\(\int x (c x^2)^{3/2} (a+b x)^p \, dx\) [424]

Optimal result

Integrand size = 18, antiderivative size = 169 \[ \int x \left (c x^2\right )^{3/2} (a+b x)^p \, dx=\frac {a^4 c \sqrt {c x^2} (a+b x)^{1+p}}{b^5 (1+p) x}-\frac {4 a^3 c \sqrt {c x^2} (a+b x)^{2+p}}{b^5 (2+p) x}+\frac {6 a^2 c \sqrt {c x^2} (a+b x)^{3+p}}{b^5 (3+p) x}-\frac {4 a c \sqrt {c x^2} (a+b x)^{4+p}}{b^5 (4+p) x}+\frac {c \sqrt {c x^2} (a+b x)^{5+p}}{b^5 (5+p) x} \] Output:

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

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.78 \[ \int x \left (c x^2\right )^{3/2} (a+b x)^p \, dx=\frac {\left (c x^2\right )^{3/2} (a+b x)^{1+p} \left (24 a^4-24 a^3 b (1+p) x+12 a^2 b^2 \left (2+3 p+p^2\right ) x^2-4 a b^3 \left (6+11 p+6 p^2+p^3\right ) x^3+b^4 \left (24+50 p+35 p^2+10 p^3+p^4\right ) x^4\right )}{b^5 (1+p) (2+p) (3+p) (4+p) (5+p) x^3} \] Input:

Integrate[x*(c*x^2)^(3/2)*(a + b*x)^p,x]
 

Output:

((c*x^2)^(3/2)*(a + b*x)^(1 + p)*(24*a^4 - 24*a^3*b*(1 + p)*x + 12*a^2*b^2 
*(2 + 3*p + p^2)*x^2 - 4*a*b^3*(6 + 11*p + 6*p^2 + p^3)*x^3 + b^4*(24 + 50 
*p + 35*p^2 + 10*p^3 + p^4)*x^4))/(b^5*(1 + p)*(2 + p)*(3 + p)*(4 + p)*(5 
+ p)*x^3)
 

Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.70, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {30, 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.

\(\displaystyle \int x \left (c x^2\right )^{3/2} (a+b x)^p \, dx\)

\(\Big \downarrow \) 30

\(\displaystyle \frac {c \sqrt {c x^2} \int x^4 (a+b x)^pdx}{x}\)

\(\Big \downarrow \) 53

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

\(\Big \downarrow \) 2009

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

Input:

Int[x*(c*x^2)^(3/2)*(a + b*x)^p,x]
 

Output:

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

Defintions of rubi rules used

rule 30

Int[(u_.)*((a_.)*(x_))^(m_.)*((b_.)*(x_)^(i_.))^(p_), x_Symbol] :> Simp[b^I 
ntPart[p]*((b*x^i)^FracPart[p]/(a^(i*IntPart[p])*(a*x)^(i*FracPart[p]))) 
Int[u*(a*x)^(m + i*p), x], x] /; FreeQ[{a, b, i, m, p}, x] && IntegerQ[i] & 
&  !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 = 0.18 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.18

method	result	size
gosper	\(\frac {\left (c \,x^{2}\right )^{\frac {3}{2}} \left (b x +a \right )^{p +1} \left (b^{4} p^{4} x^{4}+10 b^{4} p^{3} x^{4}-4 a \,b^{3} p^{3} x^{3}+35 b^{4} p^{2} x^{4}-24 a \,b^{3} p^{2} x^{3}+50 b^{4} p \,x^{4}+12 a^{2} b^{2} p^{2} x^{2}-44 x^{3} a p \,b^{3}+24 b^{4} x^{4}+36 a^{2} p \,x^{2} b^{2}-24 a \,b^{3} x^{3}-24 a^{3} b p x +24 a^{2} b^{2} x^{2}-24 b \,a^{3} x +24 a^{4}\right )}{b^{5} x^{3} \left (p^{5}+15 p^{4}+85 p^{3}+225 p^{2}+274 p +120\right )}\)	\(199\)
orering	\(\frac {\left (b x +a \right ) \left (b^{4} p^{4} x^{4}+10 b^{4} p^{3} x^{4}-4 a \,b^{3} p^{3} x^{3}+35 b^{4} p^{2} x^{4}-24 a \,b^{3} p^{2} x^{3}+50 b^{4} p \,x^{4}+12 a^{2} b^{2} p^{2} x^{2}-44 x^{3} a p \,b^{3}+24 b^{4} x^{4}+36 a^{2} p \,x^{2} b^{2}-24 a \,b^{3} x^{3}-24 a^{3} b p x +24 a^{2} b^{2} x^{2}-24 b \,a^{3} x +24 a^{4}\right ) \left (c \,x^{2}\right )^{\frac {3}{2}} \left (b x +a \right )^{p}}{x^{3} b^{5} \left (p^{5}+15 p^{4}+85 p^{3}+225 p^{2}+274 p +120\right )}\)	\(202\)
risch	\(\frac {c \sqrt {c \,x^{2}}\, \left (b^{5} p^{4} x^{5}+a \,b^{4} p^{4} x^{4}+10 b^{5} p^{3} x^{5}+6 a \,b^{4} p^{3} x^{4}+35 b^{5} p^{2} x^{5}-4 a^{2} b^{3} p^{3} x^{3}+11 a \,b^{4} p^{2} x^{4}+50 b^{5} p \,x^{5}-12 a^{2} b^{3} p^{2} x^{3}+6 x^{4} a p \,b^{4}+24 b^{5} x^{5}+12 a^{3} b^{2} p^{2} x^{2}-8 a^{2} p \,x^{3} b^{3}+12 a^{3} p \,x^{2} b^{2}-24 a^{4} b p x +24 a^{5}\right ) \left (b x +a \right )^{p}}{x \left (4+p \right ) \left (5+p \right ) \left (3+p \right ) \left (2+p \right ) \left (p +1\right ) b^{5}}\)	\(222\)

Input:

int(x*(c*x^2)^(3/2)*(b*x+a)^p,x,method=_RETURNVERBOSE)
 

Output:

1/b^5/x^3*(c*x^2)^(3/2)*(b*x+a)^(p+1)/(p^5+15*p^4+85*p^3+225*p^2+274*p+120 
)*(b^4*p^4*x^4+10*b^4*p^3*x^4-4*a*b^3*p^3*x^3+35*b^4*p^2*x^4-24*a*b^3*p^2* 
x^3+50*b^4*p*x^4+12*a^2*b^2*p^2*x^2-44*a*b^3*p*x^3+24*b^4*x^4+36*a^2*b^2*p 
*x^2-24*a*b^3*x^3-24*a^3*b*p*x+24*a^2*b^2*x^2-24*a^3*b*x+24*a^4)
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.38 \[ \int x \left (c x^2\right )^{3/2} (a+b x)^p \, dx=-\frac {{\left (24 \, a^{4} b c p x - 24 \, a^{5} c - {\left (b^{5} c p^{4} + 10 \, b^{5} c p^{3} + 35 \, b^{5} c p^{2} + 50 \, b^{5} c p + 24 \, b^{5} c\right )} x^{5} - {\left (a b^{4} c p^{4} + 6 \, a b^{4} c p^{3} + 11 \, a b^{4} c p^{2} + 6 \, a b^{4} c p\right )} x^{4} + 4 \, {\left (a^{2} b^{3} c p^{3} + 3 \, a^{2} b^{3} c p^{2} + 2 \, a^{2} b^{3} c p\right )} x^{3} - 12 \, {\left (a^{3} b^{2} c p^{2} + a^{3} b^{2} c p\right )} x^{2}\right )} \sqrt {c x^{2}} {\left (b x + a\right )}^{p}}{{\left (b^{5} p^{5} + 15 \, b^{5} p^{4} + 85 \, b^{5} p^{3} + 225 \, b^{5} p^{2} + 274 \, b^{5} p + 120 \, b^{5}\right )} x} \] Input:

integrate(x*(c*x^2)^(3/2)*(b*x+a)^p,x, algorithm="fricas")
 

Output:

-(24*a^4*b*c*p*x - 24*a^5*c - (b^5*c*p^4 + 10*b^5*c*p^3 + 35*b^5*c*p^2 + 5 
0*b^5*c*p + 24*b^5*c)*x^5 - (a*b^4*c*p^4 + 6*a*b^4*c*p^3 + 11*a*b^4*c*p^2 
+ 6*a*b^4*c*p)*x^4 + 4*(a^2*b^3*c*p^3 + 3*a^2*b^3*c*p^2 + 2*a^2*b^3*c*p)*x 
^3 - 12*(a^3*b^2*c*p^2 + a^3*b^2*c*p)*x^2)*sqrt(c*x^2)*(b*x + a)^p/((b^5*p 
^5 + 15*b^5*p^4 + 85*b^5*p^3 + 225*b^5*p^2 + 274*b^5*p + 120*b^5)*x)
 

Sympy [F]

\[ \int x \left (c x^2\right )^{3/2} (a+b x)^p \, dx=\int x \left (c x^{2}\right )^{\frac {3}{2}} \left (a + b x\right )^{p}\, dx \] Input:

integrate(x*(c*x**2)**(3/2)*(b*x+a)**p,x)
 

Output:

Integral(x*(c*x**2)**(3/2)*(a + b*x)**p, x)
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.93 \[ \int x \left (c x^2\right )^{3/2} (a+b x)^p \, dx=\frac {{\left ({\left (p^{4} + 10 \, p^{3} + 35 \, p^{2} + 50 \, p + 24\right )} b^{5} c^{\frac {3}{2}} x^{5} + {\left (p^{4} + 6 \, p^{3} + 11 \, p^{2} + 6 \, p\right )} a b^{4} c^{\frac {3}{2}} x^{4} - 4 \, {\left (p^{3} + 3 \, p^{2} + 2 \, p\right )} a^{2} b^{3} c^{\frac {3}{2}} x^{3} + 12 \, {\left (p^{2} + p\right )} a^{3} b^{2} c^{\frac {3}{2}} x^{2} - 24 \, a^{4} b c^{\frac {3}{2}} p x + 24 \, a^{5} c^{\frac {3}{2}}\right )} {\left (b x + a\right )}^{p}}{{\left (p^{5} + 15 \, p^{4} + 85 \, p^{3} + 225 \, p^{2} + 274 \, p + 120\right )} b^{5}} \] Input:

integrate(x*(c*x^2)^(3/2)*(b*x+a)^p,x, algorithm="maxima")
 

Output:

((p^4 + 10*p^3 + 35*p^2 + 50*p + 24)*b^5*c^(3/2)*x^5 + (p^4 + 6*p^3 + 11*p 
^2 + 6*p)*a*b^4*c^(3/2)*x^4 - 4*(p^3 + 3*p^2 + 2*p)*a^2*b^3*c^(3/2)*x^3 + 
12*(p^2 + p)*a^3*b^2*c^(3/2)*x^2 - 24*a^4*b*c^(3/2)*p*x + 24*a^5*c^(3/2))* 
(b*x + a)^p/((p^5 + 15*p^4 + 85*p^3 + 225*p^2 + 274*p + 120)*b^5)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 426 vs. \(2 (159) = 318\).

Time = 0.14 (sec) , antiderivative size = 426, normalized size of antiderivative = 2.52 \[ \int x \left (c x^2\right )^{3/2} (a+b x)^p \, dx=-{\left (\frac {24 \, a^{5} a^{p} \mathrm {sgn}\left (x\right )}{b^{5} p^{5} + 15 \, b^{5} p^{4} + 85 \, b^{5} p^{3} + 225 \, b^{5} p^{2} + 274 \, b^{5} p + 120 \, b^{5}} - \frac {{\left (b x + a\right )}^{p} b^{5} p^{4} x^{5} \mathrm {sgn}\left (x\right ) + {\left (b x + a\right )}^{p} a b^{4} p^{4} x^{4} \mathrm {sgn}\left (x\right ) + 10 \, {\left (b x + a\right )}^{p} b^{5} p^{3} x^{5} \mathrm {sgn}\left (x\right ) + 6 \, {\left (b x + a\right )}^{p} a b^{4} p^{3} x^{4} \mathrm {sgn}\left (x\right ) + 35 \, {\left (b x + a\right )}^{p} b^{5} p^{2} x^{5} \mathrm {sgn}\left (x\right ) - 4 \, {\left (b x + a\right )}^{p} a^{2} b^{3} p^{3} x^{3} \mathrm {sgn}\left (x\right ) + 11 \, {\left (b x + a\right )}^{p} a b^{4} p^{2} x^{4} \mathrm {sgn}\left (x\right ) + 50 \, {\left (b x + a\right )}^{p} b^{5} p x^{5} \mathrm {sgn}\left (x\right ) - 12 \, {\left (b x + a\right )}^{p} a^{2} b^{3} p^{2} x^{3} \mathrm {sgn}\left (x\right ) + 6 \, {\left (b x + a\right )}^{p} a b^{4} p x^{4} \mathrm {sgn}\left (x\right ) + 24 \, {\left (b x + a\right )}^{p} b^{5} x^{5} \mathrm {sgn}\left (x\right ) + 12 \, {\left (b x + a\right )}^{p} a^{3} b^{2} p^{2} x^{2} \mathrm {sgn}\left (x\right ) - 8 \, {\left (b x + a\right )}^{p} a^{2} b^{3} p x^{3} \mathrm {sgn}\left (x\right ) + 12 \, {\left (b x + a\right )}^{p} a^{3} b^{2} p x^{2} \mathrm {sgn}\left (x\right ) - 24 \, {\left (b x + a\right )}^{p} a^{4} b p x \mathrm {sgn}\left (x\right ) + 24 \, {\left (b x + a\right )}^{p} a^{5} \mathrm {sgn}\left (x\right )}{b^{5} p^{5} + 15 \, b^{5} p^{4} + 85 \, b^{5} p^{3} + 225 \, b^{5} p^{2} + 274 \, b^{5} p + 120 \, b^{5}}\right )} c^{\frac {3}{2}} \] Input:

integrate(x*(c*x^2)^(3/2)*(b*x+a)^p,x, algorithm="giac")
 

Output:

-(24*a^5*a^p*sgn(x)/(b^5*p^5 + 15*b^5*p^4 + 85*b^5*p^3 + 225*b^5*p^2 + 274 
*b^5*p + 120*b^5) - ((b*x + a)^p*b^5*p^4*x^5*sgn(x) + (b*x + a)^p*a*b^4*p^ 
4*x^4*sgn(x) + 10*(b*x + a)^p*b^5*p^3*x^5*sgn(x) + 6*(b*x + a)^p*a*b^4*p^3 
*x^4*sgn(x) + 35*(b*x + a)^p*b^5*p^2*x^5*sgn(x) - 4*(b*x + a)^p*a^2*b^3*p^ 
3*x^3*sgn(x) + 11*(b*x + a)^p*a*b^4*p^2*x^4*sgn(x) + 50*(b*x + a)^p*b^5*p* 
x^5*sgn(x) - 12*(b*x + a)^p*a^2*b^3*p^2*x^3*sgn(x) + 6*(b*x + a)^p*a*b^4*p 
*x^4*sgn(x) + 24*(b*x + a)^p*b^5*x^5*sgn(x) + 12*(b*x + a)^p*a^3*b^2*p^2*x 
^2*sgn(x) - 8*(b*x + a)^p*a^2*b^3*p*x^3*sgn(x) + 12*(b*x + a)^p*a^3*b^2*p* 
x^2*sgn(x) - 24*(b*x + a)^p*a^4*b*p*x*sgn(x) + 24*(b*x + a)^p*a^5*sgn(x))/ 
(b^5*p^5 + 15*b^5*p^4 + 85*b^5*p^3 + 225*b^5*p^2 + 274*b^5*p + 120*b^5))*c 
^(3/2)
 

Mupad [B] (verification not implemented)

Time = 22.99 (sec) , antiderivative size = 307, normalized size of antiderivative = 1.82 \[ \int x \left (c x^2\right )^{3/2} (a+b x)^p \, dx=\frac {{\left (a+b\,x\right )}^p\,\left (\frac {24\,a^5\,c\,\sqrt {c\,x^2}}{b^5\,\left (p^5+15\,p^4+85\,p^3+225\,p^2+274\,p+120\right )}+\frac {c\,x^5\,\sqrt {c\,x^2}\,\left (p^4+10\,p^3+35\,p^2+50\,p+24\right )}{p^5+15\,p^4+85\,p^3+225\,p^2+274\,p+120}-\frac {24\,a^4\,c\,p\,x\,\sqrt {c\,x^2}}{b^4\,\left (p^5+15\,p^4+85\,p^3+225\,p^2+274\,p+120\right )}+\frac {a\,c\,p\,x^4\,\sqrt {c\,x^2}\,\left (p^3+6\,p^2+11\,p+6\right )}{b\,\left (p^5+15\,p^4+85\,p^3+225\,p^2+274\,p+120\right )}+\frac {12\,a^3\,c\,p\,x^2\,\sqrt {c\,x^2}\,\left (p+1\right )}{b^3\,\left (p^5+15\,p^4+85\,p^3+225\,p^2+274\,p+120\right )}-\frac {4\,a^2\,c\,p\,x^3\,\sqrt {c\,x^2}\,\left (p^2+3\,p+2\right )}{b^2\,\left (p^5+15\,p^4+85\,p^3+225\,p^2+274\,p+120\right )}\right )}{x} \] Input:

int(x*(c*x^2)^(3/2)*(a + b*x)^p,x)
 

Output:

((a + b*x)^p*((24*a^5*c*(c*x^2)^(1/2))/(b^5*(274*p + 225*p^2 + 85*p^3 + 15 
*p^4 + p^5 + 120)) + (c*x^5*(c*x^2)^(1/2)*(50*p + 35*p^2 + 10*p^3 + p^4 + 
24))/(274*p + 225*p^2 + 85*p^3 + 15*p^4 + p^5 + 120) - (24*a^4*c*p*x*(c*x^ 
2)^(1/2))/(b^4*(274*p + 225*p^2 + 85*p^3 + 15*p^4 + p^5 + 120)) + (a*c*p*x 
^4*(c*x^2)^(1/2)*(11*p + 6*p^2 + p^3 + 6))/(b*(274*p + 225*p^2 + 85*p^3 + 
15*p^4 + p^5 + 120)) + (12*a^3*c*p*x^2*(c*x^2)^(1/2)*(p + 1))/(b^3*(274*p 
+ 225*p^2 + 85*p^3 + 15*p^4 + p^5 + 120)) - (4*a^2*c*p*x^3*(c*x^2)^(1/2)*( 
3*p + p^2 + 2))/(b^2*(274*p + 225*p^2 + 85*p^3 + 15*p^4 + p^5 + 120))))/x
 

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 213, normalized size of antiderivative = 1.26 \[ \int x \left (c x^2\right )^{3/2} (a+b x)^p \, dx=\frac {\sqrt {c}\, \left (b x +a \right )^{p} c \left (b^{5} p^{4} x^{5}+a \,b^{4} p^{4} x^{4}+10 b^{5} p^{3} x^{5}+6 a \,b^{4} p^{3} x^{4}+35 b^{5} p^{2} x^{5}-4 a^{2} b^{3} p^{3} x^{3}+11 a \,b^{4} p^{2} x^{4}+50 b^{5} p \,x^{5}-12 a^{2} b^{3} p^{2} x^{3}+6 a \,b^{4} p \,x^{4}+24 b^{5} x^{5}+12 a^{3} b^{2} p^{2} x^{2}-8 a^{2} b^{3} p \,x^{3}+12 a^{3} b^{2} p \,x^{2}-24 a^{4} b p x +24 a^{5}\right )}{b^{5} \left (p^{5}+15 p^{4}+85 p^{3}+225 p^{2}+274 p +120\right )} \] Input:

int(x*(c*x^2)^(3/2)*(b*x+a)^p,x)
 

Output:

(sqrt(c)*(a + b*x)**p*c*(24*a**5 - 24*a**4*b*p*x + 12*a**3*b**2*p**2*x**2 
+ 12*a**3*b**2*p*x**2 - 4*a**2*b**3*p**3*x**3 - 12*a**2*b**3*p**2*x**3 - 8 
*a**2*b**3*p*x**3 + a*b**4*p**4*x**4 + 6*a*b**4*p**3*x**4 + 11*a*b**4*p**2 
*x**4 + 6*a*b**4*p*x**4 + b**5*p**4*x**5 + 10*b**5*p**3*x**5 + 35*b**5*p** 
2*x**5 + 50*b**5*p*x**5 + 24*b**5*x**5))/(b**5*(p**5 + 15*p**4 + 85*p**3 + 
 225*p**2 + 274*p + 120))