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

Optimal. Leaf size=385 \[ \frac {\sqrt [4]{a} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right ) \left (\frac {5 \sqrt {b} (3 b c-a g)}{\sqrt {a}}-9 a i+15 b e\right )}{30 b^{7/4} \sqrt {a+b x^4}}+\frac {(2 b d-a h) \tanh ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )}{4 b^{3/2}}+\frac {x \sqrt {a+b x^4} (5 b e-3 a i)}{5 b^{3/2} \left (\sqrt {a}+\sqrt {b} x^2\right )}-\frac {\sqrt [4]{a} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} (5 b e-3 a i) E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{5 b^{7/4} \sqrt {a+b x^4}}+\frac {f \sqrt {a+b x^4}}{2 b}+\frac {g x \sqrt {a+b x^4}}{3 b}+\frac {h x^2 \sqrt {a+b x^4}}{4 b}+\frac {i x^3 \sqrt {a+b x^4}}{5 b} \]

[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)

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Rubi [A]  time = 0.42, antiderivative size = 385, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 10, integrand size = 42, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {1885, 1819, 1815, 641, 217, 206, 1888, 1198, 220, 1196} \[ \frac {\sqrt [4]{a} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right ) \left (\frac {5 \sqrt {b} (3 b c-a g)}{\sqrt {a}}-9 a i+15 b e\right )}{30 b^{7/4} \sqrt {a+b x^4}}+\frac {(2 b d-a h) \tanh ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )}{4 b^{3/2}}+\frac {x \sqrt {a+b x^4} (5 b e-3 a i)}{5 b^{3/2} \left (\sqrt {a}+\sqrt {b} x^2\right )}-\frac {\sqrt [4]{a} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} (5 b e-3 a i) E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{5 b^{7/4} \sqrt {a+b x^4}}+\frac {f \sqrt {a+b x^4}}{2 b}+\frac {g x \sqrt {a+b x^4}}{3 b}+\frac {h x^2 \sqrt {a+b x^4}}{4 b}+\frac {i x^3 \sqrt {a+b x^4}}{5 b} \]

Antiderivative was successfully verified.

[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 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 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 {c+d x+e x^2+f x^3+g x^4+h x^5+220 x^6}{\sqrt {a+b x^4}} \, dx &=\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+220 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+220 x^6}{\sqrt {a+b x^4}} \, dx\\ &=\frac {44 x^3 \sqrt {a+b x^4}}{b}+\frac {1}{2} \operatorname {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 (132 a-b e) 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 {44 x^3 \sqrt {a+b x^4}}{b}+\frac {\int \frac {5 b (3 b c-a g)-15 b (132 a-b e) x^2}{\sqrt {a+b x^4}} \, dx}{15 b^2}+\frac {\operatorname {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 {44 x^3 \sqrt {a+b x^4}}{b}+\frac {\left (\sqrt {a} (132 a-b e)\right ) \int \frac {1-\frac {\sqrt {b} x^2}{\sqrt {a}}}{\sqrt {a+b x^4}} \, dx}{b^{3/2}}-\frac {\left (396 a^{3/2}-3 b^{3/2} c-3 \sqrt {a} b e+a \sqrt {b} g\right ) \int \frac {1}{\sqrt {a+b x^4}} \, dx}{3 b^{3/2}}+\frac {(2 b d-a h) \operatorname {Subst}\left (\int \frac {1}{\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 {44 x^3 \sqrt {a+b x^4}}{b}-\frac {(132 a-b e) x \sqrt {a+b x^4}}{b^{3/2} \left (\sqrt {a}+\sqrt {b} x^2\right )}+\frac {\sqrt [4]{a} (132 a-b e) \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 )}{b^{7/4} \sqrt {a+b x^4}}-\frac {\left (396 a^{3/2}-3 b^{3/2} c-3 \sqrt {a} b e+a \sqrt {b} g\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 )}{6 \sqrt [4]{a} b^{7/4} \sqrt {a+b x^4}}+\frac {(2 b d-a h) \operatorname {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 {44 x^3 \sqrt {a+b x^4}}{b}-\frac {(132 a-b e) x \sqrt {a+b x^4}}{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} (132 a-b e) \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 )}{b^{7/4} \sqrt {a+b x^4}}-\frac {\left (396 a^{3/2}-3 b^{3/2} c-3 \sqrt {a} b e+a \sqrt {b} g\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 )}{6 \sqrt [4]{a} b^{7/4} \sqrt {a+b x^4}}\\ \end {align*}

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Mathematica [C]  time = 0.26, size = 281, normalized size = 0.73 \[ \frac {-20 \sqrt {b} x \sqrt {\frac {b x^4}{a}+1} (a g-3 b c) \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};-\frac {b x^4}{a}\right )+30 b d \sqrt {a+b x^4} \tanh ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )+4 \sqrt {b} x^3 \sqrt {\frac {b x^4}{a}+1} (5 b e-3 a i) \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-\frac {b x^4}{a}\right )+30 a \sqrt {b} f+20 a \sqrt {b} g x+15 a \sqrt {b} h x^2-15 a h \sqrt {a+b x^4} \tanh ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )+12 a \sqrt {b} i x^3+30 b^{3/2} f x^4+20 b^{3/2} g x^5+15 b^{3/2} h x^6+12 b^{3/2} i x^7}{60 b^{3/2} \sqrt {a+b x^4}} \]

Antiderivative was successfully verified.

[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])

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fricas [F]  time = 0.79, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {i x^{6} + h x^{5} + g x^{4} + f x^{3} + e x^{2} + d x + c}{\sqrt {b x^{4} + a}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[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]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {i x^{6} + h x^{5} + g x^{4} + f x^{3} + e x^{2} + d x + c}{\sqrt {b x^{4} + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [C]  time = 0.20, size = 516, normalized size = 1.34 \[ \frac {\sqrt {b \,x^{4}+a}\, i \,x^{3}}{5 b}+\frac {3 i \sqrt {-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, a^{\frac {3}{2}} i \EllipticE \left (\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, x , i\right )}{5 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}\, b^{\frac {3}{2}}}-\frac {3 i \sqrt {-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, a^{\frac {3}{2}} i \EllipticF \left (\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, x , i\right )}{5 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}\, b^{\frac {3}{2}}}-\frac {\sqrt {-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, a g \EllipticF \left (\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, x , i\right )}{3 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}\, b}+\frac {\sqrt {b \,x^{4}+a}\, h \,x^{2}}{4 b}+\frac {\sqrt {-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, c \EllipticF \left (\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, x , i\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}-\frac {a h \ln \left (\sqrt {b}\, x^{2}+\sqrt {b \,x^{4}+a}\right )}{4 b^{\frac {3}{2}}}+\frac {d \ln \left (\sqrt {b}\, x^{2}+\sqrt {b \,x^{4}+a}\right )}{2 \sqrt {b}}+\frac {i \sqrt {-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, \left (-\EllipticE \left (\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, x , i\right )+\EllipticF \left (\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, x , i\right )\right ) \sqrt {a}\, e}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}\, \sqrt {b}}+\frac {\sqrt {b \,x^{4}+a}\, g x}{3 b}+\frac {\sqrt {b \,x^{4}+a}\, f}{2 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {i x^{6} + h x^{5} + g x^{4} + f x^{3} + e x^{2} + d x + c}{\sqrt {b x^{4} + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {i\,x^6+h\,x^5+g\,x^4+f\,x^3+e\,x^2+d\,x+c}{\sqrt {b\,x^4+a}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 7.36, size = 260, normalized size = 0.68 \[ \frac {\sqrt {a} h x^{2} \sqrt {1 + \frac {b x^{4}}{a}}}{4 b} - \frac {a h \operatorname {asinh}{\left (\frac {\sqrt {b} x^{2}}{\sqrt {a}} \right )}}{4 b^{\frac {3}{2}}} + f \left (\begin {cases} \frac {x^{4}}{4 \sqrt {a}} & \text {for}\: b = 0 \\\frac {\sqrt {a + b x^{4}}}{2 b} & \text {otherwise} \end {cases}\right ) + \frac {d \operatorname {asinh}{\left (\frac {\sqrt {b} x^{2}}{\sqrt {a}} \right )}}{2 \sqrt {b}} + \frac {c x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {1}{2} \\ \frac {5}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt {a} \Gamma \left (\frac {5}{4}\right )} + \frac {e x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt {a} \Gamma \left (\frac {7}{4}\right )} + \frac {g x^{5} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt {a} \Gamma \left (\frac {9}{4}\right )} + \frac {i x^{7} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt {a} \Gamma \left (\frac {11}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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