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 \[ \frac{\sqrt [4]{a} \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac{1}{2}\right ) \left (\frac{5 \sqrt{b} (3 b c-a g)}{\sqrt{a}}-9 a i+15 b e\right )}{30 b^{7/4} \sqrt{a+b x^4}}+\frac{(2 b d-a h) \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a+b x^4}}\right )}{4 b^{3/2}}+\frac{x \sqrt{a+b x^4} (5 b e-3 a i)}{5 b^{3/2} \left (\sqrt{a}+\sqrt{b} x^2\right )}-\frac{\sqrt [4]{a} \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} (5 b e-3 a i) E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{5 b^{7/4} \sqrt{a+b x^4}}+\frac{f \sqrt{a+b x^4}}{2 b}+\frac{g x \sqrt{a+b x^4}}{3 b}+\frac{h x^2 \sqrt{a+b x^4}}{4 b}+\frac{i x^3 \sqrt{a+b x^4}}{5 b} \]

[Out]

(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])

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Rubi [A]  time = 0.423748, antiderivative size = 385, normalized size of antiderivative = 1., 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 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 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 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 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 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 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 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]

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 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 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]

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*}

Mathematica [C]  time = 0.223516, 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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Maple [C]  time = 0.017, size = 516, normalized size = 1.3 \begin{align*}{\frac{i{x}^{3}}{5\,b}\sqrt{b{x}^{4}+a}}-{{\frac{3\,i}{5}}i{a}^{{\frac{3}{2}}}\sqrt{1-{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}{\it EllipticF} \left ( x\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}},i \right ){b}^{-{\frac{3}{2}}}{\frac{1}{\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{b{x}^{4}+a}}}}+{{\frac{3\,i}{5}}i{a}^{{\frac{3}{2}}}\sqrt{1-{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}{\it EllipticE} \left ( x\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}},i \right ){b}^{-{\frac{3}{2}}}{\frac{1}{\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{b{x}^{4}+a}}}}+{\frac{h{x}^{2}}{4\,b}\sqrt{b{x}^{4}+a}}-{\frac{ah}{4}\ln \left ({x}^{2}\sqrt{b}+\sqrt{b{x}^{4}+a} \right ){b}^{-{\frac{3}{2}}}}+{\frac{gx}{3\,b}\sqrt{b{x}^{4}+a}}-{\frac{ag}{3\,b}\sqrt{1-{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}{\it EllipticF} \left ( x\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}},i \right ){\frac{1}{\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{b{x}^{4}+a}}}}+{\frac{f}{2\,b}\sqrt{b{x}^{4}+a}}+{ie\sqrt{a}\sqrt{1-{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}} \left ({\it EllipticF} \left ( x\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}},i \right ) -{\it EllipticE} \left ( x\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}},i \right ) \right ){\frac{1}{\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{b{x}^{4}+a}}}{\frac{1}{\sqrt{b}}}}+{\frac{d}{2}\ln \left ({x}^{2}\sqrt{b}+\sqrt{b{x}^{4}+a} \right ){\frac{1}{\sqrt{b}}}}+{c\sqrt{1-{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}{\it EllipticF} \left ( x\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}},i \right ){\frac{1}{\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{b{x}^{4}+a}}}} \end{align*}

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/b^(3/2)*a^(3/2)/(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)+3/5*I*i/b^(3/2)*a^(3/
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)*E
llipticE(x*(I/a^(1/2)*b^(1/2))^(1/2),I)+1/4*h*x^2*(b*x^4+a)^(1/2)/b-1/4*h*a/b^(3/2)*ln(x^2*b^(1/2)+(b*x^4+a)^(
1/2))+1/3*g*x*(b*x^4+a)^(1/2)/b-1/3*g/b*a/(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*f*(b*x^4+a)^(1/2)/b+I*e*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))+1/2*d*ln(x^2*b^(1
/2)+(b*x^4+a)^(1/2))/b^(1/2)+c/(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., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

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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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\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 ) \end{align*}

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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Sympy [A]  time = 5.73435, size = 260, normalized size = 0.68 \begin{align*} \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 )} \end{align*}

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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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

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)