3.8.42 \(\int \frac {(a+b x^2+c x^4)^{3/2}}{x^{11}} \, dx\)

Optimal. Leaf size=162 \[ \frac {3 b \left (b^2-4 a c\right )^2 \tanh ^{-1}\left (\frac {2 a+b x^2}{2 \sqrt {a} \sqrt {a+b x^2+c x^4}}\right )}{512 a^{7/2}}-\frac {3 b \left (b^2-4 a c\right ) \left (2 a+b x^2\right ) \sqrt {a+b x^2+c x^4}}{256 a^3 x^4}+\frac {b \left (2 a+b x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{32 a^2 x^8}-\frac {\left (a+b x^2+c x^4\right )^{5/2}}{10 a x^{10}} \]

________________________________________________________________________________________

Rubi [A]  time = 0.14, antiderivative size = 162, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1114, 730, 720, 724, 206} \begin {gather*} -\frac {3 b \left (b^2-4 a c\right ) \left (2 a+b x^2\right ) \sqrt {a+b x^2+c x^4}}{256 a^3 x^4}+\frac {3 b \left (b^2-4 a c\right )^2 \tanh ^{-1}\left (\frac {2 a+b x^2}{2 \sqrt {a} \sqrt {a+b x^2+c x^4}}\right )}{512 a^{7/2}}+\frac {b \left (2 a+b x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{32 a^2 x^8}-\frac {\left (a+b x^2+c x^4\right )^{5/2}}{10 a x^{10}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2 + c*x^4)^(3/2)/x^11,x]

[Out]

(-3*b*(b^2 - 4*a*c)*(2*a + b*x^2)*Sqrt[a + b*x^2 + c*x^4])/(256*a^3*x^4) + (b*(2*a + b*x^2)*(a + b*x^2 + c*x^4
)^(3/2))/(32*a^2*x^8) - (a + b*x^2 + c*x^4)^(5/2)/(10*a*x^10) + (3*b*(b^2 - 4*a*c)^2*ArcTanh[(2*a + b*x^2)/(2*
Sqrt[a]*Sqrt[a + b*x^2 + c*x^4])])/(512*a^(7/2))

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 720

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(m + 1)*
(d*b - 2*a*e + (2*c*d - b*e)*x)*(a + b*x + c*x^2)^p)/(2*(m + 1)*(c*d^2 - b*d*e + a*e^2)), x] + Dist[(p*(b^2 -
4*a*c))/(2*(m + 1)*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^(m + 2)*(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[
{a, b, c, d, e}, 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] && EqQ[m +
2*p + 2, 0] && GtQ[p, 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 730

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m + 1)
*(a + b*x + c*x^2)^(p + 1))/((m + 1)*(c*d^2 - b*d*e + a*e^2)), x] + Dist[(2*c*d - b*e)/(2*(c*d^2 - b*d*e + a*e
^2)), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, 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] && EqQ[m + 2*p + 3, 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^{11}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {\left (a+b x+c x^2\right )^{3/2}}{x^6} \, dx,x,x^2\right )\\ &=-\frac {\left (a+b x^2+c x^4\right )^{5/2}}{10 a x^{10}}-\frac {b \operatorname {Subst}\left (\int \frac {\left (a+b x+c x^2\right )^{3/2}}{x^5} \, dx,x,x^2\right )}{4 a}\\ &=\frac {b \left (2 a+b x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{32 a^2 x^8}-\frac {\left (a+b x^2+c x^4\right )^{5/2}}{10 a x^{10}}+\frac {\left (3 b \left (b^2-4 a c\right )\right ) \operatorname {Subst}\left (\int \frac {\sqrt {a+b x+c x^2}}{x^3} \, dx,x,x^2\right )}{64 a^2}\\ &=-\frac {3 b \left (b^2-4 a c\right ) \left (2 a+b x^2\right ) \sqrt {a+b x^2+c x^4}}{256 a^3 x^4}+\frac {b \left (2 a+b x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{32 a^2 x^8}-\frac {\left (a+b x^2+c x^4\right )^{5/2}}{10 a x^{10}}-\frac {\left (3 b \left (b^2-4 a c\right )^2\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+b x+c x^2}} \, dx,x,x^2\right )}{512 a^3}\\ &=-\frac {3 b \left (b^2-4 a c\right ) \left (2 a+b x^2\right ) \sqrt {a+b x^2+c x^4}}{256 a^3 x^4}+\frac {b \left (2 a+b x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{32 a^2 x^8}-\frac {\left (a+b x^2+c x^4\right )^{5/2}}{10 a x^{10}}+\frac {\left (3 b \left (b^2-4 a c\right )^2\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 )}{256 a^3}\\ &=-\frac {3 b \left (b^2-4 a c\right ) \left (2 a+b x^2\right ) \sqrt {a+b x^2+c x^4}}{256 a^3 x^4}+\frac {b \left (2 a+b x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{32 a^2 x^8}-\frac {\left (a+b x^2+c x^4\right )^{5/2}}{10 a x^{10}}+\frac {3 b \left (b^2-4 a c\right )^2 \tanh ^{-1}\left (\frac {2 a+b x^2}{2 \sqrt {a} \sqrt {a+b x^2+c x^4}}\right )}{512 a^{7/2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.14, size = 167, normalized size = 1.03 \begin {gather*} \frac {b \left (16 a^{3/2} \left (2 a+b x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}-3 x^4 \left (b^2-4 a c\right ) \left (2 \sqrt {a} \left (2 a+b x^2\right ) \sqrt {a+b x^2+c x^4}-x^4 \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 )\right )\right )}{512 a^{7/2} x^8}-\frac {\left (a+b x^2+c x^4\right )^{5/2}}{10 a x^{10}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2 + c*x^4)^(3/2)/x^11,x]

[Out]

-1/10*(a + b*x^2 + c*x^4)^(5/2)/(a*x^10) + (b*(16*a^(3/2)*(2*a + b*x^2)*(a + b*x^2 + c*x^4)^(3/2) - 3*(b^2 - 4
*a*c)*x^4*(2*Sqrt[a]*(2*a + b*x^2)*Sqrt[a + b*x^2 + c*x^4] - (b^2 - 4*a*c)*x^4*ArcTanh[(2*a + b*x^2)/(2*Sqrt[a
]*Sqrt[a + b*x^2 + c*x^4])])))/(512*a^(7/2)*x^8)

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 1.36, size = 174, normalized size = 1.07 \begin {gather*} \frac {\sqrt {a+b x^2+c x^4} \left (-128 a^4-176 a^3 b x^2-256 a^3 c x^4-8 a^2 b^2 x^4-56 a^2 b c x^6-128 a^2 c^2 x^8+10 a b^3 x^6+100 a b^2 c x^8-15 b^4 x^8\right )}{1280 a^3 x^{10}}-\frac {3 \left (16 a^2 b c^2-8 a b^3 c+b^5\right ) \tanh ^{-1}\left (\frac {\sqrt {c} x^2-\sqrt {a+b x^2+c x^4}}{\sqrt {a}}\right )}{256 a^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(a + b*x^2 + c*x^4)^(3/2)/x^11,x]

[Out]

(Sqrt[a + b*x^2 + c*x^4]*(-128*a^4 - 176*a^3*b*x^2 - 8*a^2*b^2*x^4 - 256*a^3*c*x^4 + 10*a*b^3*x^6 - 56*a^2*b*c
*x^6 - 15*b^4*x^8 + 100*a*b^2*c*x^8 - 128*a^2*c^2*x^8))/(1280*a^3*x^10) - (3*(b^5 - 8*a*b^3*c + 16*a^2*b*c^2)*
ArcTanh[(Sqrt[c]*x^2 - Sqrt[a + b*x^2 + c*x^4])/Sqrt[a]])/(256*a^(7/2))

________________________________________________________________________________________

fricas [A]  time = 1.56, size = 383, normalized size = 2.36 \begin {gather*} \left [\frac {15 \, {\left (b^{5} - 8 \, a b^{3} c + 16 \, a^{2} b c^{2}\right )} \sqrt {a} x^{10} \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 ({\left (15 \, a b^{4} - 100 \, a^{2} b^{2} c + 128 \, a^{3} c^{2}\right )} x^{8} + 176 \, a^{4} b x^{2} - 2 \, {\left (5 \, a^{2} b^{3} - 28 \, a^{3} b c\right )} x^{6} + 128 \, a^{5} + 8 \, {\left (a^{3} b^{2} + 32 \, a^{4} c\right )} x^{4}\right )} \sqrt {c x^{4} + b x^{2} + a}}{5120 \, a^{4} x^{10}}, -\frac {15 \, {\left (b^{5} - 8 \, a b^{3} c + 16 \, a^{2} b c^{2}\right )} \sqrt {-a} x^{10} \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 ({\left (15 \, a b^{4} - 100 \, a^{2} b^{2} c + 128 \, a^{3} c^{2}\right )} x^{8} + 176 \, a^{4} b x^{2} - 2 \, {\left (5 \, a^{2} b^{3} - 28 \, a^{3} b c\right )} x^{6} + 128 \, a^{5} + 8 \, {\left (a^{3} b^{2} + 32 \, a^{4} c\right )} x^{4}\right )} \sqrt {c x^{4} + b x^{2} + a}}{2560 \, a^{4} x^{10}}\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^11,x, algorithm="fricas")

[Out]

[1/5120*(15*(b^5 - 8*a*b^3*c + 16*a^2*b*c^2)*sqrt(a)*x^10*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*((15*a*b^4 - 100*a^2*b^2*c + 128*a^3*c^2)*x^8 + 176*a^4*b*
x^2 - 2*(5*a^2*b^3 - 28*a^3*b*c)*x^6 + 128*a^5 + 8*(a^3*b^2 + 32*a^4*c)*x^4)*sqrt(c*x^4 + b*x^2 + a))/(a^4*x^1
0), -1/2560*(15*(b^5 - 8*a*b^3*c + 16*a^2*b*c^2)*sqrt(-a)*x^10*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*((15*a*b^4 - 100*a^2*b^2*c + 128*a^3*c^2)*x^8 + 176*a^4*b*x^2 - 2*(5
*a^2*b^3 - 28*a^3*b*c)*x^6 + 128*a^5 + 8*(a^3*b^2 + 32*a^4*c)*x^4)*sqrt(c*x^4 + b*x^2 + a))/(a^4*x^10)]

________________________________________________________________________________________

giac [B]  time = 0.53, size = 832, normalized size = 5.14 \begin {gather*} -\frac {3 \, {\left (b^{5} - 8 \, a b^{3} c + 16 \, a^{2} b c^{2}\right )} \arctan \left (-\frac {\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}}{\sqrt {-a}}\right )}{256 \, \sqrt {-a} a^{3}} + \frac {15 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{9} b^{5} - 120 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{9} a b^{3} c + 240 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{9} a^{2} b c^{2} + 1280 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{8} a^{3} c^{\frac {5}{2}} - 70 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{7} a b^{5} + 560 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{7} a^{2} b^{3} c + 2720 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{7} a^{3} b c^{2} + 5120 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{6} a^{3} b^{2} c^{\frac {3}{2}} + 128 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{5} a^{2} b^{5} + 2560 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{5} a^{3} b^{3} c + 3840 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{5} a^{4} b c^{2} + 1280 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{4} a^{3} b^{4} \sqrt {c} + 2560 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{4} a^{4} b^{2} c^{\frac {3}{2}} + 2560 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{4} a^{5} c^{\frac {5}{2}} + 70 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{3} a^{3} b^{5} + 2000 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{3} a^{4} b^{3} c + 2400 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{3} a^{5} b c^{2} + 2560 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{2} a^{5} b^{2} c^{\frac {3}{2}} - 15 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} a^{4} b^{5} + 120 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} a^{5} b^{3} c + 1040 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} a^{6} b c^{2} + 256 \, a^{7} c^{\frac {5}{2}}}{1280 \, {\left ({\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{2} - a\right )}^{5} a^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2+a)^(3/2)/x^11,x, algorithm="giac")

[Out]

-3/256*(b^5 - 8*a*b^3*c + 16*a^2*b*c^2)*arctan(-(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))/sqrt(-a))/(sqrt(-a)*a^
3) + 1/1280*(15*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))^9*b^5 - 120*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))^9*
a*b^3*c + 240*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))^9*a^2*b*c^2 + 1280*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a
))^8*a^3*c^(5/2) - 70*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))^7*a*b^5 + 560*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2
+ a))^7*a^2*b^3*c + 2720*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))^7*a^3*b*c^2 + 5120*(sqrt(c)*x^2 - sqrt(c*x^4
+ b*x^2 + a))^6*a^3*b^2*c^(3/2) + 128*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))^5*a^2*b^5 + 2560*(sqrt(c)*x^2 -
sqrt(c*x^4 + b*x^2 + a))^5*a^3*b^3*c + 3840*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))^5*a^4*b*c^2 + 1280*(sqrt(c
)*x^2 - sqrt(c*x^4 + b*x^2 + a))^4*a^3*b^4*sqrt(c) + 2560*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))^4*a^4*b^2*c^
(3/2) + 2560*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))^4*a^5*c^(5/2) + 70*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a)
)^3*a^3*b^5 + 2000*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))^3*a^4*b^3*c + 2400*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^
2 + a))^3*a^5*b*c^2 + 2560*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))^2*a^5*b^2*c^(3/2) - 15*(sqrt(c)*x^2 - sqrt(
c*x^4 + b*x^2 + a))*a^4*b^5 + 120*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))*a^5*b^3*c + 1040*(sqrt(c)*x^2 - sqrt
(c*x^4 + b*x^2 + a))*a^6*b*c^2 + 256*a^7*c^(5/2))/(((sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))^2 - a)^5*a^3)

________________________________________________________________________________________

maple [B]  time = 0.02, size = 337, normalized size = 2.08 \begin {gather*} \frac {3 b \,c^{2} \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 {3 b^{3} c \ln \left (\frac {b \,x^{2}+2 a +2 \sqrt {c \,x^{4}+b \,x^{2}+a}\, \sqrt {a}}{x^{2}}\right )}{64 a^{\frac {5}{2}}}+\frac {3 b^{5} \ln \left (\frac {b \,x^{2}+2 a +2 \sqrt {c \,x^{4}+b \,x^{2}+a}\, \sqrt {a}}{x^{2}}\right )}{512 a^{\frac {7}{2}}}-\frac {\sqrt {c \,x^{4}+b \,x^{2}+a}\, c^{2}}{10 a \,x^{2}}+\frac {5 \sqrt {c \,x^{4}+b \,x^{2}+a}\, b^{2} c}{64 a^{2} x^{2}}-\frac {3 \sqrt {c \,x^{4}+b \,x^{2}+a}\, b^{4}}{256 a^{3} x^{2}}-\frac {7 \sqrt {c \,x^{4}+b \,x^{2}+a}\, b c}{160 a \,x^{4}}+\frac {\sqrt {c \,x^{4}+b \,x^{2}+a}\, b^{3}}{128 a^{2} x^{4}}-\frac {\sqrt {c \,x^{4}+b \,x^{2}+a}\, b^{2}}{160 a \,x^{6}}-\frac {\sqrt {c \,x^{4}+b \,x^{2}+a}\, c}{5 x^{6}}-\frac {11 \sqrt {c \,x^{4}+b \,x^{2}+a}\, b}{80 x^{8}}-\frac {\sqrt {c \,x^{4}+b \,x^{2}+a}\, a}{10 x^{10}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^4+b*x^2+a)^(3/2)/x^11,x)

[Out]

-1/160/a*b^2/x^6*(c*x^4+b*x^2+a)^(1/2)+1/128/a^2*b^3/x^4*(c*x^4+b*x^2+a)^(1/2)-3/256/a^3*b^4/x^2*(c*x^4+b*x^2+
a)^(1/2)+3/512/a^(7/2)*b^5*ln((b*x^2+2*a+2*(c*x^4+b*x^2+a)^(1/2)*a^(1/2))/x^2)-1/10/a*c^2/x^2*(c*x^4+b*x^2+a)^
(1/2)+3/32*b*c^2/a^(3/2)*ln((b*x^2+2*a+2*(c*x^4+b*x^2+a)^(1/2)*a^(1/2))/x^2)-3/64*b^3*c/a^(5/2)*ln((b*x^2+2*a+
2*(c*x^4+b*x^2+a)^(1/2)*a^(1/2))/x^2)-7/160*b*c/a/x^4*(c*x^4+b*x^2+a)^(1/2)+5/64*b^2*c/a^2/x^2*(c*x^4+b*x^2+a)
^(1/2)-1/10*a/x^10*(c*x^4+b*x^2+a)^(1/2)-11/80*b/x^8*(c*x^4+b*x^2+a)^(1/2)-1/5*c/x^6*(c*x^4+b*x^2+a)^(1/2)

________________________________________________________________________________________

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^11,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?

________________________________________________________________________________________

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^{11}} \,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^11,x)

[Out]

int((a + b*x^2 + c*x^4)^(3/2)/x^11, x)

________________________________________________________________________________________

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^{11}}\, 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**11,x)

[Out]

Integral((a + b*x**2 + c*x**4)**(3/2)/x**11, x)

________________________________________________________________________________________