\(\int \frac {(A+B x^2) (b x^2+c x^4)^{3/2}}{x^9} \, dx\) [178]

Optimal result

Integrand size = 26, antiderivative size = 104 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^9} \, dx=-\frac {B c \sqrt {b x^2+c x^4}}{x^2}-\frac {B \left (b x^2+c x^4\right )^{3/2}}{3 x^6}-\frac {A \left (b x^2+c x^4\right )^{5/2}}{5 b x^{10}}+B c^{3/2} \text {arctanh}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right ) \] Output:

-B*c*(c*x^4+b*x^2)^(1/2)/x^2-1/3*B*(c*x^4+b*x^2)^(3/2)/x^6-1/5*A*(c*x^4+b* 
x^2)^(5/2)/b/x^10+B*c^(3/2)*arctanh(c^(1/2)*x^2/(c*x^4+b*x^2)^(1/2))
 

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.06 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^9} \, dx=-\frac {\sqrt {x^2 \left (b+c x^2\right )} \left (\sqrt {b+c x^2} \left (3 A \left (b+c x^2\right )^2+5 b B x^2 \left (b+4 c x^2\right )\right )+15 b B c^{3/2} x^5 \log \left (-\sqrt {c} x+\sqrt {b+c x^2}\right )\right )}{15 b x^6 \sqrt {b+c x^2}} \] Input:

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

Output:

-1/15*(Sqrt[x^2*(b + c*x^2)]*(Sqrt[b + c*x^2]*(3*A*(b + c*x^2)^2 + 5*b*B*x 
^2*(b + 4*c*x^2)) + 15*b*B*c^(3/2)*x^5*Log[-(Sqrt[c]*x) + Sqrt[b + c*x^2]] 
))/(b*x^6*Sqrt[b + c*x^2])
 

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.07, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {1940, 1220, 1130, 1125, 25, 27, 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.

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

\(\Big \downarrow \) 1940

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

\(\Big \downarrow \) 1220

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

\(\Big \downarrow \) 1130

\(\displaystyle \frac {1}{2} \left (B \left (c \int \frac {\sqrt {c x^4+b x^2}}{x^4}dx^2-\frac {2 \left (b x^2+c x^4\right )^{3/2}}{3 x^6}\right )-\frac {2 A \left (b x^2+c x^4\right )^{5/2}}{5 b x^{10}}\right )\)

\(\Big \downarrow \) 1125

\(\displaystyle \frac {1}{2} \left (B \left (c \left (-\int -\frac {c}{\sqrt {c x^4+b x^2}}dx^2-\frac {2 \sqrt {b x^2+c x^4}}{x^2}\right )-\frac {2 \left (b x^2+c x^4\right )^{3/2}}{3 x^6}\right )-\frac {2 A \left (b x^2+c x^4\right )^{5/2}}{5 b x^{10}}\right )\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{2} \left (B \left (c \left (\int \frac {c}{\sqrt {c x^4+b x^2}}dx^2-\frac {2 \sqrt {b x^2+c x^4}}{x^2}\right )-\frac {2 \left (b x^2+c x^4\right )^{3/2}}{3 x^6}\right )-\frac {2 A \left (b x^2+c x^4\right )^{5/2}}{5 b x^{10}}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{2} \left (B \left (c \left (c \int \frac {1}{\sqrt {c x^4+b x^2}}dx^2-\frac {2 \sqrt {b x^2+c x^4}}{x^2}\right )-\frac {2 \left (b x^2+c x^4\right )^{3/2}}{3 x^6}\right )-\frac {2 A \left (b x^2+c x^4\right )^{5/2}}{5 b x^{10}}\right )\)

\(\Big \downarrow \) 1091

\(\displaystyle \frac {1}{2} \left (B \left (c \left (2 c \int \frac {1}{1-c x^4}d\frac {x^2}{\sqrt {c x^4+b x^2}}-\frac {2 \sqrt {b x^2+c x^4}}{x^2}\right )-\frac {2 \left (b x^2+c x^4\right )^{3/2}}{3 x^6}\right )-\frac {2 A \left (b x^2+c x^4\right )^{5/2}}{5 b x^{10}}\right )\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{2} \left (B \left (c \left (2 \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )-\frac {2 \sqrt {b x^2+c x^4}}{x^2}\right )-\frac {2 \left (b x^2+c x^4\right )^{3/2}}{3 x^6}\right )-\frac {2 A \left (b x^2+c x^4\right )^{5/2}}{5 b x^{10}}\right )\)

Input:

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

Output:

((-2*A*(b*x^2 + c*x^4)^(5/2))/(5*b*x^10) + B*((-2*(b*x^2 + c*x^4)^(3/2))/( 
3*x^6) + c*((-2*Sqrt[b*x^2 + c*x^4])/x^2 + 2*Sqrt[c]*ArcTanh[(Sqrt[c]*x^2) 
/Sqrt[b*x^2 + c*x^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 1125

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

rule 1130

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S 
ymbol] :> Simp[(d + e*x)^(m + 1)*((a + b*x + c*x^2)^p/(e*(m + p + 1))), x] 
- Simp[c*(p/(e^2*(m + p + 1)))   Int[(d + e*x)^(m + 2)*(a + b*x + c*x^2)^(p 
 - 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] 
 && GtQ[p, 0] && (LtQ[m, -2] || EqQ[m + 2*p + 1, 0]) && NeQ[m + p + 1, 0] & 
& IntegerQ[2*p]
 

rule 1220

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c 
_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d*g - e*f)*(d + e*x)^m*((a + b*x + c*x 
^2)^(p + 1)/((2*c*d - b*e)*(m + p + 1))), x] + Simp[(m*(g*(c*d - b*e) + c*e 
*f) + e*(p + 1)*(2*c*f - b*g))/(e*(2*c*d - b*e)*(m + p + 1))   Int[(d + e*x 
)^(m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, 
 x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && ((LtQ[m, -1] &&  !IGtQ[m + p + 1, 0 
]) || (LtQ[m, 0] && LtQ[p, -1]) || EqQ[m + 2*p + 2, 0]) && NeQ[m + p + 1, 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 = 0.47 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.03

method	result	size
pseudoelliptic	\(\frac {5 x^{6} B b \left (-\ln \left (2\right )+\ln \left (\frac {2 c \,x^{2}+2 \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \sqrt {c}+b}{\sqrt {c}}\right )\right ) c^{\frac {3}{2}}-2 \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \left (\left (\frac {5 B \,x^{2}}{3}+A \right ) b^{2}+2 c \,x^{2} \left (\frac {10 B \,x^{2}}{3}+A \right ) b +x^{4} A \,c^{2}\right )}{10 b \,x^{6}}\)	\(107\)
risch	\(-\frac {\left (3 x^{4} A \,c^{2}+20 x^{4} B b c +6 A b c \,x^{2}+5 x^{2} B \,b^{2}+3 b^{2} A \right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{15 x^{6} b}+\frac {B \,c^{\frac {3}{2}} \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{x \sqrt {c \,x^{2}+b}}\)	\(110\)
default	\(-\frac {\left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} \left (-10 B \left (c \,x^{2}+b \right )^{\frac {3}{2}} c^{\frac {5}{2}} x^{6}+10 B \left (c \,x^{2}+b \right )^{\frac {5}{2}} c^{\frac {3}{2}} x^{4}-15 B \sqrt {c \,x^{2}+b}\, c^{\frac {5}{2}} b \,x^{6}-15 B \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right ) b^{2} c^{2} x^{5}+5 B \sqrt {c}\, \left (c \,x^{2}+b \right )^{\frac {5}{2}} b \,x^{2}+3 A \sqrt {c}\, \left (c \,x^{2}+b \right )^{\frac {5}{2}} b \right )}{15 x^{8} \left (c \,x^{2}+b \right )^{\frac {3}{2}} b^{2} \sqrt {c}}\)	\(153\)

Input:

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

Output:

1/10*(5*x^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)))*c^(3/2)-2*(x^2*(c*x^2+b))^(1/2)*((5/3*B*x^2+A)*b^2+2*c*x^2*(10/3*B* 
x^2+A)*b+x^4*A*c^2))/b/x^6
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.99 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^9} \, dx=\left [\frac {15 \, B b c^{\frac {3}{2}} x^{6} \log \left (-2 \, c x^{2} - b - 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right ) - 2 \, {\left ({\left (20 \, B b c + 3 \, A c^{2}\right )} x^{4} + 3 \, A b^{2} + {\left (5 \, B b^{2} + 6 \, A b c\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{30 \, b x^{6}}, -\frac {15 \, B b \sqrt {-c} c x^{6} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-c}}{c x^{2} + b}\right ) + {\left ({\left (20 \, B b c + 3 \, A c^{2}\right )} x^{4} + 3 \, A b^{2} + {\left (5 \, B b^{2} + 6 \, A b c\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{15 \, b x^{6}}\right ] \] Input:

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

Output:

[1/30*(15*B*b*c^(3/2)*x^6*log(-2*c*x^2 - b - 2*sqrt(c*x^4 + b*x^2)*sqrt(c) 
) - 2*((20*B*b*c + 3*A*c^2)*x^4 + 3*A*b^2 + (5*B*b^2 + 6*A*b*c)*x^2)*sqrt( 
c*x^4 + b*x^2))/(b*x^6), -1/15*(15*B*b*sqrt(-c)*c*x^6*arctan(sqrt(c*x^4 + 
b*x^2)*sqrt(-c)/(c*x^2 + b)) + ((20*B*b*c + 3*A*c^2)*x^4 + 3*A*b^2 + (5*B* 
b^2 + 6*A*b*c)*x^2)*sqrt(c*x^4 + b*x^2))/(b*x^6)]
 

Sympy [F]

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

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

Output:

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 177 vs. \(2 (88) = 176\).

Time = 0.05 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.70 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^9} \, dx=\frac {1}{6} \, {\left (3 \, c^{\frac {3}{2}} \log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right ) - \frac {7 \, \sqrt {c x^{4} + b x^{2}} c}{x^{2}} - \frac {\sqrt {c x^{4} + b x^{2}} b}{x^{4}} - \frac {{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}}}{x^{6}}\right )} B - \frac {1}{10} \, A {\left (\frac {2 \, \sqrt {c x^{4} + b x^{2}} c^{2}}{b x^{2}} - \frac {\sqrt {c x^{4} + b x^{2}} c}{x^{4}} - \frac {3 \, \sqrt {c x^{4} + b x^{2}} b}{x^{6}} + \frac {5 \, {\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}}}{x^{8}}\right )} \] Input:

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

Output:

1/6*(3*c^(3/2)*log(2*c*x^2 + b + 2*sqrt(c*x^4 + b*x^2)*sqrt(c)) - 7*sqrt(c 
*x^4 + b*x^2)*c/x^2 - sqrt(c*x^4 + b*x^2)*b/x^4 - (c*x^4 + b*x^2)^(3/2)/x^ 
6)*B - 1/10*A*(2*sqrt(c*x^4 + b*x^2)*c^2/(b*x^2) - sqrt(c*x^4 + b*x^2)*c/x 
^4 - 3*sqrt(c*x^4 + b*x^2)*b/x^6 + 5*(c*x^4 + b*x^2)^(3/2)/x^8)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 254 vs. \(2 (88) = 176\).

Time = 0.70 (sec) , antiderivative size = 254, normalized size of antiderivative = 2.44 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^9} \, dx=-\frac {1}{2} \, B c^{\frac {3}{2}} \log \left ({\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2}\right ) \mathrm {sgn}\left (x\right ) + \frac {2 \, {\left (30 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{8} B b c^{\frac {3}{2}} \mathrm {sgn}\left (x\right ) + 15 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{8} A c^{\frac {5}{2}} \mathrm {sgn}\left (x\right ) - 90 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{6} B b^{2} c^{\frac {3}{2}} \mathrm {sgn}\left (x\right ) + 110 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{4} B b^{3} c^{\frac {3}{2}} \mathrm {sgn}\left (x\right ) + 30 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{4} A b^{2} c^{\frac {5}{2}} \mathrm {sgn}\left (x\right ) - 70 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} B b^{4} c^{\frac {3}{2}} \mathrm {sgn}\left (x\right ) + 20 \, B b^{5} c^{\frac {3}{2}} \mathrm {sgn}\left (x\right ) + 3 \, A b^{4} c^{\frac {5}{2}} \mathrm {sgn}\left (x\right )\right )}}{15 \, {\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} - b\right )}^{5}} \] Input:

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

Output:

-1/2*B*c^(3/2)*log((sqrt(c)*x - sqrt(c*x^2 + b))^2)*sgn(x) + 2/15*(30*(sqr 
t(c)*x - sqrt(c*x^2 + b))^8*B*b*c^(3/2)*sgn(x) + 15*(sqrt(c)*x - sqrt(c*x^ 
2 + b))^8*A*c^(5/2)*sgn(x) - 90*(sqrt(c)*x - sqrt(c*x^2 + b))^6*B*b^2*c^(3 
/2)*sgn(x) + 110*(sqrt(c)*x - sqrt(c*x^2 + b))^4*B*b^3*c^(3/2)*sgn(x) + 30 
*(sqrt(c)*x - sqrt(c*x^2 + b))^4*A*b^2*c^(5/2)*sgn(x) - 70*(sqrt(c)*x - sq 
rt(c*x^2 + b))^2*B*b^4*c^(3/2)*sgn(x) + 20*B*b^5*c^(3/2)*sgn(x) + 3*A*b^4* 
c^(5/2)*sgn(x))/((sqrt(c)*x - sqrt(c*x^2 + b))^2 - b)^5
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.36 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^9} \, dx=\frac {-3 \sqrt {c \,x^{2}+b}\, a \,b^{2}-6 \sqrt {c \,x^{2}+b}\, a b c \,x^{2}-3 \sqrt {c \,x^{2}+b}\, a \,c^{2} x^{4}-5 \sqrt {c \,x^{2}+b}\, b^{3} x^{2}-20 \sqrt {c \,x^{2}+b}\, b^{2} c \,x^{4}+15 \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}+\sqrt {c}\, x}{\sqrt {b}}\right ) b^{2} c \,x^{5}-3 \sqrt {c}\, a \,c^{2} x^{5}+8 \sqrt {c}\, b^{2} c \,x^{5}}{15 b \,x^{5}} \] Input:

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

Output:

( - 3*sqrt(b + c*x**2)*a*b**2 - 6*sqrt(b + c*x**2)*a*b*c*x**2 - 3*sqrt(b + 
 c*x**2)*a*c**2*x**4 - 5*sqrt(b + c*x**2)*b**3*x**2 - 20*sqrt(b + c*x**2)* 
b**2*c*x**4 + 15*sqrt(c)*log((sqrt(b + c*x**2) + sqrt(c)*x)/sqrt(b))*b**2* 
c*x**5 - 3*sqrt(c)*a*c**2*x**5 + 8*sqrt(c)*b**2*c*x**5)/(15*b*x**5)