\(\int \frac {(c+d x)^n}{x^2 (a+b x^2)^2} \, dx\) [227]

Optimal result

Integrand size = 20, antiderivative size = 504 \[ \int \frac {(c+d x)^n}{x^2 \left (a+b x^2\right )^2} \, dx=-\frac {b (a d+b c x) (c+d x)^{1+n}}{2 a^2 \left (b c^2+a d^2\right ) \left (a+b x^2\right )}-\frac {b (c+d x)^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt {b} (c+d x)}{\sqrt {b} c-\sqrt {-a} d}\right )}{2 a^2 \left (\sqrt {-a} \sqrt {b} c+a d\right ) (1+n)}-\frac {b \left (b c^2+a d^2 (1-n)+\sqrt {-a} \sqrt {b} c d n\right ) (c+d x)^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt {b} (c+d x)}{\sqrt {b} c-\sqrt {-a} d}\right )}{4 (-a)^{5/2} \left (\sqrt {b} c-\sqrt {-a} d\right ) \left (b c^2+a d^2\right ) (1+n)}-\frac {b (c+d x)^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt {b} (c+d x)}{\sqrt {b} c+\sqrt {-a} d}\right )}{2 a^3 \left (\frac {\sqrt {b} c}{\sqrt {-a}}+d\right ) (1+n)}+\frac {b \left (b c^2+a d^2 (1-n)-\sqrt {-a} \sqrt {b} c d n\right ) (c+d x)^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt {b} (c+d x)}{\sqrt {b} c+\sqrt {-a} d}\right )}{4 (-a)^{5/2} \left (\sqrt {b} c+\sqrt {-a} d\right ) \left (b c^2+a d^2\right ) (1+n)}+\frac {d (c+d x)^{1+n} \operatorname {Hypergeometric2F1}\left (2,1+n,2+n,1+\frac {d x}{c}\right )}{a^2 c^2 (1+n)} \] Output:

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

Mathematica [A] (verified)

Time = 0.35 (sec) , antiderivative size = 437, normalized size of antiderivative = 0.87 \[ \int \frac {(c+d x)^n}{x^2 \left (a+b x^2\right )^2} \, dx=\frac {1}{4} (c+d x)^{1+n} \left (-\frac {2 b (a d+b c x)}{a^2 \left (b c^2+a d^2\right ) \left (a+b x^2\right )}+\frac {2 b \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt {b} (c+d x)}{\sqrt {b} c-\sqrt {-a} d}\right )}{(-a)^{5/2} \left (-\sqrt {b} c+\sqrt {-a} d\right ) (1+n)}+\frac {2 b \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt {b} (c+d x)}{\sqrt {b} c+\sqrt {-a} d}\right )}{(-a)^{5/2} \left (\sqrt {b} c+\sqrt {-a} d\right ) (1+n)}+\frac {a b \left (\frac {\left (b c^2-a d^2 (-1+n)+\sqrt {-a} \sqrt {b} c d n\right ) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt {b} (c+d x)}{\sqrt {b} c-\sqrt {-a} d}\right )}{\sqrt {b} c-\sqrt {-a} d}-\frac {\left (b c^2-a d^2 (-1+n)-\sqrt {-a} \sqrt {b} c d n\right ) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt {b} (c+d x)}{\sqrt {b} c+\sqrt {-a} d}\right )}{\sqrt {b} c+\sqrt {-a} d}\right )}{(-a)^{7/2} \left (b c^2+a d^2\right ) (1+n)}+\frac {4 d \operatorname {Hypergeometric2F1}\left (2,1+n,2+n,1+\frac {d x}{c}\right )}{a^2 c^2 (1+n)}\right ) \] Input:

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

Output:

((c + d*x)^(1 + n)*((-2*b*(a*d + b*c*x))/(a^2*(b*c^2 + a*d^2)*(a + b*x^2)) 
 + (2*b*Hypergeometric2F1[1, 1 + n, 2 + n, (Sqrt[b]*(c + d*x))/(Sqrt[b]*c 
- Sqrt[-a]*d)])/((-a)^(5/2)*(-(Sqrt[b]*c) + Sqrt[-a]*d)*(1 + n)) + (2*b*Hy 
pergeometric2F1[1, 1 + n, 2 + n, (Sqrt[b]*(c + d*x))/(Sqrt[b]*c + Sqrt[-a] 
*d)])/((-a)^(5/2)*(Sqrt[b]*c + Sqrt[-a]*d)*(1 + n)) + (a*b*(((b*c^2 - a*d^ 
2*(-1 + n) + Sqrt[-a]*Sqrt[b]*c*d*n)*Hypergeometric2F1[1, 1 + n, 2 + n, (S 
qrt[b]*(c + d*x))/(Sqrt[b]*c - Sqrt[-a]*d)])/(Sqrt[b]*c - Sqrt[-a]*d) - (( 
b*c^2 - a*d^2*(-1 + n) - Sqrt[-a]*Sqrt[b]*c*d*n)*Hypergeometric2F1[1, 1 + 
n, 2 + n, (Sqrt[b]*(c + d*x))/(Sqrt[b]*c + Sqrt[-a]*d)])/(Sqrt[b]*c + Sqrt 
[-a]*d)))/((-a)^(7/2)*(b*c^2 + a*d^2)*(1 + n)) + (4*d*Hypergeometric2F1[2, 
 1 + n, 2 + n, 1 + (d*x)/c])/(a^2*c^2*(1 + n))))/4
 

Rubi [A] (verified)

Time = 0.80 (sec) , antiderivative size = 513, normalized size of antiderivative = 1.02, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {615, 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.

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

\(\Big \downarrow \) 615

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

-1/2*(b*(a*d + b*c*x)*(c + d*x)^(1 + n))/(a^2*(b*c^2 + a*d^2)*(a + b*x^2)) 
 - (b*(c + d*x)^(1 + n)*Hypergeometric2F1[1, 1 + n, 2 + n, (Sqrt[b]*(c + d 
*x))/(Sqrt[b]*c - Sqrt[-a]*d)])/(2*(-a)^(5/2)*(Sqrt[b]*c - Sqrt[-a]*d)*(1 
+ n)) - (b*(b*c^2 + a*d^2*(1 - n) + Sqrt[-a]*Sqrt[b]*c*d*n)*(c + d*x)^(1 + 
 n)*Hypergeometric2F1[1, 1 + n, 2 + n, (Sqrt[b]*(c + d*x))/(Sqrt[b]*c - Sq 
rt[-a]*d)])/(4*(-a)^(5/2)*(Sqrt[b]*c - Sqrt[-a]*d)*(b*c^2 + a*d^2)*(1 + n) 
) + (b*(c + d*x)^(1 + n)*Hypergeometric2F1[1, 1 + n, 2 + n, (Sqrt[b]*(c + 
d*x))/(Sqrt[b]*c + Sqrt[-a]*d)])/(2*(-a)^(5/2)*(Sqrt[b]*c + Sqrt[-a]*d)*(1 
 + n)) + (b*(b*c^2 + a*d^2*(1 - n) - Sqrt[-a]*Sqrt[b]*c*d*n)*(c + d*x)^(1 
+ n)*Hypergeometric2F1[1, 1 + n, 2 + n, (Sqrt[b]*(c + d*x))/(Sqrt[b]*c + S 
qrt[-a]*d)])/(4*(-a)^(5/2)*(Sqrt[b]*c + Sqrt[-a]*d)*(b*c^2 + a*d^2)*(1 + n 
)) + (d*(c + d*x)^(1 + n)*Hypergeometric2F1[2, 1 + n, 2 + n, 1 + (d*x)/c]) 
/(a^2*c^2*(1 + n))
 

Defintions of rubi rules used

rule 615

Int[((e_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), 
 x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*(c + d*x)^n*(a + b*x^2)^p, x], x] 
 /; FreeQ[{a, b, c, d, e, m, n}, x] && ILtQ[p, 0]
 

rule 2009

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

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {(c+d x)^n}{x^2 \left (a+b x^2\right )^2} \, dx=\int \frac {\left (d x +c \right )^{n}}{b^{2} x^{6}+2 a b \,x^{4}+a^{2} x^{2}}d x \] Input:

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

Output:

int((c + d*x)**n/(a**2*x**2 + 2*a*b*x**4 + b**2*x**6),x)