3.2955 \(\int (d x)^m \sqrt {a+b (c x^2)^{3/2}} \, dx\)

Optimal. Leaf size=86 \[ \frac {(d x)^{m+1} \sqrt {a+b \left (c x^2\right )^{3/2}} \, _2F_1\left (-\frac {1}{2},\frac {m+1}{3};\frac {m+4}{3};-\frac {b \left (c x^2\right )^{3/2}}{a}\right )}{d (m+1) \sqrt {\frac {b \left (c x^2\right )^{3/2}}{a}+1}} \]

[Out]

(d*x)^(1+m)*hypergeom([-1/2, 1/3+1/3*m],[4/3+1/3*m],-b*(c*x^2)^(3/2)/a)*(a+b*(c*x^2)^(3/2))^(1/2)/d/(1+m)/(1+b
*(c*x^2)^(3/2)/a)^(1/2)

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Rubi [A]  time = 0.06, antiderivative size = 86, 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, 365, 364} \[ \frac {(d x)^{m+1} \sqrt {a+b \left (c x^2\right )^{3/2}} \, _2F_1\left (-\frac {1}{2},\frac {m+1}{3};\frac {m+4}{3};-\frac {b \left (c x^2\right )^{3/2}}{a}\right )}{d (m+1) \sqrt {\frac {b \left (c x^2\right )^{3/2}}{a}+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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])

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

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Mathematica [F]  time = 0.08, size = 0, normalized size = 0.00 \[ \int (d x)^m \sqrt {a+b \left (c x^2\right )^{3/2}} \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[(d*x)^m*Sqrt[a + b*(c*x^2)^(3/2)],x]

[Out]

Integrate[(d*x)^m*Sqrt[a + b*(c*x^2)^(3/2)], x]

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fricas [F]  time = 1.02, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {\sqrt {c x^{2}} b c x^{2} + a} \left (d x\right )^{m}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.22, size = 0, normalized size = 0.00 \[ \int \sqrt {a +\left (c \,x^{2}\right )^{\frac {3}{2}} b}\, \left (d x \right )^{m}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (d\,x\right )}^m\,\sqrt {a+b\,{\left (c\,x^2\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((d*x)**m*sqrt(a + b*(c*x**2)**(3/2)), x)

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