\(\int \frac {(A+B x) (d+e x)^{1+m}}{a+c x^2} \, dx\) [300]

Optimal result

Integrand size = 24, antiderivative size = 212 \[ \int \frac {(A+B x) (d+e x)^{1+m}}{a+c x^2} \, dx=-\frac {\left (\sqrt {-a} B-A \sqrt {c}\right ) (d+e x)^{2+m} \operatorname {Hypergeometric2F1}\left (1,2+m,3+m,\frac {\sqrt {c} (d+e x)}{\sqrt {c} d-\sqrt {-a} e}\right )}{2 \sqrt {-a} \sqrt {c} \left (\sqrt {c} d-\sqrt {-a} e\right ) (2+m)}-\frac {\left (\sqrt {-a} B+A \sqrt {c}\right ) (d+e x)^{2+m} \operatorname {Hypergeometric2F1}\left (1,2+m,3+m,\frac {\sqrt {c} (d+e x)}{\sqrt {c} d+\sqrt {-a} e}\right )}{2 \sqrt {-a} \sqrt {c} \left (\sqrt {c} d+\sqrt {-a} e\right ) (2+m)} \] Output:

-1/2*((-a)^(1/2)*B-A*c^(1/2))*(e*x+d)^(2+m)*hypergeom([1, 2+m],[3+m],c^(1/ 
2)*(e*x+d)/(c^(1/2)*d-(-a)^(1/2)*e))/(-a)^(1/2)/c^(1/2)/(c^(1/2)*d-(-a)^(1 
/2)*e)/(2+m)-1/2*((-a)^(1/2)*B+A*c^(1/2))*(e*x+d)^(2+m)*hypergeom([1, 2+m] 
,[3+m],c^(1/2)*(e*x+d)/(c^(1/2)*d+(-a)^(1/2)*e))/(-a)^(1/2)/c^(1/2)/(c^(1/ 
2)*d+(-a)^(1/2)*e)/(2+m)
 

Mathematica [A] (verified)

Time = 0.29 (sec) , antiderivative size = 182, normalized size of antiderivative = 0.86 \[ \int \frac {(A+B x) (d+e x)^{1+m}}{a+c x^2} \, dx=\frac {(d+e x)^{2+m} \left (\frac {\left (a B+\sqrt {-a} A \sqrt {c}\right ) \operatorname {Hypergeometric2F1}\left (1,2+m,3+m,\frac {\sqrt {c} (d+e x)}{\sqrt {c} d-\sqrt {-a} e}\right )}{-\sqrt {c} d+\sqrt {-a} e}+\frac {\left (-a B+\sqrt {-a} A \sqrt {c}\right ) \operatorname {Hypergeometric2F1}\left (1,2+m,3+m,\frac {\sqrt {c} (d+e x)}{\sqrt {c} d+\sqrt {-a} e}\right )}{\sqrt {c} d+\sqrt {-a} e}\right )}{2 a \sqrt {c} (2+m)} \] Input:

Integrate[((A + B*x)*(d + e*x)^(1 + m))/(a + c*x^2),x]
 

Output:

((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))
 

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 202, normalized size of antiderivative = 0.95, 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

\(\displaystyle -\frac {\left (\sqrt {-a} A \sqrt {c}+a B\right ) (d+e x)^{m+2} \operatorname {Hypergeometric2F1}\left (1,m+2,m+3,\frac {\sqrt {c} (d+e x)}{\sqrt {c} d-\sqrt {-a} e}\right )}{2 a \sqrt {c} (m+2) \left (\sqrt {c} d-\sqrt {-a} e\right )}-\frac {\left (\frac {\sqrt {-a} B}{\sqrt {c}}+A\right ) (d+e x)^{m+2} \operatorname {Hypergeometric2F1}\left (1,m+2,m+3,\frac {\sqrt {c} (d+e x)}{\sqrt {c} d+\sqrt {-a} e}\right )}{2 \sqrt {-a} (m+2) \left (\sqrt {-a} e+\sqrt {c} d\right )}\)

Input:

Int[((A + B*x)*(d + e*x)^(1 + m))/(a + c*x^2),x]
 

Output:

-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))
 

Defintions of rubi rules used

rule 657

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]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

Maple [F]

\[\int \frac {\left (B x +A \right ) \left (e x +d \right )^{1+m}}{c \,x^{2}+a}d x\]

Input:

int((B*x+A)*(e*x+d)^(1+m)/(c*x^2+a),x)
 

Output:

int((B*x+A)*(e*x+d)^(1+m)/(c*x^2+a),x)
 

Fricas [F]

\[ \int \frac {(A+B x) (d+e x)^{1+m}}{a+c x^2} \, dx=\int { \frac {{\left (B x + A\right )} {\left (e x + d\right )}^{m + 1}}{c x^{2} + a} \,d x } \] Input:

integrate((B*x+A)*(e*x+d)^(1+m)/(c*x^2+a),x, algorithm="fricas")
 

Output:

integral((B*x + A)*(e*x + d)^(m + 1)/(c*x^2 + a), x)
 

Sympy [F]

\[ \int \frac {(A+B x) (d+e x)^{1+m}}{a+c x^2} \, dx=\int \frac {\left (A + B x\right ) \left (d + e x\right )^{m + 1}}{a + c x^{2}}\, dx \] Input:

integrate((B*x+A)*(e*x+d)**(1+m)/(c*x**2+a),x)
 

Output:

Integral((A + B*x)*(d + e*x)**(m + 1)/(a + c*x**2), x)
 

Maxima [F]

\[ \int \frac {(A+B x) (d+e x)^{1+m}}{a+c x^2} \, dx=\int { \frac {{\left (B x + A\right )} {\left (e x + d\right )}^{m + 1}}{c x^{2} + a} \,d x } \] Input:

integrate((B*x+A)*(e*x+d)^(1+m)/(c*x^2+a),x, algorithm="maxima")
 

Output:

integrate((B*x + A)*(e*x + d)^(m + 1)/(c*x^2 + a), x)
 

Giac [F]

\[ \int \frac {(A+B x) (d+e x)^{1+m}}{a+c x^2} \, dx=\int { \frac {{\left (B x + A\right )} {\left (e x + d\right )}^{m + 1}}{c x^{2} + a} \,d x } \] Input:

integrate((B*x+A)*(e*x+d)^(1+m)/(c*x^2+a),x, algorithm="giac")
 

Output:

integrate((B*x + A)*(e*x + d)^(m + 1)/(c*x^2 + a), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(A+B x) (d+e x)^{1+m}}{a+c x^2} \, dx=\int \frac {\left (A+B\,x\right )\,{\left (d+e\,x\right )}^{m+1}}{c\,x^2+a} \,d x \] Input:

int(((A + B*x)*(d + e*x)^(m + 1))/(a + c*x^2),x)
 

Output:

int(((A + B*x)*(d + e*x)^(m + 1))/(a + c*x^2), x)
 

Reduce [F]

\[ \int \frac {(A+B x) (d+e x)^{1+m}}{a+c x^2} \, dx=\frac {\left (e x +d \right )^{m} a e m +\left (e x +d \right )^{m} a e +2 \left (e x +d \right )^{m} b d m +\left (e x +d \right )^{m} b d +\left (e x +d \right )^{m} b e m x -\left (\int \frac {\left (e x +d \right )^{m}}{c e \,x^{3}+c d \,x^{2}+a e x +a d}d x \right ) a^{2} e^{2} m^{2}-\left (\int \frac {\left (e x +d \right )^{m}}{c e \,x^{3}+c d \,x^{2}+a e x +a d}d x \right ) a^{2} e^{2} m -2 \left (\int \frac {\left (e x +d \right )^{m}}{c e \,x^{3}+c d \,x^{2}+a e x +a d}d x \right ) a b d e \,m^{2}-2 \left (\int \frac {\left (e x +d \right )^{m}}{c e \,x^{3}+c d \,x^{2}+a e x +a d}d x \right ) a b d e m +\left (\int \frac {\left (e x +d \right )^{m}}{c e \,x^{3}+c d \,x^{2}+a e x +a d}d x \right ) a c \,d^{2} m^{2}+\left (\int \frac {\left (e x +d \right )^{m}}{c e \,x^{3}+c d \,x^{2}+a e x +a d}d x \right ) a c \,d^{2} m -\left (\int \frac {\left (e x +d \right )^{m} x}{c e \,x^{3}+c d \,x^{2}+a e x +a d}d x \right ) a b \,e^{2} m^{2}-\left (\int \frac {\left (e x +d \right )^{m} x}{c e \,x^{3}+c d \,x^{2}+a e x +a d}d x \right ) a b \,e^{2} m +2 \left (\int \frac {\left (e x +d \right )^{m} x}{c e \,x^{3}+c d \,x^{2}+a e x +a d}d x \right ) a c d e \,m^{2}+2 \left (\int \frac {\left (e x +d \right )^{m} x}{c e \,x^{3}+c d \,x^{2}+a e x +a d}d x \right ) a c d e m +\left (\int \frac {\left (e x +d \right )^{m} x}{c e \,x^{3}+c d \,x^{2}+a e x +a d}d x \right ) b c \,d^{2} m^{2}+\left (\int \frac {\left (e x +d \right )^{m} x}{c e \,x^{3}+c d \,x^{2}+a e x +a d}d x \right ) b c \,d^{2} m}{c m \left (m +1\right )} \] Input:

int((B*x+A)*(e*x+d)^(1+m)/(c*x^2+a),x)
 

Output:

((d + e*x)**m*a*e*m + (d + e*x)**m*a*e + 2*(d + e*x)**m*b*d*m + (d + e*x)* 
*m*b*d + (d + e*x)**m*b*e*m*x - int((d + e*x)**m/(a*d + a*e*x + c*d*x**2 + 
 c*e*x**3),x)*a**2*e**2*m**2 - int((d + e*x)**m/(a*d + a*e*x + c*d*x**2 + 
c*e*x**3),x)*a**2*e**2*m - 2*int((d + e*x)**m/(a*d + a*e*x + c*d*x**2 + c* 
e*x**3),x)*a*b*d*e*m**2 - 2*int((d + e*x)**m/(a*d + a*e*x + c*d*x**2 + c*e 
*x**3),x)*a*b*d*e*m + int((d + e*x)**m/(a*d + a*e*x + c*d*x**2 + c*e*x**3) 
,x)*a*c*d**2*m**2 + int((d + e*x)**m/(a*d + a*e*x + c*d*x**2 + c*e*x**3),x 
)*a*c*d**2*m - int(((d + e*x)**m*x)/(a*d + a*e*x + c*d*x**2 + c*e*x**3),x) 
*a*b*e**2*m**2 - int(((d + e*x)**m*x)/(a*d + a*e*x + c*d*x**2 + c*e*x**3), 
x)*a*b*e**2*m + 2*int(((d + e*x)**m*x)/(a*d + a*e*x + c*d*x**2 + c*e*x**3) 
,x)*a*c*d*e*m**2 + 2*int(((d + e*x)**m*x)/(a*d + a*e*x + c*d*x**2 + c*e*x* 
*3),x)*a*c*d*e*m + int(((d + e*x)**m*x)/(a*d + a*e*x + c*d*x**2 + c*e*x**3 
),x)*b*c*d**2*m**2 + int(((d + e*x)**m*x)/(a*d + a*e*x + c*d*x**2 + c*e*x* 
*3),x)*b*c*d**2*m)/(c*m*(m + 1))