\(\int \frac {(c+d x)^n (a+b x^2)^2}{x^{3/2}} \, dx\) [238]

Optimal result

Integrand size = 22, antiderivative size = 276 \[ \int \frac {(c+d x)^n \left (a+b x^2\right )^2}{x^{3/2}} \, dx=-\frac {2 a^2 (c+d x)^{1+n}}{c \sqrt {x}}+\frac {2 b \left (15 b c^2+2 a d^2 \left (35+24 n+4 n^2\right )\right ) \sqrt {x} (c+d x)^{1+n}}{d^3 (3+2 n) (5+2 n) (7+2 n)}-\frac {10 b^2 c x^{3/2} (c+d x)^{1+n}}{d^2 (5+2 n) (7+2 n)}+\frac {2 b^2 x^{5/2} (c+d x)^{1+n}}{d (7+2 n)}+\frac {\left (2 a^2 d^4 \left (35+94 n+52 n^2+8 n^3\right )-\frac {b c^2 \left (15 b c^2+2 a d^2 \left (35+24 n+4 n^2\right )\right )}{\frac {3}{2}+n}\right ) \sqrt {x} (c+d x)^n \left (\frac {c+d x}{c}\right )^{-n} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-n,\frac {3}{2},-\frac {d x}{c}\right )}{c d^3 (5+2 n) (7+2 n)} \] Output:

-2*a^2*(d*x+c)^(1+n)/c/x^(1/2)+2*b*(15*b*c^2+2*a*d^2*(4*n^2+24*n+35))*x^(1 
/2)*(d*x+c)^(1+n)/d^3/(3+2*n)/(5+2*n)/(7+2*n)-10*b^2*c*x^(3/2)*(d*x+c)^(1+ 
n)/d^2/(5+2*n)/(7+2*n)+2*b^2*x^(5/2)*(d*x+c)^(1+n)/d/(7+2*n)+(2*a^2*d^4*(8 
*n^3+52*n^2+94*n+35)-b*c^2*(15*b*c^2+2*a*d^2*(4*n^2+24*n+35))/(3/2+n))*x^( 
1/2)*(d*x+c)^n*hypergeom([1/2, -n],[3/2],-d*x/c)/c/d^3/(5+2*n)/(7+2*n)/((( 
d*x+c)/c)^n)
 

Mathematica [A] (verified)

Time = 0.37 (sec) , antiderivative size = 270, normalized size of antiderivative = 0.98 \[ \int \frac {(c+d x)^n \left (a+b x^2\right )^2}{x^{3/2}} \, dx=-\frac {2 (c+d x)^n \left (1+\frac {d x}{c}\right )^{-n} \left (b^2 c^4 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-4-n,\frac {1}{2},-\frac {d x}{c}\right )-4 b^2 c^4 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-3-n,\frac {1}{2},-\frac {d x}{c}\right )+6 b^2 c^4 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-2-n,\frac {1}{2},-\frac {d x}{c}\right )+2 a b c^2 d^2 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-2-n,\frac {1}{2},-\frac {d x}{c}\right )-4 b^2 c^4 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-1-n,\frac {1}{2},-\frac {d x}{c}\right )-4 a b c^2 d^2 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-1-n,\frac {1}{2},-\frac {d x}{c}\right )+b^2 c^4 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-n,\frac {1}{2},-\frac {d x}{c}\right )+2 a b c^2 d^2 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-n,\frac {1}{2},-\frac {d x}{c}\right )+a^2 d^4 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-n,\frac {1}{2},-\frac {d x}{c}\right )\right )}{d^4 \sqrt {x}} \] Input:

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

Output:

(-2*(c + d*x)^n*(b^2*c^4*Hypergeometric2F1[-1/2, -4 - n, 1/2, -((d*x)/c)] 
- 4*b^2*c^4*Hypergeometric2F1[-1/2, -3 - n, 1/2, -((d*x)/c)] + 6*b^2*c^4*H 
ypergeometric2F1[-1/2, -2 - n, 1/2, -((d*x)/c)] + 2*a*b*c^2*d^2*Hypergeome 
tric2F1[-1/2, -2 - n, 1/2, -((d*x)/c)] - 4*b^2*c^4*Hypergeometric2F1[-1/2, 
 -1 - n, 1/2, -((d*x)/c)] - 4*a*b*c^2*d^2*Hypergeometric2F1[-1/2, -1 - n, 
1/2, -((d*x)/c)] + b^2*c^4*Hypergeometric2F1[-1/2, -n, 1/2, -((d*x)/c)] + 
2*a*b*c^2*d^2*Hypergeometric2F1[-1/2, -n, 1/2, -((d*x)/c)] + a^2*d^4*Hyper 
geometric2F1[-1/2, -n, 1/2, -((d*x)/c)]))/(d^4*Sqrt[x]*(1 + (d*x)/c)^n)
 

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 270, normalized size of antiderivative = 0.98, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.409, Rules used = {520, 27, 2125, 27, 1194, 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.

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

\(\Big \downarrow \) 520

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 2125

\(\displaystyle \frac {\frac {2 \int \frac {(c+d x)^n \left (a^2 \left (4 n^2+16 n+7\right ) d^2+2 a b c (2 n+7) x d-5 b^2 c^2 x^2\right )}{2 \sqrt {x}}dx}{d (2 n+7)}+\frac {2 b^2 c x^{5/2} (c+d x)^{n+1}}{d (2 n+7)}}{c}-\frac {2 a^2 (c+d x)^{n+1}}{c \sqrt {x}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\int \frac {(c+d x)^n \left (a^2 \left (4 n^2+16 n+7\right ) d^2+2 a b c (2 n+7) x d-5 b^2 c^2 x^2\right )}{\sqrt {x}}dx}{d (2 n+7)}+\frac {2 b^2 c x^{5/2} (c+d x)^{n+1}}{d (2 n+7)}}{c}-\frac {2 a^2 (c+d x)^{n+1}}{c \sqrt {x}}\)

\(\Big \downarrow \) 1194

\(\displaystyle \frac {\frac {\frac {2 \int \frac {(c+d x)^n \left (a^2 \left (8 n^3+52 n^2+94 n+35\right ) d^3+b c \left (15 b c^2+2 a d^2 \left (4 n^2+24 n+35\right )\right ) x\right )}{2 \sqrt {x}}dx}{d (2 n+5)}-\frac {10 b^2 c^2 x^{3/2} (c+d x)^{n+1}}{d (2 n+5)}}{d (2 n+7)}+\frac {2 b^2 c x^{5/2} (c+d x)^{n+1}}{d (2 n+7)}}{c}-\frac {2 a^2 (c+d x)^{n+1}}{c \sqrt {x}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\frac {\int \frac {(c+d x)^n \left (a^2 \left (8 n^3+52 n^2+94 n+35\right ) d^3+b c \left (15 b c^2+2 a d^2 \left (4 n^2+24 n+35\right )\right ) x\right )}{\sqrt {x}}dx}{d (2 n+5)}-\frac {10 b^2 c^2 x^{3/2} (c+d x)^{n+1}}{d (2 n+5)}}{d (2 n+7)}+\frac {2 b^2 c x^{5/2} (c+d x)^{n+1}}{d (2 n+7)}}{c}-\frac {2 a^2 (c+d x)^{n+1}}{c \sqrt {x}}\)

\(\Big \downarrow \) 90

\(\displaystyle \frac {\frac {\frac {\frac {\left (2 a^2 d^4 \left (8 n^3+52 n^2+94 n+35\right )-\frac {b c^2 \left (2 a d^2 \left (4 n^2+24 n+35\right )+15 b c^2\right )}{n+\frac {3}{2}}\right ) \int \frac {(c+d x)^n}{\sqrt {x}}dx}{2 d}+\frac {2 b c \sqrt {x} (c+d x)^{n+1} \left (2 a d^2 \left (4 n^2+24 n+35\right )+15 b c^2\right )}{d (2 n+3)}}{d (2 n+5)}-\frac {10 b^2 c^2 x^{3/2} (c+d x)^{n+1}}{d (2 n+5)}}{d (2 n+7)}+\frac {2 b^2 c x^{5/2} (c+d x)^{n+1}}{d (2 n+7)}}{c}-\frac {2 a^2 (c+d x)^{n+1}}{c \sqrt {x}}\)

\(\Big \downarrow \) 76

\(\displaystyle \frac {\frac {\frac {\frac {(c+d x)^n \left (\frac {d x}{c}+1\right )^{-n} \left (2 a^2 d^4 \left (8 n^3+52 n^2+94 n+35\right )-\frac {b c^2 \left (2 a d^2 \left (4 n^2+24 n+35\right )+15 b c^2\right )}{n+\frac {3}{2}}\right ) \int \frac {\left (\frac {d x}{c}+1\right )^n}{\sqrt {x}}dx}{2 d}+\frac {2 b c \sqrt {x} (c+d x)^{n+1} \left (2 a d^2 \left (4 n^2+24 n+35\right )+15 b c^2\right )}{d (2 n+3)}}{d (2 n+5)}-\frac {10 b^2 c^2 x^{3/2} (c+d x)^{n+1}}{d (2 n+5)}}{d (2 n+7)}+\frac {2 b^2 c x^{5/2} (c+d x)^{n+1}}{d (2 n+7)}}{c}-\frac {2 a^2 (c+d x)^{n+1}}{c \sqrt {x}}\)

\(\Big \downarrow \) 74

\(\displaystyle \frac {\frac {\frac {\frac {\sqrt {x} (c+d x)^n \left (\frac {d x}{c}+1\right )^{-n} \left (2 a^2 d^4 \left (8 n^3+52 n^2+94 n+35\right )-\frac {b c^2 \left (2 a d^2 \left (4 n^2+24 n+35\right )+15 b c^2\right )}{n+\frac {3}{2}}\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-n,\frac {3}{2},-\frac {d x}{c}\right )}{d}+\frac {2 b c \sqrt {x} (c+d x)^{n+1} \left (2 a d^2 \left (4 n^2+24 n+35\right )+15 b c^2\right )}{d (2 n+3)}}{d (2 n+5)}-\frac {10 b^2 c^2 x^{3/2} (c+d x)^{n+1}}{d (2 n+5)}}{d (2 n+7)}+\frac {2 b^2 c x^{5/2} (c+d x)^{n+1}}{d (2 n+7)}}{c}-\frac {2 a^2 (c+d x)^{n+1}}{c \sqrt {x}}\)

Input:

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

Output:

(-2*a^2*(c + d*x)^(1 + n))/(c*Sqrt[x]) + ((2*b^2*c*x^(5/2)*(c + d*x)^(1 + 
n))/(d*(7 + 2*n)) + ((-10*b^2*c^2*x^(3/2)*(c + d*x)^(1 + n))/(d*(5 + 2*n)) 
 + ((2*b*c*(15*b*c^2 + 2*a*d^2*(35 + 24*n + 4*n^2))*Sqrt[x]*(c + d*x)^(1 + 
 n))/(d*(3 + 2*n)) + ((2*a^2*d^4*(35 + 94*n + 52*n^2 + 8*n^3) - (b*c^2*(15 
*b*c^2 + 2*a*d^2*(35 + 24*n + 4*n^2)))/(3/2 + n))*Sqrt[x]*(c + d*x)^n*Hype 
rgeometric2F1[1/2, -n, 3/2, -((d*x)/c)])/(d*(1 + (d*x)/c)^n))/(d*(5 + 2*n) 
))/(d*(7 + 2*n)))/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 1194

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) 
 + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[c^p*(d + e*x)^(m + 2*p)*((f + g*x 
)^(n + 1)/(g*e^(2*p)*(m + n + 2*p + 1))), x] + Simp[1/(g*e^(2*p)*(m + n + 2 
*p + 1))   Int[(d + e*x)^m*(f + g*x)^n*ExpandToSum[g*(m + n + 2*p + 1)*(e^( 
2*p)*(a + b*x + c*x^2)^p - c^p*(d + e*x)^(2*p)) - c^p*(e*f - d*g)*(m + 2*p) 
*(d + e*x)^(2*p - 1), x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && IGtQ 
[p, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] && NeQ[m + n + 2*p + 1, 0]
 

rule 2125

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] 
:> With[{q = Expon[Px, x], k = Coeff[Px, x, Expon[Px, x]]}, Simp[k*(a + b*x 
)^(m + q)*((c + d*x)^(n + 1)/(d*b^q*(m + n + q + 1))), x] + Simp[1/(d*b^q*( 
m + n + q + 1))   Int[(a + b*x)^m*(c + d*x)^n*ExpandToSum[d*b^q*(m + n + q 
+ 1)*Px - d*k*(m + n + q + 1)*(a + b*x)^q - k*(b*c - a*d)*(m + q)*(a + b*x) 
^(q - 1), x], x], x] /; NeQ[m + n + q + 1, 0]] /; FreeQ[{a, b, c, d, m, n}, 
 x] && PolyQ[Px, x]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(2*(16*(c + d*x)**n*a**2*d**4*n**4 + 128*(c + d*x)**n*a**2*d**4*n**3 + 344 
*(c + d*x)**n*a**2*d**4*n**2 + 352*(c + d*x)**n*a**2*d**4*n + 105*(c + d*x 
)**n*a**2*d**4 - 16*(c + d*x)**n*a*b*c**2*d**2*n**3 - 96*(c + d*x)**n*a*b* 
c**2*d**2*n**2 - 140*(c + d*x)**n*a*b*c**2*d**2*n + 32*(c + d*x)**n*a*b*c* 
d**3*n**4*x + 176*(c + d*x)**n*a*b*c*d**3*n**3*x + 184*(c + d*x)**n*a*b*c* 
d**3*n**2*x - 140*(c + d*x)**n*a*b*c*d**3*n*x + 32*(c + d*x)**n*a*b*d**4*n 
**4*x**2 + 192*(c + d*x)**n*a*b*d**4*n**3*x**2 + 272*(c + d*x)**n*a*b*d**4 
*n**2*x**2 - 48*(c + d*x)**n*a*b*d**4*n*x**2 - 70*(c + d*x)**n*a*b*d**4*x* 
*2 - 30*(c + d*x)**n*b**2*c**4*n + 60*(c + d*x)**n*b**2*c**3*d*n**2*x - 30 
*(c + d*x)**n*b**2*c**3*d*n*x - 40*(c + d*x)**n*b**2*c**2*d**2*n**3*x**2 + 
 10*(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 + 64*(c + d*x)**n*b**2*d**4*n**3*x**4 + 56*(c + d*x)**n*b**2*d** 
4*n**2*x**4 - 16*(c + d*x)**n*b**2*d**4*n*x**4 - 15*(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**2 + 240*c*n** 
4*x**2 + 560*c*n**3*x**2 + 360*c*n**2*x**2 - 142*c*n*x**2 - 105*c*x**2 + 3 
2*d*n**5*x**3 + 240*d*n**4*x**3 + 560*d*n**3*x**3 + 360*d*n**2*x**3 - 142* 
d*n*x**3 - 105*d*x**3),x)*a**2*c*d**4*n**10 + 7936*sqrt(x)*int((sqrt(x)*(c 
 + d*x)**n)/(32*c*n**5*x**2 + 240*c*n**4*x**2 + 560*c*n**3*x**2 + 360*c...