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

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

[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])

________________________________________________________________________________________

Rubi [A]  time = 0.324278, antiderivative size = 409, normalized size of antiderivative = 1., 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 = {1833, 1248, 641, 195, 217, 206, 1274, 1280, 1198, 220, 1196} \[ \frac{2 a^{9/4} \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} \left (77 \sqrt{b} d-15 \sqrt{a} f\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{1155 b^{5/4} \sqrt{a+b x^4}}-\frac{4 a^{9/4} d \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{15 b^{3/4} \sqrt{a+b x^4}}+\frac{3 a^2 c \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a+b x^4}}\right )}{16 \sqrt{b}}+\frac{4 a^2 d x \sqrt{a+b x^4}}{15 \sqrt{b} \left (\sqrt{a}+\sqrt{b} x^2\right )}+\frac{4 a^2 f x \sqrt{a+b x^4}}{77 b}+\frac{1}{8} c x^2 \left (a+b x^4\right )^{3/2}+\frac{3}{16} a c x^2 \sqrt{a+b x^4}+\frac{1}{99} x^3 \left (a+b x^4\right )^{3/2} \left (11 d+9 f x^2\right )+\frac{2 a x^3 \sqrt{a+b x^4} \left (77 d+45 f x^2\right )}{1155}+\frac{e \left (a+b x^4\right )^{5/2}}{10 b} \]

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 1833

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)*Sum[Coeff[Pq, x, j + (k*n)/2]*x^((k*n)/2), {k, 0, (2*(q - j))/n + 1}]*(a + b*x^n)^p)/c^j, {
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)]

Rule 1248

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 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 195

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 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 1274

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 1280

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 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 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} \operatorname{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 \operatorname{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) \operatorname{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 ) \operatorname{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 ) \operatorname{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*}

Mathematica [C]  time = 0.701316, size = 196, normalized size = 0.48 \[ \frac{\sqrt{a+b x^4} \left (165 c \left (\frac{3 a^{5/2} \sqrt{\frac{b x^4}{a}+1} \sinh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a}}\right )}{\sqrt{b} \left (a+b x^4\right )}+5 a x^2+2 b x^6\right )-\frac{240 a^2 f x \, _2F_1\left (-\frac{3}{2},\frac{1}{4};\frac{5}{4};-\frac{b x^4}{a}\right )}{b \sqrt{\frac{b x^4}{a}+1}}+\frac{880 a d x^3 \, _2F_1\left (-\frac{3}{2},\frac{3}{4};\frac{7}{4};-\frac{b x^4}{a}\right )}{\sqrt{\frac{b x^4}{a}+1}}+\frac{264 e \left (a+b x^4\right )^2}{b}+\frac{240 f x \left (a+b x^4\right )^2}{b}\right )}{2640} \]

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

________________________________________________________________________________________

Maple [C]  time = 0.015, size = 392, normalized size = 1. \begin{align*}{\frac{bf{x}^{9}}{11}\sqrt{b{x}^{4}+a}}+{\frac{13\,af{x}^{5}}{77}\sqrt{b{x}^{4}+a}}+{\frac{4\,{a}^{2}fx}{77\,b}\sqrt{b{x}^{4}+a}}-{\frac{4\,{a}^{3}f}{77\,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{e}{10\,b} \left ( b{x}^{4}+a \right ) ^{{\frac{5}{2}}}}+{\frac{bd{x}^{7}}{9}\sqrt{b{x}^{4}+a}}+{\frac{11\,ad{x}^{3}}{45}\sqrt{b{x}^{4}+a}}+{{\frac{4\,i}{15}}d{a}^{{\frac{5}{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 ){\frac{1}{\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{b{x}^{4}+a}}}{\frac{1}{\sqrt{b}}}}-{{\frac{4\,i}{15}}d{a}^{{\frac{5}{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 ){\frac{1}{\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{b{x}^{4}+a}}}{\frac{1}{\sqrt{b}}}}+{\frac{bc{x}^{6}}{8}\sqrt{b{x}^{4}+a}}+{\frac{5\,ac{x}^{2}}{16}\sqrt{b{x}^{4}+a}}+{\frac{3\,{a}^{2}c}{16}\ln \left ({x}^{2}\sqrt{b}+\sqrt{b{x}^{4}+a} \right ){\frac{1}{\sqrt{b}}}} \end{align*}

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)

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (b f x^{8} + b e x^{7} + b d x^{6} + b c x^{5} + a f x^{4} + a e x^{3} + a d x^{2} + a c x\right )} \sqrt{b x^{4} + a}, x\right ) \end{align*}

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]

integral((b*f*x^8 + b*e*x^7 + b*d*x^6 + b*c*x^5 + a*f*x^4 + a*e*x^3 + a*d*x^2 + a*c*x)*sqrt(b*x^4 + a), x)

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Sympy [A]  time = 10.1743, size = 396, normalized size = 0.97 \begin{align*} \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{align*}

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., size = 0, normalized size = 0. \begin{align*} \int{\left (b x^{4} + a\right )}^{\frac{3}{2}}{\left (f x^{3} + e x^{2} + d x + c\right )} x\,{d x} \end{align*}

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 + e*x^2 + d*x + c)*x, x)