3.7.64 \(\int (-\frac {b x^{1+m}}{(a+b x^2)^{3/2}}+\frac {m x^{-1+m}}{\sqrt {a+b x^2}}) \, dx\) [664]

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] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 0.05, 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 = {372, 371} \begin {gather*} \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}} \end {gather*}

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 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 372

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] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 0.04, size = 131, normalized size = 8.73 \begin {gather*} \frac {x^m \sqrt {a+b x^2} \left (a (2+m) \, _2F_1\left (-\frac {1}{2},\frac {m}{2};1+\frac {m}{2};-\frac {b x^2}{a}\right )-b x^2 \left (m \, _2F_1\left (\frac {1}{2},1+\frac {m}{2};2+\frac {m}{2};-\frac {b x^2}{a}\right )+\, _2F_1\left (\frac {3}{2},1+\frac {m}{2};2+\frac {m}{2};-\frac {b x^2}{a}\right )\right )\right )}{a^2 (2+m) \sqrt {1+\frac {b x^2}{a}}} \end {gather*}

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[a + b*x^2]*(a*(2 + m)*Hypergeometric2F1[-1/2, m/2, 1 + m/2, -((b*x^2)/a)] - b*x^2*(m*Hypergeometric2
F1[1/2, 1 + m/2, 2 + m/2, -((b*x^2)/a)] + Hypergeometric2F1[3/2, 1 + m/2, 2 + m/2, -((b*x^2)/a)])))/(a^2*(2 +
m)*Sqrt[1 + (b*x^2)/a])

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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int -\frac {b \,x^{1+m}}{\left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {m \,x^{-1+m}}{\sqrt {b \,x^{2}+a}}\, dx\]

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 = 0.33, size = 13, normalized size = 0.87 \begin {gather*} \frac {x^{m}}{\sqrt {b x^{2} + a}} \end {gather*}

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 = 1.54, size = 26, normalized size = 1.73 \begin {gather*} \frac {\sqrt {b x^{2} + a} x^{m + 1}}{b x^{3} + a x} \end {gather*}

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] Result contains complex when optimal does not.
time = 2.88, size = 94, normalized size = 6.27 \begin {gather*} \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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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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Mupad [F]
time = 0.00, size = -1, normalized size = -0.07 \begin {gather*} -\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 \end {gather*}

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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