Integrand size = 20, antiderivative size = 70 \[ \int \frac {(1-2 x) (2+3 x)^m}{(3+5 x)^3} \, dx=\frac {2 (2+3 x)^{1+m}}{15 (1-m) (3+5 x)^2}-\frac {3 (37-33 m) (2+3 x)^{1+m} \operatorname {Hypergeometric2F1}(3,1+m,2+m,5 (2+3 x))}{5 \left (1-m^2\right )} \] Output:
2/15*(2+3*x)^(1+m)/(1-m)/(3+5*x)^2-3*(37-33*m)*(2+3*x)^(1+m)*hypergeom([3, 1+m],[2+m],10+15*x)/(-5*m^2+5)
Time = 0.06 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.69 \[ \int \frac {(1-2 x) (2+3 x)^m}{(3+5 x)^3} \, dx=\frac {1}{10} (2+3 x)^{1+m} \left (-\frac {11}{(3+5 x)^2}+\frac {3 (-37+33 m) \operatorname {Hypergeometric2F1}(2,1+m,2+m,10+15 x)}{1+m}\right ) \] Input:
Integrate[((1 - 2*x)*(2 + 3*x)^m)/(3 + 5*x)^3,x]
Output:
((2 + 3*x)^(1 + m)*(-11/(3 + 5*x)^2 + (3*(-37 + 33*m)*Hypergeometric2F1[2, 1 + m, 2 + m, 10 + 15*x])/(1 + m)))/10
Time = 0.17 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.84, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {87, 78}
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.
\(\displaystyle \int \frac {(1-2 x) (3 x+2)^m}{(5 x+3)^3} \, dx\) |
\(\Big \downarrow \) 87 |
\(\displaystyle -\frac {1}{10} (37-33 m) \int \frac {(3 x+2)^m}{(5 x+3)^2}dx-\frac {11 (3 x+2)^{m+1}}{10 (5 x+3)^2}\) |
\(\Big \downarrow \) 78 |
\(\displaystyle -\frac {3 (37-33 m) (3 x+2)^{m+1} \operatorname {Hypergeometric2F1}(2,m+1,m+2,5 (3 x+2))}{10 (m+1)}-\frac {11 (3 x+2)^{m+1}}{10 (5 x+3)^2}\) |
Input:
Int[((1 - 2*x)*(2 + 3*x)^m)/(3 + 5*x)^3,x]
Output:
(-11*(2 + 3*x)^(1 + m))/(10*(3 + 5*x)^2) - (3*(37 - 33*m)*(2 + 3*x)^(1 + m )*Hypergeometric2F1[2, 1 + m, 2 + m, 5*(2 + 3*x)])/(10*(1 + m))
Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(b *c - a*d)^n*((a + b*x)^(m + 1)/(b^(n + 1)*(m + 1)))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m}, x] && !IntegerQ[m] && IntegerQ[n]
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)) Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege rQ[p] || !(IntegerQ[n] || !(EqQ[e, 0] || !(EqQ[c, 0] || LtQ[p, n]))))
\[\int \frac {\left (1-2 x \right ) \left (2+3 x \right )^{m}}{\left (3+5 x \right )^{3}}d x\]
Input:
int((1-2*x)*(2+3*x)^m/(3+5*x)^3,x)
Output:
int((1-2*x)*(2+3*x)^m/(3+5*x)^3,x)
\[ \int \frac {(1-2 x) (2+3 x)^m}{(3+5 x)^3} \, dx=\int { -\frac {{\left (3 \, x + 2\right )}^{m} {\left (2 \, x - 1\right )}}{{\left (5 \, x + 3\right )}^{3}} \,d x } \] Input:
integrate((1-2*x)*(2+3*x)^m/(3+5*x)^3,x, algorithm="fricas")
Output:
integral(-(3*x + 2)^m*(2*x - 1)/(125*x^3 + 225*x^2 + 135*x + 27), x)
\[ \int \frac {(1-2 x) (2+3 x)^m}{(3+5 x)^3} \, dx=- \int \left (- \frac {\left (3 x + 2\right )^{m}}{125 x^{3} + 225 x^{2} + 135 x + 27}\right )\, dx - \int \frac {2 x \left (3 x + 2\right )^{m}}{125 x^{3} + 225 x^{2} + 135 x + 27}\, dx \] Input:
integrate((1-2*x)*(2+3*x)**m/(3+5*x)**3,x)
Output:
-Integral(-(3*x + 2)**m/(125*x**3 + 225*x**2 + 135*x + 27), x) - Integral( 2*x*(3*x + 2)**m/(125*x**3 + 225*x**2 + 135*x + 27), x)
\[ \int \frac {(1-2 x) (2+3 x)^m}{(3+5 x)^3} \, dx=\int { -\frac {{\left (3 \, x + 2\right )}^{m} {\left (2 \, x - 1\right )}}{{\left (5 \, x + 3\right )}^{3}} \,d x } \] Input:
integrate((1-2*x)*(2+3*x)^m/(3+5*x)^3,x, algorithm="maxima")
Output:
-integrate((3*x + 2)^m*(2*x - 1)/(5*x + 3)^3, x)
\[ \int \frac {(1-2 x) (2+3 x)^m}{(3+5 x)^3} \, dx=\int { -\frac {{\left (3 \, x + 2\right )}^{m} {\left (2 \, x - 1\right )}}{{\left (5 \, x + 3\right )}^{3}} \,d x } \] Input:
integrate((1-2*x)*(2+3*x)^m/(3+5*x)^3,x, algorithm="giac")
Output:
integrate(-(3*x + 2)^m*(2*x - 1)/(5*x + 3)^3, x)
Timed out. \[ \int \frac {(1-2 x) (2+3 x)^m}{(3+5 x)^3} \, dx=-\int \frac {\left (2\,x-1\right )\,{\left (3\,x+2\right )}^m}{{\left (5\,x+3\right )}^3} \,d x \] Input:
int(-((2*x - 1)*(3*x + 2)^m)/(5*x + 3)^3,x)
Output:
-int(((2*x - 1)*(3*x + 2)^m)/(5*x + 3)^3, x)
\[ \int \frac {(1-2 x) (2+3 x)^m}{(3+5 x)^3} \, dx =\text {Too large to display} \] Input:
int((1-2*x)*(2+3*x)^m/(3+5*x)^3,x)
Output:
( - 18*(3*x + 2)**m*m*x + 10*(3*x + 2)**m*m + 40*(3*x + 2)**m*x + 2*(3*x + 2)**m - 7425*int(((3*x + 2)**m*x)/(3375*m**2*x**4 + 8325*m**2*x**3 + 7695 *m**2*x**2 + 3159*m**2*x + 486*m**2 - 10875*m*x**4 - 26825*m*x**3 - 24795* m*x**2 - 10179*m*x - 1566*m + 7500*x**4 + 18500*x**3 + 17100*x**2 + 7020*x + 1080),x)*m**4*x**2 - 8910*int(((3*x + 2)**m*x)/(3375*m**2*x**4 + 8325*m **2*x**3 + 7695*m**2*x**2 + 3159*m**2*x + 486*m**2 - 10875*m*x**4 - 26825* m*x**3 - 24795*m*x**2 - 10179*m*x - 1566*m + 7500*x**4 + 18500*x**3 + 1710 0*x**2 + 7020*x + 1080),x)*m**4*x - 2673*int(((3*x + 2)**m*x)/(3375*m**2*x **4 + 8325*m**2*x**3 + 7695*m**2*x**2 + 3159*m**2*x + 486*m**2 - 10875*m*x **4 - 26825*m*x**3 - 24795*m*x**2 - 10179*m*x - 1566*m + 7500*x**4 + 18500 *x**3 + 17100*x**2 + 7020*x + 1080),x)*m**4 + 32250*int(((3*x + 2)**m*x)/( 3375*m**2*x**4 + 8325*m**2*x**3 + 7695*m**2*x**2 + 3159*m**2*x + 486*m**2 - 10875*m*x**4 - 26825*m*x**3 - 24795*m*x**2 - 10179*m*x - 1566*m + 7500*x **4 + 18500*x**3 + 17100*x**2 + 7020*x + 1080),x)*m**3*x**2 + 38700*int((( 3*x + 2)**m*x)/(3375*m**2*x**4 + 8325*m**2*x**3 + 7695*m**2*x**2 + 3159*m* *2*x + 486*m**2 - 10875*m*x**4 - 26825*m*x**3 - 24795*m*x**2 - 10179*m*x - 1566*m + 7500*x**4 + 18500*x**3 + 17100*x**2 + 7020*x + 1080),x)*m**3*x + 11610*int(((3*x + 2)**m*x)/(3375*m**2*x**4 + 8325*m**2*x**3 + 7695*m**2*x **2 + 3159*m**2*x + 486*m**2 - 10875*m*x**4 - 26825*m*x**3 - 24795*m*x**2 - 10179*m*x - 1566*m + 7500*x**4 + 18500*x**3 + 17100*x**2 + 7020*x + 1...