\(\int (d x)^{5/2} (a^2+2 a b x^2+b^2 x^4)^2 \, dx\) [672]

Optimal result

Integrand size = 28, antiderivative size = 91 \[ \int (d x)^{5/2} \left (a^2+2 a b x^2+b^2 x^4\right )^2 \, dx=\frac {2 a^4 (d x)^{7/2}}{7 d}+\frac {8 a^3 b (d x)^{11/2}}{11 d^3}+\frac {4 a^2 b^2 (d x)^{15/2}}{5 d^5}+\frac {8 a b^3 (d x)^{19/2}}{19 d^7}+\frac {2 b^4 (d x)^{23/2}}{23 d^9} \]

[Out]

2/7*a^4*(d*x)^(7/2)/d+8/11*a^3*b*(d*x)^(11/2)/d^3+4/5*a^2*b^2*(d*x)^(15/2)/d^5+8/19*a*b^3*(d*x)^(19/2)/d^7+2/2
3*b^4*(d*x)^(23/2)/d^9

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {28, 276} \[ \int (d x)^{5/2} \left (a^2+2 a b x^2+b^2 x^4\right )^2 \, dx=\frac {2 a^4 (d x)^{7/2}}{7 d}+\frac {8 a^3 b (d x)^{11/2}}{11 d^3}+\frac {4 a^2 b^2 (d x)^{15/2}}{5 d^5}+\frac {8 a b^3 (d x)^{19/2}}{19 d^7}+\frac {2 b^4 (d x)^{23/2}}{23 d^9} \]

[In]

Int[(d*x)^(5/2)*(a^2 + 2*a*b*x^2 + b^2*x^4)^2,x]

[Out]

(2*a^4*(d*x)^(7/2))/(7*d) + (8*a^3*b*(d*x)^(11/2))/(11*d^3) + (4*a^2*b^2*(d*x)^(15/2))/(5*d^5) + (8*a*b^3*(d*x
)^(19/2))/(19*d^7) + (2*b^4*(d*x)^(23/2))/(23*d^9)

Rule 28

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/c^p, Int[u*(b/2 + c*x^n)^(2*
p), x], x] /; FreeQ[{a, b, c, n}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 276

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\int (d x)^{5/2} \left (a b+b^2 x^2\right )^4 \, dx}{b^4} \\ & = \frac {\int \left (a^4 b^4 (d x)^{5/2}+\frac {4 a^3 b^5 (d x)^{9/2}}{d^2}+\frac {6 a^2 b^6 (d x)^{13/2}}{d^4}+\frac {4 a b^7 (d x)^{17/2}}{d^6}+\frac {b^8 (d x)^{21/2}}{d^8}\right ) \, dx}{b^4} \\ & = \frac {2 a^4 (d x)^{7/2}}{7 d}+\frac {8 a^3 b (d x)^{11/2}}{11 d^3}+\frac {4 a^2 b^2 (d x)^{15/2}}{5 d^5}+\frac {8 a b^3 (d x)^{19/2}}{19 d^7}+\frac {2 b^4 (d x)^{23/2}}{23 d^9} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.60 \[ \int (d x)^{5/2} \left (a^2+2 a b x^2+b^2 x^4\right )^2 \, dx=\frac {2 x (d x)^{5/2} \left (24035 a^4+61180 a^3 b x^2+67298 a^2 b^2 x^4+35420 a b^3 x^6+7315 b^4 x^8\right )}{168245} \]

[In]

Integrate[(d*x)^(5/2)*(a^2 + 2*a*b*x^2 + b^2*x^4)^2,x]

[Out]

(2*x*(d*x)^(5/2)*(24035*a^4 + 61180*a^3*b*x^2 + 67298*a^2*b^2*x^4 + 35420*a*b^3*x^6 + 7315*b^4*x^8))/168245

Maple [A] (verified)

Time = 0.28 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.57

method	result	size
gosper	\(\frac {2 x \left (7315 b^{4} x^{8}+35420 a \,b^{3} x^{6}+67298 a^{2} b^{2} x^{4}+61180 a^{3} b \,x^{2}+24035 a^{4}\right ) \left (d x \right )^{\frac {5}{2}}}{168245}\)	\(52\)
pseudoelliptic	\(\frac {2 \sqrt {d x}\, \left (\frac {7}{23} b^{4} x^{8}+\frac {28}{19} a \,b^{3} x^{6}+\frac {14}{5} a^{2} b^{2} x^{4}+\frac {28}{11} a^{3} b \,x^{2}+a^{4}\right ) d^{2} x^{3}}{7}\)	\(55\)
trager	\(\frac {2 d^{2} x^{3} \left (7315 b^{4} x^{8}+35420 a \,b^{3} x^{6}+67298 a^{2} b^{2} x^{4}+61180 a^{3} b \,x^{2}+24035 a^{4}\right ) \sqrt {d x}}{168245}\)	\(57\)
risch	\(\frac {2 d^{3} x^{4} \left (7315 b^{4} x^{8}+35420 a \,b^{3} x^{6}+67298 a^{2} b^{2} x^{4}+61180 a^{3} b \,x^{2}+24035 a^{4}\right )}{168245 \sqrt {d x}}\)	\(57\)
derivativedivides	\(\frac {\frac {2 b^{4} \left (d x \right )^{\frac {23}{2}}}{23}+\frac {8 a \,b^{3} d^{2} \left (d x \right )^{\frac {19}{2}}}{19}+\frac {4 a^{2} d^{4} b^{2} \left (d x \right )^{\frac {15}{2}}}{5}+\frac {8 a^{3} d^{6} b \left (d x \right )^{\frac {11}{2}}}{11}+\frac {2 a^{4} d^{8} \left (d x \right )^{\frac {7}{2}}}{7}}{d^{9}}\)	\(74\)
default	\(\frac {\frac {2 b^{4} \left (d x \right )^{\frac {23}{2}}}{23}+\frac {8 a \,b^{3} d^{2} \left (d x \right )^{\frac {19}{2}}}{19}+\frac {4 a^{2} d^{4} b^{2} \left (d x \right )^{\frac {15}{2}}}{5}+\frac {8 a^{3} d^{6} b \left (d x \right )^{\frac {11}{2}}}{11}+\frac {2 a^{4} d^{8} \left (d x \right )^{\frac {7}{2}}}{7}}{d^{9}}\)	\(74\)

[In]

int((d*x)^(5/2)*(b^2*x^4+2*a*b*x^2+a^2)^2,x,method=_RETURNVERBOSE)

[Out]

2/168245*x*(7315*b^4*x^8+35420*a*b^3*x^6+67298*a^2*b^2*x^4+61180*a^3*b*x^2+24035*a^4)*(d*x)^(5/2)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.75 \[ \int (d x)^{5/2} \left (a^2+2 a b x^2+b^2 x^4\right )^2 \, dx=\frac {2}{168245} \, {\left (7315 \, b^{4} d^{2} x^{11} + 35420 \, a b^{3} d^{2} x^{9} + 67298 \, a^{2} b^{2} d^{2} x^{7} + 61180 \, a^{3} b d^{2} x^{5} + 24035 \, a^{4} d^{2} x^{3}\right )} \sqrt {d x} \]

[In]

integrate((d*x)^(5/2)*(b^2*x^4+2*a*b*x^2+a^2)^2,x, algorithm="fricas")

[Out]

2/168245*(7315*b^4*d^2*x^11 + 35420*a*b^3*d^2*x^9 + 67298*a^2*b^2*d^2*x^7 + 61180*a^3*b*d^2*x^5 + 24035*a^4*d^
2*x^3)*sqrt(d*x)

Sympy [A] (verification not implemented)

Time = 0.51 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.97 \[ \int (d x)^{5/2} \left (a^2+2 a b x^2+b^2 x^4\right )^2 \, dx=\frac {2 a^{4} x \left (d x\right )^{\frac {5}{2}}}{7} + \frac {8 a^{3} b x^{3} \left (d x\right )^{\frac {5}{2}}}{11} + \frac {4 a^{2} b^{2} x^{5} \left (d x\right )^{\frac {5}{2}}}{5} + \frac {8 a b^{3} x^{7} \left (d x\right )^{\frac {5}{2}}}{19} + \frac {2 b^{4} x^{9} \left (d x\right )^{\frac {5}{2}}}{23} \]

[In]

integrate((d*x)**(5/2)*(b**2*x**4+2*a*b*x**2+a**2)**2,x)

[Out]

2*a**4*x*(d*x)**(5/2)/7 + 8*a**3*b*x**3*(d*x)**(5/2)/11 + 4*a**2*b**2*x**5*(d*x)**(5/2)/5 + 8*a*b**3*x**7*(d*x
)**(5/2)/19 + 2*b**4*x**9*(d*x)**(5/2)/23

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.80 \[ \int (d x)^{5/2} \left (a^2+2 a b x^2+b^2 x^4\right )^2 \, dx=\frac {2 \, {\left (7315 \, \left (d x\right )^{\frac {23}{2}} b^{4} + 35420 \, \left (d x\right )^{\frac {19}{2}} a b^{3} d^{2} + 67298 \, \left (d x\right )^{\frac {15}{2}} a^{2} b^{2} d^{4} + 61180 \, \left (d x\right )^{\frac {11}{2}} a^{3} b d^{6} + 24035 \, \left (d x\right )^{\frac {7}{2}} a^{4} d^{8}\right )}}{168245 \, d^{9}} \]

[In]

integrate((d*x)^(5/2)*(b^2*x^4+2*a*b*x^2+a^2)^2,x, algorithm="maxima")

[Out]

2/168245*(7315*(d*x)^(23/2)*b^4 + 35420*(d*x)^(19/2)*a*b^3*d^2 + 67298*(d*x)^(15/2)*a^2*b^2*d^4 + 61180*(d*x)^
(11/2)*a^3*b*d^6 + 24035*(d*x)^(7/2)*a^4*d^8)/d^9

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.95 \[ \int (d x)^{5/2} \left (a^2+2 a b x^2+b^2 x^4\right )^2 \, dx=\frac {2}{23} \, \sqrt {d x} b^{4} d^{2} x^{11} + \frac {8}{19} \, \sqrt {d x} a b^{3} d^{2} x^{9} + \frac {4}{5} \, \sqrt {d x} a^{2} b^{2} d^{2} x^{7} + \frac {8}{11} \, \sqrt {d x} a^{3} b d^{2} x^{5} + \frac {2}{7} \, \sqrt {d x} a^{4} d^{2} x^{3} \]

[In]

integrate((d*x)^(5/2)*(b^2*x^4+2*a*b*x^2+a^2)^2,x, algorithm="giac")

[Out]

2/23*sqrt(d*x)*b^4*d^2*x^11 + 8/19*sqrt(d*x)*a*b^3*d^2*x^9 + 4/5*sqrt(d*x)*a^2*b^2*d^2*x^7 + 8/11*sqrt(d*x)*a^
3*b*d^2*x^5 + 2/7*sqrt(d*x)*a^4*d^2*x^3

Mupad [B] (verification not implemented)

Time = 13.32 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.78 \[ \int (d x)^{5/2} \left (a^2+2 a b x^2+b^2 x^4\right )^2 \, dx=\frac {2\,a^4\,{\left (d\,x\right )}^{7/2}}{7\,d}+\frac {2\,b^4\,{\left (d\,x\right )}^{23/2}}{23\,d^9}+\frac {4\,a^2\,b^2\,{\left (d\,x\right )}^{15/2}}{5\,d^5}+\frac {8\,a^3\,b\,{\left (d\,x\right )}^{11/2}}{11\,d^3}+\frac {8\,a\,b^3\,{\left (d\,x\right )}^{19/2}}{19\,d^7} \]

[In]

int((d*x)^(5/2)*(a^2 + b^2*x^4 + 2*a*b*x^2)^2,x)

[Out]

(2*a^4*(d*x)^(7/2))/(7*d) + (2*b^4*(d*x)^(23/2))/(23*d^9) + (4*a^2*b^2*(d*x)^(15/2))/(5*d^5) + (8*a^3*b*(d*x)^
(11/2))/(11*d^3) + (8*a*b^3*(d*x)^(19/2))/(19*d^7)