3.6.47 \(\int \frac {c+d x+e x^2+f x^3}{x (a+b x^4)^{3/2}} \, dx\) [547]

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

[Out]

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

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Rubi [A]
time = 0.20, antiderivative size = 323, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 10, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {1843, 1846, 272, 65, 214, 1899, 267, 1212, 226, 1210} \begin {gather*} \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 {b} d-\sqrt {a} f\right ) F\left (2 \text {ArcTan}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{4 a^{5/4} b^{3/4} \sqrt {a+b x^4}}+\frac {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 \text {ArcTan}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{2 a^{3/4} b^{3/4} \sqrt {a+b x^4}}-\frac {c \tanh ^{-1}\left (\frac {\sqrt {a+b x^4}}{\sqrt {a}}\right )}{2 a^{3/2}}+\frac {x \left (a d+a e x+a f x^2-b c x^3\right )}{2 a^2 \sqrt {a+b x^4}}+\frac {c \sqrt {a+b x^4}}{2 a^2}-\frac {f x \sqrt {a+b x^4}}{2 a \sqrt {b} \left (\sqrt {a}+\sqrt {b} x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 65

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 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 226

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)]/(2*q*Sqrt[a + b*x^4]))*EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 267

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rule 272

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 1210

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)]/(q*Sqrt[a + c*x^4
]))*EllipticE[2*ArcTan[q*x], 1/2], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

Rule 1212

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 1843

Int[(Pq_)*(x_)^(m_)*((a_) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> With[{q = Expon[Pq, x]}, Module[{Q = Polynomi
alQuotient[a*b^(Floor[(q - 1)/n] + 1)*x^m*Pq, a + b*x^n, x], R = PolynomialRemainder[a*b^(Floor[(q - 1)/n] + 1
)*x^m*Pq, a + b*x^n, x], i}, Dist[1/(a*n*(p + 1)*b^(Floor[(q - 1)/n] + 1)), Int[x^m*(a + b*x^n)^(p + 1)*Expand
ToSum[(n*(p + 1)*Q)/x^m + Sum[((n*(p + 1) + i + 1)/a)*Coeff[R, x, i]*x^(i - m), {i, 0, n - 1}], x], x], x] + S
imp[(-x)*R*((a + b*x^n)^(p + 1)/(a^2*n*(p + 1)*b^(Floor[(q - 1)/n] + 1))), x]]] /; FreeQ[{a, b}, x] && PolyQ[P
q, x] && IGtQ[n, 0] && LtQ[p, -1] && ILtQ[m, 0]

Rule 1846

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 1899

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

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Mathematica [C] Result contains complex when optimal does not.
time = 11.08, size = 225, normalized size = 0.70 \begin {gather*} \frac {\sqrt {a} b (c+x (d+x (e+f x)))-b c \sqrt {a+b x^4} \tanh ^{-1}\left (\frac {\sqrt {a+b x^4}}{\sqrt {a}}\right )+i a^{3/2} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}} f \sqrt {1+\frac {b x^4}{a}} E\left (\left .i \sinh ^{-1}\left (\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}} x\right )\right |-1\right )+\frac {b \left (\sqrt {b} d+i \sqrt {a} f\right ) \sqrt {1+\frac {b x^4}{a}} F\left (\left .i \sinh ^{-1}\left (\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}} x\right )\right |-1\right )}{\left (\frac {i \sqrt {b}}{\sqrt {a}}\right )^{3/2}}}{2 a^{3/2} b \sqrt {a+b x^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C] Result contains complex when optimal does not.
time = 0.37, size = 280, normalized size = 0.87

method	result	size
elliptic	\(-\frac {2 b \left (-\frac {f \,x^{3}}{4 a b}-\frac {x^{2} e}{4 a b}-\frac {d x}{4 a b}-\frac {c}{4 b a}\right )}{\sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {d \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \EllipticF \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{2 a \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}-\frac {i f \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\EllipticF \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\EllipticE \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{2 \sqrt {a}\, \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}\, \sqrt {b}}-\frac {c \arctanh \left (\frac {\sqrt {a}}{\sqrt {b \,x^{4}+a}}\right )}{2 a^{\frac {3}{2}}}\)	\(253\)
default	\(f \left (\frac {x^{3}}{2 a \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}-\frac {i \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\EllipticF \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\EllipticE \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{2 \sqrt {a}\, \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}\, \sqrt {b}}\right )+\frac {e \,x^{2}}{2 a \sqrt {b \,x^{4}+a}}+d \left (\frac {x}{2 a \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {\sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \EllipticF \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{2 a \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )+c \left (\frac {1}{2 a \sqrt {b \,x^{4}+a}}-\frac {\ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{4}+a}}{x^{2}}\right )}{2 a^{\frac {3}{2}}}\right )\)	\(280\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.13, size = 191, normalized size = 0.59 \begin {gather*} \frac {2 \, {\left (a b f x^{4} + a^{2} f\right )} \sqrt {a} \left (-\frac {b}{a}\right )^{\frac {3}{4}} E(\arcsin \left (x \left (-\frac {b}{a}\right )^{\frac {1}{4}}\right )\,|\,-1) - 2 \, {\left ({\left (a b d + a b f\right )} x^{4} + a^{2} d + a^{2} f\right )} \sqrt {a} \left (-\frac {b}{a}\right )^{\frac {3}{4}} F(\arcsin \left (x \left (-\frac {b}{a}\right )^{\frac {1}{4}}\right )\,|\,-1) + {\left (b^{2} c x^{4} + a b c\right )} \sqrt {a} \log \left (-\frac {b x^{4} - 2 \, \sqrt {b x^{4} + a} \sqrt {a} + 2 \, a}{x^{4}}\right ) + 2 \, {\left (a b f x^{3} + a b e x^{2} + a b d x + a b c\right )} \sqrt {b x^{4} + a}}{4 \, {\left (a^{2} b^{2} x^{4} + a^{3} b\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(2*(a*b*f*x^4 + a^2*f)*sqrt(a)*(-b/a)^(3/4)*elliptic_e(arcsin(x*(-b/a)^(1/4)), -1) - 2*((a*b*d + a*b*f)*x^
4 + a^2*d + a^2*f)*sqrt(a)*(-b/a)^(3/4)*elliptic_f(arcsin(x*(-b/a)^(1/4)), -1) + (b^2*c*x^4 + a*b*c)*sqrt(a)*l
og(-(b*x^4 - 2*sqrt(b*x^4 + a)*sqrt(a) + 2*a)/x^4) + 2*(a*b*f*x^3 + a*b*e*x^2 + a*b*d*x + a*b*c)*sqrt(b*x^4 +
a))/(a^2*b^2*x^4 + a^3*b)

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Sympy [C] Result contains complex when optimal does not.
time = 8.28, size = 289, normalized size = 0.89 \begin {gather*} c \left (\frac {2 a^{3} \sqrt {1 + \frac {b x^{4}}{a}}}{4 a^{\frac {9}{2}} + 4 a^{\frac {7}{2}} b x^{4}} + \frac {a^{3} \log {\left (\frac {b x^{4}}{a} \right )}}{4 a^{\frac {9}{2}} + 4 a^{\frac {7}{2}} b x^{4}} - \frac {2 a^{3} \log {\left (\sqrt {1 + \frac {b x^{4}}{a}} + 1 \right )}}{4 a^{\frac {9}{2}} + 4 a^{\frac {7}{2}} b x^{4}} + \frac {a^{2} b x^{4} \log {\left (\frac {b x^{4}}{a} \right )}}{4 a^{\frac {9}{2}} + 4 a^{\frac {7}{2}} b x^{4}} - \frac {2 a^{2} b x^{4} \log {\left (\sqrt {1 + \frac {b x^{4}}{a}} + 1 \right )}}{4 a^{\frac {9}{2}} + 4 a^{\frac {7}{2}} b x^{4}}\right ) + \frac {d x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {3}{2} \\ \frac {5}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {3}{2}} \Gamma \left (\frac {5}{4}\right )} + \frac {e x^{2}}{2 a^{\frac {3}{2}} \sqrt {1 + \frac {b x^{4}}{a}}} + \frac {f x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {3}{2} \\ \frac {7}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {3}{2}} \Gamma \left (\frac {7}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c*(2*a**3*sqrt(1 + b*x**4/a)/(4*a**(9/2) + 4*a**(7/2)*b*x**4) + a**3*log(b*x**4/a)/(4*a**(9/2) + 4*a**(7/2)*b*
x**4) - 2*a**3*log(sqrt(1 + b*x**4/a) + 1)/(4*a**(9/2) + 4*a**(7/2)*b*x**4) + a**2*b*x**4*log(b*x**4/a)/(4*a**
(9/2) + 4*a**(7/2)*b*x**4) - 2*a**2*b*x**4*log(sqrt(1 + b*x**4/a) + 1)/(4*a**(9/2) + 4*a**(7/2)*b*x**4)) + d*x
*gamma(1/4)*hyper((1/4, 3/2), (5/4,), b*x**4*exp_polar(I*pi)/a)/(4*a**(3/2)*gamma(5/4)) + e*x**2/(2*a**(3/2)*s
qrt(1 + b*x**4/a)) + f*x**3*gamma(3/4)*hyper((3/4, 3/2), (7/4,), b*x**4*exp_polar(I*pi)/a)/(4*a**(3/2)*gamma(7
/4))

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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