\(\int \sqrt {x} (b x^2+c x^4) \, dx\) [295]

Optimal result

Integrand size = 17, antiderivative size = 21 \[ \int \sqrt {x} \left (b x^2+c x^4\right ) \, dx=\frac {2}{7} b x^{7/2}+\frac {2}{11} c x^{11/2} \]

[Out]

2/7*b*x^(7/2)+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, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {14} \[ \int \sqrt {x} \left (b x^2+c x^4\right ) \, dx=\frac {2}{7} b x^{7/2}+\frac {2}{11} c x^{11/2} \]

[In]

Int[Sqrt[x]*(b*x^2 + c*x^4),x]

[Out]

(2*b*x^(7/2))/7 + (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 (b x^{5/2}+c x^{9/2}\right ) \, dx \\ & = \frac {2}{7} b x^{7/2}+\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 \[ \int \sqrt {x} \left (b x^2+c x^4\right ) \, dx=\frac {2}{77} x^{7/2} \left (11 b+7 c x^2\right ) \]

[In]

Integrate[Sqrt[x]*(b*x^2 + c*x^4),x]

[Out]

(2*x^(7/2)*(11*b + 7*c*x^2))/77

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.67

method	result	size
derivativedivides	\(\frac {2 b \,x^{\frac {7}{2}}}{7}+\frac {2 c \,x^{\frac {11}{2}}}{11}\)	\(14\)
default	\(\frac {2 b \,x^{\frac {7}{2}}}{7}+\frac {2 c \,x^{\frac {11}{2}}}{11}\)	\(14\)
gosper	\(\frac {2 x^{\frac {7}{2}} \left (7 c \,x^{2}+11 b \right )}{77}\)	\(16\)
trager	\(\frac {2 x^{\frac {7}{2}} \left (7 c \,x^{2}+11 b \right )}{77}\)	\(16\)
risch	\(\frac {2 x^{\frac {7}{2}} \left (7 c \,x^{2}+11 b \right )}{77}\)	\(16\)

[In]

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

[Out]

2/7*b*x^(7/2)+2/11*c*x^(11/2)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.86 \[ \int \sqrt {x} \left (b x^2+c x^4\right ) \, dx=\frac {2}{77} \, {\left (7 \, c x^{5} + 11 \, b x^{3}\right )} \sqrt {x} \]

[In]

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

[Out]

2/77*(7*c*x^5 + 11*b*x^3)*sqrt(x)

Sympy [A] (verification not implemented)

Time = 0.40 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \sqrt {x} \left (b x^2+c x^4\right ) \, dx=\frac {2 b x^{\frac {7}{2}}}{7} + \frac {2 c x^{\frac {11}{2}}}{11} \]

[In]

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

[Out]

2*b*x**(7/2)/7 + 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 \[ \int \sqrt {x} \left (b x^2+c x^4\right ) \, dx=\frac {2}{11} \, c x^{\frac {11}{2}} + \frac {2}{7} \, b x^{\frac {7}{2}} \]

[In]

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

[Out]

2/11*c*x^(11/2) + 2/7*b*x^(7/2)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.62 \[ \int \sqrt {x} \left (b x^2+c x^4\right ) \, dx=\frac {2}{11} \, c x^{\frac {11}{2}} + \frac {2}{7} \, b x^{\frac {7}{2}} \]

[In]

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

[Out]

2/11*c*x^(11/2) + 2/7*b*x^(7/2)

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.71 \[ \int \sqrt {x} \left (b x^2+c x^4\right ) \, dx=\frac {2\,x^{7/2}\,\left (7\,c\,x^2+11\,b\right )}{77} \]

[In]

int(x^(1/2)*(b*x^2 + c*x^4),x)

[Out]

(2*x^(7/2)*(11*b + 7*c*x^2))/77