\(\int \frac {(c+d x)^n (a+b x^2)}{x^{5/2}} \, dx\) [232]

Optimal result

Integrand size = 20, antiderivative size = 129 \[ \int \frac {(c+d x)^n \left (a+b x^2\right )}{x^{5/2}} \, dx=-\frac {2 a (c+d x)^{1+n}}{3 c x^{3/2}}+\frac {2 b (c+d x)^{1+n}}{d (1+2 n) \sqrt {x}}-\frac {2 \left (3 b c^2-a d^2 \left (1-4 n^2\right )\right ) (c+d x)^n \left (\frac {c+d x}{c}\right )^{-n} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-n,\frac {1}{2},-\frac {d x}{c}\right )}{3 c d (1+2 n) \sqrt {x}} \] Output:

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

Mathematica [A] (verified)

Time = 0.24 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.85 \[ \int \frac {(c+d x)^n \left (a+b x^2\right )}{x^{5/2}} \, dx=-\frac {2 (c+d x)^n \left (1+\frac {d x}{c}\right )^{-n} \left (b c^2 \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-2-n,-\frac {1}{2},-\frac {d x}{c}\right )-2 b c^2 \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-1-n,-\frac {1}{2},-\frac {d x}{c}\right )+\left (b c^2+a d^2\right ) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-n,-\frac {1}{2},-\frac {d x}{c}\right )\right )}{3 d^2 x^{3/2}} \] Input:

Integrate[((c + d*x)^n*(a + b*x^2))/x^(5/2),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.95, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {520, 27, 87, 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.

\(\displaystyle \int \frac {\left (a+b x^2\right ) (c+d x)^n}{x^{5/2}} \, dx\)

\(\Big \downarrow \) 520

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 87

\(\displaystyle -\frac {-\frac {\left (3 b c^2-a d^2 \left (1-4 n^2\right )\right ) \int \frac {(c+d x)^n}{\sqrt {x}}dx}{c}-\frac {2 a d (1-2 n) (c+d x)^{n+1}}{c \sqrt {x}}}{3 c}-\frac {2 a (c+d x)^{n+1}}{3 c x^{3/2}}\)

\(\Big \downarrow \) 76

\(\displaystyle -\frac {-\frac {(c+d x)^n \left (\frac {d x}{c}+1\right )^{-n} \left (3 b c^2-a d^2 \left (1-4 n^2\right )\right ) \int \frac {\left (\frac {d x}{c}+1\right )^n}{\sqrt {x}}dx}{c}-\frac {2 a d (1-2 n) (c+d x)^{n+1}}{c \sqrt {x}}}{3 c}-\frac {2 a (c+d x)^{n+1}}{3 c x^{3/2}}\)

\(\Big \downarrow \) 74

\(\displaystyle -\frac {-\frac {2 \sqrt {x} (c+d x)^n \left (\frac {d x}{c}+1\right )^{-n} \left (3 b c^2-a d^2 \left (1-4 n^2\right )\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-n,\frac {3}{2},-\frac {d x}{c}\right )}{c}-\frac {2 a d (1-2 n) (c+d x)^{n+1}}{c \sqrt {x}}}{3 c}-\frac {2 a (c+d x)^{n+1}}{3 c x^{3/2}}\)

Input:

Int[((c + d*x)^n*(a + b*x^2))/x^(5/2),x]
 

Output:

(-2*a*(c + d*x)^(1 + n))/(3*c*x^(3/2)) - ((-2*a*d*(1 - 2*n)*(c + d*x)^(1 + 
 n))/(c*Sqrt[x]) - (2*(3*b*c^2 - a*d^2*(1 - 4*n^2))*Sqrt[x]*(c + d*x)^n*Hy 
pergeometric2F1[1/2, -n, 3/2, -((d*x)/c)])/(c*(1 + (d*x)/c)^n))/(3*c)
 

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 87

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

rule 520

Int[((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_.), 
 x_Symbol] :> With[{Qx = PolynomialQuotient[(a + b*x^2)^p, e*x, x], R = Pol 
ynomialRemainder[(a + b*x^2)^p, e*x, x]}, Simp[R*(e*x)^(m + 1)*((c + d*x)^( 
n + 1)/((m + 1)*(e*c))), x] + Simp[1/((m + 1)*(e*c))   Int[(e*x)^(m + 1)*(c 
 + d*x)^n*ExpandToSum[(m + 1)*(e*c)*Qx - d*R*(m + n + 2), x], x], x]] /; Fr 
eeQ[{a, b, c, d, e, n}, x] && IGtQ[p, 0] && LtQ[m, -1] &&  !IntegerQ[n]
 

Maple [F]

\[\int \frac {\left (d x +c \right )^{n} \left (b \,x^{2}+a \right )}{x^{\frac {5}{2}}}d x\]

Input:

int((d*x+c)^n*(b*x^2+a)/x^(5/2),x)
 

Output:

int((d*x+c)^n*(b*x^2+a)/x^(5/2),x)
 

Fricas [F]

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

integrate((d*x+c)^n*(b*x^2+a)/x^(5/2),x, algorithm="fricas")
 

Output:

integral((b*x^2 + a)*(d*x + c)^n/x^(5/2), x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(c+d x)^n \left (a+b x^2\right )}{x^{5/2}} \, dx=\text {Timed out} \] Input:

integrate((d*x+c)**n*(b*x**2+a)/x**(5/2),x)
 

Output:

Timed out
 

Maxima [F]

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

integrate((d*x+c)^n*(b*x^2+a)/x^(5/2),x, algorithm="maxima")
 

Output:

integrate((b*x^2 + a)*(d*x + c)^n/x^(5/2), x)
 

Giac [F]

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

integrate((d*x+c)^n*(b*x^2+a)/x^(5/2),x, algorithm="giac")
 

Output:

integrate((b*x^2 + a)*(d*x + c)^n/x^(5/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(c+d x)^n \left (a+b x^2\right )}{x^{5/2}} \, dx=\int \frac {\left (b\,x^2+a\right )\,{\left (c+d\,x\right )}^n}{x^{5/2}} \,d x \] Input:

int(((a + b*x^2)*(c + d*x)^n)/x^(5/2),x)
 

Output:

int(((a + b*x^2)*(c + d*x)^n)/x^(5/2), x)
 

Reduce [F]

\[ \int \frac {(c+d x)^n \left (a+b x^2\right )}{x^{5/2}} \, dx =\text {Too large to display} \] Input:

int((d*x+c)^n*(b*x^2+a)/x^(5/2),x)
 

Output:

(2*(4*(c + d*x)**n*a*d**2*n**2 - (c + d*x)**n*a*d**2 + 2*(c + d*x)**n*b*c* 
*2*n + 4*(c + d*x)**n*b*c*d*n**2*x - 6*(c + d*x)**n*b*c*d*n*x + 4*(c + d*x 
)**n*b*d**2*n**2*x**2 - 8*(c + d*x)**n*b*d**2*n*x**2 + 3*(c + d*x)**n*b*d* 
*2*x**2 + 32*sqrt(x)*int((sqrt(x)*(c + d*x)**n)/(8*c*n**3*x**3 - 12*c*n**2 
*x**3 - 2*c*n*x**3 + 3*c*x**3 + 8*d*n**3*x**4 - 12*d*n**2*x**4 - 2*d*n*x** 
4 + 3*d*x**4),x)*a*c*d**2*n**6*x - 48*sqrt(x)*int((sqrt(x)*(c + d*x)**n)/( 
8*c*n**3*x**3 - 12*c*n**2*x**3 - 2*c*n*x**3 + 3*c*x**3 + 8*d*n**3*x**4 - 1 
2*d*n**2*x**4 - 2*d*n*x**4 + 3*d*x**4),x)*a*c*d**2*n**5*x - 16*sqrt(x)*int 
((sqrt(x)*(c + d*x)**n)/(8*c*n**3*x**3 - 12*c*n**2*x**3 - 2*c*n*x**3 + 3*c 
*x**3 + 8*d*n**3*x**4 - 12*d*n**2*x**4 - 2*d*n*x**4 + 3*d*x**4),x)*a*c*d** 
2*n**4*x + 24*sqrt(x)*int((sqrt(x)*(c + d*x)**n)/(8*c*n**3*x**3 - 12*c*n** 
2*x**3 - 2*c*n*x**3 + 3*c*x**3 + 8*d*n**3*x**4 - 12*d*n**2*x**4 - 2*d*n*x* 
*4 + 3*d*x**4),x)*a*c*d**2*n**3*x + 2*sqrt(x)*int((sqrt(x)*(c + d*x)**n)/( 
8*c*n**3*x**3 - 12*c*n**2*x**3 - 2*c*n*x**3 + 3*c*x**3 + 8*d*n**3*x**4 - 1 
2*d*n**2*x**4 - 2*d*n*x**4 + 3*d*x**4),x)*a*c*d**2*n**2*x - 3*sqrt(x)*int( 
(sqrt(x)*(c + d*x)**n)/(8*c*n**3*x**3 - 12*c*n**2*x**3 - 2*c*n*x**3 + 3*c* 
x**3 + 8*d*n**3*x**4 - 12*d*n**2*x**4 - 2*d*n*x**4 + 3*d*x**4),x)*a*c*d**2 
*n*x + 24*sqrt(x)*int((sqrt(x)*(c + d*x)**n)/(8*c*n**3*x**3 - 12*c*n**2*x* 
*3 - 2*c*n*x**3 + 3*c*x**3 + 8*d*n**3*x**4 - 12*d*n**2*x**4 - 2*d*n*x**4 + 
 3*d*x**4),x)*b*c**3*n**4*x - 36*sqrt(x)*int((sqrt(x)*(c + d*x)**n)/(8*...