3.523 \(\int \frac{(c+d x+e x^2+f x^3) (a+b x^4)^{3/2}}{x^9} \, dx\)

Optimal. Leaf size=377 \[ \frac{2 b^{5/4} \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} \left (21 \sqrt{a} f+5 \sqrt{b} d\right ) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac{1}{2}\right )}{35 \sqrt [4]{a} \sqrt{a+b x^4}}-\frac{3 b^2 c \tanh ^{-1}\left (\frac{\sqrt{a+b x^4}}{\sqrt{a}}\right )}{16 \sqrt{a}}+\frac{1}{2} b^{3/2} e \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a+b x^4}}\right )+\frac{12 b^{3/2} f x \sqrt{a+b x^4}}{5 \left (\sqrt{a}+\sqrt{b} x^2\right )}-\frac{12 \sqrt [4]{a} b^{5/4} f \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{5 \sqrt{a+b x^4}}-\frac{1}{560} b \sqrt{a+b x^4} \left (\frac{105 c}{x^4}+\frac{160 d}{x^3}+\frac{280 e}{x^2}+\frac{672 f}{x}\right )-\frac{1}{840} \left (a+b x^4\right )^{3/2} \left (\frac{105 c}{x^8}+\frac{120 d}{x^7}+\frac{140 e}{x^6}+\frac{168 f}{x^5}\right ) \]

[Out]

-(b*((105*c)/x^4 + (160*d)/x^3 + (280*e)/x^2 + (672*f)/x)*Sqrt[a + b*x^4])/560 + (12*b^(3/2)*f*x*Sqrt[a + b*x^
4])/(5*(Sqrt[a] + Sqrt[b]*x^2)) - (((105*c)/x^8 + (120*d)/x^7 + (140*e)/x^6 + (168*f)/x^5)*(a + b*x^4)^(3/2))/
840 + (b^(3/2)*e*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a + b*x^4]])/2 - (3*b^2*c*ArcTanh[Sqrt[a + b*x^4]/Sqrt[a]])/(16*Sq
rt[a]) - (12*a^(1/4)*b^(5/4)*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])/(5*Sqrt[a + b*x^4]) + (2*b^(5/4)*(5*Sqrt[b]*d + 21*Sqrt[a]*f)*(Sqrt[a] + S
qrt[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])/(35*a^(
1/4)*Sqrt[a + b*x^4])

________________________________________________________________________________________

Rubi [A]  time = 0.370944, antiderivative size = 377, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 13, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.433, Rules used = {14, 1825, 1832, 266, 63, 208, 1885, 275, 217, 206, 1198, 220, 1196} \[ -\frac{3 b^2 c \tanh ^{-1}\left (\frac{\sqrt{a+b x^4}}{\sqrt{a}}\right )}{16 \sqrt{a}}+\frac{2 b^{5/4} \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} \left (21 \sqrt{a} f+5 \sqrt{b} d\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{35 \sqrt [4]{a} \sqrt{a+b x^4}}+\frac{1}{2} b^{3/2} e \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a+b x^4}}\right )+\frac{12 b^{3/2} f x \sqrt{a+b x^4}}{5 \left (\sqrt{a}+\sqrt{b} x^2\right )}-\frac{12 \sqrt [4]{a} b^{5/4} f \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{5 \sqrt{a+b x^4}}-\frac{1}{560} b \sqrt{a+b x^4} \left (\frac{105 c}{x^4}+\frac{160 d}{x^3}+\frac{280 e}{x^2}+\frac{672 f}{x}\right )-\frac{1}{840} \left (a+b x^4\right )^{3/2} \left (\frac{105 c}{x^8}+\frac{120 d}{x^7}+\frac{140 e}{x^6}+\frac{168 f}{x^5}\right ) \]

Antiderivative was successfully verified.

[In]

Int[((c + d*x + e*x^2 + f*x^3)*(a + b*x^4)^(3/2))/x^9,x]

[Out]

-(b*((105*c)/x^4 + (160*d)/x^3 + (280*e)/x^2 + (672*f)/x)*Sqrt[a + b*x^4])/560 + (12*b^(3/2)*f*x*Sqrt[a + b*x^
4])/(5*(Sqrt[a] + Sqrt[b]*x^2)) - (((105*c)/x^8 + (120*d)/x^7 + (140*e)/x^6 + (168*f)/x^5)*(a + b*x^4)^(3/2))/
840 + (b^(3/2)*e*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a + b*x^4]])/2 - (3*b^2*c*ArcTanh[Sqrt[a + b*x^4]/Sqrt[a]])/(16*Sq
rt[a]) - (12*a^(1/4)*b^(5/4)*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])/(5*Sqrt[a + b*x^4]) + (2*b^(5/4)*(5*Sqrt[b]*d + 21*Sqrt[a]*f)*(Sqrt[a] + S
qrt[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])/(35*a^(
1/4)*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 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 1832

Int[(Pq_)/((x_)*Sqrt[(a_) + (b_.)*(x_)^(n_)]), x_Symbol] :> Dist[Coeff[Pq, x, 0], Int[1/(x*Sqrt[a + b*x^n]), x
], x] + Int[ExpandToSum[(Pq - Coeff[Pq, x, 0])/x, x]/Sqrt[a + b*x^n], x] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] &
& IGtQ[n, 0] && NeQ[Coeff[Pq, x, 0], 0]

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 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 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 275

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

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])

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]

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^9} \, dx &=-\frac{1}{840} \left (\frac{105 c}{x^8}+\frac{120 d}{x^7}+\frac{140 e}{x^6}+\frac{168 f}{x^5}\right ) \left (a+b x^4\right )^{3/2}-(6 b) \int \frac{\left (-\frac{c}{8}-\frac{d x}{7}-\frac{e x^2}{6}-\frac{f x^3}{5}\right ) \sqrt{a+b x^4}}{x^5} \, dx\\ &=-\frac{1}{560} b \left (\frac{105 c}{x^4}+\frac{160 d}{x^3}+\frac{280 e}{x^2}+\frac{672 f}{x}\right ) \sqrt{a+b x^4}-\frac{1}{840} \left (\frac{105 c}{x^8}+\frac{120 d}{x^7}+\frac{140 e}{x^6}+\frac{168 f}{x^5}\right ) \left (a+b x^4\right )^{3/2}+\left (12 b^2\right ) \int \frac{\frac{c}{32}+\frac{d x}{21}+\frac{e x^2}{12}+\frac{f x^3}{5}}{x \sqrt{a+b x^4}} \, dx\\ &=-\frac{1}{560} b \left (\frac{105 c}{x^4}+\frac{160 d}{x^3}+\frac{280 e}{x^2}+\frac{672 f}{x}\right ) \sqrt{a+b x^4}-\frac{1}{840} \left (\frac{105 c}{x^8}+\frac{120 d}{x^7}+\frac{140 e}{x^6}+\frac{168 f}{x^5}\right ) \left (a+b x^4\right )^{3/2}+\left (12 b^2\right ) \int \frac{\frac{d}{21}+\frac{e x}{12}+\frac{f x^2}{5}}{\sqrt{a+b x^4}} \, dx+\frac{1}{8} \left (3 b^2 c\right ) \int \frac{1}{x \sqrt{a+b x^4}} \, dx\\ &=-\frac{1}{560} b \left (\frac{105 c}{x^4}+\frac{160 d}{x^3}+\frac{280 e}{x^2}+\frac{672 f}{x}\right ) \sqrt{a+b x^4}-\frac{1}{840} \left (\frac{105 c}{x^8}+\frac{120 d}{x^7}+\frac{140 e}{x^6}+\frac{168 f}{x^5}\right ) \left (a+b x^4\right )^{3/2}+\left (12 b^2\right ) \int \left (\frac{e x}{12 \sqrt{a+b x^4}}+\frac{\frac{d}{21}+\frac{f x^2}{5}}{\sqrt{a+b x^4}}\right ) \, dx+\frac{1}{32} \left (3 b^2 c\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,x^4\right )\\ &=-\frac{1}{560} b \left (\frac{105 c}{x^4}+\frac{160 d}{x^3}+\frac{280 e}{x^2}+\frac{672 f}{x}\right ) \sqrt{a+b x^4}-\frac{1}{840} \left (\frac{105 c}{x^8}+\frac{120 d}{x^7}+\frac{140 e}{x^6}+\frac{168 f}{x^5}\right ) \left (a+b x^4\right )^{3/2}+\left (12 b^2\right ) \int \frac{\frac{d}{21}+\frac{f x^2}{5}}{\sqrt{a+b x^4}} \, dx+\frac{1}{16} (3 b c) \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b x^4}\right )+\left (b^2 e\right ) \int \frac{x}{\sqrt{a+b x^4}} \, dx\\ &=-\frac{1}{560} b \left (\frac{105 c}{x^4}+\frac{160 d}{x^3}+\frac{280 e}{x^2}+\frac{672 f}{x}\right ) \sqrt{a+b x^4}-\frac{1}{840} \left (\frac{105 c}{x^8}+\frac{120 d}{x^7}+\frac{140 e}{x^6}+\frac{168 f}{x^5}\right ) \left (a+b x^4\right )^{3/2}-\frac{3 b^2 c \tanh ^{-1}\left (\frac{\sqrt{a+b x^4}}{\sqrt{a}}\right )}{16 \sqrt{a}}+\frac{1}{2} \left (b^2 e\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^2}} \, dx,x,x^2\right )-\frac{1}{5} \left (12 \sqrt{a} b^{3/2} f\right ) \int \frac{1-\frac{\sqrt{b} x^2}{\sqrt{a}}}{\sqrt{a+b x^4}} \, dx+\frac{1}{35} \left (4 b^2 \left (5 d+\frac{21 \sqrt{a} f}{\sqrt{b}}\right )\right ) \int \frac{1}{\sqrt{a+b x^4}} \, dx\\ &=-\frac{1}{560} b \left (\frac{105 c}{x^4}+\frac{160 d}{x^3}+\frac{280 e}{x^2}+\frac{672 f}{x}\right ) \sqrt{a+b x^4}+\frac{12 b^{3/2} f x \sqrt{a+b x^4}}{5 \left (\sqrt{a}+\sqrt{b} x^2\right )}-\frac{1}{840} \left (\frac{105 c}{x^8}+\frac{120 d}{x^7}+\frac{140 e}{x^6}+\frac{168 f}{x^5}\right ) \left (a+b x^4\right )^{3/2}-\frac{3 b^2 c \tanh ^{-1}\left (\frac{\sqrt{a+b x^4}}{\sqrt{a}}\right )}{16 \sqrt{a}}-\frac{12 \sqrt [4]{a} b^{5/4} 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 )}{5 \sqrt{a+b x^4}}+\frac{2 b^{5/4} \left (5 \sqrt{b} d+21 \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 )}{35 \sqrt [4]{a} \sqrt{a+b x^4}}+\frac{1}{2} \left (b^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{1}{560} b \left (\frac{105 c}{x^4}+\frac{160 d}{x^3}+\frac{280 e}{x^2}+\frac{672 f}{x}\right ) \sqrt{a+b x^4}+\frac{12 b^{3/2} f x \sqrt{a+b x^4}}{5 \left (\sqrt{a}+\sqrt{b} x^2\right )}-\frac{1}{840} \left (\frac{105 c}{x^8}+\frac{120 d}{x^7}+\frac{140 e}{x^6}+\frac{168 f}{x^5}\right ) \left (a+b x^4\right )^{3/2}+\frac{1}{2} b^{3/2} e \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a+b x^4}}\right )-\frac{3 b^2 c \tanh ^{-1}\left (\frac{\sqrt{a+b x^4}}{\sqrt{a}}\right )}{16 \sqrt{a}}-\frac{12 \sqrt [4]{a} b^{5/4} 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 )}{5 \sqrt{a+b x^4}}+\frac{2 b^{5/4} \left (5 \sqrt{b} d+21 \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 )}{35 \sqrt [4]{a} \sqrt{a+b x^4}}\\ \end{align*}

Mathematica [C]  time = 0.293568, size = 174, normalized size = 0.46 \[ -\frac{\sqrt{a+b x^4} \left (7 \left (40 a^2 e x^2 \, _2F_1\left (-\frac{3}{2},-\frac{3}{2};-\frac{1}{2};-\frac{b x^4}{a}\right )+48 a^2 f x^3 \, _2F_1\left (-\frac{3}{2},-\frac{5}{4};-\frac{1}{4};-\frac{b x^4}{a}\right )+15 c \left (3 b^2 x^8 \tanh ^{-1}\left (\sqrt{\frac{b x^4}{a}+1}\right )+a \left (2 a+5 b x^4\right ) \sqrt{\frac{b x^4}{a}+1}\right )\right )+240 a^2 d x \, _2F_1\left (-\frac{7}{4},-\frac{3}{2};-\frac{3}{4};-\frac{b x^4}{a}\right )\right )}{1680 a x^8 \sqrt{\frac{b x^4}{a}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[((c + d*x + e*x^2 + f*x^3)*(a + b*x^4)^(3/2))/x^9,x]

[Out]

-(Sqrt[a + b*x^4]*(240*a^2*d*x*Hypergeometric2F1[-7/4, -3/2, -3/4, -((b*x^4)/a)] + 7*(15*c*(a*(2*a + 5*b*x^4)*
Sqrt[1 + (b*x^4)/a] + 3*b^2*x^8*ArcTanh[Sqrt[1 + (b*x^4)/a]]) + 40*a^2*e*x^2*Hypergeometric2F1[-3/2, -3/2, -1/
2, -((b*x^4)/a)] + 48*a^2*f*x^3*Hypergeometric2F1[-3/2, -5/4, -1/4, -((b*x^4)/a)])))/(1680*a*x^8*Sqrt[1 + (b*x
^4)/a])

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Maple [C]  time = 0.023, size = 416, normalized size = 1.1 \begin{align*} -{\frac{af}{5\,{x}^{5}}\sqrt{b{x}^{4}+a}}-{\frac{7\,fb}{5\,x}\sqrt{b{x}^{4}+a}}+{{\frac{12\,i}{5}}f{b}^{{\frac{3}{2}}}\sqrt{a}\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{12\,i}{5}}f{b}^{{\frac{3}{2}}}\sqrt{a}\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{3\,{b}^{2}c}{16}\ln \left ({\frac{1}{{x}^{2}} \left ( 2\,a+2\,\sqrt{a}\sqrt{b{x}^{4}+a} \right ) } \right ){\frac{1}{\sqrt{a}}}}-{\frac{ac}{8\,{x}^{8}}\sqrt{b{x}^{4}+a}}-{\frac{5\,bc}{16\,{x}^{4}}\sqrt{b{x}^{4}+a}}+{\frac{e}{2}{b}^{{\frac{3}{2}}}\ln \left ({x}^{2}\sqrt{b}+\sqrt{b{x}^{4}+a} \right ) }-{\frac{ae}{6\,{x}^{6}}\sqrt{b{x}^{4}+a}}-{\frac{2\,be}{3\,{x}^{2}}\sqrt{b{x}^{4}+a}}-{\frac{ad}{7\,{x}^{7}}\sqrt{b{x}^{4}+a}}-{\frac{3\,bd}{7\,{x}^{3}}\sqrt{b{x}^{4}+a}}+{\frac{4\,{b}^{2}d}{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^9,x)

[Out]

-1/5*f*a*(b*x^4+a)^(1/2)/x^5-7/5*f*b*(b*x^4+a)^(1/2)/x+12/5*I*f*b^(3/2)*a^(1/2)/(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)-12/5*I*f*b^(3/2)*a^(1/2)/(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)*EllipticE(x*(I/a^(1/2)*b^(1/2))^(1/2),I)-3/16*c*b^2/a^(1/2)*ln((2*a+2*a^(1/2)*(b*x
^4+a)^(1/2))/x^2)-1/8*c*a/x^8*(b*x^4+a)^(1/2)-5/16*c*b/x^4*(b*x^4+a)^(1/2)+1/2*e*b^(3/2)*ln(x^2*b^(1/2)+(b*x^4
+a)^(1/2))-1/6*e*a/x^6*(b*x^4+a)^(1/2)-2/3*e*b/x^2*(b*x^4+a)^(1/2)-1/7*d*a*(b*x^4+a)^(1/2)/x^7-3/7*d*b*(b*x^4+
a)^(1/2)/x^3+4/7*d*b^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)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \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^9,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\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^{9}}, 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^9,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^9, x)

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Sympy [C]  time = 12.9674, size = 444, normalized size = 1.18 \begin{align*} \frac{a^{\frac{3}{2}} d \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{a^{\frac{3}{2}} f \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{\sqrt{a} b d \Gamma \left (- \frac{3}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{3}{4}, - \frac{1}{2} \\ \frac{1}{4} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{4 x^{3} \Gamma \left (\frac{1}{4}\right )} - \frac{\sqrt{a} b e}{2 x^{2} \sqrt{1 + \frac{b x^{4}}{a}}} + \frac{\sqrt{a} b f \Gamma \left (- \frac{1}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, - \frac{1}{4} \\ \frac{3}{4} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{4 x \Gamma \left (\frac{3}{4}\right )} - \frac{a^{2} c}{8 \sqrt{b} x^{10} \sqrt{\frac{a}{b x^{4}} + 1}} - \frac{3 a \sqrt{b} c}{16 x^{6} \sqrt{\frac{a}{b x^{4}} + 1}} - \frac{a \sqrt{b} e \sqrt{\frac{a}{b x^{4}} + 1}}{6 x^{4}} - \frac{b^{\frac{3}{2}} c \sqrt{\frac{a}{b x^{4}} + 1}}{4 x^{2}} - \frac{b^{\frac{3}{2}} c}{16 x^{2} \sqrt{\frac{a}{b x^{4}} + 1}} - \frac{b^{\frac{3}{2}} e \sqrt{\frac{a}{b x^{4}} + 1}}{6} + \frac{b^{\frac{3}{2}} e \operatorname{asinh}{\left (\frac{\sqrt{b} x^{2}}{\sqrt{a}} \right )}}{2} - \frac{3 b^{2} c \operatorname{asinh}{\left (\frac{\sqrt{a}}{\sqrt{b} x^{2}} \right )}}{16 \sqrt{a}} - \frac{b^{2} e x^{2}}{2 \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**9,x)

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x^{4} + a\right )}^{\frac{3}{2}}{\left (f x^{3} + e x^{2} + d x + c\right )}}{x^{9}}\,{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^9,x, algorithm="giac")

[Out]

integrate((b*x^4 + a)^(3/2)*(f*x^3 + e*x^2 + d*x + c)/x^9, x)