3.7.74 \(\int \sqrt {d x} (a^2+2 a b x^2+b^2 x^4)^2 \, dx\) [674]

Optimal. Leaf size=91 \[ \frac {2 a^4 (d x)^{3/2}}{3 d}+\frac {8 a^3 b (d x)^{7/2}}{7 d^3}+\frac {12 a^2 b^2 (d x)^{11/2}}{11 d^5}+\frac {8 a b^3 (d x)^{15/2}}{15 d^7}+\frac {2 b^4 (d x)^{19/2}}{19 d^9} \]

[Out]

2/3*a^4*(d*x)^(3/2)/d+8/7*a^3*b*(d*x)^(7/2)/d^3+12/11*a^2*b^2*(d*x)^(11/2)/d^5+8/15*a*b^3*(d*x)^(15/2)/d^7+2/1
9*b^4*(d*x)^(19/2)/d^9

________________________________________________________________________________________

Rubi [A]
time = 0.03, antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {28, 276} \begin {gather*} \frac {2 a^4 (d x)^{3/2}}{3 d}+\frac {8 a^3 b (d x)^{7/2}}{7 d^3}+\frac {12 a^2 b^2 (d x)^{11/2}}{11 d^5}+\frac {8 a b^3 (d x)^{15/2}}{15 d^7}+\frac {2 b^4 (d x)^{19/2}}{19 d^9} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^4*(d*x)^(3/2))/(3*d) + (8*a^3*b*(d*x)^(7/2))/(7*d^3) + (12*a^2*b^2*(d*x)^(11/2))/(11*d^5) + (8*a*b^3*(d*x
)^(15/2))/(15*d^7) + (2*b^4*(d*x)^(19/2))/(19*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*} \int \sqrt {d x} \left (a^2+2 a b x^2+b^2 x^4\right )^2 \, dx &=\frac {\int \sqrt {d x} \left (a b+b^2 x^2\right )^4 \, dx}{b^4}\\ &=\frac {\int \left (a^4 b^4 \sqrt {d x}+\frac {4 a^3 b^5 (d x)^{5/2}}{d^2}+\frac {6 a^2 b^6 (d x)^{9/2}}{d^4}+\frac {4 a b^7 (d x)^{13/2}}{d^6}+\frac {b^8 (d x)^{17/2}}{d^8}\right ) \, dx}{b^4}\\ &=\frac {2 a^4 (d x)^{3/2}}{3 d}+\frac {8 a^3 b (d x)^{7/2}}{7 d^3}+\frac {12 a^2 b^2 (d x)^{11/2}}{11 d^5}+\frac {8 a b^3 (d x)^{15/2}}{15 d^7}+\frac {2 b^4 (d x)^{19/2}}{19 d^9}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.02, size = 55, normalized size = 0.60 \begin {gather*} \frac {2 x \sqrt {d x} \left (7315 a^4+12540 a^3 b x^2+11970 a^2 b^2 x^4+5852 a b^3 x^6+1155 b^4 x^8\right )}{21945} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x*Sqrt[d*x]*(7315*a^4 + 12540*a^3*b*x^2 + 11970*a^2*b^2*x^4 + 5852*a*b^3*x^6 + 1155*b^4*x^8))/21945

________________________________________________________________________________________

Maple [A]
time = 0.06, size = 74, normalized size = 0.81

method	result	size
gosper	\(\frac {2 x \left (1155 b^{4} x^{8}+5852 a \,b^{3} x^{6}+11970 a^{2} b^{2} x^{4}+12540 a^{3} b \,x^{2}+7315 a^{4}\right ) \sqrt {d x}}{21945}\)	\(52\)
trager	\(\frac {2 x \left (1155 b^{4} x^{8}+5852 a \,b^{3} x^{6}+11970 a^{2} b^{2} x^{4}+12540 a^{3} b \,x^{2}+7315 a^{4}\right ) \sqrt {d x}}{21945}\)	\(52\)
risch	\(\frac {2 d \,x^{2} \left (1155 b^{4} x^{8}+5852 a \,b^{3} x^{6}+11970 a^{2} b^{2} x^{4}+12540 a^{3} b \,x^{2}+7315 a^{4}\right )}{21945 \sqrt {d x}}\)	\(55\)
derivativedivides	\(\frac {\frac {2 b^{4} \left (d x \right )^{\frac {19}{2}}}{19}+\frac {8 a \,b^{3} d^{2} \left (d x \right )^{\frac {15}{2}}}{15}+\frac {12 a^{2} d^{4} b^{2} \left (d x \right )^{\frac {11}{2}}}{11}+\frac {8 a^{3} b \,d^{6} \left (d x \right )^{\frac {7}{2}}}{7}+\frac {2 a^{4} d^{8} \left (d x \right )^{\frac {3}{2}}}{3}}{d^{9}}\)	\(74\)
default	\(\frac {\frac {2 b^{4} \left (d x \right )^{\frac {19}{2}}}{19}+\frac {8 a \,b^{3} d^{2} \left (d x \right )^{\frac {15}{2}}}{15}+\frac {12 a^{2} d^{4} b^{2} \left (d x \right )^{\frac {11}{2}}}{11}+\frac {8 a^{3} b \,d^{6} \left (d x \right )^{\frac {7}{2}}}{7}+\frac {2 a^{4} d^{8} \left (d x \right )^{\frac {3}{2}}}{3}}{d^{9}}\)	\(74\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b^2*x^4+2*a*b*x^2+a^2)^2*(d*x)^(1/2),x,method=_RETURNVERBOSE)

[Out]

2/d^9*(1/19*b^4*(d*x)^(19/2)+4/15*a*b^3*d^2*(d*x)^(15/2)+6/11*a^2*d^4*b^2*(d*x)^(11/2)+4/7*a^3*b*d^6*(d*x)^(7/
2)+1/3*a^4*d^8*(d*x)^(3/2))

________________________________________________________________________________________

Maxima [A]
time = 0.27, size = 73, normalized size = 0.80 \begin {gather*} \frac {2 \, {\left (1155 \, \left (d x\right )^{\frac {19}{2}} b^{4} + 5852 \, \left (d x\right )^{\frac {15}{2}} a b^{3} d^{2} + 11970 \, \left (d x\right )^{\frac {11}{2}} a^{2} b^{2} d^{4} + 12540 \, \left (d x\right )^{\frac {7}{2}} a^{3} b d^{6} + 7315 \, \left (d x\right )^{\frac {3}{2}} a^{4} d^{8}\right )}}{21945 \, d^{9}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^2*(d*x)^(1/2),x, algorithm="maxima")

[Out]

2/21945*(1155*(d*x)^(19/2)*b^4 + 5852*(d*x)^(15/2)*a*b^3*d^2 + 11970*(d*x)^(11/2)*a^2*b^2*d^4 + 12540*(d*x)^(7
/2)*a^3*b*d^6 + 7315*(d*x)^(3/2)*a^4*d^8)/d^9

________________________________________________________________________________________

Fricas [A]
time = 0.33, size = 51, normalized size = 0.56 \begin {gather*} \frac {2}{21945} \, {\left (1155 \, b^{4} x^{9} + 5852 \, a b^{3} x^{7} + 11970 \, a^{2} b^{2} x^{5} + 12540 \, a^{3} b x^{3} + 7315 \, a^{4} x\right )} \sqrt {d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^2*(d*x)^(1/2),x, algorithm="fricas")

[Out]

2/21945*(1155*b^4*x^9 + 5852*a*b^3*x^7 + 11970*a^2*b^2*x^5 + 12540*a^3*b*x^3 + 7315*a^4*x)*sqrt(d*x)

________________________________________________________________________________________

Sympy [A]
time = 0.24, size = 88, normalized size = 0.97 \begin {gather*} \frac {2 a^{4} x \sqrt {d x}}{3} + \frac {8 a^{3} b x^{3} \sqrt {d x}}{7} + \frac {12 a^{2} b^{2} x^{5} \sqrt {d x}}{11} + \frac {8 a b^{3} x^{7} \sqrt {d x}}{15} + \frac {2 b^{4} x^{9} \sqrt {d x}}{19} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b**2*x**4+2*a*b*x**2+a**2)**2*(d*x)**(1/2),x)

[Out]

2*a**4*x*sqrt(d*x)/3 + 8*a**3*b*x**3*sqrt(d*x)/7 + 12*a**2*b**2*x**5*sqrt(d*x)/11 + 8*a*b**3*x**7*sqrt(d*x)/15
+ 2*b**4*x**9*sqrt(d*x)/19

________________________________________________________________________________________

Giac [A]
time = 3.45, size = 69, normalized size = 0.76 \begin {gather*} \frac {2}{19} \, \sqrt {d x} b^{4} x^{9} + \frac {8}{15} \, \sqrt {d x} a b^{3} x^{7} + \frac {12}{11} \, \sqrt {d x} a^{2} b^{2} x^{5} + \frac {8}{7} \, \sqrt {d x} a^{3} b x^{3} + \frac {2}{3} \, \sqrt {d x} a^{4} x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^2*(d*x)^(1/2),x, algorithm="giac")

[Out]

2/19*sqrt(d*x)*b^4*x^9 + 8/15*sqrt(d*x)*a*b^3*x^7 + 12/11*sqrt(d*x)*a^2*b^2*x^5 + 8/7*sqrt(d*x)*a^3*b*x^3 + 2/
3*sqrt(d*x)*a^4*x

________________________________________________________________________________________

Mupad [B]
time = 0.03, size = 71, normalized size = 0.78 \begin {gather*} \frac {2\,a^4\,{\left (d\,x\right )}^{3/2}}{3\,d}+\frac {2\,b^4\,{\left (d\,x\right )}^{19/2}}{19\,d^9}+\frac {12\,a^2\,b^2\,{\left (d\,x\right )}^{11/2}}{11\,d^5}+\frac {8\,a^3\,b\,{\left (d\,x\right )}^{7/2}}{7\,d^3}+\frac {8\,a\,b^3\,{\left (d\,x\right )}^{15/2}}{15\,d^7} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x)^(1/2)*(a^2 + b^2*x^4 + 2*a*b*x^2)^2,x)

[Out]

(2*a^4*(d*x)^(3/2))/(3*d) + (2*b^4*(d*x)^(19/2))/(19*d^9) + (12*a^2*b^2*(d*x)^(11/2))/(11*d^5) + (8*a^3*b*(d*x
)^(7/2))/(7*d^3) + (8*a*b^3*(d*x)^(15/2))/(15*d^7)

________________________________________________________________________________________