\(\int \frac {(c+d x)^n (a+b x^2)^2}{\sqrt {x}} \, dx\) [237]

Optimal result

Integrand size = 22, antiderivative size = 330 \[ \int \frac {(c+d x)^n \left (a+b x^2\right )^2}{\sqrt {x}} \, dx=-\frac {6 b c \left (35 b c^2+2 a d^2 \left (63+32 n+4 n^2\right )\right ) \sqrt {x} (c+d x)^{1+n}}{d^4 (3+2 n) (5+2 n) (7+2 n) (9+2 n)}+\frac {2 b \left (35 b c^2+2 a d^2 \left (63+32 n+4 n^2\right )\right ) x^{3/2} (c+d x)^{1+n}}{d^3 (5+2 n) (7+2 n) (9+2 n)}-\frac {14 b^2 c x^{5/2} (c+d x)^{1+n}}{d^2 (7+2 n) (9+2 n)}+\frac {2 b^2 x^{7/2} (c+d x)^{1+n}}{d (9+2 n)}+\frac {\left (2 a^2 d^4 (5+2 n) \left (63+32 n+4 n^2\right )+\frac {3 b c^2 \left (35 b c^2+2 a d^2 \left (63+32 n+4 n^2\right )\right )}{\frac {3}{2}+n}\right ) \sqrt {x} (c+d x)^n \left (\frac {c+d x}{c}\right )^{-n} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-n,\frac {3}{2},-\frac {d x}{c}\right )}{d^4 (5+2 n) (7+2 n) (9+2 n)} \] Output:

-6*b*c*(35*b*c^2+2*a*d^2*(4*n^2+32*n+63))*x^(1/2)*(d*x+c)^(1+n)/d^4/(3+2*n 
)/(5+2*n)/(7+2*n)/(9+2*n)+2*b*(35*b*c^2+2*a*d^2*(4*n^2+32*n+63))*x^(3/2)*( 
d*x+c)^(1+n)/d^3/(5+2*n)/(7+2*n)/(9+2*n)-14*b^2*c*x^(5/2)*(d*x+c)^(1+n)/d^ 
2/(7+2*n)/(9+2*n)+2*b^2*x^(7/2)*(d*x+c)^(1+n)/d/(9+2*n)+(2*a^2*d^4*(5+2*n) 
*(4*n^2+32*n+63)+3*b*c^2*(35*b*c^2+2*a*d^2*(4*n^2+32*n+63))/(3/2+n))*x^(1/ 
2)*(d*x+c)^n*hypergeom([1/2, -n],[3/2],-d*x/c)/d^4/(5+2*n)/(7+2*n)/(9+2*n) 
/(((d*x+c)/c)^n)
 

Mathematica [A] (verified)

Time = 0.31 (sec) , antiderivative size = 270, normalized size of antiderivative = 0.82 \[ \int \frac {(c+d x)^n \left (a+b x^2\right )^2}{\sqrt {x}} \, dx=\frac {2 \sqrt {x} (c+d x)^n \left (1+\frac {d x}{c}\right )^{-n} \left (b^2 c^4 \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-4-n,\frac {3}{2},-\frac {d x}{c}\right )-4 b^2 c^4 \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-3-n,\frac {3}{2},-\frac {d x}{c}\right )+6 b^2 c^4 \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-2-n,\frac {3}{2},-\frac {d x}{c}\right )+2 a b c^2 d^2 \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-2-n,\frac {3}{2},-\frac {d x}{c}\right )-4 b^2 c^4 \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-1-n,\frac {3}{2},-\frac {d x}{c}\right )-4 a b c^2 d^2 \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-1-n,\frac {3}{2},-\frac {d x}{c}\right )+b^2 c^4 \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-n,\frac {3}{2},-\frac {d x}{c}\right )+2 a b c^2 d^2 \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-n,\frac {3}{2},-\frac {d x}{c}\right )+a^2 d^4 \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-n,\frac {3}{2},-\frac {d x}{c}\right )\right )}{d^4} \] Input:

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

Output:

(2*Sqrt[x]*(c + d*x)^n*(b^2*c^4*Hypergeometric2F1[1/2, -4 - n, 3/2, -((d*x 
)/c)] - 4*b^2*c^4*Hypergeometric2F1[1/2, -3 - n, 3/2, -((d*x)/c)] + 6*b^2* 
c^4*Hypergeometric2F1[1/2, -2 - n, 3/2, -((d*x)/c)] + 2*a*b*c^2*d^2*Hyperg 
eometric2F1[1/2, -2 - n, 3/2, -((d*x)/c)] - 4*b^2*c^4*Hypergeometric2F1[1/ 
2, -1 - n, 3/2, -((d*x)/c)] - 4*a*b*c^2*d^2*Hypergeometric2F1[1/2, -1 - n, 
 3/2, -((d*x)/c)] + b^2*c^4*Hypergeometric2F1[1/2, -n, 3/2, -((d*x)/c)] + 
2*a*b*c^2*d^2*Hypergeometric2F1[1/2, -n, 3/2, -((d*x)/c)] + a^2*d^4*Hyperg 
eometric2F1[1/2, -n, 3/2, -((d*x)/c)]))/(d^4*(1 + (d*x)/c)^n)
 

Rubi [A] (verified)

Time = 0.54 (sec) , antiderivative size = 302, normalized size of antiderivative = 0.92, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.409, Rules used = {521, 27, 2125, 27, 521, 27, 90, 76, 74}

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.

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

\(\Big \downarrow \) 521

\(\displaystyle \frac {2 \int \frac {(c+d x)^n \left (-7 b^2 c x^3+2 a b d (2 n+9) x^2+a^2 d (2 n+9)\right )}{2 \sqrt {x}}dx}{d (2 n+9)}+\frac {2 b^2 x^{7/2} (c+d x)^{n+1}}{d (2 n+9)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {(c+d x)^n \left (-7 b^2 c x^3+2 a b d (2 n+9) x^2+a^2 d (2 n+9)\right )}{\sqrt {x}}dx}{d (2 n+9)}+\frac {2 b^2 x^{7/2} (c+d x)^{n+1}}{d (2 n+9)}\)

\(\Big \downarrow \) 2125

\(\displaystyle \frac {\frac {2 \int \frac {(c+d x)^n \left (a^2 \left (4 n^2+32 n+63\right ) d^2+b \left (35 b c^2+2 a d^2 \left (4 n^2+32 n+63\right )\right ) x^2\right )}{2 \sqrt {x}}dx}{d (2 n+7)}-\frac {14 b^2 c x^{5/2} (c+d x)^{n+1}}{d (2 n+7)}}{d (2 n+9)}+\frac {2 b^2 x^{7/2} (c+d x)^{n+1}}{d (2 n+9)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\int \frac {(c+d x)^n \left (a^2 \left (4 n^2+32 n+63\right ) d^2+b \left (35 b c^2+2 a d^2 \left (4 n^2+32 n+63\right )\right ) x^2\right )}{\sqrt {x}}dx}{d (2 n+7)}-\frac {14 b^2 c x^{5/2} (c+d x)^{n+1}}{d (2 n+7)}}{d (2 n+9)}+\frac {2 b^2 x^{7/2} (c+d x)^{n+1}}{d (2 n+9)}\)

\(\Big \downarrow \) 521

\(\displaystyle \frac {\frac {\frac {2 \int \frac {(c+d x)^n \left (a^2 d^3 (2 n+5) \left (4 n^2+32 n+63\right )-3 b c \left (35 b c^2+2 a d^2 \left (4 n^2+32 n+63\right )\right ) x\right )}{2 \sqrt {x}}dx}{d (2 n+5)}+\frac {2 b x^{3/2} (c+d x)^{n+1} \left (2 a d^2 \left (4 n^2+32 n+63\right )+35 b c^2\right )}{d (2 n+5)}}{d (2 n+7)}-\frac {14 b^2 c x^{5/2} (c+d x)^{n+1}}{d (2 n+7)}}{d (2 n+9)}+\frac {2 b^2 x^{7/2} (c+d x)^{n+1}}{d (2 n+9)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\frac {\int \frac {(c+d x)^n \left (a^2 d^3 (2 n+5) \left (4 n^2+32 n+63\right )-3 b c \left (35 b c^2+2 a d^2 \left (4 n^2+32 n+63\right )\right ) x\right )}{\sqrt {x}}dx}{d (2 n+5)}+\frac {2 b x^{3/2} (c+d x)^{n+1} \left (2 a d^2 \left (4 n^2+32 n+63\right )+35 b c^2\right )}{d (2 n+5)}}{d (2 n+7)}-\frac {14 b^2 c x^{5/2} (c+d x)^{n+1}}{d (2 n+7)}}{d (2 n+9)}+\frac {2 b^2 x^{7/2} (c+d x)^{n+1}}{d (2 n+9)}\)

\(\Big \downarrow \) 90

\(\displaystyle \frac {\frac {\frac {\frac {\left (2 a^2 d^4 (2 n+5) \left (4 n^2+32 n+63\right )+\frac {3 b c^2 \left (2 a d^2 \left (4 n^2+32 n+63\right )+35 b c^2\right )}{n+\frac {3}{2}}\right ) \int \frac {(c+d x)^n}{\sqrt {x}}dx}{2 d}-\frac {6 b c \sqrt {x} (c+d x)^{n+1} \left (2 a d^2 \left (4 n^2+32 n+63\right )+35 b c^2\right )}{d (2 n+3)}}{d (2 n+5)}+\frac {2 b x^{3/2} (c+d x)^{n+1} \left (2 a d^2 \left (4 n^2+32 n+63\right )+35 b c^2\right )}{d (2 n+5)}}{d (2 n+7)}-\frac {14 b^2 c x^{5/2} (c+d x)^{n+1}}{d (2 n+7)}}{d (2 n+9)}+\frac {2 b^2 x^{7/2} (c+d x)^{n+1}}{d (2 n+9)}\)

\(\Big \downarrow \) 76

\(\displaystyle \frac {\frac {\frac {\frac {(c+d x)^n \left (\frac {d x}{c}+1\right )^{-n} \left (2 a^2 d^4 (2 n+5) \left (4 n^2+32 n+63\right )+\frac {3 b c^2 \left (2 a d^2 \left (4 n^2+32 n+63\right )+35 b c^2\right )}{n+\frac {3}{2}}\right ) \int \frac {\left (\frac {d x}{c}+1\right )^n}{\sqrt {x}}dx}{2 d}-\frac {6 b c \sqrt {x} (c+d x)^{n+1} \left (2 a d^2 \left (4 n^2+32 n+63\right )+35 b c^2\right )}{d (2 n+3)}}{d (2 n+5)}+\frac {2 b x^{3/2} (c+d x)^{n+1} \left (2 a d^2 \left (4 n^2+32 n+63\right )+35 b c^2\right )}{d (2 n+5)}}{d (2 n+7)}-\frac {14 b^2 c x^{5/2} (c+d x)^{n+1}}{d (2 n+7)}}{d (2 n+9)}+\frac {2 b^2 x^{7/2} (c+d x)^{n+1}}{d (2 n+9)}\)

\(\Big \downarrow \) 74

\(\displaystyle \frac {\frac {\frac {\frac {\sqrt {x} (c+d x)^n \left (\frac {d x}{c}+1\right )^{-n} \left (2 a^2 d^4 (2 n+5) \left (4 n^2+32 n+63\right )+\frac {3 b c^2 \left (2 a d^2 \left (4 n^2+32 n+63\right )+35 b c^2\right )}{n+\frac {3}{2}}\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-n,\frac {3}{2},-\frac {d x}{c}\right )}{d}-\frac {6 b c \sqrt {x} (c+d x)^{n+1} \left (2 a d^2 \left (4 n^2+32 n+63\right )+35 b c^2\right )}{d (2 n+3)}}{d (2 n+5)}+\frac {2 b x^{3/2} (c+d x)^{n+1} \left (2 a d^2 \left (4 n^2+32 n+63\right )+35 b c^2\right )}{d (2 n+5)}}{d (2 n+7)}-\frac {14 b^2 c x^{5/2} (c+d x)^{n+1}}{d (2 n+7)}}{d (2 n+9)}+\frac {2 b^2 x^{7/2} (c+d x)^{n+1}}{d (2 n+9)}\)

Input:

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

Output:

(2*b^2*x^(7/2)*(c + d*x)^(1 + n))/(d*(9 + 2*n)) + ((-14*b^2*c*x^(5/2)*(c + 
 d*x)^(1 + n))/(d*(7 + 2*n)) + ((2*b*(35*b*c^2 + 2*a*d^2*(63 + 32*n + 4*n^ 
2))*x^(3/2)*(c + d*x)^(1 + n))/(d*(5 + 2*n)) + ((-6*b*c*(35*b*c^2 + 2*a*d^ 
2*(63 + 32*n + 4*n^2))*Sqrt[x]*(c + d*x)^(1 + n))/(d*(3 + 2*n)) + ((2*a^2* 
d^4*(5 + 2*n)*(63 + 32*n + 4*n^2) + (3*b*c^2*(35*b*c^2 + 2*a*d^2*(63 + 32* 
n + 4*n^2)))/(3/2 + n))*Sqrt[x]*(c + d*x)^n*Hypergeometric2F1[1/2, -n, 3/2 
, -((d*x)/c)])/(d*(1 + (d*x)/c)^n))/(d*(5 + 2*n)))/(d*(7 + 2*n)))/(d*(9 + 
2*n))
 

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 74

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[c^n*((b*x 
)^(m + 1)/(b*(m + 1)))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*(x/c)], x] 
/; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[m] && (IntegerQ[n] || (GtQ[c, 0] 
 &&  !(EqQ[n, -2^(-1)] && EqQ[c^2 - d^2, 0] && GtQ[-d/(b*c), 0])))
 

rule 76

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[c^IntPart 
[n]*((c + d*x)^FracPart[n]/(1 + d*(x/c))^FracPart[n])   Int[(b*x)^m*(1 + d* 
(x/c))^n, x], x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[m] &&  !Integer 
Q[n] &&  !GtQ[c, 0] &&  !GtQ[-d/(b*c), 0] && ((RationalQ[m] &&  !(EqQ[n, -2 
^(-1)] && EqQ[c^2 - d^2, 0])) ||  !RationalQ[n])
 

rule 90

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

rule 521

Int[((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_.), 
 x_Symbol] :> Simp[b^p*(e*x)^(m + 2*p)*((c + d*x)^(n + 1)/(d*e^(2*p)*(m + n 
 + 2*p + 1))), x] + Simp[1/(d*e^(2*p)*(m + n + 2*p + 1))   Int[(e*x)^m*(c + 
 d*x)^n*ExpandToSum[d*(m + n + 2*p + 1)*(e^(2*p)*(a + b*x^2)^p - b^p*(e*x)^ 
(2*p)) - b^p*(e*c)*(m + 2*p)*(e*x)^(2*p - 1), x], x], x] /; FreeQ[{a, b, c, 
 d, e}, x] && IGtQ[p, 0] && NeQ[m + n + 2*p + 1, 0] &&  !IntegerQ[m] &&  !I 
ntegerQ[n]
 

rule 2125

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] 
:> With[{q = Expon[Px, x], k = Coeff[Px, x, Expon[Px, x]]}, Simp[k*(a + b*x 
)^(m + q)*((c + d*x)^(n + 1)/(d*b^q*(m + n + q + 1))), x] + Simp[1/(d*b^q*( 
m + n + q + 1))   Int[(a + b*x)^m*(c + d*x)^n*ExpandToSum[d*b^q*(m + n + q 
+ 1)*Px - d*k*(m + n + q + 1)*(a + b*x)^q - k*(b*c - a*d)*(m + q)*(a + b*x) 
^(q - 1), x], x], x] /; NeQ[m + n + q + 1, 0]] /; FreeQ[{a, b, c, d, m, n}, 
 x] && PolyQ[Px, x]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral((b^2*x^4 + 2*a*b*x^2 + a^2)*(d*x + c)^n/sqrt(x), x)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

integrate((b*x^2 + a)^2*(d*x + c)^n/sqrt(x), x)
 

Giac [F]

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

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

Output:

integrate((b*x^2 + a)^2*(d*x + c)^n/sqrt(x), x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {(c+d x)^n \left (a+b x^2\right )^2}{\sqrt {x}} \, dx=\text {too large to display} \] Input:

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

Output:

(2*(16*sqrt(x)*(c + d*x)**n*a**2*d**4*n**4 + 192*sqrt(x)*(c + d*x)**n*a**2 
*d**4*n**3 + 824*sqrt(x)*(c + d*x)**n*a**2*d**4*n**2 + 1488*sqrt(x)*(c + d 
*x)**n*a**2*d**4*n + 945*sqrt(x)*(c + d*x)**n*a**2*d**4 - 48*sqrt(x)*(c + 
d*x)**n*a*b*c**2*d**2*n**3 - 384*sqrt(x)*(c + d*x)**n*a*b*c**2*d**2*n**2 - 
 756*sqrt(x)*(c + d*x)**n*a*b*c**2*d**2*n + 32*sqrt(x)*(c + d*x)**n*a*b*c* 
d**3*n**4*x + 272*sqrt(x)*(c + d*x)**n*a*b*c*d**3*n**3*x + 632*sqrt(x)*(c 
+ d*x)**n*a*b*c*d**3*n**2*x + 252*sqrt(x)*(c + d*x)**n*a*b*c*d**3*n*x + 32 
*sqrt(x)*(c + d*x)**n*a*b*d**4*n**4*x**2 + 320*sqrt(x)*(c + d*x)**n*a*b*d* 
*4*n**3*x**2 + 1040*sqrt(x)*(c + d*x)**n*a*b*d**4*n**2*x**2 + 1200*sqrt(x) 
*(c + d*x)**n*a*b*d**4*n*x**2 + 378*sqrt(x)*(c + d*x)**n*a*b*d**4*x**2 - 2 
10*sqrt(x)*(c + d*x)**n*b**2*c**4*n + 140*sqrt(x)*(c + d*x)**n*b**2*c**3*d 
*n**2*x + 70*sqrt(x)*(c + d*x)**n*b**2*c**3*d*n*x - 56*sqrt(x)*(c + d*x)** 
n*b**2*c**2*d**2*n**3*x**2 - 112*sqrt(x)*(c + d*x)**n*b**2*c**2*d**2*n**2* 
x**2 - 42*sqrt(x)*(c + d*x)**n*b**2*c**2*d**2*n*x**2 + 16*sqrt(x)*(c + d*x 
)**n*b**2*c*d**3*n**4*x**3 + 72*sqrt(x)*(c + d*x)**n*b**2*c*d**3*n**3*x**3 
 + 92*sqrt(x)*(c + d*x)**n*b**2*c*d**3*n**2*x**3 + 30*sqrt(x)*(c + d*x)**n 
*b**2*c*d**3*n*x**3 + 16*sqrt(x)*(c + d*x)**n*b**2*d**4*n**4*x**4 + 128*sq 
rt(x)*(c + d*x)**n*b**2*d**4*n**3*x**4 + 344*sqrt(x)*(c + d*x)**n*b**2*d** 
4*n**2*x**4 + 352*sqrt(x)*(c + d*x)**n*b**2*d**4*n*x**4 + 105*sqrt(x)*(c + 
 d*x)**n*b**2*d**4*x**4 + 512*int((sqrt(x)*(c + d*x)**n)/(32*c*n**5*x +...