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

Optimal. Leaf size=365 \[ \frac {\sqrt [4]{a} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \left (9 \sqrt {b} c-5 \sqrt {a} e\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{12 b^{9/4} \sqrt {a+b x^4}}+\frac {3 c x \sqrt {a+b x^4}}{2 b^{3/2} \left (\sqrt {a}+\sqrt {b} x^2\right )}-\frac {3 \sqrt [4]{a} c \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{2 b^{7/4} \sqrt {a+b x^4}}-\frac {3 a f \tanh ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )}{4 b^{5/2}}+\frac {x \left (a e+a f x-b c x^2-b d x^3\right )}{2 b^2 \sqrt {a+b x^4}}+\frac {d \sqrt {a+b x^4}}{b^2}+\frac {e x \sqrt {a+b x^4}}{3 b^2}+\frac {f x^2 \sqrt {a+b x^4}}{4 b^2} \]

[Out]

-3/4*a*f*arctanh(x^2*b^(1/2)/(b*x^4+a)^(1/2))/b^(5/2)+1/2*x*(-b*d*x^3-b*c*x^2+a*f*x+a*e)/b^2/(b*x^4+a)^(1/2)+d
*(b*x^4+a)^(1/2)/b^2+1/3*e*x*(b*x^4+a)^(1/2)/b^2+1/4*f*x^2*(b*x^4+a)^(1/2)/b^2+3/2*c*x*(b*x^4+a)^(1/2)/b^(3/2)
/(a^(1/2)+x^2*b^(1/2))-3/2*a^(1/4)*c*(cos(2*arctan(b^(1/4)*x/a^(1/4)))^2)^(1/2)/cos(2*arctan(b^(1/4)*x/a^(1/4)
))*EllipticE(sin(2*arctan(b^(1/4)*x/a^(1/4))),1/2*2^(1/2))*(a^(1/2)+x^2*b^(1/2))*((b*x^4+a)/(a^(1/2)+x^2*b^(1/
2))^2)^(1/2)/b^(7/4)/(b*x^4+a)^(1/2)+1/12*a^(1/4)*(cos(2*arctan(b^(1/4)*x/a^(1/4)))^2)^(1/2)/cos(2*arctan(b^(1
/4)*x/a^(1/4)))*EllipticF(sin(2*arctan(b^(1/4)*x/a^(1/4))),1/2*2^(1/2))*(-5*e*a^(1/2)+9*c*b^(1/2))*(a^(1/2)+x^
2*b^(1/2))*((b*x^4+a)/(a^(1/2)+x^2*b^(1/2))^2)^(1/2)/b^(9/4)/(b*x^4+a)^(1/2)

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Rubi [A]  time = 0.52, antiderivative size = 365, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 11, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.367, Rules used = {1828, 1885, 1888, 1198, 220, 1196, 1819, 1815, 641, 217, 206} \[ \frac {x \left (a e+a f x-b c x^2-b d x^3\right )}{2 b^2 \sqrt {a+b x^4}}+\frac {\sqrt [4]{a} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \left (9 \sqrt {b} c-5 \sqrt {a} e\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{12 b^{9/4} \sqrt {a+b x^4}}+\frac {3 c x \sqrt {a+b x^4}}{2 b^{3/2} \left (\sqrt {a}+\sqrt {b} x^2\right )}-\frac {3 \sqrt [4]{a} c \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{2 b^{7/4} \sqrt {a+b x^4}}+\frac {d \sqrt {a+b x^4}}{b^2}+\frac {e x \sqrt {a+b x^4}}{3 b^2}+\frac {f x^2 \sqrt {a+b x^4}}{4 b^2}-\frac {3 a f \tanh ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )}{4 b^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(a*e + a*f*x - b*c*x^2 - b*d*x^3))/(2*b^2*Sqrt[a + b*x^4]) + (d*Sqrt[a + b*x^4])/b^2 + (e*x*Sqrt[a + b*x^4]
)/(3*b^2) + (f*x^2*Sqrt[a + b*x^4])/(4*b^2) + (3*c*x*Sqrt[a + b*x^4])/(2*b^(3/2)*(Sqrt[a] + Sqrt[b]*x^2)) - (3
*a*f*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a + b*x^4]])/(4*b^(5/2)) - (3*a^(1/4)*c*(Sqrt[a] + Sqrt[b]*x^2)*Sqrt[(a + b*x^
4)/(Sqrt[a] + Sqrt[b]*x^2)^2]*EllipticE[2*ArcTan[(b^(1/4)*x)/a^(1/4)], 1/2])/(2*b^(7/4)*Sqrt[a + b*x^4]) + (a^
(1/4)*(9*Sqrt[b]*c - 5*Sqrt[a]*e)*(Sqrt[a] + Sqrt[b]*x^2)*Sqrt[(a + b*x^4)/(Sqrt[a] + Sqrt[b]*x^2)^2]*Elliptic
F[2*ArcTan[(b^(1/4)*x)/a^(1/4)], 1/2])/(12*b^(9/4)*Sqrt[a + b*x^4])

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 220

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[((1 + q^2*x^2)*Sqrt[(a + b*x^4)/(a*(
1 + q^2*x^2)^2)]*EllipticF[2*ArcTan[q*x], 1/2])/(2*q*Sqrt[a + b*x^4]), x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 641

Int[((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(e*(a + c*x^2)^(p + 1))/(2*c*(p + 1)),
x] + Dist[d, Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, p}, x] && NeQ[p, -1]

Rule 1196

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, -Simp[(d*x*Sqrt[a + c
*x^4])/(a*(1 + q^2*x^2)), x] + Simp[(d*(1 + q^2*x^2)*Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]*EllipticE[2*ArcTan[
q*x], 1/2])/(q*Sqrt[a + c*x^4]), x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

Rule 1198

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

Rule 1815

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], e = Coeff[Pq, x, Expon[Pq, x]]}, Si
mp[(e*x^(q - 1)*(a + b*x^2)^(p + 1))/(b*(q + 2*p + 1)), x] + Dist[1/(b*(q + 2*p + 1)), Int[(a + b*x^2)^p*Expan
dToSum[b*(q + 2*p + 1)*Pq - a*e*(q - 1)*x^(q - 2) - b*e*(q + 2*p + 1)*x^q, x], x], x]] /; FreeQ[{a, b, p}, x]
&& PolyQ[Pq, x] &&  !LeQ[p, -1]

Rule 1819

Int[(Pq_)*(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/(m + 1), Subst[Int[SubstFor[x^(m + 1)
, Pq, x]*(a + b*x^Simplify[n/(m + 1)])^p, x], x, x^(m + 1)], x] /; FreeQ[{a, b, m, n, p}, x] && NeQ[m, -1] &&
IGtQ[Simplify[n/(m + 1)], 0] && PolyQ[Pq, x^(m + 1)]

Rule 1828

Int[(Pq_)*(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> With[{q = m + Expon[Pq, x]}, Module[{Q = Pol
ynomialQuotient[b^(Floor[(q - 1)/n] + 1)*x^m*Pq, a + b*x^n, x], R = PolynomialRemainder[b^(Floor[(q - 1)/n] +
1)*x^m*Pq, a + b*x^n, x]}, Dist[1/(a*n*(p + 1)*b^(Floor[(q - 1)/n] + 1)), Int[(a + b*x^n)^(p + 1)*ExpandToSum[
a*n*(p + 1)*Q + n*(p + 1)*R + D[x*R, x], x], x], x] - Simp[(x*R*(a + b*x^n)^(p + 1))/(a*n*(p + 1)*b^(Floor[(q
- 1)/n] + 1)), x]] /; GeQ[q, n]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && IGtQ[n, 0] && LtQ[p, -1] && IGtQ[m, 0]

Rule 1885

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Module[{q = Expon[Pq, x], j, k}, Int[Sum[x^j*Sum[Coeff[P
q, x, j + (k*n)/2]*x^((k*n)/2), {k, 0, (2*(q - j))/n + 1}]*(a + b*x^n)^p, {j, 0, n/2 - 1}], x]] /; FreeQ[{a, b
, p}, x] && PolyQ[Pq, x] && IGtQ[n/2, 0] &&  !PolyQ[Pq, x^(n/2)]

Rule 1888

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{q = Expon[Pq, x]}, With[{Pqq = Coeff[Pq, x, q]}, D
ist[1/(b*(q + n*p + 1)), Int[ExpandToSum[b*(q + n*p + 1)*(Pq - Pqq*x^q) - a*Pqq*(q - n + 1)*x^(q - n), x]*(a +
b*x^n)^p, x], x] + Simp[(Pqq*x^(q - n + 1)*(a + b*x^n)^(p + 1))/(b*(q + n*p + 1)), x]] /; NeQ[q + n*p + 1, 0]
&& q - n >= 0 && (IntegerQ[2*p] || IntegerQ[p + (q + 1)/(2*n)])] /; FreeQ[{a, b, p}, x] && PolyQ[Pq, x] && IG
tQ[n, 0]

Rubi steps

\begin {align*} \int \frac {x^6 \left (c+d x+e x^2+f x^3\right )}{\left (a+b x^4\right )^{3/2}} \, dx &=\frac {x \left (a e+a f x-b c x^2-b d x^3\right )}{2 b^2 \sqrt {a+b x^4}}-\frac {\int \frac {a^2 b e+2 a^2 b f x-3 a b^2 c x^2-4 a b^2 d x^3-2 a b^2 e x^4-2 a b^2 f x^5}{\sqrt {a+b x^4}} \, dx}{2 a b^3}\\ &=\frac {x \left (a e+a f x-b c x^2-b d x^3\right )}{2 b^2 \sqrt {a+b x^4}}-\frac {\int \left (\frac {a^2 b e-3 a b^2 c x^2-2 a b^2 e x^4}{\sqrt {a+b x^4}}+\frac {x \left (2 a^2 b f-4 a b^2 d x^2-2 a b^2 f x^4\right )}{\sqrt {a+b x^4}}\right ) \, dx}{2 a b^3}\\ &=\frac {x \left (a e+a f x-b c x^2-b d x^3\right )}{2 b^2 \sqrt {a+b x^4}}-\frac {\int \frac {a^2 b e-3 a b^2 c x^2-2 a b^2 e x^4}{\sqrt {a+b x^4}} \, dx}{2 a b^3}-\frac {\int \frac {x \left (2 a^2 b f-4 a b^2 d x^2-2 a b^2 f x^4\right )}{\sqrt {a+b x^4}} \, dx}{2 a b^3}\\ &=\frac {x \left (a e+a f x-b c x^2-b d x^3\right )}{2 b^2 \sqrt {a+b x^4}}+\frac {e x \sqrt {a+b x^4}}{3 b^2}-\frac {\int \frac {5 a^2 b^2 e-9 a b^3 c x^2}{\sqrt {a+b x^4}} \, dx}{6 a b^4}-\frac {\operatorname {Subst}\left (\int \frac {2 a^2 b f-4 a b^2 d x-2 a b^2 f x^2}{\sqrt {a+b x^2}} \, dx,x,x^2\right )}{4 a b^3}\\ &=\frac {x \left (a e+a f x-b c x^2-b d x^3\right )}{2 b^2 \sqrt {a+b x^4}}+\frac {e x \sqrt {a+b x^4}}{3 b^2}+\frac {f x^2 \sqrt {a+b x^4}}{4 b^2}-\frac {\operatorname {Subst}\left (\int \frac {6 a^2 b^2 f-8 a b^3 d x}{\sqrt {a+b x^2}} \, dx,x,x^2\right )}{8 a b^4}-\frac {\left (3 \sqrt {a} c\right ) \int \frac {1-\frac {\sqrt {b} x^2}{\sqrt {a}}}{\sqrt {a+b x^4}} \, dx}{2 b^{3/2}}+\frac {\left (\sqrt {a} \left (9 \sqrt {b} c-5 \sqrt {a} e\right )\right ) \int \frac {1}{\sqrt {a+b x^4}} \, dx}{6 b^2}\\ &=\frac {x \left (a e+a f x-b c x^2-b d x^3\right )}{2 b^2 \sqrt {a+b x^4}}+\frac {d \sqrt {a+b x^4}}{b^2}+\frac {e x \sqrt {a+b x^4}}{3 b^2}+\frac {f x^2 \sqrt {a+b x^4}}{4 b^2}+\frac {3 c x \sqrt {a+b x^4}}{2 b^{3/2} \left (\sqrt {a}+\sqrt {b} x^2\right )}-\frac {3 \sqrt [4]{a} c \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{2 b^{7/4} \sqrt {a+b x^4}}+\frac {\sqrt [4]{a} \left (9 \sqrt {b} c-5 \sqrt {a} e\right ) \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{12 b^{9/4} \sqrt {a+b x^4}}-\frac {(3 a f) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,x^2\right )}{4 b^2}\\ &=\frac {x \left (a e+a f x-b c x^2-b d x^3\right )}{2 b^2 \sqrt {a+b x^4}}+\frac {d \sqrt {a+b x^4}}{b^2}+\frac {e x \sqrt {a+b x^4}}{3 b^2}+\frac {f x^2 \sqrt {a+b x^4}}{4 b^2}+\frac {3 c x \sqrt {a+b x^4}}{2 b^{3/2} \left (\sqrt {a}+\sqrt {b} x^2\right )}-\frac {3 \sqrt [4]{a} c \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{2 b^{7/4} \sqrt {a+b x^4}}+\frac {\sqrt [4]{a} \left (9 \sqrt {b} c-5 \sqrt {a} e\right ) \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{12 b^{9/4} \sqrt {a+b x^4}}-\frac {(3 a f) \operatorname {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x^2}{\sqrt {a+b x^4}}\right )}{4 b^2}\\ &=\frac {x \left (a e+a f x-b c x^2-b d x^3\right )}{2 b^2 \sqrt {a+b x^4}}+\frac {d \sqrt {a+b x^4}}{b^2}+\frac {e x \sqrt {a+b x^4}}{3 b^2}+\frac {f x^2 \sqrt {a+b x^4}}{4 b^2}+\frac {3 c x \sqrt {a+b x^4}}{2 b^{3/2} \left (\sqrt {a}+\sqrt {b} x^2\right )}-\frac {3 a f \tanh ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )}{4 b^{5/2}}-\frac {3 \sqrt [4]{a} c \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{2 b^{7/4} \sqrt {a+b x^4}}+\frac {\sqrt [4]{a} \left (9 \sqrt {b} c-5 \sqrt {a} e\right ) \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{12 b^{9/4} \sqrt {a+b x^4}}\\ \end {align*}

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Mathematica [C]  time = 0.19, size = 220, normalized size = 0.60 \[ \frac {-9 a^{3/2} f \sqrt {\frac {b x^4}{a}+1} \sinh ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )-12 b^{3/2} c x^3 \sqrt {\frac {b x^4}{a}+1} \, _2F_1\left (\frac {3}{4},\frac {3}{2};\frac {7}{4};-\frac {b x^4}{a}\right )+12 a \sqrt {b} d-10 a \sqrt {b} e x \sqrt {\frac {b x^4}{a}+1} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};-\frac {b x^4}{a}\right )+10 a \sqrt {b} e x+9 a \sqrt {b} f x^2+12 b^{3/2} c x^3+6 b^{3/2} d x^4+4 b^{3/2} e x^5+3 b^{3/2} f x^6}{12 b^{5/2} \sqrt {a+b x^4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(12*a*Sqrt[b]*d + 10*a*Sqrt[b]*e*x + 9*a*Sqrt[b]*f*x^2 + 12*b^(3/2)*c*x^3 + 6*b^(3/2)*d*x^4 + 4*b^(3/2)*e*x^5
+ 3*b^(3/2)*f*x^6 - 9*a^(3/2)*f*Sqrt[1 + (b*x^4)/a]*ArcSinh[(Sqrt[b]*x^2)/Sqrt[a]] - 10*a*Sqrt[b]*e*x*Sqrt[1 +
(b*x^4)/a]*Hypergeometric2F1[1/4, 1/2, 5/4, -((b*x^4)/a)] - 12*b^(3/2)*c*x^3*Sqrt[1 + (b*x^4)/a]*Hypergeometr
ic2F1[3/4, 3/2, 7/4, -((b*x^4)/a)])/(12*b^(5/2)*Sqrt[a + b*x^4])

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fricas [F]  time = 0.48, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (f x^{9} + e x^{8} + d x^{7} + c x^{6}\right )} \sqrt {b x^{4} + a}}{b^{2} x^{8} + 2 \, a b x^{4} + a^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((f*x^9 + e*x^8 + d*x^7 + c*x^6)*sqrt(b*x^4 + a)/(b^2*x^8 + 2*a*b*x^4 + a^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.19, size = 378, normalized size = 1.04 \[ \frac {f \,x^{6}}{4 \sqrt {b \,x^{4}+a}\, b}-\frac {c \,x^{3}}{2 \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}\, b}+\frac {3 a f \,x^{2}}{4 \sqrt {b \,x^{4}+a}\, b^{2}}+\frac {a e x}{2 \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}\, b^{2}}-\frac {5 \sqrt {-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, a e \EllipticF \left (\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, x , i\right )}{6 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}\, b^{2}}-\frac {3 i \sqrt {-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {a}\, c \EllipticE \left (\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, x , i\right )}{2 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}\, b^{\frac {3}{2}}}+\frac {3 i \sqrt {-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {a}\, c \EllipticF \left (\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, x , i\right )}{2 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}\, b^{\frac {3}{2}}}-\frac {3 a f \ln \left (\sqrt {b}\, x^{2}+\sqrt {b \,x^{4}+a}\right )}{4 b^{\frac {5}{2}}}+\frac {\sqrt {b \,x^{4}+a}\, e x}{3 b^{2}}+\frac {\left (b \,x^{4}+2 a \right ) d}{2 \sqrt {b \,x^{4}+a}\, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*f*x^6/b/(b*x^4+a)^(1/2)+3/4*f*a/b^2*x^2/(b*x^4+a)^(1/2)-3/4*f*a/b^(5/2)*ln(b^(1/2)*x^2+(b*x^4+a)^(1/2))+1/
2*e/b^2*a*x/((x^4+a/b)*b)^(1/2)+1/3*e*x*(b*x^4+a)^(1/2)/b^2-5/6*e*a/b^2/(I/a^(1/2)*b^(1/2))^(1/2)*(-I/a^(1/2)*
b^(1/2)*x^2+1)^(1/2)*(I/a^(1/2)*b^(1/2)*x^2+1)^(1/2)/(b*x^4+a)^(1/2)*EllipticF((I/a^(1/2)*b^(1/2))^(1/2)*x,I)+
1/2*d*(b*x^4+2*a)/(b*x^4+a)^(1/2)/b^2-1/2*c/b*x^3/((x^4+a/b)*b)^(1/2)+3/2*I*c/b^(3/2)*a^(1/2)/(I/a^(1/2)*b^(1/
2))^(1/2)*(-I/a^(1/2)*b^(1/2)*x^2+1)^(1/2)*(I/a^(1/2)*b^(1/2)*x^2+1)^(1/2)/(b*x^4+a)^(1/2)*EllipticF((I/a^(1/2
)*b^(1/2))^(1/2)*x,I)-3/2*I*c/b^(3/2)*a^(1/2)/(I/a^(1/2)*b^(1/2))^(1/2)*(-I/a^(1/2)*b^(1/2)*x^2+1)^(1/2)*(I/a^
(1/2)*b^(1/2)*x^2+1)^(1/2)/(b*x^4+a)^(1/2)*EllipticE((I/a^(1/2)*b^(1/2))^(1/2)*x,I)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 51.87, size = 202, normalized size = 0.55 \[ d \left (\begin {cases} \frac {a}{b^{2} \sqrt {a + b x^{4}}} + \frac {x^{4}}{2 b \sqrt {a + b x^{4}}} & \text {for}\: b \neq 0 \\\frac {x^{8}}{8 a^{\frac {3}{2}}} & \text {otherwise} \end {cases}\right ) + f \left (\frac {3 \sqrt {a} x^{2}}{4 b^{2} \sqrt {1 + \frac {b x^{4}}{a}}} - \frac {3 a \operatorname {asinh}{\left (\frac {\sqrt {b} x^{2}}{\sqrt {a}} \right )}}{4 b^{\frac {5}{2}}} + \frac {x^{6}}{4 \sqrt {a} b \sqrt {1 + \frac {b x^{4}}{a}}}\right ) + \frac {c x^{7} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {3}{2}} \Gamma \left (\frac {11}{4}\right )} + \frac {e x^{9} \Gamma \left (\frac {9}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, \frac {9}{4} \\ \frac {13}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {3}{2}} \Gamma \left (\frac {13}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d*Piecewise((a/(b**2*sqrt(a + b*x**4)) + x**4/(2*b*sqrt(a + b*x**4)), Ne(b, 0)), (x**8/(8*a**(3/2)), True)) +
f*(3*sqrt(a)*x**2/(4*b**2*sqrt(1 + b*x**4/a)) - 3*a*asinh(sqrt(b)*x**2/sqrt(a))/(4*b**(5/2)) + x**6/(4*sqrt(a)
*b*sqrt(1 + b*x**4/a))) + c*x**7*gamma(7/4)*hyper((3/2, 7/4), (11/4,), b*x**4*exp_polar(I*pi)/a)/(4*a**(3/2)*g
amma(11/4)) + e*x**9*gamma(9/4)*hyper((3/2, 9/4), (13/4,), b*x**4*exp_polar(I*pi)/a)/(4*a**(3/2)*gamma(13/4))

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