3.6.41 \(\int \frac {x^5 (c+d x+e x^2+f x^3)}{(a+b x^4)^{3/2}} \, dx\) [541]

Optimal. Leaf size=343 \[ \frac {x \left (a f-b c x-b d x^2-b e x^3\right )}{2 b^2 \sqrt {a+b x^4}}+\frac {e \sqrt {a+b x^4}}{b^2}+\frac {f x \sqrt {a+b x^4}}{3 b^2}+\frac {3 d x \sqrt {a+b x^4}}{2 b^{3/2} \left (\sqrt {a}+\sqrt {b} x^2\right )}+\frac {c \tanh ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )}{2 b^{3/2}}-\frac {3 \sqrt [4]{a} d \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{2 b^{7/4} \sqrt {a+b x^4}}+\frac {\sqrt [4]{a} \left (9 \sqrt {b} d-5 \sqrt {a} f\right ) \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{12 b^{9/4} \sqrt {a+b x^4}} \]

[Out]

1/2*c*arctanh(x^2*b^(1/2)/(b*x^4+a)^(1/2))/b^(3/2)+1/2*x*(-b*e*x^3-b*d*x^2-b*c*x+a*f)/b^2/(b*x^4+a)^(1/2)+e*(b
*x^4+a)^(1/2)/b^2+1/3*f*x*(b*x^4+a)^(1/2)/b^2+3/2*d*x*(b*x^4+a)^(1/2)/b^(3/2)/(a^(1/2)+x^2*b^(1/2))-3/2*a^(1/4
)*d*(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^(7/4)/(b*x^4+a)^(1
/2)+1/12*a^(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*a
rctan(b^(1/4)*x/a^(1/4))),1/2*2^(1/2))*(-5*f*a^(1/2)+9*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)/b^(9/4)/(b*x^4+a)^(1/2)

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Rubi [A]
time = 0.25, antiderivative size = 343, 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 = {1842, 1899, 1262, 655, 223, 212, 1902, 1212, 226, 1210} \begin {gather*} \frac {\sqrt [4]{a} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \left (9 \sqrt {b} d-5 \sqrt {a} f\right ) F\left (2 \text {ArcTan}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{12 b^{9/4} \sqrt {a+b x^4}}-\frac {3 \sqrt [4]{a} d \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 b^{7/4} \sqrt {a+b x^4}}+\frac {c \tanh ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )}{2 b^{3/2}}+\frac {3 d x \sqrt {a+b x^4}}{2 b^{3/2} \left (\sqrt {a}+\sqrt {b} x^2\right )}+\frac {x \left (a f-b c x-b d x^2-b e x^3\right )}{2 b^2 \sqrt {a+b x^4}}+\frac {e \sqrt {a+b x^4}}{b^2}+\frac {f x \sqrt {a+b x^4}}{3 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(a*f - b*c*x - b*d*x^2 - b*e*x^3))/(2*b^2*Sqrt[a + b*x^4]) + (e*Sqrt[a + b*x^4])/b^2 + (f*x*Sqrt[a + b*x^4]
)/(3*b^2) + (3*d*x*Sqrt[a + b*x^4])/(2*b^(3/2)*(Sqrt[a] + Sqrt[b]*x^2)) + (c*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a + b*
x^4]])/(2*b^(3/2)) - (3*a^(1/4)*d*(Sqrt[a] + Sqrt[b]*x^2)*Sqrt[(a + b*x^4)/(Sqrt[a] + Sqrt[b]*x^2)^2]*Elliptic
E[2*ArcTan[(b^(1/4)*x)/a^(1/4)], 1/2])/(2*b^(7/4)*Sqrt[a + b*x^4]) + (a^(1/4)*(9*Sqrt[b]*d - 5*Sqrt[a]*f)*(Sqr
t[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])
/(12*b^(9/4)*Sqrt[a + b*x^4])

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 223

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 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 655

Int[((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[e*((a + c*x^2)^(p + 1)/(2*c*(p + 1))),
x] + Dist[d, Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, p}, x] && NeQ[p, -1]

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 1262

Int[(x_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[(d + e*x)^q
*(a + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, c, d, e, p, q}, x]

Rule 1842

Int[(Pq_)*(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> With[{q = m + Expon[Pq, x]}, Module[{Q = Pol
ynomialQuotient[b^(Floor[(q - 1)/n] + 1)*x^m*Pq, a + b*x^n, x], R = PolynomialRemainder[b^(Floor[(q - 1)/n] +
1)*x^m*Pq, a + b*x^n, x]}, Dist[1/(a*n*(p + 1)*b^(Floor[(q - 1)/n] + 1)), Int[(a + b*x^n)^(p + 1)*ExpandToSum[
a*n*(p + 1)*Q + n*(p + 1)*R + D[x*R, x], x], x], x] + Simp[(-x)*R*((a + b*x^n)^(p + 1)/(a*n*(p + 1)*b^(Floor[(
q - 1)/n] + 1))), x]] /; GeQ[q, n]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && IGtQ[n, 0] && LtQ[p, -1] && IGtQ[m,
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)]

Rule 1902

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{q = Expon[Pq, x]}, With[{Pqq = Coeff[Pq, x, q]}, D
ist[1/(b*(q + n*p + 1)), Int[ExpandToSum[b*(q + n*p + 1)*(Pq - Pqq*x^q) - a*Pqq*(q - n + 1)*x^(q - n), x]*(a +
b*x^n)^p, x], x] + Simp[Pqq*x^(q - n + 1)*((a + b*x^n)^(p + 1)/(b*(q + n*p + 1))), x]] /; NeQ[q + n*p + 1, 0]
&& q - n >= 0 && (IntegerQ[2*p] || IntegerQ[p + (q + 1)/(2*n)])] /; FreeQ[{a, b, p}, x] && PolyQ[Pq, x] && IG
tQ[n, 0]

Rubi steps

\begin {align*} \int \frac {x^5 \left (c+d x+e x^2+f x^3\right )}{\left (a+b x^4\right )^{3/2}} \, dx &=\frac {x \left (a f-b c x-b d x^2-b e x^3\right )}{2 b^2 \sqrt {a+b x^4}}-\frac {\int \frac {a^2 f-2 a b c x-3 a b d x^2-4 a b e x^3-2 a b f x^4}{\sqrt {a+b x^4}} \, dx}{2 a b^2}\\ &=\frac {x \left (a f-b c x-b d x^2-b e x^3\right )}{2 b^2 \sqrt {a+b x^4}}-\frac {\int \left (\frac {x \left (-2 a b c-4 a b e x^2\right )}{\sqrt {a+b x^4}}+\frac {a^2 f-3 a b d x^2-2 a b f x^4}{\sqrt {a+b x^4}}\right ) \, dx}{2 a b^2}\\ &=\frac {x \left (a f-b c x-b d x^2-b e x^3\right )}{2 b^2 \sqrt {a+b x^4}}-\frac {\int \frac {x \left (-2 a b c-4 a b e x^2\right )}{\sqrt {a+b x^4}} \, dx}{2 a b^2}-\frac {\int \frac {a^2 f-3 a b d x^2-2 a b f x^4}{\sqrt {a+b x^4}} \, dx}{2 a b^2}\\ &=\frac {x \left (a f-b c x-b d x^2-b e x^3\right )}{2 b^2 \sqrt {a+b x^4}}+\frac {f x \sqrt {a+b x^4}}{3 b^2}-\frac {\int \frac {5 a^2 b f-9 a b^2 d x^2}{\sqrt {a+b x^4}} \, dx}{6 a b^3}-\frac {\text {Subst}\left (\int \frac {-2 a b c-4 a b e x}{\sqrt {a+b x^2}} \, dx,x,x^2\right )}{4 a b^2}\\ &=\frac {x \left (a f-b c x-b d x^2-b e x^3\right )}{2 b^2 \sqrt {a+b x^4}}+\frac {e \sqrt {a+b x^4}}{b^2}+\frac {f x \sqrt {a+b x^4}}{3 b^2}+\frac {c \text {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,x^2\right )}{2 b}-\frac {\left (3 \sqrt {a} d\right ) \int \frac {1-\frac {\sqrt {b} x^2}{\sqrt {a}}}{\sqrt {a+b x^4}} \, dx}{2 b^{3/2}}+\frac {\left (\sqrt {a} \left (9 \sqrt {b} d-5 \sqrt {a} f\right )\right ) \int \frac {1}{\sqrt {a+b x^4}} \, dx}{6 b^2}\\ &=\frac {x \left (a f-b c x-b d x^2-b e x^3\right )}{2 b^2 \sqrt {a+b x^4}}+\frac {e \sqrt {a+b x^4}}{b^2}+\frac {f x \sqrt {a+b x^4}}{3 b^2}+\frac {3 d x \sqrt {a+b x^4}}{2 b^{3/2} \left (\sqrt {a}+\sqrt {b} x^2\right )}-\frac {3 \sqrt [4]{a} d \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 b^{7/4} \sqrt {a+b x^4}}+\frac {\sqrt [4]{a} \left (9 \sqrt {b} d-5 \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 )}{12 b^{9/4} \sqrt {a+b x^4}}+\frac {c \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x^2}{\sqrt {a+b x^4}}\right )}{2 b}\\ &=\frac {x \left (a f-b c x-b d x^2-b e x^3\right )}{2 b^2 \sqrt {a+b x^4}}+\frac {e \sqrt {a+b x^4}}{b^2}+\frac {f x \sqrt {a+b x^4}}{3 b^2}+\frac {3 d x \sqrt {a+b x^4}}{2 b^{3/2} \left (\sqrt {a}+\sqrt {b} x^2\right )}+\frac {c \tanh ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )}{2 b^{3/2}}-\frac {3 \sqrt [4]{a} d \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 b^{7/4} \sqrt {a+b x^4}}+\frac {\sqrt [4]{a} \left (9 \sqrt {b} d-5 \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 )}{12 b^{9/4} \sqrt {a+b x^4}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 10.11, size = 176, normalized size = 0.51 \begin {gather*} \frac {6 a e+5 a f x-3 b c x^2+6 b d x^3+3 b e x^4+2 b f x^5+3 \sqrt {a} \sqrt {b} c \sqrt {1+\frac {b x^4}{a}} \sinh ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )-5 a f x \sqrt {1+\frac {b x^4}{a}} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};-\frac {b x^4}{a}\right )-6 b d x^3 \sqrt {1+\frac {b x^4}{a}} \, _2F_1\left (\frac {3}{4},\frac {3}{2};\frac {7}{4};-\frac {b x^4}{a}\right )}{6 b^2 \sqrt {a+b x^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(6*a*e + 5*a*f*x - 3*b*c*x^2 + 6*b*d*x^3 + 3*b*e*x^4 + 2*b*f*x^5 + 3*Sqrt[a]*Sqrt[b]*c*Sqrt[1 + (b*x^4)/a]*Arc
Sinh[(Sqrt[b]*x^2)/Sqrt[a]] - 5*a*f*x*Sqrt[1 + (b*x^4)/a]*Hypergeometric2F1[1/4, 1/2, 5/4, -((b*x^4)/a)] - 6*b
*d*x^3*Sqrt[1 + (b*x^4)/a]*Hypergeometric2F1[3/4, 3/2, 7/4, -((b*x^4)/a)])/(6*b^2*Sqrt[a + b*x^4])

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Maple [C] Result contains complex when optimal does not.
time = 0.41, size = 301, normalized size = 0.88

method	result	size
elliptic	\(-\frac {2 b \left (\frac {d \,x^{3}}{4 b^{2}}+\frac {x^{2} c}{4 b^{2}}-\frac {a f x}{4 b^{3}}-\frac {a e}{4 b^{3}}\right )}{\sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {f x \sqrt {b \,x^{4}+a}}{3 b^{2}}+\frac {e \sqrt {b \,x^{4}+a}}{2 b^{2}}-\frac {5 a f \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \EllipticF \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{6 b^{2} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}+\frac {c \ln \left (2 x^{2} \sqrt {b}+2 \sqrt {b \,x^{4}+a}\right )}{2 b^{\frac {3}{2}}}+\frac {3 i d \sqrt {a}\, \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\EllipticF \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\EllipticE \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{2 b^{\frac {3}{2}} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\)	\(282\)
default	\(f \left (\frac {a x}{2 b^{2} \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {x \sqrt {b \,x^{4}+a}}{3 b^{2}}-\frac {5 a \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \EllipticF \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{6 b^{2} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )+\frac {e \left (b \,x^{4}+2 a \right )}{2 \sqrt {b \,x^{4}+a}\, b^{2}}+d \left (-\frac {x^{3}}{2 b \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {3 i \sqrt {a}\, \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\EllipticF \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\EllipticE \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{2 b^{\frac {3}{2}} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )+c \left (-\frac {x^{2}}{2 b \sqrt {b \,x^{4}+a}}+\frac {\ln \left (x^{2} \sqrt {b}+\sqrt {b \,x^{4}+a}\right )}{2 b^{\frac {3}{2}}}\right )\)	\(301\)
risch	\(\frac {\left (2 f x +3 e \right ) \sqrt {b \,x^{4}+a}}{6 b^{2}}-\frac {d \,x^{3}}{2 b \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {3 i d \sqrt {a}\, \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \EllipticF \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{2 b^{\frac {3}{2}} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}-\frac {3 i d \sqrt {a}\, \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \EllipticE \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{2 b^{\frac {3}{2}} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}-\frac {c \,x^{2}}{2 b \sqrt {b \,x^{4}+a}}+\frac {c \ln \left (x^{2} \sqrt {b}+\sqrt {b \,x^{4}+a}\right )}{2 b^{\frac {3}{2}}}+\frac {a f x}{2 b^{2} \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}-\frac {5 a f \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \EllipticF \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{6 b^{2} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}+\frac {a e}{2 b^{2} \sqrt {b \,x^{4}+a}}\)	\(356\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

f*(1/2/b^2*a*x/((x^4+a/b)*b)^(1/2)+1/3*x*(b*x^4+a)^(1/2)/b^2-5/6*a/b^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))+1
/2*e*(b*x^4+2*a)/(b*x^4+a)^(1/2)/b^2+d*(-1/2/b*x^3/((x^4+a/b)*b)^(1/2)+3/2*I/b^(3/2)*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)-EllipticE(x*(I/a^(1/2)*b^(1/2))^(1/2),I)))+c*(-1/2*x^2/b/(b*x^4+a)^(1/2)+1/2/b^(3/2)*ln(x
^2*b^(1/2)+(b*x^4+a)^(1/2)))

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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(x^5*(f*x^3+e*x^2+d*x+c)/(b*x^4+a)^(3/2),x, algorithm="maxima")

[Out]

-1/4*c*(2*x^2/(sqrt(b*x^4 + a)*b) + log(-(sqrt(b) - sqrt(b*x^4 + a)/x^2)/(sqrt(b) + sqrt(b*x^4 + a)/x^2))/b^(3
/2)) + integrate((f*x^8 + x^7*e + d*x^6)/(b*x^4 + a)^(3/2), x)

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Fricas [A]
time = 0.12, size = 211, normalized size = 0.62 \begin {gather*} \frac {18 \, {\left (b d x^{5} + a d x\right )} \sqrt {b} \left (-\frac {a}{b}\right )^{\frac {3}{4}} E(\arcsin \left (\frac {\left (-\frac {a}{b}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) - 2 \, {\left ({\left (9 \, b d + 5 \, b f\right )} x^{5} + {\left (9 \, a d + 5 \, a f\right )} x\right )} \sqrt {b} \left (-\frac {a}{b}\right )^{\frac {3}{4}} F(\arcsin \left (\frac {\left (-\frac {a}{b}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) + 3 \, {\left (b c x^{5} + a c x\right )} \sqrt {b} \log \left (-2 \, b x^{4} - 2 \, \sqrt {b x^{4} + a} \sqrt {b} x^{2} - a\right ) + 2 \, {\left (2 \, b f x^{6} + 3 \, b e x^{5} + 6 \, b d x^{4} - 3 \, b c x^{3} + 5 \, a f x^{2} + 6 \, a e x + 9 \, a d\right )} \sqrt {b x^{4} + a}}{12 \, {\left (b^{3} x^{5} + a b^{2} x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(f*x^3+e*x^2+d*x+c)/(b*x^4+a)^(3/2),x, algorithm="fricas")

[Out]

1/12*(18*(b*d*x^5 + a*d*x)*sqrt(b)*(-a/b)^(3/4)*elliptic_e(arcsin((-a/b)^(1/4)/x), -1) - 2*((9*b*d + 5*b*f)*x^
5 + (9*a*d + 5*a*f)*x)*sqrt(b)*(-a/b)^(3/4)*elliptic_f(arcsin((-a/b)^(1/4)/x), -1) + 3*(b*c*x^5 + a*c*x)*sqrt(
b)*log(-2*b*x^4 - 2*sqrt(b*x^4 + a)*sqrt(b)*x^2 - a) + 2*(2*b*f*x^6 + 3*b*e*x^5 + 6*b*d*x^4 - 3*b*c*x^3 + 5*a*
f*x^2 + 6*a*e*x + 9*a*d)*sqrt(b*x^4 + a))/(b^3*x^5 + a*b^2*x)

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Sympy [A]
time = 10.82, size = 172, normalized size = 0.50 \begin {gather*} c \left (\frac {\operatorname {asinh}{\left (\frac {\sqrt {b} x^{2}}{\sqrt {a}} \right )}}{2 b^{\frac {3}{2}}} - \frac {x^{2}}{2 \sqrt {a} b \sqrt {1 + \frac {b x^{4}}{a}}}\right ) + e \left (\begin {cases} \frac {a}{b^{2} \sqrt {a + b x^{4}}} + \frac {x^{4}}{2 b \sqrt {a + b x^{4}}} & \text {for}\: b \neq 0 \\\frac {x^{8}}{8 a^{\frac {3}{2}}} & \text {otherwise} \end {cases}\right ) + \frac {d x^{7} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {3}{2}} \Gamma \left (\frac {11}{4}\right )} + \frac {f x^{9} \Gamma \left (\frac {9}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, \frac {9}{4} \\ \frac {13}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {3}{2}} \Gamma \left (\frac {13}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c*(asinh(sqrt(b)*x**2/sqrt(a))/(2*b**(3/2)) - x**2/(2*sqrt(a)*b*sqrt(1 + b*x**4/a))) + e*Piecewise((a/(b**2*sq
rt(a + b*x**4)) + x**4/(2*b*sqrt(a + b*x**4)), Ne(b, 0)), (x**8/(8*a**(3/2)), True)) + d*x**7*gamma(7/4)*hyper
((3/2, 7/4), (11/4,), b*x**4*exp_polar(I*pi)/a)/(4*a**(3/2)*gamma(11/4)) + f*x**9*gamma(9/4)*hyper((3/2, 9/4),
(13/4,), b*x**4*exp_polar(I*pi)/a)/(4*a**(3/2)*gamma(13/4))

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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(x^5*(f*x^3+e*x^2+d*x+c)/(b*x^4+a)^(3/2),x, algorithm="giac")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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