3.661 \(\int \frac {x^{1+m} (a (2+m)+b (3+m) x^2)}{\sqrt {a+b x^2}} \, dx\)

Optimal. Leaf size=17 \[ x^{m+2} \sqrt {a+b x^2} \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.032, Rules used = {449} \[ x^{m+2} \sqrt {a+b x^2} \]

Antiderivative was successfully verified.

[In]

Int[(x^(1 + m)*(a*(2 + m) + b*(3 + m)*x^2))/Sqrt[a + b*x^2],x]

[Out]

x^(2 + m)*Sqrt[a + b*x^2]

Rule 449

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(c*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(a*e*(m + 1)), x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0] && EqQ[
a*d*(m + 1) - b*c*(m + n*(p + 1) + 1), 0] && NeQ[m, -1]

Rubi steps

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

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Mathematica [C]  time = 0.11, size = 104, normalized size = 6.12 \[ \frac {x^{m+2} \sqrt {\frac {b x^2}{a}+1} \left (b (m+3) x^2 \, _2F_1\left (\frac {1}{2},\frac {m+4}{2};\frac {m+6}{2};-\frac {b x^2}{a}\right )+a (m+4) \, _2F_1\left (\frac {1}{2},\frac {m+2}{2};\frac {m+4}{2};-\frac {b x^2}{a}\right )\right )}{(m+4) \sqrt {a+b x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^(1 + m)*(a*(2 + m) + b*(3 + m)*x^2))/Sqrt[a + b*x^2],x]

[Out]

(x^(2 + m)*Sqrt[1 + (b*x^2)/a]*(a*(4 + m)*Hypergeometric2F1[1/2, (2 + m)/2, (4 + m)/2, -((b*x^2)/a)] + b*(3 +
m)*x^2*Hypergeometric2F1[1/2, (4 + m)/2, (6 + m)/2, -((b*x^2)/a)]))/((4 + m)*Sqrt[a + b*x^2])

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fricas [A]  time = 0.90, size = 16, normalized size = 0.94 \[ \sqrt {b x^{2} + a} x x^{m + 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(1+m)*(a*(2+m)+b*(3+m)*x^2)/(b*x^2+a)^(1/2),x, algorithm="fricas")

[Out]

sqrt(b*x^2 + a)*x*x^(m + 1)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b {\left (m + 3\right )} x^{2} + a {\left (m + 2\right )}\right )} x^{m + 1}}{\sqrt {b x^{2} + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(1+m)*(a*(2+m)+b*(3+m)*x^2)/(b*x^2+a)^(1/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.01, size = 16, normalized size = 0.94 \[ \sqrt {b \,x^{2}+a}\, x^{m +2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.88, size = 16, normalized size = 0.94 \[ \sqrt {b x^{2} + a} x^{2} x^{m} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(1+m)*(a*(2+m)+b*(3+m)*x^2)/(b*x^2+a)^(1/2),x, algorithm="maxima")

[Out]

sqrt(b*x^2 + a)*x^2*x^m

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mupad [B]  time = 5.19, size = 24, normalized size = 1.41 \[ \frac {x^{m+1}\,\left (b\,x^3+a\,x\right )}{\sqrt {b\,x^2+a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [C]  time = 10.06, size = 202, normalized size = 11.88 \[ \frac {\sqrt {a} m x^{2} x^{m} \Gamma \left (\frac {m}{2} + 1\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {m}{2} + 1 \\ \frac {m}{2} + 2 \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \Gamma \left (\frac {m}{2} + 2\right )} + \frac {\sqrt {a} x^{2} x^{m} \Gamma \left (\frac {m}{2} + 1\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {m}{2} + 1 \\ \frac {m}{2} + 2 \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{\Gamma \left (\frac {m}{2} + 2\right )} + \frac {b m x^{4} x^{m} \Gamma \left (\frac {m}{2} + 2\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {m}{2} + 2 \\ \frac {m}{2} + 3 \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \sqrt {a} \Gamma \left (\frac {m}{2} + 3\right )} + \frac {3 b x^{4} x^{m} \Gamma \left (\frac {m}{2} + 2\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {m}{2} + 2 \\ \frac {m}{2} + 3 \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \sqrt {a} \Gamma \left (\frac {m}{2} + 3\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(a)*m*x**2*x**m*gamma(m/2 + 1)*hyper((1/2, m/2 + 1), (m/2 + 2,), b*x**2*exp_polar(I*pi)/a)/(2*gamma(m/2 +
2)) + sqrt(a)*x**2*x**m*gamma(m/2 + 1)*hyper((1/2, m/2 + 1), (m/2 + 2,), b*x**2*exp_polar(I*pi)/a)/gamma(m/2 +
2) + b*m*x**4*x**m*gamma(m/2 + 2)*hyper((1/2, m/2 + 2), (m/2 + 3,), b*x**2*exp_polar(I*pi)/a)/(2*sqrt(a)*gamm
a(m/2 + 3)) + 3*b*x**4*x**m*gamma(m/2 + 2)*hyper((1/2, m/2 + 2), (m/2 + 3,), b*x**2*exp_polar(I*pi)/a)/(2*sqrt
(a)*gamma(m/2 + 3))

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