\(\int \frac {(a^2+2 a b x^2+b^2 x^4)^{5/2}}{(d x)^{5/2}} \, dx\) [747]

Optimal result

Integrand size = 30, antiderivative size = 293 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{(d x)^{5/2}} \, dx=-\frac {2 a^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{3 d (d x)^{3/2} \left (a+b x^2\right )}+\frac {10 a^4 b \sqrt {d x} \sqrt {a^2+2 a b x^2+b^2 x^4}}{d^3 \left (a+b x^2\right )}+\frac {4 a^3 b^2 (d x)^{5/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{d^5 \left (a+b x^2\right )}+\frac {20 a^2 b^3 (d x)^{9/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{9 d^7 \left (a+b x^2\right )}+\frac {10 a b^4 (d x)^{13/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{13 d^9 \left (a+b x^2\right )}+\frac {2 b^5 (d x)^{17/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{17 d^{11} \left (a+b x^2\right )} \]

[Out]

-2/3*a^5*((b*x^2+a)^2)^(1/2)/d/(d*x)^(3/2)/(b*x^2+a)+4*a^3*b^2*(d*x)^(5/2)*((b*x^2+a)^2)^(1/2)/d^5/(b*x^2+a)+2
0/9*a^2*b^3*(d*x)^(9/2)*((b*x^2+a)^2)^(1/2)/d^7/(b*x^2+a)+10/13*a*b^4*(d*x)^(13/2)*((b*x^2+a)^2)^(1/2)/d^9/(b*
x^2+a)+2/17*b^5*(d*x)^(17/2)*((b*x^2+a)^2)^(1/2)/d^11/(b*x^2+a)+10*a^4*b*(d*x)^(1/2)*((b*x^2+a)^2)^(1/2)/d^3/(
b*x^2+a)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 293, 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} \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{(d x)^{5/2}} \, dx=\frac {2 b^5 (d x)^{17/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{17 d^{11} \left (a+b x^2\right )}+\frac {10 a b^4 (d x)^{13/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{13 d^9 \left (a+b x^2\right )}+\frac {20 a^2 b^3 (d x)^{9/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{9 d^7 \left (a+b x^2\right )}-\frac {2 a^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{3 d (d x)^{3/2} \left (a+b x^2\right )}+\frac {10 a^4 b \sqrt {d x} \sqrt {a^2+2 a b x^2+b^2 x^4}}{d^3 \left (a+b x^2\right )}+\frac {4 a^3 b^2 (d x)^{5/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{d^5 \left (a+b x^2\right )} \]

[In]

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

[Out]

(-2*a^5*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(3*d*(d*x)^(3/2)*(a + b*x^2)) + (10*a^4*b*Sqrt[d*x]*Sqrt[a^2 + 2*a*b*
x^2 + b^2*x^4])/(d^3*(a + b*x^2)) + (4*a^3*b^2*(d*x)^(5/2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(d^5*(a + b*x^2))
+ (20*a^2*b^3*(d*x)^(9/2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(9*d^7*(a + b*x^2)) + (10*a*b^4*(d*x)^(13/2)*Sqrt[a
^2 + 2*a*b*x^2 + b^2*x^4])/(13*d^9*(a + b*x^2)) + (2*b^5*(d*x)^(17/2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(17*d^1
1*(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 \frac {\left (a b+b^2 x^2\right )^5}{(d x)^{5/2}} \, 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 (\frac {a^5 b^5}{(d x)^{5/2}}+\frac {5 a^4 b^6}{d^2 \sqrt {d x}}+\frac {10 a^3 b^7 (d x)^{3/2}}{d^4}+\frac {10 a^2 b^8 (d x)^{7/2}}{d^6}+\frac {5 a b^9 (d x)^{11/2}}{d^8}+\frac {b^{10} (d x)^{15/2}}{d^{10}}\right ) \, dx}{b^4 \left (a b+b^2 x^2\right )} \\ & = -\frac {2 a^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{3 d (d x)^{3/2} \left (a+b x^2\right )}+\frac {10 a^4 b \sqrt {d x} \sqrt {a^2+2 a b x^2+b^2 x^4}}{d^3 \left (a+b x^2\right )}+\frac {4 a^3 b^2 (d x)^{5/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{d^5 \left (a+b x^2\right )}+\frac {20 a^2 b^3 (d x)^{9/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{9 d^7 \left (a+b x^2\right )}+\frac {10 a b^4 (d x)^{13/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{13 d^9 \left (a+b x^2\right )}+\frac {2 b^5 (d x)^{17/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{17 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 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{(d x)^{5/2}} \, dx=-\frac {2 x \left (\left (a+b x^2\right )^2\right )^{5/2} \left (663 a^5-9945 a^4 b x^2-3978 a^3 b^2 x^4-2210 a^2 b^3 x^6-765 a b^4 x^8-117 b^5 x^{10}\right )}{1989 (d x)^{5/2} \left (a+b x^2\right )^5} \]

[In]

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

[Out]

(-2*x*((a + b*x^2)^2)^(5/2)*(663*a^5 - 9945*a^4*b*x^2 - 3978*a^3*b^2*x^4 - 2210*a^2*b^3*x^6 - 765*a*b^4*x^8 -
117*b^5*x^10))/(1989*(d*x)^(5/2)*(a + b*x^2)^5)

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.28

method	result	size
gosper	\(-\frac {2 x \left (-117 x^{10} b^{5}-765 a \,x^{8} b^{4}-2210 a^{2} x^{6} b^{3}-3978 a^{3} x^{4} b^{2}-9945 x^{2} a^{4} b +663 a^{5}\right ) {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {5}{2}}}{1989 \left (b \,x^{2}+a \right )^{5} \left (d x \right )^{\frac {5}{2}}}\)	\(83\)
default	\(-\frac {2 {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {5}{2}} \left (-117 x^{10} b^{5}-765 a \,x^{8} b^{4}-2210 a^{2} x^{6} b^{3}-3978 a^{3} x^{4} b^{2}-9945 x^{2} a^{4} b +663 a^{5}\right )}{1989 d \left (b \,x^{2}+a \right )^{5} \left (d x \right )^{\frac {3}{2}}}\)	\(85\)
risch	\(-\frac {2 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (-117 x^{10} b^{5}-765 a \,x^{8} b^{4}-2210 a^{2} x^{6} b^{3}-3978 a^{3} x^{4} b^{2}-9945 x^{2} a^{4} b +663 a^{5}\right )}{1989 d^{2} \left (b \,x^{2}+a \right ) x \sqrt {d x}}\)	\(88\)

[In]

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

[Out]

-2/1989*x*(-117*b^5*x^10-765*a*b^4*x^8-2210*a^2*b^3*x^6-3978*a^3*b^2*x^4-9945*a^4*b*x^2+663*a^5)*((b*x^2+a)^2)
^(5/2)/(b*x^2+a)^5/(d*x)^(5/2)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.23 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{(d x)^{5/2}} \, dx=\frac {2 \, {\left (117 \, b^{5} x^{10} + 765 \, a b^{4} x^{8} + 2210 \, a^{2} b^{3} x^{6} + 3978 \, a^{3} b^{2} x^{4} + 9945 \, a^{4} b x^{2} - 663 \, a^{5}\right )} \sqrt {d x}}{1989 \, d^{3} x^{2}} \]

[In]

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

[Out]

2/1989*(117*b^5*x^10 + 765*a*b^4*x^8 + 2210*a^2*b^3*x^6 + 3978*a^3*b^2*x^4 + 9945*a^4*b*x^2 - 663*a^5)*sqrt(d*
x)/(d^3*x^2)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.52 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{(d x)^{5/2}} \, dx=\frac {2 \, {\left (45 \, {\left (13 \, b^{5} \sqrt {d} x^{3} + 17 \, a b^{4} \sqrt {d} x\right )} x^{\frac {11}{2}} + 340 \, {\left (9 \, a b^{4} \sqrt {d} x^{3} + 13 \, a^{2} b^{3} \sqrt {d} x\right )} x^{\frac {7}{2}} + 1326 \, {\left (5 \, a^{2} b^{3} \sqrt {d} x^{3} + 9 \, a^{3} b^{2} \sqrt {d} x\right )} x^{\frac {3}{2}} + \frac {7956 \, {\left (a^{3} b^{2} \sqrt {d} x^{3} + 5 \, a^{4} b \sqrt {d} x\right )}}{\sqrt {x}} + \frac {3315 \, {\left (3 \, a^{4} b \sqrt {d} x^{3} - a^{5} \sqrt {d} x\right )}}{x^{\frac {5}{2}}}\right )}}{9945 \, d^{3}} \]

[In]

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

[Out]

2/9945*(45*(13*b^5*sqrt(d)*x^3 + 17*a*b^4*sqrt(d)*x)*x^(11/2) + 340*(9*a*b^4*sqrt(d)*x^3 + 13*a^2*b^3*sqrt(d)*
x)*x^(7/2) + 1326*(5*a^2*b^3*sqrt(d)*x^3 + 9*a^3*b^2*sqrt(d)*x)*x^(3/2) + 7956*(a^3*b^2*sqrt(d)*x^3 + 5*a^4*b*
sqrt(d)*x)/sqrt(x) + 3315*(3*a^4*b*sqrt(d)*x^3 - a^5*sqrt(d)*x)/x^(5/2))/d^3

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.54 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{(d x)^{5/2}} \, dx=-\frac {2 \, {\left (\frac {663 \, a^{5} d \mathrm {sgn}\left (b x^{2} + a\right )}{\sqrt {d x} x} - \frac {117 \, \sqrt {d x} b^{5} d^{136} x^{8} \mathrm {sgn}\left (b x^{2} + a\right ) + 765 \, \sqrt {d x} a b^{4} d^{136} x^{6} \mathrm {sgn}\left (b x^{2} + a\right ) + 2210 \, \sqrt {d x} a^{2} b^{3} d^{136} x^{4} \mathrm {sgn}\left (b x^{2} + a\right ) + 3978 \, \sqrt {d x} a^{3} b^{2} d^{136} x^{2} \mathrm {sgn}\left (b x^{2} + a\right ) + 9945 \, \sqrt {d x} a^{4} b d^{136} \mathrm {sgn}\left (b x^{2} + a\right )}{d^{136}}\right )}}{1989 \, d^{3}} \]

[In]

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

[Out]

-2/1989*(663*a^5*d*sgn(b*x^2 + a)/(sqrt(d*x)*x) - (117*sqrt(d*x)*b^5*d^136*x^8*sgn(b*x^2 + a) + 765*sqrt(d*x)*
a*b^4*d^136*x^6*sgn(b*x^2 + a) + 2210*sqrt(d*x)*a^2*b^3*d^136*x^4*sgn(b*x^2 + a) + 3978*sqrt(d*x)*a^3*b^2*d^13
6*x^2*sgn(b*x^2 + a) + 9945*sqrt(d*x)*a^4*b*d^136*sgn(b*x^2 + a))/d^136)/d^3

Mupad [B] (verification not implemented)

Time = 13.60 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.40 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{(d x)^{5/2}} \, dx=\frac {\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}\,\left (\frac {10\,a^4\,x^2}{d^2}-\frac {2\,a^5}{3\,b\,d^2}+\frac {2\,b^4\,x^{10}}{17\,d^2}+\frac {4\,a^3\,b\,x^4}{d^2}+\frac {10\,a\,b^3\,x^8}{13\,d^2}+\frac {20\,a^2\,b^2\,x^6}{9\,d^2}\right )}{x^3\,\sqrt {d\,x}+\frac {a\,x\,\sqrt {d\,x}}{b}} \]

[In]

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

[Out]

((a^2 + b^2*x^4 + 2*a*b*x^2)^(1/2)*((10*a^4*x^2)/d^2 - (2*a^5)/(3*b*d^2) + (2*b^4*x^10)/(17*d^2) + (4*a^3*b*x^
4)/d^2 + (10*a*b^3*x^8)/(13*d^2) + (20*a^2*b^2*x^6)/(9*d^2)))/(x^3*(d*x)^(1/2) + (a*x*(d*x)^(1/2))/b)