\(\int \frac {(A+B x) (d+e x)^m}{(a+c x^2)^2} \, dx\) [299]

Optimal result

Integrand size = 22, antiderivative size = 365 \[ \int \frac {(A+B x) (d+e x)^m}{\left (a+c x^2\right )^2} \, dx=-\frac {(d+e x)^{1+m} (a (B d-A e)-(A c d+a B e) x)}{2 a \left (c d^2+a e^2\right ) \left (a+c x^2\right )}-\frac {\left (\sqrt {-a} e (A c d+a B e) m+\sqrt {c} \left (A c d^2+a A e^2 (1-m)+a B d e m\right )\right ) (d+e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {\sqrt {c} (d+e x)}{\sqrt {c} d-\sqrt {-a} e}\right )}{4 (-a)^{3/2} \sqrt {c} \left (\sqrt {c} d-\sqrt {-a} e\right ) \left (c d^2+a e^2\right ) (1+m)}-\frac {\left (\sqrt {-a} e (A c d+a B e) m-\sqrt {c} \left (A c d^2+a A e^2 (1-m)+a B d e m\right )\right ) (d+e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {\sqrt {c} (d+e x)}{\sqrt {c} d+\sqrt {-a} e}\right )}{4 (-a)^{3/2} \sqrt {c} \left (\sqrt {c} d+\sqrt {-a} e\right ) \left (c d^2+a e^2\right ) (1+m)} \] Output:

-1/2*(e*x+d)^(1+m)*(a*(-A*e+B*d)-(A*c*d+B*a*e)*x)/a/(a*e^2+c*d^2)/(c*x^2+a 
)-1/4*((-a)^(1/2)*e*(A*c*d+B*a*e)*m+c^(1/2)*(A*c*d^2+a*A*e^2*(1-m)+a*B*d*e 
*m))*(e*x+d)^(1+m)*hypergeom([1, 1+m],[2+m],c^(1/2)*(e*x+d)/(c^(1/2)*d-(-a 
)^(1/2)*e))/(-a)^(3/2)/c^(1/2)/(c^(1/2)*d-(-a)^(1/2)*e)/(a*e^2+c*d^2)/(1+m 
)-1/4*((-a)^(1/2)*e*(A*c*d+B*a*e)*m-c^(1/2)*(A*c*d^2+a*A*e^2*(1-m)+a*B*d*e 
*m))*(e*x+d)^(1+m)*hypergeom([1, 1+m],[2+m],c^(1/2)*(e*x+d)/(c^(1/2)*d+(-a 
)^(1/2)*e))/(-a)^(3/2)/c^(1/2)/(c^(1/2)*d+(-a)^(1/2)*e)/(a*e^2+c*d^2)/(1+m 
)
 

Mathematica [A] (verified)

Time = 0.79 (sec) , antiderivative size = 310, normalized size of antiderivative = 0.85 \[ \int \frac {(A+B x) (d+e x)^m}{\left (a+c x^2\right )^2} \, dx=\frac {(d+e x)^{1+m} \left (\frac {2 c (A c d x+a (-B d+A e+B e x))}{a+c x^2}+\frac {\sqrt {c} \left (a e (A c d+a B e) m-\sqrt {-a} \sqrt {c} \left (A c d^2-a A e^2 (-1+m)+a B d e m\right )\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {\sqrt {c} (d+e x)}{\sqrt {c} d-\sqrt {-a} e}\right )}{a \left (\sqrt {c} d-\sqrt {-a} e\right ) (1+m)}+\frac {\sqrt {c} \left (a e (A c d+a B e) m+\sqrt {-a} \sqrt {c} \left (A c d^2-a A e^2 (-1+m)+a B d e m\right )\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {\sqrt {c} (d+e x)}{\sqrt {c} d+\sqrt {-a} e}\right )}{a \left (\sqrt {c} d+\sqrt {-a} e\right ) (1+m)}\right )}{4 a c \left (c d^2+a e^2\right )} \] Input:

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

Output:

((d + e*x)^(1 + m)*((2*c*(A*c*d*x + a*(-(B*d) + A*e + B*e*x)))/(a + c*x^2) 
 + (Sqrt[c]*(a*e*(A*c*d + a*B*e)*m - Sqrt[-a]*Sqrt[c]*(A*c*d^2 - a*A*e^2*( 
-1 + m) + a*B*d*e*m))*Hypergeometric2F1[1, 1 + m, 2 + m, (Sqrt[c]*(d + e*x 
))/(Sqrt[c]*d - Sqrt[-a]*e)])/(a*(Sqrt[c]*d - Sqrt[-a]*e)*(1 + m)) + (Sqrt 
[c]*(a*e*(A*c*d + a*B*e)*m + Sqrt[-a]*Sqrt[c]*(A*c*d^2 - a*A*e^2*(-1 + m) 
+ a*B*d*e*m))*Hypergeometric2F1[1, 1 + m, 2 + m, (Sqrt[c]*(d + e*x))/(Sqrt 
[c]*d + Sqrt[-a]*e)])/(a*(Sqrt[c]*d + Sqrt[-a]*e)*(1 + m))))/(4*a*c*(c*d^2 
 + a*e^2))
 

Rubi [A] (verified)

Time = 0.69 (sec) , antiderivative size = 344, normalized size of antiderivative = 0.94, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {686, 25, 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.

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

\(\Big \downarrow \) 686

\(\displaystyle -\frac {\int -\frac {c (d+e x)^m \left (A c d^2+a B e m d+a A e^2 (1-m)-e (A c d+a B e) m x\right )}{c x^2+a}dx}{2 a c \left (a e^2+c d^2\right )}-\frac {(d+e x)^{m+1} (a (B d-A e)-x (a B e+A c d))}{2 a \left (a+c x^2\right ) \left (a e^2+c d^2\right )}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\int \frac {c (d+e x)^m \left (A c d^2+a B e m d+a A e^2 (1-m)-e (A c d+a B e) m x\right )}{c x^2+a}dx}{2 a c \left (a e^2+c d^2\right )}-\frac {(d+e x)^{m+1} (a (B d-A e)-x (a B e+A c d))}{2 a \left (a+c x^2\right ) \left (a e^2+c d^2\right )}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {(d+e x)^m \left (A c d^2+a B e m d+a A e^2 (1-m)-e (A c d+a B e) m x\right )}{c x^2+a}dx}{2 a \left (a e^2+c d^2\right )}-\frac {(d+e x)^{m+1} (a (B d-A e)-x (a B e+A c d))}{2 a \left (a+c x^2\right ) \left (a e^2+c d^2\right )}\)

\(\Big \downarrow \) 657

\(\displaystyle \frac {\int \left (\frac {\left (\frac {a e (A c d+a B e) m}{\sqrt {c}}+\sqrt {-a} \left (A c d^2+a B e m d+a A e^2 (1-m)\right )\right ) (d+e x)^m}{2 a \left (\sqrt {-a}-\sqrt {c} x\right )}+\frac {\left (\sqrt {-a} \left (A c d^2+a B e m d+a A e^2 (1-m)\right )-\frac {a e (A c d+a B e) m}{\sqrt {c}}\right ) (d+e x)^m}{2 a \left (\sqrt {c} x+\sqrt {-a}\right )}\right )dx}{2 a \left (a e^2+c d^2\right )}-\frac {(d+e x)^{m+1} (a (B d-A e)-x (a B e+A c d))}{2 a \left (a+c x^2\right ) \left (a e^2+c d^2\right )}\)

\(\Big \downarrow \) 2009

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

Input:

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

Output:

-1/2*((d + e*x)^(1 + m)*(a*(B*d - A*e) - (A*c*d + a*B*e)*x))/(a*(c*d^2 + a 
*e^2)*(a + c*x^2)) + ((((a*e*(A*c*d + a*B*e)*m)/Sqrt[c] - Sqrt[-a]*(A*c*d^ 
2 + a*A*e^2*(1 - m) + a*B*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*a*(Sqrt[c]* 
d - Sqrt[-a]*e)*(1 + m)) + (((a*e*(A*c*d + a*B*e)*m)/Sqrt[c] + Sqrt[-a]*(A 
*c*d^2 + a*A*e^2*(1 - m) + a*B*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*a*(Sqr 
t[c]*d + Sqrt[-a]*e)*(1 + m)))/(2*a*(c*d^2 + a*e^2))
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

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 686

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p 
_), x_Symbol] :> Simp[(-(d + e*x)^(m + 1))*(f*a*c*e - a*g*c*d + c*(c*d*f + 
a*e*g)*x)*((a + c*x^2)^(p + 1)/(2*a*c*(p + 1)*(c*d^2 + a*e^2))), x] + Simp[ 
1/(2*a*c*(p + 1)*(c*d^2 + a*e^2))   Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*Sim 
p[f*(c^2*d^2*(2*p + 3) + a*c*e^2*(m + 2*p + 3)) - a*c*d*e*g*m + c*e*(c*d*f 
+ a*e*g)*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x] && LtQ 
[p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])
 

rule 2009

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

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral((B*x + A)*(e*x + d)^m/(c^2*x^4 + 2*a*c*x^2 + a^2), x)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

( - (d + e*x)**m*a*e - (d + e*x)**m*b*d + int((d + e*x)**m/(a**2*d + a**2* 
e*x + 2*a*c*d*x**2 + 2*a*c*e*x**3 + c**2*d*x**4 + c**2*e*x**5),x)*a**3*e** 
2*m + int((d + e*x)**m/(a**2*d + a**2*e*x + 2*a*c*d*x**2 + 2*a*c*e*x**3 + 
c**2*d*x**4 + c**2*e*x**5),x)*a**2*b*d*e*m + 2*int((d + e*x)**m/(a**2*d + 
a**2*e*x + 2*a*c*d*x**2 + 2*a*c*e*x**3 + c**2*d*x**4 + c**2*e*x**5),x)*a** 
2*c*d**2 + int((d + e*x)**m/(a**2*d + a**2*e*x + 2*a*c*d*x**2 + 2*a*c*e*x* 
*3 + c**2*d*x**4 + c**2*e*x**5),x)*a**2*c*e**2*m*x**2 + int((d + e*x)**m/( 
a**2*d + a**2*e*x + 2*a*c*d*x**2 + 2*a*c*e*x**3 + c**2*d*x**4 + c**2*e*x** 
5),x)*a*b*c*d*e*m*x**2 + 2*int((d + e*x)**m/(a**2*d + a**2*e*x + 2*a*c*d*x 
**2 + 2*a*c*e*x**3 + c**2*d*x**4 + c**2*e*x**5),x)*a*c**2*d**2*x**2 + int( 
((d + e*x)**m*x**2)/(a**2*d + a**2*e*x + 2*a*c*d*x**2 + 2*a*c*e*x**3 + c** 
2*d*x**4 + c**2*e*x**5),x)*a**2*c*e**2*m - 2*int(((d + e*x)**m*x**2)/(a**2 
*d + a**2*e*x + 2*a*c*d*x**2 + 2*a*c*e*x**3 + c**2*d*x**4 + c**2*e*x**5),x 
)*a**2*c*e**2 + int(((d + e*x)**m*x**2)/(a**2*d + a**2*e*x + 2*a*c*d*x**2 
+ 2*a*c*e*x**3 + c**2*d*x**4 + c**2*e*x**5),x)*a*b*c*d*e*m + int(((d + e*x 
)**m*x**2)/(a**2*d + a**2*e*x + 2*a*c*d*x**2 + 2*a*c*e*x**3 + c**2*d*x**4 
+ c**2*e*x**5),x)*a*c**2*e**2*m*x**2 - 2*int(((d + e*x)**m*x**2)/(a**2*d + 
 a**2*e*x + 2*a*c*d*x**2 + 2*a*c*e*x**3 + c**2*d*x**4 + c**2*e*x**5),x)*a* 
c**2*e**2*x**2 + int(((d + e*x)**m*x**2)/(a**2*d + a**2*e*x + 2*a*c*d*x**2 
 + 2*a*c*e*x**3 + c**2*d*x**4 + c**2*e*x**5),x)*b*c**2*d*e*m*x**2)/(2*c...