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

Optimal. Leaf size=309 \[ \frac{\left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} \left (\sqrt{a} e+\sqrt{b} c\right ) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac{1}{2}\right )}{2 a^{3/4} \sqrt [4]{b} \sqrt{a+b x^4}}-\frac{\sqrt [4]{b} c \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{a^{3/4} \sqrt{a+b x^4}}-\frac{c \sqrt{a+b x^4}}{a x}+\frac{\sqrt{b} c x \sqrt{a+b x^4}}{a \left (\sqrt{a}+\sqrt{b} x^2\right )}-\frac{d \tanh ^{-1}\left (\frac{\sqrt{a+b x^4}}{\sqrt{a}}\right )}{2 \sqrt{a}}+\frac{f \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a+b x^4}}\right )}{2 \sqrt{b}} \]

[Out]

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

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Rubi [A]  time = 0.224348, antiderivative size = 309, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 12, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {1833, 1282, 1198, 220, 1196, 1252, 844, 217, 206, 266, 63, 208} \[ \frac{\left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} \left (\sqrt{a} e+\sqrt{b} c\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{2 a^{3/4} \sqrt [4]{b} \sqrt{a+b x^4}}-\frac{\sqrt [4]{b} c \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{a^{3/4} \sqrt{a+b x^4}}-\frac{c \sqrt{a+b x^4}}{a x}+\frac{\sqrt{b} c x \sqrt{a+b x^4}}{a \left (\sqrt{a}+\sqrt{b} x^2\right )}-\frac{d \tanh ^{-1}\left (\frac{\sqrt{a+b x^4}}{\sqrt{a}}\right )}{2 \sqrt{a}}+\frac{f \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a+b x^4}}\right )}{2 \sqrt{b}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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)]

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 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]

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 844

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

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 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]

Rubi steps

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

Mathematica [C]  time = 0.227375, size = 157, normalized size = 0.51 \[ -\frac{c \sqrt{\frac{b x^4}{a}+1} \, _2F_1\left (-\frac{1}{4},\frac{1}{2};\frac{3}{4};-\frac{b x^4}{a}\right )}{x \sqrt{a+b x^4}}-\frac{d \tanh ^{-1}\left (\frac{\sqrt{a+b x^4}}{\sqrt{a}}\right )}{2 \sqrt{a}}+\frac{e x \sqrt{\frac{b x^4}{a}+1} \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{5}{4};-\frac{b x^4}{a}\right )}{\sqrt{a+b x^4}}+\frac{f \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a+b x^4}}\right )}{2 \sqrt{b}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(f*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a + b*x^4]])/(2*Sqrt[b]) - (d*ArcTanh[Sqrt[a + b*x^4]/Sqrt[a]])/(2*Sqrt[a]) - (c
*Sqrt[1 + (b*x^4)/a]*Hypergeometric2F1[-1/4, 1/2, 3/4, -((b*x^4)/a)])/(x*Sqrt[a + b*x^4]) + (e*x*Sqrt[1 + (b*x
^4)/a]*Hypergeometric2F1[1/4, 1/2, 5/4, -((b*x^4)/a)])/Sqrt[a + b*x^4]

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Maple [C]  time = 0.009, size = 299, normalized size = 1. \begin{align*}{\frac{f}{2}\ln \left ({x}^{2}\sqrt{b}+\sqrt{b{x}^{4}+a} \right ){\frac{1}{\sqrt{b}}}}+{e\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{d}{2}\ln \left ({\frac{1}{{x}^{2}} \left ( 2\,a+2\,\sqrt{a}\sqrt{b{x}^{4}+a} \right ) } \right ){\frac{1}{\sqrt{a}}}}-{\frac{c}{ax}\sqrt{b{x}^{4}+a}}+{ic\sqrt{b}\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{a}}}{\frac{1}{\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{b{x}^{4}+a}}}}-{ic\sqrt{b}\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{a}}}{\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)/x^2/(b*x^4+a)^(1/2),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{f x^{3} + e x^{2} + d x + c}{\sqrt{b x^{4} + a} x^{2}}\,{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)/x^2/(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^2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{b x^{4} + a}{\left (f x^{3} + e x^{2} + d x + c\right )}}{b x^{6} + a x^{2}}, 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)/x^2/(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^6 + a*x^2), x)

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Sympy [C]  time = 3.3097, size = 128, normalized size = 0.41 \begin{align*} \frac{f \operatorname{asinh}{\left (\frac{\sqrt{b} x^{2}}{\sqrt{a}} \right )}}{2 \sqrt{b}} + \frac{c \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{d \operatorname{asinh}{\left (\frac{\sqrt{a}}{\sqrt{b} x^{2}} \right )}}{2 \sqrt{a}} + \frac{e x \Gamma \left (\frac{1}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{1}{4}, \frac{1}{2} \\ \frac{5}{4} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt{a} \Gamma \left (\frac{5}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

f*asinh(sqrt(b)*x**2/sqrt(a))/(2*sqrt(b)) + c*gamma(-1/4)*hyper((-1/4, 1/2), (3/4,), b*x**4*exp_polar(I*pi)/a)
/(4*sqrt(a)*x*gamma(3/4)) - d*asinh(sqrt(a)/(sqrt(b)*x**2))/(2*sqrt(a)) + e*x*gamma(1/4)*hyper((1/4, 1/2), (5/
4,), b*x**4*exp_polar(I*pi)/a)/(4*sqrt(a)*gamma(5/4))

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