Integrand size = 22, antiderivative size = 296 \[ \int \frac {A+B x+C x^2}{\left (a+b x^2\right )^{5/3}} \, dx=-\frac {3 (a B-(A b-a C) x)}{4 a b \left (a+b x^2\right )^{2/3}}-\frac {3^{3/4} \sqrt {2-\sqrt {3}} (A b+3 a C) \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right ) \sqrt {\frac {a^{2/3}+\sqrt [3]{a} \sqrt [3]{a+b x^2}+\left (a+b x^2\right )^{2/3}}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1+\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}}{\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}}\right ),-7+4 \sqrt {3}\right )}{4 a b^2 x \sqrt {-\frac {\sqrt [3]{a} \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )^2}}} \] Output:
1/4*(-3*B*a+3*(A*b-C*a)*x)/a/b/(b*x^2+a)^(2/3)-1/4*3^(3/4)*(1/2*6^(1/2)-1/ 2*2^(1/2))*(A*b+3*C*a)*(a^(1/3)-(b*x^2+a)^(1/3))*((a^(2/3)+a^(1/3)*(b*x^2+ a)^(1/3)+(b*x^2+a)^(2/3))/((1-3^(1/2))*a^(1/3)-(b*x^2+a)^(1/3))^2)^(1/2)*E llipticF(((1+3^(1/2))*a^(1/3)-(b*x^2+a)^(1/3))/((1-3^(1/2))*a^(1/3)-(b*x^2 +a)^(1/3)),2*I-I*3^(1/2))/a/b^2/x/(-a^(1/3)*(a^(1/3)-(b*x^2+a)^(1/3))/((1- 3^(1/2))*a^(1/3)-(b*x^2+a)^(1/3))^2)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.05 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.26 \[ \int \frac {A+B x+C x^2}{\left (a+b x^2\right )^{5/3}} \, dx=\frac {3 A b x-3 a (B+C x)+(A b+3 a C) x \left (1+\frac {b x^2}{a}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {3}{2},-\frac {b x^2}{a}\right )}{4 a b \left (a+b x^2\right )^{2/3}} \] Input:
Integrate[(A + B*x + C*x^2)/(a + b*x^2)^(5/3),x]
Output:
(3*A*b*x - 3*a*(B + C*x) + (A*b + 3*a*C)*x*(1 + (b*x^2)/a)^(2/3)*Hypergeom etric2F1[1/2, 2/3, 3/2, -((b*x^2)/a)])/(4*a*b*(a + b*x^2)^(2/3))
Time = 0.32 (sec) , antiderivative size = 296, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {2345, 27, 234, 760}
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.
\(\displaystyle \int \frac {A+B x+C x^2}{\left (a+b x^2\right )^{5/3}} \, dx\) |
\(\Big \downarrow \) 2345 |
\(\displaystyle -\frac {3 \int -\frac {A b+3 a C}{3 b \left (b x^2+a\right )^{2/3}}dx}{4 a}-\frac {3 (a B-x (A b-a C))}{4 a b \left (a+b x^2\right )^{2/3}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {(3 a C+A b) \int \frac {1}{\left (b x^2+a\right )^{2/3}}dx}{4 a b}-\frac {3 (a B-x (A b-a C))}{4 a b \left (a+b x^2\right )^{2/3}}\) |
\(\Big \downarrow \) 234 |
\(\displaystyle \frac {3 \sqrt {b x^2} (3 a C+A b) \int \frac {1}{\sqrt {b x^2}}d\sqrt [3]{b x^2+a}}{8 a b^2 x}-\frac {3 (a B-x (A b-a C))}{4 a b \left (a+b x^2\right )^{2/3}}\) |
\(\Big \downarrow \) 760 |
\(\displaystyle -\frac {3^{3/4} \sqrt {2-\sqrt {3}} \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right ) \sqrt {\frac {a^{2/3}+\sqrt [3]{a} \sqrt [3]{a+b x^2}+\left (a+b x^2\right )^{2/3}}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )^2}} (3 a C+A b) \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1+\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{b x^2+a}}{\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{b x^2+a}}\right ),-7+4 \sqrt {3}\right )}{4 a b^2 x \sqrt {-\frac {\sqrt [3]{a} \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )^2}}}-\frac {3 (a B-x (A b-a C))}{4 a b \left (a+b x^2\right )^{2/3}}\) |
Input:
Int[(A + B*x + C*x^2)/(a + b*x^2)^(5/3),x]
Output:
(-3*(a*B - (A*b - a*C)*x))/(4*a*b*(a + b*x^2)^(2/3)) - (3^(3/4)*Sqrt[2 - S qrt[3]]*(A*b + 3*a*C)*(a^(1/3) - (a + b*x^2)^(1/3))*Sqrt[(a^(2/3) + a^(1/3 )*(a + b*x^2)^(1/3) + (a + b*x^2)^(2/3))/((1 - Sqrt[3])*a^(1/3) - (a + b*x ^2)^(1/3))^2]*EllipticF[ArcSin[((1 + Sqrt[3])*a^(1/3) - (a + b*x^2)^(1/3)) /((1 - Sqrt[3])*a^(1/3) - (a + b*x^2)^(1/3))], -7 + 4*Sqrt[3]])/(4*a*b^2*x *Sqrt[-((a^(1/3)*(a^(1/3) - (a + b*x^2)^(1/3)))/((1 - Sqrt[3])*a^(1/3) - ( a + b*x^2)^(1/3))^2)])
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_)^2)^(-2/3), x_Symbol] :> Simp[3*(Sqrt[b*x^2]/(2*b*x)) Subst[Int[1/Sqrt[-a + x^3], x], x, (a + b*x^2)^(1/3)], x] /; FreeQ[{a, b }, x]
Int[1/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Simp[2*Sqrt[2 - Sqrt[3]]*(s + r*x)*(Sqrt[(s^2 - r*s *x + r^2*x^2)/((1 - Sqrt[3])*s + r*x)^2]/(3^(1/4)*r*Sqrt[a + b*x^3]*Sqrt[(- s)*((s + r*x)/((1 - Sqrt[3])*s + r*x)^2)]))*EllipticF[ArcSin[((1 + Sqrt[3]) *s + r*x)/((1 - Sqrt[3])*s + r*x)], -7 + 4*Sqrt[3]], x]] /; FreeQ[{a, b}, x ] && NegQ[a]
Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuot ient[Pq, a + b*x^2, x], f = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 1]}, Simp[(a*g - b *f*x)*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*(p + 1)) In t[(a + b*x^2)^(p + 1)*ExpandToSum[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]
\[\int \frac {C \,x^{2}+B x +A}{\left (b \,x^{2}+a \right )^{\frac {5}{3}}}d x\]
Input:
int((C*x^2+B*x+A)/(b*x^2+a)^(5/3),x)
Output:
int((C*x^2+B*x+A)/(b*x^2+a)^(5/3),x)
\[ \int \frac {A+B x+C x^2}{\left (a+b x^2\right )^{5/3}} \, dx=\int { \frac {C x^{2} + B x + A}{{\left (b x^{2} + a\right )}^{\frac {5}{3}}} \,d x } \] Input:
integrate((C*x^2+B*x+A)/(b*x^2+a)^(5/3),x, algorithm="fricas")
Output:
integral((C*x^2 + B*x + A)*(b*x^2 + a)^(1/3)/(b^2*x^4 + 2*a*b*x^2 + a^2), x)
Time = 2.68 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.29 \[ \int \frac {A+B x+C x^2}{\left (a+b x^2\right )^{5/3}} \, dx=\frac {A x {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {5}{3} \\ \frac {3}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{a^{\frac {5}{3}}} + B \left (\begin {cases} - \frac {3}{4 b \left (a + b x^{2}\right )^{\frac {2}{3}}} & \text {for}\: b \neq 0 \\\frac {x^{2}}{2 a^{\frac {5}{3}}} & \text {otherwise} \end {cases}\right ) + \frac {C x^{3} {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, \frac {5}{3} \\ \frac {5}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{3 a^{\frac {5}{3}}} \] Input:
integrate((C*x**2+B*x+A)/(b*x**2+a)**(5/3),x)
Output:
A*x*hyper((1/2, 5/3), (3/2,), b*x**2*exp_polar(I*pi)/a)/a**(5/3) + B*Piece wise((-3/(4*b*(a + b*x**2)**(2/3)), Ne(b, 0)), (x**2/(2*a**(5/3)), True)) + C*x**3*hyper((3/2, 5/3), (5/2,), b*x**2*exp_polar(I*pi)/a)/(3*a**(5/3))
\[ \int \frac {A+B x+C x^2}{\left (a+b x^2\right )^{5/3}} \, dx=\int { \frac {C x^{2} + B x + A}{{\left (b x^{2} + a\right )}^{\frac {5}{3}}} \,d x } \] Input:
integrate((C*x^2+B*x+A)/(b*x^2+a)^(5/3),x, algorithm="maxima")
Output:
integrate((C*x^2 + B*x + A)/(b*x^2 + a)^(5/3), x)
\[ \int \frac {A+B x+C x^2}{\left (a+b x^2\right )^{5/3}} \, dx=\int { \frac {C x^{2} + B x + A}{{\left (b x^{2} + a\right )}^{\frac {5}{3}}} \,d x } \] Input:
integrate((C*x^2+B*x+A)/(b*x^2+a)^(5/3),x, algorithm="giac")
Output:
integrate((C*x^2 + B*x + A)/(b*x^2 + a)^(5/3), x)
Timed out. \[ \int \frac {A+B x+C x^2}{\left (a+b x^2\right )^{5/3}} \, dx=\int \frac {C\,x^2+B\,x+A}{{\left (b\,x^2+a\right )}^{5/3}} \,d x \] Input:
int((A + B*x + C*x^2)/(a + b*x^2)^(5/3),x)
Output:
int((A + B*x + C*x^2)/(a + b*x^2)^(5/3), x)
\[ \int \frac {A+B x+C x^2}{\left (a+b x^2\right )^{5/3}} \, dx=\left (\int \frac {x^{2}}{\left (b \,x^{2}+a \right )^{\frac {2}{3}} a +\left (b \,x^{2}+a \right )^{\frac {2}{3}} b \,x^{2}}d x \right ) c +\left (\int \frac {x}{\left (b \,x^{2}+a \right )^{\frac {2}{3}} a +\left (b \,x^{2}+a \right )^{\frac {2}{3}} b \,x^{2}}d x \right ) b +\left (\int \frac {1}{\left (b \,x^{2}+a \right )^{\frac {2}{3}} a +\left (b \,x^{2}+a \right )^{\frac {2}{3}} b \,x^{2}}d x \right ) a \] Input:
int((C*x^2+B*x+A)/(b*x^2+a)^(5/3),x)
Output:
int(x**2/((a + b*x**2)**(2/3)*a + (a + b*x**2)**(2/3)*b*x**2),x)*c + int(x /((a + b*x**2)**(2/3)*a + (a + b*x**2)**(2/3)*b*x**2),x)*b + int(1/((a + b *x**2)**(2/3)*a + (a + b*x**2)**(2/3)*b*x**2),x)*a