\(\int (a^2+2 a b x^3+b^2 x^6)^{5/2} \, dx\) [63]

Optimal result

Integrand size = 22, antiderivative size = 247 \[ \int \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx=\frac {a^5 x \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{\left (a+b x^3\right )^5}+\frac {5 a^4 b x^4 \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{4 \left (a+b x^3\right )^5}+\frac {10 a^3 b^2 x^7 \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{7 \left (a+b x^3\right )^5}+\frac {a^2 b^3 x^{10} \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{\left (a+b x^3\right )^5}+\frac {5 a b^4 x^{13} \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{13 \left (a+b x^3\right )^5}+\frac {b^5 x^{16} \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{16 \left (a+b x^3\right )^5} \]

[Out]

a^5*x*(b^2*x^6+2*a*b*x^3+a^2)^(5/2)/(b*x^3+a)^5+5/4*a^4*b*x^4*(b^2*x^6+2*a*b*x^3+a^2)^(5/2)/(b*x^3+a)^5+10/7*a
^3*b^2*x^7*(b^2*x^6+2*a*b*x^3+a^2)^(5/2)/(b*x^3+a)^5+a^2*b^3*x^10*(b^2*x^6+2*a*b*x^3+a^2)^(5/2)/(b*x^3+a)^5+5/
13*a*b^4*x^13*(b^2*x^6+2*a*b*x^3+a^2)^(5/2)/(b*x^3+a)^5+1/16*b^5*x^16*(b^2*x^6+2*a*b*x^3+a^2)^(5/2)/(b*x^3+a)^
5

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 247, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {1357, 200} \[ \int \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx=\frac {b^5 x^{16} \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{16 \left (a+b x^3\right )^5}+\frac {5 a b^4 x^{13} \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{13 \left (a+b x^3\right )^5}+\frac {a^2 b^3 x^{10} \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{\left (a+b x^3\right )^5}+\frac {a^5 x \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{\left (a+b x^3\right )^5}+\frac {5 a^4 b x^4 \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{4 \left (a+b x^3\right )^5}+\frac {10 a^3 b^2 x^7 \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{7 \left (a+b x^3\right )^5} \]

[In]

Int[(a^2 + 2*a*b*x^3 + b^2*x^6)^(5/2),x]

[Out]

(a^5*x*(a^2 + 2*a*b*x^3 + b^2*x^6)^(5/2))/(a + b*x^3)^5 + (5*a^4*b*x^4*(a^2 + 2*a*b*x^3 + b^2*x^6)^(5/2))/(4*(
a + b*x^3)^5) + (10*a^3*b^2*x^7*(a^2 + 2*a*b*x^3 + b^2*x^6)^(5/2))/(7*(a + b*x^3)^5) + (a^2*b^3*x^10*(a^2 + 2*
a*b*x^3 + b^2*x^6)^(5/2))/(a + b*x^3)^5 + (5*a*b^4*x^13*(a^2 + 2*a*b*x^3 + b^2*x^6)^(5/2))/(13*(a + b*x^3)^5)
+ (b^5*x^16*(a^2 + 2*a*b*x^3 + b^2*x^6)^(5/2))/(16*(a + b*x^3)^5)

Rule 200

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^n)^p, x], x] /; FreeQ[{a, b}, x]
&& IGtQ[n, 0] && IGtQ[p, 0]

Rule 1357

Int[((a_) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.))^(p_), x_Symbol] :> Dist[(a + b*x^n + c*x^(2*n))^p/(b + 2*c*x
^n)^(2*p), Int[(b + 2*c*x^n)^(2*p), x], x] /; FreeQ[{a, b, c, n, p}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \int \left (2 a b+2 b^2 x^3\right )^5 \, dx}{\left (2 a b+2 b^2 x^3\right )^5} \\ & = \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \int \left (32 a^5 b^5+160 a^4 b^6 x^3+320 a^3 b^7 x^6+320 a^2 b^8 x^9+160 a b^9 x^{12}+32 b^{10} x^{15}\right ) \, dx}{\left (2 a b+2 b^2 x^3\right )^5} \\ & = \frac {a^5 x \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{\left (a+b x^3\right )^5}+\frac {5 a^4 b x^4 \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{4 \left (a+b x^3\right )^5}+\frac {10 a^3 b^2 x^7 \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{7 \left (a+b x^3\right )^5}+\frac {a^2 b^3 x^{10} \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{\left (a+b x^3\right )^5}+\frac {5 a b^4 x^{13} \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{13 \left (a+b x^3\right )^5}+\frac {b^5 x^{16} \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{16 \left (a+b x^3\right )^5} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.01 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.33 \[ \int \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx=\frac {x \sqrt {\left (a+b x^3\right )^2} \left (1456 a^5+1820 a^4 b x^3+2080 a^3 b^2 x^6+1456 a^2 b^3 x^9+560 a b^4 x^{12}+91 b^5 x^{15}\right )}{1456 \left (a+b x^3\right )} \]

[In]

Integrate[(a^2 + 2*a*b*x^3 + b^2*x^6)^(5/2),x]

[Out]

(x*Sqrt[(a + b*x^3)^2]*(1456*a^5 + 1820*a^4*b*x^3 + 2080*a^3*b^2*x^6 + 1456*a^2*b^3*x^9 + 560*a*b^4*x^12 + 91*
b^5*x^15))/(1456*(a + b*x^3))

Maple [A] (verified)

Time = 1.47 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.32

method	result	size
gosper	\(\frac {x \left (91 b^{5} x^{15}+560 a \,b^{4} x^{12}+1456 a^{2} b^{3} x^{9}+2080 a^{3} b^{2} x^{6}+1820 a^{4} b \,x^{3}+1456 a^{5}\right ) {\left (\left (b \,x^{3}+a \right )^{2}\right )}^{\frac {5}{2}}}{1456 \left (b \,x^{3}+a \right )^{5}}\)	\(78\)
default	\(\frac {x \left (91 b^{5} x^{15}+560 a \,b^{4} x^{12}+1456 a^{2} b^{3} x^{9}+2080 a^{3} b^{2} x^{6}+1820 a^{4} b \,x^{3}+1456 a^{5}\right ) {\left (\left (b \,x^{3}+a \right )^{2}\right )}^{\frac {5}{2}}}{1456 \left (b \,x^{3}+a \right )^{5}}\)	\(78\)
risch	\(\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}\, a^{5} x}{b \,x^{3}+a}+\frac {5 \sqrt {\left (b \,x^{3}+a \right )^{2}}\, b \,a^{4} x^{4}}{4 \left (b \,x^{3}+a \right )}+\frac {10 \sqrt {\left (b \,x^{3}+a \right )^{2}}\, a^{3} b^{2} x^{7}}{7 \left (b \,x^{3}+a \right )}+\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}\, a^{2} b^{3} x^{10}}{b \,x^{3}+a}+\frac {5 \sqrt {\left (b \,x^{3}+a \right )^{2}}\, b^{4} a \,x^{13}}{13 \left (b \,x^{3}+a \right )}+\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}\, b^{5} x^{16}}{16 b \,x^{3}+16 a}\)	\(174\)

[In]

int((b^2*x^6+2*a*b*x^3+a^2)^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/1456*x*(91*b^5*x^15+560*a*b^4*x^12+1456*a^2*b^3*x^9+2080*a^3*b^2*x^6+1820*a^4*b*x^3+1456*a^5)*((b*x^3+a)^2)^
(5/2)/(b*x^3+a)^5

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.21 \[ \int \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx=\frac {1}{16} \, b^{5} x^{16} + \frac {5}{13} \, a b^{4} x^{13} + a^{2} b^{3} x^{10} + \frac {10}{7} \, a^{3} b^{2} x^{7} + \frac {5}{4} \, a^{4} b x^{4} + a^{5} x \]

[In]

integrate((b^2*x^6+2*a*b*x^3+a^2)^(5/2),x, algorithm="fricas")

[Out]

1/16*b^5*x^16 + 5/13*a*b^4*x^13 + a^2*b^3*x^10 + 10/7*a^3*b^2*x^7 + 5/4*a^4*b*x^4 + a^5*x

Sympy [F]

\[ \int \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx=\int \left (a^{2} + 2 a b x^{3} + b^{2} x^{6}\right )^{\frac {5}{2}}\, dx \]

[In]

integrate((b**2*x**6+2*a*b*x**3+a**2)**(5/2),x)

[Out]

Integral((a**2 + 2*a*b*x**3 + b**2*x**6)**(5/2), x)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.21 \[ \int \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx=\frac {1}{16} \, b^{5} x^{16} + \frac {5}{13} \, a b^{4} x^{13} + a^{2} b^{3} x^{10} + \frac {10}{7} \, a^{3} b^{2} x^{7} + \frac {5}{4} \, a^{4} b x^{4} + a^{5} x \]

[In]

integrate((b^2*x^6+2*a*b*x^3+a^2)^(5/2),x, algorithm="maxima")

[Out]

1/16*b^5*x^16 + 5/13*a*b^4*x^13 + a^2*b^3*x^10 + 10/7*a^3*b^2*x^7 + 5/4*a^4*b*x^4 + a^5*x

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.41 \[ \int \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx=\frac {1}{16} \, b^{5} x^{16} \mathrm {sgn}\left (b x^{3} + a\right ) + \frac {5}{13} \, a b^{4} x^{13} \mathrm {sgn}\left (b x^{3} + a\right ) + a^{2} b^{3} x^{10} \mathrm {sgn}\left (b x^{3} + a\right ) + \frac {10}{7} \, a^{3} b^{2} x^{7} \mathrm {sgn}\left (b x^{3} + a\right ) + \frac {5}{4} \, a^{4} b x^{4} \mathrm {sgn}\left (b x^{3} + a\right ) + a^{5} x \mathrm {sgn}\left (b x^{3} + a\right ) \]

[In]

integrate((b^2*x^6+2*a*b*x^3+a^2)^(5/2),x, algorithm="giac")

[Out]

1/16*b^5*x^16*sgn(b*x^3 + a) + 5/13*a*b^4*x^13*sgn(b*x^3 + a) + a^2*b^3*x^10*sgn(b*x^3 + a) + 10/7*a^3*b^2*x^7
*sgn(b*x^3 + a) + 5/4*a^4*b*x^4*sgn(b*x^3 + a) + a^5*x*sgn(b*x^3 + a)

Mupad [F(-1)]

Timed out. \[ \int \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx=\int {\left (a^2+2\,a\,b\,x^3+b^2\,x^6\right )}^{5/2} \,d x \]

[In]

int((a^2 + b^2*x^6 + 2*a*b*x^3)^(5/2),x)

[Out]

int((a^2 + b^2*x^6 + 2*a*b*x^3)^(5/2), x)