Integrand size = 28, antiderivative size = 89 \[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{23/3}} \, dx=\frac {3 \left (a+b x+c x^2\right )^{7/3}}{10 \left (b^2-4 a c\right ) d (b d+2 c d x)^{20/3}}+\frac {9 \left (a+b x+c x^2\right )^{7/3}}{70 \left (b^2-4 a c\right )^2 d^3 (b d+2 c d x)^{14/3}} \] Output:
3/10*(c*x^2+b*x+a)^(7/3)/(-4*a*c+b^2)/d/(2*c*d*x+b*d)^(20/3)+9/70*(c*x^2+b *x+a)^(7/3)/(-4*a*c+b^2)^2/d^3/(2*c*d*x+b*d)^(14/3)
Time = 5.00 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.83 \[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{23/3}} \, dx=\frac {3 \sqrt [3]{d (b+2 c x)} (a+x (b+c x))^{7/3} \left (5 b^2+6 b c x+2 c \left (-7 a+3 c x^2\right )\right )}{35 \left (b^2-4 a c\right )^2 d^8 (b+2 c x)^7} \] Input:
Integrate[(a + b*x + c*x^2)^(4/3)/(b*d + 2*c*d*x)^(23/3),x]
Output:
(3*(d*(b + 2*c*x))^(1/3)*(a + x*(b + c*x))^(7/3)*(5*b^2 + 6*b*c*x + 2*c*(- 7*a + 3*c*x^2)))/(35*(b^2 - 4*a*c)^2*d^8*(b + 2*c*x)^7)
Time = 0.22 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {1117, 1106}
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.
\(\displaystyle \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{23/3}} \, dx\) |
\(\Big \downarrow \) 1117 |
\(\displaystyle \frac {3 \int \frac {\left (c x^2+b x+a\right )^{4/3}}{(b d+2 c x d)^{17/3}}dx}{10 d^2 \left (b^2-4 a c\right )}+\frac {3 \left (a+b x+c x^2\right )^{7/3}}{10 d \left (b^2-4 a c\right ) (b d+2 c d x)^{20/3}}\) |
\(\Big \downarrow \) 1106 |
\(\displaystyle \frac {9 \left (a+b x+c x^2\right )^{7/3}}{70 d^3 \left (b^2-4 a c\right )^2 (b d+2 c d x)^{14/3}}+\frac {3 \left (a+b x+c x^2\right )^{7/3}}{10 d \left (b^2-4 a c\right ) (b d+2 c d x)^{20/3}}\) |
Input:
Int[(a + b*x + c*x^2)^(4/3)/(b*d + 2*c*d*x)^(23/3),x]
Output:
(3*(a + b*x + c*x^2)^(7/3))/(10*(b^2 - 4*a*c)*d*(b*d + 2*c*d*x)^(20/3)) + (9*(a + b*x + c*x^2)^(7/3))/(70*(b^2 - 4*a*c)^2*d^3*(b*d + 2*c*d*x)^(14/3) )
Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_S ymbol] :> Simp[2*c*(d + e*x)^(m + 1)*((a + b*x + c*x^2)^(p + 1)/(e*(p + 1)* (b^2 - 4*a*c))), x] /; FreeQ[{a, b, c, d, e, m, p}, x] && EqQ[2*c*d - b*e, 0] && EqQ[m + 2*p + 3, 0]
Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_S ymbol] :> Simp[-2*b*d*(d + e*x)^(m + 1)*((a + b*x + c*x^2)^(p + 1)/(d^2*(m + 1)*(b^2 - 4*a*c))), x] + Simp[b^2*((m + 2*p + 3)/(d^2*(m + 1)*(b^2 - 4*a* c))) Int[(d + e*x)^(m + 2)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[2*c*d - b*e, 0] && NeQ[m + 2*p + 3, 0] && LtQ[m, -1] & & (IntegerQ[2*p] || (IntegerQ[m] && RationalQ[p]) || IntegerQ[(m + 2*p + 3) /2])
Time = 1.00 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.85
method | result | size |
gosper | \(-\frac {3 \left (c \,x^{2}+b x +a \right )^{\frac {7}{3}} \left (2 c x +b \right ) \left (-6 c^{2} x^{2}-6 c b x +14 a c -5 b^{2}\right )}{35 \left (16 a^{2} c^{2}-8 c a \,b^{2}+b^{4}\right ) \left (2 c d x +b d \right )^{\frac {23}{3}}}\) | \(76\) |
orering | \(-\frac {3 \left (c \,x^{2}+b x +a \right )^{\frac {7}{3}} \left (2 c x +b \right ) \left (-6 c^{2} x^{2}-6 c b x +14 a c -5 b^{2}\right )}{35 \left (16 a^{2} c^{2}-8 c a \,b^{2}+b^{4}\right ) \left (2 c d x +b d \right )^{\frac {23}{3}}}\) | \(76\) |
Input:
int((c*x^2+b*x+a)^(4/3)/(2*c*d*x+b*d)^(23/3),x,method=_RETURNVERBOSE)
Output:
-3/35*(c*x^2+b*x+a)^(7/3)*(2*c*x+b)*(-6*c^2*x^2-6*b*c*x+14*a*c-5*b^2)/(16* a^2*c^2-8*a*b^2*c+b^4)/(2*c*d*x+b*d)^(23/3)
Leaf count of result is larger than twice the leaf count of optimal. 409 vs. \(2 (77) = 154\).
Time = 0.20 (sec) , antiderivative size = 409, normalized size of antiderivative = 4.60 \[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{23/3}} \, dx=\frac {3 \, {\left (6 \, c^{4} x^{6} + 18 \, b c^{3} x^{5} + {\left (23 \, b^{2} c^{2} - 2 \, a c^{3}\right )} x^{4} + 5 \, a^{2} b^{2} - 14 \, a^{3} c + 4 \, {\left (4 \, b^{3} c - a b c^{2}\right )} x^{3} + {\left (5 \, b^{4} + 8 \, a b^{2} c - 22 \, a^{2} c^{2}\right )} x^{2} + 2 \, {\left (5 \, a b^{3} - 11 \, a^{2} b c\right )} x\right )} {\left (2 \, c d x + b d\right )}^{\frac {1}{3}} {\left (c x^{2} + b x + a\right )}^{\frac {1}{3}}}{35 \, {\left (128 \, {\left (b^{4} c^{7} - 8 \, a b^{2} c^{8} + 16 \, a^{2} c^{9}\right )} d^{8} x^{7} + 448 \, {\left (b^{5} c^{6} - 8 \, a b^{3} c^{7} + 16 \, a^{2} b c^{8}\right )} d^{8} x^{6} + 672 \, {\left (b^{6} c^{5} - 8 \, a b^{4} c^{6} + 16 \, a^{2} b^{2} c^{7}\right )} d^{8} x^{5} + 560 \, {\left (b^{7} c^{4} - 8 \, a b^{5} c^{5} + 16 \, a^{2} b^{3} c^{6}\right )} d^{8} x^{4} + 280 \, {\left (b^{8} c^{3} - 8 \, a b^{6} c^{4} + 16 \, a^{2} b^{4} c^{5}\right )} d^{8} x^{3} + 84 \, {\left (b^{9} c^{2} - 8 \, a b^{7} c^{3} + 16 \, a^{2} b^{5} c^{4}\right )} d^{8} x^{2} + 14 \, {\left (b^{10} c - 8 \, a b^{8} c^{2} + 16 \, a^{2} b^{6} c^{3}\right )} d^{8} x + {\left (b^{11} - 8 \, a b^{9} c + 16 \, a^{2} b^{7} c^{2}\right )} d^{8}\right )}} \] Input:
integrate((c*x^2+b*x+a)^(4/3)/(2*c*d*x+b*d)^(23/3),x, algorithm="fricas")
Output:
3/35*(6*c^4*x^6 + 18*b*c^3*x^5 + (23*b^2*c^2 - 2*a*c^3)*x^4 + 5*a^2*b^2 - 14*a^3*c + 4*(4*b^3*c - a*b*c^2)*x^3 + (5*b^4 + 8*a*b^2*c - 22*a^2*c^2)*x^ 2 + 2*(5*a*b^3 - 11*a^2*b*c)*x)*(2*c*d*x + b*d)^(1/3)*(c*x^2 + b*x + a)^(1 /3)/(128*(b^4*c^7 - 8*a*b^2*c^8 + 16*a^2*c^9)*d^8*x^7 + 448*(b^5*c^6 - 8*a *b^3*c^7 + 16*a^2*b*c^8)*d^8*x^6 + 672*(b^6*c^5 - 8*a*b^4*c^6 + 16*a^2*b^2 *c^7)*d^8*x^5 + 560*(b^7*c^4 - 8*a*b^5*c^5 + 16*a^2*b^3*c^6)*d^8*x^4 + 280 *(b^8*c^3 - 8*a*b^6*c^4 + 16*a^2*b^4*c^5)*d^8*x^3 + 84*(b^9*c^2 - 8*a*b^7* c^3 + 16*a^2*b^5*c^4)*d^8*x^2 + 14*(b^10*c - 8*a*b^8*c^2 + 16*a^2*b^6*c^3) *d^8*x + (b^11 - 8*a*b^9*c + 16*a^2*b^7*c^2)*d^8)
Timed out. \[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{23/3}} \, dx=\text {Timed out} \] Input:
integrate((c*x**2+b*x+a)**(4/3)/(2*c*d*x+b*d)**(23/3),x)
Output:
Timed out
\[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{23/3}} \, dx=\int { \frac {{\left (c x^{2} + b x + a\right )}^{\frac {4}{3}}}{{\left (2 \, c d x + b d\right )}^{\frac {23}{3}}} \,d x } \] Input:
integrate((c*x^2+b*x+a)^(4/3)/(2*c*d*x+b*d)^(23/3),x, algorithm="maxima")
Output:
integrate((c*x^2 + b*x + a)^(4/3)/(2*c*d*x + b*d)^(23/3), x)
\[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{23/3}} \, dx=\int { \frac {{\left (c x^{2} + b x + a\right )}^{\frac {4}{3}}}{{\left (2 \, c d x + b d\right )}^{\frac {23}{3}}} \,d x } \] Input:
integrate((c*x^2+b*x+a)^(4/3)/(2*c*d*x+b*d)^(23/3),x, algorithm="giac")
Output:
integrate((c*x^2 + b*x + a)^(4/3)/(2*c*d*x + b*d)^(23/3), x)
Timed out. \[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{23/3}} \, dx=\int \frac {{\left (c\,x^2+b\,x+a\right )}^{4/3}}{{\left (b\,d+2\,c\,d\,x\right )}^{23/3}} \,d x \] Input:
int((a + b*x + c*x^2)^(4/3)/(b*d + 2*c*d*x)^(23/3),x)
Output:
int((a + b*x + c*x^2)^(4/3)/(b*d + 2*c*d*x)^(23/3), x)
\[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{23/3}} \, dx=\int \frac {\left (c \,x^{2}+b x +a \right )^{\frac {4}{3}}}{\left (2 c d x +b d \right )^{\frac {23}{3}}}d x \] Input:
int((c*x^2+b*x+a)^(4/3)/(2*c*d*x+b*d)^(23/3),x)
Output:
int((c*x^2+b*x+a)^(4/3)/(2*c*d*x+b*d)^(23/3),x)