\(\int (a+b x)^m (c+d x)^{-5-m} (e+f x)^3 \, dx\) [1752]

Optimal result

Integrand size = 26, antiderivative size = 469 \[ \int (a+b x)^m (c+d x)^{-5-m} (e+f x)^3 \, dx=\frac {(d e-c f)^3 (a+b x)^{1+m} (c+d x)^{-4-m}}{d^3 (b c-a d) (4+m)}+\frac {3 (d e-c f)^2 (b d e+b c f (3+m)-a d f (4+m)) (a+b x)^{1+m} (c+d x)^{-3-m}}{d^3 (b c-a d)^2 (3+m) (4+m)}+\frac {3 (d e-c f) \left (a^2 d^2 f^2 \left (12+7 m+m^2\right )-2 a b d f (4+m) (d e+c f (2+m))+b^2 \left (2 d^2 e^2+2 c d e f (2+m)+c^2 f^2 \left (6+5 m+m^2\right )\right )\right ) (a+b x)^{1+m} (c+d x)^{-2-m}}{d^3 (b c-a d)^3 (2+m) (3+m) (4+m)}+\frac {\left ((b c-a d)^2 f^3 (3+m) (4+m) (b c (1+m)-a d (2+m))+b \left ((b c-a d) f^2 (3 d e-2 c f) (4+m) (b c (1+m)-a d (3+m))-2 b \left (a d f \left (3 d^2 e^2-3 c d e f+c^2 f^2\right ) (4+m)-b \left (3 d^3 e^3+3 c d^2 e^2 f (1+m)-3 c^2 d e f^2 (1+m)+c^3 f^3 (1+m)\right )\right )\right )\right ) (a+b x)^{1+m} (c+d x)^{-1-m}}{d^3 (b c-a d)^4 (1+m) (2+m) (3+m) (4+m)} \] Output:

(-c*f+d*e)^3*(b*x+a)^(1+m)*(d*x+c)^(-4-m)/d^3/(-a*d+b*c)/(4+m)+3*(-c*f+d*e 
)^2*(b*d*e+b*c*f*(3+m)-a*d*f*(4+m))*(b*x+a)^(1+m)*(d*x+c)^(-3-m)/d^3/(-a*d 
+b*c)^2/(3+m)/(4+m)+3*(-c*f+d*e)*(a^2*d^2*f^2*(m^2+7*m+12)-2*a*b*d*f*(4+m) 
*(d*e+c*f*(2+m))+b^2*(2*d^2*e^2+2*c*d*e*f*(2+m)+c^2*f^2*(m^2+5*m+6)))*(b*x 
+a)^(1+m)*(d*x+c)^(-2-m)/d^3/(-a*d+b*c)^3/(2+m)/(3+m)/(4+m)+((-a*d+b*c)^2* 
f^3*(3+m)*(4+m)*(b*c*(1+m)-a*d*(2+m))+b*((-a*d+b*c)*f^2*(-2*c*f+3*d*e)*(4+ 
m)*(b*c*(1+m)-a*d*(3+m))-2*b*(a*d*f*(c^2*f^2-3*c*d*e*f+3*d^2*e^2)*(4+m)-b* 
(3*d^3*e^3+3*c*d^2*e^2*f*(1+m)-3*c^2*d*e*f^2*(1+m)+c^3*f^3*(1+m)))))*(b*x+ 
a)^(1+m)*(d*x+c)^(-1-m)/d^3/(-a*d+b*c)^4/(1+m)/(2+m)/(3+m)/(4+m)
 

Mathematica [A] (verified)

Time = 0.55 (sec) , antiderivative size = 263, normalized size of antiderivative = 0.56 \[ \int (a+b x)^m (c+d x)^{-5-m} (e+f x)^3 \, dx=\frac {(a+b x)^{1+m} (c+d x)^{-4-m} \left ((e+f x)^3-\frac {3 (b e-a f) (c+d x) \left ((b c-a d)^2 (-d e+c f) (1+m) (2+m) (b d e+b c f (2+m)-a d f (3+m))-\left (a^2 d^2 f^2 \left (6+5 m+m^2\right )-2 a b d f (3+m) (d e+c f (1+m))+b^2 \left (2 d^2 e^2+2 c d e f (1+m)+c^2 f^2 \left (2+3 m+m^2\right )\right )\right ) (c+d x) (-a d (1+m)+b c (2+m)+b d x)+d (b c-a d)^3 f (1+m) (2+m) (3+m) (e+f x)\right )}{b d^2 (b c-a d)^3 (1+m) (2+m) (3+m)}\right )}{(b c-a d) (4+m)} \] Input:

Integrate[(a + b*x)^m*(c + d*x)^(-5 - m)*(e + f*x)^3,x]
 

Output:

((a + b*x)^(1 + m)*(c + d*x)^(-4 - m)*((e + f*x)^3 - (3*(b*e - a*f)*(c + d 
*x)*((b*c - a*d)^2*(-(d*e) + c*f)*(1 + m)*(2 + m)*(b*d*e + b*c*f*(2 + m) - 
 a*d*f*(3 + m)) - (a^2*d^2*f^2*(6 + 5*m + m^2) - 2*a*b*d*f*(3 + m)*(d*e + 
c*f*(1 + m)) + b^2*(2*d^2*e^2 + 2*c*d*e*f*(1 + m) + c^2*f^2*(2 + 3*m + m^2 
)))*(c + d*x)*(-(a*d*(1 + m)) + b*c*(2 + m) + b*d*x) + d*(b*c - a*d)^3*f*( 
1 + m)*(2 + m)*(3 + m)*(e + f*x)))/(b*d^2*(b*c - a*d)^3*(1 + m)*(2 + m)*(3 
 + m))))/((b*c - a*d)*(4 + m))
 

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 339, normalized size of antiderivative = 0.72, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {105, 101, 25, 88, 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.

\(\displaystyle \int (e+f x)^3 (a+b x)^m (c+d x)^{-m-5} \, dx\)

\(\Big \downarrow \) 105

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

\(\Big \downarrow \) 101

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 88

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

\(\Big \downarrow \) 55

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

\(\Big \downarrow \) 48

\(\displaystyle \frac {3 (b e-a f) \left (\frac {\frac {\left (\frac {(a+b x)^{m+1} (c+d x)^{-m-2}}{(m+2) (b c-a d)}+\frac {b (a+b x)^{m+1} (c+d x)^{-m-1}}{(m+1) (m+2) (b c-a d)^2}\right ) \left (2 b d (a f (c f-d e (m+3))+b e (c f (m+1)+d e))+f^2 (m+2) (b c-a d) (b c (m+1)-a d (m+3))\right )}{d (m+3) (b c-a d)}+\frac {(a+b x)^{m+1} (d e-c f) (c+d x)^{-m-3} (-a d f (m+3)+b c f (m+2)+b d e)}{d (m+3) (b c-a d)}}{b d}-\frac {f (e+f x) (a+b x)^{m+1} (c+d x)^{-m-3}}{b d}\right )}{(m+4) (b c-a d)}+\frac {(e+f x)^3 (a+b x)^{m+1} (c+d x)^{-m-4}}{(m+4) (b c-a d)}\)

Input:

Int[(a + b*x)^m*(c + d*x)^(-5 - m)*(e + f*x)^3,x]
 

Output:

((a + b*x)^(1 + m)*(c + d*x)^(-4 - m)*(e + f*x)^3)/((b*c - a*d)*(4 + m)) + 
 (3*(b*e - a*f)*(-((f*(a + b*x)^(1 + m)*(c + d*x)^(-3 - m)*(e + f*x))/(b*d 
)) + (((d*e - c*f)*(b*d*e + b*c*f*(2 + m) - a*d*f*(3 + m))*(a + b*x)^(1 + 
m)*(c + d*x)^(-3 - m))/(d*(b*c - a*d)*(3 + m)) + (((b*c - a*d)*f^2*(2 + m) 
*(b*c*(1 + m) - a*d*(3 + m)) + 2*b*d*(b*e*(d*e + c*f*(1 + m)) + a*f*(c*f - 
 d*e*(3 + m))))*(((a + b*x)^(1 + m)*(c + d*x)^(-2 - m))/((b*c - a*d)*(2 + 
m)) + (b*(a + b*x)^(1 + m)*(c + d*x)^(-1 - m))/((b*c - a*d)^2*(1 + m)*(2 + 
 m))))/(d*(b*c - a*d)*(3 + m)))/(b*d)))/((b*c - a*d)*(4 + m))
 

Defintions of rubi rules used

rule 25

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

rule 48

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]
 

rule 55

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])
 

rule 88

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)^Simplify[p + 1], 
 x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] &&  !RationalQ[p] && SumSimpl 
erQ[p, 1]
 

rule 101

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^( 
p_), x_] :> Simp[b*(a + b*x)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + 
 p + 3))), x] + Simp[1/(d*f*(n + p + 3))   Int[(c + d*x)^n*(e + f*x)^p*Simp 
[a^2*d*f*(n + p + 3) - b*(b*c*e + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f 
*(n + p + 4) - b*(d*e*(n + 2) + c*f*(p + 2)))*x, x], x], x] /; FreeQ[{a, b, 
 c, d, e, f, n, p}, x] && NeQ[n + p + 3, 0]
 

rule 105

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Simp[(a + b*x)^(m + 1)*(c + d*x)^n*((e + f*x)^(p + 1)/((m + 
1)*(b*e - a*f))), x] - Simp[n*((d*e - c*f)/((m + 1)*(b*e - a*f)))   Int[(a 
+ b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, 
e, f, m, p}, x] && EqQ[m + n + p + 2, 0] && GtQ[n, 0] && (SumSimplerQ[m, 1] 
 ||  !SumSimplerQ[p, 1]) && NeQ[m, -1]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(2480\) vs. \(2(469)=938\).

Time = 1.91 (sec) , antiderivative size = 2481, normalized size of antiderivative = 5.29

method	result	size
gosper	\(\text {Expression too large to display}\)	\(2481\)
orering	\(\text {Expression too large to display}\)	\(2489\)
parallelrisch	\(\text {Expression too large to display}\)	\(9767\)

Input:

int((b*x+a)^m*(d*x+c)^(-5-m)*(f*x+e)^3,x,method=_RETURNVERBOSE)
 

Output:

-(b*x+a)^(1+m)*(d*x+c)^(-4-m)/(a^4*d^4*m^4-4*a^3*b*c*d^3*m^4+6*a^2*b^2*c^2 
*d^2*m^4-4*a*b^3*c^3*d*m^4+b^4*c^4*m^4+10*a^4*d^4*m^3-40*a^3*b*c*d^3*m^3+6 
0*a^2*b^2*c^2*d^2*m^3-40*a*b^3*c^3*d*m^3+10*b^4*c^4*m^3+35*a^4*d^4*m^2-140 
*a^3*b*c*d^3*m^2+210*a^2*b^2*c^2*d^2*m^2-140*a*b^3*c^3*d*m^2+35*b^4*c^4*m^ 
2+50*a^4*d^4*m-200*a^3*b*c*d^3*m+300*a^2*b^2*c^2*d^2*m-200*a*b^3*c^3*d*m+5 
0*b^4*c^4*m+24*a^4*d^4-96*a^3*b*c*d^3+144*a^2*b^2*c^2*d^2-96*a*b^3*c^3*d+2 
4*b^4*c^4)*(a^3*d^3*f^3*m^3*x^3-3*a^2*b*c*d^2*f^3*m^3*x^3+3*a*b^2*c^2*d*f^ 
3*m^3*x^3-b^3*c^3*f^3*m^3*x^3+3*a^3*d^3*e*f^2*m^3*x^2+9*a^3*d^3*f^3*m^2*x^ 
3-9*a^2*b*c*d^2*e*f^2*m^3*x^2-24*a^2*b*c*d^2*f^3*m^2*x^3-3*a^2*b*d^3*e*f^2 
*m^2*x^3+9*a*b^2*c^2*d*e*f^2*m^3*x^2+21*a*b^2*c^2*d*f^3*m^2*x^3+6*a*b^2*c* 
d^2*e*f^2*m^2*x^3-3*b^3*c^3*e*f^2*m^3*x^2-6*b^3*c^3*f^3*m^2*x^3-3*b^3*c^2* 
d*e*f^2*m^2*x^3+3*a^3*c*d^2*f^3*m^2*x^2+3*a^3*d^3*e^2*f*m^3*x+24*a^3*d^3*e 
*f^2*m^2*x^2+26*a^3*d^3*f^3*m*x^3-6*a^2*b*c^2*d*f^3*m^2*x^2-9*a^2*b*c*d^2* 
e^2*f*m^3*x-69*a^2*b*c*d^2*e*f^2*m^2*x^2-57*a^2*b*c*d^2*f^3*m*x^3-6*a^2*b* 
d^3*e^2*f*m^2*x^2-21*a^2*b*d^3*e*f^2*m*x^3+3*a*b^2*c^3*f^3*m^2*x^2+9*a*b^2 
*c^2*d*e^2*f*m^3*x+66*a*b^2*c^2*d*e*f^2*m^2*x^2+42*a*b^2*c^2*d*f^3*m*x^3+1 
2*a*b^2*c*d^2*e^2*f*m^2*x^2+30*a*b^2*c*d^2*e*f^2*m*x^3+6*a*b^2*d^3*e^2*f*m 
*x^3-3*b^3*c^3*e^2*f*m^3*x-21*b^3*c^3*e*f^2*m^2*x^2-11*b^3*c^3*f^3*m*x^3-6 
*b^3*c^2*d*e^2*f*m^2*x^2-9*b^3*c^2*d*e*f^2*m*x^3-6*b^3*c*d^2*e^2*f*m*x^3+6 
*a^3*c*d^2*e*f^2*m^2*x+21*a^3*c*d^2*f^3*m*x^2+a^3*d^3*e^3*m^3+21*a^3*d^...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 3421 vs. \(2 (471) = 942\).

Time = 0.30 (sec) , antiderivative size = 3421, normalized size of antiderivative = 7.29 \[ \int (a+b x)^m (c+d x)^{-5-m} (e+f x)^3 \, dx=\text {Too large to display} \] Input:

integrate((b*x+a)^m*(d*x+c)^(-5-m)*(f*x+e)^3,x, algorithm="fricas")
 

Output:

Too large to include
 

Sympy [F(-2)]

Exception generated. \[ \int (a+b x)^m (c+d x)^{-5-m} (e+f x)^3 \, dx=\text {Exception raised: HeuristicGCDFailed} \] Input:

integrate((b*x+a)**m*(d*x+c)**(-5-m)*(f*x+e)**3,x)
 

Output:

Exception raised: HeuristicGCDFailed >> no luck
 

Maxima [F]

\[ \int (a+b x)^m (c+d x)^{-5-m} (e+f x)^3 \, dx=\int { {\left (f x + e\right )}^{3} {\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m - 5} \,d x } \] Input:

integrate((b*x+a)^m*(d*x+c)^(-5-m)*(f*x+e)^3,x, algorithm="maxima")
 

Output:

integrate((f*x + e)^3*(b*x + a)^m*(d*x + c)^(-m - 5), x)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 11495 vs. \(2 (471) = 942\).

Time = 0.30 (sec) , antiderivative size = 11495, normalized size of antiderivative = 24.51 \[ \int (a+b x)^m (c+d x)^{-5-m} (e+f x)^3 \, dx=\text {Too large to display} \] Input:

integrate((b*x+a)^m*(d*x+c)^(-5-m)*(f*x+e)^3,x, algorithm="giac")
 

Output:

((b*x + a)^m*b^4*c^3*d*f^3*m^3*x^5*e^(-m*log(d*x + c) - 5*log(d*x + c)) - 
3*(b*x + a)^m*a*b^3*c^2*d^2*f^3*m^3*x^5*e^(-m*log(d*x + c) - 5*log(d*x + c 
)) + 3*(b*x + a)^m*a^2*b^2*c*d^3*f^3*m^3*x^5*e^(-m*log(d*x + c) - 5*log(d* 
x + c)) - (b*x + a)^m*a^3*b*d^4*f^3*m^3*x^5*e^(-m*log(d*x + c) - 5*log(d*x 
 + c)) + 3*(b*x + a)^m*b^4*c^3*d*e*f^2*m^3*x^4*e^(-m*log(d*x + c) - 5*log( 
d*x + c)) - 9*(b*x + a)^m*a*b^3*c^2*d^2*e*f^2*m^3*x^4*e^(-m*log(d*x + c) - 
 5*log(d*x + c)) + 9*(b*x + a)^m*a^2*b^2*c*d^3*e*f^2*m^3*x^4*e^(-m*log(d*x 
 + c) - 5*log(d*x + c)) - 3*(b*x + a)^m*a^3*b*d^4*e*f^2*m^3*x^4*e^(-m*log( 
d*x + c) - 5*log(d*x + c)) + (b*x + a)^m*b^4*c^4*f^3*m^3*x^4*e^(-m*log(d*x 
 + c) - 5*log(d*x + c)) - 2*(b*x + a)^m*a*b^3*c^3*d*f^3*m^3*x^4*e^(-m*log( 
d*x + c) - 5*log(d*x + c)) + 2*(b*x + a)^m*a^3*b*c*d^3*f^3*m^3*x^4*e^(-m*l 
og(d*x + c) - 5*log(d*x + c)) - (b*x + a)^m*a^4*d^4*f^3*m^3*x^4*e^(-m*log( 
d*x + c) - 5*log(d*x + c)) + 3*(b*x + a)^m*b^4*c^2*d^2*e*f^2*m^2*x^5*e^(-m 
*log(d*x + c) - 5*log(d*x + c)) - 6*(b*x + a)^m*a*b^3*c*d^3*e*f^2*m^2*x^5* 
e^(-m*log(d*x + c) - 5*log(d*x + c)) + 3*(b*x + a)^m*a^2*b^2*d^4*e*f^2*m^2 
*x^5*e^(-m*log(d*x + c) - 5*log(d*x + c)) + 6*(b*x + a)^m*b^4*c^3*d*f^3*m^ 
2*x^5*e^(-m*log(d*x + c) - 5*log(d*x + c)) - 21*(b*x + a)^m*a*b^3*c^2*d^2* 
f^3*m^2*x^5*e^(-m*log(d*x + c) - 5*log(d*x + c)) + 24*(b*x + a)^m*a^2*b^2* 
c*d^3*f^3*m^2*x^5*e^(-m*log(d*x + c) - 5*log(d*x + c)) - 9*(b*x + a)^m*a^3 
*b*d^4*f^3*m^2*x^5*e^(-m*log(d*x + c) - 5*log(d*x + c)) + 3*(b*x + a)^m...
 

Mupad [B] (verification not implemented)

Time = 4.40 (sec) , antiderivative size = 4265, normalized size of antiderivative = 9.09 \[ \int (a+b x)^m (c+d x)^{-5-m} (e+f x)^3 \, dx=\text {Too large to display} \] Input:

int(((e + f*x)^3*(a + b*x)^m)/(c + d*x)^(m + 5),x)
                                                                                    
                                                                                    
 

Output:

(x^3*(a + b*x)^m*(24*b^4*c^4*e*f^2 - 36*a^4*d^4*e*f^2 - 60*a^4*c*d^3*f^3 + 
 60*b^4*c^2*d^2*e^3 + 3*a^2*b^2*d^4*e^3*m^2 + 3*b^4*c^2*d^2*e^3*m^2 + 60*b 
^4*c^3*d*e^2*f + 2*a*b^3*c^4*f^3*m - 47*a^4*c*d^3*f^3*m - 57*a^4*d^4*e*f^2 
*m + 42*b^4*c^4*e*f^2*m + 3*a^2*b^2*d^4*e^3*m + 3*a*b^3*c^4*f^3*m^2 + a*b^ 
3*c^4*f^3*m^3 + 27*b^4*c^2*d^2*e^3*m - 12*a^4*c*d^3*f^3*m^2 - a^4*c*d^3*f^ 
3*m^3 - 24*a^4*d^4*e*f^2*m^2 + 21*b^4*c^4*e*f^2*m^2 - 3*a^4*d^4*e*f^2*m^3 
+ 3*b^4*c^4*e*f^2*m^3 - 240*a*b^3*c^2*d^2*e^2*f - 6*a*b^3*c*d^3*e^3*m^2 - 
15*a^2*b^2*c^3*d*f^3*m + 60*a^3*b*c^2*d^2*f^3*m - 15*a^3*b*d^4*e^2*f*m^2 - 
 3*a^3*b*d^4*e^2*f*m^3 + 30*b^4*c^3*d*e^2*f*m^2 + 3*b^4*c^3*d*e^2*f*m^3 + 
144*a^2*b^2*c^2*d^2*e*f^2 - 18*a^2*b^2*c^3*d*f^3*m^2 + 27*a^3*b*c^2*d^2*f^ 
3*m^2 - 3*a^2*b^2*c^3*d*f^3*m^3 + 3*a^3*b*c^2*d^2*f^3*m^3 - 96*a*b^3*c^3*d 
*e*f^2 + 144*a^3*b*c*d^3*e*f^2 - 30*a*b^3*c*d^3*e^3*m - 12*a^3*b*d^4*e^2*f 
*m + 87*b^4*c^3*d*e^2*f*m - 138*a*b^3*c^3*d*e*f^2*m + 108*a^3*b*c*d^3*e*f^ 
2*m + 9*a^2*b^2*c^2*d^2*e*f^2*m^2 - 198*a*b^3*c^2*d^2*e^2*f*m + 123*a^2*b^ 
2*c*d^3*e^2*f*m - 48*a*b^3*c^3*d*e*f^2*m^2 + 42*a^3*b*c*d^3*e*f^2*m^2 - 6* 
a*b^3*c^3*d*e*f^2*m^3 + 6*a^3*b*c*d^3*e*f^2*m^3 + 45*a^2*b^2*c^2*d^2*e*f^2 
*m - 75*a*b^3*c^2*d^2*e^2*f*m^2 + 60*a^2*b^2*c*d^3*e^2*f*m^2 - 9*a*b^3*c^2 
*d^2*e^2*f*m^3 + 9*a^2*b^2*c*d^3*e^2*f*m^3))/((a*d - b*c)^4*(c + d*x)^(m + 
 5)*(50*m + 35*m^2 + 10*m^3 + m^4 + 24)) - ((a + b*x)^m*(6*a^4*c^4*f^3 - 2 
4*a*b^3*c^4*e^3 + 6*a^4*c*d^3*e^3 - 24*a^3*b*c^4*e*f^2 + 6*a^4*c^3*d*e*...
 

Reduce [F]

\[ \int (a+b x)^m (c+d x)^{-5-m} (e+f x)^3 \, dx=\left (\int \frac {\left (b x +a \right )^{m}}{\left (d x +c \right )^{m} c^{5}+5 \left (d x +c \right )^{m} c^{4} d x +10 \left (d x +c \right )^{m} c^{3} d^{2} x^{2}+10 \left (d x +c \right )^{m} c^{2} d^{3} x^{3}+5 \left (d x +c \right )^{m} c \,d^{4} x^{4}+\left (d x +c \right )^{m} d^{5} x^{5}}d x \right ) e^{3}+\left (\int \frac {\left (b x +a \right )^{m} x^{3}}{\left (d x +c \right )^{m} c^{5}+5 \left (d x +c \right )^{m} c^{4} d x +10 \left (d x +c \right )^{m} c^{3} d^{2} x^{2}+10 \left (d x +c \right )^{m} c^{2} d^{3} x^{3}+5 \left (d x +c \right )^{m} c \,d^{4} x^{4}+\left (d x +c \right )^{m} d^{5} x^{5}}d x \right ) f^{3}+3 \left (\int \frac {\left (b x +a \right )^{m} x^{2}}{\left (d x +c \right )^{m} c^{5}+5 \left (d x +c \right )^{m} c^{4} d x +10 \left (d x +c \right )^{m} c^{3} d^{2} x^{2}+10 \left (d x +c \right )^{m} c^{2} d^{3} x^{3}+5 \left (d x +c \right )^{m} c \,d^{4} x^{4}+\left (d x +c \right )^{m} d^{5} x^{5}}d x \right ) e \,f^{2}+3 \left (\int \frac {\left (b x +a \right )^{m} x}{\left (d x +c \right )^{m} c^{5}+5 \left (d x +c \right )^{m} c^{4} d x +10 \left (d x +c \right )^{m} c^{3} d^{2} x^{2}+10 \left (d x +c \right )^{m} c^{2} d^{3} x^{3}+5 \left (d x +c \right )^{m} c \,d^{4} x^{4}+\left (d x +c \right )^{m} d^{5} x^{5}}d x \right ) e^{2} f \] Input:

int((b*x+a)^m*(d*x+c)^(-5-m)*(f*x+e)^3,x)
 

Output:

int((a + b*x)**m/((c + d*x)**m*c**5 + 5*(c + d*x)**m*c**4*d*x + 10*(c + d* 
x)**m*c**3*d**2*x**2 + 10*(c + d*x)**m*c**2*d**3*x**3 + 5*(c + d*x)**m*c*d 
**4*x**4 + (c + d*x)**m*d**5*x**5),x)*e**3 + int(((a + b*x)**m*x**3)/((c + 
 d*x)**m*c**5 + 5*(c + d*x)**m*c**4*d*x + 10*(c + d*x)**m*c**3*d**2*x**2 + 
 10*(c + d*x)**m*c**2*d**3*x**3 + 5*(c + d*x)**m*c*d**4*x**4 + (c + d*x)** 
m*d**5*x**5),x)*f**3 + 3*int(((a + b*x)**m*x**2)/((c + d*x)**m*c**5 + 5*(c 
 + d*x)**m*c**4*d*x + 10*(c + d*x)**m*c**3*d**2*x**2 + 10*(c + d*x)**m*c** 
2*d**3*x**3 + 5*(c + d*x)**m*c*d**4*x**4 + (c + d*x)**m*d**5*x**5),x)*e*f* 
*2 + 3*int(((a + b*x)**m*x)/((c + d*x)**m*c**5 + 5*(c + d*x)**m*c**4*d*x + 
 10*(c + d*x)**m*c**3*d**2*x**2 + 10*(c + d*x)**m*c**2*d**3*x**3 + 5*(c + 
d*x)**m*c*d**4*x**4 + (c + d*x)**m*d**5*x**5),x)*e**2*f