3.32.12 \(\int (a+b x)^m (c+d x)^{1-m} (e+f x)^3 \, dx\) [3112]

3.32.12.1 Optimal result

Integrand size = 26, antiderivative size = 445 \[ \int (a+b x)^m (c+d x)^{1-m} (e+f x)^3 \, dx=\frac {f (a+b x)^{1+m} (c+d x)^{2-m} (e+f x)^2}{5 b d}+\frac {f (a+b x)^{1+m} (c+d x)^{2-m} \left (a^2 d^2 f^2 \left (12-7 m+m^2\right )-a b d f \left (15 d e (3-m)-c f \left (9+2 m-2 m^2\right )\right )+b^2 \left (48 d^2 e^2-15 c d e f (2+m)+c^2 f^2 \left (6+5 m+m^2\right )\right )-3 b d f (a d f (4-m)-b (7 d e-c f (3+m))) x\right )}{60 b^3 d^3}-\frac {(b c-a d) \left (a^3 d^3 f^3 \left (24-26 m+9 m^2-m^3\right )-3 a^2 b d^2 f^2 \left (6-5 m+m^2\right ) (5 d e-c f (1+m))+3 a b^2 d f (2-m) \left (20 d^2 e^2-10 c d e f (1+m)+c^2 f^2 \left (2+3 m+m^2\right )\right )-b^3 \left (60 d^3 e^3-60 c d^2 e^2 f (1+m)+15 c^2 d e f^2 \left (2+3 m+m^2\right )-c^3 f^3 \left (6+11 m+6 m^2+m^3\right )\right )\right ) (a+b x)^{1+m} (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m \operatorname {Hypergeometric2F1}\left (-1+m,1+m,2+m,-\frac {d (a+b x)}{b c-a d}\right )}{60 b^5 d^3 (1+m)} \]

output

1/5*f*(b*x+a)^(1+m)*(d*x+c)^(2-m)*(f*x+e)^2/b/d+1/60*f*(b*x+a)^(1+m)*(d*x+ 
c)^(2-m)*(a^2*d^2*f^2*(m^2-7*m+12)-a*b*d*f*(15*d*e*(3-m)-c*f*(-2*m^2+2*m+9 
))+b^2*(48*d^2*e^2-15*c*d*e*f*(2+m)+c^2*f^2*(m^2+5*m+6))-3*b*d*f*(a*d*f*(4 
-m)-b*(7*d*e-c*f*(3+m)))*x)/b^3/d^3-1/60*(-a*d+b*c)*(a^3*d^3*f^3*(-m^3+9*m 
^2-26*m+24)-3*a^2*b*d^2*f^2*(m^2-5*m+6)*(5*d*e-c*f*(1+m))+3*a*b^2*d*f*(2-m 
)*(20*d^2*e^2-10*c*d*e*f*(1+m)+c^2*f^2*(m^2+3*m+2))-b^3*(60*d^3*e^3-60*c*d 
^2*e^2*f*(1+m)+15*c^2*d*e*f^2*(m^2+3*m+2)-c^3*f^3*(m^3+6*m^2+11*m+6)))*(b* 
x+a)^(1+m)*(b*(d*x+c)/(-a*d+b*c))^m*hypergeom([-1+m, 1+m],[2+m],-d*(b*x+a) 
/(-a*d+b*c))/b^5/d^3/(1+m)/((d*x+c)^m)
 

3.32.12.2 Mathematica [A] (verified)

Time = 0.56 (sec) , antiderivative size = 317, normalized size of antiderivative = 0.71 \[ \int (a+b x)^m (c+d x)^{1-m} (e+f x)^3 \, dx=\frac {(a+b x)^{1+m} (c+d x)^{-m} \left (b^4 d^2 f (1+m) (c+d x)^2 (e+f x)^2+(b c-a d)^3 f^2 (7 b d e+a d f (-4+m)-b c f (3+m)) \left (\frac {b (c+d x)}{b c-a d}\right )^m \operatorname {Hypergeometric2F1}\left (-3+m,1+m,2+m,\frac {d (a+b x)}{-b c+a d}\right )+2 b (b c-a d)^2 f (d e-c f) (6 b d e+a d f (-3+m)-b c f (3+m)) \left (\frac {b (c+d x)}{b c-a d}\right )^m \operatorname {Hypergeometric2F1}\left (-2+m,1+m,2+m,\frac {d (a+b x)}{-b c+a d}\right )+b^3 (d e-c f)^2 (5 b d e+a d f (-2+m)-b c f (3+m)) (c+d x) \left (\frac {b (c+d x)}{b c-a d}\right )^{-1+m} \operatorname {Hypergeometric2F1}\left (-1+m,1+m,2+m,\frac {d (a+b x)}{-b c+a d}\right )\right )}{5 b^5 d^3 (1+m)} \]

input

Integrate[(a + b*x)^m*(c + d*x)^(1 - m)*(e + f*x)^3,x]
 

output

((a + b*x)^(1 + m)*(b^4*d^2*f*(1 + m)*(c + d*x)^2*(e + f*x)^2 + (b*c - a*d 
)^3*f^2*(7*b*d*e + a*d*f*(-4 + m) - b*c*f*(3 + m))*((b*(c + d*x))/(b*c - a 
*d))^m*Hypergeometric2F1[-3 + m, 1 + m, 2 + m, (d*(a + b*x))/(-(b*c) + a*d 
)] + 2*b*(b*c - a*d)^2*f*(d*e - c*f)*(6*b*d*e + a*d*f*(-3 + m) - b*c*f*(3 
+ m))*((b*(c + d*x))/(b*c - a*d))^m*Hypergeometric2F1[-2 + m, 1 + m, 2 + m 
, (d*(a + b*x))/(-(b*c) + a*d)] + b^3*(d*e - c*f)^2*(5*b*d*e + a*d*f*(-2 + 
 m) - b*c*f*(3 + m))*(c + d*x)*((b*(c + d*x))/(b*c - a*d))^(-1 + m)*Hyperg 
eometric2F1[-1 + m, 1 + m, 2 + m, (d*(a + b*x))/(-(b*c) + a*d)]))/(5*b^5*d 
^3*(1 + m)*(c + d*x)^m)
 

3.32.12.3 Rubi [A] (verified)

Time = 0.63 (sec) , antiderivative size = 455, normalized size of antiderivative = 1.02, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {111, 25, 164, 80, 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.

\(\displaystyle \int (e+f x)^3 (a+b x)^m (c+d x)^{1-m} \, dx\)

\(\Big \downarrow \) 111

\(\displaystyle \frac {\int -(a+b x)^m (c+d x)^{1-m} (e+f x) (a f (2 c f+d e (2-m))-b e (5 d e-c f (m+1))-f (7 b d e-a d f (4-m)-b c f (m+3)) x)dx}{5 b d}+\frac {f (e+f x)^2 (a+b x)^{m+1} (c+d x)^{2-m}}{5 b d}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {f (e+f x)^2 (a+b x)^{m+1} (c+d x)^{2-m}}{5 b d}-\frac {\int (a+b x)^m (c+d x)^{1-m} (e+f x) (a f (2 c f+d e (2-m))-b e (5 d e-c f (m+1))-f (7 b d e-a d f (4-m)-b c f (m+3)) x)dx}{5 b d}\)

\(\Big \downarrow \) 164

\(\displaystyle \frac {f (e+f x)^2 (a+b x)^{m+1} (c+d x)^{2-m}}{5 b d}-\frac {\frac {\left (a^3 d^3 f^3 \left (-m^3+9 m^2-26 m+24\right )-3 a^2 b d^2 f^2 \left (m^2-5 m+6\right ) (5 d e-c f (m+1))+3 a b^2 d f (2-m) \left (c^2 f^2 \left (m^2+3 m+2\right )-10 c d e f (m+1)+20 d^2 e^2\right )-\left (b^3 \left (-c^3 f^3 \left (m^3+6 m^2+11 m+6\right )+15 c^2 d e f^2 \left (m^2+3 m+2\right )-60 c d^2 e^2 f (m+1)+60 d^3 e^3\right )\right )\right ) \int (a+b x)^m (c+d x)^{1-m}dx}{12 b^2 d^2}-\frac {f (a+b x)^{m+1} (c+d x)^{2-m} \left (a^2 d^2 f^2 \left (m^2-7 m+12\right )-a b d f \left (15 d e (3-m)-c f \left (-2 m^2+2 m+9\right )\right )+3 b d f x (-a d f (4-m)-b c f (m+3)+7 b d e)+b^2 \left (c^2 f^2 \left (m^2+5 m+6\right )-15 c d e f (m+2)+48 d^2 e^2\right )\right )}{12 b^2 d^2}}{5 b d}\)

\(\Big \downarrow \) 80

\(\displaystyle \frac {f (e+f x)^2 (a+b x)^{m+1} (c+d x)^{2-m}}{5 b d}-\frac {\frac {(b c-a d) (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m \left (a^3 d^3 f^3 \left (-m^3+9 m^2-26 m+24\right )-3 a^2 b d^2 f^2 \left (m^2-5 m+6\right ) (5 d e-c f (m+1))+3 a b^2 d f (2-m) \left (c^2 f^2 \left (m^2+3 m+2\right )-10 c d e f (m+1)+20 d^2 e^2\right )-\left (b^3 \left (-c^3 f^3 \left (m^3+6 m^2+11 m+6\right )+15 c^2 d e f^2 \left (m^2+3 m+2\right )-60 c d^2 e^2 f (m+1)+60 d^3 e^3\right )\right )\right ) \int (a+b x)^m \left (\frac {b c}{b c-a d}+\frac {b d x}{b c-a d}\right )^{1-m}dx}{12 b^3 d^2}-\frac {f (a+b x)^{m+1} (c+d x)^{2-m} \left (a^2 d^2 f^2 \left (m^2-7 m+12\right )-a b d f \left (15 d e (3-m)-c f \left (-2 m^2+2 m+9\right )\right )+3 b d f x (-a d f (4-m)-b c f (m+3)+7 b d e)+b^2 \left (c^2 f^2 \left (m^2+5 m+6\right )-15 c d e f (m+2)+48 d^2 e^2\right )\right )}{12 b^2 d^2}}{5 b d}\)

\(\Big \downarrow \) 79

\(\displaystyle \frac {f (e+f x)^2 (a+b x)^{m+1} (c+d x)^{2-m}}{5 b d}-\frac {\frac {(b c-a d) (a+b x)^{m+1} (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m \left (a^3 d^3 f^3 \left (-m^3+9 m^2-26 m+24\right )-3 a^2 b d^2 f^2 \left (m^2-5 m+6\right ) (5 d e-c f (m+1))+3 a b^2 d f (2-m) \left (c^2 f^2 \left (m^2+3 m+2\right )-10 c d e f (m+1)+20 d^2 e^2\right )-\left (b^3 \left (-c^3 f^3 \left (m^3+6 m^2+11 m+6\right )+15 c^2 d e f^2 \left (m^2+3 m+2\right )-60 c d^2 e^2 f (m+1)+60 d^3 e^3\right )\right )\right ) \operatorname {Hypergeometric2F1}\left (m-1,m+1,m+2,-\frac {d (a+b x)}{b c-a d}\right )}{12 b^4 d^2 (m+1)}-\frac {f (a+b x)^{m+1} (c+d x)^{2-m} \left (a^2 d^2 f^2 \left (m^2-7 m+12\right )-a b d f \left (15 d e (3-m)-c f \left (-2 m^2+2 m+9\right )\right )+3 b d f x (-a d f (4-m)-b c f (m+3)+7 b d e)+b^2 \left (c^2 f^2 \left (m^2+5 m+6\right )-15 c d e f (m+2)+48 d^2 e^2\right )\right )}{12 b^2 d^2}}{5 b d}\)

input

Int[(a + b*x)^m*(c + d*x)^(1 - m)*(e + f*x)^3,x]
 

output

(f*(a + b*x)^(1 + m)*(c + d*x)^(2 - m)*(e + f*x)^2)/(5*b*d) - (-1/12*(f*(a 
 + b*x)^(1 + m)*(c + d*x)^(2 - m)*(a^2*d^2*f^2*(12 - 7*m + m^2) - a*b*d*f* 
(15*d*e*(3 - m) - c*f*(9 + 2*m - 2*m^2)) + b^2*(48*d^2*e^2 - 15*c*d*e*f*(2 
 + m) + c^2*f^2*(6 + 5*m + m^2)) + 3*b*d*f*(7*b*d*e - a*d*f*(4 - m) - b*c* 
f*(3 + m))*x))/(b^2*d^2) + ((b*c - a*d)*(a^3*d^3*f^3*(24 - 26*m + 9*m^2 - 
m^3) - 3*a^2*b*d^2*f^2*(6 - 5*m + m^2)*(5*d*e - c*f*(1 + m)) + 3*a*b^2*d*f 
*(2 - m)*(20*d^2*e^2 - 10*c*d*e*f*(1 + m) + c^2*f^2*(2 + 3*m + m^2)) - b^3 
*(60*d^3*e^3 - 60*c*d^2*e^2*f*(1 + m) + 15*c^2*d*e*f^2*(2 + 3*m + m^2) - c 
^3*f^3*(6 + 11*m + 6*m^2 + m^3)))*(a + b*x)^(1 + m)*((b*(c + d*x))/(b*c - 
a*d))^m*Hypergeometric2F1[-1 + m, 1 + m, 2 + m, -((d*(a + b*x))/(b*c - a*d 
))])/(12*b^4*d^2*(1 + m)*(c + d*x)^m))/(5*b*d)
 

3.32.12.3.1 Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 79

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]))
 

rule 80

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(c 
 + d*x)^FracPart[n]/((b/(b*c - a*d))^IntPart[n]*(b*((c + d*x)/(b*c - a*d))) 
^FracPart[n])   Int[(a + b*x)^m*Simp[b*(c/(b*c - a*d)) + b*d*(x/(b*c - a*d) 
), x]^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] &&  !IntegerQ[m] &&  !Integ 
erQ[n] && (RationalQ[m] ||  !SimplerQ[n + 1, m + 1])
 

rule 111

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]
 

rule 164

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]
 

3.32.12.4 Maple [F]

\[\int \left (b x +a \right )^{m} \left (d x +c \right )^{1-m} \left (f x +e \right )^{3}d x\]

input

int((b*x+a)^m*(d*x+c)^(1-m)*(f*x+e)^3,x)
 

output

int((b*x+a)^m*(d*x+c)^(1-m)*(f*x+e)^3,x)
 

3.32.12.5 Fricas [F]

\[ \int (a+b x)^m (c+d x)^{1-m} (e+f x)^3 \, dx=\int { {\left (f x + e\right )}^{3} {\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m + 1} \,d x } \]

input

integrate((b*x+a)^m*(d*x+c)^(1-m)*(f*x+e)^3,x, algorithm="fricas")
 

output

integral((f^3*x^3 + 3*e*f^2*x^2 + 3*e^2*f*x + e^3)*(b*x + a)^m*(d*x + c)^( 
-m + 1), x)
 

3.32.12.6 Sympy [F(-2)]

Exception generated. \[ \int (a+b x)^m (c+d x)^{1-m} (e+f x)^3 \, dx=\text {Exception raised: HeuristicGCDFailed} \]

input

integrate((b*x+a)**m*(d*x+c)**(1-m)*(f*x+e)**3,x)
 

output

Exception raised: HeuristicGCDFailed >> no luck
 

3.32.12.7 Maxima [F]

\[ \int (a+b x)^m (c+d x)^{1-m} (e+f x)^3 \, dx=\int { {\left (f x + e\right )}^{3} {\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m + 1} \,d x } \]

input

integrate((b*x+a)^m*(d*x+c)^(1-m)*(f*x+e)^3,x, algorithm="maxima")
 

output

integrate((f*x + e)^3*(b*x + a)^m*(d*x + c)^(-m + 1), x)
 

3.32.12.8 Giac [F]

\[ \int (a+b x)^m (c+d x)^{1-m} (e+f x)^3 \, dx=\int { {\left (f x + e\right )}^{3} {\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m + 1} \,d x } \]

input

integrate((b*x+a)^m*(d*x+c)^(1-m)*(f*x+e)^3,x, algorithm="giac")
 

output

integrate((f*x + e)^3*(b*x + a)^m*(d*x + c)^(-m + 1), x)
 

3.32.12.9 Mupad [F(-1)]

Timed out. \[ \int (a+b x)^m (c+d x)^{1-m} (e+f x)^3 \, dx=\int {\left (e+f\,x\right )}^3\,{\left (a+b\,x\right )}^m\,{\left (c+d\,x\right )}^{1-m} \,d x \]

input

int((e + f*x)^3*(a + b*x)^m*(c + d*x)^(1 - m),x)
 

output

int((e + f*x)^3*(a + b*x)^m*(c + d*x)^(1 - m), x)