\(\int x^3 (a+b x^2+c x^4)^{3/2} \, dx\) [938]

Optimal result

Integrand size = 20, antiderivative size = 150 \[ \int x^3 \left (a+b x^2+c x^4\right )^{3/2} \, dx=\frac {3 b \left (b^2-4 a c\right ) \left (b+2 c x^2\right ) \sqrt {a+b x^2+c x^4}}{256 c^3}-\frac {b \left (b+2 c x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{32 c^2}+\frac {\left (a+b x^2+c x^4\right )^{5/2}}{10 c}-\frac {3 b \left (b^2-4 a c\right )^2 \text {arctanh}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )}{512 c^{7/2}} \]

[Out]

-1/32*b*(2*c*x^2+b)*(c*x^4+b*x^2+a)^(3/2)/c^2+1/10*(c*x^4+b*x^2+a)^(5/2)/c-3/512*b*(-4*a*c+b^2)^2*arctanh(1/2*
(2*c*x^2+b)/c^(1/2)/(c*x^4+b*x^2+a)^(1/2))/c^(7/2)+3/256*b*(-4*a*c+b^2)*(2*c*x^2+b)*(c*x^4+b*x^2+a)^(1/2)/c^3

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1128, 654, 626, 635, 212} \[ \int x^3 \left (a+b x^2+c x^4\right )^{3/2} \, dx=-\frac {3 b \left (b^2-4 a c\right )^2 \text {arctanh}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )}{512 c^{7/2}}+\frac {3 b \left (b^2-4 a c\right ) \left (b+2 c x^2\right ) \sqrt {a+b x^2+c x^4}}{256 c^3}-\frac {b \left (b+2 c x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{32 c^2}+\frac {\left (a+b x^2+c x^4\right )^{5/2}}{10 c} \]

[In]

Int[x^3*(a + b*x^2 + c*x^4)^(3/2),x]

[Out]

(3*b*(b^2 - 4*a*c)*(b + 2*c*x^2)*Sqrt[a + b*x^2 + c*x^4])/(256*c^3) - (b*(b + 2*c*x^2)*(a + b*x^2 + c*x^4)^(3/
2))/(32*c^2) + (a + b*x^2 + c*x^4)^(5/2)/(10*c) - (3*b*(b^2 - 4*a*c)^2*ArcTanh[(b + 2*c*x^2)/(2*Sqrt[c]*Sqrt[a
+ b*x^2 + c*x^4])])/(512*c^(7/2))

Rule 212

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] && (GtQ[a, 0] || LtQ[b, 0])

Rule 626

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] - Dist[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]
&& NeQ[b^2 - 4*a*c, 0] && GtQ[p, 0] && IntegerQ[4*p]

Rule 635

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(4*c - x^2), x], x, (b + 2*c*x)
/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 654

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] + Dist[(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[2*c*d - b*e, 0] && NeQ[p, -1]

Rule 1128

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[x^((m - 1)/2)*(a +
b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, p}, x] && IntegerQ[(m - 1)/2]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int x \left (a+b x+c x^2\right )^{3/2} \, dx,x,x^2\right ) \\ & = \frac {\left (a+b x^2+c x^4\right )^{5/2}}{10 c}-\frac {b \text {Subst}\left (\int \left (a+b x+c x^2\right )^{3/2} \, dx,x,x^2\right )}{4 c} \\ & = -\frac {b \left (b+2 c x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{32 c^2}+\frac {\left (a+b x^2+c x^4\right )^{5/2}}{10 c}+\frac {\left (3 b \left (b^2-4 a c\right )\right ) \text {Subst}\left (\int \sqrt {a+b x+c x^2} \, dx,x,x^2\right )}{64 c^2} \\ & = \frac {3 b \left (b^2-4 a c\right ) \left (b+2 c x^2\right ) \sqrt {a+b x^2+c x^4}}{256 c^3}-\frac {b \left (b+2 c x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{32 c^2}+\frac {\left (a+b x^2+c x^4\right )^{5/2}}{10 c}-\frac {\left (3 b \left (b^2-4 a c\right )^2\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+b x+c x^2}} \, dx,x,x^2\right )}{512 c^3} \\ & = \frac {3 b \left (b^2-4 a c\right ) \left (b+2 c x^2\right ) \sqrt {a+b x^2+c x^4}}{256 c^3}-\frac {b \left (b+2 c x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{32 c^2}+\frac {\left (a+b x^2+c x^4\right )^{5/2}}{10 c}-\frac {\left (3 b \left (b^2-4 a c\right )^2\right ) \text {Subst}\left (\int \frac {1}{4 c-x^2} \, dx,x,\frac {b+2 c x^2}{\sqrt {a+b x^2+c x^4}}\right )}{256 c^3} \\ & = \frac {3 b \left (b^2-4 a c\right ) \left (b+2 c x^2\right ) \sqrt {a+b x^2+c x^4}}{256 c^3}-\frac {b \left (b+2 c x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{32 c^2}+\frac {\left (a+b x^2+c x^4\right )^{5/2}}{10 c}-\frac {3 b \left (b^2-4 a c\right )^2 \tanh ^{-1}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )}{512 c^{7/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.40 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.93 \[ \int x^3 \left (a+b x^2+c x^4\right )^{3/2} \, dx=\frac {\sqrt {a+b x^2+c x^4} \left (15 b^4-10 b^3 c x^2+128 c^2 \left (a+c x^4\right )^2+4 b^2 c \left (-25 a+2 c x^4\right )+8 b c^2 x^2 \left (7 a+22 c x^4\right )\right )}{1280 c^3}+\frac {3 b \left (b^2-4 a c\right )^2 \log \left (b+2 c x^2-2 \sqrt {c} \sqrt {a+b x^2+c x^4}\right )}{512 c^{7/2}} \]

[In]

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

[Out]

(Sqrt[a + b*x^2 + c*x^4]*(15*b^4 - 10*b^3*c*x^2 + 128*c^2*(a + c*x^4)^2 + 4*b^2*c*(-25*a + 2*c*x^4) + 8*b*c^2*
x^2*(7*a + 22*c*x^4)))/(1280*c^3) + (3*b*(b^2 - 4*a*c)^2*Log[b + 2*c*x^2 - 2*Sqrt[c]*Sqrt[a + b*x^2 + c*x^4]])
/(512*c^(7/2))

Maple [A] (verified)

Time = 0.16 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.01

method	result	size
risch	\(\frac {\left (128 c^{4} x^{8}+176 c^{3} x^{6} b +256 a \,c^{3} x^{4}+8 b^{2} c^{2} x^{4}+56 a b \,c^{2} x^{2}-10 x^{2} c \,b^{3}+128 a^{2} c^{2}-100 a \,b^{2} c +15 b^{4}\right ) \sqrt {c \,x^{4}+b \,x^{2}+a}}{1280 c^{3}}-\frac {3 b \left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right ) \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{512 c^{\frac {7}{2}}}\)	\(152\)
pseudoelliptic	\(\frac {-\frac {15 b \left (a c -\frac {b^{2}}{4}\right )^{2} \ln \left (\frac {2 c \,x^{2}+2 \sqrt {c \,x^{4}+b \,x^{2}+a}\, \sqrt {c}+b}{\sqrt {c}}\right )}{16}+\left (\left (\frac {1}{16} b^{2} x^{4}+\frac {7}{16} a b \,x^{2}+a^{2}\right ) c^{\frac {5}{2}}-\frac {25 \left (\frac {b \,x^{2}}{10}+a \right ) b^{2} c^{\frac {3}{2}}}{32}+\left (\frac {11}{8} b \,x^{6}+2 a \,x^{4}\right ) c^{\frac {7}{2}}+c^{\frac {9}{2}} x^{8}+\frac {15 \sqrt {c}\, b^{4}}{128}\right ) \sqrt {c \,x^{4}+b \,x^{2}+a}+\frac {15 \ln \left (2\right ) b \left (a c -\frac {b^{2}}{4}\right )^{2}}{16}}{10 c^{\frac {7}{2}}}\)	\(156\)
default	\(-\frac {5 b^{2} a \sqrt {c \,x^{4}+b \,x^{2}+a}}{64 c^{2}}+\frac {b^{2} x^{4} \sqrt {c \,x^{4}+b \,x^{2}+a}}{160 c}-\frac {b^{3} x^{2} \sqrt {c \,x^{4}+b \,x^{2}+a}}{128 c^{2}}-\frac {3 b^{5} \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{512 c^{\frac {7}{2}}}-\frac {3 a^{2} b \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{32 c^{\frac {3}{2}}}+\frac {a^{2} \sqrt {c \,x^{4}+b \,x^{2}+a}}{10 c}+\frac {3 b^{3} a \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{64 c^{\frac {5}{2}}}+\frac {7 b a \,x^{2} \sqrt {c \,x^{4}+b \,x^{2}+a}}{160 c}+\frac {c \,x^{8} \sqrt {c \,x^{4}+b \,x^{2}+a}}{10}+\frac {11 b \,x^{6} \sqrt {c \,x^{4}+b \,x^{2}+a}}{80}+\frac {3 b^{4} \sqrt {c \,x^{4}+b \,x^{2}+a}}{256 c^{3}}+\frac {a \,x^{4} \sqrt {c \,x^{4}+b \,x^{2}+a}}{5}\)	\(316\)
elliptic	\(-\frac {5 b^{2} a \sqrt {c \,x^{4}+b \,x^{2}+a}}{64 c^{2}}+\frac {b^{2} x^{4} \sqrt {c \,x^{4}+b \,x^{2}+a}}{160 c}-\frac {b^{3} x^{2} \sqrt {c \,x^{4}+b \,x^{2}+a}}{128 c^{2}}-\frac {3 b^{5} \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{512 c^{\frac {7}{2}}}-\frac {3 a^{2} b \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{32 c^{\frac {3}{2}}}+\frac {a^{2} \sqrt {c \,x^{4}+b \,x^{2}+a}}{10 c}+\frac {3 b^{3} a \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{64 c^{\frac {5}{2}}}+\frac {7 b a \,x^{2} \sqrt {c \,x^{4}+b \,x^{2}+a}}{160 c}+\frac {c \,x^{8} \sqrt {c \,x^{4}+b \,x^{2}+a}}{10}+\frac {11 b \,x^{6} \sqrt {c \,x^{4}+b \,x^{2}+a}}{80}+\frac {3 b^{4} \sqrt {c \,x^{4}+b \,x^{2}+a}}{256 c^{3}}+\frac {a \,x^{4} \sqrt {c \,x^{4}+b \,x^{2}+a}}{5}\)	\(316\)

[In]

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

[Out]

1/1280*(128*c^4*x^8+176*b*c^3*x^6+256*a*c^3*x^4+8*b^2*c^2*x^4+56*a*b*c^2*x^2-10*b^3*c*x^2+128*a^2*c^2-100*a*b^
2*c+15*b^4)*(c*x^4+b*x^2+a)^(1/2)/c^3-3/512*b*(16*a^2*c^2-8*a*b^2*c+b^4)/c^(7/2)*ln((1/2*b+c*x^2)/c^(1/2)+(c*x
^4+b*x^2+a)^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 361, normalized size of antiderivative = 2.41 \[ \int x^3 \left (a+b x^2+c x^4\right )^{3/2} \, dx=\left [\frac {15 \, {\left (b^{5} - 8 \, a b^{3} c + 16 \, a^{2} b c^{2}\right )} \sqrt {c} \log \left (-8 \, c^{2} x^{4} - 8 \, b c x^{2} - b^{2} + 4 \, \sqrt {c x^{4} + b x^{2} + a} {\left (2 \, c x^{2} + b\right )} \sqrt {c} - 4 \, a c\right ) + 4 \, {\left (128 \, c^{5} x^{8} + 176 \, b c^{4} x^{6} + 15 \, b^{4} c - 100 \, a b^{2} c^{2} + 128 \, a^{2} c^{3} + 8 \, {\left (b^{2} c^{3} + 32 \, a c^{4}\right )} x^{4} - 2 \, {\left (5 \, b^{3} c^{2} - 28 \, a b c^{3}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2} + a}}{5120 \, c^{4}}, \frac {15 \, {\left (b^{5} - 8 \, a b^{3} c + 16 \, a^{2} b c^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2} + a} {\left (2 \, c x^{2} + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{4} + b c x^{2} + a c\right )}}\right ) + 2 \, {\left (128 \, c^{5} x^{8} + 176 \, b c^{4} x^{6} + 15 \, b^{4} c - 100 \, a b^{2} c^{2} + 128 \, a^{2} c^{3} + 8 \, {\left (b^{2} c^{3} + 32 \, a c^{4}\right )} x^{4} - 2 \, {\left (5 \, b^{3} c^{2} - 28 \, a b c^{3}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2} + a}}{2560 \, c^{4}}\right ] \]

[In]

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

[Out]

[1/5120*(15*(b^5 - 8*a*b^3*c + 16*a^2*b*c^2)*sqrt(c)*log(-8*c^2*x^4 - 8*b*c*x^2 - b^2 + 4*sqrt(c*x^4 + b*x^2 +
a)*(2*c*x^2 + b)*sqrt(c) - 4*a*c) + 4*(128*c^5*x^8 + 176*b*c^4*x^6 + 15*b^4*c - 100*a*b^2*c^2 + 128*a^2*c^3 +
8*(b^2*c^3 + 32*a*c^4)*x^4 - 2*(5*b^3*c^2 - 28*a*b*c^3)*x^2)*sqrt(c*x^4 + b*x^2 + a))/c^4, 1/2560*(15*(b^5 -
8*a*b^3*c + 16*a^2*b*c^2)*sqrt(-c)*arctan(1/2*sqrt(c*x^4 + b*x^2 + a)*(2*c*x^2 + b)*sqrt(-c)/(c^2*x^4 + b*c*x^
2 + a*c)) + 2*(128*c^5*x^8 + 176*b*c^4*x^6 + 15*b^4*c - 100*a*b^2*c^2 + 128*a^2*c^3 + 8*(b^2*c^3 + 32*a*c^4)*x
^4 - 2*(5*b^3*c^2 - 28*a*b*c^3)*x^2)*sqrt(c*x^4 + b*x^2 + a))/c^4]

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F(-2)]

Exception generated. \[ \int x^3 \left (a+b x^2+c x^4\right )^{3/2} \, dx=\text {Exception raised: ValueError} \]

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` f
or more deta

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 408 vs. \(2 (128) = 256\).

Time = 0.34 (sec) , antiderivative size = 408, normalized size of antiderivative = 2.72 \[ \int x^3 \left (a+b x^2+c x^4\right )^{3/2} \, dx=\frac {1}{96} \, {\left (2 \, \sqrt {c x^{4} + b x^{2} + a} {\left (2 \, {\left (4 \, x^{2} + \frac {b}{c}\right )} x^{2} - \frac {3 \, b^{2} - 8 \, a c}{c^{2}}\right )} - \frac {3 \, {\left (b^{3} - 4 \, a b c\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} \sqrt {c} + b \right |}\right )}{c^{\frac {5}{2}}}\right )} a + \frac {1}{768} \, {\left (2 \, \sqrt {c x^{4} + b x^{2} + a} {\left (2 \, {\left (4 \, {\left (6 \, x^{2} + \frac {b}{c}\right )} x^{2} - \frac {5 \, b^{2} c - 12 \, a c^{2}}{c^{3}}\right )} x^{2} + \frac {15 \, b^{3} - 52 \, a b c}{c^{3}}\right )} + \frac {3 \, {\left (5 \, b^{4} - 24 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} \sqrt {c} + b \right |}\right )}{c^{\frac {7}{2}}}\right )} b + \frac {1}{7680} \, {\left (2 \, \sqrt {c x^{4} + b x^{2} + a} {\left (2 \, {\left (4 \, {\left (6 \, {\left (8 \, x^{2} + \frac {b}{c}\right )} x^{2} - \frac {7 \, b^{2} c^{2} - 16 \, a c^{3}}{c^{4}}\right )} x^{2} + \frac {35 \, b^{3} c - 116 \, a b c^{2}}{c^{4}}\right )} x^{2} - \frac {105 \, b^{4} - 460 \, a b^{2} c + 256 \, a^{2} c^{2}}{c^{4}}\right )} - \frac {15 \, {\left (7 \, b^{5} - 40 \, a b^{3} c + 48 \, a^{2} b c^{2}\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} \sqrt {c} + b \right |}\right )}{c^{\frac {9}{2}}}\right )} c \]

[In]

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

[Out]

1/96*(2*sqrt(c*x^4 + b*x^2 + a)*(2*(4*x^2 + b/c)*x^2 - (3*b^2 - 8*a*c)/c^2) - 3*(b^3 - 4*a*b*c)*log(abs(2*(sqr
t(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))*sqrt(c) + b))/c^(5/2))*a + 1/768*(2*sqrt(c*x^4 + b*x^2 + a)*(2*(4*(6*x^2 +
b/c)*x^2 - (5*b^2*c - 12*a*c^2)/c^3)*x^2 + (15*b^3 - 52*a*b*c)/c^3) + 3*(5*b^4 - 24*a*b^2*c + 16*a^2*c^2)*log
(abs(2*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))*sqrt(c) + b))/c^(7/2))*b + 1/7680*(2*sqrt(c*x^4 + b*x^2 + a)*(2
*(4*(6*(8*x^2 + b/c)*x^2 - (7*b^2*c^2 - 16*a*c^3)/c^4)*x^2 + (35*b^3*c - 116*a*b*c^2)/c^4)*x^2 - (105*b^4 - 46
0*a*b^2*c + 256*a^2*c^2)/c^4) - 15*(7*b^5 - 40*a*b^3*c + 48*a^2*b*c^2)*log(abs(2*(sqrt(c)*x^2 - sqrt(c*x^4 + b
*x^2 + a))*sqrt(c) + b))/c^(9/2))*c

Mupad [B] (verification not implemented)

Time = 13.94 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.49 \[ \int x^3 \left (a+b x^2+c x^4\right )^{3/2} \, dx=\frac {{\left (c\,x^4+b\,x^2+a\right )}^{5/2}}{10\,c}-\frac {b\,\left (\frac {3\,a\,\left (\ln \left (\sqrt {c\,x^4+b\,x^2+a}+\frac {c\,x^2+\frac {b}{2}}{\sqrt {c}}\right )\,\left (\frac {a}{2\,\sqrt {c}}-\frac {b^2}{8\,c^{3/2}}\right )+\frac {\left (2\,c\,x^2+b\right )\,\sqrt {c\,x^4+b\,x^2+a}}{4\,c}\right )}{4}+\frac {x^2\,{\left (c\,x^4+b\,x^2+a\right )}^{3/2}}{4}-\frac {3\,b^2\,\left (\ln \left (\sqrt {c\,x^4+b\,x^2+a}+\frac {c\,x^2+\frac {b}{2}}{\sqrt {c}}\right )\,\left (\frac {a}{2\,\sqrt {c}}-\frac {b^2}{8\,c^{3/2}}\right )+\frac {\left (2\,c\,x^2+b\right )\,\sqrt {c\,x^4+b\,x^2+a}}{4\,c}\right )}{16\,c}+\frac {b\,{\left (c\,x^4+b\,x^2+a\right )}^{3/2}}{8\,c}\right )}{4\,c} \]

[In]

int(x^3*(a + b*x^2 + c*x^4)^(3/2),x)

[Out]

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