\(\int x^3 (b x^2+c x^4)^{3/2} \, dx\) [170]

Optimal result

Integrand size = 19, antiderivative size = 124 \[ \int x^3 \left (b x^2+c x^4\right )^{3/2} \, dx=\frac {3 b^3 \left (b+2 c x^2\right ) \sqrt {b x^2+c x^4}}{256 c^3}-\frac {b \left (b+2 c x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{32 c^2}+\frac {\left (b x^2+c x^4\right )^{5/2}}{10 c}-\frac {3 b^5 \text {arctanh}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{256 c^{7/2}} \] Output:

3/256*b^3*(2*c*x^2+b)*(c*x^4+b*x^2)^(1/2)/c^3-1/32*b*(2*c*x^2+b)*(c*x^4+b* 
x^2)^(3/2)/c^2+1/10*(c*x^4+b*x^2)^(5/2)/c-3/256*b^5*arctanh(c^(1/2)*x^2/(c 
*x^4+b*x^2)^(1/2))/c^(7/2)
 

Mathematica [A] (verified)

Time = 0.45 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.06 \[ \int x^3 \left (b x^2+c x^4\right )^{3/2} \, dx=\frac {x \sqrt {b+c x^2} \left (\sqrt {c} x \sqrt {b+c x^2} \left (15 b^4-10 b^3 c x^2+8 b^2 c^2 x^4+176 b c^3 x^6+128 c^4 x^8\right )+30 b^5 \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b}-\sqrt {b+c x^2}}\right )\right )}{1280 c^{7/2} \sqrt {x^2 \left (b+c x^2\right )}} \] Input:

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

Output:

(x*Sqrt[b + c*x^2]*(Sqrt[c]*x*Sqrt[b + c*x^2]*(15*b^4 - 10*b^3*c*x^2 + 8*b 
^2*c^2*x^4 + 176*b*c^3*x^6 + 128*c^4*x^8) + 30*b^5*ArcTanh[(Sqrt[c]*x)/(Sq 
rt[b] - Sqrt[b + c*x^2])]))/(1280*c^(7/2)*Sqrt[x^2*(b + c*x^2)])
 

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.16, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {1424, 1160, 1087, 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.

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

\(\Big \downarrow \) 1424

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

\(\Big \downarrow \) 1160

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

\(\Big \downarrow \) 1087

\(\displaystyle \frac {1}{2} \left (\frac {\left (b x^2+c x^4\right )^{5/2}}{5 c}-\frac {b \left (\frac {\left (b+2 c x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{8 c}-\frac {3 b^2 \int \sqrt {c x^4+b x^2}dx^2}{16 c}\right )}{2 c}\right )\)

\(\Big \downarrow \) 1087

\(\displaystyle \frac {1}{2} \left (\frac {\left (b x^2+c x^4\right )^{5/2}}{5 c}-\frac {b \left (\frac {\left (b+2 c x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{8 c}-\frac {3 b^2 \left (\frac {\left (b+2 c x^2\right ) \sqrt {b x^2+c x^4}}{4 c}-\frac {b^2 \int \frac {1}{\sqrt {c x^4+b x^2}}dx^2}{8 c}\right )}{16 c}\right )}{2 c}\right )\)

\(\Big \downarrow \) 1091

\(\displaystyle \frac {1}{2} \left (\frac {\left (b x^2+c x^4\right )^{5/2}}{5 c}-\frac {b \left (\frac {\left (b+2 c x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{8 c}-\frac {3 b^2 \left (\frac {\left (b+2 c x^2\right ) \sqrt {b x^2+c x^4}}{4 c}-\frac {b^2 \int \frac {1}{1-c x^4}d\frac {x^2}{\sqrt {c x^4+b x^2}}}{4 c}\right )}{16 c}\right )}{2 c}\right )\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{2} \left (\frac {\left (b x^2+c x^4\right )^{5/2}}{5 c}-\frac {b \left (\frac {\left (b+2 c x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{8 c}-\frac {3 b^2 \left (\frac {\left (b+2 c x^2\right ) \sqrt {b x^2+c x^4}}{4 c}-\frac {b^2 \text {arctanh}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{4 c^{3/2}}\right )}{16 c}\right )}{2 c}\right )\)

Input:

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

Output:

((b*x^2 + c*x^4)^(5/2)/(5*c) - (b*(((b + 2*c*x^2)*(b*x^2 + c*x^4)^(3/2))/( 
8*c) - (3*b^2*(((b + 2*c*x^2)*Sqrt[b*x^2 + c*x^4])/(4*c) - (b^2*ArcTanh[(S 
qrt[c]*x^2)/Sqrt[b*x^2 + c*x^4]])/(4*c^(3/2))))/(16*c)))/(2*c))/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 1160

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol 
] :> Simp[e*((a + b*x + c*x^2)^(p + 1)/(2*c*(p + 1))), x] + Simp[(2*c*d - b 
*e)/(2*c)   Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] 
 && NeQ[p, -1]
 

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 = 0.21 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.90

method	result	size
pseudoelliptic	\(-\frac {3 \left (\ln \left (\frac {2 c \,x^{2}+2 \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \sqrt {c}+b}{\sqrt {c}}\right ) b^{5}+\left (-\frac {256 c^{\frac {9}{2}} x^{8}}{15}-\frac {352 c^{\frac {7}{2}} b \,x^{6}}{15}-\frac {16 c^{\frac {5}{2}} b^{2} x^{4}}{15}+\frac {4 c^{\frac {3}{2}} b^{3} x^{2}}{3}-2 \sqrt {c}\, b^{4}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}-\ln \left (2\right ) b^{5}\right )}{512 c^{\frac {7}{2}}}\)	\(111\)
risch	\(\frac {\left (128 c^{4} x^{8}+176 b \,c^{3} x^{6}+8 b^{2} c^{2} x^{4}-10 x^{2} b^{3} c +15 b^{4}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{1280 c^{3}}-\frac {3 b^{5} \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{256 c^{\frac {7}{2}} x \sqrt {c \,x^{2}+b}}\)	\(112\)
default	\(\frac {\left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} \left (128 x^{5} \left (c \,x^{2}+b \right )^{\frac {5}{2}} c^{\frac {5}{2}}-80 c^{\frac {3}{2}} \left (c \,x^{2}+b \right )^{\frac {5}{2}} b \,x^{3}+40 \sqrt {c}\, \left (c \,x^{2}+b \right )^{\frac {5}{2}} b^{2} x -10 \sqrt {c}\, \left (c \,x^{2}+b \right )^{\frac {3}{2}} b^{3} x -15 \sqrt {c}\, \sqrt {c \,x^{2}+b}\, b^{4} x -15 \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right ) b^{5}\right )}{1280 x^{3} \left (c \,x^{2}+b \right )^{\frac {3}{2}} c^{\frac {7}{2}}}\)	\(142\)

Input:

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

Output:

-3/512/c^(7/2)*(ln((2*c*x^2+2*(x^2*(c*x^2+b))^(1/2)*c^(1/2)+b)/c^(1/2))*b^ 
5+(-256/15*c^(9/2)*x^8-352/15*c^(7/2)*b*x^6-16/15*c^(5/2)*b^2*x^4+4/3*c^(3 
/2)*b^3*x^2-2*c^(1/2)*b^4)*(x^2*(c*x^2+b))^(1/2)-ln(2)*b^5)
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.69 \[ \int x^3 \left (b x^2+c x^4\right )^{3/2} \, dx=\left [\frac {15 \, b^{5} \sqrt {c} \log \left (-2 \, c x^{2} - b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right ) + 2 \, {\left (128 \, c^{5} x^{8} + 176 \, b c^{4} x^{6} + 8 \, b^{2} c^{3} x^{4} - 10 \, b^{3} c^{2} x^{2} + 15 \, b^{4} c\right )} \sqrt {c x^{4} + b x^{2}}}{2560 \, c^{4}}, \frac {15 \, b^{5} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-c}}{c x^{2} + b}\right ) + {\left (128 \, c^{5} x^{8} + 176 \, b c^{4} x^{6} + 8 \, b^{2} c^{3} x^{4} - 10 \, b^{3} c^{2} x^{2} + 15 \, b^{4} c\right )} \sqrt {c x^{4} + b x^{2}}}{1280 \, c^{4}}\right ] \] Input:

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

Output:

[1/2560*(15*b^5*sqrt(c)*log(-2*c*x^2 - b + 2*sqrt(c*x^4 + b*x^2)*sqrt(c)) 
+ 2*(128*c^5*x^8 + 176*b*c^4*x^6 + 8*b^2*c^3*x^4 - 10*b^3*c^2*x^2 + 15*b^4 
*c)*sqrt(c*x^4 + b*x^2))/c^4, 1/1280*(15*b^5*sqrt(-c)*arctan(sqrt(c*x^4 + 
b*x^2)*sqrt(-c)/(c*x^2 + b)) + (128*c^5*x^8 + 176*b*c^4*x^6 + 8*b^2*c^3*x^ 
4 - 10*b^3*c^2*x^2 + 15*b^4*c)*sqrt(c*x^4 + b*x^2))/c^4]
                                                                                    
                                                                                    
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.15 \[ \int x^3 \left (b x^2+c x^4\right )^{3/2} \, dx=\frac {3 \, \sqrt {c x^{4} + b x^{2}} b^{3} x^{2}}{128 \, c^{2}} - \frac {{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} b x^{2}}{16 \, c} - \frac {3 \, b^{5} \log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right )}{512 \, c^{\frac {7}{2}}} + \frac {3 \, \sqrt {c x^{4} + b x^{2}} b^{4}}{256 \, c^{3}} - \frac {{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} b^{2}}{32 \, c^{2}} + \frac {{\left (c x^{4} + b x^{2}\right )}^{\frac {5}{2}}}{10 \, c} \] Input:

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

Output:

3/128*sqrt(c*x^4 + b*x^2)*b^3*x^2/c^2 - 1/16*(c*x^4 + b*x^2)^(3/2)*b*x^2/c 
 - 3/512*b^5*log(2*c*x^2 + b + 2*sqrt(c*x^4 + b*x^2)*sqrt(c))/c^(7/2) + 3/ 
256*sqrt(c*x^4 + b*x^2)*b^4/c^3 - 1/32*(c*x^4 + b*x^2)^(3/2)*b^2/c^2 + 1/1 
0*(c*x^4 + b*x^2)^(5/2)/c
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.93 \[ \int x^3 \left (b x^2+c x^4\right )^{3/2} \, dx=\frac {3 \, b^{5} \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + b} \right |}\right ) \mathrm {sgn}\left (x\right )}{256 \, c^{\frac {7}{2}}} - \frac {3 \, b^{5} \log \left ({\left | b \right |}\right ) \mathrm {sgn}\left (x\right )}{512 \, c^{\frac {7}{2}}} + \frac {1}{1280} \, {\left (2 \, {\left (4 \, {\left (2 \, {\left (8 \, c x^{2} \mathrm {sgn}\left (x\right ) + 11 \, b \mathrm {sgn}\left (x\right )\right )} x^{2} + \frac {b^{2} \mathrm {sgn}\left (x\right )}{c}\right )} x^{2} - \frac {5 \, b^{3} \mathrm {sgn}\left (x\right )}{c^{2}}\right )} x^{2} + \frac {15 \, b^{4} \mathrm {sgn}\left (x\right )}{c^{3}}\right )} \sqrt {c x^{2} + b} x \] Input:

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

Output:

3/256*b^5*log(abs(-sqrt(c)*x + sqrt(c*x^2 + b)))*sgn(x)/c^(7/2) - 3/512*b^ 
5*log(abs(b))*sgn(x)/c^(7/2) + 1/1280*(2*(4*(2*(8*c*x^2*sgn(x) + 11*b*sgn( 
x))*x^2 + b^2*sgn(x)/c)*x^2 - 5*b^3*sgn(x)/c^2)*x^2 + 15*b^4*sgn(x)/c^3)*s 
qrt(c*x^2 + b)*x
 

Mupad [B] (verification not implemented)

Time = 18.33 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.08 \[ \int x^3 \left (b x^2+c x^4\right )^{3/2} \, dx=\frac {{\left (c\,x^4+b\,x^2\right )}^{5/2}}{10\,c}-\frac {b\,\left (\frac {x^2\,{\left (c\,x^4+b\,x^2\right )}^{3/2}}{4}-\frac {3\,b^2\,\left (\frac {\left (2\,c\,x^2+b\right )\,\sqrt {c\,x^4+b\,x^2}}{4\,c}-\frac {b^2\,\ln \left (\frac {c\,x^2+\frac {b}{2}}{\sqrt {c}}+\sqrt {c\,x^4+b\,x^2}\right )}{8\,c^{3/2}}\right )}{16\,c}+\frac {b\,{\left (c\,x^4+b\,x^2\right )}^{3/2}}{8\,c}\right )}{4\,c} \] Input:

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

Output:

(b*x^2 + c*x^4)^(5/2)/(10*c) - (b*((x^2*(b*x^2 + c*x^4)^(3/2))/4 - (3*b^2* 
(((b + 2*c*x^2)*(b*x^2 + c*x^4)^(1/2))/(4*c) - (b^2*log((b/2 + c*x^2)/c^(1 
/2) + (b*x^2 + c*x^4)^(1/2)))/(8*c^(3/2))))/(16*c) + (b*(b*x^2 + c*x^4)^(3 
/2))/(8*c)))/(4*c)
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.95 \[ \int x^3 \left (b x^2+c x^4\right )^{3/2} \, dx=\frac {15 \sqrt {c \,x^{2}+b}\, b^{4} c x -10 \sqrt {c \,x^{2}+b}\, b^{3} c^{2} x^{3}+8 \sqrt {c \,x^{2}+b}\, b^{2} c^{3} x^{5}+176 \sqrt {c \,x^{2}+b}\, b \,c^{4} x^{7}+128 \sqrt {c \,x^{2}+b}\, c^{5} x^{9}-15 \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}+\sqrt {c}\, x}{\sqrt {b}}\right ) b^{5}}{1280 c^{4}} \] Input:

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

Output:

(15*sqrt(b + c*x**2)*b**4*c*x - 10*sqrt(b + c*x**2)*b**3*c**2*x**3 + 8*sqr 
t(b + c*x**2)*b**2*c**3*x**5 + 176*sqrt(b + c*x**2)*b*c**4*x**7 + 128*sqrt 
(b + c*x**2)*c**5*x**9 - 15*sqrt(c)*log((sqrt(b + c*x**2) + sqrt(c)*x)/sqr 
t(b))*b**5)/(1280*c**4)