\(\int \sqrt {d x} (a^2+2 a b x^2+b^2 x^4)^{5/2} \, dx\) [744]

Optimal result

Integrand size = 30, antiderivative size = 297 \[ \int \sqrt {d x} \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\frac {2 a^5 (d x)^{3/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{3 d \left (a+b x^2\right )}+\frac {10 a^4 b (d x)^{7/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{7 d^3 \left (a+b x^2\right )}+\frac {20 a^3 b^2 (d x)^{11/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{11 d^5 \left (a+b x^2\right )}+\frac {4 a^2 b^3 (d x)^{15/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{3 d^7 \left (a+b x^2\right )}+\frac {10 a b^4 (d x)^{19/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{19 d^9 \left (a+b x^2\right )}+\frac {2 b^5 (d x)^{23/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{23 d^{11} \left (a+b x^2\right )} \]

[Out]

2/3*a^5*(d*x)^(3/2)*((b*x^2+a)^2)^(1/2)/d/(b*x^2+a)+10/7*a^4*b*(d*x)^(7/2)*((b*x^2+a)^2)^(1/2)/d^3/(b*x^2+a)+2
0/11*a^3*b^2*(d*x)^(11/2)*((b*x^2+a)^2)^(1/2)/d^5/(b*x^2+a)+4/3*a^2*b^3*(d*x)^(15/2)*((b*x^2+a)^2)^(1/2)/d^7/(
b*x^2+a)+10/19*a*b^4*(d*x)^(19/2)*((b*x^2+a)^2)^(1/2)/d^9/(b*x^2+a)+2/23*b^5*(d*x)^(23/2)*((b*x^2+a)^2)^(1/2)/
d^11/(b*x^2+a)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 297, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {1126, 276} \[ \int \sqrt {d x} \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\frac {2 b^5 (d x)^{23/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{23 d^{11} \left (a+b x^2\right )}+\frac {10 a b^4 (d x)^{19/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{19 d^9 \left (a+b x^2\right )}+\frac {4 a^2 b^3 (d x)^{15/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{3 d^7 \left (a+b x^2\right )}+\frac {2 a^5 (d x)^{3/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{3 d \left (a+b x^2\right )}+\frac {10 a^4 b (d x)^{7/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{7 d^3 \left (a+b x^2\right )}+\frac {20 a^3 b^2 (d x)^{11/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{11 d^5 \left (a+b x^2\right )} \]

[In]

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

[Out]

(2*a^5*(d*x)^(3/2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(3*d*(a + b*x^2)) + (10*a^4*b*(d*x)^(7/2)*Sqrt[a^2 + 2*a*b
*x^2 + b^2*x^4])/(7*d^3*(a + b*x^2)) + (20*a^3*b^2*(d*x)^(11/2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(11*d^5*(a +
b*x^2)) + (4*a^2*b^3*(d*x)^(15/2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(3*d^7*(a + b*x^2)) + (10*a*b^4*(d*x)^(19/2
)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(19*d^9*(a + b*x^2)) + (2*b^5*(d*x)^(23/2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])
/(23*d^11*(a + b*x^2))

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]

Rule 1126

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[(a + b*x^2 + c*x^4)^FracPa
rt[p]/(c^IntPart[p]*(b/2 + c*x^2)^(2*FracPart[p])), Int[(d*x)^m*(b/2 + c*x^2)^(2*p), x], x] /; FreeQ[{a, b, c,
d, m, p}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2]

Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \sqrt {d x} \left (a b+b^2 x^2\right )^5 \, dx}{b^4 \left (a b+b^2 x^2\right )} \\ & = \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \left (a^5 b^5 \sqrt {d x}+\frac {5 a^4 b^6 (d x)^{5/2}}{d^2}+\frac {10 a^3 b^7 (d x)^{9/2}}{d^4}+\frac {10 a^2 b^8 (d x)^{13/2}}{d^6}+\frac {5 a b^9 (d x)^{17/2}}{d^8}+\frac {b^{10} (d x)^{21/2}}{d^{10}}\right ) \, dx}{b^4 \left (a b+b^2 x^2\right )} \\ & = \frac {2 a^5 (d x)^{3/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{3 d \left (a+b x^2\right )}+\frac {10 a^4 b (d x)^{7/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{7 d^3 \left (a+b x^2\right )}+\frac {20 a^3 b^2 (d x)^{11/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{11 d^5 \left (a+b x^2\right )}+\frac {4 a^2 b^3 (d x)^{15/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{3 d^7 \left (a+b x^2\right )}+\frac {10 a b^4 (d x)^{19/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{19 d^9 \left (a+b x^2\right )}+\frac {2 b^5 (d x)^{23/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{23 d^{11} \left (a+b x^2\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.30 \[ \int \sqrt {d x} \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\frac {2 x \sqrt {d x} \sqrt {\left (a+b x^2\right )^2} \left (33649 a^5+72105 a^4 b x^2+91770 a^3 b^2 x^4+67298 a^2 b^3 x^6+26565 a b^4 x^8+4389 b^5 x^{10}\right )}{100947 \left (a+b x^2\right )} \]

[In]

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

[Out]

(2*x*Sqrt[d*x]*Sqrt[(a + b*x^2)^2]*(33649*a^5 + 72105*a^4*b*x^2 + 91770*a^3*b^2*x^4 + 67298*a^2*b^3*x^6 + 2656
5*a*b^4*x^8 + 4389*b^5*x^10))/(100947*(a + b*x^2))

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.28

method	result	size
gosper	\(\frac {2 x \left (4389 x^{10} b^{5}+26565 a \,x^{8} b^{4}+67298 a^{2} x^{6} b^{3}+91770 a^{3} x^{4} b^{2}+72105 x^{2} a^{4} b +33649 a^{5}\right ) {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {5}{2}} \sqrt {d x}}{100947 \left (b \,x^{2}+a \right )^{5}}\)	\(83\)
default	\(\frac {2 {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {5}{2}} \left (d x \right )^{\frac {3}{2}} \left (4389 x^{10} b^{5}+26565 a \,x^{8} b^{4}+67298 a^{2} x^{6} b^{3}+91770 a^{3} x^{4} b^{2}+72105 x^{2} a^{4} b +33649 a^{5}\right )}{100947 d \left (b \,x^{2}+a \right )^{5}}\)	\(85\)
risch	\(\frac {2 d \sqrt {\left (b \,x^{2}+a \right )^{2}}\, x^{2} \left (4389 x^{10} b^{5}+26565 a \,x^{8} b^{4}+67298 a^{2} x^{6} b^{3}+91770 a^{3} x^{4} b^{2}+72105 x^{2} a^{4} b +33649 a^{5}\right )}{100947 \left (b \,x^{2}+a \right ) \sqrt {d x}}\)	\(86\)

[In]

int((b^2*x^4+2*a*b*x^2+a^2)^(5/2)*(d*x)^(1/2),x,method=_RETURNVERBOSE)

[Out]

2/100947*x*(4389*b^5*x^10+26565*a*b^4*x^8+67298*a^2*b^3*x^6+91770*a^3*b^2*x^4+72105*a^4*b*x^2+33649*a^5)*((b*x
^2+a)^2)^(5/2)*(d*x)^(1/2)/(b*x^2+a)^5

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.21 \[ \int \sqrt {d x} \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\frac {2}{100947} \, {\left (4389 \, b^{5} x^{11} + 26565 \, a b^{4} x^{9} + 67298 \, a^{2} b^{3} x^{7} + 91770 \, a^{3} b^{2} x^{5} + 72105 \, a^{4} b x^{3} + 33649 \, a^{5} x\right )} \sqrt {d x} \]

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^(5/2)*(d*x)^(1/2),x, algorithm="fricas")

[Out]

2/100947*(4389*b^5*x^11 + 26565*a*b^4*x^9 + 67298*a^2*b^3*x^7 + 91770*a^3*b^2*x^5 + 72105*a^4*b*x^3 + 33649*a^
5*x)*sqrt(d*x)

Sympy [F]

\[ \int \sqrt {d x} \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\int \sqrt {d x} \left (\left (a + b x^{2}\right )^{2}\right )^{\frac {5}{2}}\, dx \]

[In]

integrate((b**2*x**4+2*a*b*x**2+a**2)**(5/2)*(d*x)**(1/2),x)

[Out]

Integral(sqrt(d*x)*((a + b*x**2)**2)**(5/2), x)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.49 \[ \int \sqrt {d x} \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\frac {2}{437} \, {\left (19 \, b^{5} \sqrt {d} x^{3} + 23 \, a b^{4} \sqrt {d} x\right )} x^{\frac {17}{2}} + \frac {8}{285} \, {\left (15 \, a b^{4} \sqrt {d} x^{3} + 19 \, a^{2} b^{3} \sqrt {d} x\right )} x^{\frac {13}{2}} + \frac {4}{55} \, {\left (11 \, a^{2} b^{3} \sqrt {d} x^{3} + 15 \, a^{3} b^{2} \sqrt {d} x\right )} x^{\frac {9}{2}} + \frac {8}{77} \, {\left (7 \, a^{3} b^{2} \sqrt {d} x^{3} + 11 \, a^{4} b \sqrt {d} x\right )} x^{\frac {5}{2}} + \frac {2}{21} \, {\left (3 \, a^{4} b \sqrt {d} x^{3} + 7 \, a^{5} \sqrt {d} x\right )} \sqrt {x} \]

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^(5/2)*(d*x)^(1/2),x, algorithm="maxima")

[Out]

2/437*(19*b^5*sqrt(d)*x^3 + 23*a*b^4*sqrt(d)*x)*x^(17/2) + 8/285*(15*a*b^4*sqrt(d)*x^3 + 19*a^2*b^3*sqrt(d)*x)
*x^(13/2) + 4/55*(11*a^2*b^3*sqrt(d)*x^3 + 15*a^3*b^2*sqrt(d)*x)*x^(9/2) + 8/77*(7*a^3*b^2*sqrt(d)*x^3 + 11*a^
4*b*sqrt(d)*x)*x^(5/2) + 2/21*(3*a^4*b*sqrt(d)*x^3 + 7*a^5*sqrt(d)*x)*sqrt(x)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.45 \[ \int \sqrt {d x} \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\frac {2}{23} \, \sqrt {d x} b^{5} x^{11} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {10}{19} \, \sqrt {d x} a b^{4} x^{9} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {4}{3} \, \sqrt {d x} a^{2} b^{3} x^{7} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {20}{11} \, \sqrt {d x} a^{3} b^{2} x^{5} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {10}{7} \, \sqrt {d x} a^{4} b x^{3} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {2}{3} \, \sqrt {d x} a^{5} x \mathrm {sgn}\left (b x^{2} + a\right ) \]

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^(5/2)*(d*x)^(1/2),x, algorithm="giac")

[Out]

2/23*sqrt(d*x)*b^5*x^11*sgn(b*x^2 + a) + 10/19*sqrt(d*x)*a*b^4*x^9*sgn(b*x^2 + a) + 4/3*sqrt(d*x)*a^2*b^3*x^7*
sgn(b*x^2 + a) + 20/11*sqrt(d*x)*a^3*b^2*x^5*sgn(b*x^2 + a) + 10/7*sqrt(d*x)*a^4*b*x^3*sgn(b*x^2 + a) + 2/3*sq
rt(d*x)*a^5*x*sgn(b*x^2 + a)

Mupad [F(-1)]

Timed out. \[ \int \sqrt {d x} \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\int \sqrt {d\,x}\,{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{5/2} \,d x \]

[In]

int((d*x)^(1/2)*(a^2 + b^2*x^4 + 2*a*b*x^2)^(5/2),x)

[Out]

int((d*x)^(1/2)*(a^2 + b^2*x^4 + 2*a*b*x^2)^(5/2), x)