\(\int \frac {(c+d x)^n (A+B x+C x^2+D x^3)}{(a+b x)^{7/2}} \, dx\) [187]

Optimal result

Integrand size = 32, antiderivative size = 404 \[ \int \frac {(c+d x)^n \left (A+B x+C x^2+D x^3\right )}{(a+b x)^{7/2}} \, dx=-\frac {2 \left (A b^3-a \left (b^2 B-a b C+a^2 D\right )\right ) (c+d x)^{1+n}}{5 b^3 (b c-a d) (a+b x)^{5/2}}-\frac {2 (b c D-2 a d D (2+3 n)+b C (d+2 d n)) (c+d x)^{1+n}}{b^3 d^2 \left (1-4 n^2\right ) (a+b x)^{3/2}}+\frac {2 D (c+d x)^{1+n}}{b^3 d (1+2 n) \sqrt {a+b x}}-\frac {2 \left (8 a^3 d^3 D \left (6+11 n+6 n^2+n^3\right )-4 a^2 b d^2 \left (2+3 n+n^2\right ) (15 c D+C (d+2 d n))+2 a b^2 d (1+n) \left (45 c^2 D+10 c C d (1+2 n)-B d^2 \left (1-4 n^2\right )\right )-b^3 \left (15 c^3 D+15 c^2 C d (1+2 n)-5 B c d^2 \left (1-4 n^2\right )+A d^3 \left (3-2 n-12 n^2+8 n^3\right )\right )\right ) (c+d x)^n \left (\frac {b (c+d x)}{b c-a d}\right )^{-n} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-n,-\frac {1}{2},-\frac {d (a+b x)}{b c-a d}\right )}{15 b^4 d^2 (b c-a d) (1-2 n) (1+2 n) (a+b x)^{3/2}} \] Output:

-2/5*(A*b^3-a*(B*b^2-C*a*b+D*a^2))*(d*x+c)^(1+n)/b^3/(-a*d+b*c)/(b*x+a)^(5 
/2)-2*(D*b*c-2*a*d*D*(2+3*n)+b*C*(2*d*n+d))*(d*x+c)^(1+n)/b^3/d^2/(-4*n^2+ 
1)/(b*x+a)^(3/2)+2*D*(d*x+c)^(1+n)/b^3/d/(1+2*n)/(b*x+a)^(1/2)-2/15*(8*a^3 
*d^3*D*(n^3+6*n^2+11*n+6)-4*a^2*b*d^2*(n^2+3*n+2)*(15*D*c+C*(2*d*n+d))+2*a 
*b^2*d*(1+n)*(45*D*c^2+10*c*C*d*(1+2*n)-B*d^2*(-4*n^2+1))-b^3*(15*D*c^3+15 
*c^2*C*d*(1+2*n)-5*B*c*d^2*(-4*n^2+1)+A*d^3*(8*n^3-12*n^2-2*n+3)))*(d*x+c) 
^n*hypergeom([-3/2, -n],[-1/2],-d*(b*x+a)/(-a*d+b*c))/b^4/d^2/(-a*d+b*c)/( 
1-2*n)/(1+2*n)/(b*x+a)^(3/2)/((b*(d*x+c)/(-a*d+b*c))^n)
 

Mathematica [A] (verified)

Time = 2.96 (sec) , antiderivative size = 231, normalized size of antiderivative = 0.57 \[ \int \frac {(c+d x)^n \left (A+B x+C x^2+D x^3\right )}{(a+b x)^{7/2}} \, dx=\frac {2 (c+d x)^n \left (\frac {b (c+d x)}{b c-a d}\right )^{-n} \left (-3 \left (A b^3-a \left (b^2 B-a b C+a^2 D\right )\right ) \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},-n,-\frac {3}{2},\frac {d (a+b x)}{-b c+a d}\right )-5 (a+b x) \left (\left (b^2 B-2 a b C+3 a^2 D\right ) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-n,-\frac {1}{2},\frac {d (a+b x)}{-b c+a d}\right )-3 (a+b x) \left ((-b C+3 a D) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-n,\frac {1}{2},\frac {d (a+b x)}{-b c+a d}\right )+D (a+b x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-n,\frac {3}{2},\frac {d (a+b x)}{-b c+a d}\right )\right )\right )\right )}{15 b^4 (a+b x)^{5/2}} \] Input:

Integrate[((c + d*x)^n*(A + B*x + C*x^2 + D*x^3))/(a + b*x)^(7/2),x]
 

Output:

(2*(c + d*x)^n*(-3*(A*b^3 - a*(b^2*B - a*b*C + a^2*D))*Hypergeometric2F1[- 
5/2, -n, -3/2, (d*(a + b*x))/(-(b*c) + a*d)] - 5*(a + b*x)*((b^2*B - 2*a*b 
*C + 3*a^2*D)*Hypergeometric2F1[-3/2, -n, -1/2, (d*(a + b*x))/(-(b*c) + a* 
d)] - 3*(a + b*x)*((-(b*C) + 3*a*D)*Hypergeometric2F1[-1/2, -n, 1/2, (d*(a 
 + b*x))/(-(b*c) + a*d)] + D*(a + b*x)*Hypergeometric2F1[1/2, -n, 3/2, (d* 
(a + b*x))/(-(b*c) + a*d)]))))/(15*b^4*(a + b*x)^(5/2)*((b*(c + d*x))/(b*c 
 - a*d))^n)
 

Rubi [A] (warning: unable to verify)

Time = 1.03 (sec) , antiderivative size = 575, normalized size of antiderivative = 1.42, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.219, Rules used = {2124, 27, 1193, 27, 87, 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.

\(\displaystyle \int \frac {(c+d x)^n \left (A+B x+C x^2+D x^3\right )}{(a+b x)^{7/2}} \, dx\)

\(\Big \downarrow \) 2124

\(\displaystyle -\frac {2 \int -\frac {(c+d x)^n \left (5 \left (c-\frac {a d}{b}\right ) D x^2+\frac {5 (b c-a d) (b C-a D) x}{b^2}+\frac {-2 d D (n+1) a^3+b (5 c D+2 C d (n+1)) a^2-b^2 (5 c C+2 B d (n+1)) a+b^3 (5 B c-A d (3-2 n))}{b^3}\right )}{2 (a+b x)^{5/2}}dx}{5 (b c-a d)}-\frac {2 (c+d x)^{n+1} \left (A b^3-a \left (a^2 D-a b C+b^2 B\right )\right )}{5 b^3 (a+b x)^{5/2} (b c-a d)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {(c+d x)^n \left (-\frac {2 d D (n+1) a^3}{b^3}+\frac {(5 c D+2 C d (n+1)) a^2}{b^2}-\frac {(5 c C+2 B d (n+1)) a}{b}+5 \left (c-\frac {a d}{b}\right ) D x^2+5 B c-A d (3-2 n)+\frac {5 (b c-a d) (b C-a D) x}{b^2}\right )}{(a+b x)^{5/2}}dx}{5 (b c-a d)}-\frac {2 (c+d x)^{n+1} \left (A b^3-a \left (a^2 D-a b C+b^2 B\right )\right )}{5 b^3 (a+b x)^{5/2} (b c-a d)}\)

\(\Big \downarrow \) 1193

\(\displaystyle \frac {-\frac {2 \int \frac {(c+d x)^n \left (2 d^2 D \left (2 n^2+11 n+9\right ) a^3-b d \left (15 c D (2 n+3)+4 C d \left (n^2+3 n+2\right )\right ) a^2+2 b^2 \left (15 D c^2+10 C d (n+1) c-B d^2 \left (-2 n^2-n+1\right )\right ) a-b^3 \left (15 C c^2-5 B d (1-2 n) c+A d^2 \left (4 n^2-8 n+3\right )\right )-15 b (b c-a d)^2 D x\right )}{2 b^3 (a+b x)^{3/2}}dx}{3 (b c-a d)}-\frac {2 (c+d x)^{n+1} \left (-2 a^3 d D (n+6)+a^2 b (15 c D+C d (2 n+7))-2 a b^2 (B d (n+1)+5 c C)+b^3 (5 B c-A d (3-2 n))\right )}{3 b^3 (a+b x)^{3/2} (b c-a d)}}{5 (b c-a d)}-\frac {2 (c+d x)^{n+1} \left (A b^3-a \left (a^2 D-a b C+b^2 B\right )\right )}{5 b^3 (a+b x)^{5/2} (b c-a d)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {-\frac {\int \frac {(c+d x)^n \left (2 d^2 D \left (2 n^2+11 n+9\right ) a^3-b d \left (15 c D (2 n+3)+4 C d \left (n^2+3 n+2\right )\right ) a^2+2 b^2 \left (15 D c^2+10 C d (n+1) c-B d^2 \left (-2 n^2-n+1\right )\right ) a-b^3 \left (15 C c^2-5 B d (1-2 n) c+A d^2 \left (4 n^2-8 n+3\right )\right )-15 b (b c-a d)^2 D x\right )}{(a+b x)^{3/2}}dx}{3 b^3 (b c-a d)}-\frac {2 (c+d x)^{n+1} \left (-2 a^3 d D (n+6)+a^2 b (15 c D+C d (2 n+7))-2 a b^2 (B d (n+1)+5 c C)+b^3 (5 B c-A d (3-2 n))\right )}{3 b^3 (a+b x)^{3/2} (b c-a d)}}{5 (b c-a d)}-\frac {2 (c+d x)^{n+1} \left (A b^3-a \left (a^2 D-a b C+b^2 B\right )\right )}{5 b^3 (a+b x)^{5/2} (b c-a d)}\)

\(\Big \downarrow \) 87

\(\displaystyle \frac {-\frac {\frac {\left (8 a^3 d^3 D \left (n^3+6 n^2+11 n+6\right )-4 a^2 b d^2 \left (n^2+3 n+2\right ) (15 c D+C (2 d n+d))+2 a b^2 d (n+1) \left (-B d^2 \left (1-4 n^2\right )+45 c^2 D+10 c C d (2 n+1)\right )-\left (b^3 \left (A d^3 \left (8 n^3-12 n^2-2 n+3\right )-5 B c d^2 \left (1-4 n^2\right )+15 c^3 D+15 c^2 C d (2 n+1)\right )\right )\right ) \int \frac {(c+d x)^n}{\sqrt {a+b x}}dx}{b c-a d}-\frac {2 (c+d x)^{n+1} \left (a^3 d^2 D \left (4 n^2+22 n+33\right )-a^2 b d \left (15 c D (2 n+5)+4 C d \left (n^2+3 n+2\right )\right )+a b^2 \left (-2 B d^2 \left (-2 n^2-n+1\right )+45 c^2 D+20 c C d (n+1)\right )-b^3 \left (A d^2 \left (4 n^2-8 n+3\right )-5 B c d (1-2 n)+15 c^2 C\right )\right )}{\sqrt {a+b x} (b c-a d)}}{3 b^3 (b c-a d)}-\frac {2 (c+d x)^{n+1} \left (-2 a^3 d D (n+6)+a^2 b (15 c D+C d (2 n+7))-2 a b^2 (B d (n+1)+5 c C)+b^3 (5 B c-A d (3-2 n))\right )}{3 b^3 (a+b x)^{3/2} (b c-a d)}}{5 (b c-a d)}-\frac {2 (c+d x)^{n+1} \left (A b^3-a \left (a^2 D-a b C+b^2 B\right )\right )}{5 b^3 (a+b x)^{5/2} (b c-a d)}\)

\(\Big \downarrow \) 80

\(\displaystyle \frac {-\frac {\frac {(c+d x)^n \left (\frac {b (c+d x)}{b c-a d}\right )^{-n} \left (8 a^3 d^3 D \left (n^3+6 n^2+11 n+6\right )-4 a^2 b d^2 \left (n^2+3 n+2\right ) (15 c D+C (2 d n+d))+2 a b^2 d (n+1) \left (-B d^2 \left (1-4 n^2\right )+45 c^2 D+10 c C d (2 n+1)\right )-\left (b^3 \left (A d^3 \left (8 n^3-12 n^2-2 n+3\right )-5 B c d^2 \left (1-4 n^2\right )+15 c^3 D+15 c^2 C d (2 n+1)\right )\right )\right ) \int \frac {\left (\frac {b c}{b c-a d}+\frac {b d x}{b c-a d}\right )^n}{\sqrt {a+b x}}dx}{b c-a d}-\frac {2 (c+d x)^{n+1} \left (a^3 d^2 D \left (4 n^2+22 n+33\right )-a^2 b d \left (15 c D (2 n+5)+4 C d \left (n^2+3 n+2\right )\right )+a b^2 \left (-2 B d^2 \left (-2 n^2-n+1\right )+45 c^2 D+20 c C d (n+1)\right )-b^3 \left (A d^2 \left (4 n^2-8 n+3\right )-5 B c d (1-2 n)+15 c^2 C\right )\right )}{\sqrt {a+b x} (b c-a d)}}{3 b^3 (b c-a d)}-\frac {2 (c+d x)^{n+1} \left (-2 a^3 d D (n+6)+a^2 b (15 c D+C d (2 n+7))-2 a b^2 (B d (n+1)+5 c C)+b^3 (5 B c-A d (3-2 n))\right )}{3 b^3 (a+b x)^{3/2} (b c-a d)}}{5 (b c-a d)}-\frac {2 (c+d x)^{n+1} \left (A b^3-a \left (a^2 D-a b C+b^2 B\right )\right )}{5 b^3 (a+b x)^{5/2} (b c-a d)}\)

\(\Big \downarrow \) 79

\(\displaystyle \frac {-\frac {\frac {2 \sqrt {a+b x} (c+d x)^n \left (\frac {b (c+d x)}{b c-a d}\right )^{-n} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-n,\frac {3}{2},-\frac {d (a+b x)}{b c-a d}\right ) \left (8 a^3 d^3 D \left (n^3+6 n^2+11 n+6\right )-4 a^2 b d^2 \left (n^2+3 n+2\right ) (15 c D+C (2 d n+d))+2 a b^2 d (n+1) \left (-B d^2 \left (1-4 n^2\right )+45 c^2 D+10 c C d (2 n+1)\right )-\left (b^3 \left (A d^3 \left (8 n^3-12 n^2-2 n+3\right )-5 B c d^2 \left (1-4 n^2\right )+15 c^3 D+15 c^2 C d (2 n+1)\right )\right )\right )}{b (b c-a d)}-\frac {2 (c+d x)^{n+1} \left (a^3 d^2 D \left (4 n^2+22 n+33\right )-a^2 b d \left (15 c D (2 n+5)+4 C d \left (n^2+3 n+2\right )\right )+a b^2 \left (-2 B d^2 \left (-2 n^2-n+1\right )+45 c^2 D+20 c C d (n+1)\right )-b^3 \left (A d^2 \left (4 n^2-8 n+3\right )-5 B c d (1-2 n)+15 c^2 C\right )\right )}{\sqrt {a+b x} (b c-a d)}}{3 b^3 (b c-a d)}-\frac {2 (c+d x)^{n+1} \left (-2 a^3 d D (n+6)+a^2 b (15 c D+C d (2 n+7))-2 a b^2 (B d (n+1)+5 c C)+b^3 (5 B c-A d (3-2 n))\right )}{3 b^3 (a+b x)^{3/2} (b c-a d)}}{5 (b c-a d)}-\frac {2 (c+d x)^{n+1} \left (A b^3-a \left (a^2 D-a b C+b^2 B\right )\right )}{5 b^3 (a+b x)^{5/2} (b c-a d)}\)

Input:

Int[((c + d*x)^n*(A + B*x + C*x^2 + D*x^3))/(a + b*x)^(7/2),x]
 

Output:

(-2*(A*b^3 - a*(b^2*B - a*b*C + a^2*D))*(c + d*x)^(1 + n))/(5*b^3*(b*c - a 
*d)*(a + b*x)^(5/2)) + ((-2*(b^3*(5*B*c - A*d*(3 - 2*n)) - 2*a^3*d*D*(6 + 
n) - 2*a*b^2*(5*c*C + B*d*(1 + n)) + a^2*b*(15*c*D + C*d*(7 + 2*n)))*(c + 
d*x)^(1 + n))/(3*b^3*(b*c - a*d)*(a + b*x)^(3/2)) - ((-2*(a^3*d^2*D*(33 + 
22*n + 4*n^2) + a*b^2*(45*c^2*D + 20*c*C*d*(1 + n) - 2*B*d^2*(1 - n - 2*n^ 
2)) - a^2*b*d*(15*c*D*(5 + 2*n) + 4*C*d*(2 + 3*n + n^2)) - b^3*(15*c^2*C - 
 5*B*c*d*(1 - 2*n) + A*d^2*(3 - 8*n + 4*n^2)))*(c + d*x)^(1 + n))/((b*c - 
a*d)*Sqrt[a + b*x]) + (2*(8*a^3*d^3*D*(6 + 11*n + 6*n^2 + n^3) - 4*a^2*b*d 
^2*(2 + 3*n + n^2)*(15*c*D + C*(d + 2*d*n)) + 2*a*b^2*d*(1 + n)*(45*c^2*D 
+ 10*c*C*d*(1 + 2*n) - B*d^2*(1 - 4*n^2)) - b^3*(15*c^3*D + 15*c^2*C*d*(1 
+ 2*n) - 5*B*c*d^2*(1 - 4*n^2) + A*d^3*(3 - 2*n - 12*n^2 + 8*n^3)))*Sqrt[a 
 + b*x]*(c + d*x)^n*Hypergeometric2F1[1/2, -n, 3/2, -((d*(a + b*x))/(b*c - 
 a*d))])/(b*(b*c - a*d)*((b*(c + d*x))/(b*c - a*d))^n))/(3*b^3*(b*c - a*d) 
))/(5*(b*c - a*d))
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, 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 87

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]))))
 

rule 1193

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) 
 + (c_.)*(x_)^2)^(p_.), x_Symbol] :> With[{Qx = PolynomialQuotient[(a + b*x 
 + c*x^2)^p, d + e*x, x], R = PolynomialRemainder[(a + b*x + c*x^2)^p, d + 
e*x, x]}, Simp[R*(d + e*x)^(m + 1)*((f + g*x)^(n + 1)/((m + 1)*(e*f - d*g)) 
), x] + Simp[1/((m + 1)*(e*f - d*g))   Int[(d + e*x)^(m + 1)*(f + g*x)^n*Ex 
pandToSum[(m + 1)*(e*f - d*g)*Qx - g*R*(m + n + 2), x], x], x]] /; FreeQ[{a 
, b, c, d, e, f, g, n}, x] && IGtQ[p, 0] && ILtQ[2*m, -2] &&  !IntegerQ[n] 
&&  !(EqQ[m, -2] && EqQ[p, 1] && EqQ[2*c*d - b*e, 0])
 

rule 2124

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] : 
> With[{Qx = PolynomialQuotient[Px, a + b*x, x], R = PolynomialRemainder[Px 
, a + b*x, x]}, Simp[R*(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((m + 1)*(b*c - 
 a*d))), x] + Simp[1/((m + 1)*(b*c - a*d))   Int[(a + b*x)^(m + 1)*(c + d*x 
)^n*ExpandToSum[(m + 1)*(b*c - a*d)*Qx - d*R*(m + n + 2), x], x], x]] /; Fr 
eeQ[{a, b, c, d, n}, x] && PolyQ[Px, x] && LtQ[m, -1] && (IntegerQ[m] ||  ! 
ILtQ[n, -1])
 

Maple [F]

\[\int \frac {\left (x d +c \right )^{n} \left (D x^{3}+C \,x^{2}+B x +A \right )}{\left (b x +a \right )^{\frac {7}{2}}}d x\]

Input:

int((d*x+c)^n*(D*x^3+C*x^2+B*x+A)/(b*x+a)^(7/2),x)
 

Output:

int((d*x+c)^n*(D*x^3+C*x^2+B*x+A)/(b*x+a)^(7/2),x)
 

Fricas [F]

\[ \int \frac {(c+d x)^n \left (A+B x+C x^2+D x^3\right )}{(a+b x)^{7/2}} \, dx=\int { \frac {{\left (D x^{3} + C x^{2} + B x + A\right )} {\left (d x + c\right )}^{n}}{{\left (b x + a\right )}^{\frac {7}{2}}} \,d x } \] Input:

integrate((d*x+c)^n*(D*x^3+C*x^2+B*x+A)/(b*x+a)^(7/2),x, algorithm="fricas 
")
 

Output:

integral((D*x^3 + C*x^2 + B*x + A)*sqrt(b*x + a)*(d*x + c)^n/(b^4*x^4 + 4* 
a*b^3*x^3 + 6*a^2*b^2*x^2 + 4*a^3*b*x + a^4), x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(c+d x)^n \left (A+B x+C x^2+D x^3\right )}{(a+b x)^{7/2}} \, dx=\text {Timed out} \] Input:

integrate((d*x+c)**n*(D*x**3+C*x**2+B*x+A)/(b*x+a)**(7/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {(c+d x)^n \left (A+B x+C x^2+D x^3\right )}{(a+b x)^{7/2}} \, dx=\int { \frac {{\left (D x^{3} + C x^{2} + B x + A\right )} {\left (d x + c\right )}^{n}}{{\left (b x + a\right )}^{\frac {7}{2}}} \,d x } \] Input:

integrate((d*x+c)^n*(D*x^3+C*x^2+B*x+A)/(b*x+a)^(7/2),x, algorithm="maxima 
")
 

Output:

integrate((D*x^3 + C*x^2 + B*x + A)*(d*x + c)^n/(b*x + a)^(7/2), x)
 

Giac [F]

\[ \int \frac {(c+d x)^n \left (A+B x+C x^2+D x^3\right )}{(a+b x)^{7/2}} \, dx=\int { \frac {{\left (D x^{3} + C x^{2} + B x + A\right )} {\left (d x + c\right )}^{n}}{{\left (b x + a\right )}^{\frac {7}{2}}} \,d x } \] Input:

integrate((d*x+c)^n*(D*x^3+C*x^2+B*x+A)/(b*x+a)^(7/2),x, algorithm="giac")
 

Output:

integrate((D*x^3 + C*x^2 + B*x + A)*(d*x + c)^n/(b*x + a)^(7/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(c+d x)^n \left (A+B x+C x^2+D x^3\right )}{(a+b x)^{7/2}} \, dx=\int \frac {{\left (c+d\,x\right )}^n\,\left (A+B\,x+C\,x^2+x^3\,D\right )}{{\left (a+b\,x\right )}^{7/2}} \,d x \] Input:

int(((c + d*x)^n*(A + B*x + C*x^2 + x^3*D))/(a + b*x)^(7/2),x)
 

Output:

int(((c + d*x)^n*(A + B*x + C*x^2 + x^3*D))/(a + b*x)^(7/2), x)
 

Reduce [F]

\[ \int \frac {(c+d x)^n \left (A+B x+C x^2+D x^3\right )}{(a+b x)^{7/2}} \, dx=\text {too large to display} \] Input:

int((d*x+c)^n*(D*x^3+C*x^2+B*x+A)/(b*x+a)^(7/2),x)
                                                                                    
                                                                                    
 

Output:

(2*( - 8*(c + d*x)**n*sqrt(a + b*x)*a**3*c*d**2*n**2 - 40*(c + d*x)**n*sqr 
t(a + b*x)*a**3*c*d**2*n - 48*(c + d*x)**n*sqrt(a + b*x)*a**3*c*d**2 + 8*( 
c + d*x)**n*sqrt(a + b*x)*a**3*d**3*n**3*x + 40*(c + d*x)**n*sqrt(a + b*x) 
*a**3*d**3*n**2*x + 48*(c + d*x)**n*sqrt(a + b*x)*a**3*d**3*n*x + 16*(c + 
d*x)**n*sqrt(a + b*x)*a**2*b*c**2*d*n**2 + 64*(c + d*x)**n*sqrt(a + b*x)*a 
**2*b*c**2*d*n + 8*(c + d*x)**n*sqrt(a + b*x)*a**2*b*c**2*d - 16*(c + d*x) 
**n*sqrt(a + b*x)*a**2*b*c*d**2*n**3*x - 84*(c + d*x)**n*sqrt(a + b*x)*a** 
2*b*c*d**2*n**2*x - 108*(c + d*x)**n*sqrt(a + b*x)*a**2*b*c*d**2*n*x - 120 
*(c + d*x)**n*sqrt(a + b*x)*a**2*b*c*d**2*x - 8*(c + d*x)**n*sqrt(a + b*x) 
*a**2*b*d**3*n**3*x**2 - 12*(c + d*x)**n*sqrt(a + b*x)*a**2*b*d**3*n**2*x* 
*2 + 36*(c + d*x)**n*sqrt(a + b*x)*a**2*b*d**3*n*x**2 + 8*(c + d*x)**n*sqr 
t(a + b*x)*a*b**3*c*d*n**3 - 20*(c + d*x)**n*sqrt(a + b*x)*a*b**3*c*d*n**2 
 - 2*(c + d*x)**n*sqrt(a + b*x)*a*b**3*c*d*n + 5*(c + d*x)**n*sqrt(a + b*x 
)*a*b**3*c*d + 8*(c + d*x)**n*sqrt(a + b*x)*a*b**3*d**2*n**3*x - 2*(c + d* 
x)**n*sqrt(a + b*x)*a*b**3*d**2*n*x - 8*(c + d*x)**n*sqrt(a + b*x)*a*b**2* 
c**3*n**2 - 8*(c + d*x)**n*sqrt(a + b*x)*a*b**2*c**3*n + 8*(c + d*x)**n*sq 
rt(a + b*x)*a*b**2*c**2*d*n**3*x + 48*(c + d*x)**n*sqrt(a + b*x)*a*b**2*c* 
*2*d*n**2*x + 160*(c + d*x)**n*sqrt(a + b*x)*a*b**2*c**2*d*n*x + 20*(c + d 
*x)**n*sqrt(a + b*x)*a*b**2*c**2*d*x + 16*(c + d*x)**n*sqrt(a + b*x)*a*b** 
2*c*d**2*n**3*x**2 + 24*(c + d*x)**n*sqrt(a + b*x)*a*b**2*c*d**2*n*x**2...