Integrand size = 17, antiderivative size = 304 \[ \int \frac {(d+e x)^m}{\left (a+c x^2\right )^2} \, dx=\frac {(a e+c d x) (d+e x)^{1+m}}{2 a \left (c d^2+a e^2\right ) \left (a+c x^2\right )}-\frac {\left (c d^2+a e^2 (1-m)+\sqrt {-a} \sqrt {c} d e m\right ) (d+e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {\sqrt {c} (d+e x)}{\sqrt {c} d-\sqrt {-a} e}\right )}{4 (-a)^{3/2} \left (\sqrt {c} d-\sqrt {-a} e\right ) \left (c d^2+a e^2\right ) (1+m)}+\frac {\left (c d^2+a e^2 (1-m)-\sqrt {-a} \sqrt {c} d e m\right ) (d+e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {\sqrt {c} (d+e x)}{\sqrt {c} d+\sqrt {-a} e}\right )}{4 (-a)^{3/2} \left (\sqrt {c} d+\sqrt {-a} e\right ) \left (c d^2+a e^2\right ) (1+m)} \]
1/2*(c*d*x+a*e)*(e*x+d)^(1+m)/a/(a*e^2+c*d^2)/(c*x^2+a)+1/4*(e*x+d)^(1+m)* hypergeom([1, 1+m],[2+m],(e*x+d)*c^(1/2)/(e*(-a)^(1/2)+d*c^(1/2)))*(c*d^2+ a*e^2*(1-m)-d*e*m*(-a)^(1/2)*c^(1/2))/(-a)^(3/2)/(a*e^2+c*d^2)/(1+m)/(e*(- a)^(1/2)+d*c^(1/2))-1/4*(e*x+d)^(1+m)*hypergeom([1, 1+m],[2+m],(e*x+d)*c^( 1/2)/(-e*(-a)^(1/2)+d*c^(1/2)))*(c*d^2+a*e^2*(1-m)+d*e*m*(-a)^(1/2)*c^(1/2 ))/(-a)^(3/2)/(a*e^2+c*d^2)/(1+m)/(-e*(-a)^(1/2)+d*c^(1/2))
Time = 0.44 (sec) , antiderivative size = 253, normalized size of antiderivative = 0.83 \[ \int \frac {(d+e x)^m}{\left (a+c x^2\right )^2} \, dx=\frac {(d+e x)^{1+m} \left (\frac {2 (a e+c d x)}{a+c x^2}+\frac {\left (c d^2-a e^2 (-1+m)+\sqrt {-a} \sqrt {c} d e m\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {\sqrt {c} (d+e x)}{\sqrt {c} d-\sqrt {-a} e}\right )}{\sqrt {-a} \left (\sqrt {c} d-\sqrt {-a} e\right ) (1+m)}+\frac {\left (-c d^2+a e^2 (-1+m)+\sqrt {-a} \sqrt {c} d e m\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {\sqrt {c} (d+e x)}{\sqrt {c} d+\sqrt {-a} e}\right )}{\sqrt {-a} \left (\sqrt {c} d+\sqrt {-a} e\right ) (1+m)}\right )}{4 a \left (c d^2+a e^2\right )} \]
((d + e*x)^(1 + m)*((2*(a*e + c*d*x))/(a + c*x^2) + ((c*d^2 - a*e^2*(-1 + m) + Sqrt[-a]*Sqrt[c]*d*e*m)*Hypergeometric2F1[1, 1 + m, 2 + m, (Sqrt[c]*( d + e*x))/(Sqrt[c]*d - Sqrt[-a]*e)])/(Sqrt[-a]*(Sqrt[c]*d - Sqrt[-a]*e)*(1 + m)) + ((-(c*d^2) + a*e^2*(-1 + m) + Sqrt[-a]*Sqrt[c]*d*e*m)*Hypergeomet ric2F1[1, 1 + m, 2 + m, (Sqrt[c]*(d + e*x))/(Sqrt[c]*d + Sqrt[-a]*e)])/(Sq rt[-a]*(Sqrt[c]*d + Sqrt[-a]*e)*(1 + m))))/(4*a*(c*d^2 + a*e^2))
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.
\(\displaystyle \int \frac {(d+e x)^m}{\left (a+c x^2\right )^2} \, dx\) |
\(\Big \downarrow \) 496 |
\(\displaystyle \frac {(d+e x)^{m+1} (a e+c d x)}{2 a \left (a+c x^2\right ) \left (a e^2+c d^2\right )}-\frac {\int -\frac {(d+e x)^m \left (c d^2-c e m x d+a e^2 (1-m)\right )}{c x^2+a}dx}{2 a \left (a e^2+c d^2\right )}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int \frac {(d+e x)^m \left (c d^2-c e m x d+a e^2 (1-m)\right )}{c x^2+a}dx}{2 a \left (a e^2+c d^2\right )}+\frac {(d+e x)^{m+1} (a e+c d x)}{2 a \left (a+c x^2\right ) \left (a e^2+c d^2\right )}\) |
\(\Big \downarrow \) 657 |
\(\displaystyle \frac {\int \left (\frac {\left (\sqrt {-a} \left (c d^2+a e^2 (1-m)\right )+a \sqrt {c} d e m\right ) (d+e x)^m}{2 a \left (\sqrt {-a}-\sqrt {c} x\right )}+\frac {\left (\sqrt {-a} \left (c d^2+a e^2 (1-m)\right )-a \sqrt {c} d e m\right ) (d+e x)^m}{2 a \left (\sqrt {c} x+\sqrt {-a}\right )}\right )dx}{2 a \left (a e^2+c d^2\right )}+\frac {(d+e x)^{m+1} (a e+c d x)}{2 a \left (a+c x^2\right ) \left (a e^2+c d^2\right )}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {(d+e x)^{m+1} \left (\sqrt {-a} \sqrt {c} d e m+a e^2 (1-m)+c d^2\right ) \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,\frac {\sqrt {c} (d+e x)}{\sqrt {c} d-\sqrt {-a} e}\right )}{2 \sqrt {-a} (m+1) \left (\sqrt {c} d-\sqrt {-a} e\right )}-\frac {(d+e x)^{m+1} \left (-\sqrt {-a} \sqrt {c} d e m+a e^2 (1-m)+c d^2\right ) \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,\frac {\sqrt {c} (d+e x)}{\sqrt {c} d+\sqrt {-a} e}\right )}{2 \sqrt {-a} (m+1) \left (\sqrt {-a} e+\sqrt {c} d\right )}}{2 a \left (a e^2+c d^2\right )}+\frac {(d+e x)^{m+1} (a e+c d x)}{2 a \left (a+c x^2\right ) \left (a e^2+c d^2\right )}\) |
((a*e + c*d*x)*(d + e*x)^(1 + m))/(2*a*(c*d^2 + a*e^2)*(a + c*x^2)) + (((c *d^2 + a*e^2*(1 - m) + Sqrt[-a]*Sqrt[c]*d*e*m)*(d + e*x)^(1 + m)*Hypergeom etric2F1[1, 1 + m, 2 + m, (Sqrt[c]*(d + e*x))/(Sqrt[c]*d - Sqrt[-a]*e)])/( 2*Sqrt[-a]*(Sqrt[c]*d - Sqrt[-a]*e)*(1 + m)) - ((c*d^2 + a*e^2*(1 - m) - S qrt[-a]*Sqrt[c]*d*e*m)*(d + e*x)^(1 + m)*Hypergeometric2F1[1, 1 + m, 2 + m , (Sqrt[c]*(d + e*x))/(Sqrt[c]*d + Sqrt[-a]*e)])/(2*Sqrt[-a]*(Sqrt[c]*d + Sqrt[-a]*e)*(1 + m)))/(2*a*(c*d^2 + a*e^2))
3.8.25.3.1 Defintions of rubi rules used
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]
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]
\[\int \frac {\left (e x +d \right )^{m}}{\left (c \,x^{2}+a \right )^{2}}d x\]
\[ \int \frac {(d+e x)^m}{\left (a+c x^2\right )^2} \, dx=\int { \frac {{\left (e x + d\right )}^{m}}{{\left (c x^{2} + a\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {(d+e x)^m}{\left (a+c x^2\right )^2} \, dx=\text {Timed out} \]
\[ \int \frac {(d+e x)^m}{\left (a+c x^2\right )^2} \, dx=\int { \frac {{\left (e x + d\right )}^{m}}{{\left (c x^{2} + a\right )}^{2}} \,d x } \]
\[ \int \frac {(d+e x)^m}{\left (a+c x^2\right )^2} \, dx=\int { \frac {{\left (e x + d\right )}^{m}}{{\left (c x^{2} + a\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {(d+e x)^m}{\left (a+c x^2\right )^2} \, dx=\int \frac {{\left (d+e\,x\right )}^m}{{\left (c\,x^2+a\right )}^2} \,d x \]