\(\int \frac {(b x^2+c x^4)^{3/2}}{x} \, dx\) [172]

Optimal result

Integrand size = 19, antiderivative size = 88 \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x} \, dx=\frac {b \left (b+2 c x^2\right ) \sqrt {b x^2+c x^4}}{16 c}+\frac {1}{6} \left (b x^2+c x^4\right )^{3/2}-\frac {b^3 \text {arctanh}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{16 c^{3/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.29 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.25 \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x} \, dx=\frac {x \sqrt {b+c x^2} \left (\sqrt {c} x \sqrt {b+c x^2} \left (3 b^2+14 b c x^2+8 c^2 x^4\right )+6 b^3 \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b}-\sqrt {b+c x^2}}\right )\right )}{48 c^{3/2} \sqrt {x^2 \left (b+c x^2\right )}} \] Input:

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

Output:

(x*Sqrt[b + c*x^2]*(Sqrt[c]*x*Sqrt[b + c*x^2]*(3*b^2 + 14*b*c*x^2 + 8*c^2* 
x^4) + 6*b^3*ArcTanh[(Sqrt[c]*x)/(Sqrt[b] - Sqrt[b + c*x^2])]))/(48*c^(3/2 
)*Sqrt[x^2*(b + c*x^2)])
 

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.10, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {1424, 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.

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

\(\Big \downarrow \) 1424

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

\(\Big \downarrow \) 1131

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

\(\Big \downarrow \) 1087

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

\(\Big \downarrow \) 1091

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

((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)/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 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]
 

Maple [A] (verified)

Time = 0.17 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.02

method	result	size
risch	\(\frac {\left (8 c^{2} x^{4}+14 b c \,x^{2}+3 b^{2}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{48 c}-\frac {b^{3} \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{16 c^{\frac {3}{2}} x \sqrt {c \,x^{2}+b}}\)	\(90\)
default	\(\frac {\left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} \left (8 x \left (c \,x^{2}+b \right )^{\frac {5}{2}} \sqrt {c}-2 \sqrt {c}\, \left (c \,x^{2}+b \right )^{\frac {3}{2}} b x -3 \sqrt {c}\, \sqrt {c \,x^{2}+b}\, b^{2} x -3 \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right ) b^{3}\right )}{48 x^{3} \left (c \,x^{2}+b \right )^{\frac {3}{2}} c^{\frac {3}{2}}}\)	\(102\)
pseudoelliptic	\(\frac {16 c^{\frac {5}{2}} x^{4} \sqrt {x^{2} \left (c \,x^{2}+b \right )}+28 \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, c^{\frac {3}{2}} b \,x^{2}+3 \ln \left (2\right ) b^{3}-3 \ln \left (\frac {2 c \,x^{2}+2 \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \sqrt {c}+b}{\sqrt {c}}\right ) b^{3}+6 \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \sqrt {c}\, b^{2}}{96 c^{\frac {3}{2}}}\)	\(114\)

Input:

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

Output:

1/48*(8*c^2*x^4+14*b*c*x^2+3*b^2)/c*(x^2*(c*x^2+b))^(1/2)-1/16*b^3/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 = 0.20 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.89 \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x} \, dx=\left [\frac {3 \, b^{3} \sqrt {c} \log \left (-2 \, c x^{2} - b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right ) + 2 \, {\left (8 \, c^{3} x^{4} + 14 \, b c^{2} x^{2} + 3 \, b^{2} c\right )} \sqrt {c x^{4} + b x^{2}}}{96 \, c^{2}}, \frac {3 \, b^{3} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-c}}{c x^{2} + b}\right ) + {\left (8 \, c^{3} x^{4} + 14 \, b c^{2} x^{2} + 3 \, b^{2} c\right )} \sqrt {c x^{4} + b x^{2}}}{48 \, c^{2}}\right ] \] Input:

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

Output:

[1/96*(3*b^3*sqrt(c)*log(-2*c*x^2 - b + 2*sqrt(c*x^4 + b*x^2)*sqrt(c)) + 2 
*(8*c^3*x^4 + 14*b*c^2*x^2 + 3*b^2*c)*sqrt(c*x^4 + b*x^2))/c^2, 1/48*(3*b^ 
3*sqrt(-c)*arctan(sqrt(c*x^4 + b*x^2)*sqrt(-c)/(c*x^2 + b)) + (8*c^3*x^4 + 
 14*b*c^2*x^2 + 3*b^2*c)*sqrt(c*x^4 + b*x^2))/c^2]
                                                                                    
                                                                                    
 

Sympy [A] (verification not implemented)

Time = 1.89 (sec) , antiderivative size = 255, normalized size of antiderivative = 2.90 \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x} \, dx=\frac {b \left (\begin {cases} - \frac {b^{2} \left (\begin {cases} \frac {\log {\left (b + 2 \sqrt {c} \sqrt {b x^{2} + c x^{4}} + 2 c x^{2} \right )}}{\sqrt {c}} & \text {for}\: \frac {b^{2}}{c} \neq 0 \\\frac {\left (\frac {b}{2 c} + x^{2}\right ) \log {\left (\frac {b}{2 c} + x^{2} \right )}}{\sqrt {c \left (\frac {b}{2 c} + x^{2}\right )^{2}}} & \text {otherwise} \end {cases}\right )}{8 c} + \left (\frac {b}{4 c} + \frac {x^{2}}{2}\right ) \sqrt {b x^{2} + c x^{4}} & \text {for}\: c \neq 0 \\\frac {2 \left (b x^{2}\right )^{\frac {3}{2}}}{3 b} & \text {for}\: b \neq 0 \\0 & \text {otherwise} \end {cases}\right )}{2} + \frac {c \left (\begin {cases} \frac {b^{3} \left (\begin {cases} \frac {\log {\left (b + 2 \sqrt {c} \sqrt {b x^{2} + c x^{4}} + 2 c x^{2} \right )}}{\sqrt {c}} & \text {for}\: \frac {b^{2}}{c} \neq 0 \\\frac {\left (\frac {b}{2 c} + x^{2}\right ) \log {\left (\frac {b}{2 c} + x^{2} \right )}}{\sqrt {c \left (\frac {b}{2 c} + x^{2}\right )^{2}}} & \text {otherwise} \end {cases}\right )}{16 c^{2}} + \sqrt {b x^{2} + c x^{4}} \left (- \frac {b^{2}}{8 c^{2}} + \frac {b x^{2}}{12 c} + \frac {x^{4}}{3}\right ) & \text {for}\: c \neq 0 \\\frac {2 \left (b x^{2}\right )^{\frac {5}{2}}}{5 b^{2}} & \text {for}\: b \neq 0 \\0 & \text {otherwise} \end {cases}\right )}{2} \] Input:

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

Output:

b*Piecewise((-b**2*Piecewise((log(b + 2*sqrt(c)*sqrt(b*x**2 + c*x**4) + 2* 
c*x**2)/sqrt(c), Ne(b**2/c, 0)), ((b/(2*c) + x**2)*log(b/(2*c) + x**2)/sqr 
t(c*(b/(2*c) + x**2)**2), True))/(8*c) + (b/(4*c) + x**2/2)*sqrt(b*x**2 + 
c*x**4), Ne(c, 0)), (2*(b*x**2)**(3/2)/(3*b), Ne(b, 0)), (0, True))/2 + c* 
Piecewise((b**3*Piecewise((log(b + 2*sqrt(c)*sqrt(b*x**2 + c*x**4) + 2*c*x 
**2)/sqrt(c), Ne(b**2/c, 0)), ((b/(2*c) + x**2)*log(b/(2*c) + x**2)/sqrt(c 
*(b/(2*c) + x**2)**2), True))/(16*c**2) + sqrt(b*x**2 + c*x**4)*(-b**2/(8* 
c**2) + b*x**2/(12*c) + x**4/3), Ne(c, 0)), (2*(b*x**2)**(5/2)/(5*b**2), N 
e(b, 0)), (0, True))/2
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.03 \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x} \, dx=\frac {1}{8} \, \sqrt {c x^{4} + b x^{2}} b x^{2} - \frac {b^{3} \log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right )}{32 \, c^{\frac {3}{2}}} + \frac {1}{6} \, {\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} + \frac {\sqrt {c x^{4} + b x^{2}} b^{2}}{16 \, c} \] Input:

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

Output:

1/8*sqrt(c*x^4 + b*x^2)*b*x^2 - 1/32*b^3*log(2*c*x^2 + b + 2*sqrt(c*x^4 + 
b*x^2)*sqrt(c))/c^(3/2) + 1/6*(c*x^4 + b*x^2)^(3/2) + 1/16*sqrt(c*x^4 + b* 
x^2)*b^2/c
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.95 \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x} \, dx=\frac {b^{3} \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + b} \right |}\right ) \mathrm {sgn}\left (x\right )}{16 \, c^{\frac {3}{2}}} - \frac {b^{3} \log \left ({\left | b \right |}\right ) \mathrm {sgn}\left (x\right )}{32 \, c^{\frac {3}{2}}} + \frac {1}{48} \, {\left (2 \, {\left (4 \, c x^{2} \mathrm {sgn}\left (x\right ) + 7 \, b \mathrm {sgn}\left (x\right )\right )} x^{2} + \frac {3 \, b^{2} \mathrm {sgn}\left (x\right )}{c}\right )} \sqrt {c x^{2} + b} x \] Input:

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.91 \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x} \, dx=\frac {3 \sqrt {c \,x^{2}+b}\, b^{2} c x +14 \sqrt {c \,x^{2}+b}\, b \,c^{2} x^{3}+8 \sqrt {c \,x^{2}+b}\, c^{3} x^{5}-3 \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}+\sqrt {c}\, x}{\sqrt {b}}\right ) b^{3}}{48 c^{2}} \] Input:

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

Output:

(3*sqrt(b + c*x**2)*b**2*c*x + 14*sqrt(b + c*x**2)*b*c**2*x**3 + 8*sqrt(b 
+ c*x**2)*c**3*x**5 - 3*sqrt(c)*log((sqrt(b + c*x**2) + sqrt(c)*x)/sqrt(b) 
)*b**3)/(48*c**2)