\(\int \frac {(A+B x^2) (b x^2+c x^4)^{3/2}}{x^5} \, dx\) [176]

Optimal result

Integrand size = 26, antiderivative size = 128 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^5} \, dx=\frac {3}{8} (b B+4 A c) \sqrt {b x^2+c x^4}+\frac {(b B+4 A c) \left (b x^2+c x^4\right )^{3/2}}{4 b x^2}-\frac {A \left (b x^2+c x^4\right )^{5/2}}{b x^6}+\frac {3 b (b B+4 A c) \text {arctanh}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{8 \sqrt {c}} \] Output:

3/8*(4*A*c+B*b)*(c*x^4+b*x^2)^(1/2)+1/4*(4*A*c+B*b)*(c*x^4+b*x^2)^(3/2)/b/ 
x^2-A*(c*x^4+b*x^2)^(5/2)/b/x^6+3/8*b*(4*A*c+B*b)*arctanh(c^(1/2)*x^2/(c*x 
^4+b*x^2)^(1/2))/c^(1/2)
 

Mathematica [A] (verified)

Time = 0.32 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.91 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^5} \, dx=\frac {\sqrt {c} \left (b+c x^2\right ) \left (-8 A b+5 b B x^2+4 A c x^2+2 B c x^4\right )+6 b (b B+4 A c) x \sqrt {b+c x^2} \text {arctanh}\left (\frac {\sqrt {c} x}{-\sqrt {b}+\sqrt {b+c x^2}}\right )}{8 \sqrt {c} \sqrt {x^2 \left (b+c x^2\right )}} \] Input:

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

Output:

(Sqrt[c]*(b + c*x^2)*(-8*A*b + 5*b*B*x^2 + 4*A*c*x^2 + 2*B*c*x^4) + 6*b*(b 
*B + 4*A*c)*x*Sqrt[b + c*x^2]*ArcTanh[(Sqrt[c]*x)/(-Sqrt[b] + Sqrt[b + c*x 
^2])])/(8*Sqrt[c]*Sqrt[x^2*(b + c*x^2)])
 

Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.91, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {1940, 1220, 1131, 1131, 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.

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

\(\Big \downarrow \) 1940

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

\(\Big \downarrow \) 1220

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

\(\Big \downarrow \) 1131

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

\(\Big \downarrow \) 1131

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

\(\Big \downarrow \) 1091

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

((-2*A*(b*x^2 + c*x^4)^(5/2))/(b*x^6) + ((b*B + 4*A*c)*((b*x^2 + c*x^4)^(3 
/2)/(2*x^2) + (3*b*(Sqrt[b*x^2 + c*x^4] + (b*ArcTanh[(Sqrt[c]*x^2)/Sqrt[b* 
x^2 + c*x^4]])/Sqrt[c]))/4))/b)/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 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 1131

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 + 2*p + 1))), x 
] - Simp[p*((2*c*d - b*e)/(e^2*(m + 2*p + 1)))   Int[(d + e*x)^(m + 1)*(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] && (LeQ[-2, m, 0] || EqQ[m + p + 1, 0]) && Ne 
Q[m + 2*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 = 101, normalized size of antiderivative = 0.79

method	result	size
risch	\(-\frac {\left (-2 B c \,x^{4}-4 A c \,x^{2}-5 B b \,x^{2}+8 A b \right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{8 x^{2}}+\frac {3 b \left (4 A c +B b \right ) \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{8 \sqrt {c}\, x \sqrt {c \,x^{2}+b}}\)	\(101\)
pseudoelliptic	\(\frac {\frac {\left (\frac {B \,x^{2}}{2}+A \right ) x^{2} \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, c^{\frac {3}{2}}}{2}+\frac {3 \left (-\frac {4 \left (-\frac {5 B \,x^{2}}{8}+A \right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \sqrt {c}}{3}+x^{2} \left (-\ln \left (2\right )+\ln \left (\frac {2 c \,x^{2}+2 \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \sqrt {c}+b}{\sqrt {c}}\right )\right ) \left (A c +\frac {B b}{4}\right )\right ) b}{4}}{\sqrt {c}\, x^{2}}\)	\(116\)
default	\(\frac {\left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} \left (8 A \,c^{\frac {3}{2}} \left (c \,x^{2}+b \right )^{\frac {3}{2}} x^{2}+12 A \,c^{\frac {3}{2}} \sqrt {c \,x^{2}+b}\, b \,x^{2}+2 B \sqrt {c}\, \left (c \,x^{2}+b \right )^{\frac {3}{2}} b \,x^{2}-8 A \sqrt {c}\, \left (c \,x^{2}+b \right )^{\frac {5}{2}}+3 B \sqrt {c}\, \sqrt {c \,x^{2}+b}\, b^{2} x^{2}+12 A \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right ) b^{2} c x +3 B \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right ) b^{3} x \right )}{8 x^{4} \left (c \,x^{2}+b \right )^{\frac {3}{2}} b \sqrt {c}}\)	\(174\)

Input:

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

Output:

-1/8*(-2*B*c*x^4-4*A*c*x^2-5*B*b*x^2+8*A*b)/x^2*(x^2*(c*x^2+b))^(1/2)+3/8* 
b*(4*A*c+B*b)*ln(c^(1/2)*x+(c*x^2+b)^(1/2))/c^(1/2)*(x^2*(c*x^2+b))^(1/2)/ 
x/(c*x^2+b)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.63 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^5} \, dx=\left [\frac {3 \, {\left (B b^{2} + 4 \, A b c\right )} \sqrt {c} x^{2} \log \left (-2 \, c x^{2} - b - 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right ) + 2 \, {\left (2 \, B c^{2} x^{4} - 8 \, A b c + {\left (5 \, B b c + 4 \, A c^{2}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{16 \, c x^{2}}, -\frac {3 \, {\left (B b^{2} + 4 \, A b c\right )} \sqrt {-c} x^{2} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-c}}{c x^{2} + b}\right ) - {\left (2 \, B c^{2} x^{4} - 8 \, A b c + {\left (5 \, B b c + 4 \, A c^{2}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{8 \, c x^{2}}\right ] \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.16 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^5} \, dx=\frac {1}{4} \, {\left (3 \, b \sqrt {c} \log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right ) - \frac {6 \, \sqrt {c x^{4} + b x^{2}} b}{x^{2}} + \frac {2 \, {\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}}}{x^{4}}\right )} A + \frac {1}{16} \, {\left (\frac {3 \, b^{2} \log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right )}{\sqrt {c}} + 6 \, \sqrt {c x^{4} + b x^{2}} b + \frac {4 \, {\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}}}{x^{2}}\right )} B \] Input:

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

Output:

1/4*(3*b*sqrt(c)*log(2*c*x^2 + b + 2*sqrt(c*x^4 + b*x^2)*sqrt(c)) - 6*sqrt 
(c*x^4 + b*x^2)*b/x^2 + 2*(c*x^4 + b*x^2)^(3/2)/x^4)*A + 1/16*(3*b^2*log(2 
*c*x^2 + b + 2*sqrt(c*x^4 + b*x^2)*sqrt(c))/sqrt(c) + 6*sqrt(c*x^4 + b*x^2 
)*b + 4*(c*x^4 + b*x^2)^(3/2)/x^2)*B
 

Giac [A] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.95 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^5} \, dx=\frac {2 \, A b^{2} \sqrt {c} \mathrm {sgn}\left (x\right )}{{\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} - b} + \frac {1}{8} \, {\left (2 \, B c x^{2} \mathrm {sgn}\left (x\right ) + \frac {5 \, B b c^{2} \mathrm {sgn}\left (x\right ) + 4 \, A c^{3} \mathrm {sgn}\left (x\right )}{c^{2}}\right )} \sqrt {c x^{2} + b} x - \frac {3 \, {\left (B b^{2} \mathrm {sgn}\left (x\right ) + 4 \, A b c \mathrm {sgn}\left (x\right )\right )} \log \left ({\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2}\right )}{16 \, \sqrt {c}} \] Input:

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

Output:

2*A*b^2*sqrt(c)*sgn(x)/((sqrt(c)*x - sqrt(c*x^2 + b))^2 - b) + 1/8*(2*B*c* 
x^2*sgn(x) + (5*B*b*c^2*sgn(x) + 4*A*c^3*sgn(x))/c^2)*sqrt(c*x^2 + b)*x - 
3/16*(B*b^2*sgn(x) + 4*A*b*c*sgn(x))*log((sqrt(c)*x - sqrt(c*x^2 + b))^2)/ 
sqrt(c)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.12 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^5} \, dx=\frac {-8 \sqrt {c \,x^{2}+b}\, a b c +4 \sqrt {c \,x^{2}+b}\, a \,c^{2} x^{2}+5 \sqrt {c \,x^{2}+b}\, b^{2} c \,x^{2}+2 \sqrt {c \,x^{2}+b}\, b \,c^{2} x^{4}+12 \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}+\sqrt {c}\, x}{\sqrt {b}}\right ) a b c x +3 \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}+\sqrt {c}\, x}{\sqrt {b}}\right ) b^{3} x -9 \sqrt {c}\, a b c x -\sqrt {c}\, b^{3} x}{8 c x} \] Input:

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

Output:

( - 8*sqrt(b + c*x**2)*a*b*c + 4*sqrt(b + c*x**2)*a*c**2*x**2 + 5*sqrt(b + 
 c*x**2)*b**2*c*x**2 + 2*sqrt(b + c*x**2)*b*c**2*x**4 + 12*sqrt(c)*log((sq 
rt(b + c*x**2) + sqrt(c)*x)/sqrt(b))*a*b*c*x + 3*sqrt(c)*log((sqrt(b + c*x 
**2) + sqrt(c)*x)/sqrt(b))*b**3*x - 9*sqrt(c)*a*b*c*x - sqrt(c)*b**3*x)/(8 
*c*x)