\(\int \frac {(a^2+2 a b x^2+b^2 x^4)^2}{(d x)^{3/2}} \, dx\) [676]

Optimal result

Integrand size = 28, antiderivative size = 89 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^2}{(d x)^{3/2}} \, dx=-\frac {2 a^4}{d \sqrt {d x}}+\frac {8 a^3 b (d x)^{3/2}}{3 d^3}+\frac {12 a^2 b^2 (d x)^{7/2}}{7 d^5}+\frac {8 a b^3 (d x)^{11/2}}{11 d^7}+\frac {2 b^4 (d x)^{15/2}}{15 d^9} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 89, 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} \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^2}{(d x)^{3/2}} \, dx=-\frac {2 a^4}{d \sqrt {d x}}+\frac {8 a^3 b (d x)^{3/2}}{3 d^3}+\frac {12 a^2 b^2 (d x)^{7/2}}{7 d^5}+\frac {8 a b^3 (d x)^{11/2}}{11 d^7}+\frac {2 b^4 (d x)^{15/2}}{15 d^9} \]

[In]

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

[Out]

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.62 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^2}{(d x)^{3/2}} \, dx=-\frac {2 x \left (1155 a^4-1540 a^3 b x^2-990 a^2 b^2 x^4-420 a b^3 x^6-77 b^4 x^8\right )}{1155 (d x)^{3/2}} \]

[In]

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

[Out]

(-2*x*(1155*a^4 - 1540*a^3*b*x^2 - 990*a^2*b^2*x^4 - 420*a*b^3*x^6 - 77*b^4*x^8))/(1155*(d*x)^(3/2))

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.58

method	result	size
gosper	\(-\frac {2 \left (-77 b^{4} x^{8}-420 a \,b^{3} x^{6}-990 a^{2} b^{2} x^{4}-1540 a^{3} b \,x^{2}+1155 a^{4}\right ) x}{1155 \left (d x \right )^{\frac {3}{2}}}\)	\(52\)
pseudoelliptic	\(-\frac {2 \left (-\frac {1}{15} b^{4} x^{8}-\frac {4}{11} a \,b^{3} x^{6}-\frac {6}{7} a^{2} b^{2} x^{4}-\frac {4}{3} a^{3} b \,x^{2}+a^{4}\right )}{\sqrt {d x}\, d}\)	\(52\)
risch	\(-\frac {2 \left (-77 b^{4} x^{8}-420 a \,b^{3} x^{6}-990 a^{2} b^{2} x^{4}-1540 a^{3} b \,x^{2}+1155 a^{4}\right )}{1155 d \sqrt {d x}}\)	\(54\)
trager	\(-\frac {2 \left (-77 b^{4} x^{8}-420 a \,b^{3} x^{6}-990 a^{2} b^{2} x^{4}-1540 a^{3} b \,x^{2}+1155 a^{4}\right ) \sqrt {d x}}{1155 d^{2} x}\)	\(57\)
derivativedivides	\(\frac {\frac {2 b^{4} \left (d x \right )^{\frac {15}{2}}}{15}+\frac {8 a \,b^{3} d^{2} \left (d x \right )^{\frac {11}{2}}}{11}+\frac {12 a^{2} b^{2} d^{4} \left (d x \right )^{\frac {7}{2}}}{7}+\frac {8 a^{3} b \,d^{6} \left (d x \right )^{\frac {3}{2}}}{3}-\frac {2 a^{4} d^{8}}{\sqrt {d x}}}{d^{9}}\)	\(74\)
default	\(\frac {\frac {2 b^{4} \left (d x \right )^{\frac {15}{2}}}{15}+\frac {8 a \,b^{3} d^{2} \left (d x \right )^{\frac {11}{2}}}{11}+\frac {12 a^{2} b^{2} d^{4} \left (d x \right )^{\frac {7}{2}}}{7}+\frac {8 a^{3} b \,d^{6} \left (d x \right )^{\frac {3}{2}}}{3}-\frac {2 a^{4} d^{8}}{\sqrt {d x}}}{d^{9}}\)	\(74\)

[In]

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

[Out]

-2/1155*(-77*b^4*x^8-420*a*b^3*x^6-990*a^2*b^2*x^4-1540*a^3*b*x^2+1155*a^4)*x/(d*x)^(3/2)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.63 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^2}{(d x)^{3/2}} \, dx=\frac {2 \, {\left (77 \, b^{4} x^{8} + 420 \, a b^{3} x^{6} + 990 \, a^{2} b^{2} x^{4} + 1540 \, a^{3} b x^{2} - 1155 \, a^{4}\right )} \sqrt {d x}}{1155 \, d^{2} x} \]

[In]

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

[Out]

2/1155*(77*b^4*x^8 + 420*a*b^3*x^6 + 990*a^2*b^2*x^4 + 1540*a^3*b*x^2 - 1155*a^4)*sqrt(d*x)/(d^2*x)

Sympy [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.98 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^2}{(d x)^{3/2}} \, dx=- \frac {2 a^{4} x}{\left (d x\right )^{\frac {3}{2}}} + \frac {8 a^{3} b x^{3}}{3 \left (d x\right )^{\frac {3}{2}}} + \frac {12 a^{2} b^{2} x^{5}}{7 \left (d x\right )^{\frac {3}{2}}} + \frac {8 a b^{3} x^{7}}{11 \left (d x\right )^{\frac {3}{2}}} + \frac {2 b^{4} x^{9}}{15 \left (d x\right )^{\frac {3}{2}}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.85 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^2}{(d x)^{3/2}} \, dx=-\frac {2 \, {\left (\frac {1155 \, a^{4}}{\sqrt {d x}} - \frac {77 \, \left (d x\right )^{\frac {15}{2}} b^{4} + 420 \, \left (d x\right )^{\frac {11}{2}} a b^{3} d^{2} + 990 \, \left (d x\right )^{\frac {7}{2}} a^{2} b^{2} d^{4} + 1540 \, \left (d x\right )^{\frac {3}{2}} a^{3} b d^{6}}{d^{8}}\right )}}{1155 \, d} \]

[In]

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

[Out]

-2/1155*(1155*a^4/sqrt(d*x) - (77*(d*x)^(15/2)*b^4 + 420*(d*x)^(11/2)*a*b^3*d^2 + 990*(d*x)^(7/2)*a^2*b^2*d^4
+ 1540*(d*x)^(3/2)*a^3*b*d^6)/d^8)/d

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.00 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^2}{(d x)^{3/2}} \, dx=-\frac {2 \, {\left (\frac {1155 \, a^{4}}{\sqrt {d x}} - \frac {77 \, \sqrt {d x} b^{4} d^{119} x^{7} + 420 \, \sqrt {d x} a b^{3} d^{119} x^{5} + 990 \, \sqrt {d x} a^{2} b^{2} d^{119} x^{3} + 1540 \, \sqrt {d x} a^{3} b d^{119} x}{d^{120}}\right )}}{1155 \, d} \]

[In]

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

[Out]

-2/1155*(1155*a^4/sqrt(d*x) - (77*sqrt(d*x)*b^4*d^119*x^7 + 420*sqrt(d*x)*a*b^3*d^119*x^5 + 990*sqrt(d*x)*a^2*
b^2*d^119*x^3 + 1540*sqrt(d*x)*a^3*b*d^119*x)/d^120)/d

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.80 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^2}{(d x)^{3/2}} \, dx=\frac {2\,b^4\,{\left (d\,x\right )}^{15/2}}{15\,d^9}-\frac {2\,a^4}{d\,\sqrt {d\,x}}+\frac {12\,a^2\,b^2\,{\left (d\,x\right )}^{7/2}}{7\,d^5}+\frac {8\,a^3\,b\,{\left (d\,x\right )}^{3/2}}{3\,d^3}+\frac {8\,a\,b^3\,{\left (d\,x\right )}^{11/2}}{11\,d^7} \]

[In]

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

[Out]

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