Integrand size = 18, antiderivative size = 168 \[ \int \frac {x (c+d x)^n}{\left (a+b x^2\right )^2} \, dx=-\frac {(c+d x)^n}{2 b \left (a+b x^2\right )}+\frac {d (c+d x)^n \operatorname {Hypergeometric2F1}\left (1,n,1+n,\frac {\sqrt {b} (c+d x)}{\sqrt {b} c-\sqrt {-a} d}\right )}{4 b \left (\sqrt {-a} \sqrt {b} c+a d\right )}-\frac {d (c+d x)^n \operatorname {Hypergeometric2F1}\left (1,n,1+n,\frac {\sqrt {b} (c+d x)}{\sqrt {b} c+\sqrt {-a} d}\right )}{4 b \left (\sqrt {-a} \sqrt {b} c-a d\right )} \] Output:
-1/2*(d*x+c)^n/b/(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))/b/((-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))/b/(( -a)^(1/2)*b^(1/2)*c-a*d)
Time = 0.22 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.37 \[ \int \frac {x (c+d x)^n}{\left (a+b x^2\right )^2} \, dx=\frac {(c+d x)^{1+n} \left (-\frac {2 a b (c-d x)}{a+b x^2}-\frac {\left (\sqrt {-a} b c d n-a \sqrt {b} d^2 n\right ) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt {b} (c+d x)}{\sqrt {b} c-\sqrt {-a} d}\right )}{\left (\sqrt {b} c-\sqrt {-a} d\right ) (1+n)}+\frac {\left (\sqrt {-a} b c d n+a \sqrt {b} d^2 n\right ) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt {b} (c+d x)}{\sqrt {b} c+\sqrt {-a} d}\right )}{\left (\sqrt {b} c+\sqrt {-a} d\right ) (1+n)}\right )}{4 a b \left (b c^2+a d^2\right )} \] Input:
Integrate[(x*(c + d*x)^n)/(a + b*x^2)^2,x]
Output:
((c + d*x)^(1 + n)*((-2*a*b*(c - d*x))/(a + b*x^2) - ((Sqrt[-a]*b*c*d*n - a*Sqrt[b]*d^2*n)*Hypergeometric2F1[1, 1 + n, 2 + n, (Sqrt[b]*(c + d*x))/(S qrt[b]*c - Sqrt[-a]*d)])/((Sqrt[b]*c - Sqrt[-a]*d)*(1 + n)) + ((Sqrt[-a]*b *c*d*n + a*Sqrt[b]*d^2*n)*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))))/(4* a*b*(b*c^2 + a*d^2))
Time = 0.42 (sec) , antiderivative size = 263, normalized size of antiderivative = 1.57, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {593, 27, 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.
\(\displaystyle \int \frac {x (c+d x)^n}{\left (a+b x^2\right )^2} \, dx\) |
\(\Big \downarrow \) 593 |
\(\displaystyle \frac {d \int \frac {n (c-d x) (c+d x)^n}{b x^2+a}dx}{2 \left (a d^2+b c^2\right )}-\frac {(c-d x) (c+d x)^{n+1}}{2 \left (a+b x^2\right ) \left (a d^2+b c^2\right )}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {d n \int \frac {(c-d x) (c+d x)^n}{b x^2+a}dx}{2 \left (a d^2+b c^2\right )}-\frac {(c-d x) (c+d x)^{n+1}}{2 \left (a+b x^2\right ) \left (a d^2+b c^2\right )}\) |
\(\Big \downarrow \) 657 |
\(\displaystyle \frac {d n \int \left (\frac {\left (\sqrt {-a} c+\frac {a d}{\sqrt {b}}\right ) (c+d x)^n}{2 a \left (\sqrt {-a}-\sqrt {b} x\right )}+\frac {\left (\sqrt {-a} c-\frac {a d}{\sqrt {b}}\right ) (c+d x)^n}{2 a \left (\sqrt {b} x+\sqrt {-a}\right )}\right )dx}{2 \left (a d^2+b c^2\right )}-\frac {(c-d x) (c+d x)^{n+1}}{2 \left (a+b x^2\right ) \left (a d^2+b c^2\right )}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {d n \left (\frac {\left (\frac {\sqrt {-a} d}{\sqrt {b}}+c\right ) (c+d x)^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {\sqrt {b} (c+d x)}{\sqrt {b} c-\sqrt {-a} d}\right )}{2 \sqrt {-a} (n+1) \left (\sqrt {b} c-\sqrt {-a} d\right )}+\frac {\left (\sqrt {-a} \sqrt {b} c+a d\right ) (c+d x)^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {\sqrt {b} (c+d x)}{\sqrt {b} c+\sqrt {-a} d}\right )}{2 a \sqrt {b} (n+1) \left (\sqrt {-a} d+\sqrt {b} c\right )}\right )}{2 \left (a d^2+b c^2\right )}-\frac {(c-d x) (c+d x)^{n+1}}{2 \left (a+b x^2\right ) \left (a d^2+b c^2\right )}\) |
Input:
Int[(x*(c + d*x)^n)/(a + b*x^2)^2,x]
Output:
-1/2*((c - d*x)*(c + d*x)^(1 + n))/((b*c^2 + a*d^2)*(a + b*x^2)) + (d*n*(( (c + (Sqrt[-a]*d)/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*Sqrt[-a]*(Sqrt[b]* c - Sqrt[-a]*d)*(1 + n)) + ((Sqrt[-a]*Sqrt[b]*c + a*d)*(c + d*x)^(1 + n)*H ypergeometric2F1[1, 1 + n, 2 + n, (Sqrt[b]*(c + d*x))/(Sqrt[b]*c + Sqrt[-a ]*d)])/(2*a*Sqrt[b]*(Sqrt[b]*c + Sqrt[-a]*d)*(1 + n))))/(2*(b*c^2 + a*d^2) )
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(x_)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c + d*x)^(n + 1)*(c - d*x)*((a + b*x^2)^(p + 1)/(2*(p + 1)*(b*c^2 + a*d^2))), x] - Simp[d/(2*(p + 1)*(b*c^2 + a*d^2)) Int[(c + d*x)^n*(a + b* x^2)^(p + 1)*(c*n - d*(n + 2*p + 4)*x), x], x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[p, -1] && NeQ[b*c^2 + a*d^2, 0]
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]
\[\int \frac {x \left (d x +c \right )^{n}}{\left (b \,x^{2}+a \right )^{2}}d x\]
Input:
int(x*(d*x+c)^n/(b*x^2+a)^2,x)
Output:
int(x*(d*x+c)^n/(b*x^2+a)^2,x)
\[ \int \frac {x (c+d x)^n}{\left (a+b x^2\right )^2} \, dx=\int { \frac {{\left (d x + c\right )}^{n} x}{{\left (b x^{2} + a\right )}^{2}} \,d x } \] Input:
integrate(x*(d*x+c)^n/(b*x^2+a)^2,x, algorithm="fricas")
Output:
integral((d*x + c)^n*x/(b^2*x^4 + 2*a*b*x^2 + a^2), x)
\[ \int \frac {x (c+d x)^n}{\left (a+b x^2\right )^2} \, dx=\int \frac {x \left (c + d x\right )^{n}}{\left (a + b x^{2}\right )^{2}}\, dx \] Input:
integrate(x*(d*x+c)**n/(b*x**2+a)**2,x)
Output:
Integral(x*(c + d*x)**n/(a + b*x**2)**2, x)
\[ \int \frac {x (c+d x)^n}{\left (a+b x^2\right )^2} \, dx=\int { \frac {{\left (d x + c\right )}^{n} x}{{\left (b x^{2} + a\right )}^{2}} \,d x } \] Input:
integrate(x*(d*x+c)^n/(b*x^2+a)^2,x, algorithm="maxima")
Output:
integrate((d*x + c)^n*x/(b*x^2 + a)^2, x)
\[ \int \frac {x (c+d x)^n}{\left (a+b x^2\right )^2} \, dx=\int { \frac {{\left (d x + c\right )}^{n} x}{{\left (b x^{2} + a\right )}^{2}} \,d x } \] Input:
integrate(x*(d*x+c)^n/(b*x^2+a)^2,x, algorithm="giac")
Output:
integrate((d*x + c)^n*x/(b*x^2 + a)^2, x)
Timed out. \[ \int \frac {x (c+d x)^n}{\left (a+b x^2\right )^2} \, dx=\int \frac {x\,{\left (c+d\,x\right )}^n}{{\left (b\,x^2+a\right )}^2} \,d x \] Input:
int((x*(c + d*x)^n)/(a + b*x^2)^2,x)
Output:
int((x*(c + d*x)^n)/(a + b*x^2)^2, x)
\[ \int \frac {x (c+d x)^n}{\left (a+b x^2\right )^2} \, dx=\frac {-\left (d x +c \right )^{n}+\left (\int \frac {\left (d x +c \right )^{n}}{b d \,x^{3}+b c \,x^{2}+a d x +a c}d x \right ) a d n +\left (\int \frac {\left (d x +c \right )^{n}}{b d \,x^{3}+b c \,x^{2}+a d x +a c}d x \right ) b d n \,x^{2}}{2 b \left (b \,x^{2}+a \right )} \] Input:
int(x*(d*x+c)^n/(b*x^2+a)^2,x)
Output:
( - (c + d*x)**n + int((c + d*x)**n/(a*c + a*d*x + b*c*x**2 + b*d*x**3),x) *a*d*n + int((c + d*x)**n/(a*c + a*d*x + b*c*x**2 + b*d*x**3),x)*b*d*n*x** 2)/(2*b*(a + b*x**2))