3.6.43 \(\int \frac {x^3 (c+d x+e x^2+f x^3)}{(a+b x^4)^{3/2}} \, dx\) [543]

Optimal. Leaf size=302 \[ \frac {-c-d x-e x^2-f x^3}{2 b \sqrt {a+b x^4}}+\frac {3 f x \sqrt {a+b x^4}}{2 b^{3/2} \left (\sqrt {a}+\sqrt {b} x^2\right )}+\frac {e \tanh ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )}{2 b^{3/2}}-\frac {3 \sqrt [4]{a} f \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{2 b^{7/4} \sqrt {a+b x^4}}+\frac {\left (\sqrt {b} d+3 \sqrt {a} f\right ) \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{4 \sqrt [4]{a} b^{7/4} \sqrt {a+b x^4}} \]

[Out]

1/2*e*arctanh(x^2*b^(1/2)/(b*x^4+a)^(1/2))/b^(3/2)+1/2*(-f*x^3-e*x^2-d*x-c)/b/(b*x^4+a)^(1/2)+3/2*f*x*(b*x^4+a
)^(1/2)/b^(3/2)/(a^(1/2)+x^2*b^(1/2))-3/2*a^(1/4)*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)/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*arcta
n(b^(1/4)*x/a^(1/4)))*EllipticF(sin(2*arctan(b^(1/4)*x/a^(1/4))),1/2*2^(1/2))*(3*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^(1/4)/b^(7/4)/(b*x^4+a)^(1/2)

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Rubi [A]
time = 0.14, antiderivative size = 297, normalized size of antiderivative = 0.98, number of steps used = 9, number of rules used = 8, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {1837, 1899, 281, 223, 212, 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 (3 \sqrt {a} f+\sqrt {b} d\right ) F\left (2 \text {ArcTan}\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 {3 \sqrt [4]{a} 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 b^{7/4} \sqrt {a+b x^4}}+\frac {e \tanh ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )}{2 b^{3/2}}+\frac {3 f x \sqrt {a+b x^4}}{2 b^{3/2} \left (\sqrt {a}+\sqrt {b} x^2\right )}-\frac {c+d x+e x^2+f x^3}{2 b \sqrt {a+b x^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(c + d*x + e*x^2 + f*x^3)/(b*Sqrt[a + b*x^4]) + (3*f*x*Sqrt[a + b*x^4])/(2*b^(3/2)*(Sqrt[a] + Sqrt[b]*x^2
)) + (e*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a + b*x^4]])/(2*b^(3/2)) - (3*a^(1/4)*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*b^(7/4)*Sqrt[a + b*x^4]) +
((Sqrt[b]*d + 3*Sqrt[a]*f)*(Sqrt[a] + Sqrt[b]*x^2)*Sqrt[(a + b*x^4)/(Sqrt[a] + Sqrt[b]*x^2)^2]*EllipticF[2*Arc
Tan[(b^(1/4)*x)/a^(1/4)], 1/2])/(4*a^(1/4)*b^(7/4)*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 281

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

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 1837

Int[(Pq_)*(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[Pq*((a + b*x^n)^(p + 1)/(b*n*(p + 1))),
x] - Dist[1/(b*n*(p + 1)), Int[D[Pq, x]*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, m, n}, x] && PolyQ[Pq, x]
&& EqQ[m - n + 1, 0] && LtQ[p, -1]

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 {x^3 \left (c+d x+e x^2+f x^3\right )}{\left (a+b x^4\right )^{3/2}} \, dx &=-\frac {c+d x+e x^2+f x^3}{2 b \sqrt {a+b x^4}}+\frac {\int \frac {d+2 e x+3 f x^2}{\sqrt {a+b x^4}} \, dx}{2 b}\\ &=-\frac {c+d x+e x^2+f x^3}{2 b \sqrt {a+b x^4}}+\frac {\int \left (\frac {2 e x}{\sqrt {a+b x^4}}+\frac {d+3 f x^2}{\sqrt {a+b x^4}}\right ) \, dx}{2 b}\\ &=-\frac {c+d x+e x^2+f x^3}{2 b \sqrt {a+b x^4}}+\frac {\int \frac {d+3 f x^2}{\sqrt {a+b x^4}} \, dx}{2 b}+\frac {e \int \frac {x}{\sqrt {a+b x^4}} \, dx}{b}\\ &=-\frac {c+d x+e x^2+f x^3}{2 b \sqrt {a+b x^4}}+\frac {e \text {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,x^2\right )}{2 b}-\frac {\left (3 \sqrt {a} f\right ) \int \frac {1-\frac {\sqrt {b} x^2}{\sqrt {a}}}{\sqrt {a+b x^4}} \, dx}{2 b^{3/2}}+\frac {\left (d+\frac {3 \sqrt {a} f}{\sqrt {b}}\right ) \int \frac {1}{\sqrt {a+b x^4}} \, dx}{2 b}\\ &=-\frac {c+d x+e x^2+f x^3}{2 b \sqrt {a+b x^4}}+\frac {3 f x \sqrt {a+b x^4}}{2 b^{3/2} \left (\sqrt {a}+\sqrt {b} x^2\right )}-\frac {3 \sqrt [4]{a} 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 b^{7/4} \sqrt {a+b x^4}}+\frac {\left (\sqrt {b} d+3 \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 \sqrt [4]{a} b^{7/4} \sqrt {a+b x^4}}+\frac {e \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x^2}{\sqrt {a+b x^4}}\right )}{2 b}\\ &=-\frac {c+d x+e x^2+f x^3}{2 b \sqrt {a+b x^4}}+\frac {3 f x \sqrt {a+b x^4}}{2 b^{3/2} \left (\sqrt {a}+\sqrt {b} x^2\right )}+\frac {e \tanh ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )}{2 b^{3/2}}-\frac {3 \sqrt [4]{a} 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 b^{7/4} \sqrt {a+b x^4}}+\frac {\left (\sqrt {b} d+3 \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 \sqrt [4]{a} b^{7/4} \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 = 10.10, size = 181, normalized size = 0.60 \begin {gather*} \frac {-\sqrt {b} c-\sqrt {b} d x-\sqrt {b} e x^2+2 \sqrt {b} f x^3+\sqrt {a} e \sqrt {1+\frac {b x^4}{a}} \sinh ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )+\sqrt {b} d 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 )-2 \sqrt {b} f 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 )}{2 b^{3/2} \sqrt {a+b x^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C] Result contains complex when optimal does not.
time = 0.37, size = 275, normalized size = 0.91

method	result	size
elliptic	\(-\frac {2 b \left (\frac {f \,x^{3}}{4 b^{2}}+\frac {e \,x^{2}}{4 b^{2}}+\frac {d x}{4 b^{2}}+\frac {c}{4 b^{2}}\right )}{\sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {d \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \EllipticF \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{2 b \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}+\frac {e \ln \left (2 x^{2} \sqrt {b}+2 \sqrt {b \,x^{4}+a}\right )}{2 b^{\frac {3}{2}}}+\frac {3 i f \sqrt {a}\, \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\EllipticF \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\EllipticE \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{2 b^{\frac {3}{2}} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\)	\(248\)
default	\(f \left (-\frac {x^{3}}{2 b \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {3 i \sqrt {a}\, \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\EllipticF \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\EllipticE \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{2 b^{\frac {3}{2}} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )+e \left (-\frac {x^{2}}{2 b \sqrt {b \,x^{4}+a}}+\frac {\ln \left (x^{2} \sqrt {b}+\sqrt {b \,x^{4}+a}\right )}{2 b^{\frac {3}{2}}}\right )+d \left (-\frac {x}{2 b \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {\sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \EllipticF \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{2 b \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )-\frac {c}{2 b \sqrt {b \,x^{4}+a}}\)	\(275\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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Fricas [A]
time = 0.12, size = 215, normalized size = 0.71 \begin {gather*} \frac {6 \, {\left (a b f x^{5} + a^{2} f 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 (b^{2} d - 3 \, a b f\right )} x^{5} + {\left (a b d - 3 \, a^{2} 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) + {\left (a b e x^{5} + a^{2} e 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 \, a b f x^{4} - a b e x^{3} - a b d x^{2} - a b c x + 3 \, a^{2} f\right )} \sqrt {b x^{4} + a}}{4 \, {\left (a b^{3} x^{5} + a^{2} b^{2} x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 7.91, size = 156, normalized size = 0.52 \begin {gather*} c \left (\begin {cases} - \frac {1}{2 b \sqrt {a + b x^{4}}} & \text {for}\: b \neq 0 \\\frac {x^{4}}{4 a^{\frac {3}{2}}} & \text {otherwise} \end {cases}\right ) + e \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 ) + \frac {d 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 {f 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 )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c*Piecewise((-1/(2*b*sqrt(a + b*x**4)), Ne(b, 0)), (x**4/(4*a**(3/2)), True)) + e*(asinh(sqrt(b)*x**2/sqrt(a))
/(2*b**(3/2)) - x**2/(2*sqrt(a)*b*sqrt(1 + b*x**4/a))) + d*x**5*gamma(5/4)*hyper((5/4, 3/2), (9/4,), b*x**4*ex
p_polar(I*pi)/a)/(4*a**(3/2)*gamma(9/4)) + f*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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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^3*(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^3/(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^3\,\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^3*(c + d*x + e*x^2 + f*x^3))/(a + b*x^4)^(3/2),x)

[Out]

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

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