3.504 \(\int \frac {(c+d x+e x^2+f x^3) \sqrt {a+b x^4}}{x^5} \, dx\)

Optimal. Leaf size=329 \[ -\frac {1}{12} \sqrt {a+b x^4} \left (\frac {3 c}{x^4}+\frac {4 d}{x^3}+\frac {6 e}{x^2}+\frac {12 f}{x}\right )-\frac {b c \tanh ^{-1}\left (\frac {\sqrt {a+b x^4}}{\sqrt {a}}\right )}{4 \sqrt {a}}+\frac {\sqrt [4]{b} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \left (3 \sqrt {a} f+\sqrt {b} d\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{3 \sqrt [4]{a} \sqrt {a+b x^4}}+\frac {1}{2} \sqrt {b} e \tanh ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )+\frac {2 \sqrt {b} f x \sqrt {a+b x^4}}{\sqrt {a}+\sqrt {b} x^2}-\frac {2 \sqrt [4]{a} \sqrt [4]{b} f \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{\sqrt {a+b x^4}} \]

[Out]

-1/4*b*c*arctanh((b*x^4+a)^(1/2)/a^(1/2))/a^(1/2)+1/2*e*arctanh(x^2*b^(1/2)/(b*x^4+a)^(1/2))*b^(1/2)-1/12*(3*c
/x^4+4*d/x^3+6*e/x^2+12*f/x)*(b*x^4+a)^(1/2)+2*f*x*b^(1/2)*(b*x^4+a)^(1/2)/(a^(1/2)+x^2*b^(1/2))-2*a^(1/4)*b^(
1/4)*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)+1
/3*b^(1/4)*(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))*(3*f*a^(1/2)+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)

________________________________________________________________________________________

Rubi [A]  time = 0.28, antiderivative size = 329, normalized size of antiderivative = 1.00, number of steps used = 13, 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} \[ -\frac {1}{12} \sqrt {a+b x^4} \left (\frac {3 c}{x^4}+\frac {4 d}{x^3}+\frac {6 e}{x^2}+\frac {12 f}{x}\right )-\frac {b c \tanh ^{-1}\left (\frac {\sqrt {a+b x^4}}{\sqrt {a}}\right )}{4 \sqrt {a}}+\frac {\sqrt [4]{b} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \left (3 \sqrt {a} f+\sqrt {b} d\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{3 \sqrt [4]{a} \sqrt {a+b x^4}}+\frac {1}{2} \sqrt {b} e \tanh ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )+\frac {2 \sqrt {b} f x \sqrt {a+b x^4}}{\sqrt {a}+\sqrt {b} x^2}-\frac {2 \sqrt [4]{a} \sqrt [4]{b} f \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{\sqrt {a+b x^4}} \]

Antiderivative was successfully verified.

[In]

Int[((c + d*x + e*x^2 + f*x^3)*Sqrt[a + b*x^4])/x^5,x]

[Out]

-(((3*c)/x^4 + (4*d)/x^3 + (6*e)/x^2 + (12*f)/x)*Sqrt[a + b*x^4])/12 + (2*Sqrt[b]*f*x*Sqrt[a + b*x^4])/(Sqrt[a
] + Sqrt[b]*x^2) + (Sqrt[b]*e*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a + b*x^4]])/2 - (b*c*ArcTanh[Sqrt[a + b*x^4]/Sqrt[a]
])/(4*Sqrt[a]) - (2*a^(1/4)*b^(1/4)*f*(Sqrt[a] + Sqrt[b]*x^2)*Sqrt[(a + b*x^4)/(Sqrt[a] + Sqrt[b]*x^2)^2]*Elli
pticE[2*ArcTan[(b^(1/4)*x)/a^(1/4)], 1/2])/Sqrt[a + b*x^4] + (b^(1/4)*(Sqrt[b]*d + 3*Sqrt[a]*f)*(Sqrt[a] + Sqr
t[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])/(3*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 ) \sqrt {a+b x^4}}{x^5} \, dx &=-\frac {1}{12} \left (\frac {3 c}{x^4}+\frac {4 d}{x^3}+\frac {6 e}{x^2}+\frac {12 f}{x}\right ) \sqrt {a+b x^4}-(2 b) \int \frac {-\frac {c}{4}-\frac {d x}{3}-\frac {e x^2}{2}-f x^3}{x \sqrt {a+b x^4}} \, dx\\ &=-\frac {1}{12} \left (\frac {3 c}{x^4}+\frac {4 d}{x^3}+\frac {6 e}{x^2}+\frac {12 f}{x}\right ) \sqrt {a+b x^4}-(2 b) \int \frac {-\frac {d}{3}-\frac {e x}{2}-f x^2}{\sqrt {a+b x^4}} \, dx+\frac {1}{2} (b c) \int \frac {1}{x \sqrt {a+b x^4}} \, dx\\ &=-\frac {1}{12} \left (\frac {3 c}{x^4}+\frac {4 d}{x^3}+\frac {6 e}{x^2}+\frac {12 f}{x}\right ) \sqrt {a+b x^4}-(2 b) \int \left (-\frac {e x}{2 \sqrt {a+b x^4}}+\frac {-\frac {d}{3}-f x^2}{\sqrt {a+b x^4}}\right ) \, dx+\frac {1}{8} (b c) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,x^4\right )\\ &=-\frac {1}{12} \left (\frac {3 c}{x^4}+\frac {4 d}{x^3}+\frac {6 e}{x^2}+\frac {12 f}{x}\right ) \sqrt {a+b x^4}-(2 b) \int \frac {-\frac {d}{3}-f x^2}{\sqrt {a+b x^4}} \, dx+\frac {1}{4} c \operatorname {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b x^4}\right )+(b e) \int \frac {x}{\sqrt {a+b x^4}} \, dx\\ &=-\frac {1}{12} \left (\frac {3 c}{x^4}+\frac {4 d}{x^3}+\frac {6 e}{x^2}+\frac {12 f}{x}\right ) \sqrt {a+b x^4}-\frac {b c \tanh ^{-1}\left (\frac {\sqrt {a+b x^4}}{\sqrt {a}}\right )}{4 \sqrt {a}}+\frac {1}{2} (b e) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,x^2\right )-\left (2 \sqrt {a} \sqrt {b} f\right ) \int \frac {1-\frac {\sqrt {b} x^2}{\sqrt {a}}}{\sqrt {a+b x^4}} \, dx+\frac {1}{3} \left (2 b \left (d+\frac {3 \sqrt {a} f}{\sqrt {b}}\right )\right ) \int \frac {1}{\sqrt {a+b x^4}} \, dx\\ &=-\frac {1}{12} \left (\frac {3 c}{x^4}+\frac {4 d}{x^3}+\frac {6 e}{x^2}+\frac {12 f}{x}\right ) \sqrt {a+b x^4}+\frac {2 \sqrt {b} f x \sqrt {a+b x^4}}{\sqrt {a}+\sqrt {b} x^2}-\frac {b c \tanh ^{-1}\left (\frac {\sqrt {a+b x^4}}{\sqrt {a}}\right )}{4 \sqrt {a}}-\frac {2 \sqrt [4]{a} \sqrt [4]{b} 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 )}{\sqrt {a+b x^4}}+\frac {\sqrt [4]{b} \left (\sqrt {b} d+3 \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 )}{3 \sqrt [4]{a} \sqrt {a+b x^4}}+\frac {1}{2} (b e) \operatorname {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x^2}{\sqrt {a+b x^4}}\right )\\ &=-\frac {1}{12} \left (\frac {3 c}{x^4}+\frac {4 d}{x^3}+\frac {6 e}{x^2}+\frac {12 f}{x}\right ) \sqrt {a+b x^4}+\frac {2 \sqrt {b} f x \sqrt {a+b x^4}}{\sqrt {a}+\sqrt {b} x^2}+\frac {1}{2} \sqrt {b} e \tanh ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )-\frac {b c \tanh ^{-1}\left (\frac {\sqrt {a+b x^4}}{\sqrt {a}}\right )}{4 \sqrt {a}}-\frac {2 \sqrt [4]{a} \sqrt [4]{b} 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 )}{\sqrt {a+b x^4}}+\frac {\sqrt [4]{b} \left (\sqrt {b} d+3 \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 )}{3 \sqrt [4]{a} \sqrt {a+b x^4}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 0.27, size = 175, normalized size = 0.53 \[ -\frac {\sqrt {\frac {b x^4}{a}+1} \left (3 a c \sqrt {\frac {b x^4}{a}+1}+3 b c x^4 \tanh ^{-1}\left (\sqrt {\frac {b x^4}{a}+1}\right )+4 a d x \, _2F_1\left (-\frac {3}{4},-\frac {1}{2};\frac {1}{4};-\frac {b x^4}{a}\right )+6 a e x^2 \sqrt {\frac {b x^4}{a}+1}-6 \sqrt {a} \sqrt {b} e x^4 \sinh ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )+12 a f x^3 \, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};-\frac {b x^4}{a}\right )\right )}{12 x^4 \sqrt {a+b x^4}} \]

Antiderivative was successfully verified.

[In]

Integrate[((c + d*x + e*x^2 + f*x^3)*Sqrt[a + b*x^4])/x^5,x]

[Out]

-1/12*(Sqrt[1 + (b*x^4)/a]*(3*a*c*Sqrt[1 + (b*x^4)/a] + 6*a*e*x^2*Sqrt[1 + (b*x^4)/a] - 6*Sqrt[a]*Sqrt[b]*e*x^
4*ArcSinh[(Sqrt[b]*x^2)/Sqrt[a]] + 3*b*c*x^4*ArcTanh[Sqrt[1 + (b*x^4)/a]] + 4*a*d*x*Hypergeometric2F1[-3/4, -1
/2, 1/4, -((b*x^4)/a)] + 12*a*f*x^3*Hypergeometric2F1[-1/2, -1/4, 3/4, -((b*x^4)/a)]))/(x^4*Sqrt[a + b*x^4])

________________________________________________________________________________________

fricas [F]  time = 0.73, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {b x^{4} + a} {\left (f x^{3} + e x^{2} + d x + c\right )}}{x^{5}}, x\right ) \]

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)^(1/2)/x^5,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

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

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)^(1/2)/x^5,x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

maple [C]  time = 0.20, size = 385, normalized size = 1.17 \[ -\frac {2 i \sqrt {-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {a}\, \sqrt {b}\, f \EllipticE \left (\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, x , i\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}+\frac {2 i \sqrt {-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {a}\, \sqrt {b}\, f \EllipticF \left (\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, x , i\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}+\frac {\sqrt {b \,x^{4}+a}\, b e \,x^{2}}{2 a}+\frac {2 \sqrt {-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, b d \EllipticF \left (\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, x , i\right )}{3 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}-\frac {b c \ln \left (\frac {2 a +2 \sqrt {b \,x^{4}+a}\, \sqrt {a}}{x^{2}}\right )}{4 \sqrt {a}}+\frac {\sqrt {b}\, e \ln \left (\sqrt {b}\, x^{2}+\sqrt {b \,x^{4}+a}\right )}{2}+\frac {\sqrt {b \,x^{4}+a}\, b c}{4 a}-\frac {\sqrt {b \,x^{4}+a}\, f}{x}-\frac {\left (b \,x^{4}+a \right )^{\frac {3}{2}} e}{2 a \,x^{2}}-\frac {\sqrt {b \,x^{4}+a}\, d}{3 x^{3}}-\frac {\left (b \,x^{4}+a \right )^{\frac {3}{2}} c}{4 a \,x^{4}} \]

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)^(1/2)/x^5,x)

[Out]

-1/4*c/a/x^4*(b*x^4+a)^(3/2)-1/4*c/a^(1/2)*b*ln((2*a+2*(b*x^4+a)^(1/2)*a^(1/2))/x^2)+1/4*c/a*b*(b*x^4+a)^(1/2)
-1/3*d/x^3*(b*x^4+a)^(1/2)+2/3*d*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)-1/2*e/a/x^2*(b*x^4+a)^(3/2)+1/2*e/a*b
*x^2*(b*x^4+a)^(1/2)+1/2*e*b^(1/2)*ln(b^(1/2)*x^2+(b*x^4+a)^(1/2))-f/x*(b*x^4+a)^(1/2)+2*I*f*b^(1/2)*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)*Elli
pticF((I/a^(1/2)*b^(1/2))^(1/2)*x,I)-2*I*f*b^(1/2)*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)

________________________________________________________________________________________

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

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)^(1/2)/x^5,x, algorithm="maxima")

[Out]

1/8*(b*log((sqrt(b*x^4 + a) - sqrt(a))/(sqrt(b*x^4 + a) + sqrt(a)))/sqrt(a) - 2*sqrt(b*x^4 + a)/x^4)*c + integ
rate(sqrt(b*x^4 + a)*(f*x^2 + e*x + d)/x^4, x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [C]  time = 6.78, size = 211, normalized size = 0.64 \[ \frac {\sqrt {a} d \Gamma \left (- \frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, - \frac {1}{2} \\ \frac {1}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 x^{3} \Gamma \left (\frac {1}{4}\right )} - \frac {\sqrt {a} e}{2 x^{2} \sqrt {1 + \frac {b x^{4}}{a}}} + \frac {\sqrt {a} f \Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, - \frac {1}{4} \\ \frac {3}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 x \Gamma \left (\frac {3}{4}\right )} - \frac {\sqrt {b} c \sqrt {\frac {a}{b x^{4}} + 1}}{4 x^{2}} + \frac {\sqrt {b} e \operatorname {asinh}{\left (\frac {\sqrt {b} x^{2}}{\sqrt {a}} \right )}}{2} - \frac {b c \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x^{2}} \right )}}{4 \sqrt {a}} - \frac {b e x^{2}}{2 \sqrt {a} \sqrt {1 + \frac {b x^{4}}{a}}} \]

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)**(1/2)/x**5,x)

[Out]

sqrt(a)*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)*e/(2
*x**2*sqrt(1 + b*x**4/a)) + sqrt(a)*f*gamma(-1/4)*hyper((-1/2, -1/4), (3/4,), b*x**4*exp_polar(I*pi)/a)/(4*x*g
amma(3/4)) - sqrt(b)*c*sqrt(a/(b*x**4) + 1)/(4*x**2) + sqrt(b)*e*asinh(sqrt(b)*x**2/sqrt(a))/2 - b*c*asinh(sqr
t(a)/(sqrt(b)*x**2))/(4*sqrt(a)) - b*e*x**2/(2*sqrt(a)*sqrt(1 + b*x**4/a))

________________________________________________________________________________________