\(\int \frac {(c+d x)^n (a+b x^2)}{\sqrt {x}} \, dx\) [230]

Optimal result

Integrand size = 20, antiderivative size = 146 \[ \int \frac {(c+d x)^n \left (a+b x^2\right )}{\sqrt {x}} \, dx=-\frac {6 b c \sqrt {x} (c+d x)^{1+n}}{d^2 (3+2 n) (5+2 n)}+\frac {2 b x^{3/2} (c+d x)^{1+n}}{d (5+2 n)}+\frac {2 \left (3 b c^2+a d^2 (3+2 n) (5+2 n)\right ) \sqrt {x} (c+d x)^n \left (\frac {c+d x}{c}\right )^{-n} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-n,\frac {3}{2},-\frac {d x}{c}\right )}{d^2 (3+2 n) (5+2 n)} \] Output:

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

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.74 \[ \int \frac {(c+d x)^n \left (a+b x^2\right )}{\sqrt {x}} \, dx=\frac {2 \sqrt {x} (c+d x)^n \left (1+\frac {d x}{c}\right )^{-n} \left (b c^2 \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-2-n,\frac {3}{2},-\frac {d x}{c}\right )-2 b c^2 \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-1-n,\frac {3}{2},-\frac {d x}{c}\right )+\left (b c^2+a d^2\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-n,\frac {3}{2},-\frac {d x}{c}\right )\right )}{d^2} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.26 (sec) , antiderivative size = 143, 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.

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

\(\Big \downarrow \) 521

\(\displaystyle \frac {2 \int \frac {(a d (2 n+5)-3 b c x) (c+d x)^n}{2 \sqrt {x}}dx}{d (2 n+5)}+\frac {2 b x^{3/2} (c+d x)^{n+1}}{d (2 n+5)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {(a d (2 n+5)-3 b c x) (c+d x)^n}{\sqrt {x}}dx}{d (2 n+5)}+\frac {2 b x^{3/2} (c+d x)^{n+1}}{d (2 n+5)}\)

\(\Big \downarrow \) 90

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

\(\Big \downarrow \) 76

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

\(\Big \downarrow \) 74

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

Input:

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

Output:

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

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 72.29 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.40 \[ \int \frac {(c+d x)^n \left (a+b x^2\right )}{\sqrt {x}} \, dx=2 a c^{n} \sqrt {x} {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, - n \\ \frac {3}{2} \end {matrix}\middle | {\frac {d x e^{i \pi }}{c}} \right )} + \frac {2 b c^{n} x^{\frac {5}{2}} {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{2}, - n \\ \frac {7}{2} \end {matrix}\middle | {\frac {d x e^{i \pi }}{c}} \right )}}{5} \] Input:

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

Output:

2*a*c**n*sqrt(x)*hyper((1/2, -n), (3/2,), d*x*exp_polar(I*pi)/c) + 2*b*c** 
n*x**(5/2)*hyper((5/2, -n), (7/2,), d*x*exp_polar(I*pi)/c)/5
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(2*(4*sqrt(x)*(c + d*x)**n*a*d**2*n**2 + 16*sqrt(x)*(c + d*x)**n*a*d**2*n 
+ 15*sqrt(x)*(c + d*x)**n*a*d**2 - 6*sqrt(x)*(c + d*x)**n*b*c**2*n + 4*sqr 
t(x)*(c + d*x)**n*b*c*d*n**2*x + 2*sqrt(x)*(c + d*x)**n*b*c*d*n*x + 4*sqrt 
(x)*(c + d*x)**n*b*d**2*n**2*x**2 + 8*sqrt(x)*(c + d*x)**n*b*d**2*n*x**2 + 
 3*sqrt(x)*(c + d*x)**n*b*d**2*x**2 + 32*int((sqrt(x)*(c + d*x)**n)/(8*c*n 
**3*x + 36*c*n**2*x + 46*c*n*x + 15*c*x + 8*d*n**3*x**2 + 36*d*n**2*x**2 + 
 46*d*n*x**2 + 15*d*x**2),x)*a*c*d**2*n**6 + 272*int((sqrt(x)*(c + d*x)**n 
)/(8*c*n**3*x + 36*c*n**2*x + 46*c*n*x + 15*c*x + 8*d*n**3*x**2 + 36*d*n** 
2*x**2 + 46*d*n*x**2 + 15*d*x**2),x)*a*c*d**2*n**5 + 880*int((sqrt(x)*(c + 
 d*x)**n)/(8*c*n**3*x + 36*c*n**2*x + 46*c*n*x + 15*c*x + 8*d*n**3*x**2 + 
36*d*n**2*x**2 + 46*d*n*x**2 + 15*d*x**2),x)*a*c*d**2*n**4 + 1336*int((sqr 
t(x)*(c + d*x)**n)/(8*c*n**3*x + 36*c*n**2*x + 46*c*n*x + 15*c*x + 8*d*n** 
3*x**2 + 36*d*n**2*x**2 + 46*d*n*x**2 + 15*d*x**2),x)*a*c*d**2*n**3 + 930* 
int((sqrt(x)*(c + d*x)**n)/(8*c*n**3*x + 36*c*n**2*x + 46*c*n*x + 15*c*x + 
 8*d*n**3*x**2 + 36*d*n**2*x**2 + 46*d*n*x**2 + 15*d*x**2),x)*a*c*d**2*n** 
2 + 225*int((sqrt(x)*(c + d*x)**n)/(8*c*n**3*x + 36*c*n**2*x + 46*c*n*x + 
15*c*x + 8*d*n**3*x**2 + 36*d*n**2*x**2 + 46*d*n*x**2 + 15*d*x**2),x)*a*c* 
d**2*n + 24*int((sqrt(x)*(c + d*x)**n)/(8*c*n**3*x + 36*c*n**2*x + 46*c*n* 
x + 15*c*x + 8*d*n**3*x**2 + 36*d*n**2*x**2 + 46*d*n*x**2 + 15*d*x**2),x)* 
b*c**3*n**4 + 108*int((sqrt(x)*(c + d*x)**n)/(8*c*n**3*x + 36*c*n**2*x ...