3.6.5 \(\int \frac {(c+d x+e x^2+f x^3) \sqrt {a+b x^4}}{x^6} \, dx\) [505]

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

[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 = 0.21, 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, 1839, 1847, 1296, 1212, 226, 1210, 1266, 858, 223, 212, 272, 65, 214} \begin {gather*} \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 \text {ArcTan}\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 \text {ArcTan}\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 ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/60*(((12*c)/x^5 + (15*d)/x^4 + (20*e)/x^3 + (30*f)/x^2)*Sqrt[a + b*x^4]) - (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]*Ell
ipticF[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 65

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 212

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

Rule 214

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

Rule 223

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 226

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)]/(2*q*Sqrt[a + b*x^4]))*EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 272

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 858

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 1210

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)]/(q*Sqrt[a + c*x^4
]))*EllipticE[2*ArcTan[q*x], 1/2], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

Rule 1212

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 1266

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 1296

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 1839

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 1847

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)/c^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, {
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 \text {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) \text {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) \text {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) \text {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,x^4\right )+\frac {1}{2} (b f) \text {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 \text {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] Result contains complex when optimal does not.
time = 10.50, size = 314, normalized size = 0.87 \begin {gather*} \frac {-\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}} \left (\left (a+b x^4\right ) \left (12 a c+24 b c x^4+5 a x \left (3 d+4 e x+6 f x^2\right )\right )-30 a \sqrt {b} f x^5 \sqrt {a+b x^4} \tanh ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )+15 \sqrt {a} b d x^5 \sqrt {a+b x^4} \tanh ^{-1}\left (\frac {\sqrt {a+b x^4}}{\sqrt {a}}\right )\right )+24 \sqrt {a} b^{3/2} c x^5 \sqrt {1+\frac {b x^4}{a}} E\left (\left .i \sinh ^{-1}\left (\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}} x\right )\right |-1\right )-8 i \sqrt {a} b \left (-3 i \sqrt {b} c+5 \sqrt {a} e\right ) x^5 \sqrt {1+\frac {b x^4}{a}} F\left (\left .i \sinh ^{-1}\left (\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}} x\right )\right |-1\right )}{60 a \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}} x^5 \sqrt {a+b x^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-(Sqrt[(I*Sqrt[b])/Sqrt[a]]*((a + b*x^4)*(12*a*c + 24*b*c*x^4 + 5*a*x*(3*d + 4*e*x + 6*f*x^2)) - 30*a*Sqrt[b]
*f*x^5*Sqrt[a + b*x^4]*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a + b*x^4]] + 15*Sqrt[a]*b*d*x^5*Sqrt[a + b*x^4]*ArcTanh[Sqr
t[a + b*x^4]/Sqrt[a]])) + 24*Sqrt[a]*b^(3/2)*c*x^5*Sqrt[1 + (b*x^4)/a]*EllipticE[I*ArcSinh[Sqrt[(I*Sqrt[b])/Sq
rt[a]]*x], -1] - (8*I)*Sqrt[a]*b*((-3*I)*Sqrt[b]*c + 5*Sqrt[a]*e)*x^5*Sqrt[1 + (b*x^4)/a]*EllipticF[I*ArcSinh[
Sqrt[(I*Sqrt[b])/Sqrt[a]]*x], -1])/(60*a*Sqrt[(I*Sqrt[b])/Sqrt[a]]*x^5*Sqrt[a + b*x^4])

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Maple [C] Result contains complex when optimal does not.
time = 0.41, size = 346, normalized size = 0.96

method result size
elliptic \(-\frac {c \sqrt {b \,x^{4}+a}}{5 x^{5}}-\frac {d \sqrt {b \,x^{4}+a}}{4 x^{4}}-\frac {e \sqrt {b \,x^{4}+a}}{3 x^{3}}-\frac {f \sqrt {b \,x^{4}+a}}{2 x^{2}}-\frac {2 b c \sqrt {b \,x^{4}+a}}{5 a x}+\frac {2 b e \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \EllipticF \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{3 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}+\frac {\sqrt {b}\, f \ln \left (2 x^{2} \sqrt {b}+2 \sqrt {b \,x^{4}+a}\right )}{2}+\frac {2 i b^{\frac {3}{2}} c \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\EllipticF \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\EllipticE \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{5 \sqrt {a}\, \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}-\frac {b d \arctanh \left (\frac {\sqrt {a}}{\sqrt {b \,x^{4}+a}}\right )}{4 \sqrt {a}}\) \(298\)
risch \(-\frac {\sqrt {b \,x^{4}+a}\, \left (24 b c \,x^{4}+30 a f \,x^{3}+20 a e \,x^{2}+15 a d x +12 a c \right )}{60 x^{5} a}+\frac {2 i b^{\frac {3}{2}} c \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \EllipticF \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{5 \sqrt {a}\, \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}-\frac {2 i b^{\frac {3}{2}} c \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \EllipticE \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{5 \sqrt {a}\, \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}+\frac {\sqrt {b}\, f \ln \left (x^{2} \sqrt {b}+\sqrt {b \,x^{4}+a}\right )}{2}+\frac {2 b e \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \EllipticF \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{3 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}-\frac {b d \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{4}+a}}{x^{2}}\right )}{4 \sqrt {a}}\) \(332\)
default \(c \left (-\frac {\sqrt {b \,x^{4}+a}}{5 x^{5}}-\frac {2 b \sqrt {b \,x^{4}+a}}{5 a x}+\frac {2 i b^{\frac {3}{2}} \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\EllipticF \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\EllipticE \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{5 \sqrt {a}\, \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )+d \left (-\frac {\left (b \,x^{4}+a \right )^{\frac {3}{2}}}{4 a \,x^{4}}-\frac {b \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{4}+a}}{x^{2}}\right )}{4 \sqrt {a}}+\frac {b \sqrt {b \,x^{4}+a}}{4 a}\right )+f \left (-\frac {\left (b \,x^{4}+a \right )^{\frac {3}{2}}}{2 a \,x^{2}}+\frac {b \,x^{2} \sqrt {b \,x^{4}+a}}{2 a}+\frac {\sqrt {b}\, \ln \left (x^{2} \sqrt {b}+\sqrt {b \,x^{4}+a}\right )}{2}\right )+e \left (-\frac {\sqrt {b \,x^{4}+a}}{3 x^{3}}+\frac {2 b \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \EllipticF \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{3 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )\) \(346\)

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 + x^2*e + d*x + c)/x^6, x)

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Fricas [F]
time = 0.25, size = 30, normalized size = 0.08 \begin {gather*} {\rm integral}\left (\frac {\sqrt {b x^{4} + a} {\left (f x^{3} + e x^{2} + d x + c\right )}}{x^{6}}, x\right ) \end {gather*}

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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Sympy [C] Result contains complex when optimal does not.
time = 3.26, size = 216, normalized size = 0.60 \begin {gather*} \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}}} \end {gather*}

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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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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 + x^2*e + d*x + c)/x^6, x)

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

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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