\(\int \frac {(c+d x)^n}{(a+b x^2)^2} \, dx\) [225]

Optimal result

Integrand size = 17, antiderivative size = 304 \[ \int \frac {(c+d x)^n}{\left (a+b x^2\right )^2} \, dx=\frac {(a d+b c x) (c+d x)^{1+n}}{2 a \left (b c^2+a d^2\right ) \left (a+b x^2\right )}-\frac {\left (b c^2+a d^2 (1-n)+\sqrt {-a} \sqrt {b} c d n\right ) (c+d x)^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt {b} (c+d x)}{\sqrt {b} c-\sqrt {-a} d}\right )}{4 (-a)^{3/2} \left (\sqrt {b} c-\sqrt {-a} d\right ) \left (b c^2+a d^2\right ) (1+n)}+\frac {\left (b c^2+a d^2 (1-n)-\sqrt {-a} \sqrt {b} c d n\right ) (c+d x)^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt {b} (c+d x)}{\sqrt {b} c+\sqrt {-a} d}\right )}{4 (-a)^{3/2} \left (\sqrt {b} c+\sqrt {-a} d\right ) \left (b c^2+a d^2\right ) (1+n)} \] Output:

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

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 253, normalized size of antiderivative = 0.83 \[ \int \frac {(c+d x)^n}{\left (a+b x^2\right )^2} \, dx=\frac {(c+d x)^{1+n} \left (\frac {2 (a d+b c x)}{a+b x^2}+\frac {\left (b c^2-a d^2 (-1+n)+\sqrt {-a} \sqrt {b} c d n\right ) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt {b} (c+d x)}{\sqrt {b} c-\sqrt {-a} d}\right )}{\sqrt {-a} \left (\sqrt {b} c-\sqrt {-a} d\right ) (1+n)}+\frac {\left (-b c^2+a d^2 (-1+n)+\sqrt {-a} \sqrt {b} c d n\right ) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt {b} (c+d x)}{\sqrt {b} c+\sqrt {-a} d}\right )}{\sqrt {-a} \left (\sqrt {b} c+\sqrt {-a} d\right ) (1+n)}\right )}{4 a \left (b c^2+a d^2\right )} \] Input:

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

Output:

((c + d*x)^(1 + n)*((2*(a*d + b*c*x))/(a + b*x^2) + ((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[-a]*(Sqrt[b]*c - Sqrt[-a]*d)*(1 
 + n)) + ((-(b*c^2) + a*d^2*(-1 + n) + Sqrt[-a]*Sqrt[b]*c*d*n)*Hypergeomet 
ric2F1[1, 1 + n, 2 + n, (Sqrt[b]*(c + d*x))/(Sqrt[b]*c + Sqrt[-a]*d)])/(Sq 
rt[-a]*(Sqrt[b]*c + Sqrt[-a]*d)*(1 + n))))/(4*a*(b*c^2 + a*d^2))
 

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 299, normalized size of antiderivative = 0.98, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {496, 25, 657, 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.

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

\(\Big \downarrow \) 496

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 657

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

((a*d + b*c*x)*(c + d*x)^(1 + n))/(2*a*(b*c^2 + a*d^2)*(a + b*x^2)) + (((b 
*c^2 + a*d^2*(1 - n) + Sqrt[-a]*Sqrt[b]*c*d*n)*(c + d*x)^(1 + n)*Hypergeom 
etric2F1[1, 1 + n, 2 + n, (Sqrt[b]*(c + d*x))/(Sqrt[b]*c - Sqrt[-a]*d)])/( 
2*Sqrt[-a]*(Sqrt[b]*c - Sqrt[-a]*d)*(1 + n)) - ((b*c^2 + a*d^2*(1 - n) - S 
qrt[-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 + Sqrt[-a]*d)])/(2*Sqrt[-a]*(Sqrt[b]*c + 
Sqrt[-a]*d)*(1 + n)))/(2*a*(b*c^2 + a*d^2))
 

Defintions of rubi rules used

rule 25

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

rule 496

Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ 
(-(a*d + b*c*x))*(c + d*x)^(n + 1)*((a + b*x^2)^(p + 1)/(2*a*(p + 1)*(b*c^2 
 + a*d^2))), x] + Simp[1/(2*a*(p + 1)*(b*c^2 + a*d^2))   Int[(c + d*x)^n*(a 
 + b*x^2)^(p + 1)*Simp[b*c^2*(2*p + 3) + a*d^2*(n + 2*p + 3) + b*c*d*(n + 2 
*p + 4)*x, x], x], x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[p, -1] && IntQuad 
raticQ[a, 0, b, c, d, n, p, x]
 

rule 657

Int[(((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.))/((a_) + (c_.)*( 
x_)^2), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*((f + g*x)^n/(a + c*x^ 
2)), x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && IntegersQ[n]
 

rule 2009

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

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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