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

Optimal. Leaf size=323 \[ -\frac {\sqrt [4]{b} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \left (\sqrt {b} c-3 \sqrt {a} e\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{6 a^{5/4} \sqrt {a+b x^4}}-\frac {\sqrt [4]{b} e \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{a^{3/4} \sqrt {a+b x^4}}-\frac {c \sqrt {a+b x^4}}{3 a x^3}-\frac {d \sqrt {a+b x^4}}{2 a x^2}-\frac {e \sqrt {a+b x^4}}{a x}+\frac {\sqrt {b} e x \sqrt {a+b x^4}}{a \left (\sqrt {a}+\sqrt {b} x^2\right )}-\frac {f \tanh ^{-1}\left (\frac {\sqrt {a+b x^4}}{\sqrt {a}}\right )}{2 \sqrt {a}} \]

[Out]

-1/2*f*arctanh((b*x^4+a)^(1/2)/a^(1/2))/a^(1/2)-1/3*c*(b*x^4+a)^(1/2)/a/x^3-1/2*d*(b*x^4+a)^(1/2)/a/x^2-e*(b*x
^4+a)^(1/2)/a/x+e*x*b^(1/2)*(b*x^4+a)^(1/2)/a/(a^(1/2)+x^2*b^(1/2))-b^(1/4)*e*(cos(2*arctan(b^(1/4)*x/a^(1/4))
)^2)^(1/2)/cos(2*arctan(b^(1/4)*x/a^(1/4)))*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)/a^(3/4)/(b*x^4+a)^(1/2)-1/6*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*e*a^(1/2)+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 = 0.26, antiderivative size = 323, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 10, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {1833, 1282, 1198, 220, 1196, 1252, 807, 266, 63, 208} \[ -\frac {\sqrt [4]{b} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \left (\sqrt {b} c-3 \sqrt {a} e\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{6 a^{5/4} \sqrt {a+b x^4}}-\frac {\sqrt [4]{b} e \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{a^{3/4} \sqrt {a+b x^4}}-\frac {c \sqrt {a+b x^4}}{3 a x^3}-\frac {d \sqrt {a+b x^4}}{2 a x^2}-\frac {e \sqrt {a+b x^4}}{a x}+\frac {\sqrt {b} e x \sqrt {a+b x^4}}{a \left (\sqrt {a}+\sqrt {b} x^2\right )}-\frac {f \tanh ^{-1}\left (\frac {\sqrt {a+b x^4}}{\sqrt {a}}\right )}{2 \sqrt {a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [C]  time = 0.17, size = 149, normalized size = 0.46 \[ \frac {-2 a c \sqrt {\frac {b x^4}{a}+1} \, _2F_1\left (-\frac {3}{4},\frac {1}{2};\frac {1}{4};-\frac {b x^4}{a}\right )-3 x \left (2 a e x \sqrt {\frac {b x^4}{a}+1} \, _2F_1\left (-\frac {1}{4},\frac {1}{2};\frac {3}{4};-\frac {b x^4}{a}\right )+\sqrt {a} f x^2 \sqrt {a+b x^4} \tanh ^{-1}\left (\frac {\sqrt {a+b x^4}}{\sqrt {a}}\right )+a d+b d x^4\right )}{6 a x^3 \sqrt {a+b x^4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.49, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {b x^{4} + a} {\left (f x^{3} + e x^{2} + d x + c\right )}}{b x^{8} + a x^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x^3+e*x^2+d*x+c)/x^4/(b*x^4+a)^(1/2),x, algorithm="fricas")

[Out]

integral(sqrt(b*x^4 + a)*(f*x^3 + e*x^2 + d*x + c)/(b*x^8 + a*x^4), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x^3+e*x^2+d*x+c)/x^4/(b*x^4+a)^(1/2),x, algorithm="giac")

[Out]

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

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maple [C]  time = 0.18, size = 316, normalized size = 0.98 \[ -\frac {i \sqrt {-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {b}\, e \EllipticE \left (\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, x , i\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}\, \sqrt {a}}+\frac {i \sqrt {-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {b}\, e \EllipticF \left (\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, x , i\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}\, \sqrt {a}}-\frac {\sqrt {-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, b c \EllipticF \left (\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, x , i\right )}{3 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}\, a}-\frac {f \ln \left (\frac {2 a +2 \sqrt {b \,x^{4}+a}\, \sqrt {a}}{x^{2}}\right )}{2 \sqrt {a}}-\frac {\sqrt {b \,x^{4}+a}\, e}{a x}-\frac {\sqrt {b \,x^{4}+a}\, d}{2 a \,x^{2}}-\frac {\sqrt {b \,x^{4}+a}\, c}{3 a \,x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*c*(b*x^4+a)^(1/2)/a/x^3-1/3*c/a*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*d*(b*x^4+a)^(1/2)/a/x^2-e*(b*
x^4+a)^(1/2)/a/x+I*e/a^(1/2)*b^(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)-I*e/a^(1/2)*b^(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)-1/2*f/a^(1/2)*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 \[ \int \frac {f x^{3} + e x^{2} + d x + c}{\sqrt {b x^{4} + a} x^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x^3+e*x^2+d*x+c)/x^4/(b*x^4+a)^(1/2),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [C]  time = 6.79, size = 131, normalized size = 0.41 \[ - \frac {\sqrt {b} d \sqrt {\frac {a}{b x^{4}} + 1}}{2 a} + \frac {c \Gamma \left (- \frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, \frac {1}{2} \\ \frac {1}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt {a} x^{3} \Gamma \left (\frac {1}{4}\right )} + \frac {e \Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {1}{2} \\ \frac {3}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt {a} x \Gamma \left (\frac {3}{4}\right )} - \frac {f \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x^{2}} \right )}}{2 \sqrt {a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-sqrt(b)*d*sqrt(a/(b*x**4) + 1)/(2*a) + c*gamma(-3/4)*hyper((-3/4, 1/2), (1/4,), b*x**4*exp_polar(I*pi)/a)/(4*
sqrt(a)*x**3*gamma(1/4)) + e*gamma(-1/4)*hyper((-1/4, 1/2), (3/4,), b*x**4*exp_polar(I*pi)/a)/(4*sqrt(a)*x*gam
ma(3/4)) - f*asinh(sqrt(a)/(sqrt(b)*x**2))/(2*sqrt(a))

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