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

Optimal. Leaf size=331 \[ \frac{a^{3/4} \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} \left (3 \sqrt{a} e+5 \sqrt{b} c\right ) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac{1}{2}\right )}{15 b^{3/4} \sqrt{a+b x^4}}-\frac{2 a^{5/4} e \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{5 b^{3/4} \sqrt{a+b x^4}}+\frac{1}{15} x \sqrt{a+b x^4} \left (5 c+3 e x^2\right )+\frac{1}{4} d x^2 \sqrt{a+b x^4}+\frac{a d \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a+b x^4}}\right )}{4 \sqrt{b}}+\frac{2 a e x \sqrt{a+b x^4}}{5 \sqrt{b} \left (\sqrt{a}+\sqrt{b} x^2\right )}+\frac{f \left (a+b x^4\right )^{3/2}}{6 b} \]

[Out]

(d*x^2*Sqrt[a + b*x^4])/4 + (2*a*e*x*Sqrt[a + b*x^4])/(5*Sqrt[b]*(Sqrt[a] + Sqrt[b]*x^2)) + (x*(5*c + 3*e*x^2)
*Sqrt[a + b*x^4])/15 + (f*(a + b*x^4)^(3/2))/(6*b) + (a*d*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a + b*x^4]])/(4*Sqrt[b])
- (2*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^(3/4)*Sqrt[a + b*x^4]) + (a^(3/4)*(5*Sqrt[b]*c + 3*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])/(15*b^(3/4)*Sqrt[a
+ b*x^4])

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Rubi [A]  time = 0.194189, antiderivative size = 331, normalized size of antiderivative = 1., number of steps used = 11, 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} \[ \frac{a^{3/4} \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} \left (3 \sqrt{a} e+5 \sqrt{b} c\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{15 b^{3/4} \sqrt{a+b x^4}}-\frac{2 a^{5/4} e \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{5 b^{3/4} \sqrt{a+b x^4}}+\frac{1}{15} x \sqrt{a+b x^4} \left (5 c+3 e x^2\right )+\frac{1}{4} d x^2 \sqrt{a+b x^4}+\frac{a d \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a+b x^4}}\right )}{4 \sqrt{b}}+\frac{2 a e x \sqrt{a+b x^4}}{5 \sqrt{b} \left (\sqrt{a}+\sqrt{b} x^2\right )}+\frac{f \left (a+b x^4\right )^{3/2}}{6 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d*x^2*Sqrt[a + b*x^4])/4 + (2*a*e*x*Sqrt[a + b*x^4])/(5*Sqrt[b]*(Sqrt[a] + Sqrt[b]*x^2)) + (x*(5*c + 3*e*x^2)
*Sqrt[a + b*x^4])/15 + (f*(a + b*x^4)^(3/2))/(6*b) + (a*d*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a + b*x^4]])/(4*Sqrt[b])
- (2*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^(3/4)*Sqrt[a + b*x^4]) + (a^(3/4)*(5*Sqrt[b]*c + 3*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])/(15*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 ) \sqrt{a+b x^4} \, dx &=\int \left (\left (c+e x^2\right ) \sqrt{a+b x^4}+x \left (d+f x^2\right ) \sqrt{a+b x^4}\right ) \, dx\\ &=\int \left (c+e x^2\right ) \sqrt{a+b x^4} \, dx+\int x \left (d+f x^2\right ) \sqrt{a+b x^4} \, dx\\ &=\frac{1}{15} x \left (5 c+3 e x^2\right ) \sqrt{a+b x^4}+\frac{1}{15} \int \frac{10 a c+6 a e x^2}{\sqrt{a+b x^4}} \, dx+\frac{1}{2} \operatorname{Subst}\left (\int (d+f x) \sqrt{a+b x^2} \, dx,x,x^2\right )\\ &=\frac{1}{15} x \left (5 c+3 e x^2\right ) \sqrt{a+b x^4}+\frac{f \left (a+b x^4\right )^{3/2}}{6 b}+\frac{1}{2} d \operatorname{Subst}\left (\int \sqrt{a+b x^2} \, dx,x,x^2\right )-\frac{\left (2 a^{3/2} e\right ) \int \frac{1-\frac{\sqrt{b} x^2}{\sqrt{a}}}{\sqrt{a+b x^4}} \, dx}{5 \sqrt{b}}+\frac{1}{15} \left (2 a \left (5 c+\frac{3 \sqrt{a} e}{\sqrt{b}}\right )\right ) \int \frac{1}{\sqrt{a+b x^4}} \, dx\\ &=\frac{1}{4} d x^2 \sqrt{a+b x^4}+\frac{2 a e x \sqrt{a+b x^4}}{5 \sqrt{b} \left (\sqrt{a}+\sqrt{b} x^2\right )}+\frac{1}{15} x \left (5 c+3 e x^2\right ) \sqrt{a+b x^4}+\frac{f \left (a+b x^4\right )^{3/2}}{6 b}-\frac{2 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^{3/4} \sqrt{a+b x^4}}+\frac{a^{3/4} \left (5 \sqrt{b} c+3 \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 )}{15 b^{3/4} \sqrt{a+b x^4}}+\frac{1}{4} (a d) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^2}} \, dx,x,x^2\right )\\ &=\frac{1}{4} d x^2 \sqrt{a+b x^4}+\frac{2 a e x \sqrt{a+b x^4}}{5 \sqrt{b} \left (\sqrt{a}+\sqrt{b} x^2\right )}+\frac{1}{15} x \left (5 c+3 e x^2\right ) \sqrt{a+b x^4}+\frac{f \left (a+b x^4\right )^{3/2}}{6 b}-\frac{2 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^{3/4} \sqrt{a+b x^4}}+\frac{a^{3/4} \left (5 \sqrt{b} c+3 \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 )}{15 b^{3/4} \sqrt{a+b x^4}}+\frac{1}{4} (a d) \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{x^2}{\sqrt{a+b x^4}}\right )\\ &=\frac{1}{4} d x^2 \sqrt{a+b x^4}+\frac{2 a e x \sqrt{a+b x^4}}{5 \sqrt{b} \left (\sqrt{a}+\sqrt{b} x^2\right )}+\frac{1}{15} x \left (5 c+3 e x^2\right ) \sqrt{a+b x^4}+\frac{f \left (a+b x^4\right )^{3/2}}{6 b}+\frac{a d \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a+b x^4}}\right )}{4 \sqrt{b}}-\frac{2 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^{3/4} \sqrt{a+b x^4}}+\frac{a^{3/4} \left (5 \sqrt{b} c+3 \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 )}{15 b^{3/4} \sqrt{a+b x^4}}\\ \end{align*}

Mathematica [C]  time = 0.128262, size = 171, normalized size = 0.52 \[ \frac{\sqrt{a+b x^4} \left (12 b c x \, _2F_1\left (-\frac{1}{2},\frac{1}{4};\frac{5}{4};-\frac{b x^4}{a}\right )+3 b d x^2 \sqrt{\frac{b x^4}{a}+1}+3 \sqrt{a} \sqrt{b} d \sinh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a}}\right )+4 b e x^3 \, _2F_1\left (-\frac{1}{2},\frac{3}{4};\frac{7}{4};-\frac{b x^4}{a}\right )+2 b f x^4 \sqrt{\frac{b x^4}{a}+1}+2 a f \sqrt{\frac{b x^4}{a}+1}\right )}{12 b \sqrt{\frac{b x^4}{a}+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.007, size = 313, normalized size = 1. \begin{align*}{\frac{f}{6\,b} \left ( b{x}^{4}+a \right ) ^{{\frac{3}{2}}}}+{\frac{{x}^{3}e}{5}\sqrt{b{x}^{4}+a}}+{{\frac{2\,i}{5}}e{a}^{{\frac{3}{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{2\,i}{5}}e{a}^{{\frac{3}{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{d{x}^{2}}{4}\sqrt{b{x}^{4}+a}}+{\frac{ad}{4}\ln \left ({x}^{2}\sqrt{b}+\sqrt{b{x}^{4}+a} \right ){\frac{1}{\sqrt{b}}}}+{\frac{cx}{3}\sqrt{b{x}^{4}+a}}+{\frac{2\,ac}{3}\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)^(1/2),x)

[Out]

1/6*f*(b*x^4+a)^(3/2)/b+1/5*x^3*e*(b*x^4+a)^(1/2)+2/5*I*e*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)
-2/5*I*e*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)*EllipticE(x*(I/a^(1/2)*b^(1/2))^(1/2),I)+1/4*d*x^2*(b*x^4+a)^(1/2)+1/4*d*a/b^(1/2)*ln(x^2
*b^(1/2)+(b*x^4+a)^(1/2))+1/3*x*c*(b*x^4+a)^(1/2)+2/3*c*a/(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., size = 0, normalized size = 0. \begin{align*} \int \sqrt{b x^{4} + a}{\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)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\sqrt{b x^{4} + a}{\left (f x^{3} + e x^{2} + d x + c\right )}, 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)^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 4.14321, size = 156, normalized size = 0.47 \begin{align*} \frac{\sqrt{a} 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{\sqrt{a} d x^{2} \sqrt{1 + \frac{b x^{4}}{a}}}{4} + \frac{\sqrt{a} 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{a d \operatorname{asinh}{\left (\frac{\sqrt{b} x^{2}}{\sqrt{a}} \right )}}{4 \sqrt{b}} + 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 ) \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)**(1/2),x)

[Out]

sqrt(a)*c*x*gamma(1/4)*hyper((-1/2, 1/4), (5/4,), b*x**4*exp_polar(I*pi)/a)/(4*gamma(5/4)) + sqrt(a)*d*x**2*sq
rt(1 + b*x**4/a)/4 + sqrt(a)*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)) + a*d*asinh(sqrt(b)*x**2/sqrt(a))/(4*sqrt(b)) + f*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., size = 0, normalized size = 0. \begin{align*} \int \sqrt{b x^{4} + a}{\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)^(1/2),x, algorithm="giac")

[Out]

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