\(\int x^4 (a+b x^2)^2 (c+d x^2)^3 \, dx\) [159]

Optimal result

Integrand size = 22, antiderivative size = 127 \[ \int x^4 \left (a+b x^2\right )^2 \left (c+d x^2\right )^3 \, dx=\frac {1}{5} a^2 c^3 x^5+\frac {1}{7} a c^2 (2 b c+3 a d) x^7+\frac {1}{9} c \left (b^2 c^2+6 a b c d+3 a^2 d^2\right ) x^9+\frac {1}{11} d \left (3 b^2 c^2+6 a b c d+a^2 d^2\right ) x^{11}+\frac {1}{13} b d^2 (3 b c+2 a d) x^{13}+\frac {1}{15} b^2 d^3 x^{15} \]

[Out]

1/5*a^2*c^3*x^5+1/7*a*c^2*(3*a*d+2*b*c)*x^7+1/9*c*(3*a^2*d^2+6*a*b*c*d+b^2*c^2)*x^9+1/11*d*(a^2*d^2+6*a*b*c*d+
3*b^2*c^2)*x^11+1/13*b*d^2*(2*a*d+3*b*c)*x^13+1/15*b^2*d^3*x^15

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {459} \[ \int x^4 \left (a+b x^2\right )^2 \left (c+d x^2\right )^3 \, dx=\frac {1}{11} d x^{11} \left (a^2 d^2+6 a b c d+3 b^2 c^2\right )+\frac {1}{9} c x^9 \left (3 a^2 d^2+6 a b c d+b^2 c^2\right )+\frac {1}{5} a^2 c^3 x^5+\frac {1}{7} a c^2 x^7 (3 a d+2 b c)+\frac {1}{13} b d^2 x^{13} (2 a d+3 b c)+\frac {1}{15} b^2 d^3 x^{15} \]

[In]

Int[x^4*(a + b*x^2)^2*(c + d*x^2)^3,x]

[Out]

(a^2*c^3*x^5)/5 + (a*c^2*(2*b*c + 3*a*d)*x^7)/7 + (c*(b^2*c^2 + 6*a*b*c*d + 3*a^2*d^2)*x^9)/9 + (d*(3*b^2*c^2
+ 6*a*b*c*d + a^2*d^2)*x^11)/11 + (b*d^2*(3*b*c + 2*a*d)*x^13)/13 + (b^2*d^3*x^15)/15

Rule 459

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

Rubi steps \begin{align*} \text {integral}& = \int \left (a^2 c^3 x^4+a c^2 (2 b c+3 a d) x^6+c \left (b^2 c^2+6 a b c d+3 a^2 d^2\right ) x^8+d \left (3 b^2 c^2+6 a b c d+a^2 d^2\right ) x^{10}+b d^2 (3 b c+2 a d) x^{12}+b^2 d^3 x^{14}\right ) \, dx \\ & = \frac {1}{5} a^2 c^3 x^5+\frac {1}{7} a c^2 (2 b c+3 a d) x^7+\frac {1}{9} c \left (b^2 c^2+6 a b c d+3 a^2 d^2\right ) x^9+\frac {1}{11} d \left (3 b^2 c^2+6 a b c d+a^2 d^2\right ) x^{11}+\frac {1}{13} b d^2 (3 b c+2 a d) x^{13}+\frac {1}{15} b^2 d^3 x^{15} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.00 \[ \int x^4 \left (a+b x^2\right )^2 \left (c+d x^2\right )^3 \, dx=\frac {1}{5} a^2 c^3 x^5+\frac {1}{7} a c^2 (2 b c+3 a d) x^7+\frac {1}{9} c \left (b^2 c^2+6 a b c d+3 a^2 d^2\right ) x^9+\frac {1}{11} d \left (3 b^2 c^2+6 a b c d+a^2 d^2\right ) x^{11}+\frac {1}{13} b d^2 (3 b c+2 a d) x^{13}+\frac {1}{15} b^2 d^3 x^{15} \]

[In]

Integrate[x^4*(a + b*x^2)^2*(c + d*x^2)^3,x]

[Out]

(a^2*c^3*x^5)/5 + (a*c^2*(2*b*c + 3*a*d)*x^7)/7 + (c*(b^2*c^2 + 6*a*b*c*d + 3*a^2*d^2)*x^9)/9 + (d*(3*b^2*c^2
+ 6*a*b*c*d + a^2*d^2)*x^11)/11 + (b*d^2*(3*b*c + 2*a*d)*x^13)/13 + (b^2*d^3*x^15)/15

Maple [A] (verified)

Time = 2.72 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.99

method	result	size
norman	\(\frac {a^{2} c^{3} x^{5}}{5}+\left (\frac {3}{7} a^{2} c^{2} d +\frac {2}{7} a b \,c^{3}\right ) x^{7}+\left (\frac {1}{3} c \,a^{2} d^{2}+\frac {2}{3} a b \,c^{2} d +\frac {1}{9} b^{2} c^{3}\right ) x^{9}+\left (\frac {1}{11} a^{2} d^{3}+\frac {6}{11} a b c \,d^{2}+\frac {3}{11} b^{2} c^{2} d \right ) x^{11}+\left (\frac {2}{13} a b \,d^{3}+\frac {3}{13} b^{2} c \,d^{2}\right ) x^{13}+\frac {b^{2} d^{3} x^{15}}{15}\)	\(126\)
default	\(\frac {b^{2} d^{3} x^{15}}{15}+\frac {\left (2 a b \,d^{3}+3 b^{2} c \,d^{2}\right ) x^{13}}{13}+\frac {\left (a^{2} d^{3}+6 a b c \,d^{2}+3 b^{2} c^{2} d \right ) x^{11}}{11}+\frac {\left (3 c \,a^{2} d^{2}+6 a b \,c^{2} d +b^{2} c^{3}\right ) x^{9}}{9}+\frac {\left (3 a^{2} c^{2} d +2 a b \,c^{3}\right ) x^{7}}{7}+\frac {a^{2} c^{3} x^{5}}{5}\)	\(128\)
gosper	\(\frac {1}{5} a^{2} c^{3} x^{5}+\frac {3}{7} x^{7} a^{2} c^{2} d +\frac {2}{7} x^{7} a b \,c^{3}+\frac {1}{3} x^{9} c \,a^{2} d^{2}+\frac {2}{3} x^{9} a b \,c^{2} d +\frac {1}{9} x^{9} b^{2} c^{3}+\frac {1}{11} x^{11} a^{2} d^{3}+\frac {6}{11} x^{11} a b c \,d^{2}+\frac {3}{11} x^{11} b^{2} c^{2} d +\frac {2}{13} x^{13} a b \,d^{3}+\frac {3}{13} x^{13} b^{2} c \,d^{2}+\frac {1}{15} b^{2} d^{3} x^{15}\)	\(136\)
risch	\(\frac {1}{5} a^{2} c^{3} x^{5}+\frac {3}{7} x^{7} a^{2} c^{2} d +\frac {2}{7} x^{7} a b \,c^{3}+\frac {1}{3} x^{9} c \,a^{2} d^{2}+\frac {2}{3} x^{9} a b \,c^{2} d +\frac {1}{9} x^{9} b^{2} c^{3}+\frac {1}{11} x^{11} a^{2} d^{3}+\frac {6}{11} x^{11} a b c \,d^{2}+\frac {3}{11} x^{11} b^{2} c^{2} d +\frac {2}{13} x^{13} a b \,d^{3}+\frac {3}{13} x^{13} b^{2} c \,d^{2}+\frac {1}{15} b^{2} d^{3} x^{15}\)	\(136\)
parallelrisch	\(\frac {1}{5} a^{2} c^{3} x^{5}+\frac {3}{7} x^{7} a^{2} c^{2} d +\frac {2}{7} x^{7} a b \,c^{3}+\frac {1}{3} x^{9} c \,a^{2} d^{2}+\frac {2}{3} x^{9} a b \,c^{2} d +\frac {1}{9} x^{9} b^{2} c^{3}+\frac {1}{11} x^{11} a^{2} d^{3}+\frac {6}{11} x^{11} a b c \,d^{2}+\frac {3}{11} x^{11} b^{2} c^{2} d +\frac {2}{13} x^{13} a b \,d^{3}+\frac {3}{13} x^{13} b^{2} c \,d^{2}+\frac {1}{15} b^{2} d^{3} x^{15}\)	\(136\)

[In]

int(x^4*(b*x^2+a)^2*(d*x^2+c)^3,x,method=_RETURNVERBOSE)

[Out]

1/5*a^2*c^3*x^5+(3/7*a^2*c^2*d+2/7*a*b*c^3)*x^7+(1/3*c*a^2*d^2+2/3*a*b*c^2*d+1/9*b^2*c^3)*x^9+(1/11*a^2*d^3+6/
11*a*b*c*d^2+3/11*b^2*c^2*d)*x^11+(2/13*a*b*d^3+3/13*b^2*c*d^2)*x^13+1/15*b^2*d^3*x^15

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.00 \[ \int x^4 \left (a+b x^2\right )^2 \left (c+d x^2\right )^3 \, dx=\frac {1}{15} \, b^{2} d^{3} x^{15} + \frac {1}{13} \, {\left (3 \, b^{2} c d^{2} + 2 \, a b d^{3}\right )} x^{13} + \frac {1}{11} \, {\left (3 \, b^{2} c^{2} d + 6 \, a b c d^{2} + a^{2} d^{3}\right )} x^{11} + \frac {1}{5} \, a^{2} c^{3} x^{5} + \frac {1}{9} \, {\left (b^{2} c^{3} + 6 \, a b c^{2} d + 3 \, a^{2} c d^{2}\right )} x^{9} + \frac {1}{7} \, {\left (2 \, a b c^{3} + 3 \, a^{2} c^{2} d\right )} x^{7} \]

[In]

integrate(x^4*(b*x^2+a)^2*(d*x^2+c)^3,x, algorithm="fricas")

[Out]

1/15*b^2*d^3*x^15 + 1/13*(3*b^2*c*d^2 + 2*a*b*d^3)*x^13 + 1/11*(3*b^2*c^2*d + 6*a*b*c*d^2 + a^2*d^3)*x^11 + 1/
5*a^2*c^3*x^5 + 1/9*(b^2*c^3 + 6*a*b*c^2*d + 3*a^2*c*d^2)*x^9 + 1/7*(2*a*b*c^3 + 3*a^2*c^2*d)*x^7

Sympy [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.13 \[ \int x^4 \left (a+b x^2\right )^2 \left (c+d x^2\right )^3 \, dx=\frac {a^{2} c^{3} x^{5}}{5} + \frac {b^{2} d^{3} x^{15}}{15} + x^{13} \cdot \left (\frac {2 a b d^{3}}{13} + \frac {3 b^{2} c d^{2}}{13}\right ) + x^{11} \left (\frac {a^{2} d^{3}}{11} + \frac {6 a b c d^{2}}{11} + \frac {3 b^{2} c^{2} d}{11}\right ) + x^{9} \left (\frac {a^{2} c d^{2}}{3} + \frac {2 a b c^{2} d}{3} + \frac {b^{2} c^{3}}{9}\right ) + x^{7} \cdot \left (\frac {3 a^{2} c^{2} d}{7} + \frac {2 a b c^{3}}{7}\right ) \]

[In]

integrate(x**4*(b*x**2+a)**2*(d*x**2+c)**3,x)

[Out]

a**2*c**3*x**5/5 + b**2*d**3*x**15/15 + x**13*(2*a*b*d**3/13 + 3*b**2*c*d**2/13) + x**11*(a**2*d**3/11 + 6*a*b
*c*d**2/11 + 3*b**2*c**2*d/11) + x**9*(a**2*c*d**2/3 + 2*a*b*c**2*d/3 + b**2*c**3/9) + x**7*(3*a**2*c**2*d/7 +
2*a*b*c**3/7)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.00 \[ \int x^4 \left (a+b x^2\right )^2 \left (c+d x^2\right )^3 \, dx=\frac {1}{15} \, b^{2} d^{3} x^{15} + \frac {1}{13} \, {\left (3 \, b^{2} c d^{2} + 2 \, a b d^{3}\right )} x^{13} + \frac {1}{11} \, {\left (3 \, b^{2} c^{2} d + 6 \, a b c d^{2} + a^{2} d^{3}\right )} x^{11} + \frac {1}{5} \, a^{2} c^{3} x^{5} + \frac {1}{9} \, {\left (b^{2} c^{3} + 6 \, a b c^{2} d + 3 \, a^{2} c d^{2}\right )} x^{9} + \frac {1}{7} \, {\left (2 \, a b c^{3} + 3 \, a^{2} c^{2} d\right )} x^{7} \]

[In]

integrate(x^4*(b*x^2+a)^2*(d*x^2+c)^3,x, algorithm="maxima")

[Out]

1/15*b^2*d^3*x^15 + 1/13*(3*b^2*c*d^2 + 2*a*b*d^3)*x^13 + 1/11*(3*b^2*c^2*d + 6*a*b*c*d^2 + a^2*d^3)*x^11 + 1/
5*a^2*c^3*x^5 + 1/9*(b^2*c^3 + 6*a*b*c^2*d + 3*a^2*c*d^2)*x^9 + 1/7*(2*a*b*c^3 + 3*a^2*c^2*d)*x^7

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.06 \[ \int x^4 \left (a+b x^2\right )^2 \left (c+d x^2\right )^3 \, dx=\frac {1}{15} \, b^{2} d^{3} x^{15} + \frac {3}{13} \, b^{2} c d^{2} x^{13} + \frac {2}{13} \, a b d^{3} x^{13} + \frac {3}{11} \, b^{2} c^{2} d x^{11} + \frac {6}{11} \, a b c d^{2} x^{11} + \frac {1}{11} \, a^{2} d^{3} x^{11} + \frac {1}{9} \, b^{2} c^{3} x^{9} + \frac {2}{3} \, a b c^{2} d x^{9} + \frac {1}{3} \, a^{2} c d^{2} x^{9} + \frac {2}{7} \, a b c^{3} x^{7} + \frac {3}{7} \, a^{2} c^{2} d x^{7} + \frac {1}{5} \, a^{2} c^{3} x^{5} \]

[In]

integrate(x^4*(b*x^2+a)^2*(d*x^2+c)^3,x, algorithm="giac")

[Out]

1/15*b^2*d^3*x^15 + 3/13*b^2*c*d^2*x^13 + 2/13*a*b*d^3*x^13 + 3/11*b^2*c^2*d*x^11 + 6/11*a*b*c*d^2*x^11 + 1/11
*a^2*d^3*x^11 + 1/9*b^2*c^3*x^9 + 2/3*a*b*c^2*d*x^9 + 1/3*a^2*c*d^2*x^9 + 2/7*a*b*c^3*x^7 + 3/7*a^2*c^2*d*x^7
+ 1/5*a^2*c^3*x^5

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.94 \[ \int x^4 \left (a+b x^2\right )^2 \left (c+d x^2\right )^3 \, dx=x^9\,\left (\frac {a^2\,c\,d^2}{3}+\frac {2\,a\,b\,c^2\,d}{3}+\frac {b^2\,c^3}{9}\right )+x^{11}\,\left (\frac {a^2\,d^3}{11}+\frac {6\,a\,b\,c\,d^2}{11}+\frac {3\,b^2\,c^2\,d}{11}\right )+\frac {a^2\,c^3\,x^5}{5}+\frac {b^2\,d^3\,x^{15}}{15}+\frac {a\,c^2\,x^7\,\left (3\,a\,d+2\,b\,c\right )}{7}+\frac {b\,d^2\,x^{13}\,\left (2\,a\,d+3\,b\,c\right )}{13} \]

[In]

int(x^4*(a + b*x^2)^2*(c + d*x^2)^3,x)

[Out]

x^9*((b^2*c^3)/9 + (a^2*c*d^2)/3 + (2*a*b*c^2*d)/3) + x^11*((a^2*d^3)/11 + (3*b^2*c^2*d)/11 + (6*a*b*c*d^2)/11
) + (a^2*c^3*x^5)/5 + (b^2*d^3*x^15)/15 + (a*c^2*x^7*(3*a*d + 2*b*c))/7 + (b*d^2*x^13*(2*a*d + 3*b*c))/13