3.506 \(\int \frac {(c+d x+e x^2+f x^3) \sqrt {a+b x^4}}{x^7} \, dx\)

Optimal. Leaf size=352 \[ \frac {b^{3/4} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \left (5 \sqrt {a} f+3 \sqrt {b} d\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{15 a^{3/4} \sqrt {a+b x^4}}-\frac {2 b^{5/4} d \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{5 a^{3/4} \sqrt {a+b x^4}}+\frac {2 b^{3/2} d x \sqrt {a+b x^4}}{5 a \left (\sqrt {a}+\sqrt {b} x^2\right )}-\frac {1}{60} \sqrt {a+b x^4} \left (\frac {10 c}{x^6}+\frac {12 d}{x^5}+\frac {15 e}{x^4}+\frac {20 f}{x^3}\right )-\frac {b c \sqrt {a+b x^4}}{6 a x^2}-\frac {2 b d \sqrt {a+b x^4}}{5 a x}-\frac {b e \tanh ^{-1}\left (\frac {\sqrt {a+b x^4}}{\sqrt {a}}\right )}{4 \sqrt {a}} \]

[Out]

-1/4*b*e*arctanh((b*x^4+a)^(1/2)/a^(1/2))/a^(1/2)-1/60*(10*c/x^6+12*d/x^5+15*e/x^4+20*f/x^3)*(b*x^4+a)^(1/2)-1
/6*b*c*(b*x^4+a)^(1/2)/a/x^2-2/5*b*d*(b*x^4+a)^(1/2)/a/x+2/5*b^(3/2)*d*x*(b*x^4+a)^(1/2)/a/(a^(1/2)+x^2*b^(1/2
))-2/5*b^(5/4)*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*a
rctan(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*x^4+a)^(1/2)+1/15*b^(3/4)*(cos(2*arctan(b^(1/4)*x/a^(1/4)))^2)^(1/2)/cos(2*arctan(b^(1/4)*x/a^(1/4)))*Elli
pticF(sin(2*arctan(b^(1/4)*x/a^(1/4))),1/2*2^(1/2))*(5*f*a^(1/2)+3*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^(3/4)/(b*x^4+a)^(1/2)

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Rubi [A]  time = 0.32, antiderivative size = 352, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 12, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {14, 1825, 1833, 1252, 807, 266, 63, 208, 1282, 1198, 220, 1196} \[ \frac {b^{3/4} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \left (5 \sqrt {a} f+3 \sqrt {b} d\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{15 a^{3/4} \sqrt {a+b x^4}}-\frac {2 b^{5/4} d \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{5 a^{3/4} \sqrt {a+b x^4}}+\frac {2 b^{3/2} d x \sqrt {a+b x^4}}{5 a \left (\sqrt {a}+\sqrt {b} x^2\right )}-\frac {1}{60} \sqrt {a+b x^4} \left (\frac {10 c}{x^6}+\frac {12 d}{x^5}+\frac {15 e}{x^4}+\frac {20 f}{x^3}\right )-\frac {b c \sqrt {a+b x^4}}{6 a x^2}-\frac {2 b d \sqrt {a+b x^4}}{5 a x}-\frac {b e \tanh ^{-1}\left (\frac {\sqrt {a+b x^4}}{\sqrt {a}}\right )}{4 \sqrt {a}} \]

Antiderivative was successfully verified.

[In]

Int[((c + d*x + e*x^2 + f*x^3)*Sqrt[a + b*x^4])/x^7,x]

[Out]

-(((10*c)/x^6 + (12*d)/x^5 + (15*e)/x^4 + (20*f)/x^3)*Sqrt[a + b*x^4])/60 - (b*c*Sqrt[a + b*x^4])/(6*a*x^2) -
(2*b*d*Sqrt[a + b*x^4])/(5*a*x) + (2*b^(3/2)*d*x*Sqrt[a + b*x^4])/(5*a*(Sqrt[a] + Sqrt[b]*x^2)) - (b*e*ArcTanh
[Sqrt[a + b*x^4]/Sqrt[a]])/(4*Sqrt[a]) - (2*b^(5/4)*d*(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])/(5*a^(3/4)*Sqrt[a + b*x^4]) + (b^(3/4)*(3*Sqrt[b]*d
+ 5*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])/(15*a^(3/4)*Sqrt[a + b*x^4])

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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]

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 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 807

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Simp[((e*f - d*g
)*(d + e*x)^(m + 1)*(a + c*x^2)^(p + 1))/(2*(p + 1)*(c*d^2 + a*e^2)), x] + Dist[(c*d*f + a*e*g)/(c*d^2 + a*e^2
), Int[(d + e*x)^(m + 1)*(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]
&& EqQ[Simplify[m + 2*p + 3], 0]

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 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 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 1825

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

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)]

Rubi steps

\begin {align*} \int \frac {\left (c+d x+e x^2+f x^3\right ) \sqrt {a+b x^4}}{x^7} \, dx &=-\frac {1}{60} \left (\frac {10 c}{x^6}+\frac {12 d}{x^5}+\frac {15 e}{x^4}+\frac {20 f}{x^3}\right ) \sqrt {a+b x^4}-(2 b) \int \frac {-\frac {c}{6}-\frac {d x}{5}-\frac {e x^2}{4}-\frac {f x^3}{3}}{x^3 \sqrt {a+b x^4}} \, dx\\ &=-\frac {1}{60} \left (\frac {10 c}{x^6}+\frac {12 d}{x^5}+\frac {15 e}{x^4}+\frac {20 f}{x^3}\right ) \sqrt {a+b x^4}-(2 b) \int \left (\frac {-\frac {c}{6}-\frac {e x^2}{4}}{x^3 \sqrt {a+b x^4}}+\frac {-\frac {d}{5}-\frac {f x^2}{3}}{x^2 \sqrt {a+b x^4}}\right ) \, dx\\ &=-\frac {1}{60} \left (\frac {10 c}{x^6}+\frac {12 d}{x^5}+\frac {15 e}{x^4}+\frac {20 f}{x^3}\right ) \sqrt {a+b x^4}-(2 b) \int \frac {-\frac {c}{6}-\frac {e x^2}{4}}{x^3 \sqrt {a+b x^4}} \, dx-(2 b) \int \frac {-\frac {d}{5}-\frac {f x^2}{3}}{x^2 \sqrt {a+b x^4}} \, dx\\ &=-\frac {1}{60} \left (\frac {10 c}{x^6}+\frac {12 d}{x^5}+\frac {15 e}{x^4}+\frac {20 f}{x^3}\right ) \sqrt {a+b x^4}-\frac {2 b d \sqrt {a+b x^4}}{5 a x}-b \operatorname {Subst}\left (\int \frac {-\frac {c}{6}-\frac {e x}{4}}{x^2 \sqrt {a+b x^2}} \, dx,x,x^2\right )+\frac {(2 b) \int \frac {\frac {a f}{3}+\frac {1}{5} b d x^2}{\sqrt {a+b x^4}} \, dx}{a}\\ &=-\frac {1}{60} \left (\frac {10 c}{x^6}+\frac {12 d}{x^5}+\frac {15 e}{x^4}+\frac {20 f}{x^3}\right ) \sqrt {a+b x^4}-\frac {b c \sqrt {a+b x^4}}{6 a x^2}-\frac {2 b d \sqrt {a+b x^4}}{5 a x}-\frac {\left (2 b^{3/2} d\right ) \int \frac {1-\frac {\sqrt {b} x^2}{\sqrt {a}}}{\sqrt {a+b x^4}} \, dx}{5 \sqrt {a}}+\frac {1}{4} (b e) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+b x^2}} \, dx,x,x^2\right )+\frac {1}{15} \left (2 b \left (\frac {3 \sqrt {b} d}{\sqrt {a}}+5 f\right )\right ) \int \frac {1}{\sqrt {a+b x^4}} \, dx\\ &=-\frac {1}{60} \left (\frac {10 c}{x^6}+\frac {12 d}{x^5}+\frac {15 e}{x^4}+\frac {20 f}{x^3}\right ) \sqrt {a+b x^4}-\frac {b c \sqrt {a+b x^4}}{6 a x^2}-\frac {2 b d \sqrt {a+b x^4}}{5 a x}+\frac {2 b^{3/2} d x \sqrt {a+b x^4}}{5 a \left (\sqrt {a}+\sqrt {b} x^2\right )}-\frac {2 b^{5/4} 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 )}{5 a^{3/4} \sqrt {a+b x^4}}+\frac {b^{3/4} \left (3 \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 )}{15 a^{3/4} \sqrt {a+b x^4}}+\frac {1}{8} (b e) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,x^4\right )\\ &=-\frac {1}{60} \left (\frac {10 c}{x^6}+\frac {12 d}{x^5}+\frac {15 e}{x^4}+\frac {20 f}{x^3}\right ) \sqrt {a+b x^4}-\frac {b c \sqrt {a+b x^4}}{6 a x^2}-\frac {2 b d \sqrt {a+b x^4}}{5 a x}+\frac {2 b^{3/2} d x \sqrt {a+b x^4}}{5 a \left (\sqrt {a}+\sqrt {b} x^2\right )}-\frac {2 b^{5/4} 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 )}{5 a^{3/4} \sqrt {a+b x^4}}+\frac {b^{3/4} \left (3 \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 )}{15 a^{3/4} \sqrt {a+b x^4}}+\frac {1}{4} e \operatorname {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b x^4}\right )\\ &=-\frac {1}{60} \left (\frac {10 c}{x^6}+\frac {12 d}{x^5}+\frac {15 e}{x^4}+\frac {20 f}{x^3}\right ) \sqrt {a+b x^4}-\frac {b c \sqrt {a+b x^4}}{6 a x^2}-\frac {2 b d \sqrt {a+b x^4}}{5 a x}+\frac {2 b^{3/2} d x \sqrt {a+b x^4}}{5 a \left (\sqrt {a}+\sqrt {b} x^2\right )}-\frac {b e \tanh ^{-1}\left (\frac {\sqrt {a+b x^4}}{\sqrt {a}}\right )}{4 \sqrt {a}}-\frac {2 b^{5/4} 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 )}{5 a^{3/4} \sqrt {a+b x^4}}+\frac {b^{3/4} \left (3 \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 )}{15 a^{3/4} \sqrt {a+b x^4}}\\ \end {align*}

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Mathematica [C]  time = 0.27, size = 145, normalized size = 0.41 \[ -\frac {\sqrt {a+b x^4} \left (5 \left (\sqrt {\frac {b x^4}{a}+1} \left (2 a c+3 a e x^2+2 b c x^4\right )+3 b e x^6 \tanh ^{-1}\left (\sqrt {\frac {b x^4}{a}+1}\right )+4 a f x^3 \, _2F_1\left (-\frac {3}{4},-\frac {1}{2};\frac {1}{4};-\frac {b x^4}{a}\right )\right )+12 a d x \, _2F_1\left (-\frac {5}{4},-\frac {1}{2};-\frac {1}{4};-\frac {b x^4}{a}\right )\right )}{60 a x^6 \sqrt {\frac {b x^4}{a}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[((c + d*x + e*x^2 + f*x^3)*Sqrt[a + b*x^4])/x^7,x]

[Out]

-1/60*(Sqrt[a + b*x^4]*(12*a*d*x*Hypergeometric2F1[-5/4, -1/2, -1/4, -((b*x^4)/a)] + 5*(Sqrt[1 + (b*x^4)/a]*(2
*a*c + 3*a*e*x^2 + 2*b*c*x^4) + 3*b*e*x^6*ArcTanh[Sqrt[1 + (b*x^4)/a]] + 4*a*f*x^3*Hypergeometric2F1[-3/4, -1/
2, 1/4, -((b*x^4)/a)])))/(a*x^6*Sqrt[1 + (b*x^4)/a])

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fricas [F]  time = 0.50, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {b x^{4} + a} {\left (f x^{3} + e x^{2} + d x + c\right )}}{x^{7}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x^3+e*x^2+d*x+c)*(b*x^4+a)^(1/2)/x^7,x, algorithm="fricas")

[Out]

integral(sqrt(b*x^4 + a)*(f*x^3 + e*x^2 + d*x + c)/x^7, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {b x^{4} + a} {\left (f x^{3} + e x^{2} + d x + c\right )}}{x^{7}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x^3+e*x^2+d*x+c)*(b*x^4+a)^(1/2)/x^7,x, algorithm="giac")

[Out]

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

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maple [C]  time = 0.19, size = 361, normalized size = 1.03 \[ -\frac {2 i \sqrt {-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, b^{\frac {3}{2}} d \EllipticE \left (\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, x , i\right )}{5 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}\, \sqrt {a}}+\frac {2 i \sqrt {-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, b^{\frac {3}{2}} d \EllipticF \left (\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, x , i\right )}{5 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}\, \sqrt {a}}+\frac {2 \sqrt {-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, b f \EllipticF \left (\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, x , i\right )}{3 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}-\frac {b e \ln \left (\frac {2 a +2 \sqrt {b \,x^{4}+a}\, \sqrt {a}}{x^{2}}\right )}{4 \sqrt {a}}+\frac {\sqrt {b \,x^{4}+a}\, b e}{4 a}-\frac {2 \sqrt {b \,x^{4}+a}\, b d}{5 a x}-\frac {\sqrt {b \,x^{4}+a}\, f}{3 x^{3}}-\frac {\left (b \,x^{4}+a \right )^{\frac {3}{2}} e}{4 a \,x^{4}}-\frac {\sqrt {b \,x^{4}+a}\, d}{5 x^{5}}-\frac {\left (b \,x^{4}+a \right )^{\frac {3}{2}} c}{6 a \,x^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {{\left (b x^{4} + a\right )}^{\frac {3}{2}} c}{6 \, a x^{6}} + \int \frac {\sqrt {b x^{4} + a} {\left (f x^{2} + e x + d\right )}}{x^{6}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x^3+e*x^2+d*x+c)*(b*x^4+a)^(1/2)/x^7,x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\sqrt {b\,x^4+a}\,\left (f\,x^3+e\,x^2+d\,x+c\right )}{x^7} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [C]  time = 7.04, size = 189, normalized size = 0.54 \[ \frac {\sqrt {a} d \Gamma \left (- \frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{4}, - \frac {1}{2} \\ - \frac {1}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 x^{5} \Gamma \left (- \frac {1}{4}\right )} + \frac {\sqrt {a} f \Gamma \left (- \frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, - \frac {1}{2} \\ \frac {1}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 x^{3} \Gamma \left (\frac {1}{4}\right )} - \frac {\sqrt {b} c \sqrt {\frac {a}{b x^{4}} + 1}}{6 x^{4}} - \frac {\sqrt {b} e \sqrt {\frac {a}{b x^{4}} + 1}}{4 x^{2}} - \frac {b^{\frac {3}{2}} c \sqrt {\frac {a}{b x^{4}} + 1}}{6 a} - \frac {b e \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x^{2}} \right )}}{4 \sqrt {a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x**3+e*x**2+d*x+c)*(b*x**4+a)**(1/2)/x**7,x)

[Out]

sqrt(a)*d*gamma(-5/4)*hyper((-5/4, -1/2), (-1/4,), b*x**4*exp_polar(I*pi)/a)/(4*x**5*gamma(-1/4)) + sqrt(a)*f*
gamma(-3/4)*hyper((-3/4, -1/2), (1/4,), b*x**4*exp_polar(I*pi)/a)/(4*x**3*gamma(1/4)) - sqrt(b)*c*sqrt(a/(b*x*
*4) + 1)/(6*x**4) - sqrt(b)*e*sqrt(a/(b*x**4) + 1)/(4*x**2) - b**(3/2)*c*sqrt(a/(b*x**4) + 1)/(6*a) - b*e*asin
h(sqrt(a)/(sqrt(b)*x**2))/(4*sqrt(a))

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