\(\int \frac {c+d x+e x^2+f x^3+g x^4+h x^5+i x^6}{\sqrt {a+b x^4}} \, dx\) [220]

Optimal result

Integrand size = 42, antiderivative size = 385 \[ \int \frac {c+d x+e x^2+f x^3+g x^4+h x^5+i x^6}{\sqrt {a+b x^4}} \, dx=\frac {f \sqrt {a+b x^4}}{2 b}+\frac {g x \sqrt {a+b x^4}}{3 b}+\frac {h x^2 \sqrt {a+b x^4}}{4 b}+\frac {i x^3 \sqrt {a+b x^4}}{5 b}+\frac {(5 b e-3 a i) x \sqrt {a+b x^4}}{5 b^{3/2} \left (\sqrt {a}+\sqrt {b} x^2\right )}+\frac {(2 b d-a h) \text {arctanh}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )}{4 b^{3/2}}-\frac {\sqrt [4]{a} (5 b e-3 a i) \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{5 b^{7/4} \sqrt {a+b x^4}}+\frac {\sqrt [4]{a} \left (15 b e+\frac {5 \sqrt {b} (3 b c-a g)}{\sqrt {a}}-9 a i\right ) \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{30 b^{7/4} \sqrt {a+b x^4}} \]

[Out]

1/4*(-a*h+2*b*d)*arctanh(x^2*b^(1/2)/(b*x^4+a)^(1/2))/b^(3/2)+1/2*f*(b*x^4+a)^(1/2)/b+1/3*g*x*(b*x^4+a)^(1/2)/
b+1/4*h*x^2*(b*x^4+a)^(1/2)/b+1/5*i*x^3*(b*x^4+a)^(1/2)/b+1/5*(-3*a*i+5*b*e)*x*(b*x^4+a)^(1/2)/b^(3/2)/(a^(1/2
)+x^2*b^(1/2))-1/5*a^(1/4)*(-3*a*i+5*b*e)*(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)+1/30*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))*(a^(1/2)+x^2*b^(1/2))*(15*b*e-9*a
*i+5*(-a*g+3*b*c)*b^(1/2)/a^(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)

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 385, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {1899, 1833, 1829, 655, 223, 212, 1902, 1212, 226, 1210} \[ \int \frac {c+d x+e x^2+f x^3+g x^4+h x^5+i x^6}{\sqrt {a+b x^4}} \, dx=\frac {\sqrt [4]{a} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right ) \left (\frac {5 \sqrt {b} (3 b c-a g)}{\sqrt {a}}-9 a i+15 b e\right )}{30 b^{7/4} \sqrt {a+b x^4}}-\frac {\sqrt [4]{a} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} (5 b e-3 a i) E\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{5 b^{7/4} \sqrt {a+b x^4}}+\frac {\text {arctanh}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right ) (2 b d-a h)}{4 b^{3/2}}+\frac {x \sqrt {a+b x^4} (5 b e-3 a i)}{5 b^{3/2} \left (\sqrt {a}+\sqrt {b} x^2\right )}+\frac {f \sqrt {a+b x^4}}{2 b}+\frac {g x \sqrt {a+b x^4}}{3 b}+\frac {h x^2 \sqrt {a+b x^4}}{4 b}+\frac {i x^3 \sqrt {a+b x^4}}{5 b} \]

[In]

Int[(c + d*x + e*x^2 + f*x^3 + g*x^4 + h*x^5 + i*x^6)/Sqrt[a + b*x^4],x]

[Out]

(f*Sqrt[a + b*x^4])/(2*b) + (g*x*Sqrt[a + b*x^4])/(3*b) + (h*x^2*Sqrt[a + b*x^4])/(4*b) + (i*x^3*Sqrt[a + b*x^
4])/(5*b) + ((5*b*e - 3*a*i)*x*Sqrt[a + b*x^4])/(5*b^(3/2)*(Sqrt[a] + Sqrt[b]*x^2)) + ((2*b*d - a*h)*ArcTanh[(
Sqrt[b]*x^2)/Sqrt[a + b*x^4]])/(4*b^(3/2)) - (a^(1/4)*(5*b*e - 3*a*i)*(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*b^(7/4)*Sqrt[a + b*x^4]) + (a^(1
/4)*(15*b*e + (5*Sqrt[b]*(3*b*c - a*g))/Sqrt[a] - 9*a*i)*(Sqrt[a] + Sqrt[b]*x^2)*Sqrt[(a + b*x^4)/(Sqrt[a] + S
qrt[b]*x^2)^2]*EllipticF[2*ArcTan[(b^(1/4)*x)/a^(1/4)], 1/2])/(30*b^(7/4)*Sqrt[a + b*x^4])

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 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 655

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 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 1829

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 1833

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 1899

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 1902

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*} \text {integral}& = \int \left (\frac {x \left (d+f x^2+h x^4\right )}{\sqrt {a+b x^4}}+\frac {c+e x^2+g x^4+i x^6}{\sqrt {a+b x^4}}\right ) \, dx \\ & = \int \frac {x \left (d+f x^2+h x^4\right )}{\sqrt {a+b x^4}} \, dx+\int \frac {c+e x^2+g x^4+i x^6}{\sqrt {a+b x^4}} \, dx \\ & = \frac {i x^3 \sqrt {a+b x^4}}{5 b}+\frac {1}{2} \text {Subst}\left (\int \frac {d+f x+h x^2}{\sqrt {a+b x^2}} \, dx,x,x^2\right )+\frac {\int \frac {5 b c+(5 b e-3 a i) x^2+5 b g x^4}{\sqrt {a+b x^4}} \, dx}{5 b} \\ & = \frac {g x \sqrt {a+b x^4}}{3 b}+\frac {h x^2 \sqrt {a+b x^4}}{4 b}+\frac {i x^3 \sqrt {a+b x^4}}{5 b}+\frac {\int \frac {5 b (3 b c-a g)+3 b (5 b e-3 a i) x^2}{\sqrt {a+b x^4}} \, dx}{15 b^2}+\frac {\text {Subst}\left (\int \frac {2 b d-a h+2 b f x}{\sqrt {a+b x^2}} \, dx,x,x^2\right )}{4 b} \\ & = \frac {f \sqrt {a+b x^4}}{2 b}+\frac {g x \sqrt {a+b x^4}}{3 b}+\frac {h x^2 \sqrt {a+b x^4}}{4 b}+\frac {i x^3 \sqrt {a+b x^4}}{5 b}+\frac {(2 b d-a h) \text {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,x^2\right )}{4 b}-\frac {\left (\sqrt {a} (5 b e-3 a i)\right ) \int \frac {1-\frac {\sqrt {b} x^2}{\sqrt {a}}}{\sqrt {a+b x^4}} \, dx}{5 b^{3/2}}+\frac {\left (5 \sqrt {b} (3 b c-a g)+3 \sqrt {a} (5 b e-3 a i)\right ) \int \frac {1}{\sqrt {a+b x^4}} \, dx}{15 b^{3/2}} \\ & = \frac {f \sqrt {a+b x^4}}{2 b}+\frac {g x \sqrt {a+b x^4}}{3 b}+\frac {h x^2 \sqrt {a+b x^4}}{4 b}+\frac {i x^3 \sqrt {a+b x^4}}{5 b}+\frac {(5 b e-3 a i) x \sqrt {a+b x^4}}{5 b^{3/2} \left (\sqrt {a}+\sqrt {b} x^2\right )}-\frac {\sqrt [4]{a} (5 b e-3 a i) \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 b^{7/4} \sqrt {a+b x^4}}+\frac {\left (5 \sqrt {b} (3 b c-a g)+3 \sqrt {a} (5 b e-3 a i)\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 )}{30 \sqrt [4]{a} b^{7/4} \sqrt {a+b x^4}}+\frac {(2 b d-a h) \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x^2}{\sqrt {a+b x^4}}\right )}{4 b} \\ & = \frac {f \sqrt {a+b x^4}}{2 b}+\frac {g x \sqrt {a+b x^4}}{3 b}+\frac {h x^2 \sqrt {a+b x^4}}{4 b}+\frac {i x^3 \sqrt {a+b x^4}}{5 b}+\frac {(5 b e-3 a i) x \sqrt {a+b x^4}}{5 b^{3/2} \left (\sqrt {a}+\sqrt {b} x^2\right )}+\frac {(2 b d-a h) \tanh ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )}{4 b^{3/2}}-\frac {\sqrt [4]{a} (5 b e-3 a i) \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 b^{7/4} \sqrt {a+b x^4}}+\frac {\left (5 \sqrt {b} (3 b c-a g)+3 \sqrt {a} (5 b e-3 a i)\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 )}{30 \sqrt [4]{a} b^{7/4} \sqrt {a+b x^4}} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 10.23 (sec) , antiderivative size = 281, normalized size of antiderivative = 0.73 \[ \int \frac {c+d x+e x^2+f x^3+g x^4+h x^5+i x^6}{\sqrt {a+b x^4}} \, dx=\frac {30 a \sqrt {b} f+20 a \sqrt {b} g x+15 a \sqrt {b} h x^2+12 a \sqrt {b} i x^3+30 b^{3/2} f x^4+20 b^{3/2} g x^5+15 b^{3/2} h x^6+12 b^{3/2} i x^7+30 b d \sqrt {a+b x^4} \text {arctanh}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )-15 a h \sqrt {a+b x^4} \text {arctanh}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )-20 \sqrt {b} (-3 b c+a g) x \sqrt {1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-\frac {b x^4}{a}\right )+4 \sqrt {b} (5 b e-3 a i) x^3 \sqrt {1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-\frac {b x^4}{a}\right )}{60 b^{3/2} \sqrt {a+b x^4}} \]

[In]

Integrate[(c + d*x + e*x^2 + f*x^3 + g*x^4 + h*x^5 + i*x^6)/Sqrt[a + b*x^4],x]

[Out]

(30*a*Sqrt[b]*f + 20*a*Sqrt[b]*g*x + 15*a*Sqrt[b]*h*x^2 + 12*a*Sqrt[b]*i*x^3 + 30*b^(3/2)*f*x^4 + 20*b^(3/2)*g
*x^5 + 15*b^(3/2)*h*x^6 + 12*b^(3/2)*i*x^7 + 30*b*d*Sqrt[a + b*x^4]*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a + b*x^4]] - 1
5*a*h*Sqrt[a + b*x^4]*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a + b*x^4]] - 20*Sqrt[b]*(-3*b*c + a*g)*x*Sqrt[1 + (b*x^4)/a]
*Hypergeometric2F1[1/4, 1/2, 5/4, -((b*x^4)/a)] + 4*Sqrt[b]*(5*b*e - 3*a*i)*x^3*Sqrt[1 + (b*x^4)/a]*Hypergeome
tric2F1[1/2, 3/4, 7/4, -((b*x^4)/a)])/(60*b^(3/2)*Sqrt[a + b*x^4])

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 2.01 (sec) , antiderivative size = 287, normalized size of antiderivative = 0.75

method	result	size
elliptic	\(\frac {i \,x^{3} \sqrt {b \,x^{4}+a}}{5 b}+\frac {h \,x^{2} \sqrt {b \,x^{4}+a}}{4 b}+\frac {g x \sqrt {b \,x^{4}+a}}{3 b}+\frac {f \sqrt {b \,x^{4}+a}}{2 b}+\frac {\left (c -\frac {a g}{3 b}\right ) \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, F\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}+\frac {\left (d -\frac {a h}{2 b}\right ) \ln \left (2 x^{2} \sqrt {b}+2 \sqrt {b \,x^{4}+a}\right )}{2 \sqrt {b}}+\frac {i \left (e -\frac {3 a i}{5 b}\right ) \sqrt {a}\, \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (F\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-E\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}\, \sqrt {b}}\)	\(287\)
risch	\(\frac {\left (12 i \,x^{3}+15 h \,x^{2}+20 g x +30 f \right ) \sqrt {b \,x^{4}+a}}{60 b}-\frac {\frac {10 a g \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, F\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}-\frac {30 b c \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, F\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}+\frac {i \left (18 a i -30 b e \right ) \sqrt {a}\, \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (F\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-E\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}\, \sqrt {b}}+\frac {\left (15 a h -30 b d \right ) \ln \left (x^{2} \sqrt {b}+\sqrt {b \,x^{4}+a}\right )}{2 \sqrt {b}}}{30 b}\)	\(323\)
default	\(\frac {c \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, F\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}+i \left (\frac {x^{3} \sqrt {b \,x^{4}+a}}{5 b}-\frac {3 i a^{\frac {3}{2}} \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (F\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-E\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{5 b^{\frac {3}{2}} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )+h \left (\frac {x^{2} \sqrt {b \,x^{4}+a}}{4 b}-\frac {a \ln \left (x^{2} \sqrt {b}+\sqrt {b \,x^{4}+a}\right )}{4 b^{\frac {3}{2}}}\right )+g \left (\frac {x \sqrt {b \,x^{4}+a}}{3 b}-\frac {a \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, F\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{3 b \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )+\frac {f \sqrt {b \,x^{4}+a}}{2 b}+\frac {i e \sqrt {a}\, \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (F\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-E\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}\, \sqrt {b}}+\frac {d \ln \left (x^{2} \sqrt {b}+\sqrt {b \,x^{4}+a}\right )}{2 \sqrt {b}}\)	\(460\)

[In]

int((i*x^6+h*x^5+g*x^4+f*x^3+e*x^2+d*x+c)/(b*x^4+a)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.17 (sec) , antiderivative size = 204, normalized size of antiderivative = 0.53 \[ \int \frac {c+d x+e x^2+f x^3+g x^4+h x^5+i x^6}{\sqrt {a+b x^4}} \, dx=\frac {24 \, {\left (5 \, a b e - 3 \, a^{2} i\right )} \sqrt {b} x \left (-\frac {a}{b}\right )^{\frac {3}{4}} E(\arcsin \left (\frac {\left (-\frac {a}{b}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) + 8 \, {\left (15 \, b^{2} c - 15 \, a b e - 5 \, a b g + 9 \, a^{2} i\right )} \sqrt {b} x \left (-\frac {a}{b}\right )^{\frac {3}{4}} F(\arcsin \left (\frac {\left (-\frac {a}{b}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) - 15 \, {\left (2 \, a b d - a^{2} h\right )} \sqrt {b} x \log \left (-2 \, b x^{4} + 2 \, \sqrt {b x^{4} + a} \sqrt {b} x^{2} - a\right ) + 2 \, {\left (12 \, a b i x^{4} + 15 \, a b h x^{3} + 20 \, a b g x^{2} + 30 \, a b f x + 60 \, a b e - 36 \, a^{2} i\right )} \sqrt {b x^{4} + a}}{120 \, a b^{2} x} \]

[In]

integrate((i*x^6+h*x^5+g*x^4+f*x^3+e*x^2+d*x+c)/(b*x^4+a)^(1/2),x, algorithm="fricas")

[Out]

1/120*(24*(5*a*b*e - 3*a^2*i)*sqrt(b)*x*(-a/b)^(3/4)*elliptic_e(arcsin((-a/b)^(1/4)/x), -1) + 8*(15*b^2*c - 15
*a*b*e - 5*a*b*g + 9*a^2*i)*sqrt(b)*x*(-a/b)^(3/4)*elliptic_f(arcsin((-a/b)^(1/4)/x), -1) - 15*(2*a*b*d - a^2*
h)*sqrt(b)*x*log(-2*b*x^4 + 2*sqrt(b*x^4 + a)*sqrt(b)*x^2 - a) + 2*(12*a*b*i*x^4 + 15*a*b*h*x^3 + 20*a*b*g*x^2
+ 30*a*b*f*x + 60*a*b*e - 36*a^2*i)*sqrt(b*x^4 + a))/(a*b^2*x)

Sympy [A] (verification not implemented)

Time = 3.29 (sec) , antiderivative size = 260, normalized size of antiderivative = 0.68 \[ \int \frac {c+d x+e x^2+f x^3+g x^4+h x^5+i x^6}{\sqrt {a+b x^4}} \, dx=\frac {\sqrt {a} h x^{2} \sqrt {1 + \frac {b x^{4}}{a}}}{4 b} - \frac {a h \operatorname {asinh}{\left (\frac {\sqrt {b} x^{2}}{\sqrt {a}} \right )}}{4 b^{\frac {3}{2}}} + f \left (\begin {cases} \frac {x^{4}}{4 \sqrt {a}} & \text {for}\: b = 0 \\\frac {\sqrt {a + b x^{4}}}{2 b} & \text {otherwise} \end {cases}\right ) + \frac {d \operatorname {asinh}{\left (\frac {\sqrt {b} x^{2}}{\sqrt {a}} \right )}}{2 \sqrt {b}} + \frac {c x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {1}{2} \\ \frac {5}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt {a} \Gamma \left (\frac {5}{4}\right )} + \frac {e x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt {a} \Gamma \left (\frac {7}{4}\right )} + \frac {g x^{5} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt {a} \Gamma \left (\frac {9}{4}\right )} + \frac {i x^{7} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt {a} \Gamma \left (\frac {11}{4}\right )} \]

[In]

integrate((i*x**6+h*x**5+g*x**4+f*x**3+e*x**2+d*x+c)/(b*x**4+a)**(1/2),x)

[Out]

sqrt(a)*h*x**2*sqrt(1 + b*x**4/a)/(4*b) - a*h*asinh(sqrt(b)*x**2/sqrt(a))/(4*b**(3/2)) + f*Piecewise((x**4/(4*
sqrt(a)), Eq(b, 0)), (sqrt(a + b*x**4)/(2*b), True)) + d*asinh(sqrt(b)*x**2/sqrt(a))/(2*sqrt(b)) + c*x*gamma(1
/4)*hyper((1/4, 1/2), (5/4,), b*x**4*exp_polar(I*pi)/a)/(4*sqrt(a)*gamma(5/4)) + e*x**3*gamma(3/4)*hyper((1/2,
3/4), (7/4,), b*x**4*exp_polar(I*pi)/a)/(4*sqrt(a)*gamma(7/4)) + g*x**5*gamma(5/4)*hyper((1/2, 5/4), (9/4,),
b*x**4*exp_polar(I*pi)/a)/(4*sqrt(a)*gamma(9/4)) + i*x**7*gamma(7/4)*hyper((1/2, 7/4), (11/4,), b*x**4*exp_pol
ar(I*pi)/a)/(4*sqrt(a)*gamma(11/4))

Maxima [F]

\[ \int \frac {c+d x+e x^2+f x^3+g x^4+h x^5+i x^6}{\sqrt {a+b x^4}} \, dx=\int { \frac {i x^{6} + h x^{5} + g x^{4} + f x^{3} + e x^{2} + d x + c}{\sqrt {b x^{4} + a}} \,d x } \]

[In]

integrate((i*x^6+h*x^5+g*x^4+f*x^3+e*x^2+d*x+c)/(b*x^4+a)^(1/2),x, algorithm="maxima")

[Out]

integrate((i*x^6 + h*x^5 + g*x^4 + f*x^3 + e*x^2 + d*x + c)/sqrt(b*x^4 + a), x)

Giac [F]

\[ \int \frac {c+d x+e x^2+f x^3+g x^4+h x^5+i x^6}{\sqrt {a+b x^4}} \, dx=\int { \frac {i x^{6} + h x^{5} + g x^{4} + f x^{3} + e x^{2} + d x + c}{\sqrt {b x^{4} + a}} \,d x } \]

[In]

integrate((i*x^6+h*x^5+g*x^4+f*x^3+e*x^2+d*x+c)/(b*x^4+a)^(1/2),x, algorithm="giac")

[Out]

integrate((i*x^6 + h*x^5 + g*x^4 + f*x^3 + e*x^2 + d*x + c)/sqrt(b*x^4 + a), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {c+d x+e x^2+f x^3+g x^4+h x^5+i x^6}{\sqrt {a+b x^4}} \, dx=\int \frac {i\,x^6+h\,x^5+g\,x^4+f\,x^3+e\,x^2+d\,x+c}{\sqrt {b\,x^4+a}} \,d x \]

[In]

int((c + d*x + e*x^2 + f*x^3 + g*x^4 + h*x^5 + i*x^6)/(a + b*x^4)^(1/2),x)

[Out]

int((c + d*x + e*x^2 + f*x^3 + g*x^4 + h*x^5 + i*x^6)/(a + b*x^4)^(1/2), x)