\(\int \frac {x^m (c+d x^2)^3}{(a+b x^2)^2} \, dx\) [1601]

Optimal result

Integrand size = 22, antiderivative size = 148 \[ \int \frac {x^m \left (c+d x^2\right )^3}{\left (a+b x^2\right )^2} \, dx=\frac {d^2 (3 b c-2 a d) x^{1+m}}{b^3 (1+m)}+\frac {d^3 x^{3+m}}{b^2 (3+m)}+\frac {(b c-a d)^3 x^{1+m}}{2 a b^3 \left (a+b x^2\right )}+\frac {(b c-a d)^2 (a d (5+m)+b (c-c m)) x^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )}{2 a^2 b^3 (1+m)} \] Output:

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

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 5 in optimal.

Time = 5.62 (sec) , antiderivative size = 2524, normalized size of antiderivative = 17.05 \[ \int \frac {x^m \left (c+d x^2\right )^3}{\left (a+b x^2\right )^2} \, dx=\text {Result too large to show} \] Input:

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

Output:

-1/192*(x^(1 + m)*(a*(945 + 744*m + 206*m^2 + 24*m^3 + m^4)*(c^3*(-47 + 52 
*m + 6*m^2 + 4*m^3 + m^4) + 3*c^2*d*(1 + m)^4*x^2 + 3*c*d^2*(1 + m)^4*x^4 
+ d^3*(1 + m)^4*x^6)*HurwitzLerchPhi[-((b*x^2)/a), 1, (1 + m)/2] - 3*a*(94 
5 + 744*m + 206*m^2 + 24*m^3 + m^4)*(c^3*(3 + m)^4 + 3*c^2*d*(65 + 92*m + 
54*m^2 + 12*m^3 + m^4)*x^2 + 3*c*d^2*(3 + m)^4*x^4 + d^3*(3 + m)^4*x^6)*Hu 
rwitzLerchPhi[-((b*x^2)/a), 1, (3 + m)/2] + 1771875*a*c^3*HurwitzLerchPhi[ 
-((b*x^2)/a), 1, (5 + m)/2] + 2812500*a*c^3*m*HurwitzLerchPhi[-((b*x^2)/a) 
, 1, (5 + m)/2] + 1927500*a*c^3*m^2*HurwitzLerchPhi[-((b*x^2)/a), 1, (5 + 
m)/2] + 745500*a*c^3*m^3*HurwitzLerchPhi[-((b*x^2)/a), 1, (5 + m)/2] + 178 
050*a*c^3*m^4*HurwitzLerchPhi[-((b*x^2)/a), 1, (5 + m)/2] + 26892*a*c^3*m^ 
5*HurwitzLerchPhi[-((b*x^2)/a), 1, (5 + m)/2] + 2508*a*c^3*m^6*HurwitzLerc 
hPhi[-((b*x^2)/a), 1, (5 + m)/2] + 132*a*c^3*m^7*HurwitzLerchPhi[-((b*x^2) 
/a), 1, (5 + m)/2] + 3*a*c^3*m^8*HurwitzLerchPhi[-((b*x^2)/a), 1, (5 + m)/ 
2] + 5315625*a*c^2*d*x^2*HurwitzLerchPhi[-((b*x^2)/a), 1, (5 + m)/2] + 843 
7500*a*c^2*d*m*x^2*HurwitzLerchPhi[-((b*x^2)/a), 1, (5 + m)/2] + 5782500*a 
*c^2*d*m^2*x^2*HurwitzLerchPhi[-((b*x^2)/a), 1, (5 + m)/2] + 2236500*a*c^2 
*d*m^3*x^2*HurwitzLerchPhi[-((b*x^2)/a), 1, (5 + m)/2] + 534150*a*c^2*d*m^ 
4*x^2*HurwitzLerchPhi[-((b*x^2)/a), 1, (5 + m)/2] + 80676*a*c^2*d*m^5*x^2* 
HurwitzLerchPhi[-((b*x^2)/a), 1, (5 + m)/2] + 7524*a*c^2*d*m^6*x^2*Hurwitz 
LerchPhi[-((b*x^2)/a), 1, (5 + m)/2] + 396*a*c^2*d*m^7*x^2*HurwitzLerch...
 

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.34, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {370, 25, 437, 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.

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

\(\Big \downarrow \) 370

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 437

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 370

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ 
), x_Symbol] :> Simp[(-(b*c - a*d))*(e*x)^(m + 1)*(a + b*x^2)^(p + 1)*((c + 
 d*x^2)^(q - 1)/(a*b*e*2*(p + 1))), x] + Simp[1/(a*b*2*(p + 1))   Int[(e*x) 
^m*(a + b*x^2)^(p + 1)*(c + d*x^2)^(q - 2)*Simp[c*(b*c*2*(p + 1) + (b*c - a 
*d)*(m + 1)) + d*(b*c*2*(p + 1) + (b*c - a*d)*(m + 2*(q - 1) + 1))*x^2, x], 
 x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] 
&& GtQ[q, 1] && IntBinomialQ[a, b, c, d, e, m, 2, p, q, x]
 

rule 437

Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q 
_.)*((e_) + (f_.)*(x_)^2)^(r_.), x_Symbol] :> Int[ExpandIntegrand[(g*x)^m*( 
a + b*x^2)^p*(c + d*x^2)^q*(e + f*x^2)^r, x], x] /; FreeQ[{a, b, c, d, e, f 
, g, m}, x] && IGtQ[p, -2] && IGtQ[q, 0] && IGtQ[r, 0]
 

rule 2009

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

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral((d^3*x^6 + 3*c*d^2*x^4 + 3*c^2*d*x^2 + c^3)*x^m/(b^2*x^4 + 2*a*b* 
x^2 + a^2), x)
 

Sympy [F]

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

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

Output:

Integral(x**m*(c + d*x**2)**3/(a + b*x**2)**2, x)
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(x**m*a**2*d**3*m**2*x + 8*x**m*a**2*d**3*m*x + 15*x**m*a**2*d**3*x - 3*x* 
*m*a*b*c*d**2*m**2*x - 18*x**m*a*b*c*d**2*m*x - 27*x**m*a*b*c*d**2*x - x** 
m*a*b*d**3*m**2*x**3 - 4*x**m*a*b*d**3*m*x**3 + 5*x**m*a*b*d**3*x**3 + 3*x 
**m*b**2*c**2*d*m**2*x + 12*x**m*b**2*c**2*d*m*x + 9*x**m*b**2*c**2*d*x + 
3*x**m*b**2*c*d**2*m**2*x**3 + 6*x**m*b**2*c*d**2*m*x**3 - 9*x**m*b**2*c*d 
**2*x**3 + x**m*b**2*d**3*m**2*x**5 - x**m*b**2*d**3*x**5 - int(x**m/(a**2 
*m - a**2 + 2*a*b*m*x**2 - 2*a*b*x**2 + b**2*m*x**4 - b**2*x**4),x)*a**4*d 
**3*m**4 - 8*int(x**m/(a**2*m - a**2 + 2*a*b*m*x**2 - 2*a*b*x**2 + b**2*m* 
x**4 - b**2*x**4),x)*a**4*d**3*m**3 - 14*int(x**m/(a**2*m - a**2 + 2*a*b*m 
*x**2 - 2*a*b*x**2 + b**2*m*x**4 - b**2*x**4),x)*a**4*d**3*m**2 + 8*int(x* 
*m/(a**2*m - a**2 + 2*a*b*m*x**2 - 2*a*b*x**2 + b**2*m*x**4 - b**2*x**4),x 
)*a**4*d**3*m + 15*int(x**m/(a**2*m - a**2 + 2*a*b*m*x**2 - 2*a*b*x**2 + b 
**2*m*x**4 - b**2*x**4),x)*a**4*d**3 + 3*int(x**m/(a**2*m - a**2 + 2*a*b*m 
*x**2 - 2*a*b*x**2 + b**2*m*x**4 - b**2*x**4),x)*a**3*b*c*d**2*m**4 + 18*i 
nt(x**m/(a**2*m - a**2 + 2*a*b*m*x**2 - 2*a*b*x**2 + b**2*m*x**4 - b**2*x* 
*4),x)*a**3*b*c*d**2*m**3 + 24*int(x**m/(a**2*m - a**2 + 2*a*b*m*x**2 - 2* 
a*b*x**2 + b**2*m*x**4 - b**2*x**4),x)*a**3*b*c*d**2*m**2 - 18*int(x**m/(a 
**2*m - a**2 + 2*a*b*m*x**2 - 2*a*b*x**2 + b**2*m*x**4 - b**2*x**4),x)*a** 
3*b*c*d**2*m - 27*int(x**m/(a**2*m - a**2 + 2*a*b*m*x**2 - 2*a*b*x**2 + b* 
*2*m*x**4 - b**2*x**4),x)*a**3*b*c*d**2 - int(x**m/(a**2*m - a**2 + 2*a...