\(\int x^m \sqrt {a+b x^{2+2 m}} \, dx\) [2748]

Optimal result

Integrand size = 19, antiderivative size = 72 \[ \int x^m \sqrt {a+b x^{2+2 m}} \, dx=\frac {x^{1+m} \sqrt {a+b x^{2 (1+m)}}}{2 (1+m)}+\frac {a \text {arctanh}\left (\frac {\sqrt {b} x^{1+m}}{\sqrt {a+b x^{2 (1+m)}}}\right )}{2 \sqrt {b} (1+m)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {352, 201, 223, 212} \[ \int x^m \sqrt {a+b x^{2+2 m}} \, dx=\frac {a \text {arctanh}\left (\frac {\sqrt {b} x^{m+1}}{\sqrt {a+b x^{2 (m+1)}}}\right )}{2 \sqrt {b} (m+1)}+\frac {x^{m+1} \sqrt {a+b x^{2 (m+1)}}}{2 (m+1)} \]

[In]

Int[x^m*Sqrt[a + b*x^(2 + 2*m)],x]

[Out]

(x^(1 + m)*Sqrt[a + b*x^(2*(1 + m))])/(2*(1 + m)) + (a*ArcTanh[(Sqrt[b]*x^(1 + m))/Sqrt[a + b*x^(2*(1 + m))]])
/(2*Sqrt[b]*(1 + m))

Rule 201

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^p/(n*p + 1)), x] + Dist[a*n*(p/(n*p + 1)),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 352

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/(m + 1), Subst[Int[(a + b*x^Simplify[n/(m +
1)])^p, x], x, x^(m + 1)], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[n/(m + 1)]] &&  !IntegerQ[n]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \sqrt {a+b x^2} \, dx,x,x^{1+m}\right )}{1+m} \\ & = \frac {x^{1+m} \sqrt {a+b x^{2 (1+m)}}}{2 (1+m)}+\frac {a \text {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,x^{1+m}\right )}{2 (1+m)} \\ & = \frac {x^{1+m} \sqrt {a+b x^{2 (1+m)}}}{2 (1+m)}+\frac {a \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x^{1+m}}{\sqrt {a+b x^{2+2 m}}}\right )}{2 (1+m)} \\ & = \frac {x^{1+m} \sqrt {a+b x^{2 (1+m)}}}{2 (1+m)}+\frac {a \tanh ^{-1}\left (\frac {\sqrt {b} x^{1+m}}{\sqrt {a+b x^{2 (1+m)}}}\right )}{2 \sqrt {b} (1+m)} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.07 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.18 \[ \int x^m \sqrt {a+b x^{2+2 m}} \, dx=\frac {x^{1+m} \sqrt {a+b x^{2+2 m}} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1+m}{2+2 m},1+\frac {1+m}{2+2 m},-\frac {b x^{2+2 m}}{a}\right )}{(1+m) \sqrt {1+\frac {b x^{2+2 m}}{a}}} \]

[In]

Integrate[x^m*Sqrt[a + b*x^(2 + 2*m)],x]

[Out]

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

Maple [F]

\[\int x^{m} \sqrt {a +b \,x^{2+2 m}}d x\]

[In]

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

[Out]

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

Fricas [F(-2)]

Exception generated. \[ \int x^m \sqrt {a+b x^{2+2 m}} \, dx=\text {Exception raised: TypeError} \]

[In]

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

[Out]

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.04 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.49 \[ \int x^m \sqrt {a+b x^{2+2 m}} \, dx=\frac {\sqrt {\pi } \sqrt {a} a^{- \frac {m}{2 m + 2} + \frac {1}{2} - \frac {1}{2 m + 2}} x^{m + 1} {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {1}{2} \\ \frac {m}{2 m + 2} + 1 + \frac {1}{2 m + 2} \end {matrix}\middle | {\frac {b x^{2 m + 2} e^{i \pi }}{a}} \right )}}{2 m \Gamma \left (\frac {m}{2 m + 2} + 1 + \frac {1}{2 m + 2}\right ) + 2 \Gamma \left (\frac {m}{2 m + 2} + 1 + \frac {1}{2 m + 2}\right )} \]

[In]

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

[Out]

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

Maxima [F]

\[ \int x^m \sqrt {a+b x^{2+2 m}} \, dx=\int { \sqrt {b x^{2 \, m + 2} + a} x^{m} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int x^m \sqrt {a+b x^{2+2 m}} \, dx=\int { \sqrt {b x^{2 \, m + 2} + a} x^{m} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int x^m \sqrt {a+b x^{2+2 m}} \, dx=\int x^m\,\sqrt {a+b\,x^{2\,m+2}} \,d x \]

[In]

int(x^m*(a + b*x^(2*m + 2))^(1/2),x)

[Out]

int(x^m*(a + b*x^(2*m + 2))^(1/2), x)