Integrand size = 24, antiderivative size = 188 \[ \int (5-4 x)^4 (1+2 x)^{-m} (2+3 x)^m \, dx=-\frac {1}{45} (88-m) (5-4 x)^2 (1+2 x)^{1-m} (2+3 x)^{1+m}-\frac {2}{15} (5-4 x)^3 (1+2 x)^{1-m} (2+3 x)^{1+m}-\frac {(1+2 x)^{1-m} (2+3 x)^{1+m} \left (386850-25441 m+426 m^2-2 m^3-24 \left (4359-154 m+m^2\right ) x\right )}{1215}+\frac {2^{-1-m} \left (3528363-639760 m+29050 m^2-440 m^3+2 m^4\right ) (1+2 x)^{1-m} \operatorname {Hypergeometric2F1}(1-m,-m,2-m,-3 (1+2 x))}{1215 (1-m)} \]
-1/45*(88-m)*(5-4*x)^2*(1+2*x)^(1-m)*(2+3*x)^(1+m)-2/15*(5-4*x)^3*(1+2*x)^ (1-m)*(2+3*x)^(1+m)-1/1215*(1+2*x)^(1-m)*(2+3*x)^(1+m)*(386850-25441*m+426 *m^2-2*m^3-24*(m^2-154*m+4359)*x)+1/1215*2^(-1-m)*(2*m^4-440*m^3+29050*m^2 -639760*m+3528363)*(1+2*x)^(1-m)*hypergeom([-m, 1-m],[2-m],-3-6*x)/(1-m)
Time = 0.32 (sec) , antiderivative size = 227, normalized size of antiderivative = 1.21 \[ \int (5-4 x)^4 (1+2 x)^{-m} (2+3 x)^m \, dx=\frac {(1+2 x)^{1-m} \left (-108 (-1+m) (2+3 x)^{1+m} (-5+4 x)^3-(88-m) \left (-18 (-1+m) (5-4 x)^2 (2+3 x)^{1+m}+2^{2-m} (-66+m) \operatorname {Hypergeometric2F1}(-2-m,1-m,2-m,-3-6 x)+23\ 2^{1-m} (111-2 m) \operatorname {Hypergeometric2F1}(-1-m,1-m,2-m,-3-6 x)+529\ 2^{-m} (-45+m) \operatorname {Hypergeometric2F1}(1-m,-m,2-m,-3-6 x)\right )+483 \left (4 (-1+m) (2+3 x)^{1+m} (-59+m+12 x)+2^{-m} \left (-1453+132 m-2 m^2\right ) \operatorname {Hypergeometric2F1}(1-m,-m,2-m,-3-6 x)\right )\right )}{810 (1-m)} \]
((1 + 2*x)^(1 - m)*(-108*(-1 + m)*(2 + 3*x)^(1 + m)*(-5 + 4*x)^3 - (88 - m )*(-18*(-1 + m)*(5 - 4*x)^2*(2 + 3*x)^(1 + m) + 2^(2 - m)*(-66 + m)*Hyperg eometric2F1[-2 - m, 1 - m, 2 - m, -3 - 6*x] + 23*2^(1 - m)*(111 - 2*m)*Hyp ergeometric2F1[-1 - m, 1 - m, 2 - m, -3 - 6*x] + (529*(-45 + m)*Hypergeome tric2F1[1 - m, -m, 2 - m, -3 - 6*x])/2^m) + 483*(4*(-1 + m)*(2 + 3*x)^(1 + m)*(-59 + m + 12*x) + ((-1453 + 132*m - 2*m^2)*Hypergeometric2F1[1 - m, - m, 2 - m, -3 - 6*x])/2^m)))/(810*(1 - m))
Time = 0.35 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.05, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {111, 27, 170, 27, 164, 79}
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.
| \(\displaystyle \int (5-4 x)^4 (2 x+1)^{-m} (3 x+2)^m \, dx\) |
| \(\Big \downarrow \) 111 |
| \(\displaystyle \frac {1}{30} \int 2 (5-4 x)^2 (2 x+1)^{-m} (3 x+2)^m (-10 m-8 (88-m) x+397)dx-\frac {2}{15} (5-4 x)^3 (2 x+1)^{1-m} (3 x+2)^{m+1}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{15} \int (5-4 x)^2 (2 x+1)^{-m} (3 x+2)^m (-10 m-8 (88-m) x+397)dx-\frac {2}{15} (5-4 x)^3 (2 x+1)^{1-m} (3 x+2)^{m+1}\) |
| \(\Big \downarrow \) 170 |
| \(\displaystyle \frac {1}{15} \left (\frac {1}{24} \int 8 (5-4 x) (2 x+1)^{-m} (3 x+2)^m \left (5 m^2-609 m-4 \left (m^2-154 m+4359\right ) x+7627\right )dx-\frac {1}{3} (88-m) (5-4 x)^2 (2 x+1)^{1-m} (3 x+2)^{m+1}\right )-\frac {2}{15} (5-4 x)^3 (2 x+1)^{1-m} (3 x+2)^{m+1}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{15} \left (\frac {1}{3} \int (5-4 x) (2 x+1)^{-m} (3 x+2)^m \left (5 m^2-609 m-4 \left (m^2-154 m+4359\right ) x+7627\right )dx-\frac {1}{3} (88-m) (5-4 x)^2 (2 x+1)^{1-m} (3 x+2)^{m+1}\right )-\frac {2}{15} (5-4 x)^3 (2 x+1)^{1-m} (3 x+2)^{m+1}\) |
| \(\Big \downarrow \) 164 |
| \(\displaystyle \frac {1}{15} \left (\frac {1}{3} \left (\frac {1}{27} \left (2 m^4-440 m^3+29050 m^2-639760 m+3528363\right ) \int (2 x+1)^{-m} (3 x+2)^mdx-\frac {1}{27} (2 x+1)^{1-m} (3 x+2)^{m+1} \left (-2 m^3-24 \left (m^2-154 m+4359\right ) x+426 m^2-25441 m+386850\right )\right )-\frac {1}{3} (88-m) (5-4 x)^2 (2 x+1)^{1-m} (3 x+2)^{m+1}\right )-\frac {2}{15} (5-4 x)^3 (2 x+1)^{1-m} (3 x+2)^{m+1}\) |
| \(\Big \downarrow \) 79 |
| \(\displaystyle \frac {1}{15} \left (\frac {1}{3} \left (\frac {2^{-m-1} \left (2 m^4-440 m^3+29050 m^2-639760 m+3528363\right ) (2 x+1)^{1-m} \operatorname {Hypergeometric2F1}(1-m,-m,2-m,-3 (2 x+1))}{27 (1-m)}-\frac {1}{27} (2 x+1)^{1-m} (3 x+2)^{m+1} \left (-2 m^3-24 \left (m^2-154 m+4359\right ) x+426 m^2-25441 m+386850\right )\right )-\frac {1}{3} (88-m) (5-4 x)^2 (2 x+1)^{1-m} (3 x+2)^{m+1}\right )-\frac {2}{15} (5-4 x)^3 (2 x+1)^{1-m} (3 x+2)^{m+1}\) |
(-2*(5 - 4*x)^3*(1 + 2*x)^(1 - m)*(2 + 3*x)^(1 + m))/15 + (-1/3*((88 - m)* (5 - 4*x)^2*(1 + 2*x)^(1 - m)*(2 + 3*x)^(1 + m)) + (-1/27*((1 + 2*x)^(1 - m)*(2 + 3*x)^(1 + m)*(386850 - 25441*m + 426*m^2 - 2*m^3 - 24*(4359 - 154* m + m^2)*x)) + (2^(-1 - m)*(3528363 - 639760*m + 29050*m^2 - 440*m^3 + 2*m ^4)*(1 + 2*x)^(1 - m)*Hypergeometric2F1[1 - m, -m, 2 - m, -3*(1 + 2*x)])/( 27*(1 - m)))/3)/15
3.31.60.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(( a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c - a*d))^n))*Hypergeometric2F1[-n, m + 1 , m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m, n}, x] && !IntegerQ[m] && !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] || !(RationalQ[n] && GtQ[-d/(b*c - a*d), 0]))
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Simp[b*(a + b*x)^(m - 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1 )/(d*f*(m + n + p + 1))), x] + Simp[1/(d*f*(m + n + p + 1)) Int[(a + b*x) ^(m - 2)*(c + d*x)^n*(e + f*x)^p*Simp[a^2*d*f*(m + n + p + 1) - b*(b*c*e*(m - 1) + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f*(2*m + n + p) - b*(d*e*(m + n) + c*f*(m + p)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] & & GtQ[m, 1] && NeQ[m + n + p + 1, 0] && IntegerQ[m]
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_ ))*((g_.) + (h_.)*(x_)), x_] :> Simp[(-(a*d*f*h*(n + 2) + b*c*f*h*(m + 2) - b*d*(f*g + e*h)*(m + n + 3) - b*d*f*h*(m + n + 2)*x))*(a + b*x)^(m + 1)*(( c + d*x)^(n + 1)/(b^2*d^2*(m + n + 2)*(m + n + 3))), x] + Simp[(a^2*d^2*f*h *(n + 1)*(n + 2) + a*b*d*(n + 1)*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b^2*(c^2*f*h*(m + 1)*(m + 2) - c*d*(f*g + e*h)*(m + 1)*(m + n + 3) + d^2*e*g*(m + n + 2)*(m + n + 3)))/(b^2*d^2*(m + n + 2)*(m + n + 3)) Int[( a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && NeQ[m + n + 2, 0] && NeQ[m + n + 3, 0]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_)*((g_.) + (h_.)*(x_)), x_] :> Simp[h*(a + b*x)^m*(c + d*x)^(n + 1)*(( e + f*x)^(p + 1)/(d*f*(m + n + p + 2))), x] + Simp[1/(d*f*(m + n + p + 2)) Int[(a + b*x)^(m - 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*g*(m + n + p + 2 ) - h*(b*c*e*m + a*(d*e*(n + 1) + c*f*(p + 1))) + (b*d*f*g*(m + n + p + 2) + h*(a*d*f*m - b*(d*e*(m + n + 1) + c*f*(m + p + 1))))*x, x], x], x] /; Fre eQ[{a, b, c, d, e, f, g, h, n, p}, x] && GtQ[m, 0] && NeQ[m + n + p + 2, 0] && IntegerQ[m]
\[\int \left (5-4 x \right )^{4} \left (2+3 x \right )^{m} \left (1+2 x \right )^{-m}d x\]
\[ \int (5-4 x)^4 (1+2 x)^{-m} (2+3 x)^m \, dx=\int { \frac {{\left (3 \, x + 2\right )}^{m} {\left (4 \, x - 5\right )}^{4}}{{\left (2 \, x + 1\right )}^{m}} \,d x } \]
Timed out. \[ \int (5-4 x)^4 (1+2 x)^{-m} (2+3 x)^m \, dx=\text {Timed out} \]
\[ \int (5-4 x)^4 (1+2 x)^{-m} (2+3 x)^m \, dx=\int { \frac {{\left (3 \, x + 2\right )}^{m} {\left (4 \, x - 5\right )}^{4}}{{\left (2 \, x + 1\right )}^{m}} \,d x } \]
\[ \int (5-4 x)^4 (1+2 x)^{-m} (2+3 x)^m \, dx=\int { \frac {{\left (3 \, x + 2\right )}^{m} {\left (4 \, x - 5\right )}^{4}}{{\left (2 \, x + 1\right )}^{m}} \,d x } \]
Timed out. \[ \int (5-4 x)^4 (1+2 x)^{-m} (2+3 x)^m \, dx=\int \frac {{\left (3\,x+2\right )}^m\,{\left (4\,x-5\right )}^4}{{\left (2\,x+1\right )}^m} \,d x \]