Integrand size = 24, antiderivative size = 202 \[ \int \frac {(A+B x) (d+e x)^{1+m}}{a+c x^2} \, dx=-\frac {\left (a B+\sqrt {-a} A \sqrt {c}\right ) (d+e x)^{2+m} \operatorname {Hypergeometric2F1}\left (1,2+m,3+m,\frac {\sqrt {c} (d+e x)}{\sqrt {c} d-\sqrt {-a} e}\right )}{2 a \sqrt {c} \left (\sqrt {c} d-\sqrt {-a} e\right ) (2+m)}-\frac {\left (A+\frac {\sqrt {-a} B}{\sqrt {c}}\right ) (d+e x)^{2+m} \operatorname {Hypergeometric2F1}\left (1,2+m,3+m,\frac {\sqrt {c} (d+e x)}{\sqrt {c} d+\sqrt {-a} e}\right )}{2 \sqrt {-a} \left (\sqrt {c} d+\sqrt {-a} e\right ) (2+m)} \]
-1/2*(e*x+d)^(2+m)*hypergeom([1, 2+m],[3+m],(e*x+d)*c^(1/2)/(e*(-a)^(1/2)+ d*c^(1/2)))*(A+B*(-a)^(1/2)/c^(1/2))/(2+m)/(-a)^(1/2)/(e*(-a)^(1/2)+d*c^(1 /2))-1/2*(e*x+d)^(2+m)*hypergeom([1, 2+m],[3+m],(e*x+d)*c^(1/2)/(-e*(-a)^( 1/2)+d*c^(1/2)))*(B*a+A*(-a)^(1/2)*c^(1/2))/a/(2+m)/c^(1/2)/(-e*(-a)^(1/2) +d*c^(1/2))
Time = 0.29 (sec) , antiderivative size = 182, normalized size of antiderivative = 0.90 \[ \int \frac {(A+B x) (d+e x)^{1+m}}{a+c x^2} \, dx=\frac {(d+e x)^{2+m} \left (\frac {\left (a B+\sqrt {-a} A \sqrt {c}\right ) \operatorname {Hypergeometric2F1}\left (1,2+m,3+m,\frac {\sqrt {c} (d+e x)}{\sqrt {c} d-\sqrt {-a} e}\right )}{-\sqrt {c} d+\sqrt {-a} e}+\frac {\left (-a B+\sqrt {-a} A \sqrt {c}\right ) \operatorname {Hypergeometric2F1}\left (1,2+m,3+m,\frac {\sqrt {c} (d+e x)}{\sqrt {c} d+\sqrt {-a} e}\right )}{\sqrt {c} d+\sqrt {-a} e}\right )}{2 a \sqrt {c} (2+m)} \]
((d + e*x)^(2 + m)*(((a*B + Sqrt[-a]*A*Sqrt[c])*Hypergeometric2F1[1, 2 + m , 3 + m, (Sqrt[c]*(d + e*x))/(Sqrt[c]*d - Sqrt[-a]*e)])/(-(Sqrt[c]*d) + Sq rt[-a]*e) + ((-(a*B) + Sqrt[-a]*A*Sqrt[c])*Hypergeometric2F1[1, 2 + m, 3 + m, (Sqrt[c]*(d + e*x))/(Sqrt[c]*d + Sqrt[-a]*e)])/(Sqrt[c]*d + Sqrt[-a]*e )))/(2*a*Sqrt[c]*(2 + m))
Time = 0.36 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {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.
\(\displaystyle \int \frac {(A+B x) (d+e x)^{m+1}}{a+c x^2} \, dx\) |
\(\Big \downarrow \) 657 |
\(\displaystyle \int \left (\frac {\left (\sqrt {-a} A-\frac {a B}{\sqrt {c}}\right ) (d+e x)^{m+1}}{2 a \left (\sqrt {-a}-\sqrt {c} x\right )}+\frac {\left (\sqrt {-a} A+\frac {a B}{\sqrt {c}}\right ) (d+e x)^{m+1}}{2 a \left (\sqrt {-a}+\sqrt {c} x\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\left (\sqrt {-a} A \sqrt {c}+a B\right ) (d+e x)^{m+2} \operatorname {Hypergeometric2F1}\left (1,m+2,m+3,\frac {\sqrt {c} (d+e x)}{\sqrt {c} d-\sqrt {-a} e}\right )}{2 a \sqrt {c} (m+2) \left (\sqrt {c} d-\sqrt {-a} e\right )}-\frac {\left (\frac {\sqrt {-a} B}{\sqrt {c}}+A\right ) (d+e x)^{m+2} \operatorname {Hypergeometric2F1}\left (1,m+2,m+3,\frac {\sqrt {c} (d+e x)}{\sqrt {c} d+\sqrt {-a} e}\right )}{2 \sqrt {-a} (m+2) \left (\sqrt {-a} e+\sqrt {c} d\right )}\) |
-1/2*((a*B + Sqrt[-a]*A*Sqrt[c])*(d + e*x)^(2 + m)*Hypergeometric2F1[1, 2 + m, 3 + m, (Sqrt[c]*(d + e*x))/(Sqrt[c]*d - Sqrt[-a]*e)])/(a*Sqrt[c]*(Sqr t[c]*d - Sqrt[-a]*e)*(2 + m)) - ((A + (Sqrt[-a]*B)/Sqrt[c])*(d + e*x)^(2 + m)*Hypergeometric2F1[1, 2 + m, 3 + m, (Sqrt[c]*(d + e*x))/(Sqrt[c]*d + Sq rt[-a]*e)])/(2*Sqrt[-a]*(Sqrt[c]*d + Sqrt[-a]*e)*(2 + m))
3.15.92.3.1 Defintions of rubi rules used
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]
\[\int \frac {\left (B x +A \right ) \left (e x +d \right )^{1+m}}{c \,x^{2}+a}d x\]
\[ \int \frac {(A+B x) (d+e x)^{1+m}}{a+c x^2} \, dx=\int { \frac {{\left (B x + A\right )} {\left (e x + d\right )}^{m + 1}}{c x^{2} + a} \,d x } \]
\[ \int \frac {(A+B x) (d+e x)^{1+m}}{a+c x^2} \, dx=\int \frac {\left (A + B x\right ) \left (d + e x\right )^{m + 1}}{a + c x^{2}}\, dx \]
\[ \int \frac {(A+B x) (d+e x)^{1+m}}{a+c x^2} \, dx=\int { \frac {{\left (B x + A\right )} {\left (e x + d\right )}^{m + 1}}{c x^{2} + a} \,d x } \]
\[ \int \frac {(A+B x) (d+e x)^{1+m}}{a+c x^2} \, dx=\int { \frac {{\left (B x + A\right )} {\left (e x + d\right )}^{m + 1}}{c x^{2} + a} \,d x } \]
Timed out. \[ \int \frac {(A+B x) (d+e x)^{1+m}}{a+c x^2} \, dx=\int \frac {\left (A+B\,x\right )\,{\left (d+e\,x\right )}^{m+1}}{c\,x^2+a} \,d x \]