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

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Rubi [C]  time = 0.07, antiderivative size = 123, normalized size of antiderivative = 8.20, 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 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])

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])

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*}

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Mathematica [C]  time = 0.06, 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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fricas [A]  time = 0.95, size = 26, normalized size = 1.73 \[ \frac {\sqrt {b x^{2} + a} x^{m + 1}}{b x^{3} + a x} \]

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

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)

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maple [F]  time = 0.33, size = 0, normalized size = 0.00 \[ \int -\frac {b \,x^{m +1}}{\left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {m \,x^{m -1}}{\sqrt {b \,x^{2}+a}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.92, size = 13, normalized size = 0.87 \[ \frac {x^{m}}{\sqrt {b x^{2} + a}} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.07 \[ -\int \frac {b\,x^{m+1}}{{\left (b\,x^2+a\right )}^{3/2}}-\frac {m\,x^{m-1}}{\sqrt {b\,x^2+a}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [C]  time = 5.59, size = 94, normalized size = 6.27 \[ \frac {m x^{m} \Gamma \left (\frac {m}{2}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {m}{2} \\ \frac {m}{2} + 1 \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \sqrt {a} \Gamma \left (\frac {m}{2} + 1\right )} - \frac {b x^{2} x^{m} \Gamma \left (\frac {m}{2} + 1\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, \frac {m}{2} + 1 \\ \frac {m}{2} + 2 \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {3}{2}} \Gamma \left (\frac {m}{2} + 2\right )} \]

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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