3.6.29 \(\int \frac {x^4 (c+d x+e x^2+f x^3)}{\sqrt {a+b x^4}} \, dx\) [529]

Optimal. Leaf size=361 \[ \frac {c x \sqrt {a+b x^4}}{3 b}+\frac {e x^3 \sqrt {a+b x^4}}{5 b}+\frac {f x^4 \sqrt {a+b x^4}}{6 b}-\frac {3 a e x \sqrt {a+b x^4}}{5 b^{3/2} \left (\sqrt {a}+\sqrt {b} x^2\right )}-\frac {\left (4 a f-3 b d x^2\right ) \sqrt {a+b x^4}}{12 b^2}-\frac {a d \tanh ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )}{4 b^{3/2}}+\frac {3 a^{5/4} e \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{5 b^{7/4} \sqrt {a+b x^4}}-\frac {a^{3/4} \left (5 \sqrt {b} c+9 \sqrt {a} e\right ) \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{30 b^{7/4} \sqrt {a+b x^4}} \]

[Out]

-1/4*a*d*arctanh(x^2*b^(1/2)/(b*x^4+a)^(1/2))/b^(3/2)+1/3*c*x*(b*x^4+a)^(1/2)/b+1/5*e*x^3*(b*x^4+a)^(1/2)/b+1/
6*f*x^4*(b*x^4+a)^(1/2)/b-1/12*(-3*b*d*x^2+4*a*f)*(b*x^4+a)^(1/2)/b^2-3/5*a*e*x*(b*x^4+a)^(1/2)/b^(3/2)/(a^(1/
2)+x^2*b^(1/2))+3/5*a^(5/4)*e*(cos(2*arctan(b^(1/4)*x/a^(1/4)))^2)^(1/2)/cos(2*arctan(b^(1/4)*x/a^(1/4)))*Elli
pticE(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^(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))*(9*e*a^(1/2)+5*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)/b^(7/4)/(b*x^4+a)^(1/2)

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Rubi [A]
time = 0.45, antiderivative size = 361, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 10, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {1847, 1294, 1212, 226, 1210, 1266, 847, 794, 223, 212} \begin {gather*} -\frac {a^{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 (9 \sqrt {a} e+5 \sqrt {b} c\right ) F\left (2 \text {ArcTan}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{30 b^{7/4} \sqrt {a+b x^4}}+\frac {3 a^{5/4} 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 \text {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 {a d \tanh ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )}{4 b^{3/2}}-\frac {3 a e x \sqrt {a+b x^4}}{5 b^{3/2} \left (\sqrt {a}+\sqrt {b} x^2\right )}-\frac {\sqrt {a+b x^4} \left (4 a f-3 b d x^2\right )}{12 b^2}+\frac {c x \sqrt {a+b x^4}}{3 b}+\frac {e x^3 \sqrt {a+b x^4}}{5 b}+\frac {f x^4 \sqrt {a+b x^4}}{6 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(c*x*Sqrt[a + b*x^4])/(3*b) + (e*x^3*Sqrt[a + b*x^4])/(5*b) + (f*x^4*Sqrt[a + b*x^4])/(6*b) - (3*a*e*x*Sqrt[a
+ b*x^4])/(5*b^(3/2)*(Sqrt[a] + Sqrt[b]*x^2)) - ((4*a*f - 3*b*d*x^2)*Sqrt[a + b*x^4])/(12*b^2) - (a*d*ArcTanh[
(Sqrt[b]*x^2)/Sqrt[a + b*x^4]])/(4*b^(3/2)) + (3*a^(5/4)*e*(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^(3/4)*(5*Sqrt
[b]*c + 9*Sqrt[a]*e)*(Sqrt[a] + Sqrt[b]*x^2)*Sqrt[(a + b*x^4)/(Sqrt[a] + Sqrt[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 794

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((e*f + d*g)*(2*p
+ 3) + 2*e*g*(p + 1)*x)*((a + c*x^2)^(p + 1)/(2*c*(p + 1)*(2*p + 3))), x] - Dist[(a*e*g - c*d*f*(2*p + 3))/(c*
(2*p + 3)), Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, p}, x] &&  !LeQ[p, -1]

Rule 847

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[g*(d + e*x)^
m*((a + c*x^2)^(p + 1)/(c*(m + 2*p + 2))), x] + Dist[1/(c*(m + 2*p + 2)), Int[(d + e*x)^(m - 1)*(a + c*x^2)^p*
Simp[c*d*f*(m + 2*p + 2) - a*e*g*m + c*(e*f*(m + 2*p + 2) + d*g*m)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g, p
}, x] && NeQ[c*d^2 + a*e^2, 0] && GtQ[m, 0] && NeQ[m + 2*p + 2, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ
[2*m, 2*p]) &&  !(IGtQ[m, 0] && EqQ[f, 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 1294

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)*((a_) + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[e*f*(f*x)^(m - 1)*(
(a + c*x^4)^(p + 1)/(c*(m + 4*p + 3))), x] - Dist[f^2/(c*(m + 4*p + 3)), Int[(f*x)^(m - 2)*(a + c*x^4)^p*(a*e*
(m - 1) - c*d*(m + 4*p + 3)*x^2), x], x] /; FreeQ[{a, c, d, e, f, p}, x] && GtQ[m, 1] && NeQ[m + 4*p + 3, 0] &
& IntegerQ[2*p] && (IntegerQ[p] || IntegerQ[m])

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 {x^4 \left (c+d x+e x^2+f x^3\right )}{\sqrt {a+b x^4}} \, dx &=\int \left (\frac {x^4 \left (c+e x^2\right )}{\sqrt {a+b x^4}}+\frac {x^5 \left (d+f x^2\right )}{\sqrt {a+b x^4}}\right ) \, dx\\ &=\int \frac {x^4 \left (c+e x^2\right )}{\sqrt {a+b x^4}} \, dx+\int \frac {x^5 \left (d+f x^2\right )}{\sqrt {a+b x^4}} \, dx\\ &=\frac {e x^3 \sqrt {a+b x^4}}{5 b}+\frac {1}{2} \text {Subst}\left (\int \frac {x^2 (d+f x)}{\sqrt {a+b x^2}} \, dx,x,x^2\right )-\frac {\int \frac {x^2 \left (3 a e-5 b c x^2\right )}{\sqrt {a+b x^4}} \, dx}{5 b}\\ &=\frac {c x \sqrt {a+b x^4}}{3 b}+\frac {e x^3 \sqrt {a+b x^4}}{5 b}+\frac {f x^4 \sqrt {a+b x^4}}{6 b}+\frac {\int \frac {-5 a b c-9 a b e x^2}{\sqrt {a+b x^4}} \, dx}{15 b^2}+\frac {\text {Subst}\left (\int \frac {x (-2 a f+3 b d x)}{\sqrt {a+b x^2}} \, dx,x,x^2\right )}{6 b}\\ &=\frac {c x \sqrt {a+b x^4}}{3 b}+\frac {e x^3 \sqrt {a+b x^4}}{5 b}+\frac {f x^4 \sqrt {a+b x^4}}{6 b}-\frac {\left (4 a f-3 b d x^2\right ) \sqrt {a+b x^4}}{12 b^2}-\frac {(a d) \text {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,x^2\right )}{4 b}+\frac {\left (3 a^{3/2} e\right ) \int \frac {1-\frac {\sqrt {b} x^2}{\sqrt {a}}}{\sqrt {a+b x^4}} \, dx}{5 b^{3/2}}-\frac {\left (a \left (5 \sqrt {b} c+9 \sqrt {a} e\right )\right ) \int \frac {1}{\sqrt {a+b x^4}} \, dx}{15 b^{3/2}}\\ &=\frac {c x \sqrt {a+b x^4}}{3 b}+\frac {e x^3 \sqrt {a+b x^4}}{5 b}+\frac {f x^4 \sqrt {a+b x^4}}{6 b}-\frac {3 a e x \sqrt {a+b x^4}}{5 b^{3/2} \left (\sqrt {a}+\sqrt {b} x^2\right )}-\frac {\left (4 a f-3 b d x^2\right ) \sqrt {a+b x^4}}{12 b^2}+\frac {3 a^{5/4} 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 )}{5 b^{7/4} \sqrt {a+b x^4}}-\frac {a^{3/4} \left (5 \sqrt {b} c+9 \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 )}{30 b^{7/4} \sqrt {a+b x^4}}-\frac {(a d) \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x^2}{\sqrt {a+b x^4}}\right )}{4 b}\\ &=\frac {c x \sqrt {a+b x^4}}{3 b}+\frac {e x^3 \sqrt {a+b x^4}}{5 b}+\frac {f x^4 \sqrt {a+b x^4}}{6 b}-\frac {3 a e x \sqrt {a+b x^4}}{5 b^{3/2} \left (\sqrt {a}+\sqrt {b} x^2\right )}-\frac {\left (4 a f-3 b d x^2\right ) \sqrt {a+b x^4}}{12 b^2}-\frac {a d \tanh ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )}{4 b^{3/2}}+\frac {3 a^{5/4} 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 )}{5 b^{7/4} \sqrt {a+b x^4}}-\frac {a^{3/4} \left (5 \sqrt {b} c+9 \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 )}{30 b^{7/4} \sqrt {a+b x^4}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 10.12, size = 212, normalized size = 0.59 \begin {gather*} \frac {-20 a^2 f+20 a b c x+15 a b d x^2+12 a b e x^3-10 a b f x^4+20 b^2 c x^5+15 b^2 d x^6+12 b^2 e x^7+10 b^2 f x^8-15 a \sqrt {b} d \sqrt {a+b x^4} \tanh ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )-20 a b c x \sqrt {1+\frac {b x^4}{a}} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};-\frac {b x^4}{a}\right )-12 a b e x^3 \sqrt {1+\frac {b x^4}{a}} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-\frac {b x^4}{a}\right )}{60 b^2 \sqrt {a+b x^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-20*a^2*f + 20*a*b*c*x + 15*a*b*d*x^2 + 12*a*b*e*x^3 - 10*a*b*f*x^4 + 20*b^2*c*x^5 + 15*b^2*d*x^6 + 12*b^2*e*
x^7 + 10*b^2*f*x^8 - 15*a*Sqrt[b]*d*Sqrt[a + b*x^4]*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a + b*x^4]] - 20*a*b*c*x*Sqrt[1
+ (b*x^4)/a]*Hypergeometric2F1[1/4, 1/2, 5/4, -((b*x^4)/a)] - 12*a*b*e*x^3*Sqrt[1 + (b*x^4)/a]*Hypergeometric
2F1[1/2, 3/4, 7/4, -((b*x^4)/a)])/(60*b^2*Sqrt[a + b*x^4])

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Maple [C] Result contains complex when optimal does not.
time = 0.37, size = 279, normalized size = 0.77

method	result	size
default	\(-\frac {f \sqrt {b \,x^{4}+a}\, \left (-b \,x^{4}+2 a \right )}{6 b^{2}}+e \left (\frac {x^{3} \sqrt {b \,x^{4}+a}}{5 b}-\frac {3 i a^{\frac {3}{2}} \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\EllipticF \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\EllipticE \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{5 b^{\frac {3}{2}} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )+d \left (\frac {x^{2} \sqrt {b \,x^{4}+a}}{4 b}-\frac {a \ln \left (x^{2} \sqrt {b}+\sqrt {b \,x^{4}+a}\right )}{4 b^{\frac {3}{2}}}\right )+c \left (\frac {x \sqrt {b \,x^{4}+a}}{3 b}-\frac {a \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \EllipticF \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{3 b \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )\)	\(279\)
elliptic	\(\frac {f \,x^{4} \sqrt {b \,x^{4}+a}}{6 b}+\frac {e \,x^{3} \sqrt {b \,x^{4}+a}}{5 b}+\frac {d \,x^{2} \sqrt {b \,x^{4}+a}}{4 b}+\frac {c x \sqrt {b \,x^{4}+a}}{3 b}-\frac {a f \sqrt {b \,x^{4}+a}}{3 b^{2}}-\frac {a c \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \EllipticF \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{3 b \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}-\frac {a d \ln \left (2 x^{2} \sqrt {b}+2 \sqrt {b \,x^{4}+a}\right )}{4 b^{\frac {3}{2}}}-\frac {3 i a^{\frac {3}{2}} e \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\EllipticF \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\EllipticE \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{5 b^{\frac {3}{2}} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\)	\(288\)
risch	\(-\frac {\left (-10 b f \,x^{4}-12 b e \,x^{3}-15 b d \,x^{2}-20 b c x +20 a f \right ) \sqrt {b \,x^{4}+a}}{60 b^{2}}-\frac {3 i a^{\frac {3}{2}} e \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \EllipticF \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{5 b^{\frac {3}{2}} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}+\frac {3 i a^{\frac {3}{2}} e \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \EllipticE \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{5 b^{\frac {3}{2}} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}-\frac {a d \ln \left (x^{2} \sqrt {b}+\sqrt {b \,x^{4}+a}\right )}{4 b^{\frac {3}{2}}}-\frac {a c \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \EllipticF \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{3 b \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\)	\(303\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*f*(b*x^4+a)^(1/2)*(-b*x^4+2*a)/b^2+e*(1/5*x^3*(b*x^4+a)^(1/2)/b-3/5*I*a^(3/2)/b^(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)-EllipticE(x*(I/a^(1/2)*b^(1/2))^(1/2),I)))+d*(1/4*x^2*(b*x^4+a)^(1/2)/b-1/4*a/b^(3/2)*ln(x^
2*b^(1/2)+(b*x^4+a)^(1/2)))+c*(1/3*x*(b*x^4+a)^(1/2)/b-1/3*a/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(x^4*(f*x^3+e*x^2+d*x+c)/(b*x^4+a)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.13, size = 163, normalized size = 0.45 \begin {gather*} -\frac {72 \, a \sqrt {b} e x \left (-\frac {a}{b}\right )^{\frac {3}{4}} E(\arcsin \left (\frac {\left (-\frac {a}{b}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) - 15 \, a \sqrt {b} d x \log \left (-2 \, b x^{4} + 2 \, \sqrt {b x^{4} + a} \sqrt {b} x^{2} - a\right ) + 8 \, {\left (5 \, b c - 9 \, a e\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) - 2 \, {\left (10 \, b f x^{5} + 12 \, b e x^{4} + 15 \, b d x^{3} + 20 \, b c x^{2} - 20 \, a f x - 36 \, a e\right )} \sqrt {b x^{4} + a}}{120 \, b^{2} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/120*(72*a*sqrt(b)*e*x*(-a/b)^(3/4)*elliptic_e(arcsin((-a/b)^(1/4)/x), -1) - 15*a*sqrt(b)*d*x*log(-2*b*x^4 +
2*sqrt(b*x^4 + a)*sqrt(b)*x^2 - a) + 8*(5*b*c - 9*a*e)*sqrt(b)*x*(-a/b)^(3/4)*elliptic_f(arcsin((-a/b)^(1/4)/
x), -1) - 2*(10*b*f*x^5 + 12*b*e*x^4 + 15*b*d*x^3 + 20*b*c*x^2 - 20*a*f*x - 36*a*e)*sqrt(b*x^4 + a))/(b^2*x)

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Sympy [A]
time = 3.04, size = 177, normalized size = 0.49 \begin {gather*} \frac {\sqrt {a} d x^{2} \sqrt {1 + \frac {b x^{4}}{a}}}{4 b} - \frac {a d \operatorname {asinh}{\left (\frac {\sqrt {b} x^{2}}{\sqrt {a}} \right )}}{4 b^{\frac {3}{2}}} + f \left (\begin {cases} - \frac {a \sqrt {a + b x^{4}}}{3 b^{2}} + \frac {x^{4} \sqrt {a + b x^{4}}}{6 b} & \text {for}\: b \neq 0 \\\frac {x^{8}}{8 \sqrt {a}} & \text {otherwise} \end {cases}\right ) + \frac {c 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 {e 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 {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(a)*d*x**2*sqrt(1 + b*x**4/a)/(4*b) - a*d*asinh(sqrt(b)*x**2/sqrt(a))/(4*b**(3/2)) + f*Piecewise((-a*sqrt(
a + b*x**4)/(3*b**2) + x**4*sqrt(a + b*x**4)/(6*b), Ne(b, 0)), (x**8/(8*sqrt(a)), True)) + c*x**5*gamma(5/4)*h
yper((1/2, 5/4), (9/4,), b*x**4*exp_polar(I*pi)/a)/(4*sqrt(a)*gamma(9/4)) + e*x**7*gamma(7/4)*hyper((1/2, 7/4)
, (11/4,), b*x**4*exp_polar(I*pi)/a)/(4*sqrt(a)*gamma(11/4))

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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