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

Optimal. Leaf size=360 \[ \frac {b^{3/4} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \left (5 \sqrt {a} e+3 \sqrt {b} c\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{15 a^{3/4} \sqrt {a+b x^4}}-\frac {2 b^{5/4} c \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{5 a^{3/4} \sqrt {a+b x^4}}+\frac {2 b^{3/2} c x \sqrt {a+b x^4}}{5 a \left (\sqrt {a}+\sqrt {b} x^2\right )}-\frac {1}{60} \sqrt {a+b x^4} \left (\frac {12 c}{x^5}+\frac {15 d}{x^4}+\frac {20 e}{x^3}+\frac {30 f}{x^2}\right )-\frac {2 b c \sqrt {a+b x^4}}{5 a x}-\frac {b d \tanh ^{-1}\left (\frac {\sqrt {a+b x^4}}{\sqrt {a}}\right )}{4 \sqrt {a}}+\frac {1}{2} \sqrt {b} f \tanh ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right ) \]

[Out]

-1/4*b*d*arctanh((b*x^4+a)^(1/2)/a^(1/2))/a^(1/2)+1/2*f*arctanh(x^2*b^(1/2)/(b*x^4+a)^(1/2))*b^(1/2)-1/60*(12*
c/x^5+15*d/x^4+20*e/x^3+30*f/x^2)*(b*x^4+a)^(1/2)-2/5*b*c*(b*x^4+a)^(1/2)/a/x+2/5*b^(3/2)*c*x*(b*x^4+a)^(1/2)/
a/(a^(1/2)+x^2*b^(1/2))-2/5*b^(5/4)*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)/a^(3/4)/(b*x^4+a)^(1/2)+1/15*b^(3/4)*(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)+3*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)/a^(3/4)/(b*x^4+a)^(1/2)

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Rubi [A]  time = 0.32, antiderivative size = 360, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 14, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.467, Rules used = {14, 1825, 1833, 1282, 1198, 220, 1196, 1252, 844, 217, 206, 266, 63, 208} \[ \frac {b^{3/4} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \left (5 \sqrt {a} e+3 \sqrt {b} c\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{15 a^{3/4} \sqrt {a+b x^4}}-\frac {2 b^{5/4} c \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{5 a^{3/4} \sqrt {a+b x^4}}+\frac {2 b^{3/2} c x \sqrt {a+b x^4}}{5 a \left (\sqrt {a}+\sqrt {b} x^2\right )}-\frac {1}{60} \sqrt {a+b x^4} \left (\frac {12 c}{x^5}+\frac {15 d}{x^4}+\frac {20 e}{x^3}+\frac {30 f}{x^2}\right )-\frac {2 b c \sqrt {a+b x^4}}{5 a x}-\frac {b d \tanh ^{-1}\left (\frac {\sqrt {a+b x^4}}{\sqrt {a}}\right )}{4 \sqrt {a}}+\frac {1}{2} \sqrt {b} f \tanh ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((12*c)/x^5 + (15*d)/x^4 + (20*e)/x^3 + (30*f)/x^2)*Sqrt[a + b*x^4])/60 - (2*b*c*Sqrt[a + b*x^4])/(5*a*x) +
(2*b^(3/2)*c*x*Sqrt[a + b*x^4])/(5*a*(Sqrt[a] + Sqrt[b]*x^2)) + (Sqrt[b]*f*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a + b*x^
4]])/2 - (b*d*ArcTanh[Sqrt[a + b*x^4]/Sqrt[a]])/(4*Sqrt[a]) - (2*b^(5/4)*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])/(5*a^(3/4)*Sqrt[a + b*x^4]) +
(b^(3/4)*(3*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]*Ellip
ticF[2*ArcTan[(b^(1/4)*x)/a^(1/4)], 1/2])/(15*a^(3/4)*Sqrt[a + b*x^4])

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

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 208

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

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 266

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

Rule 844

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

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 1252

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[x^((m
- 1)/2)*(d + e*x)^q*(a + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, c, d, e, p, q}, x] && IntegerQ[(m + 1)/2]

Rule 1282

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)*((a_) + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[(d*(f*x)^(m + 1)*(a
+ c*x^4)^(p + 1))/(a*f*(m + 1)), x] + Dist[1/(a*f^2*(m + 1)), Int[(f*x)^(m + 2)*(a + c*x^4)^p*(a*e*(m + 1) -
c*d*(m + 4*p + 5)*x^2), x], x] /; FreeQ[{a, c, d, e, f, p}, x] && LtQ[m, -1] && IntegerQ[2*p] && (IntegerQ[p]
|| IntegerQ[m])

Rule 1825

Int[(Pq_)*(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Module[{u = IntHide[x^m*Pq, x]}, Simp[u*(a +
b*x^n)^p, x] - Dist[b*n*p, Int[x^(m + n)*(a + b*x^n)^(p - 1)*ExpandToSum[u/x^(m + 1), x], x], x]] /; FreeQ[{a
, b}, x] && PolyQ[Pq, x] && IGtQ[n, 0] && GtQ[p, 0] && LtQ[m + Expon[Pq, x] + 1, 0]

Rule 1833

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

Rubi steps

\begin {align*} \int \frac {\left (c+d x+e x^2+f x^3\right ) \sqrt {a+b x^4}}{x^6} \, dx &=-\frac {1}{60} \left (\frac {12 c}{x^5}+\frac {15 d}{x^4}+\frac {20 e}{x^3}+\frac {30 f}{x^2}\right ) \sqrt {a+b x^4}-(2 b) \int \frac {-\frac {c}{5}-\frac {d x}{4}-\frac {e x^2}{3}-\frac {f x^3}{2}}{x^2 \sqrt {a+b x^4}} \, dx\\ &=-\frac {1}{60} \left (\frac {12 c}{x^5}+\frac {15 d}{x^4}+\frac {20 e}{x^3}+\frac {30 f}{x^2}\right ) \sqrt {a+b x^4}-(2 b) \int \left (\frac {-\frac {c}{5}-\frac {e x^2}{3}}{x^2 \sqrt {a+b x^4}}+\frac {-\frac {d}{4}-\frac {f x^2}{2}}{x \sqrt {a+b x^4}}\right ) \, dx\\ &=-\frac {1}{60} \left (\frac {12 c}{x^5}+\frac {15 d}{x^4}+\frac {20 e}{x^3}+\frac {30 f}{x^2}\right ) \sqrt {a+b x^4}-(2 b) \int \frac {-\frac {c}{5}-\frac {e x^2}{3}}{x^2 \sqrt {a+b x^4}} \, dx-(2 b) \int \frac {-\frac {d}{4}-\frac {f x^2}{2}}{x \sqrt {a+b x^4}} \, dx\\ &=-\frac {1}{60} \left (\frac {12 c}{x^5}+\frac {15 d}{x^4}+\frac {20 e}{x^3}+\frac {30 f}{x^2}\right ) \sqrt {a+b x^4}-\frac {2 b c \sqrt {a+b x^4}}{5 a x}-b \operatorname {Subst}\left (\int \frac {-\frac {d}{4}-\frac {f x}{2}}{x \sqrt {a+b x^2}} \, dx,x,x^2\right )+\frac {(2 b) \int \frac {\frac {a e}{3}+\frac {1}{5} b c x^2}{\sqrt {a+b x^4}} \, dx}{a}\\ &=-\frac {1}{60} \left (\frac {12 c}{x^5}+\frac {15 d}{x^4}+\frac {20 e}{x^3}+\frac {30 f}{x^2}\right ) \sqrt {a+b x^4}-\frac {2 b c \sqrt {a+b x^4}}{5 a x}-\frac {\left (2 b^{3/2} c\right ) \int \frac {1-\frac {\sqrt {b} x^2}{\sqrt {a}}}{\sqrt {a+b x^4}} \, dx}{5 \sqrt {a}}+\frac {1}{4} (b d) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+b x^2}} \, dx,x,x^2\right )+\frac {1}{15} \left (2 b \left (\frac {3 \sqrt {b} c}{\sqrt {a}}+5 e\right )\right ) \int \frac {1}{\sqrt {a+b x^4}} \, dx+\frac {1}{2} (b f) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,x^2\right )\\ &=-\frac {1}{60} \left (\frac {12 c}{x^5}+\frac {15 d}{x^4}+\frac {20 e}{x^3}+\frac {30 f}{x^2}\right ) \sqrt {a+b x^4}-\frac {2 b c \sqrt {a+b x^4}}{5 a x}+\frac {2 b^{3/2} c x \sqrt {a+b x^4}}{5 a \left (\sqrt {a}+\sqrt {b} x^2\right )}-\frac {2 b^{5/4} 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 )}{5 a^{3/4} \sqrt {a+b x^4}}+\frac {b^{3/4} \left (3 \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 )}{15 a^{3/4} \sqrt {a+b x^4}}+\frac {1}{8} (b d) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,x^4\right )+\frac {1}{2} (b f) \operatorname {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x^2}{\sqrt {a+b x^4}}\right )\\ &=-\frac {1}{60} \left (\frac {12 c}{x^5}+\frac {15 d}{x^4}+\frac {20 e}{x^3}+\frac {30 f}{x^2}\right ) \sqrt {a+b x^4}-\frac {2 b c \sqrt {a+b x^4}}{5 a x}+\frac {2 b^{3/2} c x \sqrt {a+b x^4}}{5 a \left (\sqrt {a}+\sqrt {b} x^2\right )}+\frac {1}{2} \sqrt {b} f \tanh ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )-\frac {2 b^{5/4} 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 )}{5 a^{3/4} \sqrt {a+b x^4}}+\frac {b^{3/4} \left (3 \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 )}{15 a^{3/4} \sqrt {a+b x^4}}+\frac {1}{4} d \operatorname {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b x^4}\right )\\ &=-\frac {1}{60} \left (\frac {12 c}{x^5}+\frac {15 d}{x^4}+\frac {20 e}{x^3}+\frac {30 f}{x^2}\right ) \sqrt {a+b x^4}-\frac {2 b c \sqrt {a+b x^4}}{5 a x}+\frac {2 b^{3/2} c x \sqrt {a+b x^4}}{5 a \left (\sqrt {a}+\sqrt {b} x^2\right )}+\frac {1}{2} \sqrt {b} f \tanh ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )-\frac {b d \tanh ^{-1}\left (\frac {\sqrt {a+b x^4}}{\sqrt {a}}\right )}{4 \sqrt {a}}-\frac {2 b^{5/4} 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 )}{5 a^{3/4} \sqrt {a+b x^4}}+\frac {b^{3/4} \left (3 \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 )}{15 a^{3/4} \sqrt {a+b x^4}}\\ \end {align*}

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Mathematica [C]  time = 0.26, size = 179, normalized size = 0.50 \[ -\frac {\sqrt {a+b x^4} \left (12 a c \, _2F_1\left (-\frac {5}{4},-\frac {1}{2};-\frac {1}{4};-\frac {b x^4}{a}\right )+5 x \left (3 a d \sqrt {\frac {b x^4}{a}+1}+3 b d x^4 \tanh ^{-1}\left (\sqrt {\frac {b x^4}{a}+1}\right )+4 a e x \, _2F_1\left (-\frac {3}{4},-\frac {1}{2};\frac {1}{4};-\frac {b x^4}{a}\right )+6 a f x^2 \sqrt {\frac {b x^4}{a}+1}-6 \sqrt {a} \sqrt {b} f x^4 \sinh ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right )\right )}{60 a x^5 \sqrt {\frac {b x^4}{a}+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/60*(Sqrt[a + b*x^4]*(12*a*c*Hypergeometric2F1[-5/4, -1/2, -1/4, -((b*x^4)/a)] + 5*x*(3*a*d*Sqrt[1 + (b*x^4)
/a] + 6*a*f*x^2*Sqrt[1 + (b*x^4)/a] - 6*Sqrt[a]*Sqrt[b]*f*x^4*ArcSinh[(Sqrt[b]*x^2)/Sqrt[a]] + 3*b*d*x^4*ArcTa
nh[Sqrt[1 + (b*x^4)/a]] + 4*a*e*x*Hypergeometric2F1[-3/4, -1/2, 1/4, -((b*x^4)/a)])))/(a*x^5*Sqrt[1 + (b*x^4)/
a])

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fricas [F]  time = 0.71, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {b x^{4} + a} {\left (f x^{3} + e x^{2} + d x + c\right )}}{x^{6}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x^3+e*x^2+d*x+c)*(b*x^4+a)^(1/2)/x^6,x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x^3+e*x^2+d*x+c)*(b*x^4+a)^(1/2)/x^6,x, algorithm="giac")

[Out]

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

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maple [C]  time = 0.20, size = 404, normalized size = 1.12 \[ -\frac {2 i \sqrt {-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, b^{\frac {3}{2}} c \EllipticE \left (\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, x , i\right )}{5 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}\, \sqrt {a}}+\frac {2 i \sqrt {-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, b^{\frac {3}{2}} c \EllipticF \left (\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, x , i\right )}{5 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}\, \sqrt {a}}+\frac {\sqrt {b \,x^{4}+a}\, b f \,x^{2}}{2 a}+\frac {2 \sqrt {-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, b e \EllipticF \left (\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, x , i\right )}{3 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}-\frac {b d \ln \left (\frac {2 a +2 \sqrt {b \,x^{4}+a}\, \sqrt {a}}{x^{2}}\right )}{4 \sqrt {a}}+\frac {\sqrt {b}\, f \ln \left (\sqrt {b}\, x^{2}+\sqrt {b \,x^{4}+a}\right )}{2}+\frac {\sqrt {b \,x^{4}+a}\, b d}{4 a}-\frac {2 \sqrt {b \,x^{4}+a}\, b c}{5 a x}-\frac {\left (b \,x^{4}+a \right )^{\frac {3}{2}} f}{2 a \,x^{2}}-\frac {\sqrt {b \,x^{4}+a}\, e}{3 x^{3}}-\frac {\left (b \,x^{4}+a \right )^{\frac {3}{2}} d}{4 a \,x^{4}}-\frac {\sqrt {b \,x^{4}+a}\, c}{5 x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f*x^3+e*x^2+d*x+c)*(b*x^4+a)^(1/2)/x^6,x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x^3+e*x^2+d*x+c)*(b*x^4+a)^(1/2)/x^6,x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*x^4)^(1/2)*(c + d*x + e*x^2 + f*x^3))/x^6,x)

[Out]

int(((a + b*x^4)^(1/2)*(c + d*x + e*x^2 + f*x^3))/x^6, x)

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sympy [C]  time = 7.06, size = 216, normalized size = 0.60 \[ \frac {\sqrt {a} c \Gamma \left (- \frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{4}, - \frac {1}{2} \\ - \frac {1}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 x^{5} \Gamma \left (- \frac {1}{4}\right )} + \frac {\sqrt {a} e \Gamma \left (- \frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, - \frac {1}{2} \\ \frac {1}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 x^{3} \Gamma \left (\frac {1}{4}\right )} - \frac {\sqrt {a} f}{2 x^{2} \sqrt {1 + \frac {b x^{4}}{a}}} - \frac {\sqrt {b} d \sqrt {\frac {a}{b x^{4}} + 1}}{4 x^{2}} + \frac {\sqrt {b} f \operatorname {asinh}{\left (\frac {\sqrt {b} x^{2}}{\sqrt {a}} \right )}}{2} - \frac {b d \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x^{2}} \right )}}{4 \sqrt {a}} - \frac {b f x^{2}}{2 \sqrt {a} \sqrt {1 + \frac {b x^{4}}{a}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x**3+e*x**2+d*x+c)*(b*x**4+a)**(1/2)/x**6,x)

[Out]

sqrt(a)*c*gamma(-5/4)*hyper((-5/4, -1/2), (-1/4,), b*x**4*exp_polar(I*pi)/a)/(4*x**5*gamma(-1/4)) + sqrt(a)*e*
gamma(-3/4)*hyper((-3/4, -1/2), (1/4,), b*x**4*exp_polar(I*pi)/a)/(4*x**3*gamma(1/4)) - sqrt(a)*f/(2*x**2*sqrt
(1 + b*x**4/a)) - sqrt(b)*d*sqrt(a/(b*x**4) + 1)/(4*x**2) + sqrt(b)*f*asinh(sqrt(b)*x**2/sqrt(a))/2 - b*d*asin
h(sqrt(a)/(sqrt(b)*x**2))/(4*sqrt(a)) - b*f*x**2/(2*sqrt(a)*sqrt(1 + b*x**4/a))

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