3.2956 \(\int (d x)^m \sqrt {a+b \sqrt {c x^2}} \, dx\)

Optimal. Leaf size=88 \[ \frac {2 (d x)^{m+1} \left (a+b \sqrt {c x^2}\right )^{3/2} \left (-\frac {b \sqrt {c x^2}}{a}\right )^{-m} \, _2F_1\left (\frac {3}{2},-m;\frac {5}{2};\frac {\sqrt {c x^2} b}{a}+1\right )}{3 b d \sqrt {c x^2}} \]

[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)

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Rubi [A]  time = 0.04, antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {368, 67, 65} \[ \frac {2 (d x)^{m+1} \left (a+b \sqrt {c x^2}\right )^{3/2} \left (-\frac {b \sqrt {c x^2}}{a}\right )^{-m} \, _2F_1\left (\frac {3}{2},-m;\frac {5}{2};\frac {\sqrt {c x^2} b}{a}+1\right )}{3 b d \sqrt {c x^2}} \]

Antiderivative was successfully verified.

[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 65

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

Rule 67

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 368

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*} \int (d x)^m \sqrt {a+b \sqrt {c x^2}} \, dx &=\frac {\left ((d x)^{1+m} \left (c x^2\right )^{\frac {1}{2} (-1-m)}\right ) \operatorname {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 ) \operatorname {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*}

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Mathematica [A]  time = 0.08, size = 74, normalized size = 0.84 \[ \frac {x (d x)^m \sqrt {a+b \sqrt {c x^2}} \, _2F_1\left (-\frac {1}{2},m+1;m+2;-\frac {b \sqrt {c x^2}}{a}\right )}{(m+1) \sqrt {\frac {b \sqrt {c x^2}}{a}+1}} \]

Antiderivative was successfully verified.

[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])

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fricas [F]  time = 0.96, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {\sqrt {c x^{2}} b + a} \left (d x\right )^{m}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [F]  time = 0.22, size = 0, normalized size = 0.00 \[ \int \sqrt {a +\sqrt {c \,x^{2}}\, b}\, \left (d x \right )^{m}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (d\,x\right )}^m\,\sqrt {a+b\,\sqrt {c\,x^2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (d x\right )^{m} \sqrt {a + b \sqrt {c x^{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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