3.6.13 \(\int x (c+d x+e x^2+f x^3) (a+b x^4)^{3/2} \, dx\) [513]

Optimal. Leaf size=409 \[ \frac {4 a^2 f x \sqrt {a+b x^4}}{77 b}+\frac {3}{16} a c x^2 \sqrt {a+b x^4}+\frac {4 a^2 d x \sqrt {a+b x^4}}{15 \sqrt {b} \left (\sqrt {a}+\sqrt {b} x^2\right )}+\frac {2 a x^3 \left (77 d+45 f x^2\right ) \sqrt {a+b x^4}}{1155}+\frac {1}{8} c x^2 \left (a+b x^4\right )^{3/2}+\frac {1}{99} x^3 \left (11 d+9 f x^2\right ) \left (a+b x^4\right )^{3/2}+\frac {e \left (a+b x^4\right )^{5/2}}{10 b}+\frac {3 a^2 c \tanh ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )}{16 \sqrt {b}}-\frac {4 a^{9/4} d \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{15 b^{3/4} \sqrt {a+b x^4}}+\frac {2 a^{9/4} \left (77 \sqrt {b} d-15 \sqrt {a} f\right ) \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{1155 b^{5/4} \sqrt {a+b x^4}} \]

[Out]

1/8*c*x^2*(b*x^4+a)^(3/2)+1/99*x^3*(9*f*x^2+11*d)*(b*x^4+a)^(3/2)+1/10*e*(b*x^4+a)^(5/2)/b+3/16*a^2*c*arctanh(
x^2*b^(1/2)/(b*x^4+a)^(1/2))/b^(1/2)+4/77*a^2*f*x*(b*x^4+a)^(1/2)/b+3/16*a*c*x^2*(b*x^4+a)^(1/2)+2/1155*a*x^3*
(45*f*x^2+77*d)*(b*x^4+a)^(1/2)+4/15*a^2*d*x*(b*x^4+a)^(1/2)/b^(1/2)/(a^(1/2)+x^2*b^(1/2))-4/15*a^(9/4)*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)+2/11
55*a^(9/4)*(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))*(-15*f*a^(1/2)+77*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 = 0.22, antiderivative size = 409, normalized size of antiderivative = 1.00, number of steps used = 14, 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 {2 a^{9/4} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \left (77 \sqrt {b} d-15 \sqrt {a} f\right ) F\left (2 \text {ArcTan}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{1155 b^{5/4} \sqrt {a+b x^4}}-\frac {4 a^{9/4} 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 )}{15 b^{3/4} \sqrt {a+b x^4}}+\frac {3 a^2 c \tanh ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )}{16 \sqrt {b}}+\frac {4 a^2 d x \sqrt {a+b x^4}}{15 \sqrt {b} \left (\sqrt {a}+\sqrt {b} x^2\right )}+\frac {4 a^2 f x \sqrt {a+b x^4}}{77 b}+\frac {1}{8} c x^2 \left (a+b x^4\right )^{3/2}+\frac {3}{16} a c x^2 \sqrt {a+b x^4}+\frac {1}{99} x^3 \left (a+b x^4\right )^{3/2} \left (11 d+9 f x^2\right )+\frac {2 a x^3 \sqrt {a+b x^4} \left (77 d+45 f x^2\right )}{1155}+\frac {e \left (a+b x^4\right )^{5/2}}{10 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(4*a^2*f*x*Sqrt[a + b*x^4])/(77*b) + (3*a*c*x^2*Sqrt[a + b*x^4])/16 + (4*a^2*d*x*Sqrt[a + b*x^4])/(15*Sqrt[b]*
(Sqrt[a] + Sqrt[b]*x^2)) + (2*a*x^3*(77*d + 45*f*x^2)*Sqrt[a + b*x^4])/1155 + (c*x^2*(a + b*x^4)^(3/2))/8 + (x
^3*(11*d + 9*f*x^2)*(a + b*x^4)^(3/2))/99 + (e*(a + b*x^4)^(5/2))/(10*b) + (3*a^2*c*ArcTanh[(Sqrt[b]*x^2)/Sqrt
[a + b*x^4]])/(16*Sqrt[b]) - (4*a^(9/4)*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])/(15*b^(3/4)*Sqrt[a + b*x^4]) + (2*a^(9/4)*(77*Sqrt[b]*d - 15*Sq
rt[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])/(1155*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 ) \left (a+b x^4\right )^{3/2} \, dx &=\int \left (x \left (c+e x^2\right ) \left (a+b x^4\right )^{3/2}+x^2 \left (d+f x^2\right ) \left (a+b x^4\right )^{3/2}\right ) \, dx\\ &=\int x \left (c+e x^2\right ) \left (a+b x^4\right )^{3/2} \, dx+\int x^2 \left (d+f x^2\right ) \left (a+b x^4\right )^{3/2} \, dx\\ &=\frac {1}{99} x^3 \left (11 d+9 f x^2\right ) \left (a+b x^4\right )^{3/2}+\frac {1}{2} \text {Subst}\left (\int (c+e x) \left (a+b x^2\right )^{3/2} \, dx,x,x^2\right )+\frac {1}{33} (2 a) \int x^2 \left (11 d+9 f x^2\right ) \sqrt {a+b x^4} \, dx\\ &=\frac {2 a x^3 \left (77 d+45 f x^2\right ) \sqrt {a+b x^4}}{1155}+\frac {1}{99} x^3 \left (11 d+9 f x^2\right ) \left (a+b x^4\right )^{3/2}+\frac {e \left (a+b x^4\right )^{5/2}}{10 b}+\frac {\left (4 a^2\right ) \int \frac {x^2 \left (77 d+45 f x^2\right )}{\sqrt {a+b x^4}} \, dx}{1155}+\frac {1}{2} c \text {Subst}\left (\int \left (a+b x^2\right )^{3/2} \, dx,x,x^2\right )\\ &=\frac {4 a^2 f x \sqrt {a+b x^4}}{77 b}+\frac {2 a x^3 \left (77 d+45 f x^2\right ) \sqrt {a+b x^4}}{1155}+\frac {1}{8} c x^2 \left (a+b x^4\right )^{3/2}+\frac {1}{99} x^3 \left (11 d+9 f x^2\right ) \left (a+b x^4\right )^{3/2}+\frac {e \left (a+b x^4\right )^{5/2}}{10 b}-\frac {\left (4 a^2\right ) \int \frac {45 a f-231 b d x^2}{\sqrt {a+b x^4}} \, dx}{3465 b}+\frac {1}{8} (3 a c) \text {Subst}\left (\int \sqrt {a+b x^2} \, dx,x,x^2\right )\\ &=\frac {4 a^2 f x \sqrt {a+b x^4}}{77 b}+\frac {3}{16} a c x^2 \sqrt {a+b x^4}+\frac {2 a x^3 \left (77 d+45 f x^2\right ) \sqrt {a+b x^4}}{1155}+\frac {1}{8} c x^2 \left (a+b x^4\right )^{3/2}+\frac {1}{99} x^3 \left (11 d+9 f x^2\right ) \left (a+b x^4\right )^{3/2}+\frac {e \left (a+b x^4\right )^{5/2}}{10 b}+\frac {1}{16} \left (3 a^2 c\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,x^2\right )-\frac {\left (4 a^{5/2} d\right ) \int \frac {1-\frac {\sqrt {b} x^2}{\sqrt {a}}}{\sqrt {a+b x^4}} \, dx}{15 \sqrt {b}}+\frac {\left (4 a^{5/2} \left (77 \sqrt {b} d-15 \sqrt {a} f\right )\right ) \int \frac {1}{\sqrt {a+b x^4}} \, dx}{1155 b}\\ &=\frac {4 a^2 f x \sqrt {a+b x^4}}{77 b}+\frac {3}{16} a c x^2 \sqrt {a+b x^4}+\frac {4 a^2 d x \sqrt {a+b x^4}}{15 \sqrt {b} \left (\sqrt {a}+\sqrt {b} x^2\right )}+\frac {2 a x^3 \left (77 d+45 f x^2\right ) \sqrt {a+b x^4}}{1155}+\frac {1}{8} c x^2 \left (a+b x^4\right )^{3/2}+\frac {1}{99} x^3 \left (11 d+9 f x^2\right ) \left (a+b x^4\right )^{3/2}+\frac {e \left (a+b x^4\right )^{5/2}}{10 b}-\frac {4 a^{9/4} 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 )}{15 b^{3/4} \sqrt {a+b x^4}}+\frac {2 a^{9/4} \left (77 \sqrt {b} d-15 \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 )}{1155 b^{5/4} \sqrt {a+b x^4}}+\frac {1}{16} \left (3 a^2 c\right ) \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x^2}{\sqrt {a+b x^4}}\right )\\ &=\frac {4 a^2 f x \sqrt {a+b x^4}}{77 b}+\frac {3}{16} a c x^2 \sqrt {a+b x^4}+\frac {4 a^2 d x \sqrt {a+b x^4}}{15 \sqrt {b} \left (\sqrt {a}+\sqrt {b} x^2\right )}+\frac {2 a x^3 \left (77 d+45 f x^2\right ) \sqrt {a+b x^4}}{1155}+\frac {1}{8} c x^2 \left (a+b x^4\right )^{3/2}+\frac {1}{99} x^3 \left (11 d+9 f x^2\right ) \left (a+b x^4\right )^{3/2}+\frac {e \left (a+b x^4\right )^{5/2}}{10 b}+\frac {3 a^2 c \tanh ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )}{16 \sqrt {b}}-\frac {4 a^{9/4} 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 )}{15 b^{3/4} \sqrt {a+b x^4}}+\frac {2 a^{9/4} \left (77 \sqrt {b} d-15 \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 )}{1155 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 = 10.43, size = 196, normalized size = 0.48 \begin {gather*} \frac {\sqrt {a+b x^4} \left (\frac {264 e \left (a+b x^4\right )^2}{b}+\frac {240 f x \left (a+b x^4\right )^2}{b}+165 c \left (5 a x^2+2 b x^6+\frac {3 a^{5/2} \sqrt {1+\frac {b x^4}{a}} \sinh ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )}{\sqrt {b} \left (a+b x^4\right )}\right )-\frac {240 a^2 f x \, _2F_1\left (-\frac {3}{2},\frac {1}{4};\frac {5}{4};-\frac {b x^4}{a}\right )}{b \sqrt {1+\frac {b x^4}{a}}}+\frac {880 a d x^3 \, _2F_1\left (-\frac {3}{2},\frac {3}{4};\frac {7}{4};-\frac {b x^4}{a}\right )}{\sqrt {1+\frac {b x^4}{a}}}\right )}{2640} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x^4]*((264*e*(a + b*x^4)^2)/b + (240*f*x*(a + b*x^4)^2)/b + 165*c*(5*a*x^2 + 2*b*x^6 + (3*a^(5/2)*
Sqrt[1 + (b*x^4)/a]*ArcSinh[(Sqrt[b]*x^2)/Sqrt[a]])/(Sqrt[b]*(a + b*x^4))) - (240*a^2*f*x*Hypergeometric2F1[-3
/2, 1/4, 5/4, -((b*x^4)/a)])/(b*Sqrt[1 + (b*x^4)/a]) + (880*a*d*x^3*Hypergeometric2F1[-3/2, 3/4, 7/4, -((b*x^4
)/a)])/Sqrt[1 + (b*x^4)/a]))/2640

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Maple [C] Result contains complex when optimal does not.
time = 0.38, size = 332, normalized size = 0.81

method	result	size
default	\(f \left (\frac {b \,x^{9} \sqrt {b \,x^{4}+a}}{11}+\frac {13 a \,x^{5} \sqrt {b \,x^{4}+a}}{77}+\frac {4 a^{2} x \sqrt {b \,x^{4}+a}}{77 b}-\frac {4 a^{3} \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \EllipticF \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{77 b \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )+\frac {e \left (b \,x^{4}+a \right )^{\frac {5}{2}}}{10 b}+d \left (\frac {b \,x^{7} \sqrt {b \,x^{4}+a}}{9}+\frac {11 a \,x^{3} \sqrt {b \,x^{4}+a}}{45}+\frac {4 i a^{\frac {5}{2}} \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\EllipticF \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\EllipticE \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{15 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}\, \sqrt {b}}\right )+c \left (\frac {b \,x^{6} \sqrt {b \,x^{4}+a}}{8}+\frac {5 a \,x^{2} \sqrt {b \,x^{4}+a}}{16}+\frac {3 a^{2} \ln \left (x^{2} \sqrt {b}+\sqrt {b \,x^{4}+a}\right )}{16 \sqrt {b}}\right )\)	\(332\)
risch	\(\frac {\left (5040 f \,x^{9} b^{2}+5544 b^{2} e \,x^{8}+6160 b^{2} d \,x^{7}+6930 b^{2} c \,x^{6}+9360 a b f \,x^{5}+11088 a b e \,x^{4}+13552 a b d \,x^{3}+17325 a b c \,x^{2}+2880 a^{2} f x +5544 a^{2} e \right ) \sqrt {b \,x^{4}+a}}{55440 b}+\frac {4 i a^{\frac {5}{2}} d \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \EllipticF \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{15 \sqrt {b}\, \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}-\frac {4 i a^{\frac {5}{2}} d \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \EllipticE \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{15 \sqrt {b}\, \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}+\frac {3 a^{2} c \ln \left (x^{2} \sqrt {b}+\sqrt {b \,x^{4}+a}\right )}{16 \sqrt {b}}-\frac {4 a^{3} f \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \EllipticF \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{77 b \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\)	\(358\)
elliptic	\(\frac {b f \,x^{9} \sqrt {b \,x^{4}+a}}{11}+\frac {b e \,x^{8} \sqrt {b \,x^{4}+a}}{10}+\frac {b d \,x^{7} \sqrt {b \,x^{4}+a}}{9}+\frac {b c \,x^{6} \sqrt {b \,x^{4}+a}}{8}+\frac {13 a f \,x^{5} \sqrt {b \,x^{4}+a}}{77}+\frac {a e \,x^{4} \sqrt {b \,x^{4}+a}}{5}+\frac {11 a d \,x^{3} \sqrt {b \,x^{4}+a}}{45}+\frac {5 a c \,x^{2} \sqrt {b \,x^{4}+a}}{16}+\frac {4 a^{2} f x \sqrt {b \,x^{4}+a}}{77 b}+\frac {a^{2} e \sqrt {b \,x^{4}+a}}{10 b}-\frac {4 a^{3} f \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \EllipticF \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{77 b \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}+\frac {3 a^{2} c \ln \left (2 x^{2} \sqrt {b}+2 \sqrt {b \,x^{4}+a}\right )}{16 \sqrt {b}}+\frac {4 i a^{\frac {5}{2}} d \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\EllipticF \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\EllipticE \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{15 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}\, \sqrt {b}}\)	\(371\)

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)^(3/2),x,method=_RETURNVERBOSE)

[Out]

f*(1/11*b*x^9*(b*x^4+a)^(1/2)+13/77*a*x^5*(b*x^4+a)^(1/2)+4/77/b*a^2*x*(b*x^4+a)^(1/2)-4/77/b*a^3/(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))+1/10*e*(b*x^4+a)^(5/2)/b+d*(1/9*b*x^7*(b*x^4+a)^(1/2)+11/45*a*x^3*(b*x^4+a)^(1/2)+4/
15*I*a^(5/2)/(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)))+c*(1/8*b
*x^6*(b*x^4+a)^(1/2)+5/16*a*x^2*(b*x^4+a)^(1/2)+3/16*a^2*ln(x^2*b^(1/2)+(b*x^4+a)^(1/2))/b^(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)^(3/2),x, algorithm="maxima")

[Out]

-1/32*(3*a^2*log(-(sqrt(b) - sqrt(b*x^4 + a)/x^2)/(sqrt(b) + sqrt(b*x^4 + a)/x^2))/sqrt(b) + 2*(3*sqrt(b*x^4 +
a)*a^2*b/x^2 - 5*(b*x^4 + a)^(3/2)*a^2/x^6)/(b^2 - 2*(b*x^4 + a)*b/x^4 + (b*x^4 + a)^2/x^8))*c + integrate((b
*f*x^8 + b*x^7*e + b*d*x^6 + a*f*x^4 + a*x^3*e + a*d*x^2)*sqrt(b*x^4 + a), x)

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Fricas [A]
time = 0.13, size = 224, normalized size = 0.55 \begin {gather*} \frac {29568 \, a^{2} \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) + 10395 \, a^{2} \sqrt {b} c x \log \left (-2 \, b x^{4} - 2 \, \sqrt {b x^{4} + a} \sqrt {b} x^{2} - a\right ) - 384 \, {\left (77 \, a^{2} d + 15 \, a^{2} 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 (5040 \, b^{2} f x^{10} + 5544 \, b^{2} e x^{9} + 6160 \, b^{2} d x^{8} + 6930 \, b^{2} c x^{7} + 9360 \, a b f x^{6} + 11088 \, a b e x^{5} + 13552 \, a b d x^{4} + 17325 \, a b c x^{3} + 2880 \, a^{2} f x^{2} + 5544 \, a^{2} e x + 14784 \, a^{2} d\right )} \sqrt {b x^{4} + a}}{110880 \, 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)^(3/2),x, algorithm="fricas")

[Out]

1/110880*(29568*a^2*sqrt(b)*d*x*(-a/b)^(3/4)*elliptic_e(arcsin((-a/b)^(1/4)/x), -1) + 10395*a^2*sqrt(b)*c*x*lo
g(-2*b*x^4 - 2*sqrt(b*x^4 + a)*sqrt(b)*x^2 - a) - 384*(77*a^2*d + 15*a^2*f)*sqrt(b)*x*(-a/b)^(3/4)*elliptic_f(
arcsin((-a/b)^(1/4)/x), -1) + 2*(5040*b^2*f*x^10 + 5544*b^2*e*x^9 + 6160*b^2*d*x^8 + 6930*b^2*c*x^7 + 9360*a*b
*f*x^6 + 11088*a*b*e*x^5 + 13552*a*b*d*x^4 + 17325*a*b*c*x^3 + 2880*a^2*f*x^2 + 5544*a^2*e*x + 14784*a^2*d)*sq
rt(b*x^4 + a))/(b*x)

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Sympy [A]
time = 5.58, size = 396, normalized size = 0.97 \begin {gather*} \frac {a^{\frac {3}{2}} c x^{2} \sqrt {1 + \frac {b x^{4}}{a}}}{4} + \frac {a^{\frac {3}{2}} c x^{2}}{16 \sqrt {1 + \frac {b x^{4}}{a}}} + \frac {a^{\frac {3}{2}} 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 {a^{\frac {3}{2}} 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 {3 \sqrt {a} b c x^{6}}{16 \sqrt {1 + \frac {b x^{4}}{a}}} + \frac {\sqrt {a} b d 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 \Gamma \left (\frac {11}{4}\right )} + \frac {\sqrt {a} b f x^{9} \Gamma \left (\frac {9}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {9}{4} \\ \frac {13}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {13}{4}\right )} + \frac {3 a^{2} c \operatorname {asinh}{\left (\frac {\sqrt {b} x^{2}}{\sqrt {a}} \right )}}{16 \sqrt {b}} + a 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 ) + b e \left (\begin {cases} - \frac {a^{2} \sqrt {a + b x^{4}}}{15 b^{2}} + \frac {a x^{4} \sqrt {a + b x^{4}}}{30 b} + \frac {x^{8} \sqrt {a + b x^{4}}}{10} & \text {for}\: b \neq 0 \\\frac {\sqrt {a} x^{8}}{8} & \text {otherwise} \end {cases}\right ) + \frac {b^{2} c x^{10}}{8 \sqrt {a} \sqrt {1 + \frac {b x^{4}}{a}}} \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)**(3/2),x)

[Out]

a**(3/2)*c*x**2*sqrt(1 + b*x**4/a)/4 + a**(3/2)*c*x**2/(16*sqrt(1 + b*x**4/a)) + a**(3/2)*d*x**3*gamma(3/4)*hy
per((-1/2, 3/4), (7/4,), b*x**4*exp_polar(I*pi)/a)/(4*gamma(7/4)) + a**(3/2)*f*x**5*gamma(5/4)*hyper((-1/2, 5/
4), (9/4,), b*x**4*exp_polar(I*pi)/a)/(4*gamma(9/4)) + 3*sqrt(a)*b*c*x**6/(16*sqrt(1 + b*x**4/a)) + sqrt(a)*b*
d*x**7*gamma(7/4)*hyper((-1/2, 7/4), (11/4,), b*x**4*exp_polar(I*pi)/a)/(4*gamma(11/4)) + sqrt(a)*b*f*x**9*gam
ma(9/4)*hyper((-1/2, 9/4), (13/4,), b*x**4*exp_polar(I*pi)/a)/(4*gamma(13/4)) + 3*a**2*c*asinh(sqrt(b)*x**2/sq
rt(a))/(16*sqrt(b)) + a*e*Piecewise((sqrt(a)*x**4/4, Eq(b, 0)), ((a + b*x**4)**(3/2)/(6*b), True)) + b*e*Piece
wise((-a**2*sqrt(a + b*x**4)/(15*b**2) + a*x**4*sqrt(a + b*x**4)/(30*b) + x**8*sqrt(a + b*x**4)/10, Ne(b, 0)),
(sqrt(a)*x**8/8, True)) + b**2*c*x**10/(8*sqrt(a)*sqrt(1 + b*x**4/a))

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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)^(3/2),x, algorithm="giac")

[Out]

integrate((b*x^4 + a)^(3/2)*(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\,{\left (b\,x^4+a\right )}^{3/2}\,\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)^(3/2)*(c + d*x + e*x^2 + f*x^3),x)

[Out]

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

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