\(\int \frac {(a^2+2 a b x^2+b^2 x^4)^3}{(d x)^{5/2}} \, dx\) [684]

Optimal result

Integrand size = 28, antiderivative size = 127 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^3}{(d x)^{5/2}} \, dx=-\frac {2 a^6}{3 d (d x)^{3/2}}+\frac {12 a^5 b \sqrt {d x}}{d^3}+\frac {6 a^4 b^2 (d x)^{5/2}}{d^5}+\frac {40 a^3 b^3 (d x)^{9/2}}{9 d^7}+\frac {30 a^2 b^4 (d x)^{13/2}}{13 d^9}+\frac {12 a b^5 (d x)^{17/2}}{17 d^{11}}+\frac {2 b^6 (d x)^{21/2}}{21 d^{13}} \]

[Out]

-2/3*a^6/d/(d*x)^(3/2)+6*a^4*b^2*(d*x)^(5/2)/d^5+40/9*a^3*b^3*(d*x)^(9/2)/d^7+30/13*a^2*b^4*(d*x)^(13/2)/d^9+1
2/17*a*b^5*(d*x)^(17/2)/d^11+2/21*b^6*(d*x)^(21/2)/d^13+12*a^5*b*(d*x)^(1/2)/d^3

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 127, 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} \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^3}{(d x)^{5/2}} \, dx=-\frac {2 a^6}{3 d (d x)^{3/2}}+\frac {12 a^5 b \sqrt {d x}}{d^3}+\frac {6 a^4 b^2 (d x)^{5/2}}{d^5}+\frac {40 a^3 b^3 (d x)^{9/2}}{9 d^7}+\frac {30 a^2 b^4 (d x)^{13/2}}{13 d^9}+\frac {12 a b^5 (d x)^{17/2}}{17 d^{11}}+\frac {2 b^6 (d x)^{21/2}}{21 d^{13}} \]

[In]

Int[(a^2 + 2*a*b*x^2 + b^2*x^4)^3/(d*x)^(5/2),x]

[Out]

(-2*a^6)/(3*d*(d*x)^(3/2)) + (12*a^5*b*Sqrt[d*x])/d^3 + (6*a^4*b^2*(d*x)^(5/2))/d^5 + (40*a^3*b^3*(d*x)^(9/2))
/(9*d^7) + (30*a^2*b^4*(d*x)^(13/2))/(13*d^9) + (12*a*b^5*(d*x)^(17/2))/(17*d^11) + (2*b^6*(d*x)^(21/2))/(21*d
^13)

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 )^6}{(d x)^{5/2}} \, dx}{b^6} \\ & = \frac {\int \left (\frac {a^6 b^6}{(d x)^{5/2}}+\frac {6 a^5 b^7}{d^2 \sqrt {d x}}+\frac {15 a^4 b^8 (d x)^{3/2}}{d^4}+\frac {20 a^3 b^9 (d x)^{7/2}}{d^6}+\frac {15 a^2 b^{10} (d x)^{11/2}}{d^8}+\frac {6 a b^{11} (d x)^{15/2}}{d^{10}}+\frac {b^{12} (d x)^{19/2}}{d^{12}}\right ) \, dx}{b^6} \\ & = -\frac {2 a^6}{3 d (d x)^{3/2}}+\frac {12 a^5 b \sqrt {d x}}{d^3}+\frac {6 a^4 b^2 (d x)^{5/2}}{d^5}+\frac {40 a^3 b^3 (d x)^{9/2}}{9 d^7}+\frac {30 a^2 b^4 (d x)^{13/2}}{13 d^9}+\frac {12 a b^5 (d x)^{17/2}}{17 d^{11}}+\frac {2 b^6 (d x)^{21/2}}{21 d^{13}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.61 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^3}{(d x)^{5/2}} \, dx=-\frac {2 x \left (4641 a^6-83538 a^5 b x^2-41769 a^4 b^2 x^4-30940 a^3 b^3 x^6-16065 a^2 b^4 x^8-4914 a b^5 x^{10}-663 b^6 x^{12}\right )}{13923 (d x)^{5/2}} \]

[In]

Integrate[(a^2 + 2*a*b*x^2 + b^2*x^4)^3/(d*x)^(5/2),x]

[Out]

(-2*x*(4641*a^6 - 83538*a^5*b*x^2 - 41769*a^4*b^2*x^4 - 30940*a^3*b^3*x^6 - 16065*a^2*b^4*x^8 - 4914*a*b^5*x^1
0 - 663*b^6*x^12))/(13923*(d*x)^(5/2))

Maple [A] (verified)

Time = 0.22 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.58

method	result	size
gosper	\(-\frac {2 \left (-663 b^{6} x^{12}-4914 a \,b^{5} x^{10}-16065 a^{2} b^{4} x^{8}-30940 a^{3} b^{3} x^{6}-41769 a^{4} b^{2} x^{4}-83538 a^{5} b \,x^{2}+4641 a^{6}\right ) x}{13923 \left (d x \right )^{\frac {5}{2}}}\)	\(74\)
pseudoelliptic	\(-\frac {2 \left (-\frac {1}{7} b^{6} x^{12}-\frac {18}{17} a \,b^{5} x^{10}-\frac {45}{13} a^{2} b^{4} x^{8}-\frac {20}{3} a^{3} b^{3} x^{6}-9 a^{4} b^{2} x^{4}-18 a^{5} b \,x^{2}+a^{6}\right )}{3 \sqrt {d x}\, d^{2} x}\)	\(77\)
trager	\(-\frac {2 \left (-663 b^{6} x^{12}-4914 a \,b^{5} x^{10}-16065 a^{2} b^{4} x^{8}-30940 a^{3} b^{3} x^{6}-41769 a^{4} b^{2} x^{4}-83538 a^{5} b \,x^{2}+4641 a^{6}\right ) \sqrt {d x}}{13923 d^{3} x^{2}}\)	\(79\)
risch	\(-\frac {2 \left (-663 b^{6} x^{12}-4914 a \,b^{5} x^{10}-16065 a^{2} b^{4} x^{8}-30940 a^{3} b^{3} x^{6}-41769 a^{4} b^{2} x^{4}-83538 a^{5} b \,x^{2}+4641 a^{6}\right )}{13923 d^{2} x \sqrt {d x}}\)	\(79\)
derivativedivides	\(\frac {\frac {2 b^{6} \left (d x \right )^{\frac {21}{2}}}{21}+\frac {12 a \,b^{5} d^{2} \left (d x \right )^{\frac {17}{2}}}{17}+\frac {30 a^{2} b^{4} d^{4} \left (d x \right )^{\frac {13}{2}}}{13}+\frac {40 a^{3} b^{3} d^{6} \left (d x \right )^{\frac {9}{2}}}{9}+6 a^{4} b^{2} d^{8} \left (d x \right )^{\frac {5}{2}}+12 a^{5} b \,d^{10} \sqrt {d x}-\frac {2 a^{6} d^{12}}{3 \left (d x \right )^{\frac {3}{2}}}}{d^{13}}\)	\(106\)
default	\(\frac {\frac {2 b^{6} \left (d x \right )^{\frac {21}{2}}}{21}+\frac {12 a \,b^{5} d^{2} \left (d x \right )^{\frac {17}{2}}}{17}+\frac {30 a^{2} b^{4} d^{4} \left (d x \right )^{\frac {13}{2}}}{13}+\frac {40 a^{3} b^{3} d^{6} \left (d x \right )^{\frac {9}{2}}}{9}+6 a^{4} b^{2} d^{8} \left (d x \right )^{\frac {5}{2}}+12 a^{5} b \,d^{10} \sqrt {d x}-\frac {2 a^{6} d^{12}}{3 \left (d x \right )^{\frac {3}{2}}}}{d^{13}}\)	\(106\)

[In]

int((b^2*x^4+2*a*b*x^2+a^2)^3/(d*x)^(5/2),x,method=_RETURNVERBOSE)

[Out]

-2/13923*(-663*b^6*x^12-4914*a*b^5*x^10-16065*a^2*b^4*x^8-30940*a^3*b^3*x^6-41769*a^4*b^2*x^4-83538*a^5*b*x^2+
4641*a^6)*x/(d*x)^(5/2)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.61 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^3}{(d x)^{5/2}} \, dx=\frac {2 \, {\left (663 \, b^{6} x^{12} + 4914 \, a b^{5} x^{10} + 16065 \, a^{2} b^{4} x^{8} + 30940 \, a^{3} b^{3} x^{6} + 41769 \, a^{4} b^{2} x^{4} + 83538 \, a^{5} b x^{2} - 4641 \, a^{6}\right )} \sqrt {d x}}{13923 \, d^{3} x^{2}} \]

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^3/(d*x)^(5/2),x, algorithm="fricas")

[Out]

2/13923*(663*b^6*x^12 + 4914*a*b^5*x^10 + 16065*a^2*b^4*x^8 + 30940*a^3*b^3*x^6 + 41769*a^4*b^2*x^4 + 83538*a^
5*b*x^2 - 4641*a^6)*sqrt(d*x)/(d^3*x^2)

Sympy [A] (verification not implemented)

Time = 0.44 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.99 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^3}{(d x)^{5/2}} \, dx=- \frac {2 a^{6} x}{3 \left (d x\right )^{\frac {5}{2}}} + \frac {12 a^{5} b x^{3}}{\left (d x\right )^{\frac {5}{2}}} + \frac {6 a^{4} b^{2} x^{5}}{\left (d x\right )^{\frac {5}{2}}} + \frac {40 a^{3} b^{3} x^{7}}{9 \left (d x\right )^{\frac {5}{2}}} + \frac {30 a^{2} b^{4} x^{9}}{13 \left (d x\right )^{\frac {5}{2}}} + \frac {12 a b^{5} x^{11}}{17 \left (d x\right )^{\frac {5}{2}}} + \frac {2 b^{6} x^{13}}{21 \left (d x\right )^{\frac {5}{2}}} \]

[In]

integrate((b**2*x**4+2*a*b*x**2+a**2)**3/(d*x)**(5/2),x)

[Out]

-2*a**6*x/(3*(d*x)**(5/2)) + 12*a**5*b*x**3/(d*x)**(5/2) + 6*a**4*b**2*x**5/(d*x)**(5/2) + 40*a**3*b**3*x**7/(
9*(d*x)**(5/2)) + 30*a**2*b**4*x**9/(13*(d*x)**(5/2)) + 12*a*b**5*x**11/(17*(d*x)**(5/2)) + 2*b**6*x**13/(21*(
d*x)**(5/2))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.85 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^3}{(d x)^{5/2}} \, dx=-\frac {2 \, {\left (\frac {4641 \, a^{6}}{\left (d x\right )^{\frac {3}{2}}} - \frac {663 \, \left (d x\right )^{\frac {21}{2}} b^{6} + 4914 \, \left (d x\right )^{\frac {17}{2}} a b^{5} d^{2} + 16065 \, \left (d x\right )^{\frac {13}{2}} a^{2} b^{4} d^{4} + 30940 \, \left (d x\right )^{\frac {9}{2}} a^{3} b^{3} d^{6} + 41769 \, \left (d x\right )^{\frac {5}{2}} a^{4} b^{2} d^{8} + 83538 \, \sqrt {d x} a^{5} b d^{10}}{d^{12}}\right )}}{13923 \, d} \]

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^3/(d*x)^(5/2),x, algorithm="maxima")

[Out]

-2/13923*(4641*a^6/(d*x)^(3/2) - (663*(d*x)^(21/2)*b^6 + 4914*(d*x)^(17/2)*a*b^5*d^2 + 16065*(d*x)^(13/2)*a^2*
b^4*d^4 + 30940*(d*x)^(9/2)*a^3*b^3*d^6 + 41769*(d*x)^(5/2)*a^4*b^2*d^8 + 83538*sqrt(d*x)*a^5*b*d^10)/d^12)/d

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.02 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^3}{(d x)^{5/2}} \, dx=-\frac {2 \, {\left (\frac {4641 \, a^{6} d}{\sqrt {d x} x} - \frac {663 \, \sqrt {d x} b^{6} d^{210} x^{10} + 4914 \, \sqrt {d x} a b^{5} d^{210} x^{8} + 16065 \, \sqrt {d x} a^{2} b^{4} d^{210} x^{6} + 30940 \, \sqrt {d x} a^{3} b^{3} d^{210} x^{4} + 41769 \, \sqrt {d x} a^{4} b^{2} d^{210} x^{2} + 83538 \, \sqrt {d x} a^{5} b d^{210}}{d^{210}}\right )}}{13923 \, d^{3}} \]

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^3/(d*x)^(5/2),x, algorithm="giac")

[Out]

-2/13923*(4641*a^6*d/(sqrt(d*x)*x) - (663*sqrt(d*x)*b^6*d^210*x^10 + 4914*sqrt(d*x)*a*b^5*d^210*x^8 + 16065*sq
rt(d*x)*a^2*b^4*d^210*x^6 + 30940*sqrt(d*x)*a^3*b^3*d^210*x^4 + 41769*sqrt(d*x)*a^4*b^2*d^210*x^2 + 83538*sqrt
(d*x)*a^5*b*d^210)/d^210)/d^3

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.81 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^3}{(d x)^{5/2}} \, dx=\frac {2\,b^6\,{\left (d\,x\right )}^{21/2}}{21\,d^{13}}-\frac {2\,a^6}{3\,d\,{\left (d\,x\right )}^{3/2}}+\frac {6\,a^4\,b^2\,{\left (d\,x\right )}^{5/2}}{d^5}+\frac {40\,a^3\,b^3\,{\left (d\,x\right )}^{9/2}}{9\,d^7}+\frac {30\,a^2\,b^4\,{\left (d\,x\right )}^{13/2}}{13\,d^9}+\frac {12\,a^5\,b\,\sqrt {d\,x}}{d^3}+\frac {12\,a\,b^5\,{\left (d\,x\right )}^{17/2}}{17\,d^{11}} \]

[In]

int((a^2 + b^2*x^4 + 2*a*b*x^2)^3/(d*x)^(5/2),x)

[Out]

(2*b^6*(d*x)^(21/2))/(21*d^13) - (2*a^6)/(3*d*(d*x)^(3/2)) + (6*a^4*b^2*(d*x)^(5/2))/d^5 + (40*a^3*b^3*(d*x)^(
9/2))/(9*d^7) + (30*a^2*b^4*(d*x)^(13/2))/(13*d^9) + (12*a^5*b*(d*x)^(1/2))/d^3 + (12*a*b^5*(d*x)^(17/2))/(17*
d^11)