\(\int \sqrt {x} (a+c x^4) \, dx\) [719]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 13, antiderivative size = 21 \[ \int \sqrt {x} \left (a+c x^4\right ) \, dx=\frac {2}{3} a x^{3/2}+\frac {2}{11} c x^{11/2} \]

[Out]

2/3*a*x^(3/2)+2/11*c*x^(11/2)

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} \[ \int \sqrt {x} \left (a+c x^4\right ) \, dx=\frac {2}{3} a x^{3/2}+\frac {2}{11} c x^{11/2} \]

[In]

Int[Sqrt[x]*(a + c*x^4),x]

[Out]

(2*a*x^(3/2))/3 + (2*c*x^(11/2))/11

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 \sqrt {x}+c x^{9/2}\right ) \, dx \\ & = \frac {2}{3} a x^{3/2}+\frac {2}{11} c x^{11/2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \sqrt {x} \left (a+c x^4\right ) \, dx=\frac {2}{33} x^{3/2} \left (11 a+3 c x^4\right ) \]

[In]

Integrate[Sqrt[x]*(a + c*x^4),x]

[Out]

(2*x^(3/2)*(11*a + 3*c*x^4))/33

Maple [A] (verified)

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

method result size
derivativedivides \(\frac {2 a \,x^{\frac {3}{2}}}{3}+\frac {2 c \,x^{\frac {11}{2}}}{11}\) \(14\)
default \(\frac {2 a \,x^{\frac {3}{2}}}{3}+\frac {2 c \,x^{\frac {11}{2}}}{11}\) \(14\)
gosper \(\frac {2 x^{\frac {3}{2}} \left (3 x^{4} c +11 a \right )}{33}\) \(16\)
trager \(\frac {2 x^{\frac {3}{2}} \left (3 x^{4} c +11 a \right )}{33}\) \(16\)
risch \(\frac {2 x^{\frac {3}{2}} \left (3 x^{4} c +11 a \right )}{33}\) \(16\)

[In]

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

[Out]

2/3*a*x^(3/2)+2/11*c*x^(11/2)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.76 \[ \int \sqrt {x} \left (a+c x^4\right ) \, dx=\frac {2}{33} \, {\left (3 \, c x^{5} + 11 \, a x\right )} \sqrt {x} \]

[In]

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

[Out]

2/33*(3*c*x^5 + 11*a*x)*sqrt(x)

Sympy [A] (verification not implemented)

Time = 0.36 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \sqrt {x} \left (a+c x^4\right ) \, dx=\frac {2 a x^{\frac {3}{2}}}{3} + \frac {2 c x^{\frac {11}{2}}}{11} \]

[In]

integrate((c*x**4+a)*x**(1/2),x)

[Out]

2*a*x**(3/2)/3 + 2*c*x**(11/2)/11

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.62 \[ \int \sqrt {x} \left (a+c x^4\right ) \, dx=\frac {2}{11} \, c x^{\frac {11}{2}} + \frac {2}{3} \, a x^{\frac {3}{2}} \]

[In]

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

[Out]

2/11*c*x^(11/2) + 2/3*a*x^(3/2)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.62 \[ \int \sqrt {x} \left (a+c x^4\right ) \, dx=\frac {2}{11} \, c x^{\frac {11}{2}} + \frac {2}{3} \, a x^{\frac {3}{2}} \]

[In]

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

[Out]

2/11*c*x^(11/2) + 2/3*a*x^(3/2)

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.71 \[ \int \sqrt {x} \left (a+c x^4\right ) \, dx=\frac {2\,x^{3/2}\,\left (3\,c\,x^4+11\,a\right )}{33} \]

[In]

int(x^(1/2)*(a + c*x^4),x)

[Out]

(2*x^(3/2)*(11*a + 3*c*x^4))/33