\(\int x^{3/2} (a+c x^4) \, dx\) [718]

Optimal result

Integrand size = 13, antiderivative size = 21 \[ \int x^{3/2} \left (a+c x^4\right ) \, dx=\frac {2}{5} a x^{5/2}+\frac {2}{13} c x^{13/2} \]

[Out]

2/5*a*x^(5/2)+2/13*c*x^(13/2)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {14} \[ \int x^{3/2} \left (a+c x^4\right ) \, dx=\frac {2}{5} a x^{5/2}+\frac {2}{13} c x^{13/2} \]

[In]

Int[x^(3/2)*(a + c*x^4),x]

[Out]

(2*a*x^(5/2))/5 + (2*c*x^(13/2))/13

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 x^{3/2}+c x^{11/2}\right ) \, dx \\ & = \frac {2}{5} a x^{5/2}+\frac {2}{13} c x^{13/2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int x^{3/2} \left (a+c x^4\right ) \, dx=\frac {2}{65} x^{5/2} \left (13 a+5 c x^4\right ) \]

[In]

Integrate[x^(3/2)*(a + c*x^4),x]

[Out]

(2*x^(5/2)*(13*a + 5*c*x^4))/65

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.67

method	result	size
derivativedivides	\(\frac {2 a \,x^{\frac {5}{2}}}{5}+\frac {2 c \,x^{\frac {13}{2}}}{13}\)	\(14\)
default	\(\frac {2 a \,x^{\frac {5}{2}}}{5}+\frac {2 c \,x^{\frac {13}{2}}}{13}\)	\(14\)
gosper	\(\frac {2 x^{\frac {5}{2}} \left (5 x^{4} c +13 a \right )}{65}\)	\(16\)
trager	\(\frac {2 x^{\frac {5}{2}} \left (5 x^{4} c +13 a \right )}{65}\)	\(16\)
risch	\(\frac {2 x^{\frac {5}{2}} \left (5 x^{4} c +13 a \right )}{65}\)	\(16\)

[In]

int(x^(3/2)*(c*x^4+a),x,method=_RETURNVERBOSE)

[Out]

2/5*a*x^(5/2)+2/13*c*x^(13/2)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.86 \[ \int x^{3/2} \left (a+c x^4\right ) \, dx=\frac {2}{65} \, {\left (5 \, c x^{6} + 13 \, a x^{2}\right )} \sqrt {x} \]

[In]

integrate(x^(3/2)*(c*x^4+a),x, algorithm="fricas")

[Out]

2/65*(5*c*x^6 + 13*a*x^2)*sqrt(x)

Sympy [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int x^{3/2} \left (a+c x^4\right ) \, dx=\frac {2 a x^{\frac {5}{2}}}{5} + \frac {2 c x^{\frac {13}{2}}}{13} \]

[In]

integrate(x**(3/2)*(c*x**4+a),x)

[Out]

2*a*x**(5/2)/5 + 2*c*x**(13/2)/13

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.62 \[ \int x^{3/2} \left (a+c x^4\right ) \, dx=\frac {2}{13} \, c x^{\frac {13}{2}} + \frac {2}{5} \, a x^{\frac {5}{2}} \]

[In]

integrate(x^(3/2)*(c*x^4+a),x, algorithm="maxima")

[Out]

2/13*c*x^(13/2) + 2/5*a*x^(5/2)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.62 \[ \int x^{3/2} \left (a+c x^4\right ) \, dx=\frac {2}{13} \, c x^{\frac {13}{2}} + \frac {2}{5} \, a x^{\frac {5}{2}} \]

[In]

integrate(x^(3/2)*(c*x^4+a),x, algorithm="giac")

[Out]

2/13*c*x^(13/2) + 2/5*a*x^(5/2)

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.71 \[ \int x^{3/2} \left (a+c x^4\right ) \, dx=\frac {2\,x^{5/2}\,\left (5\,c\,x^4+13\,a\right )}{65} \]

[In]

int(x^(3/2)*(a + c*x^4),x)

[Out]

(2*x^(5/2)*(13*a + 5*c*x^4))/65