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

Optimal. Leaf size=15 \[ \frac{x^m}{\sqrt{a+b x^2}} \]

[Out]

x^m/Sqrt[a + b*x^2]

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Rubi [C]  time = 0.0650502, antiderivative size = 123, normalized size of antiderivative = 8.2, number of steps used = 5, number of rules used = 2, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.053, Rules used = {365, 364} \[ \frac{x^m \sqrt{\frac{b x^2}{a}+1} \, _2F_1\left (\frac{1}{2},\frac{m}{2};\frac{m+2}{2};-\frac{b x^2}{a}\right )}{\sqrt{a+b x^2}}-\frac{b x^{m+2} \sqrt{\frac{b x^2}{a}+1} \, _2F_1\left (\frac{3}{2},\frac{m+2}{2};\frac{m+4}{2};-\frac{b x^2}{a}\right )}{a (m+2) \sqrt{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 365

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^n)^FracPart[p])
/(1 + (b*x^n)/a)^FracPart[p], Int[(c*x)^m*(1 + (b*x^n)/a)^p, x], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[
p, 0] &&  !(ILtQ[p, 0] || GtQ[a, 0])

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rubi steps

\begin{align*} \int \left (-\frac{b x^{1+m}}{\left (a+b x^2\right )^{3/2}}+\frac{m x^{-1+m}}{\sqrt{a+b x^2}}\right ) \, dx &=-\left (b \int \frac{x^{1+m}}{\left (a+b x^2\right )^{3/2}} \, dx\right )+m \int \frac{x^{-1+m}}{\sqrt{a+b x^2}} \, dx\\ &=-\frac{\left (b \sqrt{1+\frac{b x^2}{a}}\right ) \int \frac{x^{1+m}}{\left (1+\frac{b x^2}{a}\right )^{3/2}} \, dx}{a \sqrt{a+b x^2}}+\frac{\left (m \sqrt{1+\frac{b x^2}{a}}\right ) \int \frac{x^{-1+m}}{\sqrt{1+\frac{b x^2}{a}}} \, dx}{\sqrt{a+b x^2}}\\ &=\frac{x^m \sqrt{1+\frac{b x^2}{a}} \, _2F_1\left (\frac{1}{2},\frac{m}{2};\frac{2+m}{2};-\frac{b x^2}{a}\right )}{\sqrt{a+b x^2}}-\frac{b x^{2+m} \sqrt{1+\frac{b x^2}{a}} \, _2F_1\left (\frac{3}{2},\frac{2+m}{2};\frac{4+m}{2};-\frac{b x^2}{a}\right )}{a (2+m) \sqrt{a+b x^2}}\\ \end{align*}

Mathematica [C]  time = 0.0398933, size = 103, normalized size = 6.87 \[ \frac{x^m \sqrt{\frac{b x^2}{a}+1} \left (b (m-1) x^2 \, _2F_1\left (\frac{3}{2},\frac{m+2}{2};\frac{m+4}{2};-\frac{b x^2}{a}\right )+a (m+2) \, _2F_1\left (\frac{3}{2},\frac{m}{2};\frac{m+2}{2};-\frac{b x^2}{a}\right )\right )}{a (m+2) \sqrt{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^m*Sqrt[1 + (b*x^2)/a]*(a*(2 + m)*Hypergeometric2F1[3/2, m/2, (2 + m)/2, -((b*x^2)/a)] + b*(-1 + m)*x^2*Hype
rgeometric2F1[3/2, (2 + m)/2, (4 + m)/2, -((b*x^2)/a)]))/(a*(2 + m)*Sqrt[a + b*x^2])

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Maple [F]  time = 0.033, size = 0, normalized size = 0. \begin{align*} \int -{b{x}^{1+m} \left ( b{x}^{2}+a \right ) ^{-{\frac{3}{2}}}}+{m{x}^{-1+m}{\frac{1}{\sqrt{b{x}^{2}+a}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 2.3452, size = 18, normalized size = 1.2 \begin{align*} \frac{x^{m}}{\sqrt{b x^{2} + a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x^m/sqrt(b*x^2 + a)

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Fricas [A]  time = 1.59138, size = 55, normalized size = 3.67 \begin{align*} \frac{\sqrt{b x^{2} + a} x^{m + 1}}{b x^{3} + a x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [C]  time = 11.0682, size = 94, normalized size = 6.27 \begin{align*} \frac{m x^{m} \Gamma \left (\frac{m}{2}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{1}{2}, \frac{m}{2} \\ \frac{m}{2} + 1 \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )}}{2 \sqrt{a} \Gamma \left (\frac{m}{2} + 1\right )} - \frac{b x^{2} x^{m} \Gamma \left (\frac{m}{2} + 1\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{3}{2}, \frac{m}{2} + 1 \\ \frac{m}{2} + 2 \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac{3}{2}} \Gamma \left (\frac{m}{2} + 2\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{m x^{m - 1}}{\sqrt{b x^{2} + a}} - \frac{b x^{m + 1}}{{\left (b x^{2} + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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