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

Optimal. Leaf size=382 \[ \frac{2 a^{7/4} \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} \left (7 \sqrt{a} e+15 \sqrt{b} c\right ) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac{1}{2}\right )}{105 b^{3/4} \sqrt{a+b x^4}}-\frac{4 a^{9/4} e \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{15 b^{3/4} \sqrt{a+b x^4}}+\frac{3 a^2 d \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a+b x^4}}\right )}{16 \sqrt{b}}+\frac{4 a^2 e x \sqrt{a+b x^4}}{15 \sqrt{b} \left (\sqrt{a}+\sqrt{b} x^2\right )}+\frac{1}{63} x \left (a+b x^4\right )^{3/2} \left (9 c+7 e x^2\right )+\frac{2}{105} a x \sqrt{a+b x^4} \left (15 c+7 e x^2\right )+\frac{1}{8} d x^2 \left (a+b x^4\right )^{3/2}+\frac{3}{16} a d x^2 \sqrt{a+b x^4}+\frac{f \left (a+b x^4\right )^{5/2}}{10 b} \]

[Out]

(3*a*d*x^2*Sqrt[a + b*x^4])/16 + (4*a^2*e*x*Sqrt[a + b*x^4])/(15*Sqrt[b]*(Sqrt[a] + Sqrt[b]*x^2)) + (2*a*x*(15
*c + 7*e*x^2)*Sqrt[a + b*x^4])/105 + (d*x^2*(a + b*x^4)^(3/2))/8 + (x*(9*c + 7*e*x^2)*(a + b*x^4)^(3/2))/63 +
(f*(a + b*x^4)^(5/2))/(10*b) + (3*a^2*d*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a + b*x^4]])/(16*Sqrt[b]) - (4*a^(9/4)*e*(S
qrt[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^(7/4)*(15*Sqrt[b]*c + 7*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])/(105*b^(3/4)*Sqrt[a + b*x^4])

________________________________________________________________________________________

Rubi [A]  time = 0.254981, antiderivative size = 382, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 10, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.37, Rules used = {1885, 1177, 1198, 220, 1196, 1248, 641, 195, 217, 206} \[ \frac{2 a^{7/4} \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} \left (7 \sqrt{a} e+15 \sqrt{b} c\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{105 b^{3/4} \sqrt{a+b x^4}}-\frac{4 a^{9/4} e \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{15 b^{3/4} \sqrt{a+b x^4}}+\frac{3 a^2 d \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a+b x^4}}\right )}{16 \sqrt{b}}+\frac{4 a^2 e x \sqrt{a+b x^4}}{15 \sqrt{b} \left (\sqrt{a}+\sqrt{b} x^2\right )}+\frac{1}{63} x \left (a+b x^4\right )^{3/2} \left (9 c+7 e x^2\right )+\frac{2}{105} a x \sqrt{a+b x^4} \left (15 c+7 e x^2\right )+\frac{1}{8} d x^2 \left (a+b x^4\right )^{3/2}+\frac{3}{16} a d x^2 \sqrt{a+b x^4}+\frac{f \left (a+b x^4\right )^{5/2}}{10 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*a*d*x^2*Sqrt[a + b*x^4])/16 + (4*a^2*e*x*Sqrt[a + b*x^4])/(15*Sqrt[b]*(Sqrt[a] + Sqrt[b]*x^2)) + (2*a*x*(15
*c + 7*e*x^2)*Sqrt[a + b*x^4])/105 + (d*x^2*(a + b*x^4)^(3/2))/8 + (x*(9*c + 7*e*x^2)*(a + b*x^4)^(3/2))/63 +
(f*(a + b*x^4)^(5/2))/(10*b) + (3*a^2*d*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a + b*x^4]])/(16*Sqrt[b]) - (4*a^(9/4)*e*(S
qrt[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^(7/4)*(15*Sqrt[b]*c + 7*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])/(105*b^(3/4)*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 1177

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

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]

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])

Rubi steps

\begin{align*} \int \left (c+d x+e x^2+f x^3\right ) \left (a+b x^4\right )^{3/2} \, dx &=\int \left (\left (c+e x^2\right ) \left (a+b x^4\right )^{3/2}+x \left (d+f x^2\right ) \left (a+b x^4\right )^{3/2}\right ) \, dx\\ &=\int \left (c+e x^2\right ) \left (a+b x^4\right )^{3/2} \, dx+\int x \left (d+f x^2\right ) \left (a+b x^4\right )^{3/2} \, dx\\ &=\frac{1}{63} x \left (9 c+7 e x^2\right ) \left (a+b x^4\right )^{3/2}+\frac{1}{21} \int \left (18 a c+14 a e x^2\right ) \sqrt{a+b x^4} \, dx+\frac{1}{2} \operatorname{Subst}\left (\int (d+f x) \left (a+b x^2\right )^{3/2} \, dx,x,x^2\right )\\ &=\frac{2}{105} a x \left (15 c+7 e x^2\right ) \sqrt{a+b x^4}+\frac{1}{63} x \left (9 c+7 e x^2\right ) \left (a+b x^4\right )^{3/2}+\frac{f \left (a+b x^4\right )^{5/2}}{10 b}+\frac{1}{315} \int \frac{180 a^2 c+84 a^2 e x^2}{\sqrt{a+b x^4}} \, dx+\frac{1}{2} d \operatorname{Subst}\left (\int \left (a+b x^2\right )^{3/2} \, dx,x,x^2\right )\\ &=\frac{2}{105} a x \left (15 c+7 e x^2\right ) \sqrt{a+b x^4}+\frac{1}{8} d x^2 \left (a+b x^4\right )^{3/2}+\frac{1}{63} x \left (9 c+7 e x^2\right ) \left (a+b x^4\right )^{3/2}+\frac{f \left (a+b x^4\right )^{5/2}}{10 b}+\frac{1}{8} (3 a d) \operatorname{Subst}\left (\int \sqrt{a+b x^2} \, dx,x,x^2\right )-\frac{\left (4 a^{5/2} e\right ) \int \frac{1-\frac{\sqrt{b} x^2}{\sqrt{a}}}{\sqrt{a+b x^4}} \, dx}{15 \sqrt{b}}+\frac{1}{105} \left (4 a^2 \left (15 c+\frac{7 \sqrt{a} e}{\sqrt{b}}\right )\right ) \int \frac{1}{\sqrt{a+b x^4}} \, dx\\ &=\frac{3}{16} a d x^2 \sqrt{a+b x^4}+\frac{4 a^2 e x \sqrt{a+b x^4}}{15 \sqrt{b} \left (\sqrt{a}+\sqrt{b} x^2\right )}+\frac{2}{105} a x \left (15 c+7 e x^2\right ) \sqrt{a+b x^4}+\frac{1}{8} d x^2 \left (a+b x^4\right )^{3/2}+\frac{1}{63} x \left (9 c+7 e x^2\right ) \left (a+b x^4\right )^{3/2}+\frac{f \left (a+b x^4\right )^{5/2}}{10 b}-\frac{4 a^{9/4} 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 )}{15 b^{3/4} \sqrt{a+b x^4}}+\frac{2 a^{7/4} \left (15 \sqrt{b} c+7 \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 )}{105 b^{3/4} \sqrt{a+b x^4}}+\frac{1}{16} \left (3 a^2 d\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^2}} \, dx,x,x^2\right )\\ &=\frac{3}{16} a d x^2 \sqrt{a+b x^4}+\frac{4 a^2 e x \sqrt{a+b x^4}}{15 \sqrt{b} \left (\sqrt{a}+\sqrt{b} x^2\right )}+\frac{2}{105} a x \left (15 c+7 e x^2\right ) \sqrt{a+b x^4}+\frac{1}{8} d x^2 \left (a+b x^4\right )^{3/2}+\frac{1}{63} x \left (9 c+7 e x^2\right ) \left (a+b x^4\right )^{3/2}+\frac{f \left (a+b x^4\right )^{5/2}}{10 b}-\frac{4 a^{9/4} 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 )}{15 b^{3/4} \sqrt{a+b x^4}}+\frac{2 a^{7/4} \left (15 \sqrt{b} c+7 \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 )}{105 b^{3/4} \sqrt{a+b x^4}}+\frac{1}{16} \left (3 a^2 d\right ) \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{x^2}{\sqrt{a+b x^4}}\right )\\ &=\frac{3}{16} a d x^2 \sqrt{a+b x^4}+\frac{4 a^2 e x \sqrt{a+b x^4}}{15 \sqrt{b} \left (\sqrt{a}+\sqrt{b} x^2\right )}+\frac{2}{105} a x \left (15 c+7 e x^2\right ) \sqrt{a+b x^4}+\frac{1}{8} d x^2 \left (a+b x^4\right )^{3/2}+\frac{1}{63} x \left (9 c+7 e x^2\right ) \left (a+b x^4\right )^{3/2}+\frac{f \left (a+b x^4\right )^{5/2}}{10 b}+\frac{3 a^2 d \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a+b x^4}}\right )}{16 \sqrt{b}}-\frac{4 a^{9/4} 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 )}{15 b^{3/4} \sqrt{a+b x^4}}+\frac{2 a^{7/4} \left (15 \sqrt{b} c+7 \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 )}{105 b^{3/4} \sqrt{a+b x^4}}\\ \end{align*}

Mathematica [C]  time = 0.552603, size = 175, normalized size = 0.46 \[ \frac{1}{240} \sqrt{a+b x^4} \left (15 d \left (\frac{3 a^{5/2} \sqrt{\frac{b x^4}{a}+1} \sinh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a}}\right )}{\sqrt{b} \left (a+b x^4\right )}+5 a x^2+2 b x^6\right )+\frac{240 a c x \, _2F_1\left (-\frac{3}{2},\frac{1}{4};\frac{5}{4};-\frac{b x^4}{a}\right )}{\sqrt{\frac{b x^4}{a}+1}}+\frac{80 a e x^3 \, _2F_1\left (-\frac{3}{2},\frac{3}{4};\frac{7}{4};-\frac{b x^4}{a}\right )}{\sqrt{\frac{b x^4}{a}+1}}+\frac{24 f \left (a+b x^4\right )^2}{b}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x^4]*((24*f*(a + b*x^4)^2)/b + 15*d*(5*a*x^2 + 2*b*x^6 + (3*a^(5/2)*Sqrt[1 + (b*x^4)/a]*ArcSinh[(S
qrt[b]*x^2)/Sqrt[a]])/(Sqrt[b]*(a + b*x^4))) + (240*a*c*x*Hypergeometric2F1[-3/2, 1/4, 5/4, -((b*x^4)/a)])/Sqr
t[1 + (b*x^4)/a] + (80*a*e*x^3*Hypergeometric2F1[-3/2, 3/4, 7/4, -((b*x^4)/a)])/Sqrt[1 + (b*x^4)/a]))/240

________________________________________________________________________________________

Maple [C]  time = 0.006, size = 368, normalized size = 1. \begin{align*}{\frac{f}{10\,b} \left ( b{x}^{4}+a \right ) ^{{\frac{5}{2}}}}+{\frac{be{x}^{7}}{9}\sqrt{b{x}^{4}+a}}+{\frac{11\,ae{x}^{3}}{45}\sqrt{b{x}^{4}+a}}+{{\frac{4\,i}{15}}e{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}}e{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{bd{x}^{6}}{8}\sqrt{b{x}^{4}+a}}+{\frac{5\,ad{x}^{2}}{16}\sqrt{b{x}^{4}+a}}+{\frac{3\,{a}^{2}d}{16}\ln \left ({x}^{2}\sqrt{b}+\sqrt{b{x}^{4}+a} \right ){\frac{1}{\sqrt{b}}}}+{\frac{bc{x}^{5}}{7}\sqrt{b{x}^{4}+a}}+{\frac{3\,acx}{7}\sqrt{b{x}^{4}+a}}+{\frac{4\,{a}^{2}c}{7}\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((f*x^3+e*x^2+d*x+c)*(b*x^4+a)^(3/2),x)

[Out]

1/10*f*(b*x^4+a)^(5/2)/b+1/9*e*b*x^7*(b*x^4+a)^(1/2)+11/45*e*a*x^3*(b*x^4+a)^(1/2)+4/15*I*e*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)*Ellipt
icF(x*(I/a^(1/2)*b^(1/2))^(1/2),I)-4/15*I*e*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*d*b*x^6*(
b*x^4+a)^(1/2)+5/16*a*d*x^2*(b*x^4+a)^(1/2)+3/16*d*a^2*ln(x^2*b^(1/2)+(b*x^4+a)^(1/2))/b^(1/2)+1/7*c*b*x^5*(b*
x^4+a)^(1/2)+3/7*c*a*x*(b*x^4+a)^(1/2)+4/7*c*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)

________________________________________________________________________________________

Maxima [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 )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [A]  time = 9.64242, size = 394, normalized size = 1.03 \begin{align*} \frac{a^{\frac{3}{2}} c x \Gamma \left (\frac{1}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{1}{4} \\ \frac{5}{4} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac{5}{4}\right )} + \frac{a^{\frac{3}{2}} d x^{2} \sqrt{1 + \frac{b x^{4}}{a}}}{4} + \frac{a^{\frac{3}{2}} d x^{2}}{16 \sqrt{1 + \frac{b x^{4}}{a}}} + \frac{a^{\frac{3}{2}} 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 \Gamma \left (\frac{7}{4}\right )} + \frac{\sqrt{a} b 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 \Gamma \left (\frac{9}{4}\right )} + \frac{3 \sqrt{a} b d x^{6}}{16 \sqrt{1 + \frac{b x^{4}}{a}}} + \frac{\sqrt{a} b 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 \Gamma \left (\frac{11}{4}\right )} + \frac{3 a^{2} d \operatorname{asinh}{\left (\frac{\sqrt{b} x^{2}}{\sqrt{a}} \right )}}{16 \sqrt{b}} + a f \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 f \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} d 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((f*x**3+e*x**2+d*x+c)*(b*x**4+a)**(3/2),x)

[Out]

a**(3/2)*c*x*gamma(1/4)*hyper((-1/2, 1/4), (5/4,), b*x**4*exp_polar(I*pi)/a)/(4*gamma(5/4)) + a**(3/2)*d*x**2*
sqrt(1 + b*x**4/a)/4 + a**(3/2)*d*x**2/(16*sqrt(1 + b*x**4/a)) + a**(3/2)*e*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)*b*c*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*d*x**6/(16*sqrt(1 + b*x**4/a)) + sqrt(a)*b*e*x**7*gamma(7/
4)*hyper((-1/2, 7/4), (11/4,), b*x**4*exp_polar(I*pi)/a)/(4*gamma(11/4)) + 3*a**2*d*asinh(sqrt(b)*x**2/sqrt(a)
)/(16*sqrt(b)) + a*f*Piecewise((sqrt(a)*x**4/4, Eq(b, 0)), ((a + b*x**4)**(3/2)/(6*b), True)) + b*f*Piecewise(
(-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)), (sqr
t(a)*x**8/8, True)) + b**2*d*x**10/(8*sqrt(a)*sqrt(1 + b*x**4/a))

________________________________________________________________________________________

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 )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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)