\(\int \frac {(A+B x^2) (b x^2+c x^4)^{3/2}}{x^3} \, dx\) [175]

Optimal result

Integrand size = 26, antiderivative size = 137 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^3} \, dx=\frac {(b B-6 A c) \left (b+2 c x^2\right ) \sqrt {b x^2+c x^4}}{16 c}+\frac {(b B-6 A c) \left (b x^2+c x^4\right )^{3/2}}{6 b}+\frac {A \left (b x^2+c x^4\right )^{5/2}}{b x^4}-\frac {b^2 (b B-6 A c) \text {arctanh}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{16 c^{3/2}} \] Output:

1/16*(-6*A*c+B*b)*(2*c*x^2+b)*(c*x^4+b*x^2)^(1/2)/c+1/6*(-6*A*c+B*b)*(c*x^ 
4+b*x^2)^(3/2)/b+A*(c*x^4+b*x^2)^(5/2)/b/x^4-1/16*b^2*(-6*A*c+B*b)*arctanh 
(c^(1/2)*x^2/(c*x^4+b*x^2)^(1/2))/c^(3/2)
 

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.91 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^3} \, dx=\frac {x \left (\sqrt {c} x \left (b+c x^2\right ) \left (3 b^2 B+4 c^2 x^2 \left (3 A+2 B x^2\right )+2 b c \left (15 A+7 B x^2\right )\right )+3 b^2 (b B-6 A c) \sqrt {b+c x^2} \log \left (-\sqrt {c} x+\sqrt {b+c x^2}\right )\right )}{48 c^{3/2} \sqrt {x^2 \left (b+c x^2\right )}} \] Input:

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

Output:

(x*(Sqrt[c]*x*(b + c*x^2)*(3*b^2*B + 4*c^2*x^2*(3*A + 2*B*x^2) + 2*b*c*(15 
*A + 7*B*x^2)) + 3*b^2*(b*B - 6*A*c)*Sqrt[b + c*x^2]*Log[-(Sqrt[c]*x) + Sq 
rt[b + c*x^2]]))/(48*c^(3/2)*Sqrt[x^2*(b + c*x^2)])
 

Rubi [A] (verified)

Time = 0.54 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.98, 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, 1087, 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.

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

\(\Big \downarrow \) 1940

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

\(\Big \downarrow \) 1220

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

\(\Big \downarrow \) 1131

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

\(\Big \downarrow \) 1087

\(\displaystyle \frac {1}{2} \left (\frac {(b B-6 A c) \left (\frac {1}{2} b \left (\frac {\left (b+2 c x^2\right ) \sqrt {b x^2+c x^4}}{4 c}-\frac {b^2 \int \frac {1}{\sqrt {c x^4+b x^2}}dx^2}{8 c}\right )+\frac {1}{3} \left (b x^2+c x^4\right )^{3/2}\right )}{b}+\frac {2 A \left (b x^2+c x^4\right )^{5/2}}{b x^4}\right )\)

\(\Big \downarrow \) 1091

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

((2*A*(b*x^2 + c*x^4)^(5/2))/(b*x^4) + ((b*B - 6*A*c)*((b*x^2 + c*x^4)^(3/ 
2)/3 + (b*(((b + 2*c*x^2)*Sqrt[b*x^2 + c*x^4])/(4*c) - (b^2*ArcTanh[(Sqrt[ 
c]*x^2)/Sqrt[b*x^2 + c*x^4]])/(4*c^(3/2))))/2))/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 1087

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(b + 2*c*x) 
*((a + b*x + c*x^2)^p/(2*c*(2*p + 1))), x] - Simp[p*((b^2 - 4*a*c)/(2*c*(2* 
p + 1)))   Int[(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && 
GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[3*p])
 

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 = 116, normalized size of antiderivative = 0.85

method	result	size
risch	\(\frac {\left (8 B \,c^{2} x^{4}+12 A \,c^{2} x^{2}+14 x^{2} B b c +30 A b c +3 B \,b^{2}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{48 c}+\frac {b^{2} \left (6 A c -B b \right ) \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{16 c^{\frac {3}{2}} x \sqrt {c \,x^{2}+b}}\)	\(116\)
default	\(\frac {\left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} \left (8 B \left (c \,x^{2}+b \right )^{\frac {5}{2}} \sqrt {c}\, x +12 A \left (c \,x^{2}+b \right )^{\frac {3}{2}} c^{\frac {3}{2}} x -2 B \left (c \,x^{2}+b \right )^{\frac {3}{2}} \sqrt {c}\, b x +18 A \sqrt {c \,x^{2}+b}\, c^{\frac {3}{2}} b x -3 B \sqrt {c \,x^{2}+b}\, \sqrt {c}\, b^{2} x +18 A \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right ) b^{2} c -3 B \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right ) b^{3}\right )}{48 x^{3} \left (c \,x^{2}+b \right )^{\frac {3}{2}} c^{\frac {3}{2}}}\)	\(162\)
pseudoelliptic	\(\frac {16 B \,c^{\frac {5}{2}} x^{4} \sqrt {x^{2} \left (c \,x^{2}+b \right )}+24 A \,c^{\frac {5}{2}} \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, x^{2}+28 B \,c^{\frac {3}{2}} \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, b \,x^{2}-18 A \ln \left (2\right ) b^{2} c +18 A \ln \left (\frac {2 c \,x^{2}+2 \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \sqrt {c}+b}{\sqrt {c}}\right ) b^{2} c +60 A \,c^{\frac {3}{2}} \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, b +3 B \ln \left (2\right ) b^{3}-3 B \ln \left (\frac {2 c \,x^{2}+2 \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \sqrt {c}+b}{\sqrt {c}}\right ) b^{3}+6 B \sqrt {c}\, \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, b^{2}}{96 c^{\frac {3}{2}}}\)	\(208\)

Input:

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

Output:

1/48/c*(8*B*c^2*x^4+12*A*c^2*x^2+14*B*b*c*x^2+30*A*b*c+3*B*b^2)*(x^2*(c*x^ 
2+b))^(1/2)+1/16*b^2*(6*A*c-B*b)/c^(3/2)*ln(c^(1/2)*x+(c*x^2+b)^(1/2))*(x^ 
2*(c*x^2+b))^(1/2)/x/(c*x^2+b)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.64 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^3} \, dx=\left [-\frac {3 \, {\left (B b^{3} - 6 \, A b^{2} c\right )} \sqrt {c} \log \left (-2 \, c x^{2} - b - 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right ) - 2 \, {\left (8 \, B c^{3} x^{4} + 3 \, B b^{2} c + 30 \, A b c^{2} + 2 \, {\left (7 \, B b c^{2} + 6 \, A c^{3}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{96 \, c^{2}}, \frac {3 \, {\left (B b^{3} - 6 \, A b^{2} c\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-c}}{c x^{2} + b}\right ) + {\left (8 \, B c^{3} x^{4} + 3 \, B b^{2} c + 30 \, A b c^{2} + 2 \, {\left (7 \, B b c^{2} + 6 \, A c^{3}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{48 \, c^{2}}\right ] \] Input:

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

Output:

[-1/96*(3*(B*b^3 - 6*A*b^2*c)*sqrt(c)*log(-2*c*x^2 - b - 2*sqrt(c*x^4 + b* 
x^2)*sqrt(c)) - 2*(8*B*c^3*x^4 + 3*B*b^2*c + 30*A*b*c^2 + 2*(7*B*b*c^2 + 6 
*A*c^3)*x^2)*sqrt(c*x^4 + b*x^2))/c^2, 1/48*(3*(B*b^3 - 6*A*b^2*c)*sqrt(-c 
)*arctan(sqrt(c*x^4 + b*x^2)*sqrt(-c)/(c*x^2 + b)) + (8*B*c^3*x^4 + 3*B*b^ 
2*c + 30*A*b*c^2 + 2*(7*B*b*c^2 + 6*A*c^3)*x^2)*sqrt(c*x^4 + b*x^2))/c^2]
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.23 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^3} \, dx=\frac {1}{16} \, {\left (\frac {3 \, b^{2} \log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right )}{\sqrt {c}} + 6 \, \sqrt {c x^{4} + b x^{2}} b + \frac {4 \, {\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}}}{x^{2}}\right )} A + \frac {1}{96} \, {\left (12 \, \sqrt {c x^{4} + b x^{2}} b x^{2} - \frac {3 \, b^{3} \log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right )}{c^{\frac {3}{2}}} + 16 \, {\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} + \frac {6 \, \sqrt {c x^{4} + b x^{2}} b^{2}}{c}\right )} B \] Input:

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

Output:

1/16*(3*b^2*log(2*c*x^2 + b + 2*sqrt(c*x^4 + b*x^2)*sqrt(c))/sqrt(c) + 6*s 
qrt(c*x^4 + b*x^2)*b + 4*(c*x^4 + b*x^2)^(3/2)/x^2)*A + 1/96*(12*sqrt(c*x^ 
4 + b*x^2)*b*x^2 - 3*b^3*log(2*c*x^2 + b + 2*sqrt(c*x^4 + b*x^2)*sqrt(c))/ 
c^(3/2) + 16*(c*x^4 + b*x^2)^(3/2) + 6*sqrt(c*x^4 + b*x^2)*b^2/c)*B
 

Giac [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.04 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^3} \, dx=\frac {1}{48} \, {\left (2 \, {\left (4 \, B c x^{2} \mathrm {sgn}\left (x\right ) + \frac {7 \, B b c^{4} \mathrm {sgn}\left (x\right ) + 6 \, A c^{5} \mathrm {sgn}\left (x\right )}{c^{4}}\right )} x^{2} + \frac {3 \, {\left (B b^{2} c^{3} \mathrm {sgn}\left (x\right ) + 10 \, A b c^{4} \mathrm {sgn}\left (x\right )\right )}}{c^{4}}\right )} \sqrt {c x^{2} + b} x + \frac {{\left (B b^{3} \mathrm {sgn}\left (x\right ) - 6 \, A b^{2} c \mathrm {sgn}\left (x\right )\right )} \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + b} \right |}\right )}{16 \, c^{\frac {3}{2}}} - \frac {{\left (B b^{3} \log \left ({\left | b \right |}\right ) - 6 \, A b^{2} c \log \left ({\left | b \right |}\right )\right )} \mathrm {sgn}\left (x\right )}{32 \, c^{\frac {3}{2}}} \] Input:

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

Output:

1/48*(2*(4*B*c*x^2*sgn(x) + (7*B*b*c^4*sgn(x) + 6*A*c^5*sgn(x))/c^4)*x^2 + 
 3*(B*b^2*c^3*sgn(x) + 10*A*b*c^4*sgn(x))/c^4)*sqrt(c*x^2 + b)*x + 1/16*(B 
*b^3*sgn(x) - 6*A*b^2*c*sgn(x))*log(abs(-sqrt(c)*x + sqrt(c*x^2 + b)))/c^( 
3/2) - 1/32*(B*b^3*log(abs(b)) - 6*A*b^2*c*log(abs(b)))*sgn(x)/c^(3/2)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.05 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^3} \, dx=\frac {30 \sqrt {c \,x^{2}+b}\, a b \,c^{2} x +12 \sqrt {c \,x^{2}+b}\, a \,c^{3} x^{3}+3 \sqrt {c \,x^{2}+b}\, b^{3} c x +14 \sqrt {c \,x^{2}+b}\, b^{2} c^{2} x^{3}+8 \sqrt {c \,x^{2}+b}\, b \,c^{3} x^{5}+18 \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}+\sqrt {c}\, x}{\sqrt {b}}\right ) a \,b^{2} c -3 \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}+\sqrt {c}\, x}{\sqrt {b}}\right ) b^{4}}{48 c^{2}} \] Input:

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

Output:

(30*sqrt(b + c*x**2)*a*b*c**2*x + 12*sqrt(b + c*x**2)*a*c**3*x**3 + 3*sqrt 
(b + c*x**2)*b**3*c*x + 14*sqrt(b + c*x**2)*b**2*c**2*x**3 + 8*sqrt(b + c* 
x**2)*b*c**3*x**5 + 18*sqrt(c)*log((sqrt(b + c*x**2) + sqrt(c)*x)/sqrt(b)) 
*a*b**2*c - 3*sqrt(c)*log((sqrt(b + c*x**2) + sqrt(c)*x)/sqrt(b))*b**4)/(4 
8*c**2)