\(\int \frac {(b x^2+c x^4)^2}{x^{7/2}} \, dx\) [307]

Optimal result

Integrand size = 19, antiderivative size = 36 \[ \int \frac {\left (b x^2+c x^4\right )^2}{x^{7/2}} \, dx=\frac {2}{3} b^2 x^{3/2}+\frac {4}{7} b c x^{7/2}+\frac {2}{11} c^2 x^{11/2} \]

[Out]

2/3*b^2*x^(3/2)+4/7*b*c*x^(7/2)+2/11*c^2*x^(11/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {1598, 276} \[ \int \frac {\left (b x^2+c x^4\right )^2}{x^{7/2}} \, dx=\frac {2}{3} b^2 x^{3/2}+\frac {4}{7} b c x^{7/2}+\frac {2}{11} c^2 x^{11/2} \]

[In]

Int[(b*x^2 + c*x^4)^2/x^(7/2),x]

[Out]

(2*b^2*x^(3/2))/3 + (4*b*c*x^(7/2))/7 + (2*c^2*x^(11/2))/11

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]

Rule 1598

Int[(u_.)*(x_)^(m_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(m + n*p)*(a + b*x^(q -
p))^n, x] /; FreeQ[{a, b, m, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rubi steps \begin{align*} \text {integral}& = \int \sqrt {x} \left (b+c x^2\right )^2 \, dx \\ & = \int \left (b^2 \sqrt {x}+2 b c x^{5/2}+c^2 x^{9/2}\right ) \, dx \\ & = \frac {2}{3} b^2 x^{3/2}+\frac {4}{7} b c x^{7/2}+\frac {2}{11} c^2 x^{11/2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.83 \[ \int \frac {\left (b x^2+c x^4\right )^2}{x^{7/2}} \, dx=\frac {2}{231} x^{3/2} \left (77 b^2+66 b c x^2+21 c^2 x^4\right ) \]

[In]

Integrate[(b*x^2 + c*x^4)^2/x^(7/2),x]

[Out]

(2*x^(3/2)*(77*b^2 + 66*b*c*x^2 + 21*c^2*x^4))/231

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.69

method	result	size
derivativedivides	\(\frac {2 b^{2} x^{\frac {3}{2}}}{3}+\frac {4 b c \,x^{\frac {7}{2}}}{7}+\frac {2 c^{2} x^{\frac {11}{2}}}{11}\)	\(25\)
default	\(\frac {2 b^{2} x^{\frac {3}{2}}}{3}+\frac {4 b c \,x^{\frac {7}{2}}}{7}+\frac {2 c^{2} x^{\frac {11}{2}}}{11}\)	\(25\)
gosper	\(\frac {2 x^{\frac {3}{2}} \left (21 c^{2} x^{4}+66 b c \,x^{2}+77 b^{2}\right )}{231}\)	\(27\)
trager	\(\frac {2 x^{\frac {3}{2}} \left (21 c^{2} x^{4}+66 b c \,x^{2}+77 b^{2}\right )}{231}\)	\(27\)
risch	\(\frac {2 x^{\frac {3}{2}} \left (21 c^{2} x^{4}+66 b c \,x^{2}+77 b^{2}\right )}{231}\)	\(27\)

[In]

int((c*x^4+b*x^2)^2/x^(7/2),x,method=_RETURNVERBOSE)

[Out]

2/3*b^2*x^(3/2)+4/7*b*c*x^(7/2)+2/11*c^2*x^(11/2)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.75 \[ \int \frac {\left (b x^2+c x^4\right )^2}{x^{7/2}} \, dx=\frac {2}{231} \, {\left (21 \, c^{2} x^{5} + 66 \, b c x^{3} + 77 \, b^{2} x\right )} \sqrt {x} \]

[In]

integrate((c*x^4+b*x^2)^2/x^(7/2),x, algorithm="fricas")

[Out]

2/231*(21*c^2*x^5 + 66*b*c*x^3 + 77*b^2*x)*sqrt(x)

Sympy [A] (verification not implemented)

Time = 0.51 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.94 \[ \int \frac {\left (b x^2+c x^4\right )^2}{x^{7/2}} \, dx=\frac {2 b^{2} x^{\frac {3}{2}}}{3} + \frac {4 b c x^{\frac {7}{2}}}{7} + \frac {2 c^{2} x^{\frac {11}{2}}}{11} \]

[In]

integrate((c*x**4+b*x**2)**2/x**(7/2),x)

[Out]

2*b**2*x**(3/2)/3 + 4*b*c*x**(7/2)/7 + 2*c**2*x**(11/2)/11

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.67 \[ \int \frac {\left (b x^2+c x^4\right )^2}{x^{7/2}} \, dx=\frac {2}{11} \, c^{2} x^{\frac {11}{2}} + \frac {4}{7} \, b c x^{\frac {7}{2}} + \frac {2}{3} \, b^{2} x^{\frac {3}{2}} \]

[In]

integrate((c*x^4+b*x^2)^2/x^(7/2),x, algorithm="maxima")

[Out]

2/11*c^2*x^(11/2) + 4/7*b*c*x^(7/2) + 2/3*b^2*x^(3/2)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.67 \[ \int \frac {\left (b x^2+c x^4\right )^2}{x^{7/2}} \, dx=\frac {2}{11} \, c^{2} x^{\frac {11}{2}} + \frac {4}{7} \, b c x^{\frac {7}{2}} + \frac {2}{3} \, b^{2} x^{\frac {3}{2}} \]

[In]

integrate((c*x^4+b*x^2)^2/x^(7/2),x, algorithm="giac")

[Out]

2/11*c^2*x^(11/2) + 4/7*b*c*x^(7/2) + 2/3*b^2*x^(3/2)

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.72 \[ \int \frac {\left (b x^2+c x^4\right )^2}{x^{7/2}} \, dx=\frac {2\,x^{3/2}\,\left (77\,b^2+66\,b\,c\,x^2+21\,c^2\,x^4\right )}{231} \]

[In]

int((b*x^2 + c*x^4)^2/x^(7/2),x)

[Out]

(2*x^(3/2)*(77*b^2 + 21*c^2*x^4 + 66*b*c*x^2))/231