3.394 \(\int \frac {x^6}{(d+e x^2)^{3/2} (a+b x^2+c x^4)} \, dx\)

Optimal. Leaf size=350 \[ \frac {2 \left (-\frac {b \left (b^2-3 a c\right )}{\sqrt {b^2-4 a c}}-a c+b^2\right ) \tan ^{-1}\left (\frac {x \sqrt {2 c d-e \left (b-\sqrt {b^2-4 a c}\right )}}{\sqrt {b-\sqrt {b^2-4 a c}} \sqrt {d+e x^2}}\right )}{c \sqrt {b-\sqrt {b^2-4 a c}} \left (2 c d-e \left (b-\sqrt {b^2-4 a c}\right )\right )^{3/2}}+\frac {2 \left (\frac {b \left (b^2-3 a c\right )}{\sqrt {b^2-4 a c}}-a c+b^2\right ) \tan ^{-1}\left (\frac {x \sqrt {2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}}{\sqrt {\sqrt {b^2-4 a c}+b} \sqrt {d+e x^2}}\right )}{c \sqrt {\sqrt {b^2-4 a c}+b} \left (2 c d-e \left (\sqrt {b^2-4 a c}+b\right )\right )^{3/2}}-\frac {d^2 x}{e \sqrt {d+e x^2} \left (a e^2-b d e+c d^2\right )}+\frac {\tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{c e^{3/2}} \]

[Out]

arctanh(x*e^(1/2)/(e*x^2+d)^(1/2))/c/e^(3/2)-d^2*x/e/(a*e^2-b*d*e+c*d^2)/(e*x^2+d)^(1/2)+2*arctan(x*(2*c*d-e*(
b-(-4*a*c+b^2)^(1/2)))^(1/2)/(e*x^2+d)^(1/2)/(b-(-4*a*c+b^2)^(1/2))^(1/2))*(b^2-a*c-b*(-3*a*c+b^2)/(-4*a*c+b^2
)^(1/2))/c/(2*c*d-e*(b-(-4*a*c+b^2)^(1/2)))^(3/2)/(b-(-4*a*c+b^2)^(1/2))^(1/2)+2*arctan(x*(2*c*d-e*(b+(-4*a*c+
b^2)^(1/2)))^(1/2)/(e*x^2+d)^(1/2)/(b+(-4*a*c+b^2)^(1/2))^(1/2))*(b^2-a*c+b*(-3*a*c+b^2)/(-4*a*c+b^2)^(1/2))/c
/(2*c*d-e*(b+(-4*a*c+b^2)^(1/2)))^(3/2)/(b+(-4*a*c+b^2)^(1/2))^(1/2)

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Rubi [A]  time = 4.33, antiderivative size = 507, normalized size of antiderivative = 1.45, number of steps used = 14, number of rules used = 7, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {1297, 288, 217, 206, 1692, 377, 205} \[ \frac {\left (-\frac {2 a^2 c e-a b^2 e-3 a b c d+b^3 d}{\sqrt {b^2-4 a c}}-a b e-a c d+b^2 d\right ) \tan ^{-1}\left (\frac {x \sqrt {2 c d-e \left (b-\sqrt {b^2-4 a c}\right )}}{\sqrt {b-\sqrt {b^2-4 a c}} \sqrt {d+e x^2}}\right )}{c \sqrt {b-\sqrt {b^2-4 a c}} \sqrt {2 c d-e \left (b-\sqrt {b^2-4 a c}\right )} \left (a e^2-b d e+c d^2\right )}+\frac {\left (\frac {2 a^2 c e-a b^2 e-3 a b c d+b^3 d}{\sqrt {b^2-4 a c}}-a b e-a c d+b^2 d\right ) \tan ^{-1}\left (\frac {x \sqrt {2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}}{\sqrt {\sqrt {b^2-4 a c}+b} \sqrt {d+e x^2}}\right )}{c \sqrt {\sqrt {b^2-4 a c}+b} \sqrt {2 c d-e \left (\sqrt {b^2-4 a c}+b\right )} \left (a e^2-b d e+c d^2\right )}-\frac {d^2 x}{e \sqrt {d+e x^2} \left (a e^2-b d e+c d^2\right )}+\frac {d^2 \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{e^{3/2} \left (a e^2-b d e+c d^2\right )}-\frac {(b d-a e) \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{c \sqrt {e} \left (a e^2-b d e+c d^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[x^6/((d + e*x^2)^(3/2)*(a + b*x^2 + c*x^4)),x]

[Out]

-((d^2*x)/(e*(c*d^2 - b*d*e + a*e^2)*Sqrt[d + e*x^2])) + ((b^2*d - a*c*d - a*b*e - (b^3*d - 3*a*b*c*d - a*b^2*
e + 2*a^2*c*e)/Sqrt[b^2 - 4*a*c])*ArcTan[(Sqrt[2*c*d - (b - Sqrt[b^2 - 4*a*c])*e]*x)/(Sqrt[b - Sqrt[b^2 - 4*a*
c]]*Sqrt[d + e*x^2])])/(c*Sqrt[b - Sqrt[b^2 - 4*a*c]]*Sqrt[2*c*d - (b - Sqrt[b^2 - 4*a*c])*e]*(c*d^2 - b*d*e +
a*e^2)) + ((b^2*d - a*c*d - a*b*e + (b^3*d - 3*a*b*c*d - a*b^2*e + 2*a^2*c*e)/Sqrt[b^2 - 4*a*c])*ArcTan[(Sqrt
[2*c*d - (b + Sqrt[b^2 - 4*a*c])*e]*x)/(Sqrt[b + Sqrt[b^2 - 4*a*c]]*Sqrt[d + e*x^2])])/(c*Sqrt[b + Sqrt[b^2 -
4*a*c]]*Sqrt[2*c*d - (b + Sqrt[b^2 - 4*a*c])*e]*(c*d^2 - b*d*e + a*e^2)) + (d^2*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e
*x^2]])/(e^(3/2)*(c*d^2 - b*d*e + a*e^2)) - ((b*d - a*e)*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]])/(c*Sqrt[e]*(c*d
^2 - b*d*e + a*e^2))

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

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 217

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

Rule 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 377

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 1297

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

Rule 1692

Int[(Px_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandInteg
rand[Px*(d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, q}, x] && PolyQ[Px, x^2] && NeQ[b
^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[p]

Rubi steps

\begin {align*} \int \frac {x^6}{\left (d+e x^2\right )^{3/2} \left (a+b x^2+c x^4\right )} \, dx &=-\frac {\int \frac {x^2 \left (a d+(b d-a e) x^2\right )}{\sqrt {d+e x^2} \left (a+b x^2+c x^4\right )} \, dx}{c d^2-b d e+a e^2}+\frac {d^2 \int \frac {x^2}{\left (d+e x^2\right )^{3/2}} \, dx}{c d^2-b d e+a e^2}\\ &=-\frac {d^2 x}{e \left (c d^2-b d e+a e^2\right ) \sqrt {d+e x^2}}-\frac {\int \left (\frac {b d-a e}{c \sqrt {d+e x^2}}-\frac {a (b d-a e)+\left (b^2 d-a c d-a b e\right ) x^2}{c \sqrt {d+e x^2} \left (a+b x^2+c x^4\right )}\right ) \, dx}{c d^2-b d e+a e^2}+\frac {d^2 \int \frac {1}{\sqrt {d+e x^2}} \, dx}{e \left (c d^2-b d e+a e^2\right )}\\ &=-\frac {d^2 x}{e \left (c d^2-b d e+a e^2\right ) \sqrt {d+e x^2}}+\frac {\int \frac {a (b d-a e)+\left (b^2 d-a c d-a b e\right ) x^2}{\sqrt {d+e x^2} \left (a+b x^2+c x^4\right )} \, dx}{c \left (c d^2-b d e+a e^2\right )}+\frac {d^2 \operatorname {Subst}\left (\int \frac {1}{1-e x^2} \, dx,x,\frac {x}{\sqrt {d+e x^2}}\right )}{e \left (c d^2-b d e+a e^2\right )}-\frac {(b d-a e) \int \frac {1}{\sqrt {d+e x^2}} \, dx}{c \left (c d^2-b d e+a e^2\right )}\\ &=-\frac {d^2 x}{e \left (c d^2-b d e+a e^2\right ) \sqrt {d+e x^2}}+\frac {d^2 \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{e^{3/2} \left (c d^2-b d e+a e^2\right )}+\frac {\int \left (\frac {b^2 d-a c d-a b e+\frac {-b^3 d+3 a b c d+a b^2 e-2 a^2 c e}{\sqrt {b^2-4 a c}}}{\left (b-\sqrt {b^2-4 a c}+2 c x^2\right ) \sqrt {d+e x^2}}+\frac {b^2 d-a c d-a b e-\frac {-b^3 d+3 a b c d+a b^2 e-2 a^2 c e}{\sqrt {b^2-4 a c}}}{\left (b+\sqrt {b^2-4 a c}+2 c x^2\right ) \sqrt {d+e x^2}}\right ) \, dx}{c \left (c d^2-b d e+a e^2\right )}-\frac {(b d-a e) \operatorname {Subst}\left (\int \frac {1}{1-e x^2} \, dx,x,\frac {x}{\sqrt {d+e x^2}}\right )}{c \left (c d^2-b d e+a e^2\right )}\\ &=-\frac {d^2 x}{e \left (c d^2-b d e+a e^2\right ) \sqrt {d+e x^2}}+\frac {d^2 \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{e^{3/2} \left (c d^2-b d e+a e^2\right )}-\frac {(b d-a e) \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{c \sqrt {e} \left (c d^2-b d e+a e^2\right )}+\frac {\left (b^2 d-a c d-a b e-\frac {b^3 d-3 a b c d-a b^2 e+2 a^2 c e}{\sqrt {b^2-4 a c}}\right ) \int \frac {1}{\left (b-\sqrt {b^2-4 a c}+2 c x^2\right ) \sqrt {d+e x^2}} \, dx}{c \left (c d^2-b d e+a e^2\right )}+\frac {\left (b^2 d-a c d-a b e+\frac {b^3 d-3 a b c d-a b^2 e+2 a^2 c e}{\sqrt {b^2-4 a c}}\right ) \int \frac {1}{\left (b+\sqrt {b^2-4 a c}+2 c x^2\right ) \sqrt {d+e x^2}} \, dx}{c \left (c d^2-b d e+a e^2\right )}\\ &=-\frac {d^2 x}{e \left (c d^2-b d e+a e^2\right ) \sqrt {d+e x^2}}+\frac {d^2 \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{e^{3/2} \left (c d^2-b d e+a e^2\right )}-\frac {(b d-a e) \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{c \sqrt {e} \left (c d^2-b d e+a e^2\right )}+\frac {\left (b^2 d-a c d-a b e-\frac {b^3 d-3 a b c d-a b^2 e+2 a^2 c e}{\sqrt {b^2-4 a c}}\right ) \operatorname {Subst}\left (\int \frac {1}{b-\sqrt {b^2-4 a c}-\left (-2 c d+\left (b-\sqrt {b^2-4 a c}\right ) e\right ) x^2} \, dx,x,\frac {x}{\sqrt {d+e x^2}}\right )}{c \left (c d^2-b d e+a e^2\right )}+\frac {\left (b^2 d-a c d-a b e+\frac {b^3 d-3 a b c d-a b^2 e+2 a^2 c e}{\sqrt {b^2-4 a c}}\right ) \operatorname {Subst}\left (\int \frac {1}{b+\sqrt {b^2-4 a c}-\left (-2 c d+\left (b+\sqrt {b^2-4 a c}\right ) e\right ) x^2} \, dx,x,\frac {x}{\sqrt {d+e x^2}}\right )}{c \left (c d^2-b d e+a e^2\right )}\\ &=-\frac {d^2 x}{e \left (c d^2-b d e+a e^2\right ) \sqrt {d+e x^2}}+\frac {\left (b^2 d-a c d-a b e-\frac {b^3 d-3 a b c d-a b^2 e+2 a^2 c e}{\sqrt {b^2-4 a c}}\right ) \tan ^{-1}\left (\frac {\sqrt {2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e} x}{\sqrt {b-\sqrt {b^2-4 a c}} \sqrt {d+e x^2}}\right )}{c \sqrt {b-\sqrt {b^2-4 a c}} \sqrt {2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e} \left (c d^2-b d e+a e^2\right )}+\frac {\left (b^2 d-a c d-a b e+\frac {b^3 d-3 a b c d-a b^2 e+2 a^2 c e}{\sqrt {b^2-4 a c}}\right ) \tan ^{-1}\left (\frac {\sqrt {2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e} x}{\sqrt {b+\sqrt {b^2-4 a c}} \sqrt {d+e x^2}}\right )}{c \sqrt {b+\sqrt {b^2-4 a c}} \sqrt {2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e} \left (c d^2-b d e+a e^2\right )}+\frac {d^2 \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{e^{3/2} \left (c d^2-b d e+a e^2\right )}-\frac {(b d-a e) \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{c \sqrt {e} \left (c d^2-b d e+a e^2\right )}\\ \end {align*}

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Mathematica [B]  time = 11.26, size = 10968, normalized size = 31.34 \[ \text {Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[x^6/((d + e*x^2)^(3/2)*(a + b*x^2 + c*x^4)),x]

[Out]

Result too large to show

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6/(e*x^2+d)^(3/2)/(c*x^4+b*x^2+a),x, algorithm="fricas")

[Out]

Timed out

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giac [A]  time = 2.25, size = 75, normalized size = 0.21 \[ -\frac {c^{2} d^{2} x}{{\left (c^{3} d^{2} e - b c^{2} d e^{2} + a c^{2} e^{3}\right )} \sqrt {x^{2} e + d}} - \frac {e^{\left (-\frac {3}{2}\right )} \log \left ({\left (x e^{\frac {1}{2}} - \sqrt {x^{2} e + d}\right )}^{2}\right )}{2 \, c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6/(e*x^2+d)^(3/2)/(c*x^4+b*x^2+a),x, algorithm="giac")

[Out]

-c^2*d^2*x/((c^3*d^2*e - b*c^2*d*e^2 + a*c^2*e^3)*sqrt(x^2*e + d)) - 1/2*e^(-3/2)*log((x*e^(1/2) - sqrt(x^2*e
+ d))^2)/c

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maple [C]  time = 0.04, size = 480, normalized size = 1.37 \[ \frac {8 a b \,e^{\frac {3}{2}}}{\left (4 a \,e^{2}-4 d e b +4 c \,d^{2}\right ) \left (2 e \,x^{2}-2 \sqrt {e \,x^{2}+d}\, \sqrt {e}\, x +2 d \right ) c^{2}}+\frac {8 a d \sqrt {e}}{\left (4 a \,e^{2}-4 d e b +4 c \,d^{2}\right ) \left (2 e \,x^{2}-2 \sqrt {e \,x^{2}+d}\, \sqrt {e}\, x +2 d \right ) c}-\frac {8 b^{2} d \sqrt {e}}{\left (4 a \,e^{2}-4 d e b +4 c \,d^{2}\right ) \left (2 e \,x^{2}-2 \sqrt {e \,x^{2}+d}\, \sqrt {e}\, x +2 d \right ) c^{2}}+\frac {2 \sqrt {e}\, \left (a b \,d^{2} e +a c \,d^{3}-b^{2} d^{3}+\left (a b e +a c d -b^{2} d \right ) \RootOf \left (\textit {\_Z}^{4} c +c \,d^{4}+\left (4 b e -4 c d \right ) \textit {\_Z}^{3}+\left (16 a \,e^{2}-8 d e b +6 c \,d^{2}\right ) \textit {\_Z}^{2}+\left (4 b \,d^{2} e -4 c \,d^{3}\right ) \textit {\_Z} \right )^{2}+2 \left (2 a^{2} e^{2}-3 a b d e -a c \,d^{2}+b^{2} d^{2}\right ) \RootOf \left (\textit {\_Z}^{4} c +c \,d^{4}+\left (4 b e -4 c d \right ) \textit {\_Z}^{3}+\left (16 a \,e^{2}-8 d e b +6 c \,d^{2}\right ) \textit {\_Z}^{2}+\left (4 b \,d^{2} e -4 c \,d^{3}\right ) \textit {\_Z} \right )\right ) \ln \left (-\RootOf \left (\textit {\_Z}^{4} c +c \,d^{4}+\left (4 b e -4 c d \right ) \textit {\_Z}^{3}+\left (16 a \,e^{2}-8 d e b +6 c \,d^{2}\right ) \textit {\_Z}^{2}+\left (4 b \,d^{2} e -4 c \,d^{3}\right ) \textit {\_Z} \right )+\left (-\sqrt {e}\, x +\sqrt {e \,x^{2}+d}\right )^{2}\right )}{\left (4 a \,e^{2}-4 d e b +4 c \,d^{2}\right ) c \left (\RootOf \left (\textit {\_Z}^{4} c +c \,d^{4}+\left (4 b e -4 c d \right ) \textit {\_Z}^{3}+\left (16 a \,e^{2}-8 d e b +6 c \,d^{2}\right ) \textit {\_Z}^{2}+\left (4 b \,d^{2} e -4 c \,d^{3}\right ) \textit {\_Z} \right )^{3} c +3 \RootOf \left (\textit {\_Z}^{4} c +c \,d^{4}+\left (4 b e -4 c d \right ) \textit {\_Z}^{3}+\left (16 a \,e^{2}-8 d e b +6 c \,d^{2}\right ) \textit {\_Z}^{2}+\left (4 b \,d^{2} e -4 c \,d^{3}\right ) \textit {\_Z} \right )^{2} b e -3 \RootOf \left (\textit {\_Z}^{4} c +c \,d^{4}+\left (4 b e -4 c d \right ) \textit {\_Z}^{3}+\left (16 a \,e^{2}-8 d e b +6 c \,d^{2}\right ) \textit {\_Z}^{2}+\left (4 b \,d^{2} e -4 c \,d^{3}\right ) \textit {\_Z} \right )^{2} c d +8 \RootOf \left (\textit {\_Z}^{4} c +c \,d^{4}+\left (4 b e -4 c d \right ) \textit {\_Z}^{3}+\left (16 a \,e^{2}-8 d e b +6 c \,d^{2}\right ) \textit {\_Z}^{2}+\left (4 b \,d^{2} e -4 c \,d^{3}\right ) \textit {\_Z} \right ) a \,e^{2}-4 \RootOf \left (\textit {\_Z}^{4} c +c \,d^{4}+\left (4 b e -4 c d \right ) \textit {\_Z}^{3}+\left (16 a \,e^{2}-8 d e b +6 c \,d^{2}\right ) \textit {\_Z}^{2}+\left (4 b \,d^{2} e -4 c \,d^{3}\right ) \textit {\_Z} \right ) b d e +3 \RootOf \left (\textit {\_Z}^{4} c +c \,d^{4}+\left (4 b e -4 c d \right ) \textit {\_Z}^{3}+\left (16 a \,e^{2}-8 d e b +6 c \,d^{2}\right ) \textit {\_Z}^{2}+\left (4 b \,d^{2} e -4 c \,d^{3}\right ) \textit {\_Z} \right ) c \,d^{2}+b \,d^{2} e -c \,d^{3}\right )}-\frac {b x}{\sqrt {e \,x^{2}+d}\, c^{2} d}-\frac {x}{\sqrt {e \,x^{2}+d}\, c e}+\frac {\ln \left (\sqrt {e}\, x +\sqrt {e \,x^{2}+d}\right )}{c \,e^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^6/(e*x^2+d)^(3/2)/(c*x^4+b*x^2+a),x)

[Out]

-1/c*x/e/(e*x^2+d)^(1/2)+1/c/e^(3/2)*ln(e^(1/2)*x+(e*x^2+d)^(1/2))-1/c^2*b*x/d/(e*x^2+d)^(1/2)+2/c*e^(1/2)/(4*
a*e^2-4*b*d*e+4*c*d^2)*sum(((a*b*e+a*c*d-b^2*d)*_R^2+2*(2*a^2*e^2-3*a*b*d*e-a*c*d^2+b^2*d^2)*_R+a*b*d^2*e+a*c*
d^3-b^2*d^3)/(_R^3*c+3*_R^2*b*e-3*_R^2*c*d+8*_R*a*e^2-4*_R*b*d*e+3*_R*c*d^2+b*d^2*e-c*d^3)*ln(-_R+(-e^(1/2)*x+
(e*x^2+d)^(1/2))^2),_R=RootOf(_Z^4*c+c*d^4+(4*b*e-4*c*d)*_Z^3+(16*a*e^2-8*b*d*e+6*c*d^2)*_Z^2+(4*b*d^2*e-4*c*d
^3)*_Z))+8/c^2*e^(3/2)/(4*a*e^2-4*b*d*e+4*c*d^2)/(2*e*x^2-2*e^(1/2)*(e*x^2+d)^(1/2)*x+2*d)*a*b+8/c*e^(1/2)/(4*
a*e^2-4*b*d*e+4*c*d^2)/(2*e*x^2-2*e^(1/2)*(e*x^2+d)^(1/2)*x+2*d)*a*d-8/c^2*e^(1/2)/(4*a*e^2-4*b*d*e+4*c*d^2)/(
2*e*x^2-2*e^(1/2)*(e*x^2+d)^(1/2)*x+2*d)*b^2*d

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{6}}{{\left (c x^{4} + b x^{2} + a\right )} {\left (e x^{2} + d\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6/(e*x^2+d)^(3/2)/(c*x^4+b*x^2+a),x, algorithm="maxima")

[Out]

integrate(x^6/((c*x^4 + b*x^2 + a)*(e*x^2 + d)^(3/2)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^6}{{\left (e\,x^2+d\right )}^{3/2}\,\left (c\,x^4+b\,x^2+a\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^6/((d + e*x^2)^(3/2)*(a + b*x^2 + c*x^4)),x)

[Out]

int(x^6/((d + e*x^2)^(3/2)*(a + b*x^2 + c*x^4)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{6}}{\left (d + e x^{2}\right )^{\frac {3}{2}} \left (a + b x^{2} + c x^{4}\right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**6/(e*x**2+d)**(3/2)/(c*x**4+b*x**2+a),x)

[Out]

Integral(x**6/((d + e*x**2)**(3/2)*(a + b*x**2 + c*x**4)), x)

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