\(\int \sqrt {x} (c+d x)^n (a+b x^2) \, dx\) [229]

Optimal result

Integrand size = 20, antiderivative size = 148 \[ \int \sqrt {x} (c+d x)^n \left (a+b x^2\right ) \, dx=-\frac {10 b c x^{3/2} (c+d x)^{1+n}}{d^2 (5+2 n) (7+2 n)}+\frac {2 b x^{5/2} (c+d x)^{1+n}}{d (7+2 n)}+\frac {2 \left (15 b c^2+a d^2 (5+2 n) (7+2 n)\right ) x^{3/2} (c+d x)^n \left (\frac {c+d x}{c}\right )^{-n} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-n,\frac {5}{2},-\frac {d x}{c}\right )}{3 d^2 (5+2 n) (7+2 n)} \] Output:

-10*b*c*x^(3/2)*(d*x+c)^(1+n)/d^2/(5+2*n)/(7+2*n)+2*b*x^(5/2)*(d*x+c)^(1+n 
)/d/(7+2*n)+2/3*(15*b*c^2+a*d^2*(5+2*n)*(7+2*n))*x^(3/2)*(d*x+c)^n*hyperge 
om([3/2, -n],[5/2],-d*x/c)/d^2/(5+2*n)/(7+2*n)/(((d*x+c)/c)^n)
 

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.74 \[ \int \sqrt {x} (c+d x)^n \left (a+b x^2\right ) \, dx=\frac {2 x^{3/2} (c+d x)^n \left (1+\frac {d x}{c}\right )^{-n} \left (b c^2 \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-2-n,\frac {5}{2},-\frac {d x}{c}\right )-2 b c^2 \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-1-n,\frac {5}{2},-\frac {d x}{c}\right )+\left (b c^2+a d^2\right ) \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-n,\frac {5}{2},-\frac {d x}{c}\right )\right )}{3 d^2} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.26 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.98, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {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.

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

\(\Big \downarrow \) 521

\(\displaystyle \frac {2 \int \frac {1}{2} \sqrt {x} (a d (2 n+7)-5 b c x) (c+d x)^ndx}{d (2 n+7)}+\frac {2 b x^{5/2} (c+d x)^{n+1}}{d (2 n+7)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \sqrt {x} (a d (2 n+7)-5 b c x) (c+d x)^ndx}{d (2 n+7)}+\frac {2 b x^{5/2} (c+d x)^{n+1}}{d (2 n+7)}\)

\(\Big \downarrow \) 90

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

\(\Big \downarrow \) 76

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

\(\Big \downarrow \) 74

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

Input:

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

Output:

(2*b*x^(5/2)*(c + d*x)^(1 + n))/(d*(7 + 2*n)) + ((-10*b*c*x^(3/2)*(c + d*x 
)^(1 + n))/(d*(5 + 2*n)) + (2*(15*b*c^2 + a*d^2*(5 + 2*n)*(7 + 2*n))*x^(3/ 
2)*(c + d*x)^n*Hypergeometric2F1[3/2, -n, 5/2, -((d*x)/c)])/(3*d*(5 + 2*n) 
*(1 + (d*x)/c)^n))/(d*(7 + 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]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(2*(8*sqrt(x)*(c + d*x)**n*a*c*d**2*n**3 + 48*sqrt(x)*(c + d*x)**n*a*c*d** 
2*n**2 + 70*sqrt(x)*(c + d*x)**n*a*c*d**2*n + 8*sqrt(x)*(c + d*x)**n*a*d** 
3*n**3*x + 52*sqrt(x)*(c + d*x)**n*a*d**3*n**2*x + 94*sqrt(x)*(c + d*x)**n 
*a*d**3*n*x + 35*sqrt(x)*(c + d*x)**n*a*d**3*x + 30*sqrt(x)*(c + d*x)**n*b 
*c**3*n - 20*sqrt(x)*(c + d*x)**n*b*c**2*d*n**2*x - 10*sqrt(x)*(c + d*x)** 
n*b*c**2*d*n*x + 8*sqrt(x)*(c + d*x)**n*b*c*d**2*n**3*x**2 + 16*sqrt(x)*(c 
 + d*x)**n*b*c*d**2*n**2*x**2 + 6*sqrt(x)*(c + d*x)**n*b*c*d**2*n*x**2 + 8 
*sqrt(x)*(c + d*x)**n*b*d**3*n**3*x**3 + 36*sqrt(x)*(c + d*x)**n*b*d**3*n* 
*2*x**3 + 46*sqrt(x)*(c + d*x)**n*b*d**3*n*x**3 + 15*sqrt(x)*(c + d*x)**n* 
b*d**3*x**3 - 64*int((sqrt(x)*(c + d*x)**n)/(16*c*n**4*x + 128*c*n**3*x + 
344*c*n**2*x + 352*c*n*x + 105*c*x + 16*d*n**4*x**2 + 128*d*n**3*x**2 + 34 
4*d*n**2*x**2 + 352*d*n*x**2 + 105*d*x**2),x)*a*c**2*d**2*n**7 - 896*int(( 
sqrt(x)*(c + d*x)**n)/(16*c*n**4*x + 128*c*n**3*x + 344*c*n**2*x + 352*c*n 
*x + 105*c*x + 16*d*n**4*x**2 + 128*d*n**3*x**2 + 344*d*n**2*x**2 + 352*d* 
n*x**2 + 105*d*x**2),x)*a*c**2*d**2*n**6 - 5008*int((sqrt(x)*(c + d*x)**n) 
/(16*c*n**4*x + 128*c*n**3*x + 344*c*n**2*x + 352*c*n*x + 105*c*x + 16*d*n 
**4*x**2 + 128*d*n**3*x**2 + 344*d*n**2*x**2 + 352*d*n*x**2 + 105*d*x**2), 
x)*a*c**2*d**2*n**5 - 14144*int((sqrt(x)*(c + d*x)**n)/(16*c*n**4*x + 128* 
c*n**3*x + 344*c*n**2*x + 352*c*n*x + 105*c*x + 16*d*n**4*x**2 + 128*d*n** 
3*x**2 + 344*d*n**2*x**2 + 352*d*n*x**2 + 105*d*x**2),x)*a*c**2*d**2*n*...