\(\int \frac {(c+d x)^n (a+b x^2)^2}{x^{5/2}} \, dx\) [239]

Optimal result

Integrand size = 22, antiderivative size = 218 \[ \int \frac {(c+d x)^n \left (a+b x^2\right )^2}{x^{5/2}} \, dx=-\frac {2 a^2 (c+d x)^{1+n}}{3 c x^{3/2}}+\frac {2 a^2 d (1-2 n) (c+d x)^{1+n}}{3 c^2 \sqrt {x}}-\frac {6 b^2 c \sqrt {x} (c+d x)^{1+n}}{d^2 (3+2 n) (5+2 n)}+\frac {2 b^2 x^{3/2} (c+d x)^{1+n}}{d (5+2 n)}+\frac {2 \left (6 a b c^2+\frac {9 b^2 c^4}{d^2 (3+2 n) (5+2 n)}-a^2 d^2 \left (1-4 n^2\right )\right ) \sqrt {x} (c+d x)^n \left (\frac {c+d x}{c}\right )^{-n} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-n,\frac {3}{2},-\frac {d x}{c}\right )}{3 c^2} \] Output:

-2/3*a^2*(d*x+c)^(1+n)/c/x^(3/2)+2/3*a^2*d*(1-2*n)*(d*x+c)^(1+n)/c^2/x^(1/ 
2)-6*b^2*c*x^(1/2)*(d*x+c)^(1+n)/d^2/(3+2*n)/(5+2*n)+2*b^2*x^(3/2)*(d*x+c) 
^(1+n)/d/(5+2*n)+2/3*(6*a*b*c^2+9*b^2*c^4/d^2/(3+2*n)/(5+2*n)-a^2*d^2*(-4* 
n^2+1))*x^(1/2)*(d*x+c)^n*hypergeom([1/2, -n],[3/2],-d*x/c)/c^2/(((d*x+c)/ 
c)^n)
 

Mathematica [A] (verified)

Time = 0.30 (sec) , antiderivative size = 272, normalized size of antiderivative = 1.25 \[ \int \frac {(c+d x)^n \left (a+b x^2\right )^2}{x^{5/2}} \, dx=-\frac {2 (c+d x)^n \left (1+\frac {d x}{c}\right )^{-n} \left (b^2 c^4 \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-4-n,-\frac {1}{2},-\frac {d x}{c}\right )-4 b^2 c^4 \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-3-n,-\frac {1}{2},-\frac {d x}{c}\right )+6 b^2 c^4 \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-2-n,-\frac {1}{2},-\frac {d x}{c}\right )+2 a b c^2 d^2 \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-2-n,-\frac {1}{2},-\frac {d x}{c}\right )-4 b^2 c^4 \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-1-n,-\frac {1}{2},-\frac {d x}{c}\right )-4 a b c^2 d^2 \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-1-n,-\frac {1}{2},-\frac {d x}{c}\right )+b^2 c^4 \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-n,-\frac {1}{2},-\frac {d x}{c}\right )+2 a b c^2 d^2 \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-n,-\frac {1}{2},-\frac {d x}{c}\right )+a^2 d^4 \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-n,-\frac {1}{2},-\frac {d x}{c}\right )\right )}{3 d^4 x^{3/2}} \] Input:

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

Output:

(-2*(c + d*x)^n*(b^2*c^4*Hypergeometric2F1[-3/2, -4 - n, -1/2, -((d*x)/c)] 
 - 4*b^2*c^4*Hypergeometric2F1[-3/2, -3 - n, -1/2, -((d*x)/c)] + 6*b^2*c^4 
*Hypergeometric2F1[-3/2, -2 - n, -1/2, -((d*x)/c)] + 2*a*b*c^2*d^2*Hyperge 
ometric2F1[-3/2, -2 - n, -1/2, -((d*x)/c)] - 4*b^2*c^4*Hypergeometric2F1[- 
3/2, -1 - n, -1/2, -((d*x)/c)] - 4*a*b*c^2*d^2*Hypergeometric2F1[-3/2, -1 
- n, -1/2, -((d*x)/c)] + b^2*c^4*Hypergeometric2F1[-3/2, -n, -1/2, -((d*x) 
/c)] + 2*a*b*c^2*d^2*Hypergeometric2F1[-3/2, -n, -1/2, -((d*x)/c)] + a^2*d 
^4*Hypergeometric2F1[-3/2, -n, -1/2, -((d*x)/c)]))/(3*d^4*x^(3/2)*(1 + (d* 
x)/c)^n)
 

Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 240, normalized size of antiderivative = 1.10, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.409, Rules used = {520, 27, 2124, 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.

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

\(\Big \downarrow \) 520

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 2124

\(\displaystyle -\frac {-\frac {2 \int \frac {(c+d x)^n \left (3 b^2 c^2 x^2+a \left (6 b c^2-a d^2 \left (1-4 n^2\right )\right )\right )}{2 \sqrt {x}}dx}{c}-\frac {2 a^2 d (1-2 n) (c+d x)^{n+1}}{c \sqrt {x}}}{3 c}-\frac {2 a^2 (c+d x)^{n+1}}{3 c x^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {-\frac {\int \frac {(c+d x)^n \left (3 b^2 c^2 x^2+a \left (6 b c^2-a d^2 \left (1-4 n^2\right )\right )\right )}{\sqrt {x}}dx}{c}-\frac {2 a^2 d (1-2 n) (c+d x)^{n+1}}{c \sqrt {x}}}{3 c}-\frac {2 a^2 (c+d x)^{n+1}}{3 c x^{3/2}}\)

\(\Big \downarrow \) 521

\(\displaystyle -\frac {-\frac {\frac {2 \int \frac {\left (a d (2 n+5) \left (6 b c^2-a d^2 \left (1-4 n^2\right )\right )-9 b^2 c^3 x\right ) (c+d x)^n}{2 \sqrt {x}}dx}{d (2 n+5)}+\frac {6 b^2 c^2 x^{3/2} (c+d x)^{n+1}}{d (2 n+5)}}{c}-\frac {2 a^2 d (1-2 n) (c+d x)^{n+1}}{c \sqrt {x}}}{3 c}-\frac {2 a^2 (c+d x)^{n+1}}{3 c x^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {-\frac {\frac {\int \frac {\left (a d (2 n+5) \left (6 b c^2-a d^2 \left (1-4 n^2\right )\right )-9 b^2 c^3 x\right ) (c+d x)^n}{\sqrt {x}}dx}{d (2 n+5)}+\frac {6 b^2 c^2 x^{3/2} (c+d x)^{n+1}}{d (2 n+5)}}{c}-\frac {2 a^2 d (1-2 n) (c+d x)^{n+1}}{c \sqrt {x}}}{3 c}-\frac {2 a^2 (c+d x)^{n+1}}{3 c x^{3/2}}\)

\(\Big \downarrow \) 90

\(\displaystyle -\frac {-\frac {\frac {\frac {\left (a d^2 (2 n+3) (2 n+5) \left (6 b c^2-a d^2 \left (1-4 n^2\right )\right )+9 b^2 c^4\right ) \int \frac {(c+d x)^n}{\sqrt {x}}dx}{d (2 n+3)}-\frac {18 b^2 c^3 \sqrt {x} (c+d x)^{n+1}}{d (2 n+3)}}{d (2 n+5)}+\frac {6 b^2 c^2 x^{3/2} (c+d x)^{n+1}}{d (2 n+5)}}{c}-\frac {2 a^2 d (1-2 n) (c+d x)^{n+1}}{c \sqrt {x}}}{3 c}-\frac {2 a^2 (c+d x)^{n+1}}{3 c x^{3/2}}\)

\(\Big \downarrow \) 76

\(\displaystyle -\frac {-\frac {\frac {\frac {(c+d x)^n \left (\frac {d x}{c}+1\right )^{-n} \left (a d^2 (2 n+3) (2 n+5) \left (6 b c^2-a d^2 \left (1-4 n^2\right )\right )+9 b^2 c^4\right ) \int \frac {\left (\frac {d x}{c}+1\right )^n}{\sqrt {x}}dx}{d (2 n+3)}-\frac {18 b^2 c^3 \sqrt {x} (c+d x)^{n+1}}{d (2 n+3)}}{d (2 n+5)}+\frac {6 b^2 c^2 x^{3/2} (c+d x)^{n+1}}{d (2 n+5)}}{c}-\frac {2 a^2 d (1-2 n) (c+d x)^{n+1}}{c \sqrt {x}}}{3 c}-\frac {2 a^2 (c+d x)^{n+1}}{3 c x^{3/2}}\)

\(\Big \downarrow \) 74

\(\displaystyle -\frac {-\frac {2 a^2 d (1-2 n) (c+d x)^{n+1}}{c \sqrt {x}}-\frac {\frac {\frac {2 \sqrt {x} (c+d x)^n \left (\frac {d x}{c}+1\right )^{-n} \left (a d^2 (2 n+3) (2 n+5) \left (6 b c^2-a d^2 \left (1-4 n^2\right )\right )+9 b^2 c^4\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-n,\frac {3}{2},-\frac {d x}{c}\right )}{d (2 n+3)}-\frac {18 b^2 c^3 \sqrt {x} (c+d x)^{n+1}}{d (2 n+3)}}{d (2 n+5)}+\frac {6 b^2 c^2 x^{3/2} (c+d x)^{n+1}}{d (2 n+5)}}{c}}{3 c}-\frac {2 a^2 (c+d x)^{n+1}}{3 c x^{3/2}}\)

Input:

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

Output:

(-2*a^2*(c + d*x)^(1 + n))/(3*c*x^(3/2)) - ((-2*a^2*d*(1 - 2*n)*(c + d*x)^ 
(1 + n))/(c*Sqrt[x]) - ((6*b^2*c^2*x^(3/2)*(c + d*x)^(1 + n))/(d*(5 + 2*n) 
) + ((-18*b^2*c^3*Sqrt[x]*(c + d*x)^(1 + n))/(d*(3 + 2*n)) + (2*(9*b^2*c^4 
 + a*d^2*(3 + 2*n)*(5 + 2*n)*(6*b*c^2 - a*d^2*(1 - 4*n^2)))*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)))/c)/(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 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 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]
 

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 2124

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] : 
> With[{Qx = PolynomialQuotient[Px, a + b*x, x], R = PolynomialRemainder[Px 
, a + b*x, x]}, Simp[R*(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((m + 1)*(b*c - 
 a*d))), x] + Simp[1/((m + 1)*(b*c - a*d))   Int[(a + b*x)^(m + 1)*(c + d*x 
)^n*ExpandToSum[(m + 1)*(b*c - a*d)*Qx - d*R*(m + n + 2), x], x], x]] /; Fr 
eeQ[{a, b, c, d, n}, x] && PolyQ[Px, x] && LtQ[m, -1] && (IntegerQ[m] ||  ! 
ILtQ[n, -1])
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral((b^2*x^4 + 2*a*b*x^2 + a^2)*(d*x + c)^n/x^(5/2), x)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(2*(16*(c + d*x)**n*a**2*d**4*n**4 + 64*(c + d*x)**n*a**2*d**4*n**3 + 56*( 
c + d*x)**n*a**2*d**4*n**2 - 16*(c + d*x)**n*a**2*d**4*n - 15*(c + d*x)**n 
*a**2*d**4 + 16*(c + d*x)**n*a*b*c**2*d**2*n**3 + 64*(c + d*x)**n*a*b*c**2 
*d**2*n**2 + 60*(c + d*x)**n*a*b*c**2*d**2*n + 32*(c + d*x)**n*a*b*c*d**3* 
n**4*x + 80*(c + d*x)**n*a*b*c*d**3*n**3*x - 72*(c + d*x)**n*a*b*c*d**3*n* 
*2*x - 180*(c + d*x)**n*a*b*c*d**3*n*x + 32*(c + d*x)**n*a*b*d**4*n**4*x** 
2 + 64*(c + d*x)**n*a*b*d**4*n**3*x**2 - 112*(c + d*x)**n*a*b*d**4*n**2*x* 
*2 - 144*(c + d*x)**n*a*b*d**4*n*x**2 + 90*(c + d*x)**n*a*b*d**4*x**2 + 6* 
(c + d*x)**n*b**2*c**4*n + 12*(c + d*x)**n*b**2*c**3*d*n**2*x - 18*(c + d* 
x)**n*b**2*c**3*d*n*x - 24*(c + d*x)**n*b**2*c**2*d**2*n**3*x**2 + 48*(c + 
 d*x)**n*b**2*c**2*d**2*n**2*x**2 - 18*(c + d*x)**n*b**2*c**2*d**2*n*x**2 
+ 16*(c + d*x)**n*b**2*c*d**3*n**4*x**3 - 24*(c + d*x)**n*b**2*c*d**3*n**3 
*x**3 - 4*(c + d*x)**n*b**2*c*d**3*n**2*x**3 + 6*(c + d*x)**n*b**2*c*d**3* 
n*x**3 + 16*(c + d*x)**n*b**2*d**4*n**4*x**4 - 40*(c + d*x)**n*b**2*d**4*n 
**2*x**4 + 9*(c + d*x)**n*b**2*d**4*x**4 + 512*sqrt(x)*int((sqrt(x)*(c + d 
*x)**n)/(32*c*n**5*x**3 + 80*c*n**4*x**3 - 80*c*n**3*x**3 - 200*c*n**2*x** 
3 + 18*c*n*x**3 + 45*c*x**3 + 32*d*n**5*x**4 + 80*d*n**4*x**4 - 80*d*n**3* 
x**4 - 200*d*n**2*x**4 + 18*d*n*x**4 + 45*d*x**4),x)*a**2*c*d**4*n**10*x + 
 3328*sqrt(x)*int((sqrt(x)*(c + d*x)**n)/(32*c*n**5*x**3 + 80*c*n**4*x**3 
- 80*c*n**3*x**3 - 200*c*n**2*x**3 + 18*c*n*x**3 + 45*c*x**3 + 32*d*n**...