3.27.97 \(\int (\frac {6 a x^2}{b (4+m) \sqrt {a+b x^{-2+m}}}+\frac {x^m}{\sqrt {a+b x^{-2+m}}}) \, dx\) [2697]

Optimal. Leaf size=26 \[ \frac {2 x^3 \sqrt {a+b x^{-2+m}}}{b (4+m)} \]

[Out]

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

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Rubi [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 0.06, antiderivative size = 160, normalized size of antiderivative = 6.15, number of steps used = 5, number of rules used = 2, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.044, Rules used = {372, 371} \begin {gather*} \frac {x^{m+1} \sqrt {\frac {b x^{m-2}}{a}+1} \, _2F_1\left (\frac {1}{2},-\frac {m+1}{2-m};\frac {1-2 m}{2-m};-\frac {b x^{m-2}}{a}\right )}{(m+1) \sqrt {a+b x^{m-2}}}+\frac {2 a x^3 \sqrt {\frac {b x^{m-2}}{a}+1} \, _2F_1\left (\frac {1}{2},-\frac {3}{2-m};-\frac {m+1}{2-m};-\frac {b x^{m-2}}{a}\right )}{b (m+4) \sqrt {a+b x^{m-2}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 {6 a x^2}{b (4+m) \sqrt {a+b x^{-2+m}}}+\frac {x^m}{\sqrt {a+b x^{-2+m}}}\right ) \, dx &=\frac {(6 a) \int \frac {x^2}{\sqrt {a+b x^{-2+m}}} \, dx}{b (4+m)}+\int \frac {x^m}{\sqrt {a+b x^{-2+m}}} \, dx\\ &=\frac {\sqrt {1+\frac {b x^{-2+m}}{a}} \int \frac {x^m}{\sqrt {1+\frac {b x^{-2+m}}{a}}} \, dx}{\sqrt {a+b x^{-2+m}}}+\frac {\left (6 a \sqrt {1+\frac {b x^{-2+m}}{a}}\right ) \int \frac {x^2}{\sqrt {1+\frac {b x^{-2+m}}{a}}} \, dx}{b (4+m) \sqrt {a+b x^{-2+m}}}\\ &=\frac {2 a x^3 \sqrt {1+\frac {b x^{-2+m}}{a}} \, _2F_1\left (\frac {1}{2},-\frac {3}{2-m};-\frac {1+m}{2-m};-\frac {b x^{-2+m}}{a}\right )}{b (4+m) \sqrt {a+b x^{-2+m}}}+\frac {x^{1+m} \sqrt {1+\frac {b x^{-2+m}}{a}} \, _2F_1\left (\frac {1}{2},-\frac {1+m}{2-m};\frac {1-2 m}{2-m};-\frac {b x^{-2+m}}{a}\right )}{(1+m) \sqrt {a+b x^{-2+m}}}\\ \end {align*}

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Mathematica [A]
time = 0.18, size = 26, normalized size = 1.00 \begin {gather*} \frac {2 x^3 \sqrt {a+b x^{-2+m}}}{b (4+m)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.29, size = 40, normalized size = 1.54

method	result	size
risch	\(\frac {2 x \left (a \,x^{2}+b \,x^{m}\right )}{b \left (4+m \right ) \sqrt {\frac {a \,x^{2}+b \,x^{m}}{x^{2}}}}\)	\(40\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(6*a*x^2/b/(4+m)/(a+b*x^(-2+m))^(1/2)+x^m/(a+b*x^(-2+m))^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.34, size = 37, normalized size = 1.42 \begin {gather*} \frac {2 \, {\left (a x^{4} + b x^{2} x^{m}\right )}}{\sqrt {a x^{2} + b x^{m}} b {\left (m + 4\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(6*a*x^2/b/(4+m)/(a+b*x^(-2+m))^(1/2)+x^m/(a+b*x^(-2+m))^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(6*a*x^2/b/(4+m)/(a+b*x^(-2+m))^(1/2)+x^m/(a+b*x^(-2+m))^(1/2),x, algorithm="fricas")

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (ha
s polynomial part)

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Sympy [C] Result contains complex when optimal does not.
time = 17.00, size = 170, normalized size = 6.54 \begin {gather*} \frac {6 a x^{3} \Gamma \left (\frac {3}{m - 2}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {3}{m - 2} \\ 1 + \frac {3}{m - 2} \end {matrix}\middle | {\frac {b x^{m} e^{i \pi }}{a x^{2}}} \right )}}{b \left (m + 4\right ) \left (\sqrt {a} m \Gamma \left (1 + \frac {3}{m - 2}\right ) - 2 \sqrt {a} \Gamma \left (1 + \frac {3}{m - 2}\right )\right )} + \frac {x x^{m} \Gamma \left (\frac {m}{m - 2} + \frac {1}{m - 2}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {m}{m - 2} + \frac {1}{m - 2} \\ \frac {m}{m - 2} + 1 + \frac {1}{m - 2} \end {matrix}\middle | {\frac {b x^{m} e^{i \pi }}{a x^{2}}} \right )}}{\sqrt {a} m \Gamma \left (\frac {m}{m - 2} + 1 + \frac {1}{m - 2}\right ) - 2 \sqrt {a} \Gamma \left (\frac {m}{m - 2} + 1 + \frac {1}{m - 2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(6*a*x**2/b/(4+m)/(a+b*x**(-2+m))**(1/2)+x**m/(a+b*x**(-2+m))**(1/2),x)

[Out]

6*a*x**3*gamma(3/(m - 2))*hyper((1/2, 3/(m - 2)), (1 + 3/(m - 2),), b*x**m*exp_polar(I*pi)/(a*x**2))/(b*(m + 4
)*(sqrt(a)*m*gamma(1 + 3/(m - 2)) - 2*sqrt(a)*gamma(1 + 3/(m - 2)))) + x*x**m*gamma(m/(m - 2) + 1/(m - 2))*hyp
er((1/2, m/(m - 2) + 1/(m - 2)), (m/(m - 2) + 1 + 1/(m - 2),), b*x**m*exp_polar(I*pi)/(a*x**2))/(sqrt(a)*m*gam
ma(m/(m - 2) + 1 + 1/(m - 2)) - 2*sqrt(a)*gamma(m/(m - 2) + 1 + 1/(m - 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(6*a*x^2/b/(4+m)/(a+b*x^(-2+m))^(1/2)+x^m/(a+b*x^(-2+m))^(1/2),x, algorithm="giac")

[Out]

integrate(x^m/sqrt(b*x^(m - 2) + a) + 6*a*x^2/(sqrt(b*x^(m - 2) + a)*b*(m + 4)), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {x^m}{\sqrt {a+b\,x^{m-2}}}+\frac {6\,a\,x^2}{b\,\sqrt {a+b\,x^{m-2}}\,\left (m+4\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m/(a + b*x^(m - 2))^(1/2) + (6*a*x^2)/(b*(a + b*x^(m - 2))^(1/2)*(m + 4)),x)

[Out]

int(x^m/(a + b*x^(m - 2))^(1/2) + (6*a*x^2)/(b*(a + b*x^(m - 2))^(1/2)*(m + 4)), x)

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