3.4.71 \(\int \frac {x^3 (d+e x)^n}{(a+c x^2)^2} \, dx\) [371]

3.4.71.1 Optimal result

Integrand size = 20, antiderivative size = 297 \[ \int \frac {x^3 (d+e x)^n}{\left (a+c x^2\right )^2} \, dx=\frac {a (d-e x) (d+e x)^{1+n}}{2 c \left (c d^2+a e^2\right ) \left (a+c x^2\right )}+\frac {\left (\sqrt {-a} d e n-\frac {2 c d^2+a e^2 (2+n)}{\sqrt {c}}\right ) (d+e x)^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt {c} (d+e x)}{\sqrt {c} d-\sqrt {-a} e}\right )}{4 c \left (\sqrt {c} d-\sqrt {-a} e\right ) \left (c d^2+a e^2\right ) (1+n)}-\frac {\left (2 c d^2+\sqrt {-a} \sqrt {c} d e n+a e^2 (2+n)\right ) (d+e x)^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt {c} (d+e x)}{\sqrt {c} d+\sqrt {-a} e}\right )}{4 c^{3/2} \left (\sqrt {c} d+\sqrt {-a} e\right ) \left (c d^2+a e^2\right ) (1+n)} \]

output

1/2*a*(-e*x+d)*(e*x+d)^(1+n)/c/(a*e^2+c*d^2)/(c*x^2+a)+1/4*(e*x+d)^(1+n)*h 
ypergeom([1, 1+n],[2+n],(e*x+d)*c^(1/2)/(-e*(-a)^(1/2)+d*c^(1/2)))*(d*e*n* 
(-a)^(1/2)+(-2*c*d^2-a*e^2*(2+n))/c^(1/2))/c/(a*e^2+c*d^2)/(1+n)/(-e*(-a)^ 
(1/2)+d*c^(1/2))-1/4*(e*x+d)^(1+n)*hypergeom([1, 1+n],[2+n],(e*x+d)*c^(1/2 
)/(e*(-a)^(1/2)+d*c^(1/2)))*(2*c*d^2+a*e^2*(2+n)+d*e*n*(-a)^(1/2)*c^(1/2)) 
/c^(3/2)/(a*e^2+c*d^2)/(1+n)/(e*(-a)^(1/2)+d*c^(1/2))
 

3.4.71.2 Mathematica [A] (verified)

Time = 0.59 (sec) , antiderivative size = 247, normalized size of antiderivative = 0.83 \[ \int \frac {x^3 (d+e x)^n}{\left (a+c x^2\right )^2} \, dx=-\frac {(d+e x)^{1+n} \left (\frac {2 a \sqrt {c} (-d+e x)}{a+c x^2}+\frac {\left (2 c d^2-\sqrt {-a} \sqrt {c} d e n+a e^2 (2+n)\right ) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt {c} (d+e x)}{\sqrt {c} d-\sqrt {-a} e}\right )}{\left (\sqrt {c} d-\sqrt {-a} e\right ) (1+n)}+\frac {\left (2 c d^2+\sqrt {-a} \sqrt {c} d e n+a e^2 (2+n)\right ) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt {c} (d+e x)}{\sqrt {c} d+\sqrt {-a} e}\right )}{\left (\sqrt {c} d+\sqrt {-a} e\right ) (1+n)}\right )}{4 c^{3/2} \left (c d^2+a e^2\right )} \]

input

Integrate[(x^3*(d + e*x)^n)/(a + c*x^2)^2,x]
 

output

-1/4*((d + e*x)^(1 + n)*((2*a*Sqrt[c]*(-d + e*x))/(a + c*x^2) + ((2*c*d^2 
- Sqrt[-a]*Sqrt[c]*d*e*n + a*e^2*(2 + n))*Hypergeometric2F1[1, 1 + n, 2 + 
n, (Sqrt[c]*(d + e*x))/(Sqrt[c]*d - Sqrt[-a]*e)])/((Sqrt[c]*d - Sqrt[-a]*e 
)*(1 + n)) + ((2*c*d^2 + Sqrt[-a]*Sqrt[c]*d*e*n + a*e^2*(2 + n))*Hypergeom 
etric2F1[1, 1 + n, 2 + n, (Sqrt[c]*(d + e*x))/(Sqrt[c]*d + Sqrt[-a]*e)])/( 
(Sqrt[c]*d + Sqrt[-a]*e)*(1 + n))))/(c^(3/2)*(c*d^2 + a*e^2))
 

3.4.71.3 Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 292, normalized size of antiderivative = 0.98, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {602, 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.

\(\displaystyle \int \frac {x^3 (d+e x)^n}{\left (a+c x^2\right )^2} \, dx\)

\(\Big \downarrow \) 602

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 657

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

\(\Big \downarrow \) 2009

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

input

Int[(x^3*(d + e*x)^n)/(a + c*x^2)^2,x]
 

output

(a*(d - e*x)*(d + e*x)^(1 + n))/(2*c*(c*d^2 + a*e^2)*(a + c*x^2)) - (((2*c 
*d^2 - Sqrt[-a]*Sqrt[c]*d*e*n + a*e^2*(2 + n))*(d + e*x)^(1 + n)*Hypergeom 
etric2F1[1, 1 + n, 2 + n, (Sqrt[c]*(d + e*x))/(Sqrt[c]*d - Sqrt[-a]*e)])/( 
2*Sqrt[c]*(Sqrt[c]*d - Sqrt[-a]*e)*(1 + n)) + ((2*c*d^2 + Sqrt[-a]*Sqrt[c] 
*d*e*n + a*e^2*(2 + n))*(d + e*x)^(1 + n)*Hypergeometric2F1[1, 1 + n, 2 + 
n, (Sqrt[c]*(d + e*x))/(Sqrt[c]*d + Sqrt[-a]*e)])/(2*Sqrt[c]*(Sqrt[c]*d + 
Sqrt[-a]*e)*(1 + n)))/(2*c*(c*d^2 + a*e^2))
 

3.4.71.3.1 Defintions of rubi rules used

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 602

Int[(x_)^(m_)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo 
l] :> With[{Qx = PolynomialQuotient[x^m, a + b*x^2, x], e = Coeff[Polynomia 
lRemainder[x^m, a + b*x^2, x], x, 0], f = Coeff[PolynomialRemainder[x^m, a 
+ b*x^2, x], x, 1]}, Simp[(-(c + d*x)^(n + 1))*(a + b*x^2)^(p + 1)*((a*(d*e 
 - c*f) + (b*c*e + a*d*f)*x)/(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)*ExpandToS 
um[2*a*(p + 1)*(b*c^2 + a*d^2)*Qx + e*(b*c^2*(2*p + 3) + a*d^2*(n + 2*p + 3 
)) - a*c*d*f*n + d*(b*c*e + a*d*f)*(n + 2*p + 4)*x, x], x], x]] /; FreeQ[{a 
, b, c, d, n}, x] && IGtQ[m, 1] && LtQ[p, -1] && NeQ[b*c^2 + a*d^2, 0]
 

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]
 

3.4.71.4 Maple [F]

\[\int \frac {x^{3} \left (e x +d \right )^{n}}{\left (c \,x^{2}+a \right )^{2}}d x\]

input

int(x^3*(e*x+d)^n/(c*x^2+a)^2,x)
 

output

int(x^3*(e*x+d)^n/(c*x^2+a)^2,x)
 

3.4.71.5 Fricas [F]

\[ \int \frac {x^3 (d+e x)^n}{\left (a+c x^2\right )^2} \, dx=\int { \frac {{\left (e x + d\right )}^{n} x^{3}}{{\left (c x^{2} + a\right )}^{2}} \,d x } \]

input

integrate(x^3*(e*x+d)^n/(c*x^2+a)^2,x, algorithm="fricas")
 

output

integral((e*x + d)^n*x^3/(c^2*x^4 + 2*a*c*x^2 + a^2), x)
 

3.4.71.6 Sympy [F(-1)]

Timed out. \[ \int \frac {x^3 (d+e x)^n}{\left (a+c x^2\right )^2} \, dx=\text {Timed out} \]

input

integrate(x**3*(e*x+d)**n/(c*x**2+a)**2,x)
 

output

Timed out
 

3.4.71.7 Maxima [F]

\[ \int \frac {x^3 (d+e x)^n}{\left (a+c x^2\right )^2} \, dx=\int { \frac {{\left (e x + d\right )}^{n} x^{3}}{{\left (c x^{2} + a\right )}^{2}} \,d x } \]

input

integrate(x^3*(e*x+d)^n/(c*x^2+a)^2,x, algorithm="maxima")
 

output

integrate((e*x + d)^n*x^3/(c*x^2 + a)^2, x)
 

3.4.71.8 Giac [F]

\[ \int \frac {x^3 (d+e x)^n}{\left (a+c x^2\right )^2} \, dx=\int { \frac {{\left (e x + d\right )}^{n} x^{3}}{{\left (c x^{2} + a\right )}^{2}} \,d x } \]

input

integrate(x^3*(e*x+d)^n/(c*x^2+a)^2,x, algorithm="giac")
 

output

integrate((e*x + d)^n*x^3/(c*x^2 + a)^2, x)
 

3.4.71.9 Mupad [F(-1)]

Timed out. \[ \int \frac {x^3 (d+e x)^n}{\left (a+c x^2\right )^2} \, dx=\int \frac {x^3\,{\left (d+e\,x\right )}^n}{{\left (c\,x^2+a\right )}^2} \,d x \]

input

int((x^3*(d + e*x)^n)/(a + c*x^2)^2,x)
 

output

int((x^3*(d + e*x)^n)/(a + c*x^2)^2, x)