\(\int x^{5/2} (a+b x^2)^2 (c+d x^2) \, dx\) [392]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 22, antiderivative size = 63 \[ \int x^{5/2} \left (a+b x^2\right )^2 \left (c+d x^2\right ) \, dx=\frac {2}{7} a^2 c x^{7/2}+\frac {2}{11} a (2 b c+a d) x^{11/2}+\frac {2}{15} b (b c+2 a d) x^{15/2}+\frac {2}{19} b^2 d x^{19/2} \]

[Out]

2/7*a^2*c*x^(7/2)+2/11*a*(a*d+2*b*c)*x^(11/2)+2/15*b*(2*a*d+b*c)*x^(15/2)+2/19*b^2*d*x^(19/2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 63, 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} \[ \int x^{5/2} \left (a+b x^2\right )^2 \left (c+d x^2\right ) \, dx=\frac {2}{7} a^2 c x^{7/2}+\frac {2}{15} b x^{15/2} (2 a d+b c)+\frac {2}{11} a x^{11/2} (a d+2 b c)+\frac {2}{19} b^2 d x^{19/2} \]

[In]

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

[Out]

(2*a^2*c*x^(7/2))/7 + (2*a*(2*b*c + a*d)*x^(11/2))/11 + (2*b*(b*c + 2*a*d)*x^(15/2))/15 + (2*b^2*d*x^(19/2))/1
9

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 x^{5/2}+a (2 b c+a d) x^{9/2}+b (b c+2 a d) x^{13/2}+b^2 d x^{17/2}\right ) \, dx \\ & = \frac {2}{7} a^2 c x^{7/2}+\frac {2}{11} a (2 b c+a d) x^{11/2}+\frac {2}{15} b (b c+2 a d) x^{15/2}+\frac {2}{19} b^2 d x^{19/2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.95 \[ \int x^{5/2} \left (a+b x^2\right )^2 \left (c+d x^2\right ) \, dx=\frac {2 x^{7/2} \left (285 a^2 \left (11 c+7 d x^2\right )+266 a b x^2 \left (15 c+11 d x^2\right )+77 b^2 x^4 \left (19 c+15 d x^2\right )\right )}{21945} \]

[In]

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

[Out]

(2*x^(7/2)*(285*a^2*(11*c + 7*d*x^2) + 266*a*b*x^2*(15*c + 11*d*x^2) + 77*b^2*x^4*(19*c + 15*d*x^2)))/21945

Maple [A] (verified)

Time = 2.75 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.83

method result size
derivativedivides \(\frac {2 b^{2} d \,x^{\frac {19}{2}}}{19}+\frac {2 \left (2 a b d +b^{2} c \right ) x^{\frac {15}{2}}}{15}+\frac {2 \left (a^{2} d +2 a b c \right ) x^{\frac {11}{2}}}{11}+\frac {2 a^{2} c \,x^{\frac {7}{2}}}{7}\) \(52\)
default \(\frac {2 b^{2} d \,x^{\frac {19}{2}}}{19}+\frac {2 \left (2 a b d +b^{2} c \right ) x^{\frac {15}{2}}}{15}+\frac {2 \left (a^{2} d +2 a b c \right ) x^{\frac {11}{2}}}{11}+\frac {2 a^{2} c \,x^{\frac {7}{2}}}{7}\) \(52\)
gosper \(\frac {2 x^{\frac {7}{2}} \left (1155 b^{2} d \,x^{6}+2926 a b d \,x^{4}+1463 b^{2} c \,x^{4}+1995 a^{2} d \,x^{2}+3990 a b c \,x^{2}+3135 a^{2} c \right )}{21945}\) \(56\)
trager \(\frac {2 x^{\frac {7}{2}} \left (1155 b^{2} d \,x^{6}+2926 a b d \,x^{4}+1463 b^{2} c \,x^{4}+1995 a^{2} d \,x^{2}+3990 a b c \,x^{2}+3135 a^{2} c \right )}{21945}\) \(56\)
risch \(\frac {2 x^{\frac {7}{2}} \left (1155 b^{2} d \,x^{6}+2926 a b d \,x^{4}+1463 b^{2} c \,x^{4}+1995 a^{2} d \,x^{2}+3990 a b c \,x^{2}+3135 a^{2} c \right )}{21945}\) \(56\)

[In]

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

[Out]

2/19*b^2*d*x^(19/2)+2/15*(2*a*b*d+b^2*c)*x^(15/2)+2/11*(a^2*d+2*a*b*c)*x^(11/2)+2/7*a^2*c*x^(7/2)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.89 \[ \int x^{5/2} \left (a+b x^2\right )^2 \left (c+d x^2\right ) \, dx=\frac {2}{21945} \, {\left (1155 \, b^{2} d x^{9} + 1463 \, {\left (b^{2} c + 2 \, a b d\right )} x^{7} + 3135 \, a^{2} c x^{3} + 1995 \, {\left (2 \, a b c + a^{2} d\right )} x^{5}\right )} \sqrt {x} \]

[In]

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

[Out]

2/21945*(1155*b^2*d*x^9 + 1463*(b^2*c + 2*a*b*d)*x^7 + 3135*a^2*c*x^3 + 1995*(2*a*b*c + a^2*d)*x^5)*sqrt(x)

Sympy [A] (verification not implemented)

Time = 0.61 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.27 \[ \int x^{5/2} \left (a+b x^2\right )^2 \left (c+d x^2\right ) \, dx=\frac {2 a^{2} c x^{\frac {7}{2}}}{7} + \frac {2 a^{2} d x^{\frac {11}{2}}}{11} + \frac {4 a b c x^{\frac {11}{2}}}{11} + \frac {4 a b d x^{\frac {15}{2}}}{15} + \frac {2 b^{2} c x^{\frac {15}{2}}}{15} + \frac {2 b^{2} d x^{\frac {19}{2}}}{19} \]

[In]

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

[Out]

2*a**2*c*x**(7/2)/7 + 2*a**2*d*x**(11/2)/11 + 4*a*b*c*x**(11/2)/11 + 4*a*b*d*x**(15/2)/15 + 2*b**2*c*x**(15/2)
/15 + 2*b**2*d*x**(19/2)/19

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.81 \[ \int x^{5/2} \left (a+b x^2\right )^2 \left (c+d x^2\right ) \, dx=\frac {2}{19} \, b^{2} d x^{\frac {19}{2}} + \frac {2}{15} \, {\left (b^{2} c + 2 \, a b d\right )} x^{\frac {15}{2}} + \frac {2}{7} \, a^{2} c x^{\frac {7}{2}} + \frac {2}{11} \, {\left (2 \, a b c + a^{2} d\right )} x^{\frac {11}{2}} \]

[In]

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

[Out]

2/19*b^2*d*x^(19/2) + 2/15*(b^2*c + 2*a*b*d)*x^(15/2) + 2/7*a^2*c*x^(7/2) + 2/11*(2*a*b*c + a^2*d)*x^(11/2)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.84 \[ \int x^{5/2} \left (a+b x^2\right )^2 \left (c+d x^2\right ) \, dx=\frac {2}{19} \, b^{2} d x^{\frac {19}{2}} + \frac {2}{15} \, b^{2} c x^{\frac {15}{2}} + \frac {4}{15} \, a b d x^{\frac {15}{2}} + \frac {4}{11} \, a b c x^{\frac {11}{2}} + \frac {2}{11} \, a^{2} d x^{\frac {11}{2}} + \frac {2}{7} \, a^{2} c x^{\frac {7}{2}} \]

[In]

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

[Out]

2/19*b^2*d*x^(19/2) + 2/15*b^2*c*x^(15/2) + 4/15*a*b*d*x^(15/2) + 4/11*a*b*c*x^(11/2) + 2/11*a^2*d*x^(11/2) +
2/7*a^2*c*x^(7/2)

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.81 \[ \int x^{5/2} \left (a+b x^2\right )^2 \left (c+d x^2\right ) \, dx=x^{11/2}\,\left (\frac {2\,d\,a^2}{11}+\frac {4\,b\,c\,a}{11}\right )+x^{15/2}\,\left (\frac {2\,c\,b^2}{15}+\frac {4\,a\,d\,b}{15}\right )+\frac {2\,a^2\,c\,x^{7/2}}{7}+\frac {2\,b^2\,d\,x^{19/2}}{19} \]

[In]

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

[Out]

x^(11/2)*((2*a^2*d)/11 + (4*a*b*c)/11) + x^(15/2)*((2*b^2*c)/15 + (4*a*b*d)/15) + (2*a^2*c*x^(7/2))/7 + (2*b^2
*d*x^(19/2))/19