\(\int (c x^2)^{5/2} (a+b x)^p \, dx\) [432]

Optimal result

Integrand size = 17, antiderivative size = 217 \[ \int \left (c x^2\right )^{5/2} (a+b x)^p \, dx=-\frac {a^5 c^2 \sqrt {c x^2} (a+b x)^{1+p}}{b^6 (1+p) x}+\frac {5 a^4 c^2 \sqrt {c x^2} (a+b x)^{2+p}}{b^6 (2+p) x}-\frac {10 a^3 c^2 \sqrt {c x^2} (a+b x)^{3+p}}{b^6 (3+p) x}+\frac {10 a^2 c^2 \sqrt {c x^2} (a+b x)^{4+p}}{b^6 (4+p) x}-\frac {5 a c^2 \sqrt {c x^2} (a+b x)^{5+p}}{b^6 (5+p) x}+\frac {c^2 \sqrt {c x^2} (a+b x)^{6+p}}{b^6 (6+p) x} \] Output:

-a^5*c^2*(c*x^2)^(1/2)*(b*x+a)^(p+1)/b^6/(p+1)/x+5*a^4*c^2*(c*x^2)^(1/2)*( 
b*x+a)^(2+p)/b^6/(2+p)/x-10*a^3*c^2*(c*x^2)^(1/2)*(b*x+a)^(3+p)/b^6/(3+p)/ 
x+10*a^2*c^2*(c*x^2)^(1/2)*(b*x+a)^(4+p)/b^6/(4+p)/x-5*a*c^2*(c*x^2)^(1/2) 
*(b*x+a)^(5+p)/b^6/(5+p)/x+c^2*(c*x^2)^(1/2)*(b*x+a)^(6+p)/b^6/(6+p)/x
 

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 172, normalized size of antiderivative = 0.79 \[ \int \left (c x^2\right )^{5/2} (a+b x)^p \, dx=\frac {c^3 x (a+b x)^{1+p} \left (-120 a^5+120 a^4 b (1+p) x-60 a^3 b^2 \left (2+3 p+p^2\right ) x^2+20 a^2 b^3 \left (6+11 p+6 p^2+p^3\right ) x^3-5 a b^4 \left (24+50 p+35 p^2+10 p^3+p^4\right ) x^4+b^5 \left (120+274 p+225 p^2+85 p^3+15 p^4+p^5\right ) x^5\right )}{b^6 (1+p) (2+p) (3+p) (4+p) (5+p) (6+p) \sqrt {c x^2}} \] Input:

Integrate[(c*x^2)^(5/2)*(a + b*x)^p,x]
 

Output:

(c^3*x*(a + b*x)^(1 + p)*(-120*a^5 + 120*a^4*b*(1 + p)*x - 60*a^3*b^2*(2 + 
 3*p + p^2)*x^2 + 20*a^2*b^3*(6 + 11*p + 6*p^2 + p^3)*x^3 - 5*a*b^4*(24 + 
50*p + 35*p^2 + 10*p^3 + p^4)*x^4 + b^5*(120 + 274*p + 225*p^2 + 85*p^3 + 
15*p^4 + p^5)*x^5))/(b^6*(1 + p)*(2 + p)*(3 + p)*(4 + p)*(5 + p)*(6 + p)*S 
qrt[c*x^2])
 

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.66, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {34, 53, 2009}

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.

\(\displaystyle \int \left (c x^2\right )^{5/2} (a+b x)^p \, dx\)

\(\Big \downarrow \) 34

\(\displaystyle \frac {c^2 \sqrt {c x^2} \int x^5 (a+b x)^pdx}{x}\)

\(\Big \downarrow \) 53

\(\displaystyle \frac {c^2 \sqrt {c x^2} \int \left (-\frac {a^5 (a+b x)^p}{b^5}+\frac {5 a^4 (a+b x)^{p+1}}{b^5}-\frac {10 a^3 (a+b x)^{p+2}}{b^5}+\frac {10 a^2 (a+b x)^{p+3}}{b^5}-\frac {5 a (a+b x)^{p+4}}{b^5}+\frac {(a+b x)^{p+5}}{b^5}\right )dx}{x}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {c^2 \sqrt {c x^2} \left (-\frac {a^5 (a+b x)^{p+1}}{b^6 (p+1)}+\frac {5 a^4 (a+b x)^{p+2}}{b^6 (p+2)}-\frac {10 a^3 (a+b x)^{p+3}}{b^6 (p+3)}+\frac {10 a^2 (a+b x)^{p+4}}{b^6 (p+4)}-\frac {5 a (a+b x)^{p+5}}{b^6 (p+5)}+\frac {(a+b x)^{p+6}}{b^6 (p+6)}\right )}{x}\)

Input:

Int[(c*x^2)^(5/2)*(a + b*x)^p,x]
 

Output:

(c^2*Sqrt[c*x^2]*(-((a^5*(a + b*x)^(1 + p))/(b^6*(1 + p))) + (5*a^4*(a + b 
*x)^(2 + p))/(b^6*(2 + p)) - (10*a^3*(a + b*x)^(3 + p))/(b^6*(3 + p)) + (1 
0*a^2*(a + b*x)^(4 + p))/(b^6*(4 + p)) - (5*a*(a + b*x)^(5 + p))/(b^6*(5 + 
 p)) + (a + b*x)^(6 + p)/(b^6*(6 + p))))/x
 

Defintions of rubi rules used

rule 34

Int[(u_.)*((a_.)*(x_)^(m_))^(p_), x_Symbol] :> Simp[a^IntPart[p]*((a*x^m)^F 
racPart[p]/x^(m*FracPart[p]))   Int[u*x^(m*p), x], x] /; FreeQ[{a, m, p}, x 
] &&  !IntegerQ[p]
 

rule 53

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int 
[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, 
x] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0] && LeQ[7*m + 4*n + 4, 0]) 
|| LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

Maple [A] (verified)

Time = 0.20 (sec) , antiderivative size = 280, normalized size of antiderivative = 1.29

method	result	size
gosper	\(-\frac {\left (c \,x^{2}\right )^{\frac {5}{2}} \left (b x +a \right )^{p +1} \left (-b^{5} p^{5} x^{5}-15 b^{5} p^{4} x^{5}+5 a \,b^{4} p^{4} x^{4}-85 b^{5} p^{3} x^{5}+50 a \,b^{4} p^{3} x^{4}-225 b^{5} p^{2} x^{5}-20 a^{2} b^{3} p^{3} x^{3}+175 a \,b^{4} p^{2} x^{4}-274 b^{5} p \,x^{5}-120 a^{2} b^{3} p^{2} x^{3}+250 x^{4} a p \,b^{4}-120 b^{5} x^{5}+60 a^{3} b^{2} p^{2} x^{2}-220 a^{2} p \,x^{3} b^{3}+120 b^{4} a \,x^{4}+180 a^{3} p \,x^{2} b^{2}-120 a^{2} b^{3} x^{3}-120 a^{4} b p x +120 b^{2} x^{2} a^{3}-120 a^{4} b x +120 a^{5}\right )}{b^{6} x^{5} \left (p^{6}+21 p^{5}+175 p^{4}+735 p^{3}+1624 p^{2}+1764 p +720\right )}\)	\(280\)
orering	\(-\frac {\left (b x +a \right )^{p} \left (c \,x^{2}\right )^{\frac {5}{2}} \left (-b^{5} p^{5} x^{5}-15 b^{5} p^{4} x^{5}+5 a \,b^{4} p^{4} x^{4}-85 b^{5} p^{3} x^{5}+50 a \,b^{4} p^{3} x^{4}-225 b^{5} p^{2} x^{5}-20 a^{2} b^{3} p^{3} x^{3}+175 a \,b^{4} p^{2} x^{4}-274 b^{5} p \,x^{5}-120 a^{2} b^{3} p^{2} x^{3}+250 x^{4} a p \,b^{4}-120 b^{5} x^{5}+60 a^{3} b^{2} p^{2} x^{2}-220 a^{2} p \,x^{3} b^{3}+120 b^{4} a \,x^{4}+180 a^{3} p \,x^{2} b^{2}-120 a^{2} b^{3} x^{3}-120 a^{4} b p x +120 b^{2} x^{2} a^{3}-120 a^{4} b x +120 a^{5}\right ) \left (b x +a \right )}{x^{5} b^{6} \left (p^{6}+21 p^{5}+175 p^{4}+735 p^{3}+1624 p^{2}+1764 p +720\right )}\)	\(283\)
risch	\(-\frac {c^{2} \sqrt {c \,x^{2}}\, \left (-b^{6} p^{5} x^{6}-a \,b^{5} p^{5} x^{5}-15 b^{6} p^{4} x^{6}-10 a \,b^{5} p^{4} x^{5}-85 b^{6} p^{3} x^{6}+5 a^{2} b^{4} p^{4} x^{4}-35 a \,b^{5} p^{3} x^{5}-225 b^{6} p^{2} x^{6}+30 a^{2} b^{4} p^{3} x^{4}-50 a \,b^{5} p^{2} x^{5}-274 b^{6} p \,x^{6}-20 a^{3} b^{3} p^{3} x^{3}+55 a^{2} b^{4} p^{2} x^{4}-24 x^{5} a p \,b^{5}-120 x^{6} b^{6}-60 a^{3} b^{3} p^{2} x^{3}+30 a^{2} p \,x^{4} b^{4}+60 a^{4} b^{2} p^{2} x^{2}-40 a^{3} p \,x^{3} b^{3}+60 a^{4} b^{2} p \,x^{2}-120 a^{5} b p x +120 a^{6}\right ) \left (b x +a \right )^{p}}{x \left (5+p \right ) \left (6+p \right ) \left (4+p \right ) \left (3+p \right ) \left (2+p \right ) \left (p +1\right ) b^{6}}\)	\(309\)

Input:

int((c*x^2)^(5/2)*(b*x+a)^p,x,method=_RETURNVERBOSE)
 

Output:

-1/b^6/x^5*(c*x^2)^(5/2)*(b*x+a)^(p+1)/(p^6+21*p^5+175*p^4+735*p^3+1624*p^ 
2+1764*p+720)*(-b^5*p^5*x^5-15*b^5*p^4*x^5+5*a*b^4*p^4*x^4-85*b^5*p^3*x^5+ 
50*a*b^4*p^3*x^4-225*b^5*p^2*x^5-20*a^2*b^3*p^3*x^3+175*a*b^4*p^2*x^4-274* 
b^5*p*x^5-120*a^2*b^3*p^2*x^3+250*a*b^4*p*x^4-120*b^5*x^5+60*a^3*b^2*p^2*x 
^2-220*a^2*b^3*p*x^3+120*a*b^4*x^4+180*a^3*b^2*p*x^2-120*a^2*b^3*x^3-120*a 
^4*b*p*x+120*a^3*b^2*x^2-120*a^4*b*x+120*a^5)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 352, normalized size of antiderivative = 1.62 \[ \int \left (c x^2\right )^{5/2} (a+b x)^p \, dx=\frac {{\left (120 \, a^{5} b c^{2} p x - 120 \, a^{6} c^{2} + {\left (b^{6} c^{2} p^{5} + 15 \, b^{6} c^{2} p^{4} + 85 \, b^{6} c^{2} p^{3} + 225 \, b^{6} c^{2} p^{2} + 274 \, b^{6} c^{2} p + 120 \, b^{6} c^{2}\right )} x^{6} + {\left (a b^{5} c^{2} p^{5} + 10 \, a b^{5} c^{2} p^{4} + 35 \, a b^{5} c^{2} p^{3} + 50 \, a b^{5} c^{2} p^{2} + 24 \, a b^{5} c^{2} p\right )} x^{5} - 5 \, {\left (a^{2} b^{4} c^{2} p^{4} + 6 \, a^{2} b^{4} c^{2} p^{3} + 11 \, a^{2} b^{4} c^{2} p^{2} + 6 \, a^{2} b^{4} c^{2} p\right )} x^{4} + 20 \, {\left (a^{3} b^{3} c^{2} p^{3} + 3 \, a^{3} b^{3} c^{2} p^{2} + 2 \, a^{3} b^{3} c^{2} p\right )} x^{3} - 60 \, {\left (a^{4} b^{2} c^{2} p^{2} + a^{4} b^{2} c^{2} p\right )} x^{2}\right )} \sqrt {c x^{2}} {\left (b x + a\right )}^{p}}{{\left (b^{6} p^{6} + 21 \, b^{6} p^{5} + 175 \, b^{6} p^{4} + 735 \, b^{6} p^{3} + 1624 \, b^{6} p^{2} + 1764 \, b^{6} p + 720 \, b^{6}\right )} x} \] Input:

integrate((c*x^2)^(5/2)*(b*x+a)^p,x, algorithm="fricas")
 

Output:

(120*a^5*b*c^2*p*x - 120*a^6*c^2 + (b^6*c^2*p^5 + 15*b^6*c^2*p^4 + 85*b^6* 
c^2*p^3 + 225*b^6*c^2*p^2 + 274*b^6*c^2*p + 120*b^6*c^2)*x^6 + (a*b^5*c^2* 
p^5 + 10*a*b^5*c^2*p^4 + 35*a*b^5*c^2*p^3 + 50*a*b^5*c^2*p^2 + 24*a*b^5*c^ 
2*p)*x^5 - 5*(a^2*b^4*c^2*p^4 + 6*a^2*b^4*c^2*p^3 + 11*a^2*b^4*c^2*p^2 + 6 
*a^2*b^4*c^2*p)*x^4 + 20*(a^3*b^3*c^2*p^3 + 3*a^3*b^3*c^2*p^2 + 2*a^3*b^3* 
c^2*p)*x^3 - 60*(a^4*b^2*c^2*p^2 + a^4*b^2*c^2*p)*x^2)*sqrt(c*x^2)*(b*x + 
a)^p/((b^6*p^6 + 21*b^6*p^5 + 175*b^6*p^4 + 735*b^6*p^3 + 1624*b^6*p^2 + 1 
764*b^6*p + 720*b^6)*x)
 

Sympy [F]

\[ \int \left (c x^2\right )^{5/2} (a+b x)^p \, dx=\int \left (c x^{2}\right )^{\frac {5}{2}} \left (a + b x\right )^{p}\, dx \] Input:

integrate((c*x**2)**(5/2)*(b*x+a)**p,x)
 

Output:

Integral((c*x**2)**(5/2)*(a + b*x)**p, x)
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 203, normalized size of antiderivative = 0.94 \[ \int \left (c x^2\right )^{5/2} (a+b x)^p \, dx=\frac {{\left ({\left (p^{5} + 15 \, p^{4} + 85 \, p^{3} + 225 \, p^{2} + 274 \, p + 120\right )} b^{6} c^{\frac {5}{2}} x^{6} + {\left (p^{5} + 10 \, p^{4} + 35 \, p^{3} + 50 \, p^{2} + 24 \, p\right )} a b^{5} c^{\frac {5}{2}} x^{5} - 5 \, {\left (p^{4} + 6 \, p^{3} + 11 \, p^{2} + 6 \, p\right )} a^{2} b^{4} c^{\frac {5}{2}} x^{4} + 20 \, {\left (p^{3} + 3 \, p^{2} + 2 \, p\right )} a^{3} b^{3} c^{\frac {5}{2}} x^{3} - 60 \, {\left (p^{2} + p\right )} a^{4} b^{2} c^{\frac {5}{2}} x^{2} + 120 \, a^{5} b c^{\frac {5}{2}} p x - 120 \, a^{6} c^{\frac {5}{2}}\right )} {\left (b x + a\right )}^{p}}{{\left (p^{6} + 21 \, p^{5} + 175 \, p^{4} + 735 \, p^{3} + 1624 \, p^{2} + 1764 \, p + 720\right )} b^{6}} \] Input:

integrate((c*x^2)^(5/2)*(b*x+a)^p,x, algorithm="maxima")
 

Output:

((p^5 + 15*p^4 + 85*p^3 + 225*p^2 + 274*p + 120)*b^6*c^(5/2)*x^6 + (p^5 + 
10*p^4 + 35*p^3 + 50*p^2 + 24*p)*a*b^5*c^(5/2)*x^5 - 5*(p^4 + 6*p^3 + 11*p 
^2 + 6*p)*a^2*b^4*c^(5/2)*x^4 + 20*(p^3 + 3*p^2 + 2*p)*a^3*b^3*c^(5/2)*x^3 
 - 60*(p^2 + p)*a^4*b^2*c^(5/2)*x^2 + 120*a^5*b*c^(5/2)*p*x - 120*a^6*c^(5 
/2))*(b*x + a)^p/((p^6 + 21*p^5 + 175*p^4 + 735*p^3 + 1624*p^2 + 1764*p + 
720)*b^6)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 571 vs. \(2 (205) = 410\).

Time = 0.12 (sec) , antiderivative size = 571, normalized size of antiderivative = 2.63 \[ \int \left (c x^2\right )^{5/2} (a+b x)^p \, dx =\text {Too large to display} \] Input:

integrate((c*x^2)^(5/2)*(b*x+a)^p,x, algorithm="giac")
 

Output:

(120*a^6*a^p*sgn(x)/(b^6*p^6 + 21*b^6*p^5 + 175*b^6*p^4 + 735*b^6*p^3 + 16 
24*b^6*p^2 + 1764*b^6*p + 720*b^6) + ((b*x + a)^p*b^6*p^5*x^6*sgn(x) + (b* 
x + a)^p*a*b^5*p^5*x^5*sgn(x) + 15*(b*x + a)^p*b^6*p^4*x^6*sgn(x) + 10*(b* 
x + a)^p*a*b^5*p^4*x^5*sgn(x) + 85*(b*x + a)^p*b^6*p^3*x^6*sgn(x) - 5*(b*x 
 + a)^p*a^2*b^4*p^4*x^4*sgn(x) + 35*(b*x + a)^p*a*b^5*p^3*x^5*sgn(x) + 225 
*(b*x + a)^p*b^6*p^2*x^6*sgn(x) - 30*(b*x + a)^p*a^2*b^4*p^3*x^4*sgn(x) + 
50*(b*x + a)^p*a*b^5*p^2*x^5*sgn(x) + 274*(b*x + a)^p*b^6*p*x^6*sgn(x) + 2 
0*(b*x + a)^p*a^3*b^3*p^3*x^3*sgn(x) - 55*(b*x + a)^p*a^2*b^4*p^2*x^4*sgn( 
x) + 24*(b*x + a)^p*a*b^5*p*x^5*sgn(x) + 120*(b*x + a)^p*b^6*x^6*sgn(x) + 
60*(b*x + a)^p*a^3*b^3*p^2*x^3*sgn(x) - 30*(b*x + a)^p*a^2*b^4*p*x^4*sgn(x 
) - 60*(b*x + a)^p*a^4*b^2*p^2*x^2*sgn(x) + 40*(b*x + a)^p*a^3*b^3*p*x^3*s 
gn(x) - 60*(b*x + a)^p*a^4*b^2*p*x^2*sgn(x) + 120*(b*x + a)^p*a^5*b*p*x*sg 
n(x) - 120*(b*x + a)^p*a^6*sgn(x))/(b^6*p^6 + 21*b^6*p^5 + 175*b^6*p^4 + 7 
35*b^6*p^3 + 1624*b^6*p^2 + 1764*b^6*p + 720*b^6))*c^(5/2)
 

Mupad [B] (verification not implemented)

Time = 24.33 (sec) , antiderivative size = 424, normalized size of antiderivative = 1.95 \[ \int \left (c x^2\right )^{5/2} (a+b x)^p \, dx=\frac {{\left (a+b\,x\right )}^p\,\left (\frac {c^2\,x^6\,\sqrt {c\,x^2}\,\left (p^5+15\,p^4+85\,p^3+225\,p^2+274\,p+120\right )}{p^6+21\,p^5+175\,p^4+735\,p^3+1624\,p^2+1764\,p+720}-\frac {120\,a^6\,c^2\,\sqrt {c\,x^2}}{b^6\,\left (p^6+21\,p^5+175\,p^4+735\,p^3+1624\,p^2+1764\,p+720\right )}+\frac {120\,a^5\,c^2\,p\,x\,\sqrt {c\,x^2}}{b^5\,\left (p^6+21\,p^5+175\,p^4+735\,p^3+1624\,p^2+1764\,p+720\right )}-\frac {5\,a^2\,c^2\,p\,x^4\,\sqrt {c\,x^2}\,\left (p^3+6\,p^2+11\,p+6\right )}{b^2\,\left (p^6+21\,p^5+175\,p^4+735\,p^3+1624\,p^2+1764\,p+720\right )}-\frac {60\,a^4\,c^2\,p\,x^2\,\sqrt {c\,x^2}\,\left (p+1\right )}{b^4\,\left (p^6+21\,p^5+175\,p^4+735\,p^3+1624\,p^2+1764\,p+720\right )}+\frac {a\,c^2\,p\,x^5\,\sqrt {c\,x^2}\,\left (p^4+10\,p^3+35\,p^2+50\,p+24\right )}{b\,\left (p^6+21\,p^5+175\,p^4+735\,p^3+1624\,p^2+1764\,p+720\right )}+\frac {20\,a^3\,c^2\,p\,x^3\,\sqrt {c\,x^2}\,\left (p^2+3\,p+2\right )}{b^3\,\left (p^6+21\,p^5+175\,p^4+735\,p^3+1624\,p^2+1764\,p+720\right )}\right )}{x} \] Input:

int((c*x^2)^(5/2)*(a + b*x)^p,x)
 

Output:

((a + b*x)^p*((c^2*x^6*(c*x^2)^(1/2)*(274*p + 225*p^2 + 85*p^3 + 15*p^4 + 
p^5 + 120))/(1764*p + 1624*p^2 + 735*p^3 + 175*p^4 + 21*p^5 + p^6 + 720) - 
 (120*a^6*c^2*(c*x^2)^(1/2))/(b^6*(1764*p + 1624*p^2 + 735*p^3 + 175*p^4 + 
 21*p^5 + p^6 + 720)) + (120*a^5*c^2*p*x*(c*x^2)^(1/2))/(b^5*(1764*p + 162 
4*p^2 + 735*p^3 + 175*p^4 + 21*p^5 + p^6 + 720)) - (5*a^2*c^2*p*x^4*(c*x^2 
)^(1/2)*(11*p + 6*p^2 + p^3 + 6))/(b^2*(1764*p + 1624*p^2 + 735*p^3 + 175* 
p^4 + 21*p^5 + p^6 + 720)) - (60*a^4*c^2*p*x^2*(c*x^2)^(1/2)*(p + 1))/(b^4 
*(1764*p + 1624*p^2 + 735*p^3 + 175*p^4 + 21*p^5 + p^6 + 720)) + (a*c^2*p* 
x^5*(c*x^2)^(1/2)*(50*p + 35*p^2 + 10*p^3 + p^4 + 24))/(b*(1764*p + 1624*p 
^2 + 735*p^3 + 175*p^4 + 21*p^5 + p^6 + 720)) + (20*a^3*c^2*p*x^3*(c*x^2)^ 
(1/2)*(3*p + p^2 + 2))/(b^3*(1764*p + 1624*p^2 + 735*p^3 + 175*p^4 + 21*p^ 
5 + p^6 + 720))))/x
 

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 297, normalized size of antiderivative = 1.37 \[ \int \left (c x^2\right )^{5/2} (a+b x)^p \, dx=\frac {\sqrt {c}\, \left (b x +a \right )^{p} c^{2} \left (b^{6} p^{5} x^{6}+a \,b^{5} p^{5} x^{5}+15 b^{6} p^{4} x^{6}+10 a \,b^{5} p^{4} x^{5}+85 b^{6} p^{3} x^{6}-5 a^{2} b^{4} p^{4} x^{4}+35 a \,b^{5} p^{3} x^{5}+225 b^{6} p^{2} x^{6}-30 a^{2} b^{4} p^{3} x^{4}+50 a \,b^{5} p^{2} x^{5}+274 b^{6} p \,x^{6}+20 a^{3} b^{3} p^{3} x^{3}-55 a^{2} b^{4} p^{2} x^{4}+24 a \,b^{5} p \,x^{5}+120 b^{6} x^{6}+60 a^{3} b^{3} p^{2} x^{3}-30 a^{2} b^{4} p \,x^{4}-60 a^{4} b^{2} p^{2} x^{2}+40 a^{3} b^{3} p \,x^{3}-60 a^{4} b^{2} p \,x^{2}+120 a^{5} b p x -120 a^{6}\right )}{b^{6} \left (p^{6}+21 p^{5}+175 p^{4}+735 p^{3}+1624 p^{2}+1764 p +720\right )} \] Input:

int((c*x^2)^(5/2)*(b*x+a)^p,x)
 

Output:

(sqrt(c)*(a + b*x)**p*c**2*( - 120*a**6 + 120*a**5*b*p*x - 60*a**4*b**2*p* 
*2*x**2 - 60*a**4*b**2*p*x**2 + 20*a**3*b**3*p**3*x**3 + 60*a**3*b**3*p**2 
*x**3 + 40*a**3*b**3*p*x**3 - 5*a**2*b**4*p**4*x**4 - 30*a**2*b**4*p**3*x* 
*4 - 55*a**2*b**4*p**2*x**4 - 30*a**2*b**4*p*x**4 + a*b**5*p**5*x**5 + 10* 
a*b**5*p**4*x**5 + 35*a*b**5*p**3*x**5 + 50*a*b**5*p**2*x**5 + 24*a*b**5*p 
*x**5 + b**6*p**5*x**6 + 15*b**6*p**4*x**6 + 85*b**6*p**3*x**6 + 225*b**6* 
p**2*x**6 + 274*b**6*p*x**6 + 120*b**6*x**6))/(b**6*(p**6 + 21*p**5 + 175* 
p**4 + 735*p**3 + 1624*p**2 + 1764*p + 720))