3.943 \(\int \frac {(a+b x^2+c x^4)^{3/2}}{x^7} \, dx\)

Optimal. Leaf size=163 \[ \frac {b \left (b^2-12 a c\right ) \tanh ^{-1}\left (\frac {2 a+b x^2}{2 \sqrt {a} \sqrt {a+b x^2+c x^4}}\right )}{32 a^{3/2}}-\frac {\left (x^2 \left (8 a c+b^2\right )+2 a b\right ) \sqrt {a+b x^2+c x^4}}{16 a x^4}+\frac {1}{2} c^{3/2} \tanh ^{-1}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )-\frac {\left (a+b x^2+c x^4\right )^{3/2}}{6 x^6} \]

[Out]

-1/6*(c*x^4+b*x^2+a)^(3/2)/x^6+1/32*b*(-12*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^(
3/2)+1/2*c^(3/2)*arctanh(1/2*(2*c*x^2+b)/c^(1/2)/(c*x^4+b*x^2+a)^(1/2))-1/16*(2*a*b+(8*a*c+b^2)*x^2)*(c*x^4+b*
x^2+a)^(1/2)/a/x^4

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Rubi [A]  time = 0.18, antiderivative size = 163, 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 = {1114, 732, 810, 843, 621, 206, 724} \[ \frac {b \left (b^2-12 a c\right ) \tanh ^{-1}\left (\frac {2 a+b x^2}{2 \sqrt {a} \sqrt {a+b x^2+c x^4}}\right )}{32 a^{3/2}}-\frac {\left (x^2 \left (8 a c+b^2\right )+2 a b\right ) \sqrt {a+b x^2+c x^4}}{16 a x^4}+\frac {1}{2} c^{3/2} \tanh ^{-1}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )-\frac {\left (a+b x^2+c x^4\right )^{3/2}}{6 x^6} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2 + c*x^4)^(3/2)/x^7,x]

[Out]

-((2*a*b + (b^2 + 8*a*c)*x^2)*Sqrt[a + b*x^2 + c*x^4])/(16*a*x^4) - (a + b*x^2 + c*x^4)^(3/2)/(6*x^6) + (b*(b^
2 - 12*a*c)*ArcTanh[(2*a + b*x^2)/(2*Sqrt[a]*Sqrt[a + b*x^2 + c*x^4])])/(32*a^(3/2)) + (c^(3/2)*ArcTanh[(b + 2
*c*x^2)/(2*Sqrt[c]*Sqrt[a + b*x^2 + c*x^4])])/2

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 621

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 724

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 732

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 810

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Si
mp[((d + e*x)^(m + 1)*(a + b*x + c*x^2)^p*((d*g - e*f*(m + 2))*(c*d^2 - b*d*e + a*e^2) - d*p*(2*c*d - b*e)*(e*
f - d*g) - e*(g*(m + 1)*(c*d^2 - b*d*e + a*e^2) + p*(2*c*d - b*e)*(e*f - d*g))*x))/(e^2*(m + 1)*(m + 2)*(c*d^2
- b*d*e + a*e^2)), x] - Dist[p/(e^2*(m + 1)*(m + 2)*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^(m + 2)*(a + b*x
+ c*x^2)^(p - 1)*Simp[2*a*c*e*(e*f - d*g)*(m + 2) + b^2*e*(d*g*(p + 1) - e*f*(m + p + 2)) + b*(a*e^2*g*(m + 1)
- c*d*(d*g*(2*p + 1) - e*f*(m + 2*p + 2))) - c*(2*c*d*(d*g*(2*p + 1) - e*f*(m + 2*p + 2)) - e*(2*a*e*g*(m + 1
) - b*(d*g*(m - 2*p) + e*f*(m + 2*p + 2))))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 - 4*a*
c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && GtQ[p, 0] && LtQ[m, -2] && LtQ[m + 2*p, 0] &&  !ILtQ[m + 2*p + 3, 0]

Rule 843

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 1114

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^7} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {\left (a+b x+c x^2\right )^{3/2}}{x^4} \, dx,x,x^2\right )\\ &=-\frac {\left (a+b x^2+c x^4\right )^{3/2}}{6 x^6}+\frac {1}{4} \operatorname {Subst}\left (\int \frac {(b+2 c x) \sqrt {a+b x+c x^2}}{x^3} \, dx,x,x^2\right )\\ &=-\frac {\left (2 a b+\left (b^2+8 a c\right ) x^2\right ) \sqrt {a+b x^2+c x^4}}{16 a x^4}-\frac {\left (a+b x^2+c x^4\right )^{3/2}}{6 x^6}-\frac {\operatorname {Subst}\left (\int \frac {\frac {1}{2} b \left (b^2-12 a c\right )-8 a c^2 x}{x \sqrt {a+b x+c x^2}} \, dx,x,x^2\right )}{16 a}\\ &=-\frac {\left (2 a b+\left (b^2+8 a c\right ) x^2\right ) \sqrt {a+b x^2+c x^4}}{16 a x^4}-\frac {\left (a+b x^2+c x^4\right )^{3/2}}{6 x^6}+\frac {1}{2} c^2 \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b x+c x^2}} \, dx,x,x^2\right )-\frac {\left (b \left (b^2-12 a c\right )\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+b x+c x^2}} \, dx,x,x^2\right )}{32 a}\\ &=-\frac {\left (2 a b+\left (b^2+8 a c\right ) x^2\right ) \sqrt {a+b x^2+c x^4}}{16 a x^4}-\frac {\left (a+b x^2+c x^4\right )^{3/2}}{6 x^6}+c^2 \operatorname {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 {\left (b \left (b^2-12 a c\right )\right ) \operatorname {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 )}{16 a}\\ &=-\frac {\left (2 a b+\left (b^2+8 a c\right ) x^2\right ) \sqrt {a+b x^2+c x^4}}{16 a x^4}-\frac {\left (a+b x^2+c x^4\right )^{3/2}}{6 x^6}+\frac {b \left (b^2-12 a c\right ) \tanh ^{-1}\left (\frac {2 a+b x^2}{2 \sqrt {a} \sqrt {a+b x^2+c x^4}}\right )}{32 a^{3/2}}+\frac {1}{2} c^{3/2} \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 = 0.22, size = 149, normalized size = 0.91 \[ \frac {1}{96} \left (\frac {3 b \left (b^2-12 a c\right ) \tanh ^{-1}\left (\frac {2 a+b x^2}{2 \sqrt {a} \sqrt {a+b x^2+c x^4}}\right )}{a^{3/2}}-\frac {2 \sqrt {a+b x^2+c x^4} \left (8 a^2+14 a b x^2+32 a c x^4+3 b^2 x^4\right )}{a x^6}+48 c^{3/2} \tanh ^{-1}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2 + c*x^4)^(3/2)/x^7,x]

[Out]

((-2*Sqrt[a + b*x^2 + c*x^4]*(8*a^2 + 14*a*b*x^2 + 3*b^2*x^4 + 32*a*c*x^4))/(a*x^6) + (3*b*(b^2 - 12*a*c)*ArcT
anh[(2*a + b*x^2)/(2*Sqrt[a]*Sqrt[a + b*x^2 + c*x^4])])/a^(3/2) + 48*c^(3/2)*ArcTanh[(b + 2*c*x^2)/(2*Sqrt[c]*
Sqrt[a + b*x^2 + c*x^4])])/96

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fricas [A]  time = 1.16, size = 771, normalized size = 4.73 \[ \left [\frac {48 \, a^{2} c^{\frac {3}{2}} x^{6} \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^{3} - 12 \, a b c\right )} \sqrt {a} x^{6} \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 (14 \, a^{2} b x^{2} + {\left (3 \, a b^{2} + 32 \, a^{2} c\right )} x^{4} + 8 \, a^{3}\right )} \sqrt {c x^{4} + b x^{2} + a}}{192 \, a^{2} x^{6}}, -\frac {96 \, a^{2} \sqrt {-c} c x^{6} \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^{3} - 12 \, a b c\right )} \sqrt {a} x^{6} \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 (14 \, a^{2} b x^{2} + {\left (3 \, a b^{2} + 32 \, a^{2} c\right )} x^{4} + 8 \, a^{3}\right )} \sqrt {c x^{4} + b x^{2} + a}}{192 \, a^{2} x^{6}}, \frac {24 \, a^{2} c^{\frac {3}{2}} x^{6} \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^{3} - 12 \, a b c\right )} \sqrt {-a} x^{6} \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 (14 \, a^{2} b x^{2} + {\left (3 \, a b^{2} + 32 \, a^{2} c\right )} x^{4} + 8 \, a^{3}\right )} \sqrt {c x^{4} + b x^{2} + a}}{96 \, a^{2} x^{6}}, -\frac {48 \, a^{2} \sqrt {-c} c x^{6} \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^{3} - 12 \, a b c\right )} \sqrt {-a} x^{6} \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 (14 \, a^{2} b x^{2} + {\left (3 \, a b^{2} + 32 \, a^{2} c\right )} x^{4} + 8 \, a^{3}\right )} \sqrt {c x^{4} + b x^{2} + a}}{96 \, a^{2} x^{6}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2+a)^(3/2)/x^7,x, algorithm="fricas")

[Out]

[1/192*(48*a^2*c^(3/2)*x^6*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^3 - 12*a*b*c)*sqrt(a)*x^6*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*(14*a^2*b*x^2 + (3*a*b^2 + 32*a^2*c)*x^4 + 8*a^3)*sqrt(c*x^4 + b*x^2 + a))
/(a^2*x^6), -1/192*(96*a^2*sqrt(-c)*c*x^6*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^3 - 12*a*b*c)*sqrt(a)*x^6*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*(14*a^2*b*x^2 + (3*a*b^2 + 32*a^2*c)*x^4 + 8*a^3)*sqrt(c*x^4 + b*x
^2 + a))/(a^2*x^6), 1/96*(24*a^2*c^(3/2)*x^6*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^3 - 12*a*b*c)*sqrt(-a)*x^6*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*(14*a^2*b*x^2 + (3*a*b^2 + 32*a^2*c)*x^4 + 8*a^3)*sqrt(c*x^4 + b*x^2 +
a))/(a^2*x^6), -1/96*(48*a^2*sqrt(-c)*c*x^6*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^3 - 12*a*b*c)*sqrt(-a)*x^6*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*(14*a^2*b*x^2 + (3*a*b^2 + 32*a^2*c)*x^4 + 8*a^3)*sqrt(c*x^4 + b*x^2 + a))/(a
^2*x^6)]

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giac [B]  time = 0.68, size = 412, normalized size = 2.53 \[ -\frac {1}{2} \, c^{\frac {3}{2}} \log \left ({\left | -2 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} \sqrt {c} - b \right |}\right ) - \frac {{\left (b^{3} - 12 \, a b c\right )} \arctan \left (-\frac {\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}}{\sqrt {-a}}\right )}{16 \, \sqrt {-a} a} + \frac {3 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{5} b^{3} \sqrt {c} + 60 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{5} a b c^{\frac {3}{2}} + 48 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{4} a b^{2} c + 96 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{4} a^{2} c^{2} + 8 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{3} a b^{3} \sqrt {c} - 96 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{2} a^{3} c^{2} - 3 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} a^{2} b^{3} \sqrt {c} + 36 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} a^{3} b c^{\frac {3}{2}} + 64 \, a^{4} c^{2}}{48 \, {\left ({\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{2} - a\right )}^{3} a \sqrt {c}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2+a)^(3/2)/x^7,x, algorithm="giac")

[Out]

-1/2*c^(3/2)*log(abs(-2*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))*sqrt(c) - b)) - 1/16*(b^3 - 12*a*b*c)*arctan(-
(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))/sqrt(-a))/(sqrt(-a)*a) + 1/48*(3*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a
))^5*b^3*sqrt(c) + 60*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))^5*a*b*c^(3/2) + 48*(sqrt(c)*x^2 - sqrt(c*x^4 + b
*x^2 + a))^4*a*b^2*c + 96*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))^4*a^2*c^2 + 8*(sqrt(c)*x^2 - sqrt(c*x^4 + b*
x^2 + a))^3*a*b^3*sqrt(c) - 96*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))^2*a^3*c^2 - 3*(sqrt(c)*x^2 - sqrt(c*x^4
+ b*x^2 + a))*a^2*b^3*sqrt(c) + 36*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))*a^3*b*c^(3/2) + 64*a^4*c^2)/(((sqr
t(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))^2 - a)^3*a*sqrt(c))

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maple [A]  time = 0.02, size = 202, normalized size = 1.24 \[ -\frac {3 b c \ln \left (\frac {b \,x^{2}+2 a +2 \sqrt {c \,x^{4}+b \,x^{2}+a}\, \sqrt {a}}{x^{2}}\right )}{8 \sqrt {a}}+\frac {b^{3} \ln \left (\frac {b \,x^{2}+2 a +2 \sqrt {c \,x^{4}+b \,x^{2}+a}\, \sqrt {a}}{x^{2}}\right )}{32 a^{\frac {3}{2}}}+\frac {c^{\frac {3}{2}} \ln \left (\frac {c \,x^{2}+\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{2}-\frac {\sqrt {c \,x^{4}+b \,x^{2}+a}\, b^{2}}{16 a \,x^{2}}-\frac {2 \sqrt {c \,x^{4}+b \,x^{2}+a}\, c}{3 x^{2}}-\frac {7 \sqrt {c \,x^{4}+b \,x^{2}+a}\, b}{24 x^{4}}-\frac {\sqrt {c \,x^{4}+b \,x^{2}+a}\, a}{6 x^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^4+b*x^2+a)^(3/2)/x^7,x)

[Out]

1/2*c^(3/2)*ln((c*x^2+1/2*b)/c^(1/2)+(c*x^4+b*x^2+a)^(1/2))-7/24*b/x^4*(c*x^4+b*x^2+a)^(1/2)-1/16/a*b^2/x^2*(c
*x^4+b*x^2+a)^(1/2)+1/32/a^(3/2)*b^3*ln((b*x^2+2*a+2*(c*x^4+b*x^2+a)^(1/2)*a^(1/2))/x^2)-3/8*b*c/a^(1/2)*ln((b
*x^2+2*a+2*(c*x^4+b*x^2+a)^(1/2)*a^(1/2))/x^2)-2/3*c/x^2*(c*x^4+b*x^2+a)^(1/2)-1/6*a/x^6*(c*x^4+b*x^2+a)^(1/2)

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2+a)^(3/2)/x^7,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 details)Is 4*a*c-b^2 positive, negative or zero?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (c\,x^4+b\,x^2+a\right )}^{3/2}}{x^7} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2 + c*x^4)^(3/2)/x^7,x)

[Out]

int((a + b*x^2 + c*x^4)^(3/2)/x^7, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b x^{2} + c x^{4}\right )^{\frac {3}{2}}}{x^{7}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**4+b*x**2+a)**(3/2)/x**7,x)

[Out]

Integral((a + b*x**2 + c*x**4)**(3/2)/x**7, x)

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