3.343 \(\int x^m (a+b x^2) \, dx\)

Optimal. Leaf size=25 \[ \frac {a x^{m+1}}{m+1}+\frac {b x^{m+3}}{m+3} \]

[Out]

a*x^(1+m)/(1+m)+b*x^(3+m)/(3+m)

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Rubi [A]  time = 0.01, antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {14} \[ \frac {a x^{m+1}}{m+1}+\frac {b x^{m+3}}{m+3} \]

Antiderivative was successfully verified.

[In]

Int[x^m*(a + b*x^2),x]

[Out]

(a*x^(1 + m))/(1 + m) + (b*x^(3 + m))/(3 + m)

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*} \int x^m \left (a+b x^2\right ) \, dx &=\int \left (a x^m+b x^{2+m}\right ) \, dx\\ &=\frac {a x^{1+m}}{1+m}+\frac {b x^{3+m}}{3+m}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 25, normalized size = 1.00 \[ \frac {a x^{m+1}}{m+1}+\frac {b x^{m+3}}{m+3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^m*(a + b*x^2),x]

[Out]

(a*x^(1 + m))/(1 + m) + (b*x^(3 + m))/(3 + m)

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fricas [A]  time = 0.94, size = 33, normalized size = 1.32 \[ \frac {{\left ({\left (b m + b\right )} x^{3} + {\left (a m + 3 \, a\right )} x\right )} x^{m}}{m^{2} + 4 \, m + 3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((b*m + b)*x^3 + (a*m + 3*a)*x)*x^m/(m^2 + 4*m + 3)

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giac [A]  time = 0.69, size = 43, normalized size = 1.72 \[ \frac {b m x^{3} x^{m} + b x^{3} x^{m} + a m x x^{m} + 3 \, a x x^{m}}{m^{2} + 4 \, m + 3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b*m*x^3*x^m + b*x^3*x^m + a*m*x*x^m + 3*a*x*x^m)/(m^2 + 4*m + 3)

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maple [A]  time = 0.00, size = 35, normalized size = 1.40 \[ \frac {\left (b m \,x^{2}+b \,x^{2}+a m +3 a \right ) x^{m +1}}{\left (m +3\right ) \left (m +1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m*(b*x^2+a),x)

[Out]

x^(m+1)*(b*m*x^2+b*x^2+a*m+3*a)/(m+3)/(m+1)

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maxima [A]  time = 1.30, size = 25, normalized size = 1.00 \[ \frac {b x^{m + 3}}{m + 3} + \frac {a x^{m + 1}}{m + 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*x^(m + 3)/(m + 3) + a*x^(m + 1)/(m + 1)

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mupad [B]  time = 4.80, size = 34, normalized size = 1.36 \[ \frac {x^{m+1}\,\left (3\,a+a\,m+b\,x^2+b\,m\,x^2\right )}{m^2+4\,m+3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m*(a + b*x^2),x)

[Out]

(x^(m + 1)*(3*a + a*m + b*x^2 + b*m*x^2))/(4*m + m^2 + 3)

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sympy [A]  time = 0.41, size = 94, normalized size = 3.76 \[ \begin {cases} - \frac {a}{2 x^{2}} + b \log {\relax (x )} & \text {for}\: m = -3 \\a \log {\relax (x )} + \frac {b x^{2}}{2} & \text {for}\: m = -1 \\\frac {a m x x^{m}}{m^{2} + 4 m + 3} + \frac {3 a x x^{m}}{m^{2} + 4 m + 3} + \frac {b m x^{3} x^{m}}{m^{2} + 4 m + 3} + \frac {b x^{3} x^{m}}{m^{2} + 4 m + 3} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-a/(2*x**2) + b*log(x), Eq(m, -3)), (a*log(x) + b*x**2/2, Eq(m, -1)), (a*m*x*x**m/(m**2 + 4*m + 3)
+ 3*a*x*x**m/(m**2 + 4*m + 3) + b*m*x**3*x**m/(m**2 + 4*m + 3) + b*x**3*x**m/(m**2 + 4*m + 3), True))

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