\(\int \sqrt {x} (c+d x)^n (a+b x^2)^2 \, dx\) [236]

Optimal result

Integrand size = 22, antiderivative size = 333 \[ \int \sqrt {x} (c+d x)^n \left (a+b x^2\right )^2 \, dx=-\frac {10 b c \left (63 b c^2+2 a d^2 \left (99+40 n+4 n^2\right )\right ) x^{3/2} (c+d x)^{1+n}}{d^4 (5+2 n) (7+2 n) (9+2 n) (11+2 n)}+\frac {2 b \left (63 b c^2+2 a d^2 \left (99+40 n+4 n^2\right )\right ) x^{5/2} (c+d x)^{1+n}}{d^3 (7+2 n) (9+2 n) (11+2 n)}-\frac {18 b^2 c x^{7/2} (c+d x)^{1+n}}{d^2 (9+2 n) (11+2 n)}+\frac {2 b^2 x^{9/2} (c+d x)^{1+n}}{d (11+2 n)}+\frac {\left (2 a^2 d^4 (7+2 n) \left (99+40 n+4 n^2\right )+\frac {15 b c^2 \left (63 b c^2+2 a d^2 \left (99+40 n+4 n^2\right )\right )}{\frac {5}{2}+n}\right ) x^{3/2} (c+d x)^n \left (\frac {c+d x}{c}\right )^{-n} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-n,\frac {5}{2},-\frac {d x}{c}\right )}{3 d^4 (7+2 n) (9+2 n) (11+2 n)} \] Output:

-10*b*c*(63*b*c^2+2*a*d^2*(4*n^2+40*n+99))*x^(3/2)*(d*x+c)^(1+n)/d^4/(5+2* 
n)/(7+2*n)/(9+2*n)/(11+2*n)+2*b*(63*b*c^2+2*a*d^2*(4*n^2+40*n+99))*x^(5/2) 
*(d*x+c)^(1+n)/d^3/(7+2*n)/(9+2*n)/(11+2*n)-18*b^2*c*x^(7/2)*(d*x+c)^(1+n) 
/d^2/(9+2*n)/(11+2*n)+2*b^2*x^(9/2)*(d*x+c)^(1+n)/d/(11+2*n)+1/3*(2*a^2*d^ 
4*(7+2*n)*(4*n^2+40*n+99)+15*b*c^2*(63*b*c^2+2*a*d^2*(4*n^2+40*n+99))/(5/2 
+n))*x^(3/2)*(d*x+c)^n*hypergeom([3/2, -n],[5/2],-d*x/c)/d^4/(7+2*n)/(9+2* 
n)/(11+2*n)/(((d*x+c)/c)^n)
 

Mathematica [A] (verified)

Time = 0.26 (sec) , antiderivative size = 272, normalized size of antiderivative = 0.82 \[ \int \sqrt {x} (c+d x)^n \left (a+b x^2\right )^2 \, dx=\frac {2 x^{3/2} (c+d x)^n \left (1+\frac {d x}{c}\right )^{-n} \left (b^2 c^4 \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-4-n,\frac {5}{2},-\frac {d x}{c}\right )-4 b^2 c^4 \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-3-n,\frac {5}{2},-\frac {d x}{c}\right )+6 b^2 c^4 \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-2-n,\frac {5}{2},-\frac {d x}{c}\right )+2 a b c^2 d^2 \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-2-n,\frac {5}{2},-\frac {d x}{c}\right )-4 b^2 c^4 \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-1-n,\frac {5}{2},-\frac {d x}{c}\right )-4 a b c^2 d^2 \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-1-n,\frac {5}{2},-\frac {d x}{c}\right )+b^2 c^4 \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-n,\frac {5}{2},-\frac {d x}{c}\right )+2 a b c^2 d^2 \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-n,\frac {5}{2},-\frac {d x}{c}\right )+a^2 d^4 \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-n,\frac {5}{2},-\frac {d x}{c}\right )\right )}{3 d^4} \] Input:

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

Output:

(2*x^(3/2)*(c + d*x)^n*(b^2*c^4*Hypergeometric2F1[3/2, -4 - n, 5/2, -((d*x 
)/c)] - 4*b^2*c^4*Hypergeometric2F1[3/2, -3 - n, 5/2, -((d*x)/c)] + 6*b^2* 
c^4*Hypergeometric2F1[3/2, -2 - n, 5/2, -((d*x)/c)] + 2*a*b*c^2*d^2*Hyperg 
eometric2F1[3/2, -2 - n, 5/2, -((d*x)/c)] - 4*b^2*c^4*Hypergeometric2F1[3/ 
2, -1 - n, 5/2, -((d*x)/c)] - 4*a*b*c^2*d^2*Hypergeometric2F1[3/2, -1 - n, 
 5/2, -((d*x)/c)] + b^2*c^4*Hypergeometric2F1[3/2, -n, 5/2, -((d*x)/c)] + 
2*a*b*c^2*d^2*Hypergeometric2F1[3/2, -n, 5/2, -((d*x)/c)] + a^2*d^4*Hyperg 
eometric2F1[3/2, -n, 5/2, -((d*x)/c)]))/(3*d^4*(1 + (d*x)/c)^n)
 

Rubi [A] (verified)

Time = 0.54 (sec) , antiderivative size = 305, 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.

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

\(\Big \downarrow \) 521

\(\displaystyle \frac {2 \int \frac {1}{2} \sqrt {x} (c+d x)^n \left (-9 b^2 c x^3+2 a b d (2 n+11) x^2+a^2 d (2 n+11)\right )dx}{d (2 n+11)}+\frac {2 b^2 x^{9/2} (c+d x)^{n+1}}{d (2 n+11)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \sqrt {x} (c+d x)^n \left (-9 b^2 c x^3+2 a b d (2 n+11) x^2+a^2 d (2 n+11)\right )dx}{d (2 n+11)}+\frac {2 b^2 x^{9/2} (c+d x)^{n+1}}{d (2 n+11)}\)

\(\Big \downarrow \) 2125

\(\displaystyle \frac {\frac {2 \int \frac {1}{2} \sqrt {x} (c+d x)^n \left (a^2 \left (4 n^2+40 n+99\right ) d^2+b \left (63 b c^2+2 a d^2 \left (4 n^2+40 n+99\right )\right ) x^2\right )dx}{d (2 n+9)}-\frac {18 b^2 c x^{7/2} (c+d x)^{n+1}}{d (2 n+9)}}{d (2 n+11)}+\frac {2 b^2 x^{9/2} (c+d x)^{n+1}}{d (2 n+11)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\int \sqrt {x} (c+d x)^n \left (a^2 \left (4 n^2+40 n+99\right ) d^2+b \left (63 b c^2+2 a d^2 \left (4 n^2+40 n+99\right )\right ) x^2\right )dx}{d (2 n+9)}-\frac {18 b^2 c x^{7/2} (c+d x)^{n+1}}{d (2 n+9)}}{d (2 n+11)}+\frac {2 b^2 x^{9/2} (c+d x)^{n+1}}{d (2 n+11)}\)

\(\Big \downarrow \) 521

\(\displaystyle \frac {\frac {\frac {2 \int \frac {1}{2} \sqrt {x} (c+d x)^n \left (a^2 d^3 (2 n+7) \left (4 n^2+40 n+99\right )-5 b c \left (63 b c^2+2 a d^2 \left (4 n^2+40 n+99\right )\right ) x\right )dx}{d (2 n+7)}+\frac {2 b x^{5/2} (c+d x)^{n+1} \left (2 a d^2 \left (4 n^2+40 n+99\right )+63 b c^2\right )}{d (2 n+7)}}{d (2 n+9)}-\frac {18 b^2 c x^{7/2} (c+d x)^{n+1}}{d (2 n+9)}}{d (2 n+11)}+\frac {2 b^2 x^{9/2} (c+d x)^{n+1}}{d (2 n+11)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\frac {\int \sqrt {x} (c+d x)^n \left (a^2 d^3 (2 n+7) \left (4 n^2+40 n+99\right )-5 b c \left (63 b c^2+2 a d^2 \left (4 n^2+40 n+99\right )\right ) x\right )dx}{d (2 n+7)}+\frac {2 b x^{5/2} (c+d x)^{n+1} \left (2 a d^2 \left (4 n^2+40 n+99\right )+63 b c^2\right )}{d (2 n+7)}}{d (2 n+9)}-\frac {18 b^2 c x^{7/2} (c+d x)^{n+1}}{d (2 n+9)}}{d (2 n+11)}+\frac {2 b^2 x^{9/2} (c+d x)^{n+1}}{d (2 n+11)}\)

\(\Big \downarrow \) 90

\(\displaystyle \frac {\frac {\frac {\frac {\left (2 a^2 d^4 (2 n+7) \left (4 n^2+40 n+99\right )+\frac {15 b c^2 \left (2 a d^2 \left (4 n^2+40 n+99\right )+63 b c^2\right )}{n+\frac {5}{2}}\right ) \int \sqrt {x} (c+d x)^ndx}{2 d}-\frac {10 b c x^{3/2} (c+d x)^{n+1} \left (2 a d^2 \left (4 n^2+40 n+99\right )+63 b c^2\right )}{d (2 n+5)}}{d (2 n+7)}+\frac {2 b x^{5/2} (c+d x)^{n+1} \left (2 a d^2 \left (4 n^2+40 n+99\right )+63 b c^2\right )}{d (2 n+7)}}{d (2 n+9)}-\frac {18 b^2 c x^{7/2} (c+d x)^{n+1}}{d (2 n+9)}}{d (2 n+11)}+\frac {2 b^2 x^{9/2} (c+d x)^{n+1}}{d (2 n+11)}\)

\(\Big \downarrow \) 76

\(\displaystyle \frac {\frac {\frac {\frac {(c+d x)^n \left (\frac {d x}{c}+1\right )^{-n} \left (2 a^2 d^4 (2 n+7) \left (4 n^2+40 n+99\right )+\frac {15 b c^2 \left (2 a d^2 \left (4 n^2+40 n+99\right )+63 b c^2\right )}{n+\frac {5}{2}}\right ) \int \sqrt {x} \left (\frac {d x}{c}+1\right )^ndx}{2 d}-\frac {10 b c x^{3/2} (c+d x)^{n+1} \left (2 a d^2 \left (4 n^2+40 n+99\right )+63 b c^2\right )}{d (2 n+5)}}{d (2 n+7)}+\frac {2 b x^{5/2} (c+d x)^{n+1} \left (2 a d^2 \left (4 n^2+40 n+99\right )+63 b c^2\right )}{d (2 n+7)}}{d (2 n+9)}-\frac {18 b^2 c x^{7/2} (c+d x)^{n+1}}{d (2 n+9)}}{d (2 n+11)}+\frac {2 b^2 x^{9/2} (c+d x)^{n+1}}{d (2 n+11)}\)

\(\Big \downarrow \) 74

\(\displaystyle \frac {\frac {\frac {\frac {x^{3/2} (c+d x)^n \left (\frac {d x}{c}+1\right )^{-n} \left (2 a^2 d^4 (2 n+7) \left (4 n^2+40 n+99\right )+\frac {15 b c^2 \left (2 a d^2 \left (4 n^2+40 n+99\right )+63 b c^2\right )}{n+\frac {5}{2}}\right ) \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-n,\frac {5}{2},-\frac {d x}{c}\right )}{3 d}-\frac {10 b c x^{3/2} (c+d x)^{n+1} \left (2 a d^2 \left (4 n^2+40 n+99\right )+63 b c^2\right )}{d (2 n+5)}}{d (2 n+7)}+\frac {2 b x^{5/2} (c+d x)^{n+1} \left (2 a d^2 \left (4 n^2+40 n+99\right )+63 b c^2\right )}{d (2 n+7)}}{d (2 n+9)}-\frac {18 b^2 c x^{7/2} (c+d x)^{n+1}}{d (2 n+9)}}{d (2 n+11)}+\frac {2 b^2 x^{9/2} (c+d x)^{n+1}}{d (2 n+11)}\)

Input:

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

Output:

(2*b^2*x^(9/2)*(c + d*x)^(1 + n))/(d*(11 + 2*n)) + ((-18*b^2*c*x^(7/2)*(c 
+ d*x)^(1 + n))/(d*(9 + 2*n)) + ((2*b*(63*b*c^2 + 2*a*d^2*(99 + 40*n + 4*n 
^2))*x^(5/2)*(c + d*x)^(1 + n))/(d*(7 + 2*n)) + ((-10*b*c*(63*b*c^2 + 2*a* 
d^2*(99 + 40*n + 4*n^2))*x^(3/2)*(c + d*x)^(1 + n))/(d*(5 + 2*n)) + ((2*a^ 
2*d^4*(7 + 2*n)*(99 + 40*n + 4*n^2) + (15*b*c^2*(63*b*c^2 + 2*a*d^2*(99 + 
40*n + 4*n^2)))/(5/2 + n))*x^(3/2)*(c + d*x)^n*Hypergeometric2F1[3/2, -n, 
5/2, -((d*x)/c)])/(3*d*(1 + (d*x)/c)^n))/(d*(7 + 2*n)))/(d*(9 + 2*n)))/(d* 
(11 + 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]

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

Input:

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

Output:

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

Fricas [F]

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

integrate(x^(1/2)*(d*x+c)^n*(b*x^2+a)^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. \[ \int \sqrt {x} (c+d x)^n \left (a+b x^2\right )^2 \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(2*(32*sqrt(x)*(c + d*x)**n*a**2*c*d**4*n**5 + 512*sqrt(x)*(c + d*x)**n*a* 
*2*c*d**4*n**4 + 2992*sqrt(x)*(c + d*x)**n*a**2*c*d**4*n**3 + 7552*sqrt(x) 
*(c + d*x)**n*a**2*c*d**4*n**2 + 6930*sqrt(x)*(c + d*x)**n*a**2*c*d**4*n + 
 32*sqrt(x)*(c + d*x)**n*a**2*d**5*n**5*x + 528*sqrt(x)*(c + d*x)**n*a**2* 
d**5*n**4*x + 3248*sqrt(x)*(c + d*x)**n*a**2*d**5*n**3*x + 9048*sqrt(x)*(c 
 + d*x)**n*a**2*d**5*n**2*x + 10706*sqrt(x)*(c + d*x)**n*a**2*d**5*n*x + 3 
465*sqrt(x)*(c + d*x)**n*a**2*d**5*x + 240*sqrt(x)*(c + d*x)**n*a*b*c**3*d 
**2*n**3 + 2400*sqrt(x)*(c + d*x)**n*a*b*c**3*d**2*n**2 + 5940*sqrt(x)*(c 
+ d*x)**n*a*b*c**3*d**2*n - 160*sqrt(x)*(c + d*x)**n*a*b*c**2*d**3*n**4*x 
- 1680*sqrt(x)*(c + d*x)**n*a*b*c**2*d**3*n**3*x - 4760*sqrt(x)*(c + d*x)* 
*n*a*b*c**2*d**3*n**2*x - 1980*sqrt(x)*(c + d*x)**n*a*b*c**2*d**3*n*x + 64 
*sqrt(x)*(c + d*x)**n*a*b*c*d**4*n**5*x**2 + 768*sqrt(x)*(c + d*x)**n*a*b* 
c*d**4*n**4*x**2 + 2912*sqrt(x)*(c + d*x)**n*a*b*c*d**4*n**3*x**2 + 3648*s 
qrt(x)*(c + d*x)**n*a*b*c*d**4*n**2*x**2 + 1188*sqrt(x)*(c + d*x)**n*a*b*c 
*d**4*n*x**2 + 64*sqrt(x)*(c + d*x)**n*a*b*d**5*n**5*x**3 + 928*sqrt(x)*(c 
 + d*x)**n*a*b*d**5*n**4*x**3 + 4832*sqrt(x)*(c + d*x)**n*a*b*d**5*n**3*x* 
*3 + 10928*sqrt(x)*(c + d*x)**n*a*b*d**5*n**2*x**3 + 10308*sqrt(x)*(c + d* 
x)**n*a*b*d**5*n*x**3 + 2970*sqrt(x)*(c + d*x)**n*a*b*d**5*x**3 + 1890*sqr 
t(x)*(c + d*x)**n*b**2*c**5*n - 1260*sqrt(x)*(c + d*x)**n*b**2*c**4*d*n**2 
*x - 630*sqrt(x)*(c + d*x)**n*b**2*c**4*d*n*x + 504*sqrt(x)*(c + d*x)**...