\(\int \frac {(c+d x)^n (a+b x^2)^3}{x^2} \, dx\) [1777]

Optimal result

Integrand size = 20, antiderivative size = 198 \[ \int \frac {(c+d x)^n \left (a+b x^2\right )^3}{x^2} \, dx=\frac {b \left (b^2 c^4+3 a b c^2 d^2+3 a^2 d^4\right ) (c+d x)^{1+n}}{d^5 (1+n)}-\frac {2 b^2 c \left (2 b c^2+3 a d^2\right ) (c+d x)^{2+n}}{d^5 (2+n)}+\frac {3 b^2 \left (2 b c^2+a d^2\right ) (c+d x)^{3+n}}{d^5 (3+n)}-\frac {4 b^3 c (c+d x)^{4+n}}{d^5 (4+n)}+\frac {b^3 (c+d x)^{5+n}}{d^5 (5+n)}+\frac {a^3 d (c+d x)^{1+n} \operatorname {Hypergeometric2F1}\left (2,1+n,2+n,1+\frac {d x}{c}\right )}{c^2 (1+n)} \] Output:

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

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.86 \[ \int \frac {(c+d x)^n \left (a+b x^2\right )^3}{x^2} \, dx=\frac {(c+d x)^{1+n} \left (\frac {b \left (b^2 c^4+3 a b c^2 d^2+3 a^2 d^4\right )}{1+n}-\frac {2 b^2 c \left (2 b c^2+3 a d^2\right ) (c+d x)}{2+n}+\frac {3 b^2 \left (2 b c^2+a d^2\right ) (c+d x)^2}{3+n}-\frac {4 b^3 c (c+d x)^3}{4+n}+\frac {b^3 (c+d x)^4}{5+n}+\frac {a^3 d^6 \operatorname {Hypergeometric2F1}\left (2,1+n,2+n,1+\frac {d x}{c}\right )}{c^2 (1+n)}\right )}{d^5} \] Input:

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

Output:

((c + d*x)^(1 + n)*((b*(b^2*c^4 + 3*a*b*c^2*d^2 + 3*a^2*d^4))/(1 + n) - (2 
*b^2*c*(2*b*c^2 + 3*a*d^2)*(c + d*x))/(2 + n) + (3*b^2*(2*b*c^2 + a*d^2)*( 
c + d*x)^2)/(3 + n) - (4*b^3*c*(c + d*x)^3)/(4 + n) + (b^3*(c + d*x)^4)/(5 
 + n) + (a^3*d^6*Hypergeometric2F1[2, 1 + n, 2 + n, 1 + (d*x)/c])/(c^2*(1 
+ n))))/d^5
 

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 232, normalized size of antiderivative = 1.17, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {520, 25, 2123, 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.

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

\(\Big \downarrow \) 520

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 2123

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

-((a^3*(c + d*x)^(1 + n))/(c*x)) + ((b*c*(b^2*c^4 + 3*a*b*c^2*d^2 + 3*a^2* 
d^4)*(c + d*x)^(1 + n))/(d^5*(1 + n)) - (2*b^2*c^2*(2*b*c^2 + 3*a*d^2)*(c 
+ d*x)^(2 + n))/(d^5*(2 + n)) + (3*b^2*c*(2*b*c^2 + a*d^2)*(c + d*x)^(3 + 
n))/(d^5*(3 + n)) - (4*b^3*c^2*(c + d*x)^(4 + n))/(d^5*(4 + n)) + (b^3*c*( 
c + d*x)^(5 + n))/(d^5*(5 + n)) - (a^3*d*n*(c + d*x)^(1 + n)*Hypergeometri 
c2F1[1, 1 + n, 2 + n, 1 + (d*x)/c])/(c*(1 + n)))/c
 

Defintions of rubi rules used

rule 25

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

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 2009

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

rule 2123

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] 
:> Int[ExpandIntegrand[Px*(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c 
, d, m, n}, x] && PolyQ[Px, x] && (IntegersQ[m, n] || IGtQ[m, -2])
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2421 vs. \(2 (184) = 368\).

Time = 3.46 (sec) , antiderivative size = 3213, normalized size of antiderivative = 16.23 \[ \int \frac {(c+d x)^n \left (a+b x^2\right )^3}{x^2} \, dx=\text {Too large to display} \] Input:

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

Output:

-a**3*d**(n + 2)*n*(c/d + x)**(n + 1)*gamma(n + 1)/(c*d*x*gamma(n + 2)) - 
a**3*d**(n + 2)*(c/d + x)**(n + 1)*gamma(n + 1)/(c*d*x*gamma(n + 2)) - a** 
3*d**(n + 2)*n**2*(c/d + x)**(n + 1)*lerchphi(1 + d*x/c, 1, n + 1)*gamma(n 
 + 1)/(c**2*gamma(n + 2)) - a**3*d**(n + 2)*n*(c/d + x)**(n + 1)*lerchphi( 
1 + d*x/c, 1, n + 1)*gamma(n + 1)/(c**2*gamma(n + 2)) + 3*a**2*b*Piecewise 
((c**n*x, Eq(d, 0)), (Piecewise(((c + d*x)**(n + 1)/(n + 1), Ne(n, -1)), ( 
log(c + d*x), True))/d, True)) + 3*a*b**2*Piecewise((c**n*x**3/3, Eq(d, 0) 
), (2*c**2*log(c/d + x)/(2*c**2*d**3 + 4*c*d**4*x + 2*d**5*x**2) + 3*c**2/ 
(2*c**2*d**3 + 4*c*d**4*x + 2*d**5*x**2) + 4*c*d*x*log(c/d + x)/(2*c**2*d* 
*3 + 4*c*d**4*x + 2*d**5*x**2) + 4*c*d*x/(2*c**2*d**3 + 4*c*d**4*x + 2*d** 
5*x**2) + 2*d**2*x**2*log(c/d + x)/(2*c**2*d**3 + 4*c*d**4*x + 2*d**5*x**2 
), Eq(n, -3)), (-2*c**2*log(c/d + x)/(c*d**3 + d**4*x) - 2*c**2/(c*d**3 + 
d**4*x) - 2*c*d*x*log(c/d + x)/(c*d**3 + d**4*x) + d**2*x**2/(c*d**3 + d** 
4*x), Eq(n, -2)), (c**2*log(c/d + x)/d**3 - c*x/d**2 + x**2/(2*d), Eq(n, - 
1)), (2*c**3*(c + d*x)**n/(d**3*n**3 + 6*d**3*n**2 + 11*d**3*n + 6*d**3) - 
 2*c**2*d*n*x*(c + d*x)**n/(d**3*n**3 + 6*d**3*n**2 + 11*d**3*n + 6*d**3) 
+ c*d**2*n**2*x**2*(c + d*x)**n/(d**3*n**3 + 6*d**3*n**2 + 11*d**3*n + 6*d 
**3) + c*d**2*n*x**2*(c + d*x)**n/(d**3*n**3 + 6*d**3*n**2 + 11*d**3*n + 6 
*d**3) + d**3*n**2*x**3*(c + d*x)**n/(d**3*n**3 + 6*d**3*n**2 + 11*d**3*n 
+ 6*d**3) + 3*d**3*n*x**3*(c + d*x)**n/(d**3*n**3 + 6*d**3*n**2 + 11*d*...
 

Maxima [F]

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

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

Output:

3*(d*x + c)^(n + 1)*a^2*b/(d*(n + 1)) + integrate((b^3*x^6 + 3*a*b^2*x^4 + 
 a^3)*(d*x + c)^n/x^2, x)
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

( - (c + d*x)**n*a**3*d**5*n**5 - 15*(c + d*x)**n*a**3*d**5*n**4 - 85*(c + 
 d*x)**n*a**3*d**5*n**3 - 225*(c + d*x)**n*a**3*d**5*n**2 - 274*(c + d*x)* 
*n*a**3*d**5*n - 120*(c + d*x)**n*a**3*d**5 + 3*(c + d*x)**n*a**2*b*c*d**4 
*n**4*x + 42*(c + d*x)**n*a**2*b*c*d**4*n**3*x + 213*(c + d*x)**n*a**2*b*c 
*d**4*n**2*x + 462*(c + d*x)**n*a**2*b*c*d**4*n*x + 360*(c + d*x)**n*a**2* 
b*c*d**4*x + 3*(c + d*x)**n*a**2*b*d**5*n**4*x**2 + 42*(c + d*x)**n*a**2*b 
*d**5*n**3*x**2 + 213*(c + d*x)**n*a**2*b*d**5*n**2*x**2 + 462*(c + d*x)** 
n*a**2*b*d**5*n*x**2 + 360*(c + d*x)**n*a**2*b*d**5*x**2 + 6*(c + d*x)**n* 
a*b**2*c**3*d**2*n**2*x + 54*(c + d*x)**n*a*b**2*c**3*d**2*n*x + 120*(c + 
d*x)**n*a*b**2*c**3*d**2*x - 6*(c + d*x)**n*a*b**2*c**2*d**3*n**3*x**2 - 5 
4*(c + d*x)**n*a*b**2*c**2*d**3*n**2*x**2 - 120*(c + d*x)**n*a*b**2*c**2*d 
**3*n*x**2 + 3*(c + d*x)**n*a*b**2*c*d**4*n**4*x**3 + 30*(c + d*x)**n*a*b* 
*2*c*d**4*n**3*x**3 + 87*(c + d*x)**n*a*b**2*c*d**4*n**2*x**3 + 60*(c + d* 
x)**n*a*b**2*c*d**4*n*x**3 + 3*(c + d*x)**n*a*b**2*d**5*n**4*x**4 + 36*(c 
+ d*x)**n*a*b**2*d**5*n**3*x**4 + 147*(c + d*x)**n*a*b**2*d**5*n**2*x**4 + 
 234*(c + d*x)**n*a*b**2*d**5*n*x**4 + 120*(c + d*x)**n*a*b**2*d**5*x**4 + 
 24*(c + d*x)**n*b**3*c**5*x - 24*(c + d*x)**n*b**3*c**4*d*n*x**2 + 12*(c 
+ d*x)**n*b**3*c**3*d**2*n**2*x**3 + 12*(c + d*x)**n*b**3*c**3*d**2*n*x**3 
 - 4*(c + d*x)**n*b**3*c**2*d**3*n**3*x**4 - 12*(c + d*x)**n*b**3*c**2*d** 
3*n**2*x**4 - 8*(c + d*x)**n*b**3*c**2*d**3*n*x**4 + (c + d*x)**n*b**3*...