\(\int \frac {(A+B x^2) (b x^2+c x^4)^{3/2}}{x^4} \, dx\) [188]

Optimal result

Integrand size = 26, antiderivative size = 102 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^4} \, dx=\frac {A b \sqrt {b x^2+c x^4}}{x}+\frac {A \left (b x^2+c x^4\right )^{3/2}}{3 x^3}+\frac {B \left (b x^2+c x^4\right )^{5/2}}{5 c x^5}-A b^{3/2} \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right ) \] Output:

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

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.06 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^4} \, dx=\frac {x \left (\left (b+c x^2\right ) \left (3 b^2 B+c^2 x^2 \left (5 A+3 B x^2\right )+b \left (20 A c+6 B c x^2\right )\right )-15 A b^{3/2} c \sqrt {b+c x^2} \text {arctanh}\left (\frac {\sqrt {b+c x^2}}{\sqrt {b}}\right )\right )}{15 c \sqrt {x^2 \left (b+c x^2\right )}} \] Input:

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

Output:

(x*((b + c*x^2)*(3*b^2*B + c^2*x^2*(5*A + 3*B*x^2) + b*(20*A*c + 6*B*c*x^2 
)) - 15*A*b^(3/2)*c*Sqrt[b + c*x^2]*ArcTanh[Sqrt[b + c*x^2]/Sqrt[b]]))/(15 
*c*Sqrt[x^2*(b + c*x^2)])
 

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.02, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {1945, 1426, 1426, 1400, 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.

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

\(\Big \downarrow \) 1945

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

\(\Big \downarrow \) 1426

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

\(\Big \downarrow \) 1426

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

\(\Big \downarrow \) 1400

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

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

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 1400

Int[1/Sqrt[(b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> -Subst[Int[1/(1 - b*x 
^2), x], x, x/Sqrt[b*x^2 + c*x^4]] /; FreeQ[{b, c}, x]
 

rule 1426

Int[((d_.)*(x_))^(m_)*((b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp 
[(d*x)^(m + 1)*((b*x^2 + c*x^4)^p/(d*(m + 4*p + 1))), x] + Simp[2*b*(p/(d^2 
*(m + 4*p + 1)))   Int[(d*x)^(m + 2)*(b*x^2 + c*x^4)^(p - 1), x], x] /; Fre 
eQ[{b, c, d, m, p}, x] &&  !IntegerQ[p] && GtQ[p, 0] && NeQ[m + 4*p + 1, 0]
 

rule 1945

Int[((e_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(jn_.))^(p_)*((c_) + 
(d_.)*(x_)^(n_.)), x_Symbol] :> Simp[d*e^(j - 1)*(e*x)^(m - j + 1)*((a*x^j 
+ b*x^(j + n))^(p + 1)/(b*(m + n + p*(j + n) + 1))), x] - Simp[(a*d*(m + j* 
p + 1) - b*c*(m + n + p*(j + n) + 1))/(b*(m + n + p*(j + n) + 1))   Int[(e* 
x)^m*(a*x^j + b*x^(j + n))^p, x], x] /; FreeQ[{a, b, c, d, e, j, m, n, p}, 
x] && EqQ[jn, j + n] &&  !IntegerQ[p] && NeQ[b*c - a*d, 0] && NeQ[m + n + p 
*(j + n) + 1, 0] && (GtQ[e, 0] || IntegerQ[j])
 

Maple [A] (verified)

Time = 0.75 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.97

method	result	size
default	\(\frac {\left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} \left (3 B \left (c \,x^{2}+b \right )^{\frac {5}{2}}+5 A \left (c \,x^{2}+b \right )^{\frac {3}{2}} c -15 A \ln \left (\frac {2 b +2 \sqrt {b}\, \sqrt {c \,x^{2}+b}}{x}\right ) b^{\frac {3}{2}} c +15 A \sqrt {c \,x^{2}+b}\, b c \right )}{15 x^{3} \left (c \,x^{2}+b \right )^{\frac {3}{2}} c}\)	\(99\)

Input:

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

Output:

1/15*(c*x^4+b*x^2)^(3/2)*(3*B*(c*x^2+b)^(5/2)+5*A*(c*x^2+b)^(3/2)*c-15*A*l 
n(2*(b^(1/2)*(c*x^2+b)^(1/2)+b)/x)*b^(3/2)*c+15*A*(c*x^2+b)^(1/2)*b*c)/x^3 
/(c*x^2+b)^(3/2)/c
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.97 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^4} \, dx=\left [\frac {15 \, A b^{\frac {3}{2}} c x \log \left (-\frac {c x^{3} + 2 \, b x - 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {b}}{x^{3}}\right ) + 2 \, {\left (3 \, B c^{2} x^{4} + 3 \, B b^{2} + 20 \, A b c + {\left (6 \, B b c + 5 \, A c^{2}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{30 \, c x}, \frac {15 \, A \sqrt {-b} b c x \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-b}}{b x}\right ) + {\left (3 \, B c^{2} x^{4} + 3 \, B b^{2} + 20 \, A b c + {\left (6 \, B b c + 5 \, A c^{2}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{15 \, c x}\right ] \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.37 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^4} \, dx=\frac {A b^{2} \arctan \left (\frac {\sqrt {c x^{2} + b}}{\sqrt {-b}}\right ) \mathrm {sgn}\left (x\right )}{\sqrt {-b}} - \frac {{\left (15 \, A b^{2} c \arctan \left (\frac {\sqrt {b}}{\sqrt {-b}}\right ) + 3 \, B \sqrt {-b} b^{\frac {5}{2}} + 20 \, A \sqrt {-b} b^{\frac {3}{2}} c\right )} \mathrm {sgn}\left (x\right )}{15 \, \sqrt {-b} c} + \frac {3 \, {\left (c x^{2} + b\right )}^{\frac {5}{2}} B c^{4} \mathrm {sgn}\left (x\right ) + 5 \, {\left (c x^{2} + b\right )}^{\frac {3}{2}} A c^{5} \mathrm {sgn}\left (x\right ) + 15 \, \sqrt {c x^{2} + b} A b c^{5} \mathrm {sgn}\left (x\right )}{15 \, c^{5}} \] Input:

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

Output:

A*b^2*arctan(sqrt(c*x^2 + b)/sqrt(-b))*sgn(x)/sqrt(-b) - 1/15*(15*A*b^2*c* 
arctan(sqrt(b)/sqrt(-b)) + 3*B*sqrt(-b)*b^(5/2) + 20*A*sqrt(-b)*b^(3/2)*c) 
*sgn(x)/(sqrt(-b)*c) + 1/15*(3*(c*x^2 + b)^(5/2)*B*c^4*sgn(x) + 5*(c*x^2 + 
 b)^(3/2)*A*c^5*sgn(x) + 15*sqrt(c*x^2 + b)*A*b*c^5*sgn(x))/c^5
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.38 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^4} \, dx=\frac {20 \sqrt {c \,x^{2}+b}\, a b c +5 \sqrt {c \,x^{2}+b}\, a \,c^{2} x^{2}+3 \sqrt {c \,x^{2}+b}\, b^{3}+6 \sqrt {c \,x^{2}+b}\, b^{2} c \,x^{2}+3 \sqrt {c \,x^{2}+b}\, b \,c^{2} x^{4}+15 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}-\sqrt {b}+\sqrt {c}\, x}{\sqrt {b}}\right ) a b c -15 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}+\sqrt {b}+\sqrt {c}\, x}{\sqrt {b}}\right ) a b c}{15 c} \] Input:

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

Output:

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