\(\int \frac {x^{3/2} (c+d x)^n}{(a+b x^2)^2} \, dx\) [250]

Optimal result

Integrand size = 22, antiderivative size = 416 \[ \int \frac {x^{3/2} (c+d x)^n}{\left (a+b x^2\right )^2} \, dx=-\frac {c \sqrt {x} (c+d x)^{1+n}}{2 a \left (b c^2+a d^2\right )}+\frac {d x^{3/2} (c+d x)^{1+n}}{2 a \left (b c^2+a d^2\right )}+\frac {b x^{5/2} (c-d x) (c+d x)^{1+n}}{2 a \left (b c^2+a d^2\right ) \left (a+b x^2\right )}+\frac {\left (b c^2-2 \sqrt {-a} \sqrt {b} c d n+a d^2 (1+2 n)\right ) \sqrt {x} (c+d x)^n \left (1+\frac {d x}{c}\right )^{-n} \operatorname {AppellF1}\left (\frac {1}{2},-n,1,\frac {3}{2},-\frac {d x}{c},-\frac {\sqrt {b} x}{\sqrt {-a}}\right )}{4 a b \left (b c^2+a d^2\right )}+\frac {\left (b c^2+2 \sqrt {-a} \sqrt {b} c d n+a d^2 (1+2 n)\right ) \sqrt {x} (c+d x)^n \left (1+\frac {d x}{c}\right )^{-n} \operatorname {AppellF1}\left (\frac {1}{2},-n,1,\frac {3}{2},-\frac {d x}{c},\frac {\sqrt {b} x}{\sqrt {-a}}\right )}{4 a b \left (b c^2+a d^2\right )}-\frac {d^2 (1+2 n) \sqrt {x} (c+d x)^n \left (\frac {c+d x}{c}\right )^{-n} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-n,\frac {3}{2},-\frac {d x}{c}\right )}{2 b \left (b c^2+a d^2\right )} \] Output:

-1/2*c*x^(1/2)*(d*x+c)^(1+n)/a/(a*d^2+b*c^2)+1/2*d*x^(3/2)*(d*x+c)^(1+n)/a 
/(a*d^2+b*c^2)+1/2*b*x^(5/2)*(-d*x+c)*(d*x+c)^(1+n)/a/(a*d^2+b*c^2)/(b*x^2 
+a)+1/4*(b*c^2-2*(-a)^(1/2)*b^(1/2)*c*d*n+a*d^2*(1+2*n))*x^(1/2)*(d*x+c)^n 
*AppellF1(1/2,1,-n,3/2,-b^(1/2)*x/(-a)^(1/2),-d*x/c)/a/b/(a*d^2+b*c^2)/((1 
+d*x/c)^n)+1/4*(b*c^2+2*(-a)^(1/2)*b^(1/2)*c*d*n+a*d^2*(1+2*n))*x^(1/2)*(d 
*x+c)^n*AppellF1(1/2,1,-n,3/2,b^(1/2)*x/(-a)^(1/2),-d*x/c)/a/b/(a*d^2+b*c^ 
2)/((1+d*x/c)^n)-1/2*d^2*(1+2*n)*x^(1/2)*(d*x+c)^n*hypergeom([1/2, -n],[3/ 
2],-d*x/c)/b/(a*d^2+b*c^2)/(((d*x+c)/c)^n)
 

Mathematica [F]

\[ \int \frac {x^{3/2} (c+d x)^n}{\left (a+b x^2\right )^2} \, dx=\int \frac {x^{3/2} (c+d x)^n}{\left (a+b x^2\right )^2} \, dx \] Input:

Integrate[(x^(3/2)*(c + d*x)^n)/(a + b*x^2)^2,x]
 

Output:

Integrate[(x^(3/2)*(c + d*x)^n)/(a + b*x^2)^2, x]
 

Rubi [A] (verified)

Time = 0.69 (sec) , antiderivative size = 255, normalized size of antiderivative = 0.61, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {615, 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.

\(\displaystyle \int \frac {x^{3/2} (c+d x)^n}{\left (a+b x^2\right )^2} \, dx\)

\(\Big \downarrow \) 615

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {x^{5/2} (c+d x)^n \left (\frac {d x}{c}+1\right )^{-n} \operatorname {AppellF1}\left (\frac {5}{2},1,-n,\frac {7}{2},-\frac {\sqrt {b} x}{\sqrt {-a}},-\frac {d x}{c}\right )}{10 a^2}+\frac {x^{5/2} (c+d x)^n \left (\frac {d x}{c}+1\right )^{-n} \operatorname {AppellF1}\left (\frac {5}{2},1,-n,\frac {7}{2},\frac {\sqrt {b} x}{\sqrt {-a}},-\frac {d x}{c}\right )}{10 a^2}+\frac {x^{5/2} (c+d x)^n \left (\frac {d x}{c}+1\right )^{-n} \operatorname {AppellF1}\left (\frac {5}{2},2,-n,\frac {7}{2},-\frac {\sqrt {b} x}{\sqrt {-a}},-\frac {d x}{c}\right )}{10 a^2}+\frac {x^{5/2} (c+d x)^n \left (\frac {d x}{c}+1\right )^{-n} \operatorname {AppellF1}\left (\frac {5}{2},2,-n,\frac {7}{2},\frac {\sqrt {b} x}{\sqrt {-a}},-\frac {d x}{c}\right )}{10 a^2}\)

Input:

Int[(x^(3/2)*(c + d*x)^n)/(a + b*x^2)^2,x]
 

Output:

(x^(5/2)*(c + d*x)^n*AppellF1[5/2, 1, -n, 7/2, -((Sqrt[b]*x)/Sqrt[-a]), -( 
(d*x)/c)])/(10*a^2*(1 + (d*x)/c)^n) + (x^(5/2)*(c + d*x)^n*AppellF1[5/2, 1 
, -n, 7/2, (Sqrt[b]*x)/Sqrt[-a], -((d*x)/c)])/(10*a^2*(1 + (d*x)/c)^n) + ( 
x^(5/2)*(c + d*x)^n*AppellF1[5/2, 2, -n, 7/2, -((Sqrt[b]*x)/Sqrt[-a]), -(( 
d*x)/c)])/(10*a^2*(1 + (d*x)/c)^n) + (x^(5/2)*(c + d*x)^n*AppellF1[5/2, 2, 
 -n, 7/2, (Sqrt[b]*x)/Sqrt[-a], -((d*x)/c)])/(10*a^2*(1 + (d*x)/c)^n)
 

Defintions of rubi rules used

rule 615

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] && ILtQ[p, 0]
 

rule 2009

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

Maple [F]

\[\int \frac {x^{\frac {3}{2}} \left (d x +c \right )^{n}}{\left (b \,x^{2}+a \right )^{2}}d x\]

Input:

int(x^(3/2)*(d*x+c)^n/(b*x^2+a)^2,x)
 

Output:

int(x^(3/2)*(d*x+c)^n/(b*x^2+a)^2,x)
 

Fricas [F]

\[ \int \frac {x^{3/2} (c+d x)^n}{\left (a+b x^2\right )^2} \, dx=\int { \frac {{\left (d x + c\right )}^{n} x^{\frac {3}{2}}}{{\left (b x^{2} + a\right )}^{2}} \,d x } \] Input:

integrate(x^(3/2)*(d*x+c)^n/(b*x^2+a)^2,x, algorithm="fricas")
 

Output:

integral((d*x + c)^n*x^(3/2)/(b^2*x^4 + 2*a*b*x^2 + a^2), x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {x^{3/2} (c+d x)^n}{\left (a+b x^2\right )^2} \, dx=\text {Timed out} \] Input:

integrate(x**(3/2)*(d*x+c)**n/(b*x**2+a)**2,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {x^{3/2} (c+d x)^n}{\left (a+b x^2\right )^2} \, dx=\int { \frac {{\left (d x + c\right )}^{n} x^{\frac {3}{2}}}{{\left (b x^{2} + a\right )}^{2}} \,d x } \] Input:

integrate(x^(3/2)*(d*x+c)^n/(b*x^2+a)^2,x, algorithm="maxima")
 

Output:

integrate((d*x + c)^n*x^(3/2)/(b*x^2 + a)^2, x)
 

Giac [F]

\[ \int \frac {x^{3/2} (c+d x)^n}{\left (a+b x^2\right )^2} \, dx=\int { \frac {{\left (d x + c\right )}^{n} x^{\frac {3}{2}}}{{\left (b x^{2} + a\right )}^{2}} \,d x } \] Input:

integrate(x^(3/2)*(d*x+c)^n/(b*x^2+a)^2,x, algorithm="giac")
 

Output:

integrate((d*x + c)^n*x^(3/2)/(b*x^2 + a)^2, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^{3/2} (c+d x)^n}{\left (a+b x^2\right )^2} \, dx=\int \frac {x^{3/2}\,{\left (c+d\,x\right )}^n}{{\left (b\,x^2+a\right )}^2} \,d x \] Input:

int((x^(3/2)*(c + d*x)^n)/(a + b*x^2)^2,x)
 

Output:

int((x^(3/2)*(c + d*x)^n)/(a + b*x^2)^2, x)
 

Reduce [F]

\[ \int \frac {x^{3/2} (c+d x)^n}{\left (a+b x^2\right )^2} \, dx=\frac {-2 \sqrt {x}\, \left (d x +c \right )^{n}+\left (\int \frac {\left (d x +c \right )^{n}}{\sqrt {x}\, a^{2} c +\sqrt {x}\, a^{2} d x +2 \sqrt {x}\, a b c \,x^{2}+2 \sqrt {x}\, a b d \,x^{3}+\sqrt {x}\, b^{2} c \,x^{4}+\sqrt {x}\, b^{2} d \,x^{5}}d x \right ) a^{2} c +\left (\int \frac {\left (d x +c \right )^{n}}{\sqrt {x}\, a^{2} c +\sqrt {x}\, a^{2} d x +2 \sqrt {x}\, a b c \,x^{2}+2 \sqrt {x}\, a b d \,x^{3}+\sqrt {x}\, b^{2} c \,x^{4}+\sqrt {x}\, b^{2} d \,x^{5}}d x \right ) a b c \,x^{2}+2 \left (\int \frac {\sqrt {x}\, \left (d x +c \right )^{n} x^{2}}{b^{2} d \,x^{5}+b^{2} c \,x^{4}+2 a b d \,x^{3}+2 a b c \,x^{2}+a^{2} d x +a^{2} c}d x \right ) a b d n +2 \left (\int \frac {\sqrt {x}\, \left (d x +c \right )^{n} x^{2}}{b^{2} d \,x^{5}+b^{2} c \,x^{4}+2 a b d \,x^{3}+2 a b c \,x^{2}+a^{2} d x +a^{2} c}d x \right ) b^{2} d n \,x^{2}+2 \left (\int \frac {\sqrt {x}\, \left (d x +c \right )^{n}}{b^{2} d \,x^{5}+b^{2} c \,x^{4}+2 a b d \,x^{3}+2 a b c \,x^{2}+a^{2} d x +a^{2} c}d x \right ) a^{2} d n +\left (\int \frac {\sqrt {x}\, \left (d x +c \right )^{n}}{b^{2} d \,x^{5}+b^{2} c \,x^{4}+2 a b d \,x^{3}+2 a b c \,x^{2}+a^{2} d x +a^{2} c}d x \right ) a^{2} d +2 \left (\int \frac {\sqrt {x}\, \left (d x +c \right )^{n}}{b^{2} d \,x^{5}+b^{2} c \,x^{4}+2 a b d \,x^{3}+2 a b c \,x^{2}+a^{2} d x +a^{2} c}d x \right ) a b d n \,x^{2}+\left (\int \frac {\sqrt {x}\, \left (d x +c \right )^{n}}{b^{2} d \,x^{5}+b^{2} c \,x^{4}+2 a b d \,x^{3}+2 a b c \,x^{2}+a^{2} d x +a^{2} c}d x \right ) a b d \,x^{2}}{3 b \left (b \,x^{2}+a \right )} \] Input:

int(x^(3/2)*(d*x+c)^n/(b*x^2+a)^2,x)
 

Output:

( - 2*sqrt(x)*(c + d*x)**n + int((c + d*x)**n/(sqrt(x)*a**2*c + sqrt(x)*a* 
*2*d*x + 2*sqrt(x)*a*b*c*x**2 + 2*sqrt(x)*a*b*d*x**3 + sqrt(x)*b**2*c*x**4 
 + sqrt(x)*b**2*d*x**5),x)*a**2*c + int((c + d*x)**n/(sqrt(x)*a**2*c + sqr 
t(x)*a**2*d*x + 2*sqrt(x)*a*b*c*x**2 + 2*sqrt(x)*a*b*d*x**3 + sqrt(x)*b**2 
*c*x**4 + sqrt(x)*b**2*d*x**5),x)*a*b*c*x**2 + 2*int((sqrt(x)*(c + d*x)**n 
*x**2)/(a**2*c + a**2*d*x + 2*a*b*c*x**2 + 2*a*b*d*x**3 + b**2*c*x**4 + b* 
*2*d*x**5),x)*a*b*d*n + 2*int((sqrt(x)*(c + d*x)**n*x**2)/(a**2*c + a**2*d 
*x + 2*a*b*c*x**2 + 2*a*b*d*x**3 + b**2*c*x**4 + b**2*d*x**5),x)*b**2*d*n* 
x**2 + 2*int((sqrt(x)*(c + d*x)**n)/(a**2*c + a**2*d*x + 2*a*b*c*x**2 + 2* 
a*b*d*x**3 + b**2*c*x**4 + b**2*d*x**5),x)*a**2*d*n + int((sqrt(x)*(c + d* 
x)**n)/(a**2*c + a**2*d*x + 2*a*b*c*x**2 + 2*a*b*d*x**3 + b**2*c*x**4 + b* 
*2*d*x**5),x)*a**2*d + 2*int((sqrt(x)*(c + d*x)**n)/(a**2*c + a**2*d*x + 2 
*a*b*c*x**2 + 2*a*b*d*x**3 + b**2*c*x**4 + b**2*d*x**5),x)*a*b*d*n*x**2 + 
int((sqrt(x)*(c + d*x)**n)/(a**2*c + a**2*d*x + 2*a*b*c*x**2 + 2*a*b*d*x** 
3 + b**2*c*x**4 + b**2*d*x**5),x)*a*b*d*x**2)/(3*b*(a + b*x**2))