3.526 \(\int \frac {(c+d x+e x^2+f x^3) (a+b x^4)^{3/2}}{x^{12}} \, dx\)

Optimal. Leaf size=424 \[ -\frac {2 b^{9/4} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \left (15 \sqrt {b} c-77 \sqrt {a} e\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{1155 a^{5/4} \sqrt {a+b x^4}}-\frac {4 b^{9/4} e \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{15 a^{3/4} \sqrt {a+b x^4}}+\frac {4 b^{5/2} e x \sqrt {a+b x^4}}{15 a \left (\sqrt {a}+\sqrt {b} x^2\right )}-\frac {4 b^2 c \sqrt {a+b x^4}}{77 a x^3}-\frac {b^2 d \sqrt {a+b x^4}}{10 a x^2}-\frac {4 b^2 e \sqrt {a+b x^4}}{15 a x}-\frac {3 b^2 f \tanh ^{-1}\left (\frac {\sqrt {a+b x^4}}{\sqrt {a}}\right )}{16 \sqrt {a}}-\frac {b \sqrt {a+b x^4} \left (\frac {1440 c}{x^7}+\frac {1848 d}{x^6}+\frac {2464 e}{x^5}+\frac {3465 f}{x^4}\right )}{18480}-\frac {\left (a+b x^4\right )^{3/2} \left (\frac {360 c}{x^{11}}+\frac {396 d}{x^{10}}+\frac {440 e}{x^9}+\frac {495 f}{x^8}\right )}{3960} \]

[Out]

-1/3960*(360*c/x^11+396*d/x^10+440*e/x^9+495*f/x^8)*(b*x^4+a)^(3/2)-3/16*b^2*f*arctanh((b*x^4+a)^(1/2)/a^(1/2)
)/a^(1/2)-1/18480*b*(1440*c/x^7+1848*d/x^6+2464*e/x^5+3465*f/x^4)*(b*x^4+a)^(1/2)-4/77*b^2*c*(b*x^4+a)^(1/2)/a
/x^3-1/10*b^2*d*(b*x^4+a)^(1/2)/a/x^2-4/15*b^2*e*(b*x^4+a)^(1/2)/a/x+4/15*b^(5/2)*e*x*(b*x^4+a)^(1/2)/a/(a^(1/
2)+x^2*b^(1/2))-4/15*b^(9/4)*e*(cos(2*arctan(b^(1/4)*x/a^(1/4)))^2)^(1/2)/cos(2*arctan(b^(1/4)*x/a^(1/4)))*Ell
ipticE(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)/a^(3/4)/(b*x^4+a)^(1/2)-2/1155*b^(9/4)*(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))*(-77*e*a^(1/2)+15*c*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)/a^(5/4)/(b*x^4+a)^(1/2)

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Rubi [A]  time = 0.46, antiderivative size = 424, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 12, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {14, 1825, 1833, 1282, 1198, 220, 1196, 1252, 807, 266, 63, 208} \[ -\frac {2 b^{9/4} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \left (15 \sqrt {b} c-77 \sqrt {a} e\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{1155 a^{5/4} \sqrt {a+b x^4}}-\frac {4 b^{9/4} e \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{15 a^{3/4} \sqrt {a+b x^4}}-\frac {4 b^2 c \sqrt {a+b x^4}}{77 a x^3}-\frac {b^2 d \sqrt {a+b x^4}}{10 a x^2}+\frac {4 b^{5/2} e x \sqrt {a+b x^4}}{15 a \left (\sqrt {a}+\sqrt {b} x^2\right )}-\frac {4 b^2 e \sqrt {a+b x^4}}{15 a x}-\frac {3 b^2 f \tanh ^{-1}\left (\frac {\sqrt {a+b x^4}}{\sqrt {a}}\right )}{16 \sqrt {a}}-\frac {b \sqrt {a+b x^4} \left (\frac {1440 c}{x^7}+\frac {1848 d}{x^6}+\frac {2464 e}{x^5}+\frac {3465 f}{x^4}\right )}{18480}-\frac {\left (a+b x^4\right )^{3/2} \left (\frac {360 c}{x^{11}}+\frac {396 d}{x^{10}}+\frac {440 e}{x^9}+\frac {495 f}{x^8}\right )}{3960} \]

Antiderivative was successfully verified.

[In]

Int[((c + d*x + e*x^2 + f*x^3)*(a + b*x^4)^(3/2))/x^12,x]

[Out]

-(b*((1440*c)/x^7 + (1848*d)/x^6 + (2464*e)/x^5 + (3465*f)/x^4)*Sqrt[a + b*x^4])/18480 - (4*b^2*c*Sqrt[a + b*x
^4])/(77*a*x^3) - (b^2*d*Sqrt[a + b*x^4])/(10*a*x^2) - (4*b^2*e*Sqrt[a + b*x^4])/(15*a*x) + (4*b^(5/2)*e*x*Sqr
t[a + b*x^4])/(15*a*(Sqrt[a] + Sqrt[b]*x^2)) - (((360*c)/x^11 + (396*d)/x^10 + (440*e)/x^9 + (495*f)/x^8)*(a +
b*x^4)^(3/2))/3960 - (3*b^2*f*ArcTanh[Sqrt[a + b*x^4]/Sqrt[a]])/(16*Sqrt[a]) - (4*b^(9/4)*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])/(15*a^(3/4)*
Sqrt[a + b*x^4]) - (2*b^(9/4)*(15*Sqrt[b]*c - 77*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])/(1155*a^(5/4)*Sqrt[a + b*x^4])

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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 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 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 807

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Simp[((e*f - d*g
)*(d + e*x)^(m + 1)*(a + c*x^2)^(p + 1))/(2*(p + 1)*(c*d^2 + a*e^2)), x] + Dist[(c*d*f + a*e*g)/(c*d^2 + a*e^2
), Int[(d + e*x)^(m + 1)*(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]
&& EqQ[Simplify[m + 2*p + 3], 0]

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 1282

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)*((a_) + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[(d*(f*x)^(m + 1)*(a
+ c*x^4)^(p + 1))/(a*f*(m + 1)), x] + Dist[1/(a*f^2*(m + 1)), Int[(f*x)^(m + 2)*(a + c*x^4)^p*(a*e*(m + 1) -
c*d*(m + 4*p + 5)*x^2), x], x] /; FreeQ[{a, c, d, e, f, p}, x] && LtQ[m, -1] && IntegerQ[2*p] && (IntegerQ[p]
|| IntegerQ[m])

Rule 1825

Int[(Pq_)*(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Module[{u = IntHide[x^m*Pq, x]}, Simp[u*(a +
b*x^n)^p, x] - Dist[b*n*p, Int[x^(m + n)*(a + b*x^n)^(p - 1)*ExpandToSum[u/x^(m + 1), x], x], x]] /; FreeQ[{a
, b}, x] && PolyQ[Pq, x] && IGtQ[n, 0] && GtQ[p, 0] && LtQ[m + Expon[Pq, x] + 1, 0]

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^{12}} \, dx &=-\frac {\left (\frac {360 c}{x^{11}}+\frac {396 d}{x^{10}}+\frac {440 e}{x^9}+\frac {495 f}{x^8}\right ) \left (a+b x^4\right )^{3/2}}{3960}-(6 b) \int \frac {\left (-\frac {c}{11}-\frac {d x}{10}-\frac {e x^2}{9}-\frac {f x^3}{8}\right ) \sqrt {a+b x^4}}{x^8} \, dx\\ &=-\frac {b \left (\frac {1440 c}{x^7}+\frac {1848 d}{x^6}+\frac {2464 e}{x^5}+\frac {3465 f}{x^4}\right ) \sqrt {a+b x^4}}{18480}-\frac {\left (\frac {360 c}{x^{11}}+\frac {396 d}{x^{10}}+\frac {440 e}{x^9}+\frac {495 f}{x^8}\right ) \left (a+b x^4\right )^{3/2}}{3960}+\left (12 b^2\right ) \int \frac {\frac {c}{77}+\frac {d x}{60}+\frac {e x^2}{45}+\frac {f x^3}{32}}{x^4 \sqrt {a+b x^4}} \, dx\\ &=-\frac {b \left (\frac {1440 c}{x^7}+\frac {1848 d}{x^6}+\frac {2464 e}{x^5}+\frac {3465 f}{x^4}\right ) \sqrt {a+b x^4}}{18480}-\frac {\left (\frac {360 c}{x^{11}}+\frac {396 d}{x^{10}}+\frac {440 e}{x^9}+\frac {495 f}{x^8}\right ) \left (a+b x^4\right )^{3/2}}{3960}+\left (12 b^2\right ) \int \left (\frac {\frac {c}{77}+\frac {e x^2}{45}}{x^4 \sqrt {a+b x^4}}+\frac {\frac {d}{60}+\frac {f x^2}{32}}{x^3 \sqrt {a+b x^4}}\right ) \, dx\\ &=-\frac {b \left (\frac {1440 c}{x^7}+\frac {1848 d}{x^6}+\frac {2464 e}{x^5}+\frac {3465 f}{x^4}\right ) \sqrt {a+b x^4}}{18480}-\frac {\left (\frac {360 c}{x^{11}}+\frac {396 d}{x^{10}}+\frac {440 e}{x^9}+\frac {495 f}{x^8}\right ) \left (a+b x^4\right )^{3/2}}{3960}+\left (12 b^2\right ) \int \frac {\frac {c}{77}+\frac {e x^2}{45}}{x^4 \sqrt {a+b x^4}} \, dx+\left (12 b^2\right ) \int \frac {\frac {d}{60}+\frac {f x^2}{32}}{x^3 \sqrt {a+b x^4}} \, dx\\ &=-\frac {b \left (\frac {1440 c}{x^7}+\frac {1848 d}{x^6}+\frac {2464 e}{x^5}+\frac {3465 f}{x^4}\right ) \sqrt {a+b x^4}}{18480}-\frac {4 b^2 c \sqrt {a+b x^4}}{77 a x^3}-\frac {\left (\frac {360 c}{x^{11}}+\frac {396 d}{x^{10}}+\frac {440 e}{x^9}+\frac {495 f}{x^8}\right ) \left (a+b x^4\right )^{3/2}}{3960}+\left (6 b^2\right ) \operatorname {Subst}\left (\int \frac {\frac {d}{60}+\frac {f x}{32}}{x^2 \sqrt {a+b x^2}} \, dx,x,x^2\right )-\frac {\left (4 b^2\right ) \int \frac {-\frac {a e}{15}+\frac {1}{77} b c x^2}{x^2 \sqrt {a+b x^4}} \, dx}{a}\\ &=-\frac {b \left (\frac {1440 c}{x^7}+\frac {1848 d}{x^6}+\frac {2464 e}{x^5}+\frac {3465 f}{x^4}\right ) \sqrt {a+b x^4}}{18480}-\frac {4 b^2 c \sqrt {a+b x^4}}{77 a x^3}-\frac {b^2 d \sqrt {a+b x^4}}{10 a x^2}-\frac {4 b^2 e \sqrt {a+b x^4}}{15 a x}-\frac {\left (\frac {360 c}{x^{11}}+\frac {396 d}{x^{10}}+\frac {440 e}{x^9}+\frac {495 f}{x^8}\right ) \left (a+b x^4\right )^{3/2}}{3960}+\frac {\left (4 b^2\right ) \int \frac {-\frac {1}{77} a b c+\frac {1}{15} a b e x^2}{\sqrt {a+b x^4}} \, dx}{a^2}+\frac {1}{16} \left (3 b^2 f\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+b x^2}} \, dx,x,x^2\right )\\ &=-\frac {b \left (\frac {1440 c}{x^7}+\frac {1848 d}{x^6}+\frac {2464 e}{x^5}+\frac {3465 f}{x^4}\right ) \sqrt {a+b x^4}}{18480}-\frac {4 b^2 c \sqrt {a+b x^4}}{77 a x^3}-\frac {b^2 d \sqrt {a+b x^4}}{10 a x^2}-\frac {4 b^2 e \sqrt {a+b x^4}}{15 a x}-\frac {\left (\frac {360 c}{x^{11}}+\frac {396 d}{x^{10}}+\frac {440 e}{x^9}+\frac {495 f}{x^8}\right ) \left (a+b x^4\right )^{3/2}}{3960}-\frac {\left (4 b^{5/2} e\right ) \int \frac {1-\frac {\sqrt {b} x^2}{\sqrt {a}}}{\sqrt {a+b x^4}} \, dx}{15 \sqrt {a}}-\frac {\left (4 b^{5/2} \left (15 \sqrt {b} c-77 \sqrt {a} e\right )\right ) \int \frac {1}{\sqrt {a+b x^4}} \, dx}{1155 a}+\frac {1}{32} \left (3 b^2 f\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,x^4\right )\\ &=-\frac {b \left (\frac {1440 c}{x^7}+\frac {1848 d}{x^6}+\frac {2464 e}{x^5}+\frac {3465 f}{x^4}\right ) \sqrt {a+b x^4}}{18480}-\frac {4 b^2 c \sqrt {a+b x^4}}{77 a x^3}-\frac {b^2 d \sqrt {a+b x^4}}{10 a x^2}-\frac {4 b^2 e \sqrt {a+b x^4}}{15 a x}+\frac {4 b^{5/2} e x \sqrt {a+b x^4}}{15 a \left (\sqrt {a}+\sqrt {b} x^2\right )}-\frac {\left (\frac {360 c}{x^{11}}+\frac {396 d}{x^{10}}+\frac {440 e}{x^9}+\frac {495 f}{x^8}\right ) \left (a+b x^4\right )^{3/2}}{3960}-\frac {4 b^{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 a^{3/4} \sqrt {a+b x^4}}-\frac {2 b^{9/4} \left (15 \sqrt {b} c-77 \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 )}{1155 a^{5/4} \sqrt {a+b x^4}}+\frac {1}{16} (3 b f) \operatorname {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b x^4}\right )\\ &=-\frac {b \left (\frac {1440 c}{x^7}+\frac {1848 d}{x^6}+\frac {2464 e}{x^5}+\frac {3465 f}{x^4}\right ) \sqrt {a+b x^4}}{18480}-\frac {4 b^2 c \sqrt {a+b x^4}}{77 a x^3}-\frac {b^2 d \sqrt {a+b x^4}}{10 a x^2}-\frac {4 b^2 e \sqrt {a+b x^4}}{15 a x}+\frac {4 b^{5/2} e x \sqrt {a+b x^4}}{15 a \left (\sqrt {a}+\sqrt {b} x^2\right )}-\frac {\left (\frac {360 c}{x^{11}}+\frac {396 d}{x^{10}}+\frac {440 e}{x^9}+\frac {495 f}{x^8}\right ) \left (a+b x^4\right )^{3/2}}{3960}-\frac {3 b^2 f \tanh ^{-1}\left (\frac {\sqrt {a+b x^4}}{\sqrt {a}}\right )}{16 \sqrt {a}}-\frac {4 b^{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 a^{3/4} \sqrt {a+b x^4}}-\frac {2 b^{9/4} \left (15 \sqrt {b} c-77 \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 )}{1155 a^{5/4} \sqrt {a+b x^4}}\\ \end {align*}

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Mathematica [C]  time = 0.42, size = 172, normalized size = 0.41 \[ -\frac {\sqrt {a+b x^4} \left (11 x \left (9 \sqrt {\frac {b x^4}{a}+1} \left (2 a^2 \left (4 d+5 f x^2\right )+a b x^4 \left (16 d+25 f x^2\right )+8 b^2 d x^8\right )+80 a^2 e x \, _2F_1\left (-\frac {9}{4},-\frac {3}{2};-\frac {5}{4};-\frac {b x^4}{a}\right )+135 b^2 f x^{10} \tanh ^{-1}\left (\sqrt {\frac {b x^4}{a}+1}\right )\right )+720 a^2 c \, _2F_1\left (-\frac {11}{4},-\frac {3}{2};-\frac {7}{4};-\frac {b x^4}{a}\right )\right )}{7920 a x^{11} \sqrt {\frac {b x^4}{a}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[((c + d*x + e*x^2 + f*x^3)*(a + b*x^4)^(3/2))/x^12,x]

[Out]

-1/7920*(Sqrt[a + b*x^4]*(720*a^2*c*Hypergeometric2F1[-11/4, -3/2, -7/4, -((b*x^4)/a)] + 11*x*(9*Sqrt[1 + (b*x
^4)/a]*(8*b^2*d*x^8 + 2*a^2*(4*d + 5*f*x^2) + a*b*x^4*(16*d + 25*f*x^2)) + 135*b^2*f*x^10*ArcTanh[Sqrt[1 + (b*
x^4)/a]] + 80*a^2*e*x*Hypergeometric2F1[-9/4, -3/2, -5/4, -((b*x^4)/a)])))/(a*x^11*Sqrt[1 + (b*x^4)/a])

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fricas [F]  time = 0.48, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b f x^{7} + b e x^{6} + b d x^{5} + b c x^{4} + a f x^{3} + a e x^{2} + a d x + a c\right )} \sqrt {b x^{4} + a}}{x^{12}}, x\right ) \]

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^12,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^12, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x^{4} + a\right )}^{\frac {3}{2}} {\left (f x^{3} + e x^{2} + d x + c\right )}}{x^{12}}\,{d x} \]

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^12,x, algorithm="giac")

[Out]

integrate((b*x^4 + a)^(3/2)*(f*x^3 + e*x^2 + d*x + c)/x^12, x)

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maple [C]  time = 0.22, size = 441, normalized size = 1.04 \[ -\frac {4 i \sqrt {-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, b^{\frac {5}{2}} e \EllipticE \left (\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, x , i\right )}{15 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}\, \sqrt {a}}+\frac {4 i \sqrt {-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, b^{\frac {5}{2}} e \EllipticF \left (\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, x , i\right )}{15 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}\, \sqrt {a}}-\frac {4 \sqrt {-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, b^{3} c \EllipticF \left (\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, x , i\right )}{77 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}\, a}-\frac {3 b^{2} f \ln \left (\frac {2 a +2 \sqrt {b \,x^{4}+a}\, \sqrt {a}}{x^{2}}\right )}{16 \sqrt {a}}-\frac {4 \sqrt {b \,x^{4}+a}\, b^{2} e}{15 a x}-\frac {4 \sqrt {b \,x^{4}+a}\, b^{2} c}{77 a \,x^{3}}-\frac {5 \sqrt {b \,x^{4}+a}\, b f}{16 x^{4}}-\frac {11 \sqrt {b \,x^{4}+a}\, b e}{45 x^{5}}-\frac {13 \sqrt {b \,x^{4}+a}\, b c}{77 x^{7}}-\frac {\sqrt {b \,x^{4}+a}\, a f}{8 x^{8}}-\frac {\sqrt {b \,x^{4}+a}\, a e}{9 x^{9}}-\frac {\sqrt {b \,x^{4}+a}\, a c}{11 x^{11}}-\frac {\sqrt {b \,x^{4}+a}\, \left (b^{2} x^{8}+2 a b \,x^{4}+a^{2}\right ) d}{10 a \,x^{10}} \]

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^12,x)

[Out]

-1/8*f*a/x^8*(b*x^4+a)^(1/2)-5/16*f*b/x^4*(b*x^4+a)^(1/2)-3/16*f*b^2/a^(1/2)*ln((2*a+2*(b*x^4+a)^(1/2)*a^(1/2)
)/x^2)-1/9*e*a*(b*x^4+a)^(1/2)/x^9-11/45*e*b*(b*x^4+a)^(1/2)/x^5-4/15*b^2*e*(b*x^4+a)^(1/2)/a/x+4/15*I*e/a^(1/
2)*b^(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)*EllipticF((I/a^(1/2)*b^(1/2))^(1/2)*x,I)-4/15*I*e/a^(1/2)*b^(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)*EllipticE((I/a^(1/2)*b^(1/2))^(1/2)*x,I
)-1/11*c*a*(b*x^4+a)^(1/2)/x^11-13/77*c*b*(b*x^4+a)^(1/2)/x^7-4/77*b^2*c*(b*x^4+a)^(1/2)/a/x^3-4/77*c/a*b^3/(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)*Ellip
ticF((I/a^(1/2)*b^(1/2))^(1/2)*x,I)-1/10*d*(b*x^4+a)^(1/2)/x^10/a*(b^2*x^8+2*a*b*x^4+a^2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x^{4} + a\right )}^{\frac {3}{2}} {\left (f x^{3} + e x^{2} + d x + c\right )}}{x^{12}}\,{d x} \]

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^12,x, algorithm="maxima")

[Out]

integrate((b*x^4 + a)^(3/2)*(f*x^3 + e*x^2 + d*x + c)/x^12, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (b\,x^4+a\right )}^{3/2}\,\left (f\,x^3+e\,x^2+d\,x+c\right )}{x^{12}} \,d x \]

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^12,x)

[Out]

int(((a + b*x^4)^(3/2)*(c + d*x + e*x^2 + f*x^3))/x^12, x)

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sympy [C]  time = 22.52, size = 401, normalized size = 0.95 \[ \frac {a^{\frac {3}{2}} c \Gamma \left (- \frac {11}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {11}{4}, - \frac {1}{2} \\ - \frac {7}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 x^{11} \Gamma \left (- \frac {7}{4}\right )} + \frac {a^{\frac {3}{2}} e \Gamma \left (- \frac {9}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {9}{4}, - \frac {1}{2} \\ - \frac {5}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 x^{9} \Gamma \left (- \frac {5}{4}\right )} + \frac {\sqrt {a} b c \Gamma \left (- \frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {7}{4}, - \frac {1}{2} \\ - \frac {3}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 x^{7} \Gamma \left (- \frac {3}{4}\right )} + \frac {\sqrt {a} b e \Gamma \left (- \frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{4}, - \frac {1}{2} \\ - \frac {1}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 x^{5} \Gamma \left (- \frac {1}{4}\right )} - \frac {a^{2} f}{8 \sqrt {b} x^{10} \sqrt {\frac {a}{b x^{4}} + 1}} - \frac {a \sqrt {b} d \sqrt {\frac {a}{b x^{4}} + 1}}{10 x^{8}} - \frac {3 a \sqrt {b} f}{16 x^{6} \sqrt {\frac {a}{b x^{4}} + 1}} - \frac {b^{\frac {3}{2}} d \sqrt {\frac {a}{b x^{4}} + 1}}{5 x^{4}} - \frac {b^{\frac {3}{2}} f \sqrt {\frac {a}{b x^{4}} + 1}}{4 x^{2}} - \frac {b^{\frac {3}{2}} f}{16 x^{2} \sqrt {\frac {a}{b x^{4}} + 1}} - \frac {b^{\frac {5}{2}} d \sqrt {\frac {a}{b x^{4}} + 1}}{10 a} - \frac {3 b^{2} f \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x^{2}} \right )}}{16 \sqrt {a}} \]

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**12,x)

[Out]

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

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