3.673 \(\int (d x)^{3/2} (a^2+2 a b x^2+b^2 x^4)^2 \, dx\)

Optimal. Leaf size=91 \[ \frac {2 a^4 (d x)^{5/2}}{5 d}+\frac {8 a^3 b (d x)^{9/2}}{9 d^3}+\frac {12 a^2 b^2 (d x)^{13/2}}{13 d^5}+\frac {8 a b^3 (d x)^{17/2}}{17 d^7}+\frac {2 b^4 (d x)^{21/2}}{21 d^9} \]

[Out]

2/5*a^4*(d*x)^(5/2)/d+8/9*a^3*b*(d*x)^(9/2)/d^3+12/13*a^2*b^2*(d*x)^(13/2)/d^5+8/17*a*b^3*(d*x)^(17/2)/d^7+2/2
1*b^4*(d*x)^(21/2)/d^9

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Rubi [A]  time = 0.04, 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, 270} \[ \frac {12 a^2 b^2 (d x)^{13/2}}{13 d^5}+\frac {8 a^3 b (d x)^{9/2}}{9 d^3}+\frac {2 a^4 (d x)^{5/2}}{5 d}+\frac {8 a b^3 (d x)^{17/2}}{17 d^7}+\frac {2 b^4 (d x)^{21/2}}{21 d^9} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^4*(d*x)^(5/2))/(5*d) + (8*a^3*b*(d*x)^(9/2))/(9*d^3) + (12*a^2*b^2*(d*x)^(13/2))/(13*d^5) + (8*a*b^3*(d*x
)^(17/2))/(17*d^7) + (2*b^4*(d*x)^(21/2))/(21*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 270

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 (d x)^{3/2} \left (a^2+2 a b x^2+b^2 x^4\right )^2 \, dx &=\frac {\int (d x)^{3/2} \left (a b+b^2 x^2\right )^4 \, dx}{b^4}\\ &=\frac {\int \left (a^4 b^4 (d x)^{3/2}+\frac {4 a^3 b^5 (d x)^{7/2}}{d^2}+\frac {6 a^2 b^6 (d x)^{11/2}}{d^4}+\frac {4 a b^7 (d x)^{15/2}}{d^6}+\frac {b^8 (d x)^{19/2}}{d^8}\right ) \, dx}{b^4}\\ &=\frac {2 a^4 (d x)^{5/2}}{5 d}+\frac {8 a^3 b (d x)^{9/2}}{9 d^3}+\frac {12 a^2 b^2 (d x)^{13/2}}{13 d^5}+\frac {8 a b^3 (d x)^{17/2}}{17 d^7}+\frac {2 b^4 (d x)^{21/2}}{21 d^9}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 55, normalized size = 0.60 \[ \frac {2 x (d x)^{3/2} \left (13923 a^4+30940 a^3 b x^2+32130 a^2 b^2 x^4+16380 a b^3 x^6+3315 b^4 x^8\right )}{69615} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x*(d*x)^(3/2)*(13923*a^4 + 30940*a^3*b*x^2 + 32130*a^2*b^2*x^4 + 16380*a*b^3*x^6 + 3315*b^4*x^8))/69615

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fricas [A]  time = 0.97, size = 58, normalized size = 0.64 \[ \frac {2}{69615} \, {\left (3315 \, b^{4} d x^{10} + 16380 \, a b^{3} d x^{8} + 32130 \, a^{2} b^{2} d x^{6} + 30940 \, a^{3} b d x^{4} + 13923 \, a^{4} d x^{2}\right )} \sqrt {d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/69615*(3315*b^4*d*x^10 + 16380*a*b^3*d*x^8 + 32130*a^2*b^2*d*x^6 + 30940*a^3*b*d*x^4 + 13923*a^4*d*x^2)*sqrt
(d*x)

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giac [A]  time = 0.17, size = 74, normalized size = 0.81 \[ \frac {2}{69615} \, {\left (3315 \, \sqrt {d x} b^{4} x^{10} + 16380 \, \sqrt {d x} a b^{3} x^{8} + 32130 \, \sqrt {d x} a^{2} b^{2} x^{6} + 30940 \, \sqrt {d x} a^{3} b x^{4} + 13923 \, \sqrt {d x} a^{4} x^{2}\right )} d \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/69615*(3315*sqrt(d*x)*b^4*x^10 + 16380*sqrt(d*x)*a*b^3*x^8 + 32130*sqrt(d*x)*a^2*b^2*x^6 + 30940*sqrt(d*x)*a
^3*b*x^4 + 13923*sqrt(d*x)*a^4*x^2)*d

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maple [A]  time = 0.01, size = 52, normalized size = 0.57 \[ \frac {2 \left (3315 b^{4} x^{8}+16380 a \,b^{3} x^{6}+32130 a^{2} b^{2} x^{4}+30940 a^{3} b \,x^{2}+13923 a^{4}\right ) \left (d x \right )^{\frac {3}{2}} x}{69615} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/69615*x*(3315*b^4*x^8+16380*a*b^3*x^6+32130*a^2*b^2*x^4+30940*a^3*b*x^2+13923*a^4)*(d*x)^(3/2)

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maxima [A]  time = 1.35, size = 73, normalized size = 0.80 \[ \frac {2 \, {\left (3315 \, \left (d x\right )^{\frac {21}{2}} b^{4} + 16380 \, \left (d x\right )^{\frac {17}{2}} a b^{3} d^{2} + 32130 \, \left (d x\right )^{\frac {13}{2}} a^{2} b^{2} d^{4} + 30940 \, \left (d x\right )^{\frac {9}{2}} a^{3} b d^{6} + 13923 \, \left (d x\right )^{\frac {5}{2}} a^{4} d^{8}\right )}}{69615 \, d^{9}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/69615*(3315*(d*x)^(21/2)*b^4 + 16380*(d*x)^(17/2)*a*b^3*d^2 + 32130*(d*x)^(13/2)*a^2*b^2*d^4 + 30940*(d*x)^(
9/2)*a^3*b*d^6 + 13923*(d*x)^(5/2)*a^4*d^8)/d^9

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mupad [B]  time = 0.03, size = 71, normalized size = 0.78 \[ \frac {2\,a^4\,{\left (d\,x\right )}^{5/2}}{5\,d}+\frac {2\,b^4\,{\left (d\,x\right )}^{21/2}}{21\,d^9}+\frac {12\,a^2\,b^2\,{\left (d\,x\right )}^{13/2}}{13\,d^5}+\frac {8\,a^3\,b\,{\left (d\,x\right )}^{9/2}}{9\,d^3}+\frac {8\,a\,b^3\,{\left (d\,x\right )}^{17/2}}{17\,d^7} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*a^4*(d*x)^(5/2))/(5*d) + (2*b^4*(d*x)^(21/2))/(21*d^9) + (12*a^2*b^2*(d*x)^(13/2))/(13*d^5) + (8*a^3*b*(d*x
)^(9/2))/(9*d^3) + (8*a*b^3*(d*x)^(17/2))/(17*d^7)

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sympy [A]  time = 2.69, size = 90, normalized size = 0.99 \[ \frac {2 a^{4} d^{\frac {3}{2}} x^{\frac {5}{2}}}{5} + \frac {8 a^{3} b d^{\frac {3}{2}} x^{\frac {9}{2}}}{9} + \frac {12 a^{2} b^{2} d^{\frac {3}{2}} x^{\frac {13}{2}}}{13} + \frac {8 a b^{3} d^{\frac {3}{2}} x^{\frac {17}{2}}}{17} + \frac {2 b^{4} d^{\frac {3}{2}} x^{\frac {21}{2}}}{21} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*a**4*d**(3/2)*x**(5/2)/5 + 8*a**3*b*d**(3/2)*x**(9/2)/9 + 12*a**2*b**2*d**(3/2)*x**(13/2)/13 + 8*a*b**3*d**(
3/2)*x**(17/2)/17 + 2*b**4*d**(3/2)*x**(21/2)/21

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