Integrand size = 19, antiderivative size = 66 \[ \int \frac {(a+b x)^{7/6}}{(c+d x)^{25/6}} \, dx=\frac {6 (a+b x)^{13/6}}{19 (b c-a d) (c+d x)^{19/6}}+\frac {36 b (a+b x)^{13/6}}{247 (b c-a d)^2 (c+d x)^{13/6}} \] Output:
6/19*(b*x+a)^(13/6)/(-a*d+b*c)/(d*x+c)^(19/6)+36/247*b*(b*x+a)^(13/6)/(-a* d+b*c)^2/(d*x+c)^(13/6)
Time = 0.24 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.70 \[ \int \frac {(a+b x)^{7/6}}{(c+d x)^{25/6}} \, dx=\frac {6 (a+b x)^{13/6} (19 b c-13 a d+6 b d x)}{247 (b c-a d)^2 (c+d x)^{19/6}} \] Input:
Integrate[(a + b*x)^(7/6)/(c + d*x)^(25/6),x]
Output:
(6*(a + b*x)^(13/6)*(19*b*c - 13*a*d + 6*b*d*x))/(247*(b*c - a*d)^2*(c + d *x)^(19/6))
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.
\(\displaystyle \int \frac {(a+b x)^{7/6}}{(c+d x)^{25/6}} \, dx\) |
\(\Big \downarrow \) 55 |
\(\displaystyle \frac {6 b \int \frac {(a+b x)^{7/6}}{(c+d x)^{19/6}}dx}{19 (b c-a d)}+\frac {6 (a+b x)^{13/6}}{19 (c+d x)^{19/6} (b c-a d)}\) |
\(\Big \downarrow \) 48 |
\(\displaystyle \frac {36 b (a+b x)^{13/6}}{247 (c+d x)^{13/6} (b c-a d)^2}+\frac {6 (a+b x)^{13/6}}{19 (c+d x)^{19/6} (b c-a d)}\) |
Input:
Int[(a + b*x)^(7/6)/(c + d*x)^(25/6),x]
Output:
(6*(a + b*x)^(13/6))/(19*(b*c - a*d)*(c + d*x)^(19/6)) + (36*b*(a + b*x)^( 13/6))/(247*(b*c - a*d)^2*(c + d*x)^(13/6))
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 = 54, normalized size of antiderivative = 0.82
method | result | size |
gosper | \(-\frac {6 \left (b x +a \right )^{\frac {13}{6}} \left (-6 b d x +13 a d -19 b c \right )}{247 \left (x d +c \right )^{\frac {19}{6}} \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}\) | \(54\) |
orering | \(-\frac {6 \left (b x +a \right )^{\frac {13}{6}} \left (-6 b d x +13 a d -19 b c \right )}{247 \left (x d +c \right )^{\frac {19}{6}} \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}\) | \(54\) |
Input:
int((b*x+a)^(7/6)/(d*x+c)^(25/6),x,method=_RETURNVERBOSE)
Output:
-6/247*(b*x+a)^(13/6)*(-6*b*d*x+13*a*d-19*b*c)/(d*x+c)^(19/6)/(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. 235 vs. \(2 (54) = 108\).
Time = 0.11 (sec) , antiderivative size = 235, normalized size of antiderivative = 3.56 \[ \int \frac {(a+b x)^{7/6}}{(c+d x)^{25/6}} \, dx=\frac {6 \, {\left (6 \, b^{3} d x^{3} + 19 \, a^{2} b c - 13 \, a^{3} d + {\left (19 \, b^{3} c - a b^{2} d\right )} x^{2} + 2 \, {\left (19 \, a b^{2} c - 10 \, a^{2} b d\right )} x\right )} {\left (b x + a\right )}^{\frac {1}{6}} {\left (d x + c\right )}^{\frac {5}{6}}}{247 \, {\left (b^{2} c^{6} - 2 \, a b c^{5} d + a^{2} c^{4} d^{2} + {\left (b^{2} c^{2} d^{4} - 2 \, a b c d^{5} + a^{2} d^{6}\right )} x^{4} + 4 \, {\left (b^{2} c^{3} d^{3} - 2 \, a b c^{2} d^{4} + a^{2} c d^{5}\right )} x^{3} + 6 \, {\left (b^{2} c^{4} d^{2} - 2 \, a b c^{3} d^{3} + a^{2} c^{2} d^{4}\right )} x^{2} + 4 \, {\left (b^{2} c^{5} d - 2 \, a b c^{4} d^{2} + a^{2} c^{3} d^{3}\right )} x\right )}} \] Input:
integrate((b*x+a)^(7/6)/(d*x+c)^(25/6),x, algorithm="fricas")
Output:
6/247*(6*b^3*d*x^3 + 19*a^2*b*c - 13*a^3*d + (19*b^3*c - a*b^2*d)*x^2 + 2* (19*a*b^2*c - 10*a^2*b*d)*x)*(b*x + a)^(1/6)*(d*x + c)^(5/6)/(b^2*c^6 - 2* a*b*c^5*d + a^2*c^4*d^2 + (b^2*c^2*d^4 - 2*a*b*c*d^5 + a^2*d^6)*x^4 + 4*(b ^2*c^3*d^3 - 2*a*b*c^2*d^4 + a^2*c*d^5)*x^3 + 6*(b^2*c^4*d^2 - 2*a*b*c^3*d ^3 + a^2*c^2*d^4)*x^2 + 4*(b^2*c^5*d - 2*a*b*c^4*d^2 + a^2*c^3*d^3)*x)
Timed out. \[ \int \frac {(a+b x)^{7/6}}{(c+d x)^{25/6}} \, dx=\text {Timed out} \] Input:
integrate((b*x+a)**(7/6)/(d*x+c)**(25/6),x)
Output:
Timed out
\[ \int \frac {(a+b x)^{7/6}}{(c+d x)^{25/6}} \, dx=\int { \frac {{\left (b x + a\right )}^{\frac {7}{6}}}{{\left (d x + c\right )}^{\frac {25}{6}}} \,d x } \] Input:
integrate((b*x+a)^(7/6)/(d*x+c)^(25/6),x, algorithm="maxima")
Output:
integrate((b*x + a)^(7/6)/(d*x + c)^(25/6), x)
\[ \int \frac {(a+b x)^{7/6}}{(c+d x)^{25/6}} \, dx=\int { \frac {{\left (b x + a\right )}^{\frac {7}{6}}}{{\left (d x + c\right )}^{\frac {25}{6}}} \,d x } \] Input:
integrate((b*x+a)^(7/6)/(d*x+c)^(25/6),x, algorithm="giac")
Output:
integrate((b*x + a)^(7/6)/(d*x + c)^(25/6), x)
Time = 0.52 (sec) , antiderivative size = 189, normalized size of antiderivative = 2.86 \[ \int \frac {(a+b x)^{7/6}}{(c+d x)^{25/6}} \, dx=-\frac {{\left (c+d\,x\right )}^{5/6}\,\left (\frac {\left (78\,a^3\,d-114\,a^2\,b\,c\right )\,{\left (a+b\,x\right )}^{1/6}}{247\,d^4\,{\left (a\,d-b\,c\right )}^2}-\frac {36\,b^3\,x^3\,{\left (a+b\,x\right )}^{1/6}}{247\,d^3\,{\left (a\,d-b\,c\right )}^2}-\frac {x^2\,\left (114\,b^3\,c-6\,a\,b^2\,d\right )\,{\left (a+b\,x\right )}^{1/6}}{247\,d^4\,{\left (a\,d-b\,c\right )}^2}+\frac {12\,a\,b\,x\,\left (10\,a\,d-19\,b\,c\right )\,{\left (a+b\,x\right )}^{1/6}}{247\,d^4\,{\left (a\,d-b\,c\right )}^2}\right )}{x^4+\frac {c^4}{d^4}+\frac {4\,c\,x^3}{d}+\frac {4\,c^3\,x}{d^3}+\frac {6\,c^2\,x^2}{d^2}} \] Input:
int((a + b*x)^(7/6)/(c + d*x)^(25/6),x)
Output:
-((c + d*x)^(5/6)*(((78*a^3*d - 114*a^2*b*c)*(a + b*x)^(1/6))/(247*d^4*(a* d - b*c)^2) - (36*b^3*x^3*(a + b*x)^(1/6))/(247*d^3*(a*d - b*c)^2) - (x^2* (114*b^3*c - 6*a*b^2*d)*(a + b*x)^(1/6))/(247*d^4*(a*d - b*c)^2) + (12*a*b *x*(10*a*d - 19*b*c)*(a + b*x)^(1/6))/(247*d^4*(a*d - b*c)^2)))/(x^4 + c^4 /d^4 + (4*c*x^3)/d + (4*c^3*x)/d^3 + (6*c^2*x^2)/d^2)
Time = 14.40 (sec) , antiderivative size = 1312, normalized size of antiderivative = 19.88 \[ \int \frac {(a+b x)^{7/6}}{(c+d x)^{25/6}} \, dx =\text {Too large to display} \] Input:
int((b*x+a)^(7/6)/(d*x+c)^(25/6),x)
Output:
(2*( - 2470*(c + d*x)*(a + b*x)**(3/2)*a**3*b*d**3 + 9633*(c + d*x)*(a + b *x)**(3/2)*a**2*b**2*c*d**2 + 2223*(c + d*x)*(a + b*x)**(3/2)*a**2*b**2*d* *3*x - 16302*(c + d*x)*(a + b*x)**(3/2)*a*b**3*c**2*d - 13338*(c + d*x)*(a + b*x)**(3/2)*a*b**3*c*d**2*x - 4446*(c + d*x)*(a + b*x)**(3/2)*a*b**3*d* *3*x**2 + 9139*(c + d*x)*(a + b*x)**(3/2)*b**4*c**3 + 11115*(c + d*x)*(a + b*x)**(3/2)*b**4*c**2*d*x + 4446*(c + d*x)*(a + b*x)**(3/2)*b**4*c*d**2*x **2 - 1365*sqrt(a + b*x)*a**5*d**4 + 7350*sqrt(a + b*x)*a**4*b*c*d**3 + 52 5*sqrt(a + b*x)*a**4*b*d**4*x - 17100*sqrt(a + b*x)*a**3*b**2*c**2*d**2 - 4800*sqrt(a + b*x)*a**3*b**2*c*d**3*x - 1350*sqrt(a + b*x)*a**3*b**2*d**4* x**2 + 37050*sqrt(a + b*x)*a**2*b**3*c**3*d + 59850*sqrt(a + b*x)*a**2*b** 3*c**2*d**2*x + 52650*sqrt(a + b*x)*a**2*b**3*c*d**3*x**2 + 16200*sqrt(a + b*x)*a**2*b**3*d**4*x**3 - 25935*sqrt(a + b*x)*a*b**4*c**4 - 29640*sqrt(a + b*x)*a*b**4*c**3*d*x + 15390*sqrt(a + b*x)*a*b**4*c**2*d**2*x**2 + 4536 0*sqrt(a + b*x)*a*b**4*c*d**3*x**3 + 19440*sqrt(a + b*x)*a*b**4*d**4*x**4 - 25935*sqrt(a + b*x)*b**5*c**4*x - 66690*sqrt(a + b*x)*b**5*c**3*d*x**2 - 61560*sqrt(a + b*x)*b**5*c**2*d**2*x**3 - 19440*sqrt(a + b*x)*b**5*c*d**3 *x**4 + 53352*(c + d*x)*sqrt(a + b*x)*log((a + b*x)**(1/6))*a*b**4*c**3 + 160056*(c + d*x)*sqrt(a + b*x)*log((a + b*x)**(1/6))*a*b**4*c**2*d*x + 160 056*(c + d*x)*sqrt(a + b*x)*log((a + b*x)**(1/6))*a*b**4*c*d**2*x**2 + 533 52*(c + d*x)*sqrt(a + b*x)*log((a + b*x)**(1/6))*a*b**4*d**3*x**3 + 533...