3.10.42 \(\int \frac {(a+b x^2+c x^4)^{3/2}}{x^5} \, dx\) [942]

Optimal. Leaf size=151 \[ -\frac {3 \left (b-2 c x^2\right ) \sqrt {a+b x^2+c x^4}}{8 x^2}-\frac {\left (a+b x^2+c x^4\right )^{3/2}}{4 x^4}-\frac {3 \left (b^2+4 a c\right ) \tanh ^{-1}\left (\frac {2 a+b x^2}{2 \sqrt {a} \sqrt {a+b x^2+c x^4}}\right )}{16 \sqrt {a}}+\frac {3}{4} b \sqrt {c} \tanh ^{-1}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right ) \]

[Out]

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

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Rubi [A]
time = 0.11, antiderivative size = 151, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {1128, 746, 826, 857, 635, 212, 738} \begin {gather*} -\frac {3 \left (4 a c+b^2\right ) \tanh ^{-1}\left (\frac {2 a+b x^2}{2 \sqrt {a} \sqrt {a+b x^2+c x^4}}\right )}{16 \sqrt {a}}-\frac {\left (a+b x^2+c x^4\right )^{3/2}}{4 x^4}-\frac {3 \left (b-2 c x^2\right ) \sqrt {a+b x^2+c x^4}}{8 x^2}+\frac {3}{4} b \sqrt {c} \tanh ^{-1}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2 + c*x^4)^(3/2)/x^5,x]

[Out]

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

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 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 738

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

Rule 746

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d + e*x)^(m + 1)*((
a + b*x + c*x^2)^p/(e*(m + 1))), x] - Dist[p/(e*(m + 1)), Int[(d + e*x)^(m + 1)*(b + 2*c*x)*(a + b*x + c*x^2)^
(p - 1), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ
[2*c*d - b*e, 0] && GtQ[p, 0] && (IntegerQ[p] || LtQ[m, -1]) && NeQ[m, -1] &&  !ILtQ[m + 2*p + 1, 0] && IntQua
draticQ[a, b, c, d, e, m, p, x]

Rule 826

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Sim
p[(d + e*x)^(m + 1)*(e*f*(m + 2*p + 2) - d*g*(2*p + 1) + e*g*(m + 1)*x)*((a + b*x + c*x^2)^p/(e^2*(m + 1)*(m +
2*p + 2))), x] + Dist[p/(e^2*(m + 1)*(m + 2*p + 2)), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^(p - 1)*Simp[g*(
b*d + 2*a*e + 2*a*e*m + 2*b*d*p) - f*b*e*(m + 2*p + 2) + (g*(2*c*d + b*e + b*e*m + 4*c*d*p) - 2*c*e*f*(m + 2*p
+ 2))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2
, 0] && RationalQ[p] && p > 0 && (LtQ[m, -1] || EqQ[p, 1] || (IntegerQ[p] &&  !RationalQ[m])) && NeQ[m, -1] &&
!ILtQ[m + 2*p + 1, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 857

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dis
t[g/e, Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + b*x + c*x^
2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0]
&&  !IGtQ[m, 0]

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*} \int \frac {\left (a+b x^2+c x^4\right )^{3/2}}{x^5} \, dx &=\frac {1}{2} \text {Subst}\left (\int \frac {\left (a+b x+c x^2\right )^{3/2}}{x^3} \, dx,x,x^2\right )\\ &=-\frac {\left (a+b x^2+c x^4\right )^{3/2}}{4 x^4}+\frac {3}{8} \text {Subst}\left (\int \frac {(b+2 c x) \sqrt {a+b x+c x^2}}{x^2} \, dx,x,x^2\right )\\ &=-\frac {3 \left (b-2 c x^2\right ) \sqrt {a+b x^2+c x^4}}{8 x^2}-\frac {\left (a+b x^2+c x^4\right )^{3/2}}{4 x^4}-\frac {3}{16} \text {Subst}\left (\int \frac {-b^2-4 a c-4 b c x}{x \sqrt {a+b x+c x^2}} \, dx,x,x^2\right )\\ &=-\frac {3 \left (b-2 c x^2\right ) \sqrt {a+b x^2+c x^4}}{8 x^2}-\frac {\left (a+b x^2+c x^4\right )^{3/2}}{4 x^4}+\frac {1}{4} (3 b c) \text {Subst}\left (\int \frac {1}{\sqrt {a+b x+c x^2}} \, dx,x,x^2\right )+\frac {1}{16} \left (3 \left (b^2+4 a c\right )\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {a+b x+c x^2}} \, dx,x,x^2\right )\\ &=-\frac {3 \left (b-2 c x^2\right ) \sqrt {a+b x^2+c x^4}}{8 x^2}-\frac {\left (a+b x^2+c x^4\right )^{3/2}}{4 x^4}+\frac {1}{2} (3 b c) \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 )-\frac {1}{8} \left (3 \left (b^2+4 a c\right )\right ) \text {Subst}\left (\int \frac {1}{4 a-x^2} \, dx,x,\frac {2 a+b x^2}{\sqrt {a+b x^2+c x^4}}\right )\\ &=-\frac {3 \left (b-2 c x^2\right ) \sqrt {a+b x^2+c x^4}}{8 x^2}-\frac {\left (a+b x^2+c x^4\right )^{3/2}}{4 x^4}-\frac {3 \left (b^2+4 a c\right ) \tanh ^{-1}\left (\frac {2 a+b x^2}{2 \sqrt {a} \sqrt {a+b x^2+c x^4}}\right )}{16 \sqrt {a}}+\frac {3}{4} b \sqrt {c} \tanh ^{-1}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )\\ \end {align*}

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Mathematica [A]
time = 0.51, size = 131, normalized size = 0.87 \begin {gather*} \frac {1}{8} \left (\frac {\sqrt {a+b x^2+c x^4} \left (-2 a-5 b x^2+4 c x^4\right )}{x^4}+\frac {3 \left (b^2+4 a c\right ) \tanh ^{-1}\left (\frac {\sqrt {c} x^2-\sqrt {a+b x^2+c x^4}}{\sqrt {a}}\right )}{\sqrt {a}}-6 b \sqrt {c} \log \left (b+2 c x^2-2 \sqrt {c} \sqrt {a+b x^2+c x^4}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2 + c*x^4)^(3/2)/x^5,x]

[Out]

((Sqrt[a + b*x^2 + c*x^4]*(-2*a - 5*b*x^2 + 4*c*x^4))/x^4 + (3*(b^2 + 4*a*c)*ArcTanh[(Sqrt[c]*x^2 - Sqrt[a + b
*x^2 + c*x^4])/Sqrt[a]])/Sqrt[a] - 6*b*Sqrt[c]*Log[b + 2*c*x^2 - 2*Sqrt[c]*Sqrt[a + b*x^2 + c*x^4]])/8

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Maple [A]
time = 0.07, size = 174, normalized size = 1.15

method	result	size
risch	\(-\frac {\sqrt {c \,x^{4}+b \,x^{2}+a}\, \left (5 b \,x^{2}+2 a \right )}{8 x^{4}}+\frac {c \sqrt {c \,x^{4}+b \,x^{2}+a}}{2}+\frac {3 b \sqrt {c}\, \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{4}-\frac {3 \sqrt {a}\, c \ln \left (\frac {2 a +b \,x^{2}+2 \sqrt {a}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{x^{2}}\right )}{4}-\frac {3 b^{2} \ln \left (\frac {2 a +b \,x^{2}+2 \sqrt {a}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{x^{2}}\right )}{16 \sqrt {a}}\)	\(163\)
default	\(\frac {c \sqrt {c \,x^{4}+b \,x^{2}+a}}{2}+\frac {3 b \sqrt {c}\, \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{4}-\frac {a \sqrt {c \,x^{4}+b \,x^{2}+a}}{4 x^{4}}-\frac {5 b \sqrt {c \,x^{4}+b \,x^{2}+a}}{8 x^{2}}-\frac {3 b^{2} \ln \left (\frac {2 a +b \,x^{2}+2 \sqrt {a}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{x^{2}}\right )}{16 \sqrt {a}}-\frac {3 \sqrt {a}\, c \ln \left (\frac {2 a +b \,x^{2}+2 \sqrt {a}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{x^{2}}\right )}{4}\)	\(174\)
elliptic	\(\frac {c \sqrt {c \,x^{4}+b \,x^{2}+a}}{2}+\frac {3 b \sqrt {c}\, \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{4}-\frac {a \sqrt {c \,x^{4}+b \,x^{2}+a}}{4 x^{4}}-\frac {5 b \sqrt {c \,x^{4}+b \,x^{2}+a}}{8 x^{2}}-\frac {3 b^{2} \ln \left (\frac {2 a +b \,x^{2}+2 \sqrt {a}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{x^{2}}\right )}{16 \sqrt {a}}-\frac {3 \sqrt {a}\, c \ln \left (\frac {2 a +b \,x^{2}+2 \sqrt {a}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{x^{2}}\right )}{4}\)	\(174\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^4+b*x^2+a)^(3/2)/x^5,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2+a)^(3/2)/x^5,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

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Fricas [A]
time = 0.43, size = 713, normalized size = 4.72 \begin {gather*} \left [\frac {12 \, a b \sqrt {c} x^{4} \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 ) + 3 \, {\left (b^{2} + 4 \, a c\right )} \sqrt {a} x^{4} \log \left (-\frac {{\left (b^{2} + 4 \, a c\right )} x^{4} + 8 \, a b x^{2} - 4 \, \sqrt {c x^{4} + b x^{2} + a} {\left (b x^{2} + 2 \, a\right )} \sqrt {a} + 8 \, a^{2}}{x^{4}}\right ) + 4 \, {\left (4 \, a c x^{4} - 5 \, a b x^{2} - 2 \, a^{2}\right )} \sqrt {c x^{4} + b x^{2} + a}}{32 \, a x^{4}}, -\frac {24 \, a b \sqrt {-c} x^{4} \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 ) - 3 \, {\left (b^{2} + 4 \, a c\right )} \sqrt {a} x^{4} \log \left (-\frac {{\left (b^{2} + 4 \, a c\right )} x^{4} + 8 \, a b x^{2} - 4 \, \sqrt {c x^{4} + b x^{2} + a} {\left (b x^{2} + 2 \, a\right )} \sqrt {a} + 8 \, a^{2}}{x^{4}}\right ) - 4 \, {\left (4 \, a c x^{4} - 5 \, a b x^{2} - 2 \, a^{2}\right )} \sqrt {c x^{4} + b x^{2} + a}}{32 \, a x^{4}}, \frac {6 \, a b \sqrt {c} x^{4} \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 ) + 3 \, {\left (b^{2} + 4 \, a c\right )} \sqrt {-a} x^{4} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2} + a} {\left (b x^{2} + 2 \, a\right )} \sqrt {-a}}{2 \, {\left (a c x^{4} + a b x^{2} + a^{2}\right )}}\right ) + 2 \, {\left (4 \, a c x^{4} - 5 \, a b x^{2} - 2 \, a^{2}\right )} \sqrt {c x^{4} + b x^{2} + a}}{16 \, a x^{4}}, -\frac {12 \, a b \sqrt {-c} x^{4} \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 ) - 3 \, {\left (b^{2} + 4 \, a c\right )} \sqrt {-a} x^{4} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2} + a} {\left (b x^{2} + 2 \, a\right )} \sqrt {-a}}{2 \, {\left (a c x^{4} + a b x^{2} + a^{2}\right )}}\right ) - 2 \, {\left (4 \, a c x^{4} - 5 \, a b x^{2} - 2 \, a^{2}\right )} \sqrt {c x^{4} + b x^{2} + a}}{16 \, a x^{4}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2+a)^(3/2)/x^5,x, algorithm="fricas")

[Out]

[1/32*(12*a*b*sqrt(c)*x^4*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) + 3*(b^2 + 4*a*c)*sqrt(a)*x^4*log(-((b^2 + 4*a*c)*x^4 + 8*a*b*x^2 - 4*sqrt(c*x^4 + b*x^2 + a)*(b*x^2 +
2*a)*sqrt(a) + 8*a^2)/x^4) + 4*(4*a*c*x^4 - 5*a*b*x^2 - 2*a^2)*sqrt(c*x^4 + b*x^2 + a))/(a*x^4), -1/32*(24*a*
b*sqrt(-c)*x^4*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)) - 3*(b^2 +
4*a*c)*sqrt(a)*x^4*log(-((b^2 + 4*a*c)*x^4 + 8*a*b*x^2 - 4*sqrt(c*x^4 + b*x^2 + a)*(b*x^2 + 2*a)*sqrt(a) + 8*
a^2)/x^4) - 4*(4*a*c*x^4 - 5*a*b*x^2 - 2*a^2)*sqrt(c*x^4 + b*x^2 + a))/(a*x^4), 1/16*(6*a*b*sqrt(c)*x^4*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) + 3*(b^2 + 4*a*c)*sqrt(-
a)*x^4*arctan(1/2*sqrt(c*x^4 + b*x^2 + a)*(b*x^2 + 2*a)*sqrt(-a)/(a*c*x^4 + a*b*x^2 + a^2)) + 2*(4*a*c*x^4 - 5
*a*b*x^2 - 2*a^2)*sqrt(c*x^4 + b*x^2 + a))/(a*x^4), -1/16*(12*a*b*sqrt(-c)*x^4*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)) - 3*(b^2 + 4*a*c)*sqrt(-a)*x^4*arctan(1/2*sqrt(c*x^4 + b
*x^2 + a)*(b*x^2 + 2*a)*sqrt(-a)/(a*c*x^4 + a*b*x^2 + a^2)) - 2*(4*a*c*x^4 - 5*a*b*x^2 - 2*a^2)*sqrt(c*x^4 + b
*x^2 + a))/(a*x^4)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a + b x^{2} + c x^{4}\right )^{\frac {3}{2}}}{x^{5}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**4+b*x**2+a)**(3/2)/x**5,x)

[Out]

Integral((a + b*x**2 + c*x**4)**(3/2)/x**5, x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 302 vs. \(2 (125) = 250\).
time = 4.87, size = 302, normalized size = 2.00 \begin {gather*} -\frac {3}{4} \, b \sqrt {c} \log \left ({\left | 2 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} \sqrt {c} + b \right |}\right ) + \frac {1}{2} \, \sqrt {c x^{4} + b x^{2} + a} c + \frac {3 \, {\left (b^{2} + 4 \, a c\right )} \arctan \left (-\frac {\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}}{\sqrt {-a}}\right )}{8 \, \sqrt {-a}} + \frac {5 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{3} b^{2} + 4 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{3} a c + 16 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{2} a b \sqrt {c} - 3 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} a b^{2} + 4 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} a^{2} c - 8 \, a^{2} b \sqrt {c}}{8 \, {\left ({\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{2} - a\right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2+a)^(3/2)/x^5,x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (c\,x^4+b\,x^2+a\right )}^{3/2}}{x^5} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2 + c*x^4)^(3/2)/x^5,x)

[Out]

int((a + b*x^2 + c*x^4)^(3/2)/x^5, x)

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