Integrand size = 30, antiderivative size = 343 \[ \int \frac {\left (c+d x+e x^2+f x^3\right ) \sqrt {a+b x^4}}{x^5} \, dx=-\frac {2 \sqrt {a} f \sqrt {a+b x^4}}{x \left (\sqrt {a}+\sqrt {b} x^2\right )}-\frac {\left (c+2 e x^2\right ) \sqrt {a+b x^4}}{4 x^4}-\frac {\left (d-3 f x^2\right ) \sqrt {a+b x^4}}{3 x^3}+\frac {1}{2} \sqrt {b} e \text {arctanh}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )-\frac {b c \text {arctanh}\left (\frac {\sqrt {a+b x^4}}{\sqrt {a}}\right )}{4 \sqrt {a}}-\frac {2 \sqrt [4]{a} \sqrt [4]{b} f \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{\sqrt {a+b x^4}}+\frac {\sqrt [4]{b} \left (\sqrt {b} d+3 \sqrt {a} f\right ) \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{3 \sqrt [4]{a} \sqrt {a+b x^4}} \] Output:
-2*a^(1/2)*f*(b*x^4+a)^(1/2)/x/(a^(1/2)+b^(1/2)*x^2)-1/4*(2*e*x^2+c)*(b*x^ 4+a)^(1/2)/x^4-1/3*(-3*f*x^2+d)*(b*x^4+a)^(1/2)/x^3+1/2*b^(1/2)*e*arctanh( b^(1/2)*x^2/(b*x^4+a)^(1/2))-1/4*b*c*arctanh((b*x^4+a)^(1/2)/a^(1/2))/a^(1 /2)-2*a^(1/4)*b^(1/4)*f*(a^(1/2)+b^(1/2)*x^2)*((b*x^4+a)/(a^(1/2)+b^(1/2)* x^2)^2)^(1/2)*EllipticE(sin(2*arctan(b^(1/4)*x/a^(1/4))),1/2*2^(1/2))/(b*x ^4+a)^(1/2)+1/3*b^(1/4)*(b^(1/2)*d+3*a^(1/2)*f)*(a^(1/2)+b^(1/2)*x^2)*((b* x^4+a)/(a^(1/2)+b^(1/2)*x^2)^2)^(1/2)*InverseJacobiAM(2*arctan(b^(1/4)*x/a ^(1/4)),1/2*2^(1/2))/a^(1/4)/(b*x^4+a)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.16 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.51 \[ \int \frac {\left (c+d x+e x^2+f x^3\right ) \sqrt {a+b x^4}}{x^5} \, dx=-\frac {\sqrt {1+\frac {b x^4}{a}} \left (3 a c \sqrt {1+\frac {b x^4}{a}}+6 a e x^2 \sqrt {1+\frac {b x^4}{a}}-6 \sqrt {a} \sqrt {b} e x^4 \text {arcsinh}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )+3 b c x^4 \text {arctanh}\left (\sqrt {1+\frac {b x^4}{a}}\right )+4 a d x \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},-\frac {1}{2},\frac {1}{4},-\frac {b x^4}{a}\right )+12 a f x^3 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-\frac {1}{4},\frac {3}{4},-\frac {b x^4}{a}\right )\right )}{12 x^4 \sqrt {a+b x^4}} \] Input:
Integrate[((c + d*x + e*x^2 + f*x^3)*Sqrt[a + b*x^4])/x^5,x]
Output:
-1/12*(Sqrt[1 + (b*x^4)/a]*(3*a*c*Sqrt[1 + (b*x^4)/a] + 6*a*e*x^2*Sqrt[1 + (b*x^4)/a] - 6*Sqrt[a]*Sqrt[b]*e*x^4*ArcSinh[(Sqrt[b]*x^2)/Sqrt[a]] + 3*b *c*x^4*ArcTanh[Sqrt[1 + (b*x^4)/a]] + 4*a*d*x*Hypergeometric2F1[-3/4, -1/2 , 1/4, -((b*x^4)/a)] + 12*a*f*x^3*Hypergeometric2F1[-1/2, -1/4, 3/4, -((b* x^4)/a)]))/(x^4*Sqrt[a + b*x^4])
Time = 1.09 (sec) , antiderivative size = 330, normalized size of antiderivative = 0.96, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {2364, 27, 2371, 798, 73, 221, 2424, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {\sqrt {a+b x^4} \left (c+d x+e x^2+f x^3\right )}{x^5} \, dx\) |
| \(\Big \downarrow \) 2364 |
| \(\displaystyle -2 b \int -\frac {12 f x^3+6 e x^2+4 d x+3 c}{12 x \sqrt {b x^4+a}}dx-\frac {1}{12} \sqrt {a+b x^4} \left (\frac {3 c}{x^4}+\frac {4 d}{x^3}+\frac {6 e}{x^2}+\frac {12 f}{x}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{6} b \int \frac {12 f x^3+6 e x^2+4 d x+3 c}{x \sqrt {b x^4+a}}dx-\frac {1}{12} \sqrt {a+b x^4} \left (\frac {3 c}{x^4}+\frac {4 d}{x^3}+\frac {6 e}{x^2}+\frac {12 f}{x}\right )\) |
| \(\Big \downarrow \) 2371 |
| \(\displaystyle \frac {1}{6} b \left (3 c \int \frac {1}{x \sqrt {b x^4+a}}dx+\int \frac {12 f x^2+6 e x+4 d}{\sqrt {b x^4+a}}dx\right )-\frac {1}{12} \sqrt {a+b x^4} \left (\frac {3 c}{x^4}+\frac {4 d}{x^3}+\frac {6 e}{x^2}+\frac {12 f}{x}\right )\) |
| \(\Big \downarrow \) 798 |
| \(\displaystyle \frac {1}{6} b \left (\frac {3}{4} c \int \frac {1}{x^4 \sqrt {b x^4+a}}dx^4+\int \frac {12 f x^2+6 e x+4 d}{\sqrt {b x^4+a}}dx\right )-\frac {1}{12} \sqrt {a+b x^4} \left (\frac {3 c}{x^4}+\frac {4 d}{x^3}+\frac {6 e}{x^2}+\frac {12 f}{x}\right )\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {1}{6} b \left (\frac {3 c \int \frac {1}{\frac {x^8}{b}-\frac {a}{b}}d\sqrt {b x^4+a}}{2 b}+\int \frac {12 f x^2+6 e x+4 d}{\sqrt {b x^4+a}}dx\right )-\frac {1}{12} \sqrt {a+b x^4} \left (\frac {3 c}{x^4}+\frac {4 d}{x^3}+\frac {6 e}{x^2}+\frac {12 f}{x}\right )\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {1}{6} b \left (\int \frac {12 f x^2+6 e x+4 d}{\sqrt {b x^4+a}}dx-\frac {3 c \text {arctanh}\left (\frac {\sqrt {a+b x^4}}{\sqrt {a}}\right )}{2 \sqrt {a}}\right )-\frac {1}{12} \sqrt {a+b x^4} \left (\frac {3 c}{x^4}+\frac {4 d}{x^3}+\frac {6 e}{x^2}+\frac {12 f}{x}\right )\) |
| \(\Big \downarrow \) 2424 |
| \(\displaystyle \frac {1}{6} b \left (\int \left (\frac {6 e x}{\sqrt {b x^4+a}}+\frac {12 f x^2+4 d}{\sqrt {b x^4+a}}\right )dx-\frac {3 c \text {arctanh}\left (\frac {\sqrt {a+b x^4}}{\sqrt {a}}\right )}{2 \sqrt {a}}\right )-\frac {1}{12} \sqrt {a+b x^4} \left (\frac {3 c}{x^4}+\frac {4 d}{x^3}+\frac {6 e}{x^2}+\frac {12 f}{x}\right )\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{6} b \left (\frac {2 \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \left (3 \sqrt {a} f+\sqrt {b} d\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{\sqrt [4]{a} b^{3/4} \sqrt {a+b x^4}}-\frac {12 \sqrt [4]{a} f \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{b^{3/4} \sqrt {a+b x^4}}-\frac {3 c \text {arctanh}\left (\frac {\sqrt {a+b x^4}}{\sqrt {a}}\right )}{2 \sqrt {a}}+\frac {3 e \text {arctanh}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )}{\sqrt {b}}+\frac {12 f x \sqrt {a+b x^4}}{\sqrt {b} \left (\sqrt {a}+\sqrt {b} x^2\right )}\right )-\frac {1}{12} \sqrt {a+b x^4} \left (\frac {3 c}{x^4}+\frac {4 d}{x^3}+\frac {6 e}{x^2}+\frac {12 f}{x}\right )\) |
Input:
Int[((c + d*x + e*x^2 + f*x^3)*Sqrt[a + b*x^4])/x^5,x]
Output:
-1/12*(((3*c)/x^4 + (4*d)/x^3 + (6*e)/x^2 + (12*f)/x)*Sqrt[a + b*x^4]) + ( b*((12*f*x*Sqrt[a + b*x^4])/(Sqrt[b]*(Sqrt[a] + Sqrt[b]*x^2)) + (3*e*ArcTa nh[(Sqrt[b]*x^2)/Sqrt[a + b*x^4]])/Sqrt[b] - (3*c*ArcTanh[Sqrt[a + b*x^4]/ Sqrt[a]])/(2*Sqrt[a]) - (12*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])/(b^(3/4)*Sqrt[a + b*x^4]) + (2*(Sqrt[b]*d + 3*Sqrt[a]*f)*(Sqrt[a] + S qrt[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])/(a^(1/4)*b^(3/4)*Sqrt[a + b*x^4])))/6
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[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] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
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]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[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]]
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] - Simp[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]
Int[(Pq_)/((x_)*Sqrt[(a_) + (b_.)*(x_)^(n_)]), x_Symbol] :> Simp[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] && IG tQ[n, 0] && NeQ[Coeff[Pq, x, 0], 0]
Int[(Pq_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Module[{q = Expon[Pq, x], j, k}, Int[Sum[x^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, {j, 0, n/2 - 1}], x]] /; FreeQ[{a, b, p}, x] && PolyQ[Pq, x] && IGtQ[n/2, 0] && !PolyQ[Pq, x^(n/2)]
Result contains complex when optimal does not.
Time = 1.63 (sec) , antiderivative size = 261, normalized size of antiderivative = 0.76
| method | result | size |
| risch | \(-\frac {\sqrt {b \,x^{4}+a}\, \left (12 f \,x^{3}+6 e \,x^{2}+4 d x +3 c \right )}{12 x^{4}}+\frac {b \left (\frac {4 d \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}-\frac {3 c \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{4}+a}}{x^{2}}\right )}{2 \sqrt {a}}+\frac {3 e \ln \left (\sqrt {b}\, x^{2}+\sqrt {b \,x^{4}+a}\right )}{\sqrt {b}}+\frac {12 i f \sqrt {a}\, \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}\, \sqrt {b}}\right )}{6}\) | \(261\) |
| elliptic | \(-\frac {c \sqrt {b \,x^{4}+a}}{4 x^{4}}-\frac {d \sqrt {b \,x^{4}+a}}{3 x^{3}}-\frac {e \sqrt {b \,x^{4}+a}}{2 x^{2}}-\frac {f \sqrt {b \,x^{4}+a}}{x}+\frac {2 b d \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{3 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}+\frac {\sqrt {b}\, e \ln \left (2 \sqrt {b}\, x^{2}+2 \sqrt {b \,x^{4}+a}\right )}{2}+\frac {2 i f \sqrt {b}\, \sqrt {a}\, \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}-\frac {c b \,\operatorname {arctanh}\left (\frac {\sqrt {a}}{\sqrt {b \,x^{4}+a}}\right )}{4 \sqrt {a}}\) | \(279\) |
| default | \(c \left (-\frac {\left (b \,x^{4}+a \right )^{\frac {3}{2}}}{4 a \,x^{4}}-\frac {b \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{4}+a}}{x^{2}}\right )}{4 \sqrt {a}}+\frac {b \sqrt {b \,x^{4}+a}}{4 a}\right )+d \left (-\frac {\sqrt {b \,x^{4}+a}}{3 x^{3}}+\frac {2 b \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{3 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )+e \left (-\frac {\left (b \,x^{4}+a \right )^{\frac {3}{2}}}{2 a \,x^{2}}+\frac {b \,x^{2} \sqrt {b \,x^{4}+a}}{2 a}+\frac {\sqrt {b}\, \ln \left (\sqrt {b}\, x^{2}+\sqrt {b \,x^{4}+a}\right )}{2}\right )+f \left (-\frac {\sqrt {b \,x^{4}+a}}{x}+\frac {2 i \sqrt {b}\, \sqrt {a}\, \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )\) | \(328\) |
Input:
int((f*x^3+e*x^2+d*x+c)*(b*x^4+a)^(1/2)/x^5,x,method=_RETURNVERBOSE)
Output:
-1/12*(b*x^4+a)^(1/2)*(12*f*x^3+6*e*x^2+4*d*x+3*c)/x^4+1/6*b*(4*d/(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)-3/2*c/a^( 1/2)*ln((2*a+2*a^(1/2)*(b*x^4+a)^(1/2))/x^2)+3*e*ln(b^(1/2)*x^2+(b*x^4+a)^ (1/2))/b^(1/2)+12*I*f*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)*(Ell ipticF(x*(I/a^(1/2)*b^(1/2))^(1/2),I)-EllipticE(x*(I/a^(1/2)*b^(1/2))^(1/2 ),I)))
\[ \int \frac {\left (c+d x+e x^2+f x^3\right ) \sqrt {a+b x^4}}{x^5} \, dx=\int { \frac {\sqrt {b x^{4} + a} {\left (f x^{3} + e x^{2} + d x + c\right )}}{x^{5}} \,d x } \] Input:
integrate((f*x^3+e*x^2+d*x+c)*(b*x^4+a)^(1/2)/x^5,x, algorithm="fricas")
Output:
integral(sqrt(b*x^4 + a)*(f*x^3 + e*x^2 + d*x + c)/x^5, x)
Result contains complex when optimal does not.
Time = 2.85 (sec) , antiderivative size = 211, normalized size of antiderivative = 0.62 \[ \int \frac {\left (c+d x+e x^2+f x^3\right ) \sqrt {a+b x^4}}{x^5} \, dx=\frac {\sqrt {a} d \Gamma \left (- \frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, - \frac {1}{2} \\ \frac {1}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 x^{3} \Gamma \left (\frac {1}{4}\right )} - \frac {\sqrt {a} e}{2 x^{2} \sqrt {1 + \frac {b x^{4}}{a}}} + \frac {\sqrt {a} f \Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, - \frac {1}{4} \\ \frac {3}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 x \Gamma \left (\frac {3}{4}\right )} - \frac {\sqrt {b} c \sqrt {\frac {a}{b x^{4}} + 1}}{4 x^{2}} + \frac {\sqrt {b} e \operatorname {asinh}{\left (\frac {\sqrt {b} x^{2}}{\sqrt {a}} \right )}}{2} - \frac {b c \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x^{2}} \right )}}{4 \sqrt {a}} - \frac {b e x^{2}}{2 \sqrt {a} \sqrt {1 + \frac {b x^{4}}{a}}} \] Input:
integrate((f*x**3+e*x**2+d*x+c)*(b*x**4+a)**(1/2)/x**5,x)
Output:
sqrt(a)*d*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(a)*e/(2*x**2*sqrt(1 + b*x**4/a)) + sqrt(a)*f* gamma(-1/4)*hyper((-1/2, -1/4), (3/4,), b*x**4*exp_polar(I*pi)/a)/(4*x*gam ma(3/4)) - sqrt(b)*c*sqrt(a/(b*x**4) + 1)/(4*x**2) + sqrt(b)*e*asinh(sqrt( b)*x**2/sqrt(a))/2 - b*c*asinh(sqrt(a)/(sqrt(b)*x**2))/(4*sqrt(a)) - b*e*x **2/(2*sqrt(a)*sqrt(1 + b*x**4/a))
\[ \int \frac {\left (c+d x+e x^2+f x^3\right ) \sqrt {a+b x^4}}{x^5} \, dx=\int { \frac {\sqrt {b x^{4} + a} {\left (f x^{3} + e x^{2} + d x + c\right )}}{x^{5}} \,d x } \] Input:
integrate((f*x^3+e*x^2+d*x+c)*(b*x^4+a)^(1/2)/x^5,x, algorithm="maxima")
Output:
1/8*(b*log((sqrt(b*x^4 + a) - sqrt(a))/(sqrt(b*x^4 + a) + sqrt(a)))/sqrt(a ) - 2*sqrt(b*x^4 + a)/x^4)*c + integrate(sqrt(b*x^4 + a)*(f*x^2 + e*x + d) /x^4, x)
\[ \int \frac {\left (c+d x+e x^2+f x^3\right ) \sqrt {a+b x^4}}{x^5} \, dx=\int { \frac {\sqrt {b x^{4} + a} {\left (f x^{3} + e x^{2} + d x + c\right )}}{x^{5}} \,d x } \] Input:
integrate((f*x^3+e*x^2+d*x+c)*(b*x^4+a)^(1/2)/x^5,x, algorithm="giac")
Output:
integrate(sqrt(b*x^4 + a)*(f*x^3 + e*x^2 + d*x + c)/x^5, x)
Timed out. \[ \int \frac {\left (c+d x+e x^2+f x^3\right ) \sqrt {a+b x^4}}{x^5} \, dx=\int \frac {\sqrt {b\,x^4+a}\,\left (f\,x^3+e\,x^2+d\,x+c\right )}{x^5} \,d x \] Input:
int(((a + b*x^4)^(1/2)*(c + d*x + e*x^2 + f*x^3))/x^5,x)
Output:
int(((a + b*x^4)^(1/2)*(c + d*x + e*x^2 + f*x^3))/x^5, x)
\[ \int \frac {\left (c+d x+e x^2+f x^3\right ) \sqrt {a+b x^4}}{x^5} \, dx=\frac {-2 \sqrt {b \,x^{4}+a}\, a c -8 \sqrt {b \,x^{4}+a}\, a d x -4 \sqrt {b \,x^{4}+a}\, a e \,x^{2}+8 \sqrt {b \,x^{4}+a}\, a f \,x^{3}+\sqrt {a}\, \mathrm {log}\left (\sqrt {b \,x^{4}+a}-\sqrt {a}\right ) b c \,x^{4}-\sqrt {a}\, \mathrm {log}\left (\sqrt {b \,x^{4}+a}+\sqrt {a}\right ) b c \,x^{4}-2 \sqrt {b}\, \mathrm {log}\left (\sqrt {b \,x^{4}+a}-\sqrt {b}\, x^{2}\right ) a e \,x^{4}+2 \sqrt {b}\, \mathrm {log}\left (\sqrt {b \,x^{4}+a}+\sqrt {b}\, x^{2}\right ) a e \,x^{4}-16 \left (\int \frac {\sqrt {b \,x^{4}+a}}{b \,x^{8}+a \,x^{4}}d x \right ) a^{2} d \,x^{4}+16 \left (\int \frac {\sqrt {b \,x^{4}+a}}{b \,x^{6}+a \,x^{2}}d x \right ) a^{2} f \,x^{4}}{8 a \,x^{4}} \] Input:
int((f*x^3+e*x^2+d*x+c)*(b*x^4+a)^(1/2)/x^5,x)
Output:
( - 2*sqrt(a + b*x**4)*a*c - 8*sqrt(a + b*x**4)*a*d*x - 4*sqrt(a + b*x**4) *a*e*x**2 + 8*sqrt(a + b*x**4)*a*f*x**3 + sqrt(a)*log(sqrt(a + b*x**4) - s qrt(a))*b*c*x**4 - sqrt(a)*log(sqrt(a + b*x**4) + sqrt(a))*b*c*x**4 - 2*sq rt(b)*log(sqrt(a + b*x**4) - sqrt(b)*x**2)*a*e*x**4 + 2*sqrt(b)*log(sqrt(a + b*x**4) + sqrt(b)*x**2)*a*e*x**4 - 16*int(sqrt(a + b*x**4)/(a*x**4 + b* x**8),x)*a**2*d*x**4 + 16*int(sqrt(a + b*x**4)/(a*x**2 + b*x**6),x)*a**2*f *x**4)/(8*a*x**4)