Integrand size = 19, antiderivative size = 66 \[ \int \frac {1}{(a+b x)^{5/6} (c+d x)^{13/6}} \, dx=\frac {6 \sqrt [6]{a+b x}}{7 (b c-a d) (c+d x)^{7/6}}+\frac {36 b \sqrt [6]{a+b x}}{7 (b c-a d)^2 \sqrt [6]{c+d x}} \] Output:
6/7*(b*x+a)^(1/6)/(-a*d+b*c)/(d*x+c)^(7/6)+36/7*b*(b*x+a)^(1/6)/(-a*d+b*c) ^2/(d*x+c)^(1/6)
Time = 0.16 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.70 \[ \int \frac {1}{(a+b x)^{5/6} (c+d x)^{13/6}} \, dx=\frac {6 \sqrt [6]{a+b x} (7 b c-a d+6 b d x)}{7 (b c-a d)^2 (c+d x)^{7/6}} \] Input:
Integrate[1/((a + b*x)^(5/6)*(c + d*x)^(13/6)),x]
Output:
(6*(a + b*x)^(1/6)*(7*b*c - a*d + 6*b*d*x))/(7*(b*c - a*d)^2*(c + d*x)^(7/ 6))
Time = 0.16 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {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.
\(\displaystyle \int \frac {1}{(a+b x)^{5/6} (c+d x)^{13/6}} \, dx\) |
\(\Big \downarrow \) 55 |
\(\displaystyle \frac {6 b \int \frac {1}{(a+b x)^{5/6} (c+d x)^{7/6}}dx}{7 (b c-a d)}+\frac {6 \sqrt [6]{a+b x}}{7 (c+d x)^{7/6} (b c-a d)}\) |
\(\Big \downarrow \) 48 |
\(\displaystyle \frac {36 b \sqrt [6]{a+b x}}{7 \sqrt [6]{c+d x} (b c-a d)^2}+\frac {6 \sqrt [6]{a+b x}}{7 (c+d x)^{7/6} (b c-a d)}\) |
Input:
Int[1/((a + b*x)^(5/6)*(c + d*x)^(13/6)),x]
Output:
(6*(a + b*x)^(1/6))/(7*(b*c - a*d)*(c + d*x)^(7/6)) + (36*b*(a + b*x)^(1/6 ))/(7*(b*c - a*d)^2*(c + d*x)^(1/6))
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 = 0.18 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.80
method | result | size |
gosper | \(-\frac {6 \left (b x +a \right )^{\frac {1}{6}} \left (-6 b d x +a d -7 b c \right )}{7 \left (x d +c \right )^{\frac {7}{6}} \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}\) | \(53\) |
orering | \(-\frac {6 \left (b x +a \right )^{\frac {1}{6}} \left (-6 b d x +a d -7 b c \right )}{7 \left (x d +c \right )^{\frac {7}{6}} \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}\) | \(53\) |
Input:
int(1/(b*x+a)^(5/6)/(d*x+c)^(13/6),x,method=_RETURNVERBOSE)
Output:
-6/7*(b*x+a)^(1/6)*(-6*b*d*x+a*d-7*b*c)/(d*x+c)^(7/6)/(a^2*d^2-2*a*b*c*d+b ^2*c^2)
Leaf count of result is larger than twice the leaf count of optimal. 118 vs. \(2 (54) = 108\).
Time = 0.08 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.79 \[ \int \frac {1}{(a+b x)^{5/6} (c+d x)^{13/6}} \, dx=\frac {6 \, {\left (6 \, b d x + 7 \, b c - a d\right )} {\left (b x + a\right )}^{\frac {1}{6}} {\left (d x + c\right )}^{\frac {5}{6}}}{7 \, {\left (b^{2} c^{4} - 2 \, a b c^{3} d + a^{2} c^{2} d^{2} + {\left (b^{2} c^{2} d^{2} - 2 \, a b c d^{3} + a^{2} d^{4}\right )} x^{2} + 2 \, {\left (b^{2} c^{3} d - 2 \, a b c^{2} d^{2} + a^{2} c d^{3}\right )} x\right )}} \] Input:
integrate(1/(b*x+a)^(5/6)/(d*x+c)^(13/6),x, algorithm="fricas")
Output:
6/7*(6*b*d*x + 7*b*c - a*d)*(b*x + a)^(1/6)*(d*x + c)^(5/6)/(b^2*c^4 - 2*a *b*c^3*d + a^2*c^2*d^2 + (b^2*c^2*d^2 - 2*a*b*c*d^3 + a^2*d^4)*x^2 + 2*(b^ 2*c^3*d - 2*a*b*c^2*d^2 + a^2*c*d^3)*x)
\[ \int \frac {1}{(a+b x)^{5/6} (c+d x)^{13/6}} \, dx=\int \frac {1}{\left (a + b x\right )^{\frac {5}{6}} \left (c + d x\right )^{\frac {13}{6}}}\, dx \] Input:
integrate(1/(b*x+a)**(5/6)/(d*x+c)**(13/6),x)
Output:
Integral(1/((a + b*x)**(5/6)*(c + d*x)**(13/6)), x)
\[ \int \frac {1}{(a+b x)^{5/6} (c+d x)^{13/6}} \, dx=\int { \frac {1}{{\left (b x + a\right )}^{\frac {5}{6}} {\left (d x + c\right )}^{\frac {13}{6}}} \,d x } \] Input:
integrate(1/(b*x+a)^(5/6)/(d*x+c)^(13/6),x, algorithm="maxima")
Output:
integrate(1/((b*x + a)^(5/6)*(d*x + c)^(13/6)), x)
\[ \int \frac {1}{(a+b x)^{5/6} (c+d x)^{13/6}} \, dx=\int { \frac {1}{{\left (b x + a\right )}^{\frac {5}{6}} {\left (d x + c\right )}^{\frac {13}{6}}} \,d x } \] Input:
integrate(1/(b*x+a)^(5/6)/(d*x+c)^(13/6),x, algorithm="giac")
Output:
integrate(1/((b*x + a)^(5/6)*(d*x + c)^(13/6)), x)
Timed out. \[ \int \frac {1}{(a+b x)^{5/6} (c+d x)^{13/6}} \, dx=\int \frac {1}{{\left (a+b\,x\right )}^{5/6}\,{\left (c+d\,x\right )}^{13/6}} \,d x \] Input:
int(1/((a + b*x)^(5/6)*(c + d*x)^(13/6)),x)
Output:
int(1/((a + b*x)^(5/6)*(c + d*x)^(13/6)), x)
Time = 0.68 (sec) , antiderivative size = 340, normalized size of antiderivative = 5.15 \[ \int \frac {1}{(a+b x)^{5/6} (c+d x)^{13/6}} \, dx=\frac {2 \left (d x +c \right )^{\frac {5}{6}} \left (-\left (b x +a \right )^{\frac {3}{2}} a d +\left (b x +a \right )^{\frac {3}{2}} b c +12 \sqrt {b x +a}\, \mathrm {log}\left (\left (b x +a \right )^{\frac {1}{6}}\right ) a b c +12 \sqrt {b x +a}\, \mathrm {log}\left (\left (b x +a \right )^{\frac {1}{6}}\right ) a b d x +12 \sqrt {b x +a}\, \mathrm {log}\left (\left (b x +a \right )^{\frac {1}{6}}\right ) b^{2} c x +12 \sqrt {b x +a}\, \mathrm {log}\left (\left (b x +a \right )^{\frac {1}{6}}\right ) b^{2} d \,x^{2}-12 \sqrt {b x +a}\, \mathrm {log}\left (\left (d x +c \right )^{\frac {1}{6}}\right ) a b c -12 \sqrt {b x +a}\, \mathrm {log}\left (\left (d x +c \right )^{\frac {1}{6}}\right ) a b d x -12 \sqrt {b x +a}\, \mathrm {log}\left (\left (d x +c \right )^{\frac {1}{6}}\right ) b^{2} c x -12 \sqrt {b x +a}\, \mathrm {log}\left (\left (d x +c \right )^{\frac {1}{6}}\right ) b^{2} d \,x^{2}\right )}{3 \left (b x +a \right )^{\frac {1}{3}} \left (a^{2} b \,d^{4} x^{3}-2 a \,b^{2} c \,d^{3} x^{3}+b^{3} c^{2} d^{2} x^{3}+a^{3} d^{4} x^{2}-3 a \,b^{2} c^{2} d^{2} x^{2}+2 b^{3} c^{3} d \,x^{2}+2 a^{3} c \,d^{3} x -3 a^{2} b \,c^{2} d^{2} x +b^{3} c^{4} x +a^{3} c^{2} d^{2}-2 a^{2} b \,c^{3} d +a \,b^{2} c^{4}\right )} \] Input:
int(1/(b*x+a)^(5/6)/(d*x+c)^(13/6),x)
Output:
(2*(c + d*x)**(5/6)*( - (a + b*x)**(3/2)*a*d + (a + b*x)**(3/2)*b*c + 12*s qrt(a + b*x)*log((a + b*x)**(1/6))*a*b*c + 12*sqrt(a + b*x)*log((a + b*x)* *(1/6))*a*b*d*x + 12*sqrt(a + b*x)*log((a + b*x)**(1/6))*b**2*c*x + 12*sqr t(a + b*x)*log((a + b*x)**(1/6))*b**2*d*x**2 - 12*sqrt(a + b*x)*log((c + d *x)**(1/6))*a*b*c - 12*sqrt(a + b*x)*log((c + d*x)**(1/6))*a*b*d*x - 12*sq rt(a + b*x)*log((c + d*x)**(1/6))*b**2*c*x - 12*sqrt(a + b*x)*log((c + d*x )**(1/6))*b**2*d*x**2))/(3*(a + b*x)**(1/3)*(a**3*c**2*d**2 + 2*a**3*c*d** 3*x + a**3*d**4*x**2 - 2*a**2*b*c**3*d - 3*a**2*b*c**2*d**2*x + a**2*b*d** 4*x**3 + a*b**2*c**4 - 3*a*b**2*c**2*d**2*x**2 - 2*a*b**2*c*d**3*x**3 + b* *3*c**4*x + 2*b**3*c**3*d*x**2 + b**3*c**2*d**2*x**3))