Integrand size = 30, antiderivative size = 363 \[ \int \frac {x^5 \left (c+d x+e x^2+f x^3\right )}{\left (a+b x^4\right )^{3/2}} \, dx=\frac {a e-b c x^2}{2 b^2 \sqrt {a+b x^4}}+\frac {x \left (a f-b d x^2\right )}{2 b^2 \sqrt {a+b x^4}}+\frac {e \sqrt {a+b x^4}}{2 b^2}+\frac {f x \sqrt {a+b x^4}}{3 b^2}+\frac {3 d x \sqrt {a+b x^4}}{2 b^{3/2} \left (\sqrt {a}+\sqrt {b} x^2\right )}+\frac {c \text {arctanh}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )}{2 b^{3/2}}-\frac {3 \sqrt [4]{a} d \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{2 b^{7/4} \sqrt {a+b x^4}}+\frac {\sqrt [4]{a} \left (9 \sqrt {b} d-5 \sqrt {a} f\right ) \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{12 b^{9/4} \sqrt {a+b x^4}} \] Output:
1/2*(-b*c*x^2+a*e)/b^2/(b*x^4+a)^(1/2)+1/2*x*(-b*d*x^2+a*f)/b^2/(b*x^4+a)^ (1/2)+1/2*e*(b*x^4+a)^(1/2)/b^2+1/3*f*x*(b*x^4+a)^(1/2)/b^2+3/2*d*x*(b*x^4 +a)^(1/2)/b^(3/2)/(a^(1/2)+b^(1/2)*x^2)+1/2*c*arctanh(b^(1/2)*x^2/(b*x^4+a )^(1/2))/b^(3/2)-3/2*a^(1/4)*d*(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)+1/12*a^(1/4)*(9*b^(1/2)*d-5*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))/b^(9/4)/(b*x^4+a)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.14 (sec) , antiderivative size = 176, normalized size of antiderivative = 0.48 \[ \int \frac {x^5 \left (c+d x+e x^2+f x^3\right )}{\left (a+b x^4\right )^{3/2}} \, dx=\frac {6 a e+5 a f x-3 b c x^2+6 b d x^3+3 b e x^4+2 b f x^5+3 \sqrt {a} \sqrt {b} c \sqrt {1+\frac {b x^4}{a}} \text {arcsinh}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )-5 a f x \sqrt {1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-\frac {b x^4}{a}\right )-6 b d x^3 \sqrt {1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {3}{2},\frac {7}{4},-\frac {b x^4}{a}\right )}{6 b^2 \sqrt {a+b x^4}} \] Input:
Integrate[(x^5*(c + d*x + e*x^2 + f*x^3))/(a + b*x^4)^(3/2),x]
Output:
(6*a*e + 5*a*f*x - 3*b*c*x^2 + 6*b*d*x^3 + 3*b*e*x^4 + 2*b*f*x^5 + 3*Sqrt[ a]*Sqrt[b]*c*Sqrt[1 + (b*x^4)/a]*ArcSinh[(Sqrt[b]*x^2)/Sqrt[a]] - 5*a*f*x* Sqrt[1 + (b*x^4)/a]*Hypergeometric2F1[1/4, 1/2, 5/4, -((b*x^4)/a)] - 6*b*d *x^3*Sqrt[1 + (b*x^4)/a]*Hypergeometric2F1[3/4, 3/2, 7/4, -((b*x^4)/a)])/( 6*b^2*Sqrt[a + b*x^4])
Time = 1.14 (sec) , antiderivative size = 347, normalized size of antiderivative = 0.96, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {2367, 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.
| \(\displaystyle \int \frac {x^5 \left (c+d x+e x^2+f x^3\right )}{\left (a+b x^4\right )^{3/2}} \, dx\) |
| \(\Big \downarrow \) 2367 |
| \(\displaystyle \frac {x \left (a f-b c x-b d x^2-b e x^3\right )}{2 b^2 \sqrt {a+b x^4}}-\frac {\int \frac {-2 a b f x^4-4 a b e x^3-3 a b d x^2-2 a b c x+a^2 f}{\sqrt {b x^4+a}}dx}{2 a b^2}\) |
| \(\Big \downarrow \) 2424 |
| \(\displaystyle \frac {x \left (a f-b c x-b d x^2-b e x^3\right )}{2 b^2 \sqrt {a+b x^4}}-\frac {\int \left (\frac {x \left (-4 a b e x^2-2 a b c\right )}{\sqrt {b x^4+a}}+\frac {-2 a b f x^4-3 a b d x^2+a^2 f}{\sqrt {b x^4+a}}\right )dx}{2 a b^2}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {x \left (a f-b c x-b d x^2-b e x^3\right )}{2 b^2 \sqrt {a+b x^4}}-\frac {-\frac {a^{5/4} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \left (9 \sqrt {b} d-5 \sqrt {a} f\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{6 \sqrt [4]{b} \sqrt {a+b x^4}}+\frac {3 a^{5/4} \sqrt [4]{b} d \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{\sqrt {a+b x^4}}-a \sqrt {b} c \text {arctanh}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )-\frac {3 a \sqrt {b} d x \sqrt {a+b x^4}}{\sqrt {a}+\sqrt {b} x^2}-2 a e \sqrt {a+b x^4}-\frac {2}{3} a f x \sqrt {a+b x^4}}{2 a b^2}\) |
Input:
Int[(x^5*(c + d*x + e*x^2 + f*x^3))/(a + b*x^4)^(3/2),x]
Output:
(x*(a*f - b*c*x - b*d*x^2 - b*e*x^3))/(2*b^2*Sqrt[a + b*x^4]) - (-2*a*e*Sq rt[a + b*x^4] - (2*a*f*x*Sqrt[a + b*x^4])/3 - (3*a*Sqrt[b]*d*x*Sqrt[a + b* x^4])/(Sqrt[a] + Sqrt[b]*x^2) - a*Sqrt[b]*c*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a + b*x^4]] + (3*a^(5/4)*b^(1/4)*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])/S qrt[a + b*x^4] - (a^(5/4)*(9*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])/(6*b^(1/4)*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 = 4.23 (sec) , antiderivative size = 282, normalized size of antiderivative = 0.78
| method | result | size |
| elliptic | \(-\frac {2 b \left (\frac {d \,x^{3}}{4 b^{2}}+\frac {c \,x^{2}}{4 b^{2}}-\frac {a f x}{4 b^{3}}-\frac {a e}{4 b^{3}}\right )}{\sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {f x \sqrt {b \,x^{4}+a}}{3 b^{2}}+\frac {e \sqrt {b \,x^{4}+a}}{2 b^{2}}-\frac {5 f a \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{6 b^{2} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}+\frac {c \ln \left (2 \sqrt {b}\, x^{2}+2 \sqrt {b \,x^{4}+a}\right )}{2 b^{\frac {3}{2}}}+\frac {3 i d \sqrt {a}\, \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{2 b^{\frac {3}{2}} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\) | \(282\) |
| default | \(c \left (-\frac {x^{2}}{2 b \sqrt {b \,x^{4}+a}}+\frac {\ln \left (\sqrt {b}\, x^{2}+\sqrt {b \,x^{4}+a}\right )}{2 b^{\frac {3}{2}}}\right )+d \left (-\frac {x^{3}}{2 b \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {3 i \sqrt {a}\, \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{2 b^{\frac {3}{2}} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )+\frac {e \left (b \,x^{4}+2 a \right )}{2 \sqrt {b \,x^{4}+a}\, b^{2}}+f \left (\frac {a x}{2 b^{2} \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {x \sqrt {b \,x^{4}+a}}{3 b^{2}}-\frac {5 a \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{6 b^{2} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )\) | \(301\) |
| risch | \(\frac {\left (2 f x +3 e \right ) \sqrt {b \,x^{4}+a}}{6 b^{2}}+\frac {f a x}{2 b^{2} \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}-\frac {5 f a \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{6 b^{2} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}-\frac {c \,x^{2}}{2 b \sqrt {b \,x^{4}+a}}+\frac {c \ln \left (\sqrt {b}\, x^{2}+\sqrt {b \,x^{4}+a}\right )}{2 b^{\frac {3}{2}}}-\frac {d \,x^{3}}{2 b \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {3 i d \sqrt {a}\, \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{2 b^{\frac {3}{2}} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}-\frac {3 i d \sqrt {a}\, \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{2 b^{\frac {3}{2}} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}+\frac {a e}{2 b^{2} \sqrt {b \,x^{4}+a}}\) | \(356\) |
Input:
int(x^5*(f*x^3+e*x^2+d*x+c)/(b*x^4+a)^(3/2),x,method=_RETURNVERBOSE)
Output:
-2*b*(1/4*d/b^2*x^3+1/4/b^2*c*x^2-1/4/b^3*a*f*x-1/4*a*e/b^3)/((x^4+a/b)*b) ^(1/2)+1/3*f*x*(b*x^4+a)^(1/2)/b^2+1/2*e*(b*x^4+a)^(1/2)/b^2-5/6/b^2*f*a/( 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/b^(3/2)*c*ln(2*b^(1/2)*x^2+2*(b*x^4+a)^(1/2))+3/2*I*d/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)-El lipticE(x*(I/a^(1/2)*b^(1/2))^(1/2),I))
Time = 0.14 (sec) , antiderivative size = 211, normalized size of antiderivative = 0.58 \[ \int \frac {x^5 \left (c+d x+e x^2+f x^3\right )}{\left (a+b x^4\right )^{3/2}} \, dx=\frac {18 \, {\left (b d x^{5} + a d x\right )} \sqrt {b} \left (-\frac {a}{b}\right )^{\frac {3}{4}} E(\arcsin \left (\frac {\left (-\frac {a}{b}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) - 2 \, {\left ({\left (9 \, b d + 5 \, b f\right )} x^{5} + {\left (9 \, a d + 5 \, a f\right )} x\right )} \sqrt {b} \left (-\frac {a}{b}\right )^{\frac {3}{4}} F(\arcsin \left (\frac {\left (-\frac {a}{b}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) + 3 \, {\left (b c x^{5} + a c x\right )} \sqrt {b} \log \left (-2 \, b x^{4} - 2 \, \sqrt {b x^{4} + a} \sqrt {b} x^{2} - a\right ) + 2 \, {\left (2 \, b f x^{6} + 3 \, b e x^{5} + 6 \, b d x^{4} - 3 \, b c x^{3} + 5 \, a f x^{2} + 6 \, a e x + 9 \, a d\right )} \sqrt {b x^{4} + a}}{12 \, {\left (b^{3} x^{5} + a b^{2} x\right )}} \] Input:
integrate(x^5*(f*x^3+e*x^2+d*x+c)/(b*x^4+a)^(3/2),x, algorithm="fricas")
Output:
1/12*(18*(b*d*x^5 + a*d*x)*sqrt(b)*(-a/b)^(3/4)*elliptic_e(arcsin((-a/b)^( 1/4)/x), -1) - 2*((9*b*d + 5*b*f)*x^5 + (9*a*d + 5*a*f)*x)*sqrt(b)*(-a/b)^ (3/4)*elliptic_f(arcsin((-a/b)^(1/4)/x), -1) + 3*(b*c*x^5 + a*c*x)*sqrt(b) *log(-2*b*x^4 - 2*sqrt(b*x^4 + a)*sqrt(b)*x^2 - a) + 2*(2*b*f*x^6 + 3*b*e* x^5 + 6*b*d*x^4 - 3*b*c*x^3 + 5*a*f*x^2 + 6*a*e*x + 9*a*d)*sqrt(b*x^4 + a) )/(b^3*x^5 + a*b^2*x)
Time = 8.94 (sec) , antiderivative size = 172, normalized size of antiderivative = 0.47 \[ \int \frac {x^5 \left (c+d x+e x^2+f x^3\right )}{\left (a+b x^4\right )^{3/2}} \, dx=c \left (\frac {\operatorname {asinh}{\left (\frac {\sqrt {b} x^{2}}{\sqrt {a}} \right )}}{2 b^{\frac {3}{2}}} - \frac {x^{2}}{2 \sqrt {a} b \sqrt {1 + \frac {b x^{4}}{a}}}\right ) + e \left (\begin {cases} \frac {a}{b^{2} \sqrt {a + b x^{4}}} + \frac {x^{4}}{2 b \sqrt {a + b x^{4}}} & \text {for}\: b \neq 0 \\\frac {x^{8}}{8 a^{\frac {3}{2}}} & \text {otherwise} \end {cases}\right ) + \frac {d x^{7} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {3}{2}} \Gamma \left (\frac {11}{4}\right )} + \frac {f x^{9} \Gamma \left (\frac {9}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, \frac {9}{4} \\ \frac {13}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {3}{2}} \Gamma \left (\frac {13}{4}\right )} \] Input:
integrate(x**5*(f*x**3+e*x**2+d*x+c)/(b*x**4+a)**(3/2),x)
Output:
c*(asinh(sqrt(b)*x**2/sqrt(a))/(2*b**(3/2)) - x**2/(2*sqrt(a)*b*sqrt(1 + b *x**4/a))) + e*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)) + d*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)) + f *x**9*gamma(9/4)*hyper((3/2, 9/4), (13/4,), b*x**4*exp_polar(I*pi)/a)/(4*a **(3/2)*gamma(13/4))
\[ \int \frac {x^5 \left (c+d x+e x^2+f x^3\right )}{\left (a+b x^4\right )^{3/2}} \, dx=\int { \frac {{\left (f x^{3} + e x^{2} + d x + c\right )} x^{5}}{{\left (b x^{4} + a\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate(x^5*(f*x^3+e*x^2+d*x+c)/(b*x^4+a)^(3/2),x, algorithm="maxima")
Output:
-1/4*c*(2*x^2/(sqrt(b*x^4 + a)*b) + log(-(sqrt(b) - sqrt(b*x^4 + a)/x^2)/( sqrt(b) + sqrt(b*x^4 + a)/x^2))/b^(3/2)) + integrate((f*x^8 + e*x^7 + d*x^ 6)/(b*x^4 + a)^(3/2), x)
\[ \int \frac {x^5 \left (c+d x+e x^2+f x^3\right )}{\left (a+b x^4\right )^{3/2}} \, dx=\int { \frac {{\left (f x^{3} + e x^{2} + d x + c\right )} x^{5}}{{\left (b x^{4} + a\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate(x^5*(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^5/(b*x^4 + a)^(3/2), x)
Timed out. \[ \int \frac {x^5 \left (c+d x+e x^2+f x^3\right )}{\left (a+b x^4\right )^{3/2}} \, dx=\int \frac {x^5\,\left (f\,x^3+e\,x^2+d\,x+c\right )}{{\left (b\,x^4+a\right )}^{3/2}} \,d x \] Input:
int((x^5*(c + d*x + e*x^2 + f*x^3))/(a + b*x^4)^(3/2),x)
Output:
int((x^5*(c + d*x + e*x^2 + f*x^3))/(a + b*x^4)^(3/2), x)
\[ \int \frac {x^5 \left (c+d x+e x^2+f x^3\right )}{\left (a+b x^4\right )^{3/2}} \, dx=\frac {12 \sqrt {b \,x^{4}+a}\, a e +20 \sqrt {b \,x^{4}+a}\, a f x -6 \sqrt {b \,x^{4}+a}\, b c \,x^{2}+12 \sqrt {b \,x^{4}+a}\, b d \,x^{3}+6 \sqrt {b \,x^{4}+a}\, b e \,x^{4}+4 \sqrt {b \,x^{4}+a}\, b f \,x^{5}-3 \sqrt {b}\, \mathrm {log}\left (\sqrt {b \,x^{4}+a}-\sqrt {b}\, x^{2}\right ) a c -3 \sqrt {b}\, \mathrm {log}\left (\sqrt {b \,x^{4}+a}-\sqrt {b}\, x^{2}\right ) b c \,x^{4}+3 \sqrt {b}\, \mathrm {log}\left (\sqrt {b \,x^{4}+a}+\sqrt {b}\, x^{2}\right ) a c +3 \sqrt {b}\, \mathrm {log}\left (\sqrt {b \,x^{4}+a}+\sqrt {b}\, x^{2}\right ) b c \,x^{4}-20 \left (\int \frac {\sqrt {b \,x^{4}+a}}{b^{2} x^{8}+2 a b \,x^{4}+a^{2}}d x \right ) a^{3} f -20 \left (\int \frac {\sqrt {b \,x^{4}+a}}{b^{2} x^{8}+2 a b \,x^{4}+a^{2}}d x \right ) a^{2} b f \,x^{4}-36 \left (\int \frac {\sqrt {b \,x^{4}+a}\, x^{2}}{b^{2} x^{8}+2 a b \,x^{4}+a^{2}}d x \right ) a^{2} b d -36 \left (\int \frac {\sqrt {b \,x^{4}+a}\, x^{2}}{b^{2} x^{8}+2 a b \,x^{4}+a^{2}}d x \right ) a \,b^{2} d \,x^{4}}{12 b^{2} \left (b \,x^{4}+a \right )} \] Input:
int(x^5*(f*x^3+e*x^2+d*x+c)/(b*x^4+a)^(3/2),x)
Output:
(12*sqrt(a + b*x**4)*a*e + 20*sqrt(a + b*x**4)*a*f*x - 6*sqrt(a + b*x**4)* b*c*x**2 + 12*sqrt(a + b*x**4)*b*d*x**3 + 6*sqrt(a + b*x**4)*b*e*x**4 + 4* sqrt(a + b*x**4)*b*f*x**5 - 3*sqrt(b)*log(sqrt(a + b*x**4) - sqrt(b)*x**2) *a*c - 3*sqrt(b)*log(sqrt(a + b*x**4) - sqrt(b)*x**2)*b*c*x**4 + 3*sqrt(b) *log(sqrt(a + b*x**4) + sqrt(b)*x**2)*a*c + 3*sqrt(b)*log(sqrt(a + b*x**4) + sqrt(b)*x**2)*b*c*x**4 - 20*int(sqrt(a + b*x**4)/(a**2 + 2*a*b*x**4 + b **2*x**8),x)*a**3*f - 20*int(sqrt(a + b*x**4)/(a**2 + 2*a*b*x**4 + b**2*x* *8),x)*a**2*b*f*x**4 - 36*int((sqrt(a + b*x**4)*x**2)/(a**2 + 2*a*b*x**4 + b**2*x**8),x)*a**2*b*d - 36*int((sqrt(a + b*x**4)*x**2)/(a**2 + 2*a*b*x** 4 + b**2*x**8),x)*a*b**2*d*x**4)/(12*b**2*(a + b*x**4))