\(\int \frac {(c+d x)^n}{x (a+b x^2)^2} \, dx\) [226]

Optimal result

Integrand size = 20, antiderivative size = 350 \[ \int \frac {(c+d x)^n}{x \left (a+b x^2\right )^2} \, dx=\frac {(c+d x)^n}{2 a \left (a+b x^2\right )}-\frac {d (c+d x)^n \operatorname {Hypergeometric2F1}\left (1,n,1+n,\frac {\sqrt {b} (c+d x)}{\sqrt {b} c-\sqrt {-a} d}\right )}{4 a \left (\sqrt {-a} \sqrt {b} c+a d\right )}+\frac {d (c+d x)^n \operatorname {Hypergeometric2F1}\left (1,n,1+n,\frac {\sqrt {b} (c+d x)}{\sqrt {b} c+\sqrt {-a} d}\right )}{4 a \left (\sqrt {-a} \sqrt {b} c-a d\right )}+\frac {(c+d x)^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {c+d x}{c-\frac {\sqrt {-a} d}{\sqrt {b}}}\right )}{2 a^2 \left (c-\frac {\sqrt {-a} d}{\sqrt {b}}\right ) (1+n)}+\frac {(c+d x)^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {c+d x}{c+\frac {\sqrt {-a} d}{\sqrt {b}}}\right )}{2 a^2 \left (c+\frac {\sqrt {-a} d}{\sqrt {b}}\right ) (1+n)}-\frac {(c+d x)^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,1+\frac {d x}{c}\right )}{a^2 c (1+n)} \] Output:

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

Mathematica [A] (verified)

Time = 0.52 (sec) , antiderivative size = 391, normalized size of antiderivative = 1.12 \[ \int \frac {(c+d x)^n}{x \left (a+b x^2\right )^2} \, dx=\frac {(c+d x)^{1+n} \left (\frac {2 a b (c-d x)}{\left (b c^2+a d^2\right ) \left (a+b x^2\right )}-\frac {4 \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {c+d x}{c}\right )}{c+c n}+\frac {2 \sqrt {b} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt {b} (c+d x)}{\sqrt {b} c-\sqrt {-a} d}\right )}{\left (\sqrt {b} c-\sqrt {-a} d\right ) (1+n)}+\frac {2 \sqrt {b} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt {b} (c+d x)}{\sqrt {b} c+\sqrt {-a} d}\right )}{\left (\sqrt {b} c+\sqrt {-a} d\right ) (1+n)}+\frac {\sqrt {b} d n \left (\left (\sqrt {-a} b c^2-2 a \sqrt {b} c d+(-a)^{3/2} d^2\right ) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt {b} (c+d x)}{\sqrt {b} c-\sqrt {-a} d}\right )+\left (-\sqrt {-a} b c^2-2 a \sqrt {b} c d+\sqrt {-a} a d^2\right ) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt {b} (c+d x)}{\sqrt {b} c+\sqrt {-a} d}\right )\right )}{\left (b c^2+a d^2\right )^2 (1+n)}\right )}{4 a^2} \] Input:

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

Output:

((c + d*x)^(1 + n)*((2*a*b*(c - d*x))/((b*c^2 + a*d^2)*(a + b*x^2)) - (4*H 
ypergeometric2F1[1, 1 + n, 2 + n, (c + d*x)/c])/(c + c*n) + (2*Sqrt[b]*Hyp 
ergeometric2F1[1, 1 + n, 2 + n, (Sqrt[b]*(c + d*x))/(Sqrt[b]*c - Sqrt[-a]* 
d)])/((Sqrt[b]*c - Sqrt[-a]*d)*(1 + n)) + (2*Sqrt[b]*Hypergeometric2F1[1, 
1 + n, 2 + n, (Sqrt[b]*(c + d*x))/(Sqrt[b]*c + Sqrt[-a]*d)])/((Sqrt[b]*c + 
 Sqrt[-a]*d)*(1 + n)) + (Sqrt[b]*d*n*((Sqrt[-a]*b*c^2 - 2*a*Sqrt[b]*c*d + 
(-a)^(3/2)*d^2)*Hypergeometric2F1[1, 1 + n, 2 + n, (Sqrt[b]*(c + d*x))/(Sq 
rt[b]*c - Sqrt[-a]*d)] + (-(Sqrt[-a]*b*c^2) - 2*a*Sqrt[b]*c*d + Sqrt[-a]*a 
*d^2)*Hypergeometric2F1[1, 1 + n, 2 + n, (Sqrt[b]*(c + d*x))/(Sqrt[b]*c + 
Sqrt[-a]*d)]))/((b*c^2 + a*d^2)^2*(1 + n))))/(4*a^2)
 

Rubi [A] (verified)

Time = 0.81 (sec) , antiderivative size = 484, normalized size of antiderivative = 1.38, 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.

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

\(\Big \downarrow \) 615

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

(b*(c - d*x)*(c + d*x)^(1 + n))/(2*a*(b*c^2 + a*d^2)*(a + b*x^2)) + (Sqrt[ 
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^2*(Sqrt[b]*c - Sqrt[-a]*d)*(1 + n)) + (b 
*d*(c + (Sqrt[-a]*d)/Sqrt[b])*n*(c + d*x)^(1 + n)*Hypergeometric2F1[1, 1 + 
 n, 2 + n, (Sqrt[b]*(c + d*x))/(Sqrt[b]*c - Sqrt[-a]*d)])/(4*(-a)^(3/2)*(S 
qrt[b]*c - Sqrt[-a]*d)*(b*c^2 + a*d^2)*(1 + n)) + (Sqrt[b]*(c + d*x)^(1 + 
n)*Hypergeometric2F1[1, 1 + n, 2 + n, (Sqrt[b]*(c + d*x))/(Sqrt[b]*c + Sqr 
t[-a]*d)])/(2*a^2*(Sqrt[b]*c + Sqrt[-a]*d)*(1 + n)) - (Sqrt[b]*d*(Sqrt[-a] 
*Sqrt[b]*c + a*d)*n*(c + d*x)^(1 + n)*Hypergeometric2F1[1, 1 + n, 2 + n, ( 
Sqrt[b]*(c + d*x))/(Sqrt[b]*c + Sqrt[-a]*d)])/(4*a^2*(Sqrt[b]*c + Sqrt[-a] 
*d)*(b*c^2 + a*d^2)*(1 + n)) - ((c + d*x)^(1 + n)*Hypergeometric2F1[1, 1 + 
 n, 2 + n, 1 + (d*x)/c])/(a^2*c*(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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral((d*x + c)^n/(b^2*x^5 + 2*a*b*x^3 + a^2*x), x)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {(c+d x)^n}{x \left (a+b x^2\right )^2} \, dx=\int \frac {\left (d x +c \right )^{n}}{b^{2} x^{5}+2 a b \,x^{3}+a^{2} x}d x \] Input:

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

Output:

int((c + d*x)**n/(a**2*x + 2*a*b*x**3 + b**2*x**5),x)