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

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

[Out]

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

________________________________________________________________________________________

Rubi [C]  time = 0.07, antiderivative size = 127, normalized size of antiderivative = 7.47, number of steps used = 5, number of rules used = 2, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.047, Rules used = {365, 364} \[ \frac {a x^{m+2} \sqrt {\frac {b x^2}{a}+1} \, _2F_1\left (\frac {1}{2},\frac {m+2}{2};\frac {m+4}{2};-\frac {b x^2}{a}\right )}{\sqrt {a+b x^2}}+\frac {b (m+3) x^{m+4} \sqrt {\frac {b x^2}{a}+1} \, _2F_1\left (\frac {1}{2},\frac {m+4}{2};\frac {m+6}{2};-\frac {b x^2}{a}\right )}{(m+4) \sqrt {a+b x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*x^(2 + m)*Sqrt[1 + (b*x^2)/a]*Hypergeometric2F1[1/2, (2 + m)/2, (4 + m)/2, -((b*x^2)/a)])/Sqrt[a + b*x^2] +
(b*(3 + m)*x^(4 + m)*Sqrt[1 + (b*x^2)/a]*Hypergeometric2F1[1/2, (4 + m)/2, (6 + m)/2, -((b*x^2)/a)])/((4 + 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 {a (2+m) x^{1+m}}{\sqrt {a+b x^2}}+\frac {b (3+m) x^{3+m}}{\sqrt {a+b x^2}}\right ) \, dx &=(a (2+m)) \int \frac {x^{1+m}}{\sqrt {a+b x^2}} \, dx+(b (3+m)) \int \frac {x^{3+m}}{\sqrt {a+b x^2}} \, dx\\ &=\frac {\left (a (2+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 {\left (b (3+m) \sqrt {1+\frac {b x^2}{a}}\right ) \int \frac {x^{3+m}}{\sqrt {1+\frac {b x^2}{a}}} \, dx}{\sqrt {a+b x^2}}\\ &=\frac {a x^{2+m} \sqrt {1+\frac {b x^2}{a}} \, _2F_1\left (\frac {1}{2},\frac {2+m}{2};\frac {4+m}{2};-\frac {b x^2}{a}\right )}{\sqrt {a+b x^2}}+\frac {b (3+m) x^{4+m} \sqrt {1+\frac {b x^2}{a}} \, _2F_1\left (\frac {1}{2},\frac {4+m}{2};\frac {6+m}{2};-\frac {b x^2}{a}\right )}{(4+m) \sqrt {a+b x^2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 0.07, 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[(a*(2 + m)*x^(1 + m))/Sqrt[a + b*x^2] + (b*(3 + m)*x^(3 + m))/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])

________________________________________________________________________________________

fricas [A]  time = 0.71, size = 18, normalized size = 1.06 \[ \frac {\sqrt {b x^{2} + a} x^{m + 3}}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(b*x^2 + a)*x^(m + 3)/x

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [F]  time = 0.28, size = 0, normalized size = 0.00 \[ \int \frac {\left (m +2\right ) a \,x^{m +1}}{\sqrt {b \,x^{2}+a}}+\frac {\left (m +3\right ) b \,x^{m +3}}{\sqrt {b \,x^{2}+a}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 1.89, 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(a*(2+m)*x^(1+m)/(b*x^2+a)^(1/2)+b*(3+m)*x^(3+m)/(b*x^2+a)^(1/2),x, algorithm="maxima")

[Out]

sqrt(b*x^2 + a)*x^2*x^m

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.06 \[ \int \frac {a\,x^{m+1}\,\left (m+2\right )}{\sqrt {b\,x^2+a}}+\frac {b\,x^{m+3}\,\left (m+3\right )}{\sqrt {b\,x^2+a}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [C]  time = 5.29, size = 105, normalized size = 6.18 \[ \frac {\sqrt {a} x^{2} x^{m} \left (m + 2\right ) \Gamma \left (\frac {m}{2} + 1\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {m}{2} + 1 \\ \frac {m}{2} + 2 \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \Gamma \left (\frac {m}{2} + 2\right )} + \frac {b x^{4} x^{m} \left (m + 3\right ) \Gamma \left (\frac {m}{2} + 2\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {m}{2} + 2 \\ \frac {m}{2} + 3 \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \sqrt {a} \Gamma \left (\frac {m}{2} + 3\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(a)*x**2*x**m*(m + 2)*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)) + b*x**4*x**m*(m + 3)*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))

________________________________________________________________________________________