\(\int x^5 (c+d \sqrt {a+b x^2})^p \, dx\) [290]

Optimal result

Integrand size = 21, antiderivative size = 252 \[ \int x^5 \left (c+d \sqrt {a+b x^2}\right )^p \, dx=-\frac {c \left (c^2-a d^2\right )^2 \left (c+d \sqrt {a+b x^2}\right )^{1+p}}{b^3 d^6 (1+p)}+\frac {\left (5 c^4-6 a c^2 d^2+a^2 d^4\right ) \left (c+d \sqrt {a+b x^2}\right )^{2+p}}{b^3 d^6 (2+p)}-\frac {2 c \left (5 c^2-3 a d^2\right ) \left (c+d \sqrt {a+b x^2}\right )^{3+p}}{b^3 d^6 (3+p)}+\frac {2 \left (5 c^2-a d^2\right ) \left (c+d \sqrt {a+b x^2}\right )^{4+p}}{b^3 d^6 (4+p)}-\frac {5 c \left (c+d \sqrt {a+b x^2}\right )^{5+p}}{b^3 d^6 (5+p)}+\frac {\left (c+d \sqrt {a+b x^2}\right )^{6+p}}{b^3 d^6 (6+p)} \] Output:

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

Mathematica [A] (verified)

Time = 0.64 (sec) , antiderivative size = 299, normalized size of antiderivative = 1.19 \[ \int x^5 \left (c+d \sqrt {a+b x^2}\right )^p \, dx=\frac {\left (c+d \sqrt {a+b x^2}\right )^{1+p} \left (-120 c^5+120 c^4 d (1+p) \sqrt {a+b x^2}-12 c^3 d^2 \left (4 a \left (-5+p+p^2\right )+5 b \left (2+3 p+p^2\right ) x^2\right )+4 c^2 d^3 (1+p) \sqrt {a+b x^2} \left (2 a \left (-30-4 p+p^2\right )+5 b \left (6+5 p+p^2\right ) x^2\right )+d^5 \left (15+23 p+9 p^2+p^3\right ) \sqrt {a+b x^2} \left (8 a^2-4 a b (2+p) x^2+b^2 \left (8+6 p+p^2\right ) x^4\right )-c d^4 \left (-8 a^2 \left (-15+10 p+12 p^2+2 p^3\right )+4 a b \left (-30-43 p-10 p^2+4 p^3+p^4\right ) x^2+5 b^2 \left (24+50 p+35 p^2+10 p^3+p^4\right ) x^4\right )\right )}{b^3 d^6 (1+p) (2+p) (3+p) (4+p) (5+p) (6+p)} \] Input:

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

Output:

((c + d*Sqrt[a + b*x^2])^(1 + p)*(-120*c^5 + 120*c^4*d*(1 + p)*Sqrt[a + b* 
x^2] - 12*c^3*d^2*(4*a*(-5 + p + p^2) + 5*b*(2 + 3*p + p^2)*x^2) + 4*c^2*d 
^3*(1 + p)*Sqrt[a + b*x^2]*(2*a*(-30 - 4*p + p^2) + 5*b*(6 + 5*p + p^2)*x^ 
2) + d^5*(15 + 23*p + 9*p^2 + p^3)*Sqrt[a + b*x^2]*(8*a^2 - 4*a*b*(2 + p)* 
x^2 + b^2*(8 + 6*p + p^2)*x^4) - c*d^4*(-8*a^2*(-15 + 10*p + 12*p^2 + 2*p^ 
3) + 4*a*b*(-30 - 43*p - 10*p^2 + 4*p^3 + p^4)*x^2 + 5*b^2*(24 + 50*p + 35 
*p^2 + 10*p^3 + p^4)*x^4)))/(b^3*d^6*(1 + p)*(2 + p)*(3 + p)*(4 + p)*(5 + 
p)*(6 + p))
 

Rubi [A] (verified)

Time = 0.92 (sec) , antiderivative size = 238, normalized size of antiderivative = 0.94, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {7283, 896, 1732, 522, 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.

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

\(\Big \downarrow \) 7283

\(\displaystyle \frac {1}{2} \int x^4 \left (c+d \sqrt {b x^2+a}\right )^pdx^2\)

\(\Big \downarrow \) 896

\(\displaystyle \frac {\int b^2 x^4 \left (c+d \sqrt {b x^2+a}\right )^pd\left (b x^2+a\right )}{2 b^3}\)

\(\Big \downarrow \) 1732

\(\displaystyle \frac {\int \sqrt {b x^2+a} \left (a-x^4\right )^2 \left (c+d \sqrt {b x^2+a}\right )^pd\sqrt {b x^2+a}}{b^3}\)

\(\Big \downarrow \) 522

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 522

Int[((e_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_. 
), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*(c + d*x)^n*(a + b*x^2)^p, x], 
x] /; FreeQ[{a, b, c, d, e, m, n}, x] && IGtQ[p, 0]
 

rule 896

Int[((a_) + (b_.)*(v_)^(n_))^(p_.)*(x_)^(m_.), x_Symbol] :> With[{c = Coeff 
icient[v, x, 0], d = Coefficient[v, x, 1]}, Simp[1/d^(m + 1)   Subst[Int[Si 
mplifyIntegrand[(x - c)^m*(a + b*x^n)^p, x], x], x, v], x] /; NeQ[c, 0]] /; 
 FreeQ[{a, b, n, p}, x] && LinearQ[v, x] && IntegerQ[m]
 

rule 1732

Int[((a_) + (c_.)*(x_)^(n2_.))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q_.), x_Symb 
ol] :> With[{g = Denominator[n]}, Simp[g   Subst[Int[x^(g - 1)*(d + e*x^(g* 
n))^q*(a + c*x^(2*g*n))^p, x], x, x^(1/g)], x]] /; FreeQ[{a, c, d, e, p, q} 
, x] && EqQ[n2, 2*n] && FractionQ[n]
 

rule 2009

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

rule 7283

Int[(u_)*(x_)^(m_.), x_Symbol] :> With[{lst = PowerVariableExpn[u, m + 1, x 
]}, Simp[1/lst[[2]]   Subst[Int[NormalizeIntegrand[Simplify[lst[[1]]/x], x] 
, x], x, (lst[[3]]*x)^lst[[2]]], x] /;  !FalseQ[lst] && NeQ[lst[[2]], m + 1 
]] /; IntegerQ[m] && NeQ[m, -1] && NonsumQ[u] && (GtQ[m, 0] ||  !AlgebraicF 
unctionQ[u, x])
 

Maple [F]

\[\int x^{5} \left (c +d \sqrt {b \,x^{2}+a}\right )^{p}d x\]

Input:

int(x^5*(c+d*(b*x^2+a)^(1/2))^p,x)
 

Output:

int(x^5*(c+d*(b*x^2+a)^(1/2))^p,x)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 740 vs. \(2 (240) = 480\).

Time = 24.36 (sec) , antiderivative size = 740, normalized size of antiderivative = 2.94 \[ \int x^5 \left (c+d \sqrt {a+b x^2}\right )^p \, dx=\frac {{\left (120 \, a^{3} d^{6} - 360 \, a^{2} c^{2} d^{4} + 360 \, a c^{4} d^{2} + {\left (b^{3} d^{6} p^{5} + 15 \, b^{3} d^{6} p^{4} + 85 \, b^{3} d^{6} p^{3} + 225 \, b^{3} d^{6} p^{2} + 274 \, b^{3} d^{6} p + 120 \, b^{3} d^{6}\right )} x^{6} - 120 \, c^{6} + {\left (a b^{2} d^{6} p^{5} + {\left (11 \, a b^{2} d^{6} - 5 \, b^{2} c^{2} d^{4}\right )} p^{4} + {\left (41 \, a b^{2} d^{6} - 30 \, b^{2} c^{2} d^{4}\right )} p^{3} + {\left (61 \, a b^{2} d^{6} - 55 \, b^{2} c^{2} d^{4}\right )} p^{2} + 30 \, {\left (a b^{2} d^{6} - b^{2} c^{2} d^{4}\right )} p\right )} x^{4} + 8 \, {\left (a^{3} d^{6} + 3 \, a^{2} c^{2} d^{4}\right )} p^{3} + 24 \, {\left (3 \, a^{3} d^{6} + 3 \, a^{2} c^{2} d^{4} - 2 \, a c^{4} d^{2}\right )} p^{2} - 4 \, {\left ({\left (a^{2} b d^{6} + a b c^{2} d^{4}\right )} p^{4} + 3 \, {\left (3 \, a^{2} b d^{6} - a b c^{2} d^{4}\right )} p^{3} + {\left (23 \, a^{2} b d^{6} - 34 \, a b c^{2} d^{4} + 15 \, b c^{4} d^{2}\right )} p^{2} + 15 \, {\left (a^{2} b d^{6} - 2 \, a b c^{2} d^{4} + b c^{4} d^{2}\right )} p\right )} x^{2} + 8 \, {\left (23 \, a^{3} d^{6} - 24 \, a^{2} c^{2} d^{4} + 9 \, a c^{4} d^{2}\right )} p + {\left ({\left (b^{2} c d^{5} p^{5} + 10 \, b^{2} c d^{5} p^{4} + 35 \, b^{2} c d^{5} p^{3} + 50 \, b^{2} c d^{5} p^{2} + 24 \, b^{2} c d^{5} p\right )} x^{4} + 8 \, {\left (3 \, a^{2} c d^{5} + a c^{3} d^{3}\right )} p^{3} + 24 \, {\left (7 \, a^{2} c d^{5} - 3 \, a c^{3} d^{3}\right )} p^{2} - 4 \, {\left (2 \, a b c d^{5} p^{4} + 5 \, {\left (3 \, a b c d^{5} - b c^{3} d^{3}\right )} p^{3} + {\left (31 \, a b c d^{5} - 15 \, b c^{3} d^{3}\right )} p^{2} + 2 \, {\left (9 \, a b c d^{5} - 5 \, b c^{3} d^{3}\right )} p\right )} x^{2} + 8 \, {\left (33 \, a^{2} c d^{5} - 40 \, a c^{3} d^{3} + 15 \, c^{5} d\right )} p\right )} \sqrt {b x^{2} + a}\right )} {\left (\sqrt {b x^{2} + a} d + c\right )}^{p}}{b^{3} d^{6} p^{6} + 21 \, b^{3} d^{6} p^{5} + 175 \, b^{3} d^{6} p^{4} + 735 \, b^{3} d^{6} p^{3} + 1624 \, b^{3} d^{6} p^{2} + 1764 \, b^{3} d^{6} p + 720 \, b^{3} d^{6}} \] Input:

integrate(x^5*(c+d*(b*x^2+a)^(1/2))^p,x, algorithm="fricas")
 

Output:

(120*a^3*d^6 - 360*a^2*c^2*d^4 + 360*a*c^4*d^2 + (b^3*d^6*p^5 + 15*b^3*d^6 
*p^4 + 85*b^3*d^6*p^3 + 225*b^3*d^6*p^2 + 274*b^3*d^6*p + 120*b^3*d^6)*x^6 
 - 120*c^6 + (a*b^2*d^6*p^5 + (11*a*b^2*d^6 - 5*b^2*c^2*d^4)*p^4 + (41*a*b 
^2*d^6 - 30*b^2*c^2*d^4)*p^3 + (61*a*b^2*d^6 - 55*b^2*c^2*d^4)*p^2 + 30*(a 
*b^2*d^6 - b^2*c^2*d^4)*p)*x^4 + 8*(a^3*d^6 + 3*a^2*c^2*d^4)*p^3 + 24*(3*a 
^3*d^6 + 3*a^2*c^2*d^4 - 2*a*c^4*d^2)*p^2 - 4*((a^2*b*d^6 + a*b*c^2*d^4)*p 
^4 + 3*(3*a^2*b*d^6 - a*b*c^2*d^4)*p^3 + (23*a^2*b*d^6 - 34*a*b*c^2*d^4 + 
15*b*c^4*d^2)*p^2 + 15*(a^2*b*d^6 - 2*a*b*c^2*d^4 + b*c^4*d^2)*p)*x^2 + 8* 
(23*a^3*d^6 - 24*a^2*c^2*d^4 + 9*a*c^4*d^2)*p + ((b^2*c*d^5*p^5 + 10*b^2*c 
*d^5*p^4 + 35*b^2*c*d^5*p^3 + 50*b^2*c*d^5*p^2 + 24*b^2*c*d^5*p)*x^4 + 8*( 
3*a^2*c*d^5 + a*c^3*d^3)*p^3 + 24*(7*a^2*c*d^5 - 3*a*c^3*d^3)*p^2 - 4*(2*a 
*b*c*d^5*p^4 + 5*(3*a*b*c*d^5 - b*c^3*d^3)*p^3 + (31*a*b*c*d^5 - 15*b*c^3* 
d^3)*p^2 + 2*(9*a*b*c*d^5 - 5*b*c^3*d^3)*p)*x^2 + 8*(33*a^2*c*d^5 - 40*a*c 
^3*d^3 + 15*c^5*d)*p)*sqrt(b*x^2 + a))*(sqrt(b*x^2 + a)*d + c)^p/(b^3*d^6* 
p^6 + 21*b^3*d^6*p^5 + 175*b^3*d^6*p^4 + 735*b^3*d^6*p^3 + 1624*b^3*d^6*p^ 
2 + 1764*b^3*d^6*p + 720*b^3*d^6)
 

Sympy [F]

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

integrate(x**5*(c+d*(b*x**2+a)**(1/2))**p,x)
 

Output:

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

Maxima [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 431, normalized size of antiderivative = 1.71 \[ \int x^5 \left (c+d \sqrt {a+b x^2}\right )^p \, dx=\frac {\frac {{\left ({\left (b x^{2} + a\right )} d^{2} {\left (p + 1\right )} + \sqrt {b x^{2} + a} c d p - c^{2}\right )} {\left (\sqrt {b x^{2} + a} d + c\right )}^{p} a^{2}}{{\left (p^{2} + 3 \, p + 2\right )} d^{2}} - \frac {2 \, {\left ({\left (p^{3} + 6 \, p^{2} + 11 \, p + 6\right )} {\left (b x^{2} + a\right )}^{2} d^{4} + {\left (p^{3} + 3 \, p^{2} + 2 \, p\right )} {\left (b x^{2} + a\right )}^{\frac {3}{2}} c d^{3} - 3 \, {\left (b x^{2} + a\right )} {\left (p^{2} + p\right )} c^{2} d^{2} + 6 \, \sqrt {b x^{2} + a} c^{3} d p - 6 \, c^{4}\right )} {\left (\sqrt {b x^{2} + a} d + c\right )}^{p} a}{{\left (p^{4} + 10 \, p^{3} + 35 \, p^{2} + 50 \, p + 24\right )} d^{4}} + \frac {{\left ({\left (p^{5} + 15 \, p^{4} + 85 \, p^{3} + 225 \, p^{2} + 274 \, p + 120\right )} {\left (b x^{2} + a\right )}^{3} d^{6} + {\left (p^{5} + 10 \, p^{4} + 35 \, p^{3} + 50 \, p^{2} + 24 \, p\right )} {\left (b x^{2} + a\right )}^{\frac {5}{2}} c d^{5} - 5 \, {\left (p^{4} + 6 \, p^{3} + 11 \, p^{2} + 6 \, p\right )} {\left (b x^{2} + a\right )}^{2} c^{2} d^{4} + 20 \, {\left (p^{3} + 3 \, p^{2} + 2 \, p\right )} {\left (b x^{2} + a\right )}^{\frac {3}{2}} c^{3} d^{3} - 60 \, {\left (b x^{2} + a\right )} {\left (p^{2} + p\right )} c^{4} d^{2} + 120 \, \sqrt {b x^{2} + a} c^{5} d p - 120 \, c^{6}\right )} {\left (\sqrt {b x^{2} + a} d + c\right )}^{p}}{{\left (p^{6} + 21 \, p^{5} + 175 \, p^{4} + 735 \, p^{3} + 1624 \, p^{2} + 1764 \, p + 720\right )} d^{6}}}{b^{3}} \] Input:

integrate(x^5*(c+d*(b*x^2+a)^(1/2))^p,x, algorithm="maxima")
 

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 4289 vs. \(2 (240) = 480\).

Time = 0.35 (sec) , antiderivative size = 4289, normalized size of antiderivative = 17.02 \[ \int x^5 \left (c+d \sqrt {a+b x^2}\right )^p \, dx=\text {Too large to display} \] Input:

integrate(x^5*(c+d*(b*x^2+a)^(1/2))^p,x, algorithm="giac")
 

Output:

((sqrt(b*x^2 + a)*d + c)^2*(sqrt(b*x^2 + a)*d + c)^p*a^2*d^4*p^5*sgn((sqrt 
(b*x^2 + a)*d + c)*d - c*d) - (sqrt(b*x^2 + a)*d + c)*(sqrt(b*x^2 + a)*d + 
 c)^p*a^2*c*d^4*p^5*sgn((sqrt(b*x^2 + a)*d + c)*d - c*d) + 19*(sqrt(b*x^2 
+ a)*d + c)^2*(sqrt(b*x^2 + a)*d + c)^p*a^2*d^4*p^4*sgn((sqrt(b*x^2 + a)*d 
 + c)*d - c*d) - 20*(sqrt(b*x^2 + a)*d + c)*(sqrt(b*x^2 + a)*d + c)^p*a^2* 
c*d^4*p^4*sgn((sqrt(b*x^2 + a)*d + c)*d - c*d) - 2*(sqrt(b*x^2 + a)*d + c) 
^4*(sqrt(b*x^2 + a)*d + c)^p*a*d^2*p^5*sgn((sqrt(b*x^2 + a)*d + c)*d - c*d 
) + 6*(sqrt(b*x^2 + a)*d + c)^3*(sqrt(b*x^2 + a)*d + c)^p*a*c*d^2*p^5*sgn( 
(sqrt(b*x^2 + a)*d + c)*d - c*d) - 6*(sqrt(b*x^2 + a)*d + c)^2*(sqrt(b*x^2 
 + a)*d + c)^p*a*c^2*d^2*p^5*sgn((sqrt(b*x^2 + a)*d + c)*d - c*d) + 2*(sqr 
t(b*x^2 + a)*d + c)*(sqrt(b*x^2 + a)*d + c)^p*a*c^3*d^2*p^5*sgn((sqrt(b*x^ 
2 + a)*d + c)*d - c*d) + 137*(sqrt(b*x^2 + a)*d + c)^2*(sqrt(b*x^2 + a)*d 
+ c)^p*a^2*d^4*p^3*sgn((sqrt(b*x^2 + a)*d + c)*d - c*d) - 155*(sqrt(b*x^2 
+ a)*d + c)*(sqrt(b*x^2 + a)*d + c)^p*a^2*c*d^4*p^3*sgn((sqrt(b*x^2 + a)*d 
 + c)*d - c*d) - 34*(sqrt(b*x^2 + a)*d + c)^4*(sqrt(b*x^2 + a)*d + c)^p*a* 
d^2*p^4*sgn((sqrt(b*x^2 + a)*d + c)*d - c*d) + 108*(sqrt(b*x^2 + a)*d + c) 
^3*(sqrt(b*x^2 + a)*d + c)^p*a*c*d^2*p^4*sgn((sqrt(b*x^2 + a)*d + c)*d - c 
*d) - 114*(sqrt(b*x^2 + a)*d + c)^2*(sqrt(b*x^2 + a)*d + c)^p*a*c^2*d^2*p^ 
4*sgn((sqrt(b*x^2 + a)*d + c)*d - c*d) + 40*(sqrt(b*x^2 + a)*d + c)*(sqrt( 
b*x^2 + a)*d + c)^p*a*c^3*d^2*p^4*sgn((sqrt(b*x^2 + a)*d + c)*d - c*d) ...
 

Mupad [F(-1)]

Timed out. \[ \int x^5 \left (c+d \sqrt {a+b x^2}\right )^p \, dx=\int x^5\,{\left (c+d\,\sqrt {b\,x^2+a}\right )}^p \,d x \] Input:

int(x^5*(c + d*(a + b*x^2)^(1/2))^p,x)
 

Output:

int(x^5*(c + d*(a + b*x^2)^(1/2))^p, x)
 

Reduce [F]

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

int(x^5*(c+d*(b*x^2+a)^(1/2))^p,x)
 

Output:

int(x^5*(c+d*(b*x^2+a)^(1/2))^p,x)