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

Optimal. Leaf size=394 \[ -\frac {a^{7/4} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \left (7 \sqrt {a} f+5 \sqrt {b} d\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{105 b^{7/4} \sqrt {a+b x^4}}+\frac {2 a^{9/4} f \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{15 b^{7/4} \sqrt {a+b x^4}}-\frac {a^2 e \tanh ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )}{16 b^{3/2}}-\frac {2 a^2 f x \sqrt {a+b x^4}}{15 b^{3/2} \left (\sqrt {a}+\sqrt {b} x^2\right )}+\frac {\left (a+b x^4\right )^{3/2} \left (4 c+3 e x^2\right )}{24 b}+\frac {1}{63} x^5 \sqrt {a+b x^4} \left (9 d+7 f x^2\right )+\frac {2 a d x \sqrt {a+b x^4}}{21 b}-\frac {a e x^2 \sqrt {a+b x^4}}{16 b}+\frac {2 a f x^3 \sqrt {a+b x^4}}{45 b} \]

[Out]

1/24*(3*e*x^2+4*c)*(b*x^4+a)^(3/2)/b-1/16*a^2*e*arctanh(x^2*b^(1/2)/(b*x^4+a)^(1/2))/b^(3/2)+2/21*a*d*x*(b*x^4
+a)^(1/2)/b-1/16*a*e*x^2*(b*x^4+a)^(1/2)/b+2/45*a*f*x^3*(b*x^4+a)^(1/2)/b+1/63*x^5*(7*f*x^2+9*d)*(b*x^4+a)^(1/
2)-2/15*a^2*f*x*(b*x^4+a)^(1/2)/b^(3/2)/(a^(1/2)+x^2*b^(1/2))+2/15*a^(9/4)*f*(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^(7/4)/(b*x^4+a)^(1/2)-1/105*a^(7/4)*(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
))*(7*f*a^(1/2)+5*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^(7/4)/(b*x^4+a)
^(1/2)

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Rubi [A]  time = 0.33, antiderivative size = 394, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 11, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.367, Rules used = {1833, 1252, 780, 195, 217, 206, 1274, 1280, 1198, 220, 1196} \[ -\frac {a^{7/4} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \left (7 \sqrt {a} f+5 \sqrt {b} d\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{105 b^{7/4} \sqrt {a+b x^4}}-\frac {a^2 e \tanh ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )}{16 b^{3/2}}-\frac {2 a^2 f x \sqrt {a+b x^4}}{15 b^{3/2} \left (\sqrt {a}+\sqrt {b} x^2\right )}+\frac {2 a^{9/4} f \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{15 b^{7/4} \sqrt {a+b x^4}}+\frac {\left (a+b x^4\right )^{3/2} \left (4 c+3 e x^2\right )}{24 b}+\frac {1}{63} x^5 \sqrt {a+b x^4} \left (9 d+7 f x^2\right )+\frac {2 a d x \sqrt {a+b x^4}}{21 b}-\frac {a e x^2 \sqrt {a+b x^4}}{16 b}+\frac {2 a f x^3 \sqrt {a+b x^4}}{45 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*d*x*Sqrt[a + b*x^4])/(21*b) - (a*e*x^2*Sqrt[a + b*x^4])/(16*b) + (2*a*f*x^3*Sqrt[a + b*x^4])/(45*b) - (2*
a^2*f*x*Sqrt[a + b*x^4])/(15*b^(3/2)*(Sqrt[a] + Sqrt[b]*x^2)) + (x^5*(9*d + 7*f*x^2)*Sqrt[a + b*x^4])/63 + ((4
*c + 3*e*x^2)*(a + b*x^4)^(3/2))/(24*b) - (a^2*e*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a + b*x^4]])/(16*b^(3/2)) + (2*a^(
9/4)*f*(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^(7/4)*Sqrt[a + b*x^4]) - (a^(7/4)*(5*Sqrt[b]*d + 7*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^(7/4)*Sqrt[a + b*x
^4])

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 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 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 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 780

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(((e*f + d*g)*(2*p
+ 3) + 2*e*g*(p + 1)*x)*(a + c*x^2)^(p + 1))/(2*c*(p + 1)*(2*p + 3)), x] - Dist[(a*e*g - c*d*f*(2*p + 3))/(c*
(2*p + 3)), Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, p}, x] &&  !LeQ[p, -1]

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 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 1252

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

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 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)]

Rubi steps

\begin {align*} \int x^3 \left (c+d x+e x^2+f x^3\right ) \sqrt {a+b x^4} \, dx &=\int \left (x^3 \left (c+e x^2\right ) \sqrt {a+b x^4}+x^4 \left (d+f x^2\right ) \sqrt {a+b x^4}\right ) \, dx\\ &=\int x^3 \left (c+e x^2\right ) \sqrt {a+b x^4} \, dx+\int x^4 \left (d+f x^2\right ) \sqrt {a+b x^4} \, dx\\ &=\frac {1}{63} x^5 \left (9 d+7 f x^2\right ) \sqrt {a+b x^4}+\frac {1}{2} \operatorname {Subst}\left (\int x (c+e x) \sqrt {a+b x^2} \, dx,x,x^2\right )+\frac {1}{63} (2 a) \int \frac {x^4 \left (9 d+7 f x^2\right )}{\sqrt {a+b x^4}} \, dx\\ &=\frac {2 a f x^3 \sqrt {a+b x^4}}{45 b}+\frac {1}{63} x^5 \left (9 d+7 f x^2\right ) \sqrt {a+b x^4}+\frac {\left (4 c+3 e x^2\right ) \left (a+b x^4\right )^{3/2}}{24 b}-\frac {(2 a) \int \frac {x^2 \left (21 a f-45 b d x^2\right )}{\sqrt {a+b x^4}} \, dx}{315 b}-\frac {(a e) \operatorname {Subst}\left (\int \sqrt {a+b x^2} \, dx,x,x^2\right )}{8 b}\\ &=\frac {2 a d x \sqrt {a+b x^4}}{21 b}-\frac {a e x^2 \sqrt {a+b x^4}}{16 b}+\frac {2 a f x^3 \sqrt {a+b x^4}}{45 b}+\frac {1}{63} x^5 \left (9 d+7 f x^2\right ) \sqrt {a+b x^4}+\frac {\left (4 c+3 e x^2\right ) \left (a+b x^4\right )^{3/2}}{24 b}+\frac {(2 a) \int \frac {-45 a b d-63 a b f x^2}{\sqrt {a+b x^4}} \, dx}{945 b^2}-\frac {\left (a^2 e\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,x^2\right )}{16 b}\\ &=\frac {2 a d x \sqrt {a+b x^4}}{21 b}-\frac {a e x^2 \sqrt {a+b x^4}}{16 b}+\frac {2 a f x^3 \sqrt {a+b x^4}}{45 b}+\frac {1}{63} x^5 \left (9 d+7 f x^2\right ) \sqrt {a+b x^4}+\frac {\left (4 c+3 e x^2\right ) \left (a+b x^4\right )^{3/2}}{24 b}-\frac {\left (a^2 e\right ) \operatorname {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x^2}{\sqrt {a+b x^4}}\right )}{16 b}+\frac {\left (2 a^{5/2} f\right ) \int \frac {1-\frac {\sqrt {b} x^2}{\sqrt {a}}}{\sqrt {a+b x^4}} \, dx}{15 b^{3/2}}-\frac {\left (2 a^2 \left (5 \sqrt {b} d+7 \sqrt {a} f\right )\right ) \int \frac {1}{\sqrt {a+b x^4}} \, dx}{105 b^{3/2}}\\ &=\frac {2 a d x \sqrt {a+b x^4}}{21 b}-\frac {a e x^2 \sqrt {a+b x^4}}{16 b}+\frac {2 a f x^3 \sqrt {a+b x^4}}{45 b}-\frac {2 a^2 f x \sqrt {a+b x^4}}{15 b^{3/2} \left (\sqrt {a}+\sqrt {b} x^2\right )}+\frac {1}{63} x^5 \left (9 d+7 f x^2\right ) \sqrt {a+b x^4}+\frac {\left (4 c+3 e x^2\right ) \left (a+b x^4\right )^{3/2}}{24 b}-\frac {a^2 e \tanh ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )}{16 b^{3/2}}+\frac {2 a^{9/4} f \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^{7/4} \sqrt {a+b x^4}}-\frac {a^{7/4} \left (5 \sqrt {b} d+7 \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^{7/4} \sqrt {a+b x^4}}\\ \end {align*}

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Mathematica [C]  time = 0.66, size = 215, normalized size = 0.55 \[ \frac {\sqrt {a+b x^4} \left (63 e \left (\sqrt {b} x^2 \left (a+2 b x^4\right )-\frac {a^{3/2} \sinh ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )}{\sqrt {\frac {b x^4}{a}+1}}\right )+168 \sqrt {b} c \left (a+b x^4\right )-\frac {144 a \sqrt {b} d x \, _2F_1\left (-\frac {1}{2},\frac {1}{4};\frac {5}{4};-\frac {b x^4}{a}\right )}{\sqrt {\frac {b x^4}{a}+1}}+144 \sqrt {b} d x \left (a+b x^4\right )-\frac {112 a \sqrt {b} f x^3 \, _2F_1\left (-\frac {1}{2},\frac {3}{4};\frac {7}{4};-\frac {b x^4}{a}\right )}{\sqrt {\frac {b x^4}{a}+1}}+112 \sqrt {b} f x^3 \left (a+b x^4\right )\right )}{1008 b^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x^4]*(168*Sqrt[b]*c*(a + b*x^4) + 144*Sqrt[b]*d*x*(a + b*x^4) + 112*Sqrt[b]*f*x^3*(a + b*x^4) + 63
*e*(Sqrt[b]*x^2*(a + 2*b*x^4) - (a^(3/2)*ArcSinh[(Sqrt[b]*x^2)/Sqrt[a]])/Sqrt[1 + (b*x^4)/a]) - (144*a*Sqrt[b]
*d*x*Hypergeometric2F1[-1/2, 1/4, 5/4, -((b*x^4)/a)])/Sqrt[1 + (b*x^4)/a] - (112*a*Sqrt[b]*f*x^3*Hypergeometri
c2F1[-1/2, 3/4, 7/4, -((b*x^4)/a)])/Sqrt[1 + (b*x^4)/a]))/(1008*b^(3/2))

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fricas [F]  time = 0.49, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (f x^{6} + e x^{5} + d x^{4} + c x^{3}\right )} \sqrt {b x^{4} + a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {b x^{4} + a} {\left (f x^{3} + e x^{2} + d x + c\right )} x^{3}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(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^3, x)

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maple [C]  time = 0.19, size = 380, normalized size = 0.96 \[ \frac {\sqrt {b \,x^{4}+a}\, f \,x^{7}}{9}+\frac {\sqrt {b \,x^{4}+a}\, d \,x^{5}}{7}+\frac {2 \sqrt {b \,x^{4}+a}\, a f \,x^{3}}{45 b}+\frac {2 i \sqrt {-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, a^{\frac {5}{2}} f \EllipticE \left (\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, x , i\right )}{15 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}\, b^{\frac {3}{2}}}-\frac {2 i \sqrt {-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, a^{\frac {5}{2}} f \EllipticF \left (\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, x , i\right )}{15 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}\, b^{\frac {3}{2}}}-\frac {2 \sqrt {-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, a^{2} d \EllipticF \left (\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, x , i\right )}{21 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}\, b}-\frac {\sqrt {b \,x^{4}+a}\, a e \,x^{2}}{16 b}-\frac {a^{2} e \ln \left (\sqrt {b}\, x^{2}+\sqrt {b \,x^{4}+a}\right )}{16 b^{\frac {3}{2}}}+\frac {2 \sqrt {b \,x^{4}+a}\, a d x}{21 b}+\frac {\left (b \,x^{4}+a \right )^{\frac {3}{2}} e \,x^{2}}{8 b}+\frac {\left (b \,x^{4}+a \right )^{\frac {3}{2}} c}{6 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/9*f*x^7*(b*x^4+a)^(1/2)+2/45*a*f*x^3*(b*x^4+a)^(1/2)/b-2/15*I*f*a^(5/2)/b^(3/2)/(I/a^(1/2)*b^(1/2))^(1/2)*(-
I/a^(1/2)*b^(1/2)*x^2+1)^(1/2)*(I/a^(1/2)*b^(1/2)*x^2+1)^(1/2)/(b*x^4+a)^(1/2)*EllipticF((I/a^(1/2)*b^(1/2))^(
1/2)*x,I)+2/15*I*f*a^(5/2)/b^(3/2)/(I/a^(1/2)*b^(1/2))^(1/2)*(-I/a^(1/2)*b^(1/2)*x^2+1)^(1/2)*(I/a^(1/2)*b^(1/
2)*x^2+1)^(1/2)/(b*x^4+a)^(1/2)*EllipticE((I/a^(1/2)*b^(1/2))^(1/2)*x,I)+1/8*e*x^2*(b*x^4+a)^(3/2)/b-1/16*a*e*
x^2*(b*x^4+a)^(1/2)/b-1/16*e*a^2/b^(3/2)*ln(b^(1/2)*x^2+(b*x^4+a)^(1/2))+1/7*d*x^5*(b*x^4+a)^(1/2)+2/21*a*d*x*
(b*x^4+a)^(1/2)/b-2/21*d*a^2/b/(I/a^(1/2)*b^(1/2))^(1/2)*(-I/a^(1/2)*b^(1/2)*x^2+1)^(1/2)*(I/a^(1/2)*b^(1/2)*x
^2+1)^(1/2)/(b*x^4+a)^(1/2)*EllipticF((I/a^(1/2)*b^(1/2))^(1/2)*x,I)+1/6*c/b*(b*x^4+a)^(3/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {{\left (b x^{4} + a\right )}^{\frac {3}{2}} c}{6 \, b} + \int {\left (f x^{6} + e x^{5} + d x^{4}\right )} \sqrt {b x^{4} + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 7.51, size = 212, normalized size = 0.54 \[ \frac {a^{\frac {3}{2}} e x^{2}}{16 b \sqrt {1 + \frac {b x^{4}}{a}}} + \frac {\sqrt {a} d x^{5} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {9}{4}\right )} + \frac {3 \sqrt {a} e x^{6}}{16 \sqrt {1 + \frac {b x^{4}}{a}}} + \frac {\sqrt {a} f x^{7} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {11}{4}\right )} - \frac {a^{2} e \operatorname {asinh}{\left (\frac {\sqrt {b} x^{2}}{\sqrt {a}} \right )}}{16 b^{\frac {3}{2}}} + c \left (\begin {cases} \frac {\sqrt {a} x^{4}}{4} & \text {for}\: b = 0 \\\frac {\left (a + b x^{4}\right )^{\frac {3}{2}}}{6 b} & \text {otherwise} \end {cases}\right ) + \frac {b e x^{10}}{8 \sqrt {a} \sqrt {1 + \frac {b x^{4}}{a}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**(3/2)*e*x**2/(16*b*sqrt(1 + b*x**4/a)) + sqrt(a)*d*x**5*gamma(5/4)*hyper((-1/2, 5/4), (9/4,), b*x**4*exp_po
lar(I*pi)/a)/(4*gamma(9/4)) + 3*sqrt(a)*e*x**6/(16*sqrt(1 + b*x**4/a)) + sqrt(a)*f*x**7*gamma(7/4)*hyper((-1/2
, 7/4), (11/4,), b*x**4*exp_polar(I*pi)/a)/(4*gamma(11/4)) - a**2*e*asinh(sqrt(b)*x**2/sqrt(a))/(16*b**(3/2))
+ c*Piecewise((sqrt(a)*x**4/4, Eq(b, 0)), ((a + b*x**4)**(3/2)/(6*b), True)) + b*e*x**10/(8*sqrt(a)*sqrt(1 + b
*x**4/a))

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