\(\int \sqrt {d x} (a^2+2 a b x^2+b^2 x^4) \, dx\) [667]

Optimal result

Integrand size = 26, antiderivative size = 51 \[ \int \sqrt {d x} \left (a^2+2 a b x^2+b^2 x^4\right ) \, dx=\frac {2 a^2 (d x)^{3/2}}{3 d}+\frac {4 a b (d x)^{7/2}}{7 d^3}+\frac {2 b^2 (d x)^{11/2}}{11 d^5} \]

[Out]

2/3*a^2*(d*x)^(3/2)/d+4/7*a*b*(d*x)^(7/2)/d^3+2/11*b^2*(d*x)^(11/2)/d^5

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.038, Rules used = {14} \[ \int \sqrt {d x} \left (a^2+2 a b x^2+b^2 x^4\right ) \, dx=\frac {2 a^2 (d x)^{3/2}}{3 d}+\frac {4 a b (d x)^{7/2}}{7 d^3}+\frac {2 b^2 (d x)^{11/2}}{11 d^5} \]

[In]

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

[Out]

(2*a^2*(d*x)^(3/2))/(3*d) + (4*a*b*(d*x)^(7/2))/(7*d^3) + (2*b^2*(d*x)^(11/2))/(11*d^5)

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rubi steps \begin{align*} \text {integral}& = \int \left (a^2 \sqrt {d x}+\frac {2 a b (d x)^{5/2}}{d^2}+\frac {b^2 (d x)^{9/2}}{d^4}\right ) \, dx \\ & = \frac {2 a^2 (d x)^{3/2}}{3 d}+\frac {4 a b (d x)^{7/2}}{7 d^3}+\frac {2 b^2 (d x)^{11/2}}{11 d^5} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.65 \[ \int \sqrt {d x} \left (a^2+2 a b x^2+b^2 x^4\right ) \, dx=\frac {2}{231} x \sqrt {d x} \left (77 a^2+66 a b x^2+21 b^2 x^4\right ) \]

[In]

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

[Out]

(2*x*Sqrt[d*x]*(77*a^2 + 66*a*b*x^2 + 21*b^2*x^4))/231

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.59

method	result	size
gosper	\(\frac {2 x \left (21 b^{2} x^{4}+66 a b \,x^{2}+77 a^{2}\right ) \sqrt {d x}}{231}\)	\(30\)
trager	\(\frac {2 x \left (21 b^{2} x^{4}+66 a b \,x^{2}+77 a^{2}\right ) \sqrt {d x}}{231}\)	\(30\)
pseudoelliptic	\(\frac {2 x \left (21 b^{2} x^{4}+66 a b \,x^{2}+77 a^{2}\right ) \sqrt {d x}}{231}\)	\(30\)
risch	\(\frac {2 d \,x^{2} \left (21 b^{2} x^{4}+66 a b \,x^{2}+77 a^{2}\right )}{231 \sqrt {d x}}\)	\(33\)
derivativedivides	\(\frac {\frac {2 b^{2} \left (d x \right )^{\frac {11}{2}}}{11}+\frac {4 a b \,d^{2} \left (d x \right )^{\frac {7}{2}}}{7}+\frac {2 a^{2} d^{4} \left (d x \right )^{\frac {3}{2}}}{3}}{d^{5}}\)	\(42\)
default	\(\frac {\frac {2 b^{2} \left (d x \right )^{\frac {11}{2}}}{11}+\frac {4 a b \,d^{2} \left (d x \right )^{\frac {7}{2}}}{7}+\frac {2 a^{2} d^{4} \left (d x \right )^{\frac {3}{2}}}{3}}{d^{5}}\)	\(42\)

[In]

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

[Out]

2/231*x*(21*b^2*x^4+66*a*b*x^2+77*a^2)*(d*x)^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.57 \[ \int \sqrt {d x} \left (a^2+2 a b x^2+b^2 x^4\right ) \, dx=\frac {2}{231} \, {\left (21 \, b^{2} x^{5} + 66 \, a b x^{3} + 77 \, a^{2} x\right )} \sqrt {d x} \]

[In]

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

[Out]

2/231*(21*b^2*x^5 + 66*a*b*x^3 + 77*a^2*x)*sqrt(d*x)

Sympy [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.94 \[ \int \sqrt {d x} \left (a^2+2 a b x^2+b^2 x^4\right ) \, dx=\frac {2 a^{2} x \sqrt {d x}}{3} + \frac {4 a b x^{3} \sqrt {d x}}{7} + \frac {2 b^{2} x^{5} \sqrt {d x}}{11} \]

[In]

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

[Out]

2*a**2*x*sqrt(d*x)/3 + 4*a*b*x**3*sqrt(d*x)/7 + 2*b**2*x**5*sqrt(d*x)/11

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.80 \[ \int \sqrt {d x} \left (a^2+2 a b x^2+b^2 x^4\right ) \, dx=\frac {2 \, {\left (21 \, \left (d x\right )^{\frac {11}{2}} b^{2} + 66 \, \left (d x\right )^{\frac {7}{2}} a b d^{2} + 77 \, \left (d x\right )^{\frac {3}{2}} a^{2} d^{4}\right )}}{231 \, d^{5}} \]

[In]

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

[Out]

2/231*(21*(d*x)^(11/2)*b^2 + 66*(d*x)^(7/2)*a*b*d^2 + 77*(d*x)^(3/2)*a^2*d^4)/d^5

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.73 \[ \int \sqrt {d x} \left (a^2+2 a b x^2+b^2 x^4\right ) \, dx=\frac {2}{11} \, \sqrt {d x} b^{2} x^{5} + \frac {4}{7} \, \sqrt {d x} a b x^{3} + \frac {2}{3} \, \sqrt {d x} a^{2} x \]

[In]

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

[Out]

2/11*sqrt(d*x)*b^2*x^5 + 4/7*sqrt(d*x)*a*b*x^3 + 2/3*sqrt(d*x)*a^2*x

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.80 \[ \int \sqrt {d x} \left (a^2+2 a b x^2+b^2 x^4\right ) \, dx=\frac {42\,b^2\,{\left (d\,x\right )}^{11/2}+154\,a^2\,d^4\,{\left (d\,x\right )}^{3/2}+132\,a\,b\,d^2\,{\left (d\,x\right )}^{7/2}}{231\,d^5} \]

[In]

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

[Out]

(42*b^2*(d*x)^(11/2) + 154*a^2*d^4*(d*x)^(3/2) + 132*a*b*d^2*(d*x)^(7/2))/(231*d^5)