\(\int \frac {x^5 (c+d x^2)^2}{a+b x^2} \, dx\) [208]

Optimal result

Integrand size = 22, antiderivative size = 103 \[ \int \frac {x^5 \left (c+d x^2\right )^2}{a+b x^2} \, dx=-\frac {a (b c-a d)^2 x^2}{2 b^4}+\frac {(b c-a d)^2 x^4}{4 b^3}+\frac {d (2 b c-a d) x^6}{6 b^2}+\frac {d^2 x^8}{8 b}+\frac {a^2 (b c-a d)^2 \log \left (a+b x^2\right )}{2 b^5} \]

[Out]

-1/2*a*(-a*d+b*c)^2*x^2/b^4+1/4*(-a*d+b*c)^2*x^4/b^3+1/6*d*(-a*d+2*b*c)*x^6/b^2+1/8*d^2*x^8/b+1/2*a^2*(-a*d+b*
c)^2*ln(b*x^2+a)/b^5

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {457, 90} \[ \int \frac {x^5 \left (c+d x^2\right )^2}{a+b x^2} \, dx=\frac {a^2 (b c-a d)^2 \log \left (a+b x^2\right )}{2 b^5}-\frac {a x^2 (b c-a d)^2}{2 b^4}+\frac {x^4 (b c-a d)^2}{4 b^3}+\frac {d x^6 (2 b c-a d)}{6 b^2}+\frac {d^2 x^8}{8 b} \]

[In]

Int[(x^5*(c + d*x^2)^2)/(a + b*x^2),x]

[Out]

-1/2*(a*(b*c - a*d)^2*x^2)/b^4 + ((b*c - a*d)^2*x^4)/(4*b^3) + (d*(2*b*c - a*d)*x^6)/(6*b^2) + (d^2*x^8)/(8*b)
+ (a^2*(b*c - a*d)^2*Log[a + b*x^2])/(2*b^5)

Rule 90

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 457

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {x^2 (c+d x)^2}{a+b x} \, dx,x,x^2\right ) \\ & = \frac {1}{2} \text {Subst}\left (\int \left (-\frac {a (-b c+a d)^2}{b^4}+\frac {(b c-a d)^2 x}{b^3}+\frac {d (2 b c-a d) x^2}{b^2}+\frac {d^2 x^3}{b}+\frac {a^2 (-b c+a d)^2}{b^4 (a+b x)}\right ) \, dx,x,x^2\right ) \\ & = -\frac {a (b c-a d)^2 x^2}{2 b^4}+\frac {(b c-a d)^2 x^4}{4 b^3}+\frac {d (2 b c-a d) x^6}{6 b^2}+\frac {d^2 x^8}{8 b}+\frac {a^2 (b c-a d)^2 \log \left (a+b x^2\right )}{2 b^5} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.13 \[ \int \frac {x^5 \left (c+d x^2\right )^2}{a+b x^2} \, dx=-\frac {a (-b c+a d)^2 x^2}{2 b^4}+\frac {(b c-a d)^2 x^4}{4 b^3}+\frac {d (2 b c-a d) x^6}{6 b^2}+\frac {d^2 x^8}{8 b}+\frac {\left (a^2 b^2 c^2-2 a^3 b c d+a^4 d^2\right ) \log \left (a+b x^2\right )}{2 b^5} \]

[In]

Integrate[(x^5*(c + d*x^2)^2)/(a + b*x^2),x]

[Out]

-1/2*(a*(-(b*c) + a*d)^2*x^2)/b^4 + ((b*c - a*d)^2*x^4)/(4*b^3) + (d*(2*b*c - a*d)*x^6)/(6*b^2) + (d^2*x^8)/(8
*b) + ((a^2*b^2*c^2 - 2*a^3*b*c*d + a^4*d^2)*Log[a + b*x^2])/(2*b^5)

Maple [A] (verified)

Time = 2.73 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.22

method	result	size
norman	\(\frac {d^{2} x^{8}}{8 b}+\frac {\left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) x^{4}}{4 b^{3}}-\frac {a \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) x^{2}}{2 b^{4}}-\frac {d \left (a d -2 b c \right ) x^{6}}{6 b^{2}}+\frac {a^{2} \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \ln \left (b \,x^{2}+a \right )}{2 b^{5}}\)	\(126\)
default	\(-\frac {-\frac {d^{2} x^{8} b^{3}}{4}+\frac {\left (\left (a d -b c \right ) b^{2} d -b^{3} d c \right ) x^{6}}{3}+\frac {\left (\left (a d -b c \right ) b^{2} c -b d \left (a^{2} d -a b c \right )\right ) x^{4}}{2}+\left (a d -b c \right ) \left (a^{2} d -a b c \right ) x^{2}}{2 b^{4}}+\frac {a^{2} \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \ln \left (b \,x^{2}+a \right )}{2 b^{5}}\)	\(139\)
parallelrisch	\(\frac {3 d^{2} x^{8} b^{4}-4 x^{6} a \,b^{3} d^{2}+8 x^{6} b^{4} c d +6 x^{4} a^{2} b^{2} d^{2}-12 x^{4} a \,b^{3} c d +6 x^{4} b^{4} c^{2}-12 a^{3} b \,d^{2} x^{2}+24 a^{2} b^{2} c d \,x^{2}-12 a \,b^{3} c^{2} x^{2}+12 \ln \left (b \,x^{2}+a \right ) a^{4} d^{2}-24 \ln \left (b \,x^{2}+a \right ) a^{3} b c d +12 \ln \left (b \,x^{2}+a \right ) a^{2} b^{2} c^{2}}{24 b^{5}}\)	\(164\)
risch	\(-\frac {d^{2} a \,x^{6}}{6 b^{2}}+\frac {d c \,x^{6}}{3 b}+\frac {d^{2} a^{2} x^{4}}{4 b^{3}}+\frac {c^{2} x^{4}}{4 b}-\frac {d a c \,x^{4}}{2 b^{2}}+\frac {7 d^{2} a^{4}}{24 b^{5}}-\frac {5 d \,a^{3} c}{6 b^{4}}+\frac {3 a^{2} c^{2}}{4 b^{3}}-\frac {a \,c^{3}}{6 b^{2} d}-\frac {d^{2} a^{3} x^{2}}{2 b^{4}}+\frac {d \,a^{2} c \,x^{2}}{b^{3}}-\frac {a \,c^{2} x^{2}}{2 b^{2}}-\frac {c^{4}}{24 b \,d^{2}}+\frac {d^{2} x^{8}}{8 b}+\frac {a^{4} \ln \left (b \,x^{2}+a \right ) d^{2}}{2 b^{5}}-\frac {a^{3} \ln \left (b \,x^{2}+a \right ) c d}{b^{4}}+\frac {a^{2} \ln \left (b \,x^{2}+a \right ) c^{2}}{2 b^{3}}\)	\(220\)

[In]

int(x^5*(d*x^2+c)^2/(b*x^2+a),x,method=_RETURNVERBOSE)

[Out]

1/8*d^2*x^8/b+1/4*(a^2*d^2-2*a*b*c*d+b^2*c^2)/b^3*x^4-1/2*a*(a^2*d^2-2*a*b*c*d+b^2*c^2)/b^4*x^2-1/6*d*(a*d-2*b
*c)/b^2*x^6+1/2*a^2*(a^2*d^2-2*a*b*c*d+b^2*c^2)/b^5*ln(b*x^2+a)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.34 \[ \int \frac {x^5 \left (c+d x^2\right )^2}{a+b x^2} \, dx=\frac {3 \, b^{4} d^{2} x^{8} + 4 \, {\left (2 \, b^{4} c d - a b^{3} d^{2}\right )} x^{6} + 6 \, {\left (b^{4} c^{2} - 2 \, a b^{3} c d + a^{2} b^{2} d^{2}\right )} x^{4} - 12 \, {\left (a b^{3} c^{2} - 2 \, a^{2} b^{2} c d + a^{3} b d^{2}\right )} x^{2} + 12 \, {\left (a^{2} b^{2} c^{2} - 2 \, a^{3} b c d + a^{4} d^{2}\right )} \log \left (b x^{2} + a\right )}{24 \, b^{5}} \]

[In]

integrate(x^5*(d*x^2+c)^2/(b*x^2+a),x, algorithm="fricas")

[Out]

1/24*(3*b^4*d^2*x^8 + 4*(2*b^4*c*d - a*b^3*d^2)*x^6 + 6*(b^4*c^2 - 2*a*b^3*c*d + a^2*b^2*d^2)*x^4 - 12*(a*b^3*
c^2 - 2*a^2*b^2*c*d + a^3*b*d^2)*x^2 + 12*(a^2*b^2*c^2 - 2*a^3*b*c*d + a^4*d^2)*log(b*x^2 + a))/b^5

Sympy [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.18 \[ \int \frac {x^5 \left (c+d x^2\right )^2}{a+b x^2} \, dx=\frac {a^{2} \left (a d - b c\right )^{2} \log {\left (a + b x^{2} \right )}}{2 b^{5}} + x^{6} \left (- \frac {a d^{2}}{6 b^{2}} + \frac {c d}{3 b}\right ) + x^{4} \left (\frac {a^{2} d^{2}}{4 b^{3}} - \frac {a c d}{2 b^{2}} + \frac {c^{2}}{4 b}\right ) + x^{2} \left (- \frac {a^{3} d^{2}}{2 b^{4}} + \frac {a^{2} c d}{b^{3}} - \frac {a c^{2}}{2 b^{2}}\right ) + \frac {d^{2} x^{8}}{8 b} \]

[In]

integrate(x**5*(d*x**2+c)**2/(b*x**2+a),x)

[Out]

a**2*(a*d - b*c)**2*log(a + b*x**2)/(2*b**5) + x**6*(-a*d**2/(6*b**2) + c*d/(3*b)) + x**4*(a**2*d**2/(4*b**3)
- a*c*d/(2*b**2) + c**2/(4*b)) + x**2*(-a**3*d**2/(2*b**4) + a**2*c*d/b**3 - a*c**2/(2*b**2)) + d**2*x**8/(8*b
)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.33 \[ \int \frac {x^5 \left (c+d x^2\right )^2}{a+b x^2} \, dx=\frac {3 \, b^{3} d^{2} x^{8} + 4 \, {\left (2 \, b^{3} c d - a b^{2} d^{2}\right )} x^{6} + 6 \, {\left (b^{3} c^{2} - 2 \, a b^{2} c d + a^{2} b d^{2}\right )} x^{4} - 12 \, {\left (a b^{2} c^{2} - 2 \, a^{2} b c d + a^{3} d^{2}\right )} x^{2}}{24 \, b^{4}} + \frac {{\left (a^{2} b^{2} c^{2} - 2 \, a^{3} b c d + a^{4} d^{2}\right )} \log \left (b x^{2} + a\right )}{2 \, b^{5}} \]

[In]

integrate(x^5*(d*x^2+c)^2/(b*x^2+a),x, algorithm="maxima")

[Out]

1/24*(3*b^3*d^2*x^8 + 4*(2*b^3*c*d - a*b^2*d^2)*x^6 + 6*(b^3*c^2 - 2*a*b^2*c*d + a^2*b*d^2)*x^4 - 12*(a*b^2*c^
2 - 2*a^2*b*c*d + a^3*d^2)*x^2)/b^4 + 1/2*(a^2*b^2*c^2 - 2*a^3*b*c*d + a^4*d^2)*log(b*x^2 + a)/b^5

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.44 \[ \int \frac {x^5 \left (c+d x^2\right )^2}{a+b x^2} \, dx=\frac {3 \, b^{3} d^{2} x^{8} + 8 \, b^{3} c d x^{6} - 4 \, a b^{2} d^{2} x^{6} + 6 \, b^{3} c^{2} x^{4} - 12 \, a b^{2} c d x^{4} + 6 \, a^{2} b d^{2} x^{4} - 12 \, a b^{2} c^{2} x^{2} + 24 \, a^{2} b c d x^{2} - 12 \, a^{3} d^{2} x^{2}}{24 \, b^{4}} + \frac {{\left (a^{2} b^{2} c^{2} - 2 \, a^{3} b c d + a^{4} d^{2}\right )} \log \left ({\left | b x^{2} + a \right |}\right )}{2 \, b^{5}} \]

[In]

integrate(x^5*(d*x^2+c)^2/(b*x^2+a),x, algorithm="giac")

[Out]

1/24*(3*b^3*d^2*x^8 + 8*b^3*c*d*x^6 - 4*a*b^2*d^2*x^6 + 6*b^3*c^2*x^4 - 12*a*b^2*c*d*x^4 + 6*a^2*b*d^2*x^4 - 1
2*a*b^2*c^2*x^2 + 24*a^2*b*c*d*x^2 - 12*a^3*d^2*x^2)/b^4 + 1/2*(a^2*b^2*c^2 - 2*a^3*b*c*d + a^4*d^2)*log(abs(b
*x^2 + a))/b^5

Mupad [B] (verification not implemented)

Time = 5.09 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.42 \[ \int \frac {x^5 \left (c+d x^2\right )^2}{a+b x^2} \, dx=x^4\,\left (\frac {c^2}{4\,b}+\frac {a\,\left (\frac {a\,d^2}{b^2}-\frac {2\,c\,d}{b}\right )}{4\,b}\right )-x^6\,\left (\frac {a\,d^2}{6\,b^2}-\frac {c\,d}{3\,b}\right )+\frac {\ln \left (b\,x^2+a\right )\,\left (a^4\,d^2-2\,a^3\,b\,c\,d+a^2\,b^2\,c^2\right )}{2\,b^5}+\frac {d^2\,x^8}{8\,b}-\frac {a\,x^2\,\left (\frac {c^2}{b}+\frac {a\,\left (\frac {a\,d^2}{b^2}-\frac {2\,c\,d}{b}\right )}{b}\right )}{2\,b} \]

[In]

int((x^5*(c + d*x^2)^2)/(a + b*x^2),x)

[Out]

x^4*(c^2/(4*b) + (a*((a*d^2)/b^2 - (2*c*d)/b))/(4*b)) - x^6*((a*d^2)/(6*b^2) - (c*d)/(3*b)) + (log(a + b*x^2)*
(a^4*d^2 + a^2*b^2*c^2 - 2*a^3*b*c*d))/(2*b^5) + (d^2*x^8)/(8*b) - (a*x^2*(c^2/b + (a*((a*d^2)/b^2 - (2*c*d)/b
))/b))/(2*b)