\(\int x^{5/2} (a+c x^4)^2 \, dx\) [724]

Optimal result

Integrand size = 15, antiderivative size = 36 \[ \int x^{5/2} \left (a+c x^4\right )^2 \, dx=\frac {2}{7} a^2 x^{7/2}+\frac {4}{15} a c x^{15/2}+\frac {2}{23} c^2 x^{23/2} \]

[Out]

2/7*a^2*x^(7/2)+4/15*a*c*x^(15/2)+2/23*c^2*x^(23/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {276} \[ \int x^{5/2} \left (a+c x^4\right )^2 \, dx=\frac {2}{7} a^2 x^{7/2}+\frac {4}{15} a c x^{15/2}+\frac {2}{23} c^2 x^{23/2} \]

[In]

Int[x^(5/2)*(a + c*x^4)^2,x]

[Out]

(2*a^2*x^(7/2))/7 + (4*a*c*x^(15/2))/15 + (2*c^2*x^(23/2))/23

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}& = \int \left (a^2 x^{5/2}+2 a c x^{13/2}+c^2 x^{21/2}\right ) \, dx \\ & = \frac {2}{7} a^2 x^{7/2}+\frac {4}{15} a c x^{15/2}+\frac {2}{23} c^2 x^{23/2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.83 \[ \int x^{5/2} \left (a+c x^4\right )^2 \, dx=\frac {2 x^{7/2} \left (345 a^2+322 a c x^4+105 c^2 x^8\right )}{2415} \]

[In]

Integrate[x^(5/2)*(a + c*x^4)^2,x]

[Out]

(2*x^(7/2)*(345*a^2 + 322*a*c*x^4 + 105*c^2*x^8))/2415

Maple [A] (verified)

Time = 3.99 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.69

method	result	size
derivativedivides	\(\frac {2 a^{2} x^{\frac {7}{2}}}{7}+\frac {4 a c \,x^{\frac {15}{2}}}{15}+\frac {2 c^{2} x^{\frac {23}{2}}}{23}\)	\(25\)
default	\(\frac {2 a^{2} x^{\frac {7}{2}}}{7}+\frac {4 a c \,x^{\frac {15}{2}}}{15}+\frac {2 c^{2} x^{\frac {23}{2}}}{23}\)	\(25\)
gosper	\(\frac {2 x^{\frac {7}{2}} \left (105 c^{2} x^{8}+322 a \,x^{4} c +345 a^{2}\right )}{2415}\)	\(27\)
trager	\(\frac {2 x^{\frac {7}{2}} \left (105 c^{2} x^{8}+322 a \,x^{4} c +345 a^{2}\right )}{2415}\)	\(27\)
risch	\(\frac {2 x^{\frac {7}{2}} \left (105 c^{2} x^{8}+322 a \,x^{4} c +345 a^{2}\right )}{2415}\)	\(27\)

[In]

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

[Out]

2/7*a^2*x^(7/2)+4/15*a*c*x^(15/2)+2/23*c^2*x^(23/2)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.81 \[ \int x^{5/2} \left (a+c x^4\right )^2 \, dx=\frac {2}{2415} \, {\left (105 \, c^{2} x^{11} + 322 \, a c x^{7} + 345 \, a^{2} x^{3}\right )} \sqrt {x} \]

[In]

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

[Out]

2/2415*(105*c^2*x^11 + 322*a*c*x^7 + 345*a^2*x^3)*sqrt(x)

Sympy [A] (verification not implemented)

Time = 0.88 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.94 \[ \int x^{5/2} \left (a+c x^4\right )^2 \, dx=\frac {2 a^{2} x^{\frac {7}{2}}}{7} + \frac {4 a c x^{\frac {15}{2}}}{15} + \frac {2 c^{2} x^{\frac {23}{2}}}{23} \]

[In]

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

[Out]

2*a**2*x**(7/2)/7 + 4*a*c*x**(15/2)/15 + 2*c**2*x**(23/2)/23

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.67 \[ \int x^{5/2} \left (a+c x^4\right )^2 \, dx=\frac {2}{23} \, c^{2} x^{\frac {23}{2}} + \frac {4}{15} \, a c x^{\frac {15}{2}} + \frac {2}{7} \, a^{2} x^{\frac {7}{2}} \]

[In]

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

[Out]

2/23*c^2*x^(23/2) + 4/15*a*c*x^(15/2) + 2/7*a^2*x^(7/2)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.67 \[ \int x^{5/2} \left (a+c x^4\right )^2 \, dx=\frac {2}{23} \, c^{2} x^{\frac {23}{2}} + \frac {4}{15} \, a c x^{\frac {15}{2}} + \frac {2}{7} \, a^{2} x^{\frac {7}{2}} \]

[In]

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

[Out]

2/23*c^2*x^(23/2) + 4/15*a*c*x^(15/2) + 2/7*a^2*x^(7/2)

Mupad [B] (verification not implemented)

Time = 5.59 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.69 \[ \int x^{5/2} \left (a+c x^4\right )^2 \, dx=x^{7/2}\,\left (\frac {2\,a^2}{7}+\frac {4\,a\,c\,x^4}{15}+\frac {2\,c^2\,x^8}{23}\right ) \]

[In]

int(x^(5/2)*(a + c*x^4)^2,x)

[Out]

x^(7/2)*((2*a^2)/7 + (2*c^2*x^8)/23 + (4*a*c*x^4)/15)