Integrand size = 19, antiderivative size = 101 \[ \int \frac {(a+b x)^{5/6}}{(c+d x)^{29/6}} \, dx=\frac {6 (a+b x)^{11/6}}{23 (b c-a d) (c+d x)^{23/6}}+\frac {72 b (a+b x)^{11/6}}{391 (b c-a d)^2 (c+d x)^{17/6}}+\frac {432 b^2 (a+b x)^{11/6}}{4301 (b c-a d)^3 (c+d x)^{11/6}} \]
6/23*(b*x+a)^(11/6)/(-a*d+b*c)/(d*x+c)^(23/6)+72/391*b*(b*x+a)^(11/6)/(-a* d+b*c)^2/(d*x+c)^(17/6)+432/4301*b^2*(b*x+a)^(11/6)/(-a*d+b*c)^3/(d*x+c)^( 11/6)
Time = 0.43 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.76 \[ \int \frac {(a+b x)^{5/6}}{(c+d x)^{29/6}} \, dx=\frac {6 (a+b x)^{11/6} \left (187 a^2 d^2-22 a b d (23 c+6 d x)+b^2 \left (391 c^2+276 c d x+72 d^2 x^2\right )\right )}{4301 (b c-a d)^3 (c+d x)^{23/6}} \]
(6*(a + b*x)^(11/6)*(187*a^2*d^2 - 22*a*b*d*(23*c + 6*d*x) + b^2*(391*c^2 + 276*c*d*x + 72*d^2*x^2)))/(4301*(b*c - a*d)^3*(c + d*x)^(23/6))
Time = 0.19 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.13, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {55, 55, 48}
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.
\(\displaystyle \int \frac {(a+b x)^{5/6}}{(c+d x)^{29/6}} \, dx\) |
\(\Big \downarrow \) 55 |
\(\displaystyle \frac {12 b \int \frac {(a+b x)^{5/6}}{(c+d x)^{23/6}}dx}{23 (b c-a d)}+\frac {6 (a+b x)^{11/6}}{23 (c+d x)^{23/6} (b c-a d)}\) |
\(\Big \downarrow \) 55 |
\(\displaystyle \frac {12 b \left (\frac {6 b \int \frac {(a+b x)^{5/6}}{(c+d x)^{17/6}}dx}{17 (b c-a d)}+\frac {6 (a+b x)^{11/6}}{17 (c+d x)^{17/6} (b c-a d)}\right )}{23 (b c-a d)}+\frac {6 (a+b x)^{11/6}}{23 (c+d x)^{23/6} (b c-a d)}\) |
\(\Big \downarrow \) 48 |
\(\displaystyle \frac {6 (a+b x)^{11/6}}{23 (c+d x)^{23/6} (b c-a d)}+\frac {12 b \left (\frac {36 b (a+b x)^{11/6}}{187 (c+d x)^{11/6} (b c-a d)^2}+\frac {6 (a+b x)^{11/6}}{17 (c+d x)^{17/6} (b c-a d)}\right )}{23 (b c-a d)}\) |
(6*(a + b*x)^(11/6))/(23*(b*c - a*d)*(c + d*x)^(23/6)) + (12*b*((6*(a + b* x)^(11/6))/(17*(b*c - a*d)*(c + d*x)^(17/6)) + (36*b*(a + b*x)^(11/6))/(18 7*(b*c - a*d)^2*(c + d*x)^(11/6))))/(23*(b*c - a*d))
3.18.84.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp [(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] /; FreeQ[{ a, b, c, d, m, n}, x] && EqQ[m + n + 2, 0] && NeQ[m, -1]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(S implify[m + n + 2]/((b*c - a*d)*(m + 1))) Int[(a + b*x)^Simplify[m + 1]*( c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && ILtQ[Simplify[m + n + 2], 0] && NeQ[m, -1] && !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[ c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (SumSimplerQ[m, 1] || !SumSimp lerQ[n, 1])
Time = 0.75 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.04
method | result | size |
gosper | \(-\frac {6 \left (b x +a \right )^{\frac {11}{6}} \left (72 d^{2} x^{2} b^{2}-132 x a b \,d^{2}+276 x \,b^{2} c d +187 a^{2} d^{2}-506 a b c d +391 b^{2} c^{2}\right )}{4301 \left (d x +c \right )^{\frac {23}{6}} \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right )}\) | \(105\) |
-6/4301*(b*x+a)^(11/6)*(72*b^2*d^2*x^2-132*a*b*d^2*x+276*b^2*c*d*x+187*a^2 *d^2-506*a*b*c*d+391*b^2*c^2)/(d*x+c)^(23/6)/(a^3*d^3-3*a^2*b*c*d^2+3*a*b^ 2*c^2*d-b^3*c^3)
Leaf count of result is larger than twice the leaf count of optimal. 338 vs. \(2 (83) = 166\).
Time = 0.24 (sec) , antiderivative size = 338, normalized size of antiderivative = 3.35 \[ \int \frac {(a+b x)^{5/6}}{(c+d x)^{29/6}} \, dx=\frac {6 \, {\left (72 \, b^{3} d^{2} x^{3} + 391 \, a b^{2} c^{2} - 506 \, a^{2} b c d + 187 \, a^{3} d^{2} + 12 \, {\left (23 \, b^{3} c d - 5 \, a b^{2} d^{2}\right )} x^{2} + {\left (391 \, b^{3} c^{2} - 230 \, a b^{2} c d + 55 \, a^{2} b d^{2}\right )} x\right )} {\left (b x + a\right )}^{\frac {5}{6}} {\left (d x + c\right )}^{\frac {1}{6}}}{4301 \, {\left (b^{3} c^{7} - 3 \, a b^{2} c^{6} d + 3 \, a^{2} b c^{5} d^{2} - a^{3} c^{4} d^{3} + {\left (b^{3} c^{3} d^{4} - 3 \, a b^{2} c^{2} d^{5} + 3 \, a^{2} b c d^{6} - a^{3} d^{7}\right )} x^{4} + 4 \, {\left (b^{3} c^{4} d^{3} - 3 \, a b^{2} c^{3} d^{4} + 3 \, a^{2} b c^{2} d^{5} - a^{3} c d^{6}\right )} x^{3} + 6 \, {\left (b^{3} c^{5} d^{2} - 3 \, a b^{2} c^{4} d^{3} + 3 \, a^{2} b c^{3} d^{4} - a^{3} c^{2} d^{5}\right )} x^{2} + 4 \, {\left (b^{3} c^{6} d - 3 \, a b^{2} c^{5} d^{2} + 3 \, a^{2} b c^{4} d^{3} - a^{3} c^{3} d^{4}\right )} x\right )}} \]
6/4301*(72*b^3*d^2*x^3 + 391*a*b^2*c^2 - 506*a^2*b*c*d + 187*a^3*d^2 + 12* (23*b^3*c*d - 5*a*b^2*d^2)*x^2 + (391*b^3*c^2 - 230*a*b^2*c*d + 55*a^2*b*d ^2)*x)*(b*x + a)^(5/6)*(d*x + c)^(1/6)/(b^3*c^7 - 3*a*b^2*c^6*d + 3*a^2*b* c^5*d^2 - a^3*c^4*d^3 + (b^3*c^3*d^4 - 3*a*b^2*c^2*d^5 + 3*a^2*b*c*d^6 - a ^3*d^7)*x^4 + 4*(b^3*c^4*d^3 - 3*a*b^2*c^3*d^4 + 3*a^2*b*c^2*d^5 - a^3*c*d ^6)*x^3 + 6*(b^3*c^5*d^2 - 3*a*b^2*c^4*d^3 + 3*a^2*b*c^3*d^4 - a^3*c^2*d^5 )*x^2 + 4*(b^3*c^6*d - 3*a*b^2*c^5*d^2 + 3*a^2*b*c^4*d^3 - a^3*c^3*d^4)*x)
Timed out. \[ \int \frac {(a+b x)^{5/6}}{(c+d x)^{29/6}} \, dx=\text {Timed out} \]
\[ \int \frac {(a+b x)^{5/6}}{(c+d x)^{29/6}} \, dx=\int { \frac {{\left (b x + a\right )}^{\frac {5}{6}}}{{\left (d x + c\right )}^{\frac {29}{6}}} \,d x } \]
\[ \int \frac {(a+b x)^{5/6}}{(c+d x)^{29/6}} \, dx=\int { \frac {{\left (b x + a\right )}^{\frac {5}{6}}}{{\left (d x + c\right )}^{\frac {29}{6}}} \,d x } \]
Time = 0.90 (sec) , antiderivative size = 214, normalized size of antiderivative = 2.12 \[ \int \frac {(a+b x)^{5/6}}{(c+d x)^{29/6}} \, dx=-\frac {{\left (c+d\,x\right )}^{1/6}\,\left (\frac {{\left (a+b\,x\right )}^{5/6}\,\left (1122\,a^3\,d^2-3036\,a^2\,b\,c\,d+2346\,a\,b^2\,c^2\right )}{4301\,d^4\,{\left (a\,d-b\,c\right )}^3}+\frac {432\,b^3\,x^3\,{\left (a+b\,x\right )}^{5/6}}{4301\,d^2\,{\left (a\,d-b\,c\right )}^3}+\frac {x\,{\left (a+b\,x\right )}^{5/6}\,\left (330\,a^2\,b\,d^2-1380\,a\,b^2\,c\,d+2346\,b^3\,c^2\right )}{4301\,d^4\,{\left (a\,d-b\,c\right )}^3}-\frac {72\,b^2\,x^2\,\left (5\,a\,d-23\,b\,c\right )\,{\left (a+b\,x\right )}^{5/6}}{4301\,d^3\,{\left (a\,d-b\,c\right )}^3}\right )}{x^4+\frac {c^4}{d^4}+\frac {4\,c\,x^3}{d}+\frac {4\,c^3\,x}{d^3}+\frac {6\,c^2\,x^2}{d^2}} \]
-((c + d*x)^(1/6)*(((a + b*x)^(5/6)*(1122*a^3*d^2 + 2346*a*b^2*c^2 - 3036* a^2*b*c*d))/(4301*d^4*(a*d - b*c)^3) + (432*b^3*x^3*(a + b*x)^(5/6))/(4301 *d^2*(a*d - b*c)^3) + (x*(a + b*x)^(5/6)*(2346*b^3*c^2 + 330*a^2*b*d^2 - 1 380*a*b^2*c*d))/(4301*d^4*(a*d - b*c)^3) - (72*b^2*x^2*(5*a*d - 23*b*c)*(a + b*x)^(5/6))/(4301*d^3*(a*d - b*c)^3)))/(x^4 + c^4/d^4 + (4*c*x^3)/d + ( 4*c^3*x)/d^3 + (6*c^2*x^2)/d^2)
\[ \int \frac {(a+b x)^{5/6}}{(c+d x)^{29/6}} \, dx=\int \frac {\left (b x +a \right )^{\frac {5}{6}}}{\left (d x +c \right )^{\frac {5}{6}} \left (d^{4} x^{4}+4 c \,d^{3} x^{3}+6 c^{2} d^{2} x^{2}+4 c^{3} d x +c^{4}\right )}d x \]
int((a + b*x)**(5/6)/((c + d*x)**(5/6)*(c**4 + 4*c**3*d*x + 6*c**2*d**2*x* *2 + 4*c*d**3*x**3 + d**4*x**4)),x)