3.5.1 \(\int \frac {x^7 (d+e x^2)^q}{a+b x^2+c x^4} \, dx\) [401]

3.5.1.1 Optimal result

Integrand size = 27, antiderivative size = 313 \[ \int \frac {x^7 \left (d+e x^2\right )^q}{a+b x^2+c x^4} \, dx=-\frac {(c d+b e) \left (d+e x^2\right )^{1+q}}{2 c^2 e^2 (1+q)}+\frac {\left (d+e x^2\right )^{2+q}}{2 c e^2 (2+q)}+\frac {\left (a-\frac {b^2}{c}+\frac {b \left (b^2-3 a c\right )}{c \sqrt {b^2-4 a c}}\right ) \left (d+e x^2\right )^{1+q} \operatorname {Hypergeometric2F1}\left (1,1+q,2+q,\frac {2 c \left (d+e x^2\right )}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}\right )}{2 c \left (2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e\right ) (1+q)}+\frac {\left (a-\frac {b^2}{c}-\frac {b \left (b^2-3 a c\right )}{c \sqrt {b^2-4 a c}}\right ) \left (d+e x^2\right )^{1+q} \operatorname {Hypergeometric2F1}\left (1,1+q,2+q,\frac {2 c \left (d+e x^2\right )}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{2 c \left (2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e\right ) (1+q)} \]

output

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

3.5.1.2 Mathematica [A] (verified)

Time = 1.10 (sec) , antiderivative size = 272, normalized size of antiderivative = 0.87 \[ \int \frac {x^7 \left (d+e x^2\right )^q}{a+b x^2+c x^4} \, dx=\frac {\left (d+e x^2\right )^{1+q} \left (-\frac {c d+b e}{e^2 (1+q)}+\frac {c \left (d+e x^2\right )}{e^2 (2+q)}+\frac {c \left (a-\frac {b^2}{c}+\frac {b \left (b^2-3 a c\right )}{c \sqrt {b^2-4 a c}}\right ) \operatorname {Hypergeometric2F1}\left (1,1+q,2+q,\frac {2 c \left (d+e x^2\right )}{2 c d+\left (-b+\sqrt {b^2-4 a c}\right ) e}\right )}{\left (2 c d+\left (-b+\sqrt {b^2-4 a c}\right ) e\right ) (1+q)}+\frac {c \left (a-\frac {b^2}{c}-\frac {b \left (b^2-3 a c\right )}{c \sqrt {b^2-4 a c}}\right ) \operatorname {Hypergeometric2F1}\left (1,1+q,2+q,\frac {2 c \left (d+e x^2\right )}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{\left (2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e\right ) (1+q)}\right )}{2 c^2} \]

input

Integrate[(x^7*(d + e*x^2)^q)/(a + b*x^2 + c*x^4),x]
 

output

((d + e*x^2)^(1 + q)*(-((c*d + b*e)/(e^2*(1 + q))) + (c*(d + e*x^2))/(e^2* 
(2 + q)) + (c*(a - b^2/c + (b*(b^2 - 3*a*c))/(c*Sqrt[b^2 - 4*a*c]))*Hyperg 
eometric2F1[1, 1 + q, 2 + q, (2*c*(d + e*x^2))/(2*c*d + (-b + Sqrt[b^2 - 4 
*a*c])*e)])/((2*c*d + (-b + Sqrt[b^2 - 4*a*c])*e)*(1 + q)) + (c*(a - b^2/c 
 - (b*(b^2 - 3*a*c))/(c*Sqrt[b^2 - 4*a*c]))*Hypergeometric2F1[1, 1 + q, 2 
+ q, (2*c*(d + e*x^2))/(2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)])/((2*c*d - (b 
+ Sqrt[b^2 - 4*a*c])*e)*(1 + q))))/(2*c^2)
 

3.5.1.3 Rubi [A] (verified)

Time = 0.71 (sec) , antiderivative size = 306, normalized size of antiderivative = 0.98, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {1578, 1200, 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.

\(\displaystyle \int \frac {x^7 \left (d+e x^2\right )^q}{a+b x^2+c x^4} \, dx\)

\(\Big \downarrow \) 1578

\(\displaystyle \frac {1}{2} \int \frac {x^6 \left (e x^2+d\right )^q}{c x^4+b x^2+a}dx^2\)

\(\Big \downarrow \) 1200

\(\displaystyle \frac {1}{2} \int \left (\frac {(-c d-b e) \left (e x^2+d\right )^q}{c^2 e}+\frac {\left (\frac {b^2}{c^2}-\frac {\left (b^2-3 a c\right ) b}{c^2 \sqrt {b^2-4 a c}}-\frac {a}{c}\right ) \left (e x^2+d\right )^q}{2 c x^2+b-\sqrt {b^2-4 a c}}+\frac {\left (\frac {b^2}{c^2}+\frac {\left (b^2-3 a c\right ) b}{c^2 \sqrt {b^2-4 a c}}-\frac {a}{c}\right ) \left (e x^2+d\right )^q}{2 c x^2+b+\sqrt {b^2-4 a c}}+\frac {\left (e x^2+d\right )^{q+1}}{c e}\right )dx^2\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{2} \left (\frac {\left (\frac {b \left (b^2-3 a c\right )}{c \sqrt {b^2-4 a c}}+a-\frac {b^2}{c}\right ) \left (d+e x^2\right )^{q+1} \operatorname {Hypergeometric2F1}\left (1,q+1,q+2,\frac {2 c \left (e x^2+d\right )}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}\right )}{c (q+1) \left (2 c d-e \left (b-\sqrt {b^2-4 a c}\right )\right )}+\frac {\left (-\frac {b \left (b^2-3 a c\right )}{c \sqrt {b^2-4 a c}}+a-\frac {b^2}{c}\right ) \left (d+e x^2\right )^{q+1} \operatorname {Hypergeometric2F1}\left (1,q+1,q+2,\frac {2 c \left (e x^2+d\right )}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{c (q+1) \left (2 c d-e \left (\sqrt {b^2-4 a c}+b\right )\right )}-\frac {(b e+c d) \left (d+e x^2\right )^{q+1}}{c^2 e^2 (q+1)}+\frac {\left (d+e x^2\right )^{q+2}}{c e^2 (q+2)}\right )\)

input

Int[(x^7*(d + e*x^2)^q)/(a + b*x^2 + c*x^4),x]
 

output

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

3.5.1.3.1 Defintions of rubi rules used

rule 1200

Int[(((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.))/((a_.) + (b_.)* 
(x_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*((f + g* 
x)^n/(a + b*x + c*x^2)), x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && In 
tegersQ[n]
 

rule 1578

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_ 
)^4)^(p_.), x_Symbol] :> Simp[1/2   Subst[Int[x^((m - 1)/2)*(d + e*x)^q*(a 
+ b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x] && Int 
egerQ[(m - 1)/2]
 

rule 2009

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

3.5.1.4 Maple [F]

\[\int \frac {x^{7} \left (e \,x^{2}+d \right )^{q}}{c \,x^{4}+b \,x^{2}+a}d x\]

input

int(x^7*(e*x^2+d)^q/(c*x^4+b*x^2+a),x)
 

output

int(x^7*(e*x^2+d)^q/(c*x^4+b*x^2+a),x)
 

3.5.1.5 Fricas [F]

\[ \int \frac {x^7 \left (d+e x^2\right )^q}{a+b x^2+c x^4} \, dx=\int { \frac {{\left (e x^{2} + d\right )}^{q} x^{7}}{c x^{4} + b x^{2} + a} \,d x } \]

input

integrate(x^7*(e*x^2+d)^q/(c*x^4+b*x^2+a),x, algorithm="fricas")
 

output

integral((e*x^2 + d)^q*x^7/(c*x^4 + b*x^2 + a), x)
 

3.5.1.6 Sympy [F(-1)]

Timed out. \[ \int \frac {x^7 \left (d+e x^2\right )^q}{a+b x^2+c x^4} \, dx=\text {Timed out} \]

input

integrate(x**7*(e*x**2+d)**q/(c*x**4+b*x**2+a),x)
 

output

Timed out
 

3.5.1.7 Maxima [F]

\[ \int \frac {x^7 \left (d+e x^2\right )^q}{a+b x^2+c x^4} \, dx=\int { \frac {{\left (e x^{2} + d\right )}^{q} x^{7}}{c x^{4} + b x^{2} + a} \,d x } \]

input

integrate(x^7*(e*x^2+d)^q/(c*x^4+b*x^2+a),x, algorithm="maxima")
 

output

integrate((e*x^2 + d)^q*x^7/(c*x^4 + b*x^2 + a), x)
 

3.5.1.8 Giac [F]

\[ \int \frac {x^7 \left (d+e x^2\right )^q}{a+b x^2+c x^4} \, dx=\int { \frac {{\left (e x^{2} + d\right )}^{q} x^{7}}{c x^{4} + b x^{2} + a} \,d x } \]

input

integrate(x^7*(e*x^2+d)^q/(c*x^4+b*x^2+a),x, algorithm="giac")
 

output

integrate((e*x^2 + d)^q*x^7/(c*x^4 + b*x^2 + a), x)
 

3.5.1.9 Mupad [F(-1)]

Timed out. \[ \int \frac {x^7 \left (d+e x^2\right )^q}{a+b x^2+c x^4} \, dx=\int \frac {x^7\,{\left (e\,x^2+d\right )}^q}{c\,x^4+b\,x^2+a} \,d x \]

input

int((x^7*(d + e*x^2)^q)/(a + b*x^2 + c*x^4),x)
 

output

int((x^7*(d + e*x^2)^q)/(a + b*x^2 + c*x^4), x)