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

Optimal. Leaf size=403 \[ \frac {2 a^{7/4} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \left (7 \sqrt {a} f+15 \sqrt {b} d\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{105 b^{3/4} \sqrt {a+b x^4}}-\frac {4 a^{9/4} f \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{15 b^{3/4} \sqrt {a+b x^4}}-\frac {1}{2} a^{3/2} c \tanh ^{-1}\left (\frac {\sqrt {a+b x^4}}{\sqrt {a}}\right )+\frac {3 a^2 e \tanh ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )}{16 \sqrt {b}}+\frac {4 a^2 f x \sqrt {a+b x^4}}{15 \sqrt {b} \left (\sqrt {a}+\sqrt {b} x^2\right )}+\frac {1}{16} a \sqrt {a+b x^4} \left (8 c+3 e x^2\right )+\frac {1}{24} \left (a+b x^4\right )^{3/2} \left (4 c+3 e x^2\right )+\frac {2}{105} a x \sqrt {a+b x^4} \left (15 d+7 f x^2\right )+\frac {1}{63} x \left (a+b x^4\right )^{3/2} \left (9 d+7 f x^2\right ) \]

[Out]

1/24*(3*e*x^2+4*c)*(b*x^4+a)^(3/2)+1/63*x*(7*f*x^2+9*d)*(b*x^4+a)^(3/2)-1/2*a^(3/2)*c*arctanh((b*x^4+a)^(1/2)/
a^(1/2))+3/16*a^2*e*arctanh(x^2*b^(1/2)/(b*x^4+a)^(1/2))/b^(1/2)+1/16*a*(3*e*x^2+8*c)*(b*x^4+a)^(1/2)+2/105*a*
x*(7*f*x^2+15*d)*(b*x^4+a)^(1/2)+4/15*a^2*f*x*(b*x^4+a)^(1/2)/b^(1/2)/(a^(1/2)+x^2*b^(1/2))-4/15*a^(9/4)*f*(co
s(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)+2/1
05*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)+15*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^(3/4)/(b*x^4+a)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.35, antiderivative size = 403, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 13, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.433, Rules used = {1833, 1252, 815, 844, 217, 206, 266, 63, 208, 1177, 1198, 220, 1196} \[ \frac {2 a^{7/4} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \left (7 \sqrt {a} f+15 \sqrt {b} d\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{105 b^{3/4} \sqrt {a+b x^4}}-\frac {4 a^{9/4} f \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{15 b^{3/4} \sqrt {a+b x^4}}-\frac {1}{2} a^{3/2} c \tanh ^{-1}\left (\frac {\sqrt {a+b x^4}}{\sqrt {a}}\right )+\frac {3 a^2 e \tanh ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )}{16 \sqrt {b}}+\frac {4 a^2 f x \sqrt {a+b x^4}}{15 \sqrt {b} \left (\sqrt {a}+\sqrt {b} x^2\right )}+\frac {1}{16} a \sqrt {a+b x^4} \left (8 c+3 e x^2\right )+\frac {1}{24} \left (a+b x^4\right )^{3/2} \left (4 c+3 e x^2\right )+\frac {2}{105} a x \sqrt {a+b x^4} \left (15 d+7 f x^2\right )+\frac {1}{63} x \left (a+b x^4\right )^{3/2} \left (9 d+7 f x^2\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*a^2*f*x*Sqrt[a + b*x^4])/(15*Sqrt[b]*(Sqrt[a] + Sqrt[b]*x^2)) + (a*(8*c + 3*e*x^2)*Sqrt[a + b*x^4])/16 + (2
*a*x*(15*d + 7*f*x^2)*Sqrt[a + b*x^4])/105 + ((4*c + 3*e*x^2)*(a + b*x^4)^(3/2))/24 + (x*(9*d + 7*f*x^2)*(a +
b*x^4)^(3/2))/63 + (3*a^2*e*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a + b*x^4]])/(16*Sqrt[b]) - (a^(3/2)*c*ArcTanh[Sqrt[a +
b*x^4]/Sqrt[a]])/2 - (4*a^(9/4)*f*(Sqrt[a] + Sqrt[b]*x^2)*Sqrt[(a + b*x^4)/(Sqrt[a] + Sqrt[b]*x^2)^2]*Ellipti
cE[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]*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^(3/4)*Sqrt[a + b*x^4])

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

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 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

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 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 815

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x)^(
m + 1)*(c*e*f*(m + 2*p + 2) - g*c*d*(2*p + 1) + g*c*e*(m + 2*p + 1)*x)*(a + c*x^2)^p)/(c*e^2*(m + 2*p + 1)*(m
+ 2*p + 2)), x] + Dist[(2*p)/(c*e^2*(m + 2*p + 1)*(m + 2*p + 2)), Int[(d + e*x)^m*(a + c*x^2)^(p - 1)*Simp[f*a
*c*e^2*(m + 2*p + 2) + a*c*d*e*g*m - (c^2*f*d*e*(m + 2*p + 2) - g*(c^2*d^2*(2*p + 1) + a*c*e^2*(m + 2*p + 1)))
*x, x], x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && NeQ[c*d^2 + a*e^2, 0] && GtQ[p, 0] && (IntegerQ[p] ||  !R
ationalQ[m] || (GeQ[m, -1] && LtQ[m, 0])) &&  !ILtQ[m + 2*p, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*
m, 2*p])

Rule 844

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[g/e, Int[(d
+ e*x)^(m + 1)*(a + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + c*x^2)^p, x], x] /; FreeQ[{a,
c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !IGtQ[m, 0]

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 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 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 \frac {\left (c+d x+e x^2+f x^3\right ) \left (a+b x^4\right )^{3/2}}{x} \, dx &=\int \left (\frac {\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 \frac {\left (c+e x^2\right ) \left (a+b x^4\right )^{3/2}}{x} \, dx+\int \left (d+f x^2\right ) \left (a+b x^4\right )^{3/2} \, dx\\ &=\frac {1}{63} x \left (9 d+7 f x^2\right ) \left (a+b x^4\right )^{3/2}+\frac {1}{21} \int \left (18 a d+14 a f x^2\right ) \sqrt {a+b x^4} \, dx+\frac {1}{2} \operatorname {Subst}\left (\int \frac {(c+e x) \left (a+b x^2\right )^{3/2}}{x} \, dx,x,x^2\right )\\ &=\frac {2}{105} a x \left (15 d+7 f x^2\right ) \sqrt {a+b x^4}+\frac {1}{24} \left (4 c+3 e x^2\right ) \left (a+b x^4\right )^{3/2}+\frac {1}{63} x \left (9 d+7 f x^2\right ) \left (a+b x^4\right )^{3/2}+\frac {1}{315} \int \frac {180 a^2 d+84 a^2 f x^2}{\sqrt {a+b x^4}} \, dx+\frac {\operatorname {Subst}\left (\int \frac {(4 a b c+3 a b e x) \sqrt {a+b x^2}}{x} \, dx,x,x^2\right )}{8 b}\\ &=\frac {1}{16} a \left (8 c+3 e x^2\right ) \sqrt {a+b x^4}+\frac {2}{105} a x \left (15 d+7 f x^2\right ) \sqrt {a+b x^4}+\frac {1}{24} \left (4 c+3 e x^2\right ) \left (a+b x^4\right )^{3/2}+\frac {1}{63} x \left (9 d+7 f x^2\right ) \left (a+b x^4\right )^{3/2}+\frac {\operatorname {Subst}\left (\int \frac {8 a^2 b^2 c+3 a^2 b^2 e x}{x \sqrt {a+b x^2}} \, dx,x,x^2\right )}{16 b^2}-\frac {\left (4 a^{5/2} f\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 d+\frac {7 \sqrt {a} f}{\sqrt {b}}\right )\right ) \int \frac {1}{\sqrt {a+b x^4}} \, dx\\ &=\frac {4 a^2 f x \sqrt {a+b x^4}}{15 \sqrt {b} \left (\sqrt {a}+\sqrt {b} x^2\right )}+\frac {1}{16} a \left (8 c+3 e x^2\right ) \sqrt {a+b x^4}+\frac {2}{105} a x \left (15 d+7 f x^2\right ) \sqrt {a+b x^4}+\frac {1}{24} \left (4 c+3 e x^2\right ) \left (a+b x^4\right )^{3/2}+\frac {1}{63} x \left (9 d+7 f x^2\right ) \left (a+b x^4\right )^{3/2}-\frac {4 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^{3/4} \sqrt {a+b x^4}}+\frac {2 a^{7/4} \left (15 \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^{3/4} \sqrt {a+b x^4}}+\frac {1}{2} \left (a^2 c\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+b x^2}} \, dx,x,x^2\right )+\frac {1}{16} \left (3 a^2 e\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,x^2\right )\\ &=\frac {4 a^2 f x \sqrt {a+b x^4}}{15 \sqrt {b} \left (\sqrt {a}+\sqrt {b} x^2\right )}+\frac {1}{16} a \left (8 c+3 e x^2\right ) \sqrt {a+b x^4}+\frac {2}{105} a x \left (15 d+7 f x^2\right ) \sqrt {a+b x^4}+\frac {1}{24} \left (4 c+3 e x^2\right ) \left (a+b x^4\right )^{3/2}+\frac {1}{63} x \left (9 d+7 f x^2\right ) \left (a+b x^4\right )^{3/2}-\frac {4 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^{3/4} \sqrt {a+b x^4}}+\frac {2 a^{7/4} \left (15 \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^{3/4} \sqrt {a+b x^4}}+\frac {1}{4} \left (a^2 c\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,x^4\right )+\frac {1}{16} \left (3 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 )\\ &=\frac {4 a^2 f x \sqrt {a+b x^4}}{15 \sqrt {b} \left (\sqrt {a}+\sqrt {b} x^2\right )}+\frac {1}{16} a \left (8 c+3 e x^2\right ) \sqrt {a+b x^4}+\frac {2}{105} a x \left (15 d+7 f x^2\right ) \sqrt {a+b x^4}+\frac {1}{24} \left (4 c+3 e x^2\right ) \left (a+b x^4\right )^{3/2}+\frac {1}{63} x \left (9 d+7 f x^2\right ) \left (a+b x^4\right )^{3/2}+\frac {3 a^2 e \tanh ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )}{16 \sqrt {b}}-\frac {4 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^{3/4} \sqrt {a+b x^4}}+\frac {2 a^{7/4} \left (15 \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^{3/4} \sqrt {a+b x^4}}+\frac {\left (a^2 c\right ) \operatorname {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b x^4}\right )}{2 b}\\ &=\frac {4 a^2 f x \sqrt {a+b x^4}}{15 \sqrt {b} \left (\sqrt {a}+\sqrt {b} x^2\right )}+\frac {1}{16} a \left (8 c+3 e x^2\right ) \sqrt {a+b x^4}+\frac {2}{105} a x \left (15 d+7 f x^2\right ) \sqrt {a+b x^4}+\frac {1}{24} \left (4 c+3 e x^2\right ) \left (a+b x^4\right )^{3/2}+\frac {1}{63} x \left (9 d+7 f x^2\right ) \left (a+b x^4\right )^{3/2}+\frac {3 a^2 e \tanh ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )}{16 \sqrt {b}}-\frac {1}{2} a^{3/2} c \tanh ^{-1}\left (\frac {\sqrt {a+b x^4}}{\sqrt {a}}\right )-\frac {4 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^{3/4} \sqrt {a+b x^4}}+\frac {2 a^{7/4} \left (15 \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^{3/4} \sqrt {a+b x^4}}\\ \end {align*}

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Mathematica [C]  time = 0.57, size = 224, normalized size = 0.56 \[ \frac {1}{6} c \left (\sqrt {a+b x^4} \left (4 a+b x^4\right )-3 a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+b x^4}}{\sqrt {a}}\right )\right )+\frac {1}{16} e \sqrt {a+b x^4} \left (\frac {3 a^{3/2} \sinh ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )}{\sqrt {b} \sqrt {\frac {b x^4}{a}+1}}+5 a x^2+2 b x^6\right )+\frac {a d x \sqrt {a+b x^4} \, _2F_1\left (-\frac {3}{2},\frac {1}{4};\frac {5}{4};-\frac {b x^4}{a}\right )}{\sqrt {\frac {b x^4}{a}+1}}+\frac {a f x^3 \sqrt {a+b x^4} \, _2F_1\left (-\frac {3}{2},\frac {3}{4};\frac {7}{4};-\frac {b x^4}{a}\right )}{3 \sqrt {\frac {b x^4}{a}+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(e*Sqrt[a + b*x^4]*(5*a*x^2 + 2*b*x^6 + (3*a^(3/2)*ArcSinh[(Sqrt[b]*x^2)/Sqrt[a]])/(Sqrt[b]*Sqrt[1 + (b*x^4)/a
])))/16 + (c*(Sqrt[a + b*x^4]*(4*a + b*x^4) - 3*a^(3/2)*ArcTanh[Sqrt[a + b*x^4]/Sqrt[a]]))/6 + (a*d*x*Sqrt[a +
b*x^4]*Hypergeometric2F1[-3/2, 1/4, 5/4, -((b*x^4)/a)])/Sqrt[1 + (b*x^4)/a] + (a*f*x^3*Sqrt[a + b*x^4]*Hyperg
eometric2F1[-3/2, 3/4, 7/4, -((b*x^4)/a)])/(3*Sqrt[1 + (b*x^4)/a])

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fricas [F]  time = 0.75, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b f x^{7} + b e x^{6} + b d x^{5} + b c x^{4} + a f x^{3} + a e x^{2} + a d x + a c\right )} \sqrt {b x^{4} + a}}{x}, x\right ) \]

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,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, x)

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

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,x, algorithm="giac")

[Out]

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

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maple [C]  time = 0.18, size = 411, normalized size = 1.02 \[ \frac {\sqrt {b \,x^{4}+a}\, b f \,x^{7}}{9}+\frac {\sqrt {b \,x^{4}+a}\, b e \,x^{6}}{8}+\frac {\sqrt {b \,x^{4}+a}\, b d \,x^{5}}{7}+\frac {\sqrt {b \,x^{4}+a}\, b c \,x^{4}}{6}+\frac {11 \sqrt {b \,x^{4}+a}\, a f \,x^{3}}{45}-\frac {4 i \sqrt {-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, a^{\frac {5}{2}} f \EllipticE \left (\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, x , i\right )}{15 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}\, \sqrt {b}}+\frac {4 i \sqrt {-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, a^{\frac {5}{2}} f \EllipticF \left (\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, x , i\right )}{15 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}\, \sqrt {b}}+\frac {4 \sqrt {-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, a^{2} d \EllipticF \left (\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, x , i\right )}{7 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}+\frac {5 \sqrt {b \,x^{4}+a}\, a e \,x^{2}}{16}+\frac {3 a^{2} e \ln \left (\sqrt {b}\, x^{2}+\sqrt {b \,x^{4}+a}\right )}{16 \sqrt {b}}-\frac {a^{\frac {3}{2}} c \ln \left (\frac {2 a +2 \sqrt {b \,x^{4}+a}\, \sqrt {a}}{x^{2}}\right )}{2}+\frac {3 \sqrt {b \,x^{4}+a}\, a d x}{7}+\frac {2 \sqrt {b \,x^{4}+a}\, a c}{3} \]

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,x)

[Out]

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

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

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,x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 31.32, size = 405, normalized size = 1.00 \[ - \frac {a^{\frac {3}{2}} c \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x^{2}} \right )}}{2} + \frac {a^{\frac {3}{2}} d x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {1}{4} \\ \frac {5}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {5}{4}\right )} + \frac {a^{\frac {3}{2}} e x^{2} \sqrt {1 + \frac {b x^{4}}{a}}}{4} + \frac {a^{\frac {3}{2}} e x^{2}}{16 \sqrt {1 + \frac {b x^{4}}{a}}} + \frac {a^{\frac {3}{2}} f x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {7}{4}\right )} + \frac {\sqrt {a} b d x^{5} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {9}{4}\right )} + \frac {3 \sqrt {a} b e x^{6}}{16 \sqrt {1 + \frac {b x^{4}}{a}}} + \frac {\sqrt {a} b f x^{7} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {11}{4}\right )} + \frac {a^{2} c}{2 \sqrt {b} x^{2} \sqrt {\frac {a}{b x^{4}} + 1}} + \frac {3 a^{2} e \operatorname {asinh}{\left (\frac {\sqrt {b} x^{2}}{\sqrt {a}} \right )}}{16 \sqrt {b}} + \frac {a \sqrt {b} c x^{2}}{2 \sqrt {\frac {a}{b x^{4}} + 1}} + b c \left (\begin {cases} \frac {\sqrt {a} x^{4}}{4} & \text {for}\: b = 0 \\\frac {\left (a + b x^{4}\right )^{\frac {3}{2}}}{6 b} & \text {otherwise} \end {cases}\right ) + \frac {b^{2} e x^{10}}{8 \sqrt {a} \sqrt {1 + \frac {b x^{4}}{a}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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