\(\int \frac {(a^2+2 a b x^2+b^2 x^4)^3}{(d x)^{7/2}} \, dx\) [685]

Optimal result

Integrand size = 28, antiderivative size = 127 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^3}{(d x)^{7/2}} \, dx=-\frac {2 a^6}{5 d (d x)^{5/2}}-\frac {12 a^5 b}{d^3 \sqrt {d x}}+\frac {10 a^4 b^2 (d x)^{3/2}}{d^5}+\frac {40 a^3 b^3 (d x)^{7/2}}{7 d^7}+\frac {30 a^2 b^4 (d x)^{11/2}}{11 d^9}+\frac {4 a b^5 (d x)^{15/2}}{5 d^{11}}+\frac {2 b^6 (d x)^{19/2}}{19 d^{13}} \]

[Out]

-2/5*a^6/d/(d*x)^(5/2)+10*a^4*b^2*(d*x)^(3/2)/d^5+40/7*a^3*b^3*(d*x)^(7/2)/d^7+30/11*a^2*b^4*(d*x)^(11/2)/d^9+
4/5*a*b^5*(d*x)^(15/2)/d^11+2/19*b^6*(d*x)^(19/2)/d^13-12*a^5*b/d^3/(d*x)^(1/2)

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} \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^3}{(d x)^{7/2}} \, dx=-\frac {2 a^6}{5 d (d x)^{5/2}}-\frac {12 a^5 b}{d^3 \sqrt {d x}}+\frac {10 a^4 b^2 (d x)^{3/2}}{d^5}+\frac {40 a^3 b^3 (d x)^{7/2}}{7 d^7}+\frac {30 a^2 b^4 (d x)^{11/2}}{11 d^9}+\frac {4 a b^5 (d x)^{15/2}}{5 d^{11}}+\frac {2 b^6 (d x)^{19/2}}{19 d^{13}} \]

[In]

Int[(a^2 + 2*a*b*x^2 + b^2*x^4)^3/(d*x)^(7/2),x]

[Out]

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.65 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^3}{(d x)^{7/2}} \, dx=-\frac {2 \sqrt {d x} \left (1463 a^6+43890 a^5 b x^2-36575 a^4 b^2 x^4-20900 a^3 b^3 x^6-9975 a^2 b^4 x^8-2926 a b^5 x^{10}-385 b^6 x^{12}\right )}{7315 d^4 x^3} \]

[In]

Integrate[(a^2 + 2*a*b*x^2 + b^2*x^4)^3/(d*x)^(7/2),x]

[Out]

(-2*Sqrt[d*x]*(1463*a^6 + 43890*a^5*b*x^2 - 36575*a^4*b^2*x^4 - 20900*a^3*b^3*x^6 - 9975*a^2*b^4*x^8 - 2926*a*
b^5*x^10 - 385*b^6*x^12))/(7315*d^4*x^3)

Maple [A] (verified)

Time = 0.23 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.58

method	result	size
gosper	\(-\frac {2 \left (-385 b^{6} x^{12}-2926 a \,b^{5} x^{10}-9975 a^{2} b^{4} x^{8}-20900 a^{3} b^{3} x^{6}-36575 a^{4} b^{2} x^{4}+43890 a^{5} b \,x^{2}+1463 a^{6}\right ) x}{7315 \left (d x \right )^{\frac {7}{2}}}\)	\(74\)
pseudoelliptic	\(-\frac {2 \left (-\frac {5}{19} b^{6} x^{12}-2 a \,b^{5} x^{10}-\frac {75}{11} a^{2} b^{4} x^{8}-\frac {100}{7} a^{3} b^{3} x^{6}-25 a^{4} b^{2} x^{4}+30 a^{5} b \,x^{2}+a^{6}\right )}{5 \sqrt {d x}\, d^{3} x^{2}}\)	\(77\)
trager	\(-\frac {2 \left (-385 b^{6} x^{12}-2926 a \,b^{5} x^{10}-9975 a^{2} b^{4} x^{8}-20900 a^{3} b^{3} x^{6}-36575 a^{4} b^{2} x^{4}+43890 a^{5} b \,x^{2}+1463 a^{6}\right ) \sqrt {d x}}{7315 d^{4} x^{3}}\)	\(79\)
risch	\(-\frac {2 \left (-385 b^{6} x^{12}-2926 a \,b^{5} x^{10}-9975 a^{2} b^{4} x^{8}-20900 a^{3} b^{3} x^{6}-36575 a^{4} b^{2} x^{4}+43890 a^{5} b \,x^{2}+1463 a^{6}\right )}{7315 d^{3} x^{2} \sqrt {d x}}\)	\(79\)
derivativedivides	\(\frac {\frac {2 b^{6} \left (d x \right )^{\frac {19}{2}}}{19}+\frac {4 a \,b^{5} d^{2} \left (d x \right )^{\frac {15}{2}}}{5}+\frac {30 a^{2} b^{4} d^{4} \left (d x \right )^{\frac {11}{2}}}{11}+\frac {40 a^{3} b^{3} d^{6} \left (d x \right )^{\frac {7}{2}}}{7}+10 a^{4} b^{2} d^{8} \left (d x \right )^{\frac {3}{2}}-\frac {12 a^{5} b \,d^{10}}{\sqrt {d x}}-\frac {2 a^{6} d^{12}}{5 \left (d x \right )^{\frac {5}{2}}}}{d^{13}}\)	\(106\)
default	\(\frac {\frac {2 b^{6} \left (d x \right )^{\frac {19}{2}}}{19}+\frac {4 a \,b^{5} d^{2} \left (d x \right )^{\frac {15}{2}}}{5}+\frac {30 a^{2} b^{4} d^{4} \left (d x \right )^{\frac {11}{2}}}{11}+\frac {40 a^{3} b^{3} d^{6} \left (d x \right )^{\frac {7}{2}}}{7}+10 a^{4} b^{2} d^{8} \left (d x \right )^{\frac {3}{2}}-\frac {12 a^{5} b \,d^{10}}{\sqrt {d x}}-\frac {2 a^{6} d^{12}}{5 \left (d x \right )^{\frac {5}{2}}}}{d^{13}}\)	\(106\)

[In]

int((b^2*x^4+2*a*b*x^2+a^2)^3/(d*x)^(7/2),x,method=_RETURNVERBOSE)

[Out]

-2/7315*(-385*b^6*x^12-2926*a*b^5*x^10-9975*a^2*b^4*x^8-20900*a^3*b^3*x^6-36575*a^4*b^2*x^4+43890*a^5*b*x^2+14
63*a^6)*x/(d*x)^(7/2)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.61 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^3}{(d x)^{7/2}} \, dx=\frac {2 \, {\left (385 \, b^{6} x^{12} + 2926 \, a b^{5} x^{10} + 9975 \, a^{2} b^{4} x^{8} + 20900 \, a^{3} b^{3} x^{6} + 36575 \, a^{4} b^{2} x^{4} - 43890 \, a^{5} b x^{2} - 1463 \, a^{6}\right )} \sqrt {d x}}{7315 \, d^{4} x^{3}} \]

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^3/(d*x)^(7/2),x, algorithm="fricas")

[Out]

2/7315*(385*b^6*x^12 + 2926*a*b^5*x^10 + 9975*a^2*b^4*x^8 + 20900*a^3*b^3*x^6 + 36575*a^4*b^2*x^4 - 43890*a^5*
b*x^2 - 1463*a^6)*sqrt(d*x)/(d^4*x^3)

Sympy [A] (verification not implemented)

Time = 0.55 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.99 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^3}{(d x)^{7/2}} \, dx=- \frac {2 a^{6} x}{5 \left (d x\right )^{\frac {7}{2}}} - \frac {12 a^{5} b x^{3}}{\left (d x\right )^{\frac {7}{2}}} + \frac {10 a^{4} b^{2} x^{5}}{\left (d x\right )^{\frac {7}{2}}} + \frac {40 a^{3} b^{3} x^{7}}{7 \left (d x\right )^{\frac {7}{2}}} + \frac {30 a^{2} b^{4} x^{9}}{11 \left (d x\right )^{\frac {7}{2}}} + \frac {4 a b^{5} x^{11}}{5 \left (d x\right )^{\frac {7}{2}}} + \frac {2 b^{6} x^{13}}{19 \left (d x\right )^{\frac {7}{2}}} \]

[In]

integrate((b**2*x**4+2*a*b*x**2+a**2)**3/(d*x)**(7/2),x)

[Out]

-2*a**6*x/(5*(d*x)**(7/2)) - 12*a**5*b*x**3/(d*x)**(7/2) + 10*a**4*b**2*x**5/(d*x)**(7/2) + 40*a**3*b**3*x**7/
(7*(d*x)**(7/2)) + 30*a**2*b**4*x**9/(11*(d*x)**(7/2)) + 4*a*b**5*x**11/(5*(d*x)**(7/2)) + 2*b**6*x**13/(19*(d
*x)**(7/2))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.90 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^3}{(d x)^{7/2}} \, dx=-\frac {2 \, {\left (\frac {1463 \, {\left (30 \, a^{5} b d^{2} x^{2} + a^{6} d^{2}\right )}}{\left (d x\right )^{\frac {5}{2}} d^{2}} - \frac {385 \, \left (d x\right )^{\frac {19}{2}} b^{6} + 2926 \, \left (d x\right )^{\frac {15}{2}} a b^{5} d^{2} + 9975 \, \left (d x\right )^{\frac {11}{2}} a^{2} b^{4} d^{4} + 20900 \, \left (d x\right )^{\frac {7}{2}} a^{3} b^{3} d^{6} + 36575 \, \left (d x\right )^{\frac {3}{2}} a^{4} b^{2} d^{8}}{d^{12}}\right )}}{7315 \, d} \]

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^3/(d*x)^(7/2),x, algorithm="maxima")

[Out]

-2/7315*(1463*(30*a^5*b*d^2*x^2 + a^6*d^2)/((d*x)^(5/2)*d^2) - (385*(d*x)^(19/2)*b^6 + 2926*(d*x)^(15/2)*a*b^5
*d^2 + 9975*(d*x)^(11/2)*a^2*b^4*d^4 + 20900*(d*x)^(7/2)*a^3*b^3*d^6 + 36575*(d*x)^(3/2)*a^4*b^2*d^8)/d^12)/d

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.05 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^3}{(d x)^{7/2}} \, dx=-\frac {2 \, {\left (\frac {1463 \, {\left (30 \, a^{5} b d^{3} x^{2} + a^{6} d^{3}\right )}}{\sqrt {d x} d^{2} x^{2}} - \frac {385 \, \sqrt {d x} b^{6} d^{171} x^{9} + 2926 \, \sqrt {d x} a b^{5} d^{171} x^{7} + 9975 \, \sqrt {d x} a^{2} b^{4} d^{171} x^{5} + 20900 \, \sqrt {d x} a^{3} b^{3} d^{171} x^{3} + 36575 \, \sqrt {d x} a^{4} b^{2} d^{171} x}{d^{171}}\right )}}{7315 \, d^{4}} \]

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^3/(d*x)^(7/2),x, algorithm="giac")

[Out]

-2/7315*(1463*(30*a^5*b*d^3*x^2 + a^6*d^3)/(sqrt(d*x)*d^2*x^2) - (385*sqrt(d*x)*b^6*d^171*x^9 + 2926*sqrt(d*x)
*a*b^5*d^171*x^7 + 9975*sqrt(d*x)*a^2*b^4*d^171*x^5 + 20900*sqrt(d*x)*a^3*b^3*d^171*x^3 + 36575*sqrt(d*x)*a^4*
b^2*d^171*x)/d^171)/d^4

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.84 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^3}{(d x)^{7/2}} \, dx=\frac {2\,b^6\,{\left (d\,x\right )}^{19/2}}{19\,d^{13}}-\frac {\frac {2\,a^6\,d^2}{5}+12\,b\,a^5\,d^2\,x^2}{d^3\,{\left (d\,x\right )}^{5/2}}+\frac {10\,a^4\,b^2\,{\left (d\,x\right )}^{3/2}}{d^5}+\frac {40\,a^3\,b^3\,{\left (d\,x\right )}^{7/2}}{7\,d^7}+\frac {30\,a^2\,b^4\,{\left (d\,x\right )}^{11/2}}{11\,d^9}+\frac {4\,a\,b^5\,{\left (d\,x\right )}^{15/2}}{5\,d^{11}} \]

[In]

int((a^2 + b^2*x^4 + 2*a*b*x^2)^3/(d*x)^(7/2),x)

[Out]

(2*b^6*(d*x)^(19/2))/(19*d^13) - ((2*a^6*d^2)/5 + 12*a^5*b*d^2*x^2)/(d^3*(d*x)^(5/2)) + (10*a^4*b^2*(d*x)^(3/2
))/d^5 + (40*a^3*b^3*(d*x)^(7/2))/(7*d^7) + (30*a^2*b^4*(d*x)^(11/2))/(11*d^9) + (4*a*b^5*(d*x)^(15/2))/(5*d^1
1)