right ) \sqrt {c \,x^{2}+b}\, b^{2} c \right )}{8 \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} c^{\frac {9}{2}}}\)	\(87\)
risch	\(-\frac {x^{2} \left (-2 c \,x^{2}+7 b \right ) \left (c \,x^{2}+b \right )}{8 c^{3} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}+\frac {\left (-\frac {b^{2} x}{c^{3} \sqrt {c \,x^{2}+b}}+\frac {15 b^{2} \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right )}{8 c^{\frac {7}{2}}}\right ) x \sqrt {c \,x^{2}+b}}{\sqrt {x^{2} \left (c \,x^{2}+b \right )}}\)	\(107\)
pseudoelliptic	\(\frac {\frac {c^{\frac {5}{2}} x^{6}}{4}-\frac {5 b \,c^{\frac {3}{2}} x^{4}}{8}-\frac {15 b^{2} x^{2} \sqrt {c}}{8}+\frac {15 \ln \left (\frac {2 c \,x^{2}+2 \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \sqrt {c}+b}{\sqrt {c}}\right ) b^{2} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{16}-\frac {15 \ln \left (2\right ) b^{2} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{16}}{c^{\frac {7}{2}} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}\)	\(116\)

Input:

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

Output:

1/8*x^3*(c*x^2+b)*(2*c^(7/2)*x^5-5*c^(5/2)*b*x^3-15*c^(3/2)*b^2*x+15*ln(c^ 
(1/2)*x+(c*x^2+b)^(1/2))*(c*x^2+b)^(1/2)*b^2*c)/(c*x^4+b*x^2)^(3/2)/c^(9/2 
)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.87 \[ \int \frac {x^9}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\left [\frac {15 \, {\left (b^{2} c x^{2} + b^{3}\right )} \sqrt {c} \log \left (-2 \, c x^{2} - b - 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right ) + 2 \, {\left (2 \, c^{3} x^{4} - 5 \, b c^{2} x^{2} - 15 \, b^{2} c\right )} \sqrt {c x^{4} + b x^{2}}}{16 \, {\left (c^{5} x^{2} + b c^{4}\right )}}, -\frac {15 \, {\left (b^{2} c x^{2} + b^{3}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-c}}{c x^{2} + b}\right ) - {\left (2 \, c^{3} x^{4} - 5 \, b c^{2} x^{2} - 15 \, b^{2} c\right )} \sqrt {c x^{4} + b x^{2}}}{8 \, {\left (c^{5} x^{2} + b c^{4}\right )}}\right ] \] Input:

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

Output:

[1/16*(15*(b^2*c*x^2 + b^3)*sqrt(c)*log(-2*c*x^2 - b - 2*sqrt(c*x^4 + b*x^ 
2)*sqrt(c)) + 2*(2*c^3*x^4 - 5*b*c^2*x^2 - 15*b^2*c)*sqrt(c*x^4 + b*x^2))/ 
(c^5*x^2 + b*c^4), -1/8*(15*(b^2*c*x^2 + b^3)*sqrt(-c)*arctan(sqrt(c*x^4 + 
 b*x^2)*sqrt(-c)/(c*x^2 + b)) - (2*c^3*x^4 - 5*b*c^2*x^2 - 15*b^2*c)*sqrt( 
c*x^4 + b*x^2))/(c^5*x^2 + b*c^4)]
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.92 \[ \int \frac {x^9}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {x^{6}}{4 \, \sqrt {c x^{4} + b x^{2}} c} - \frac {5 \, b x^{4}}{8 \, \sqrt {c x^{4} + b x^{2}} c^{2}} - \frac {15 \, b^{2} x^{2}}{8 \, \sqrt {c x^{4} + b x^{2}} c^{3}} + \frac {15 \, b^{2} \log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right )}{16 \, c^{\frac {7}{2}}} \] Input:

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

Output:

1/4*x^6/(sqrt(c*x^4 + b*x^2)*c) - 5/8*b*x^4/(sqrt(c*x^4 + b*x^2)*c^2) - 15 
/8*b^2*x^2/(sqrt(c*x^4 + b*x^2)*c^3) + 15/16*b^2*log(2*c*x^2 + b + 2*sqrt( 
c*x^4 + b*x^2)*sqrt(c))/c^(7/2)
 

Giac [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.86 \[ \int \frac {x^9}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {{\left (x^{2} {\left (\frac {2 \, x^{2}}{c \mathrm {sgn}\left (x\right )} - \frac {5 \, b}{c^{2} \mathrm {sgn}\left (x\right )}\right )} - \frac {15 \, b^{2}}{c^{3} \mathrm {sgn}\left (x\right )}\right )} x}{8 \, \sqrt {c x^{2} + b}} + \frac {15 \, b^{2} \log \left ({\left | b \right |}\right ) \mathrm {sgn}\left (x\right )}{16 \, c^{\frac {7}{2}}} - \frac {15 \, b^{2} \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + b} \right |}\right )}{8 \, c^{\frac {7}{2}} \mathrm {sgn}\left (x\right )} \] Input:

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

Output:

1/8*(x^2*(2*x^2/(c*sgn(x)) - 5*b/(c^2*sgn(x))) - 15*b^2/(c^3*sgn(x)))*x/sq 
rt(c*x^2 + b) + 15/16*b^2*log(abs(b))*sgn(x)/c^(7/2) - 15/8*b^2*log(abs(-s 
qrt(c)*x + sqrt(c*x^2 + b)))/(c^(7/2)*sgn(x))
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.22 \[ \int \frac {x^9}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {-15 \sqrt {c \,x^{2}+b}\, b^{2} c x -5 \sqrt {c \,x^{2}+b}\, b \,c^{2} x^{3}+2 \sqrt {c \,x^{2}+b}\, c^{3} x^{5}+15 \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}+\sqrt {c}\, x}{\sqrt {b}}\right ) b^{3}+15 \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}+\sqrt {c}\, x}{\sqrt {b}}\right ) b^{2} c \,x^{2}-10 \sqrt {c}\, b^{3}-10 \sqrt {c}\, b^{2} c \,x^{2}}{8 c^{4} \left (c \,x^{2}+b \right )} \] Input:

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

Output:

( - 15*sqrt(b + c*x**2)*b**2*c*x - 5*sqrt(b + c*x**2)*b*c**2*x**3 + 2*sqrt 
(b + c*x**2)*c**3*x**5 + 15*sqrt(c)*log((sqrt(b + c*x**2) + sqrt(c)*x)/sqr 
t(b))*b**3 + 15*sqrt(c)*log((sqrt(b + c*x**2) + sqrt(c)*x)/sqrt(b))*b**2*c 
*x**2 - 10*sqrt(c)*b**3 - 10*sqrt(c)*b**2*c*x**2)/(8*c**4*(b + c*x**2))