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

[Out]

-1/840*(105*c/x^8+120*d/x^7+140*e/x^6+168*f/x^5)*(b*x^4+a)^(3/2)+1/2*b^(3/2)*e*arctanh(x^2*b^(1/2)/(b*x^4+a)^(
1/2))-3/16*b^2*c*arctanh((b*x^4+a)^(1/2)/a^(1/2))/a^(1/2)-1/560*b*(105*c/x^4+160*d/x^3+280*e/x^2+672*f/x)*(b*x
^4+a)^(1/2)+12/5*b^(3/2)*f*x*(b*x^4+a)^(1/2)/(a^(1/2)+x^2*b^(1/2))-12/5*a^(1/4)*b^(5/4)*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*x^4+a)^(1/2)+2/35*b^(5/4)*(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))*(21*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)/a^(1/4)/(b*x
^4+a)^(1/2)

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Rubi [A]  time = 0.37, antiderivative size = 377, normalized size of antiderivative = 1.00, 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 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 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 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 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 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)]

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*}

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Mathematica [C]  time = 0.32, 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]

-1/1680*(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)])))/(a*x^8*Sqrt[1 + (b
*x^4)/a])

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fricas [F]  time = 0.79, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b f x^{7} + b e x^{6} + b d x^{5} + b c x^{4} + a f x^{3} + a e x^{2} + a d x + a c\right )} \sqrt {b x^{4} + a}}{x^{9}}, x\right ) \]

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^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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x^{4} + a\right )}^{\frac {3}{2}} {\left (f x^{3} + e x^{2} + d x + c\right )}}{x^{9}}\,{d x} \]

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^9,x, algorithm="giac")

[Out]

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

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maple [C]  time = 0.21, size = 416, normalized size = 1.10 \[ -\frac {12 i \sqrt {-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {a}\, b^{\frac {3}{2}} f \EllipticE \left (\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, x , i\right )}{5 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}+\frac {12 i \sqrt {-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {a}\, b^{\frac {3}{2}} f \EllipticF \left (\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, x , i\right )}{5 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}+\frac {4 \sqrt {-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, b^{2} d \EllipticF \left (\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, x , i\right )}{7 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}-\frac {3 b^{2} c \ln \left (\frac {2 a +2 \sqrt {b \,x^{4}+a}\, \sqrt {a}}{x^{2}}\right )}{16 \sqrt {a}}+\frac {b^{\frac {3}{2}} e \ln \left (\sqrt {b}\, x^{2}+\sqrt {b \,x^{4}+a}\right )}{2}-\frac {7 \sqrt {b \,x^{4}+a}\, b f}{5 x}-\frac {2 \sqrt {b \,x^{4}+a}\, b e}{3 x^{2}}-\frac {3 \sqrt {b \,x^{4}+a}\, b d}{7 x^{3}}-\frac {5 \sqrt {b \,x^{4}+a}\, b c}{16 x^{4}}-\frac {\sqrt {b \,x^{4}+a}\, a f}{5 x^{5}}-\frac {\sqrt {b \,x^{4}+a}\, a e}{6 x^{6}}-\frac {\sqrt {b \,x^{4}+a}\, a d}{7 x^{7}}-\frac {\sqrt {b \,x^{4}+a}\, a c}{8 x^{8}} \]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{32} \, {\left (\frac {3 \, b^{2} \log \left (\frac {\sqrt {b x^{4} + a} - \sqrt {a}}{\sqrt {b x^{4} + a} + \sqrt {a}}\right )}{\sqrt {a}} - \frac {2 \, {\left (5 \, {\left (b x^{4} + a\right )}^{\frac {3}{2}} b^{2} - 3 \, \sqrt {b x^{4} + a} a b^{2}\right )}}{{\left (b x^{4} + a\right )}^{2} - 2 \, {\left (b x^{4} + a\right )} a + a^{2}}\right )} c + \int \frac {{\left (b f x^{6} + b e x^{5} + b d x^{4} + a f x^{2} + a e x + a d\right )} \sqrt {b x^{4} + a}}{x^{8}}\,{d x} \]

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^9,x, algorithm="maxima")

[Out]

1/32*(3*b^2*log((sqrt(b*x^4 + a) - sqrt(a))/(sqrt(b*x^4 + a) + sqrt(a)))/sqrt(a) - 2*(5*(b*x^4 + a)^(3/2)*b^2
- 3*sqrt(b*x^4 + a)*a*b^2)/((b*x^4 + a)^2 - 2*(b*x^4 + a)*a + a^2))*c + integrate((b*f*x^6 + b*e*x^5 + b*d*x^4
+ a*f*x^2 + a*e*x + a*d)*sqrt(b*x^4 + a)/x^8, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (b\,x^4+a\right )}^{3/2}\,\left (f\,x^3+e\,x^2+d\,x+c\right )}{x^9} \,d x \]

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^9,x)

[Out]

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

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sympy [C]  time = 18.59, size = 444, normalized size = 1.18 \[ \frac {a^{\frac {3}{2}} d \Gamma \left (- \frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {7}{4}, - \frac {1}{2} \\ - \frac {3}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 x^{7} \Gamma \left (- \frac {3}{4}\right )} + \frac {a^{\frac {3}{2}} f \Gamma \left (- \frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{4}, - \frac {1}{2} \\ - \frac {1}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 x^{5} \Gamma \left (- \frac {1}{4}\right )} + \frac {\sqrt {a} b d \Gamma \left (- \frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, - \frac {1}{2} \\ \frac {1}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 x^{3} \Gamma \left (\frac {1}{4}\right )} - \frac {\sqrt {a} b e}{2 x^{2} \sqrt {1 + \frac {b x^{4}}{a}}} + \frac {\sqrt {a} b f \Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, - \frac {1}{4} \\ \frac {3}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 x \Gamma \left (\frac {3}{4}\right )} - \frac {a^{2} c}{8 \sqrt {b} x^{10} \sqrt {\frac {a}{b x^{4}} + 1}} - \frac {3 a \sqrt {b} c}{16 x^{6} \sqrt {\frac {a}{b x^{4}} + 1}} - \frac {a \sqrt {b} e \sqrt {\frac {a}{b x^{4}} + 1}}{6 x^{4}} - \frac {b^{\frac {3}{2}} c \sqrt {\frac {a}{b x^{4}} + 1}}{4 x^{2}} - \frac {b^{\frac {3}{2}} c}{16 x^{2} \sqrt {\frac {a}{b x^{4}} + 1}} - \frac {b^{\frac {3}{2}} e \sqrt {\frac {a}{b x^{4}} + 1}}{6} + \frac {b^{\frac {3}{2}} e \operatorname {asinh}{\left (\frac {\sqrt {b} x^{2}}{\sqrt {a}} \right )}}{2} - \frac {3 b^{2} c \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x^{2}} \right )}}{16 \sqrt {a}} - \frac {b^{2} e x^{2}}{2 \sqrt {a} \sqrt {1 + \frac {b x^{4}}{a}}} \]

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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