\(\int \frac {(c+d x)^n}{x (a+b x^2)} \, dx\) [219]

Optimal result

Integrand size = 20, antiderivative size = 183 \[ \int \frac {(c+d x)^n}{x \left (a+b x^2\right )} \, dx=\frac {(c+d x)^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {c+d x}{c-\frac {\sqrt {-a} d}{\sqrt {b}}}\right )}{2 a \left (c-\frac {\sqrt {-a} d}{\sqrt {b}}\right ) (1+n)}+\frac {(c+d x)^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {c+d x}{c+\frac {\sqrt {-a} d}{\sqrt {b}}}\right )}{2 a \left (c+\frac {\sqrt {-a} d}{\sqrt {b}}\right ) (1+n)}-\frac {(c+d x)^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,1+\frac {d x}{c}\right )}{a c (1+n)} \] Output:

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/(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/(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/c/(1+n)
 

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.03 \[ \int \frac {(c+d x)^n}{x \left (a+b x^2\right )} \, dx=\frac {(c+d x)^{1+n} \left (\left (b c^2+\sqrt {-a} \sqrt {b} c d\right ) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt {b} (c+d x)}{\sqrt {b} c-\sqrt {-a} d}\right )+\left (b c^2-\sqrt {-a} \sqrt {b} c d\right ) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt {b} (c+d x)}{\sqrt {b} c+\sqrt {-a} d}\right )-2 \left (b c^2+a d^2\right ) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,1+\frac {d x}{c}\right )\right )}{2 a c \left (b c^2+a d^2\right ) (1+n)} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.13, 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.

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

\(\Big \downarrow \) 615

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

(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*(Sqrt[b]*c - Sqrt[-a]*d)*(1 + n)) 
+ (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*(Sqrt[b]*c + Sqrt[-a]*d)*(1 + n) 
) - ((c + d*x)^(1 + n)*Hypergeometric2F1[1, 1 + n, 2 + n, 1 + (d*x)/c])/(a 
*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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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