3.5.98 \(\int x (c+d x+e x^2+f x^3) \sqrt {a+b x^4} \, dx\) [498]

Optimal. Leaf size=354 \[ \frac {2 a f x \sqrt {a+b x^4}}{21 b}+\frac {1}{4} c x^2 \sqrt {a+b x^4}+\frac {2 a d x \sqrt {a+b x^4}}{5 \sqrt {b} \left (\sqrt {a}+\sqrt {b} x^2\right )}+\frac {1}{35} x^3 \left (7 d+5 f x^2\right ) \sqrt {a+b x^4}+\frac {e \left (a+b x^4\right )^{3/2}}{6 b}+\frac {a c \tanh ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )}{4 \sqrt {b}}-\frac {2 a^{5/4} d \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{5 b^{3/4} \sqrt {a+b x^4}}+\frac {a^{5/4} \left (21 \sqrt {b} d-5 \sqrt {a} f\right ) \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{105 b^{5/4} \sqrt {a+b x^4}} \]

[Out]

1/6*e*(b*x^4+a)^(3/2)/b+1/4*a*c*arctanh(x^2*b^(1/2)/(b*x^4+a)^(1/2))/b^(1/2)+2/21*a*f*x*(b*x^4+a)^(1/2)/b+1/4*
c*x^2*(b*x^4+a)^(1/2)+1/35*x^3*(5*f*x^2+7*d)*(b*x^4+a)^(1/2)+2/5*a*d*x*(b*x^4+a)^(1/2)/b^(1/2)/(a^(1/2)+x^2*b^
(1/2))-2/5*a^(5/4)*d*(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^(
3/4)/(b*x^4+a)^(1/2)+1/105*a^(5/4)*(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*f*a^(1/2)+21*d*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^(5/4)/(b*x^4+a)^(1/2)

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Rubi [A]
time = 0.18, antiderivative size = 354, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 11, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.393, Rules used = {1847, 1262, 655, 201, 223, 212, 1288, 1294, 1212, 226, 1210} \begin {gather*} \frac {a^{5/4} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \left (21 \sqrt {b} d-5 \sqrt {a} f\right ) F\left (2 \text {ArcTan}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{105 b^{5/4} \sqrt {a+b x^4}}-\frac {2 a^{5/4} d \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^{3/4} \sqrt {a+b x^4}}+\frac {1}{4} c x^2 \sqrt {a+b x^4}+\frac {a c \tanh ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )}{4 \sqrt {b}}+\frac {1}{35} x^3 \sqrt {a+b x^4} \left (7 d+5 f x^2\right )+\frac {2 a d x \sqrt {a+b x^4}}{5 \sqrt {b} \left (\sqrt {a}+\sqrt {b} x^2\right )}+\frac {e \left (a+b x^4\right )^{3/2}}{6 b}+\frac {2 a f x \sqrt {a+b x^4}}{21 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*f*x*Sqrt[a + b*x^4])/(21*b) + (c*x^2*Sqrt[a + b*x^4])/4 + (2*a*d*x*Sqrt[a + b*x^4])/(5*Sqrt[b]*(Sqrt[a] +
Sqrt[b]*x^2)) + (x^3*(7*d + 5*f*x^2)*Sqrt[a + b*x^4])/35 + (e*(a + b*x^4)^(3/2))/(6*b) + (a*c*ArcTanh[(Sqrt[b
]*x^2)/Sqrt[a + b*x^4]])/(4*Sqrt[b]) - (2*a^(5/4)*d*(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^(3/4)*Sqrt[a + b*x^4]) + (a^(5/4)*(21*Sqrt[b]*d
- 5*Sqrt[a]*f)*(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])/(105*b^(5/4)*Sqrt[a + b*x^4])

Rule 201

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^p/(n*p + 1)), x] + Dist[a*n*(p/(n*p + 1)),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

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 1262

Int[(x_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[(d + e*x)^q
*(a + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, c, d, e, p, q}, x]

Rule 1288

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

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 x \left (c+d x+e x^2+f x^3\right ) \sqrt {a+b x^4} \, dx &=\int \left (x \left (c+e x^2\right ) \sqrt {a+b x^4}+x^2 \left (d+f x^2\right ) \sqrt {a+b x^4}\right ) \, dx\\ &=\int x \left (c+e x^2\right ) \sqrt {a+b x^4} \, dx+\int x^2 \left (d+f x^2\right ) \sqrt {a+b x^4} \, dx\\ &=\frac {1}{35} x^3 \left (7 d+5 f x^2\right ) \sqrt {a+b x^4}+\frac {1}{2} \text {Subst}\left (\int (c+e x) \sqrt {a+b x^2} \, dx,x,x^2\right )+\frac {1}{35} (2 a) \int \frac {x^2 \left (7 d+5 f x^2\right )}{\sqrt {a+b x^4}} \, dx\\ &=\frac {2 a f x \sqrt {a+b x^4}}{21 b}+\frac {1}{35} x^3 \left (7 d+5 f x^2\right ) \sqrt {a+b x^4}+\frac {e \left (a+b x^4\right )^{3/2}}{6 b}-\frac {(2 a) \int \frac {5 a f-21 b d x^2}{\sqrt {a+b x^4}} \, dx}{105 b}+\frac {1}{2} c \text {Subst}\left (\int \sqrt {a+b x^2} \, dx,x,x^2\right )\\ &=\frac {2 a f x \sqrt {a+b x^4}}{21 b}+\frac {1}{4} c x^2 \sqrt {a+b x^4}+\frac {1}{35} x^3 \left (7 d+5 f x^2\right ) \sqrt {a+b x^4}+\frac {e \left (a+b x^4\right )^{3/2}}{6 b}+\frac {1}{4} (a c) \text {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,x^2\right )-\frac {\left (2 a^{3/2} d\right ) \int \frac {1-\frac {\sqrt {b} x^2}{\sqrt {a}}}{\sqrt {a+b x^4}} \, dx}{5 \sqrt {b}}+\frac {\left (2 a^{3/2} \left (21 \sqrt {b} d-5 \sqrt {a} f\right )\right ) \int \frac {1}{\sqrt {a+b x^4}} \, dx}{105 b}\\ &=\frac {2 a f x \sqrt {a+b x^4}}{21 b}+\frac {1}{4} c x^2 \sqrt {a+b x^4}+\frac {2 a d x \sqrt {a+b x^4}}{5 \sqrt {b} \left (\sqrt {a}+\sqrt {b} x^2\right )}+\frac {1}{35} x^3 \left (7 d+5 f x^2\right ) \sqrt {a+b x^4}+\frac {e \left (a+b x^4\right )^{3/2}}{6 b}-\frac {2 a^{5/4} d \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^{3/4} \sqrt {a+b x^4}}+\frac {a^{5/4} \left (21 \sqrt {b} d-5 \sqrt {a} f\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 )}{105 b^{5/4} \sqrt {a+b x^4}}+\frac {1}{4} (a c) \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x^2}{\sqrt {a+b x^4}}\right )\\ &=\frac {2 a f x \sqrt {a+b x^4}}{21 b}+\frac {1}{4} c x^2 \sqrt {a+b x^4}+\frac {2 a d x \sqrt {a+b x^4}}{5 \sqrt {b} \left (\sqrt {a}+\sqrt {b} x^2\right )}+\frac {1}{35} x^3 \left (7 d+5 f x^2\right ) \sqrt {a+b x^4}+\frac {e \left (a+b x^4\right )^{3/2}}{6 b}+\frac {a c \tanh ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )}{4 \sqrt {b}}-\frac {2 a^{5/4} d \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^{3/4} \sqrt {a+b x^4}}+\frac {a^{5/4} \left (21 \sqrt {b} d-5 \sqrt {a} f\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 )}{105 b^{5/4} \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 = 10.13, size = 211, normalized size = 0.60 \begin {gather*} \frac {\sqrt {a+b x^4} \left (14 a e \sqrt {1+\frac {b x^4}{a}}+12 a f x \sqrt {1+\frac {b x^4}{a}}+21 b c x^2 \sqrt {1+\frac {b x^4}{a}}+14 b e x^4 \sqrt {1+\frac {b x^4}{a}}+12 b f x^5 \sqrt {1+\frac {b x^4}{a}}+21 \sqrt {a} \sqrt {b} c \sinh ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )-12 a f x \, _2F_1\left (-\frac {1}{2},\frac {1}{4};\frac {5}{4};-\frac {b x^4}{a}\right )+28 b d x^3 \, _2F_1\left (-\frac {1}{2},\frac {3}{4};\frac {7}{4};-\frac {b x^4}{a}\right )\right )}{84 b \sqrt {1+\frac {b x^4}{a}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x^4]*(14*a*e*Sqrt[1 + (b*x^4)/a] + 12*a*f*x*Sqrt[1 + (b*x^4)/a] + 21*b*c*x^2*Sqrt[1 + (b*x^4)/a] +
14*b*e*x^4*Sqrt[1 + (b*x^4)/a] + 12*b*f*x^5*Sqrt[1 + (b*x^4)/a] + 21*Sqrt[a]*Sqrt[b]*c*ArcSinh[(Sqrt[b]*x^2)/
Sqrt[a]] - 12*a*f*x*Hypergeometric2F1[-1/2, 1/4, 5/4, -((b*x^4)/a)] + 28*b*d*x^3*Hypergeometric2F1[-1/2, 3/4,
7/4, -((b*x^4)/a)]))/(84*b*Sqrt[1 + (b*x^4)/a])

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Maple [C] Result contains complex when optimal does not.
time = 0.44, size = 280, normalized size = 0.79

method	result	size
default	\(f \left (\frac {x^{5} \sqrt {b \,x^{4}+a}}{7}+\frac {2 a x \sqrt {b \,x^{4}+a}}{21 b}-\frac {2 a^{2} \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \EllipticF \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{21 b \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )+\frac {e \left (b \,x^{4}+a \right )^{\frac {3}{2}}}{6 b}+d \left (\frac {x^{3} \sqrt {b \,x^{4}+a}}{5}+\frac {2 i a^{\frac {3}{2}} \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\EllipticF \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\EllipticE \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{5 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}\, \sqrt {b}}\right )+c \left (\frac {x^{2} \sqrt {b \,x^{4}+a}}{4}+\frac {a \ln \left (x^{2} \sqrt {b}+\sqrt {b \,x^{4}+a}\right )}{4 \sqrt {b}}\right )\)	\(280\)
elliptic	\(\frac {f \,x^{5} \sqrt {b \,x^{4}+a}}{7}+\frac {e \,x^{4} \sqrt {b \,x^{4}+a}}{6}+\frac {d \,x^{3} \sqrt {b \,x^{4}+a}}{5}+\frac {c \,x^{2} \sqrt {b \,x^{4}+a}}{4}+\frac {2 a f x \sqrt {b \,x^{4}+a}}{21 b}+\frac {a e \sqrt {b \,x^{4}+a}}{6 b}-\frac {2 a^{2} f \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \EllipticF \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{21 b \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}+\frac {a c \ln \left (2 x^{2} \sqrt {b}+2 \sqrt {b \,x^{4}+a}\right )}{4 \sqrt {b}}+\frac {2 i a^{\frac {3}{2}} d \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\EllipticF \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\EllipticE \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{5 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}\, \sqrt {b}}\)	\(297\)
risch	\(\frac {\left (60 b f \,x^{5}+70 b e \,x^{4}+84 b d \,x^{3}+105 c \,x^{2} b +40 a f x +70 a e \right ) \sqrt {b \,x^{4}+a}}{420 b}+\frac {2 i a^{\frac {3}{2}} d \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \EllipticF \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{5 \sqrt {b}\, \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}-\frac {2 i a^{\frac {3}{2}} d \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \EllipticE \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{5 \sqrt {b}\, \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}+\frac {a c \ln \left (x^{2} \sqrt {b}+\sqrt {b \,x^{4}+a}\right )}{4 \sqrt {b}}-\frac {2 a^{2} f \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \EllipticF \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{21 b \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\)	\(312\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

f*(1/7*x^5*(b*x^4+a)^(1/2)+2/21*a/b*x*(b*x^4+a)^(1/2)-2/21/b*a^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))+1/6*e*(
b*x^4+a)^(3/2)/b+d*(1/5*x^3*(b*x^4+a)^(1/2)+2/5*I*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)/b^(1/2)*(EllipticF(x*(I/a^(1/2)*b^(1/2))^(1/2),I)-Ellipt
icE(x*(I/a^(1/2)*b^(1/2))^(1/2),I)))+c*(1/4*x^2*(b*x^4+a)^(1/2)+1/4*a/b^(1/2)*ln(x^2*b^(1/2)+(b*x^4+a)^(1/2)))

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

[Out]

-1/8*(a*log(-(sqrt(b) - sqrt(b*x^4 + a)/x^2)/(sqrt(b) + sqrt(b*x^4 + a)/x^2))/sqrt(b) + 2*sqrt(b*x^4 + a)*a/((
b - (b*x^4 + a)/x^4)*x^2))*c + integrate(sqrt(b*x^4 + a)*(f*x^4 + x^3*e + d*x^2), x)

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Fricas [A]
time = 0.13, size = 170, normalized size = 0.48 \begin {gather*} \frac {336 \, a \sqrt {b} d x \left (-\frac {a}{b}\right )^{\frac {3}{4}} E(\arcsin \left (\frac {\left (-\frac {a}{b}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) + 105 \, a \sqrt {b} c x \log \left (-2 \, b x^{4} - 2 \, \sqrt {b x^{4} + a} \sqrt {b} x^{2} - a\right ) - 16 \, {\left (21 \, a d + 5 \, a f\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 (60 \, b f x^{6} + 70 \, b e x^{5} + 84 \, b d x^{4} + 105 \, b c x^{3} + 40 \, a f x^{2} + 70 \, a e x + 168 \, a d\right )} \sqrt {b x^{4} + a}}{840 \, b x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/840*(336*a*sqrt(b)*d*x*(-a/b)^(3/4)*elliptic_e(arcsin((-a/b)^(1/4)/x), -1) + 105*a*sqrt(b)*c*x*log(-2*b*x^4
- 2*sqrt(b*x^4 + a)*sqrt(b)*x^2 - a) - 16*(21*a*d + 5*a*f)*sqrt(b)*x*(-a/b)^(3/4)*elliptic_f(arcsin((-a/b)^(1/
4)/x), -1) + 2*(60*b*f*x^6 + 70*b*e*x^5 + 84*b*d*x^4 + 105*b*c*x^3 + 40*a*f*x^2 + 70*a*e*x + 168*a*d)*sqrt(b*x
^4 + a))/(b*x)

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Sympy [A]
time = 2.46, size = 158, normalized size = 0.45 \begin {gather*} \frac {\sqrt {a} c x^{2} \sqrt {1 + \frac {b x^{4}}{a}}}{4} + \frac {\sqrt {a} d 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 \Gamma \left (\frac {7}{4}\right )} + \frac {\sqrt {a} f 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 \Gamma \left (\frac {9}{4}\right )} + \frac {a c \operatorname {asinh}{\left (\frac {\sqrt {b} x^{2}}{\sqrt {a}} \right )}}{4 \sqrt {b}} + e \left (\begin {cases} \frac {\sqrt {a} x^{4}}{4} & \text {for}\: b = 0 \\\frac {\left (a + b x^{4}\right )^{\frac {3}{2}}}{6 b} & \text {otherwise} \end {cases}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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