\(\int (d x)^m \sqrt {a+b \sqrt {c x^2}} \, dx\) [2956]

Optimal result

Integrand size = 23, antiderivative size = 88 \[ \int (d x)^m \sqrt {a+b \sqrt {c x^2}} \, dx=\frac {2 (d x)^{1+m} \left (-\frac {b \sqrt {c x^2}}{a}\right )^{-m} \left (a+b \sqrt {c x^2}\right )^{3/2} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-m,\frac {5}{2},1+\frac {b \sqrt {c x^2}}{a}\right )}{3 b d \sqrt {c x^2}} \]

[Out]

2/3*(d*x)^(1+m)*hypergeom([3/2, -m],[5/2],1+b*(c*x^2)^(1/2)/a)*(a+b*(c*x^2)^(1/2))^(3/2)/b/d/(c*x^2)^(1/2)/((-
b*(c*x^2)^(1/2)/a)^m)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {375, 69, 67} \[ \int (d x)^m \sqrt {a+b \sqrt {c x^2}} \, dx=\frac {2 (d x)^{m+1} \left (a+b \sqrt {c x^2}\right )^{3/2} \left (-\frac {b \sqrt {c x^2}}{a}\right )^{-m} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-m,\frac {5}{2},\frac {\sqrt {c x^2} b}{a}+1\right )}{3 b d \sqrt {c x^2}} \]

[In]

Int[(d*x)^m*Sqrt[a + b*Sqrt[c*x^2]],x]

[Out]

(2*(d*x)^(1 + m)*(a + b*Sqrt[c*x^2])^(3/2)*Hypergeometric2F1[3/2, -m, 5/2, 1 + (b*Sqrt[c*x^2])/a])/(3*b*d*Sqrt
[c*x^2]*(-((b*Sqrt[c*x^2])/a))^m)

Rule 67

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((c + d*x)^(n + 1)/(d*(n + 1)*(-d/(b*c))^m))
*Hypergeometric2F1[-m, n + 1, n + 2, 1 + d*(x/c)], x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[n] && (Intege
rQ[m] || GtQ[-d/(b*c), 0])

Rule 69

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Dist[((-b)*(c/d))^IntPart[m]*((b*x)^FracPart[m]/(
(-d)*(x/c))^FracPart[m]), Int[((-d)*(x/c))^m*(c + d*x)^n, x], x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[m]
&&  !IntegerQ[n] &&  !GtQ[c, 0] &&  !GtQ[-d/(b*c), 0]

Rule 375

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*((c_.)*(x_)^(q_))^(n_))^(p_.), x_Symbol] :> Dist[(d*x)^(m + 1)/(d*((c*x^q
)^(1/q))^(m + 1)), Subst[Int[x^m*(a + b*x^(n*q))^p, x], x, (c*x^q)^(1/q)], x] /; FreeQ[{a, b, c, d, m, n, p, q
}, x] && IntegerQ[n*q] && NeQ[x, (c*x^q)^(1/q)]

Rubi steps \begin{align*} \text {integral}& = \frac {\left ((d x)^{1+m} \left (c x^2\right )^{\frac {1}{2} (-1-m)}\right ) \text {Subst}\left (\int x^m \sqrt {a+b x} \, dx,x,\sqrt {c x^2}\right )}{d} \\ & = \frac {\left ((d x)^{1+m} \left (c x^2\right )^{\frac {1}{2} (-1-m)+\frac {m}{2}} \left (-\frac {b \sqrt {c x^2}}{a}\right )^{-m}\right ) \text {Subst}\left (\int \left (-\frac {b x}{a}\right )^m \sqrt {a+b x} \, dx,x,\sqrt {c x^2}\right )}{d} \\ & = \frac {2 (d x)^{1+m} \left (-\frac {b \sqrt {c x^2}}{a}\right )^{-m} \left (a+b \sqrt {c x^2}\right )^{3/2} \, _2F_1\left (\frac {3}{2},-m;\frac {5}{2};1+\frac {b \sqrt {c x^2}}{a}\right )}{3 b d \sqrt {c x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.60 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.84 \[ \int (d x)^m \sqrt {a+b \sqrt {c x^2}} \, dx=\frac {x (d x)^m \sqrt {a+b \sqrt {c x^2}} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1+m,2+m,-\frac {b \sqrt {c x^2}}{a}\right )}{(1+m) \sqrt {1+\frac {b \sqrt {c x^2}}{a}}} \]

[In]

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

[Out]

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

Maple [F]

\[\int \left (d x \right )^{m} \sqrt {a +b \sqrt {c \,x^{2}}}d x\]

[In]

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

[Out]

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

Fricas [F]

\[ \int (d x)^m \sqrt {a+b \sqrt {c x^2}} \, dx=\int { \sqrt {\sqrt {c x^{2}} b + a} \left (d x\right )^{m} \,d x } \]

[In]

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

[Out]

integral(sqrt(sqrt(c*x^2)*b + a)*(d*x)^m, x)

Sympy [F]

\[ \int (d x)^m \sqrt {a+b \sqrt {c x^2}} \, dx=\int \left (d x\right )^{m} \sqrt {a + b \sqrt {c x^{2}}}\, dx \]

[In]

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

[Out]

Integral((d*x)**m*sqrt(a + b*sqrt(c*x**2)), x)

Maxima [F]

\[ \int (d x)^m \sqrt {a+b \sqrt {c x^2}} \, dx=\int { \sqrt {\sqrt {c x^{2}} b + a} \left (d x\right )^{m} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int (d x)^m \sqrt {a+b \sqrt {c x^2}} \, dx=\int { \sqrt {\sqrt {c x^{2}} b + a} \left (d x\right )^{m} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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