Integrand size = 30, antiderivative size = 338 \[ \int \frac {x^4 \left (c+d x+e x^2+f x^3\right )}{\left (a+b x^4\right )^{3/2}} \, dx=\frac {a f-b d x^2}{2 b^2 \sqrt {a+b x^4}}-\frac {x \left (c+e x^2\right )}{2 b \sqrt {a+b x^4}}+\frac {f \sqrt {a+b x^4}}{2 b^2}+\frac {3 e x \sqrt {a+b x^4}}{2 b^{3/2} \left (\sqrt {a}+\sqrt {b} x^2\right )}+\frac {d \text {arctanh}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )}{2 b^{3/2}}-\frac {3 \sqrt [4]{a} e \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{2 b^{7/4} \sqrt {a+b x^4}}+\frac {\left (\sqrt {b} c+3 \sqrt {a} e\right ) \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{4 \sqrt [4]{a} b^{7/4} \sqrt {a+b x^4}} \] Output:
1/2*(-b*d*x^2+a*f)/b^2/(b*x^4+a)^(1/2)-1/2*x*(e*x^2+c)/b/(b*x^4+a)^(1/2)+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)+b^(1/2)* x^2)+1/2*d*arctanh(b^(1/2)*x^2/(b*x^4+a)^(1/2))/b^(3/2)-3/2*a^(1/4)*e*(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^(7/4)/(b*x^4+a)^(1/2)+1/4*(b^( 1/2)*c+3*a^(1/2)*e)*(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^(7/4)/(b*x^4+a)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.12 (sec) , antiderivative size = 166, normalized size of antiderivative = 0.49 \[ \int \frac {x^4 \left (c+d x+e x^2+f x^3\right )}{\left (a+b x^4\right )^{3/2}} \, dx=\frac {2 a f-b c x-b d x^2+2 b e x^3+b f x^4+\sqrt {a} \sqrt {b} d \sqrt {1+\frac {b x^4}{a}} \text {arcsinh}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )+b c x \sqrt {1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-\frac {b x^4}{a}\right )-2 b e x^3 \sqrt {1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {3}{2},\frac {7}{4},-\frac {b x^4}{a}\right )}{2 b^2 \sqrt {a+b x^4}} \] Input:
Integrate[(x^4*(c + d*x + e*x^2 + f*x^3))/(a + b*x^4)^(3/2),x]
Output:
(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)/Sqrt[a]] + b*c*x*Sqrt[1 + (b*x^4)/a]*Hy pergeometric2F1[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])
Time = 0.91 (sec) , antiderivative size = 319, normalized size of antiderivative = 0.94, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {2367, 25, 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.
| \(\displaystyle \int \frac {x^4 \left (c+d x+e x^2+f x^3\right )}{\left (a+b x^4\right )^{3/2}} \, dx\) |
| \(\Big \downarrow \) 2367 |
| \(\displaystyle -\frac {\int -\frac {4 a b f x^3+3 a b e x^2+2 a b d x+a b c}{\sqrt {b x^4+a}}dx}{2 a b^2}-\frac {x \left (c+d x+e x^2+f x^3\right )}{2 b \sqrt {a+b x^4}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {4 a b f x^3+3 a b e x^2+2 a b d x+a b c}{\sqrt {b x^4+a}}dx}{2 a b^2}-\frac {x \left (c+d x+e x^2+f x^3\right )}{2 b \sqrt {a+b x^4}}\) |
| \(\Big \downarrow \) 2424 |
| \(\displaystyle \frac {\int \left (\frac {3 a b e x^2+a b c}{\sqrt {b x^4+a}}+\frac {x \left (4 a b f x^2+2 a b d\right )}{\sqrt {b x^4+a}}\right )dx}{2 a b^2}-\frac {x \left (c+d x+e x^2+f x^3\right )}{2 b \sqrt {a+b x^4}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {a^{3/4} \sqrt [4]{b} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \left (3 \sqrt {a} e+\sqrt {b} c\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{2 \sqrt {a+b x^4}}-\frac {3 a^{5/4} \sqrt [4]{b} e \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{\sqrt {a+b x^4}}+a \sqrt {b} d \text {arctanh}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )+\frac {3 a \sqrt {b} e x \sqrt {a+b x^4}}{\sqrt {a}+\sqrt {b} x^2}+2 a f \sqrt {a+b x^4}}{2 a b^2}-\frac {x \left (c+d x+e x^2+f x^3\right )}{2 b \sqrt {a+b x^4}}\) |
Input:
Int[(x^4*(c + d*x + e*x^2 + f*x^3))/(a + b*x^4)^(3/2),x]
Output:
-1/2*(x*(c + d*x + e*x^2 + f*x^3))/(b*Sqrt[a + b*x^4]) + (2*a*f*Sqrt[a + b *x^4] + (3*a*Sqrt[b]*e*x*Sqrt[a + b*x^4])/(Sqrt[a] + Sqrt[b]*x^2) + a*Sqrt [b]*d*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a + b*x^4]] - (3*a^(5/4)*b^(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])/Sqrt[a + b*x^4] + (a^(3/4)*b^(1/4)*(Sqr t[b]*c + 3*Sqrt[a]*e)*(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])/(2*Sqrt[a + b*x^4]))/(2*a*b^2)
Int[(Pq_)*(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> With[{q = m + Expon[Pq, x]}, Module[{Q = PolynomialQuotient[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]}, Simp[(-x)*R*((a + b*x^n)^(p + 1)/(a*n*(p + 1)*b^(Floo r[(q - 1)/n] + 1))), x] + Simp[1/(a*n*(p + 1)*b^(Floor[(q - 1)/n] + 1)) I nt[(a + b*x^n)^(p + 1)*ExpandToSum[a*n*(p + 1)*Q + n*(p + 1)*R + D[x*R, x], x], x], x]] /; GeQ[q, n]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && IGtQ[n, 0 ] && LtQ[p, -1] && IGtQ[m, 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 = 3.76 (sec) , antiderivative size = 264, normalized size of antiderivative = 0.78
| method | result | size |
| elliptic | \(-\frac {2 b \left (\frac {e \,x^{3}}{4 b^{2}}+\frac {d \,x^{2}}{4 b^{2}}+\frac {c x}{4 b^{2}}-\frac {a f}{4 b^{3}}\right )}{\sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {f \sqrt {b \,x^{4}+a}}{2 b^{2}}+\frac {c \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{2 b \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}+\frac {d \ln \left (2 \sqrt {b}\, x^{2}+2 \sqrt {b \,x^{4}+a}\right )}{2 b^{\frac {3}{2}}}+\frac {3 i e \sqrt {a}\, \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{2 b^{\frac {3}{2}} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\) | \(264\) |
| default | \(c \left (-\frac {x}{2 b \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {\sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{2 b \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )+d \left (-\frac {x^{2}}{2 b \sqrt {b \,x^{4}+a}}+\frac {\ln \left (\sqrt {b}\, x^{2}+\sqrt {b \,x^{4}+a}\right )}{2 b^{\frac {3}{2}}}\right )+e \left (-\frac {x^{3}}{2 b \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {3 i \sqrt {a}\, \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{2 b^{\frac {3}{2}} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )+\frac {f \left (b \,x^{4}+2 a \right )}{2 \sqrt {b \,x^{4}+a}\, b^{2}}\) | \(284\) |
| risch | \(\frac {f \sqrt {b \,x^{4}+a}}{2 b^{2}}+\frac {c b \left (-\frac {x}{2 b \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {\sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{2 b \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )+b d \left (-\frac {x^{2}}{2 b \sqrt {b \,x^{4}+a}}+\frac {\ln \left (\sqrt {b}\, x^{2}+\sqrt {b \,x^{4}+a}\right )}{2 b^{\frac {3}{2}}}\right )+b e \left (-\frac {x^{3}}{2 b \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {3 i \sqrt {a}\, \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{2 b^{\frac {3}{2}} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )+\frac {a f}{2 b \sqrt {b \,x^{4}+a}}}{b}\) | \(299\) |
Input:
int(x^4*(f*x^3+e*x^2+d*x+c)/(b*x^4+a)^(3/2),x,method=_RETURNVERBOSE)
Output:
-2*b*(1/4*e/b^2*x^3+1/4*d/b^2*x^2+1/4/b^2*c*x-1/4/b^3*a*f)/((x^4+a/b)*b)^( 1/2)+1/2*f*(b*x^4+a)^(1/2)/b^2+1/2/b*c/(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)*Ell ipticF(x*(I/a^(1/2)*b^(1/2))^(1/2),I)+1/2*d/b^(3/2)*ln(2*b^(1/2)*x^2+2*(b* x^4+a)^(1/2))+3/2*I*e/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)*(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))
Time = 0.12 (sec) , antiderivative size = 223, normalized size of antiderivative = 0.66 \[ \int \frac {x^4 \left (c+d x+e x^2+f x^3\right )}{\left (a+b x^4\right )^{3/2}} \, dx=\frac {6 \, {\left (a b e x^{5} + a^{2} e x\right )} \sqrt {b} \left (-\frac {a}{b}\right )^{\frac {3}{4}} E(\arcsin \left (\frac {\left (-\frac {a}{b}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) + 2 \, {\left ({\left (b^{2} c - 3 \, a b e\right )} x^{5} + {\left (a b c - 3 \, a^{2} e\right )} x\right )} \sqrt {b} \left (-\frac {a}{b}\right )^{\frac {3}{4}} F(\arcsin \left (\frac {\left (-\frac {a}{b}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) + {\left (a b d x^{5} + a^{2} d x\right )} \sqrt {b} \log \left (-2 \, b x^{4} - 2 \, \sqrt {b x^{4} + a} \sqrt {b} x^{2} - a\right ) + 2 \, {\left (a b f x^{5} + 2 \, a b e x^{4} - a b d x^{3} - a b c x^{2} + 2 \, a^{2} f x + 3 \, a^{2} e\right )} \sqrt {b x^{4} + a}}{4 \, {\left (a b^{3} x^{5} + a^{2} b^{2} x\right )}} \] Input:
integrate(x^4*(f*x^3+e*x^2+d*x+c)/(b*x^4+a)^(3/2),x, algorithm="fricas")
Output:
1/4*(6*(a*b*e*x^5 + a^2*e*x)*sqrt(b)*(-a/b)^(3/4)*elliptic_e(arcsin((-a/b) ^(1/4)/x), -1) + 2*((b^2*c - 3*a*b*e)*x^5 + (a*b*c - 3*a^2*e)*x)*sqrt(b)*( -a/b)^(3/4)*elliptic_f(arcsin((-a/b)^(1/4)/x), -1) + (a*b*d*x^5 + a^2*d*x) *sqrt(b)*log(-2*b*x^4 - 2*sqrt(b*x^4 + a)*sqrt(b)*x^2 - a) + 2*(a*b*f*x^5 + 2*a*b*e*x^4 - a*b*d*x^3 - a*b*c*x^2 + 2*a^2*f*x + 3*a^2*e)*sqrt(b*x^4 + a))/(a*b^3*x^5 + a^2*b^2*x)
Time = 7.37 (sec) , antiderivative size = 172, normalized size of antiderivative = 0.51 \[ \int \frac {x^4 \left (c+d x+e x^2+f x^3\right )}{\left (a+b x^4\right )^{3/2}} \, dx=d \left (\frac {\operatorname {asinh}{\left (\frac {\sqrt {b} x^{2}}{\sqrt {a}} \right )}}{2 b^{\frac {3}{2}}} - \frac {x^{2}}{2 \sqrt {a} b \sqrt {1 + \frac {b x^{4}}{a}}}\right ) + f \left (\begin {cases} \frac {a}{b^{2} \sqrt {a + b x^{4}}} + \frac {x^{4}}{2 b \sqrt {a + b x^{4}}} & \text {for}\: b \neq 0 \\\frac {x^{8}}{8 a^{\frac {3}{2}}} & \text {otherwise} \end {cases}\right ) + \frac {c x^{5} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{4}, \frac {3}{2} \\ \frac {9}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {3}{2}} \Gamma \left (\frac {9}{4}\right )} + \frac {e x^{7} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {3}{2}} \Gamma \left (\frac {11}{4}\right )} \] Input:
integrate(x**4*(f*x**3+e*x**2+d*x+c)/(b*x**4+a)**(3/2),x)
Output:
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*sqrt(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))
\[ \int \frac {x^4 \left (c+d x+e x^2+f x^3\right )}{\left (a+b x^4\right )^{3/2}} \, dx=\int { \frac {{\left (f x^{3} + e x^{2} + d x + c\right )} x^{4}}{{\left (b x^{4} + a\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate(x^4*(f*x^3+e*x^2+d*x+c)/(b*x^4+a)^(3/2),x, algorithm="maxima")
Output:
integrate((f*x^3 + e*x^2 + d*x + c)*x^4/(b*x^4 + a)^(3/2), x)
\[ \int \frac {x^4 \left (c+d x+e x^2+f x^3\right )}{\left (a+b x^4\right )^{3/2}} \, dx=\int { \frac {{\left (f x^{3} + e x^{2} + d x + c\right )} x^{4}}{{\left (b x^{4} + a\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate(x^4*(f*x^3+e*x^2+d*x+c)/(b*x^4+a)^(3/2),x, algorithm="giac")
Output:
integrate((f*x^3 + e*x^2 + d*x + c)*x^4/(b*x^4 + a)^(3/2), x)
Timed out. \[ \int \frac {x^4 \left (c+d x+e x^2+f x^3\right )}{\left (a+b x^4\right )^{3/2}} \, dx=\int \frac {x^4\,\left (f\,x^3+e\,x^2+d\,x+c\right )}{{\left (b\,x^4+a\right )}^{3/2}} \,d x \] Input:
int((x^4*(c + d*x + e*x^2 + f*x^3))/(a + b*x^4)^(3/2),x)
Output:
int((x^4*(c + d*x + e*x^2 + f*x^3))/(a + b*x^4)^(3/2), x)
\[ \int \frac {x^4 \left (c+d x+e x^2+f x^3\right )}{\left (a+b x^4\right )^{3/2}} \, dx=\frac {4 \sqrt {b \,x^{4}+a}\, a f -4 \sqrt {b \,x^{4}+a}\, b c x -2 \sqrt {b \,x^{4}+a}\, b d \,x^{2}+4 \sqrt {b \,x^{4}+a}\, b e \,x^{3}+2 \sqrt {b \,x^{4}+a}\, b f \,x^{4}-\sqrt {b}\, \mathrm {log}\left (\sqrt {b \,x^{4}+a}-\sqrt {b}\, x^{2}\right ) a d -\sqrt {b}\, \mathrm {log}\left (\sqrt {b \,x^{4}+a}-\sqrt {b}\, x^{2}\right ) b d \,x^{4}+\sqrt {b}\, \mathrm {log}\left (\sqrt {b \,x^{4}+a}+\sqrt {b}\, x^{2}\right ) a d +\sqrt {b}\, \mathrm {log}\left (\sqrt {b \,x^{4}+a}+\sqrt {b}\, x^{2}\right ) b d \,x^{4}+4 \left (\int \frac {\sqrt {b \,x^{4}+a}}{b^{2} x^{8}+2 a b \,x^{4}+a^{2}}d x \right ) a^{2} b c +4 \left (\int \frac {\sqrt {b \,x^{4}+a}}{b^{2} x^{8}+2 a b \,x^{4}+a^{2}}d x \right ) a \,b^{2} c \,x^{4}-12 \left (\int \frac {\sqrt {b \,x^{4}+a}\, x^{2}}{b^{2} x^{8}+2 a b \,x^{4}+a^{2}}d x \right ) a^{2} b e -12 \left (\int \frac {\sqrt {b \,x^{4}+a}\, x^{2}}{b^{2} x^{8}+2 a b \,x^{4}+a^{2}}d x \right ) a \,b^{2} e \,x^{4}}{4 b^{2} \left (b \,x^{4}+a \right )} \] Input:
int(x^4*(f*x^3+e*x^2+d*x+c)/(b*x^4+a)^(3/2),x)
Output:
(4*sqrt(a + b*x**4)*a*f - 4*sqrt(a + b*x**4)*b*c*x - 2*sqrt(a + b*x**4)*b* d*x**2 + 4*sqrt(a + b*x**4)*b*e*x**3 + 2*sqrt(a + b*x**4)*b*f*x**4 - sqrt( b)*log(sqrt(a + b*x**4) - sqrt(b)*x**2)*a*d - sqrt(b)*log(sqrt(a + b*x**4) - sqrt(b)*x**2)*b*d*x**4 + sqrt(b)*log(sqrt(a + b*x**4) + sqrt(b)*x**2)*a *d + sqrt(b)*log(sqrt(a + b*x**4) + sqrt(b)*x**2)*b*d*x**4 + 4*int(sqrt(a + b*x**4)/(a**2 + 2*a*b*x**4 + b**2*x**8),x)*a**2*b*c + 4*int(sqrt(a + b*x **4)/(a**2 + 2*a*b*x**4 + b**2*x**8),x)*a*b**2*c*x**4 - 12*int((sqrt(a + b *x**4)*x**2)/(a**2 + 2*a*b*x**4 + b**2*x**8),x)*a**2*b*e - 12*int((sqrt(a + b*x**4)*x**2)/(a**2 + 2*a*b*x**4 + b**2*x**8),x)*a*b**2*e*x**4)/(4*b**2* (a + b*x**4))