\(\int (a x+b x^3)^2 \, dx\) [8]

Optimal result

Integrand size = 11, antiderivative size = 30 \[ \int \left (a x+b x^3\right )^2 \, dx=\frac {a^2 x^3}{3}+\frac {2}{5} a b x^5+\frac {b^2 x^7}{7} \]

[Out]

1/3*a^2*x^3+2/5*a*b*x^5+1/7*b^2*x^7

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {1607, 276} \[ \int \left (a x+b x^3\right )^2 \, dx=\frac {a^2 x^3}{3}+\frac {2}{5} a b x^5+\frac {b^2 x^7}{7} \]

[In]

Int[(a*x + b*x^3)^2,x]

[Out]

(a^2*x^3)/3 + (2*a*b*x^5)/5 + (b^2*x^7)/7

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 1607

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

Rubi steps \begin{align*} \text {integral}& = \int x^2 \left (a+b x^2\right )^2 \, dx \\ & = \int \left (a^2 x^2+2 a b x^4+b^2 x^6\right ) \, dx \\ & = \frac {a^2 x^3}{3}+\frac {2}{5} a b x^5+\frac {b^2 x^7}{7} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00 \[ \int \left (a x+b x^3\right )^2 \, dx=\frac {a^2 x^3}{3}+\frac {2}{5} a b x^5+\frac {b^2 x^7}{7} \]

[In]

Integrate[(a*x + b*x^3)^2,x]

[Out]

(a^2*x^3)/3 + (2*a*b*x^5)/5 + (b^2*x^7)/7

Maple [A] (verified)

Time = 2.00 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.83

method	result	size
default	\(\frac {1}{3} a^{2} x^{3}+\frac {2}{5} a b \,x^{5}+\frac {1}{7} b^{2} x^{7}\)	\(25\)
norman	\(\frac {1}{3} a^{2} x^{3}+\frac {2}{5} a b \,x^{5}+\frac {1}{7} b^{2} x^{7}\)	\(25\)
risch	\(\frac {1}{3} a^{2} x^{3}+\frac {2}{5} a b \,x^{5}+\frac {1}{7} b^{2} x^{7}\)	\(25\)
parallelrisch	\(\frac {1}{3} a^{2} x^{3}+\frac {2}{5} a b \,x^{5}+\frac {1}{7} b^{2} x^{7}\)	\(25\)
gosper	\(\frac {x^{3} \left (15 b^{2} x^{4}+42 a b \,x^{2}+35 a^{2}\right )}{105}\)	\(27\)

[In]

int((b*x^3+a*x)^2,x,method=_RETURNVERBOSE)

[Out]

1/3*a^2*x^3+2/5*a*b*x^5+1/7*b^2*x^7

Fricas [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int \left (a x+b x^3\right )^2 \, dx=\frac {1}{7} \, b^{2} x^{7} + \frac {2}{5} \, a b x^{5} + \frac {1}{3} \, a^{2} x^{3} \]

[In]

integrate((b*x^3+a*x)^2,x, algorithm="fricas")

[Out]

1/7*b^2*x^7 + 2/5*a*b*x^5 + 1/3*a^2*x^3

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.87 \[ \int \left (a x+b x^3\right )^2 \, dx=\frac {a^{2} x^{3}}{3} + \frac {2 a b x^{5}}{5} + \frac {b^{2} x^{7}}{7} \]

[In]

integrate((b*x**3+a*x)**2,x)

[Out]

a**2*x**3/3 + 2*a*b*x**5/5 + b**2*x**7/7

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int \left (a x+b x^3\right )^2 \, dx=\frac {1}{7} \, b^{2} x^{7} + \frac {2}{5} \, a b x^{5} + \frac {1}{3} \, a^{2} x^{3} \]

[In]

integrate((b*x^3+a*x)^2,x, algorithm="maxima")

[Out]

1/7*b^2*x^7 + 2/5*a*b*x^5 + 1/3*a^2*x^3

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int \left (a x+b x^3\right )^2 \, dx=\frac {1}{7} \, b^{2} x^{7} + \frac {2}{5} \, a b x^{5} + \frac {1}{3} \, a^{2} x^{3} \]

[In]

integrate((b*x^3+a*x)^2,x, algorithm="giac")

[Out]

1/7*b^2*x^7 + 2/5*a*b*x^5 + 1/3*a^2*x^3

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int \left (a x+b x^3\right )^2 \, dx=\frac {a^2\,x^3}{3}+\frac {2\,a\,b\,x^5}{5}+\frac {b^2\,x^7}{7} \]

[In]

int((a*x + b*x^3)^2,x)

[Out]

(a^2*x^3)/3 + (b^2*x^7)/7 + (2*a*b*x^5)/5