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

Optimal. Leaf size=314 \[ \frac {\left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \left (3 \sqrt {a} e+\sqrt {b} c\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{4 \sqrt [4]{a} b^{7/4} \sqrt {a+b x^4}}+\frac {d \tanh ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )}{2 b^{3/2}}+\frac {3 e x \sqrt {a+b x^4}}{2 b^{3/2} \left (\sqrt {a}+\sqrt {b} x^2\right )}-\frac {3 \sqrt [4]{a} e \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{2 b^{7/4} \sqrt {a+b x^4}}+\frac {f \sqrt {a+b x^4}}{b^2}-\frac {x \left (c+d x+e x^2+f x^3\right )}{2 b \sqrt {a+b x^4}} \]

[Out]

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

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Rubi [A]  time = 0.27, antiderivative size = 314, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {1828, 1885, 1198, 220, 1196, 1248, 641, 217, 206} \[ \frac {\left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \left (3 \sqrt {a} e+\sqrt {b} c\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{4 \sqrt [4]{a} b^{7/4} \sqrt {a+b x^4}}+\frac {d \tanh ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )}{2 b^{3/2}}+\frac {3 e x \sqrt {a+b x^4}}{2 b^{3/2} \left (\sqrt {a}+\sqrt {b} x^2\right )}-\frac {3 \sqrt [4]{a} e \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{2 b^{7/4} \sqrt {a+b x^4}}+\frac {f \sqrt {a+b x^4}}{b^2}-\frac {x \left (c+d x+e x^2+f x^3\right )}{2 b \sqrt {a+b x^4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 641

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 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 1248

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 1828

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

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Mathematica [C]  time = 0.16, size = 166, normalized size = 0.53 \[ \frac {b c x \sqrt {\frac {b x^4}{a}+1} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};-\frac {b x^4}{a}\right )+\sqrt {a} \sqrt {b} d \sqrt {\frac {b x^4}{a}+1} \sinh ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )-2 b e x^3 \sqrt {\frac {b x^4}{a}+1} \, _2F_1\left (\frac {3}{4},\frac {3}{2};\frac {7}{4};-\frac {b x^4}{a}\right )+2 a f-b c x-b d x^2+2 b e x^3+b f x^4}{2 b^2 \sqrt {a+b x^4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.50, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (f x^{7} + e x^{6} + d x^{5} + c x^{4}\right )} \sqrt {b x^{4} + a}}{b^{2} x^{8} + 2 \, a b x^{4} + a^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((f*x^7 + e*x^6 + d*x^5 + c*x^4)*sqrt(b*x^4 + a)/(b^2*x^8 + 2*a*b*x^4 + a^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.17, size = 340, normalized size = 1.08 \[ -\frac {e \,x^{3}}{2 \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}\, b}-\frac {d \,x^{2}}{2 \sqrt {b \,x^{4}+a}\, b}-\frac {3 i \sqrt {-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {a}\, e \EllipticE \left (\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, x , i\right )}{2 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}\, b^{\frac {3}{2}}}+\frac {3 i \sqrt {-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {a}\, e \EllipticF \left (\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, x , i\right )}{2 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}\, b^{\frac {3}{2}}}-\frac {c x}{2 \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}\, b}+\frac {\sqrt {-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, c \EllipticF \left (\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, x , i\right )}{2 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}\, b}+\frac {d \ln \left (\sqrt {b}\, x^{2}+\sqrt {b \,x^{4}+a}\right )}{2 b^{\frac {3}{2}}}+\frac {\left (b \,x^{4}+2 a \right ) f}{2 \sqrt {b \,x^{4}+a}\, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 26.11, size = 172, normalized size = 0.55 \[ d \left (\frac {\operatorname {asinh}{\left (\frac {\sqrt {b} x^{2}}{\sqrt {a}} \right )}}{2 b^{\frac {3}{2}}} - \frac {x^{2}}{2 \sqrt {a} b \sqrt {1 + \frac {b x^{4}}{a}}}\right ) + f \left (\begin {cases} \frac {a}{b^{2} \sqrt {a + b x^{4}}} + \frac {x^{4}}{2 b \sqrt {a + b x^{4}}} & \text {for}\: b \neq 0 \\\frac {x^{8}}{8 a^{\frac {3}{2}}} & \text {otherwise} \end {cases}\right ) + \frac {c x^{5} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{4}, \frac {3}{2} \\ \frac {9}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {3}{2}} \Gamma \left (\frac {9}{4}\right )} + \frac {e x^{7} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {3}{2}} \Gamma \left (\frac {11}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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