3.2919 \(\int (d x)^m \sqrt{a+b (c x)^{3/2}} \, dx\)

Optimal. Leaf size=78 \[ \frac{x (d x)^m \sqrt{a+b (c x)^{3/2}} \, _2F_1\left (-\frac{1}{2},\frac{2 (m+1)}{3};\frac{1}{3} (2 m+5);-\frac{b (c x)^{3/2}}{a}\right )}{(m+1) \sqrt{\frac{b (c x)^{3/2}}{a}+1}} \]

[Out]

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

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Rubi [A]  time = 0.0827579, antiderivative size = 78, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {367, 343, 341, 365, 364} \[ \frac{x (d x)^m \sqrt{a+b (c x)^{3/2}} \, _2F_1\left (-\frac{1}{2},\frac{2 (m+1)}{3};\frac{1}{3} (2 m+5);-\frac{b (c x)^{3/2}}{a}\right )}{(m+1) \sqrt{\frac{b (c x)^{3/2}}{a}+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 367

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

Rule 343

Int[((c_)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[(c^IntPart[m]*(c*x)^FracPart[m])/x^FracP
art[m], Int[x^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, p}, x] && FractionQ[n]

Rule 341

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[n]}, Dist[k, Subst[Int[x^(k*(
m + 1) - 1)*(a + b*x^(k*n))^p, x], x, x^(1/k)], x]] /; FreeQ[{a, b, m, p}, x] && FractionQ[n]

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 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])

Rubi steps

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

Mathematica [A]  time = 0.0997827, size = 79, normalized size = 1.01 \[ \frac{x (d x)^m \sqrt{a+b (c x)^{3/2}} \, _2F_1\left (-\frac{1}{2},\frac{2 (m+1)}{3};\frac{1}{3} (2 m+5);-\frac{b (c x)^{3/2}}{a}\right )}{(m+1) \sqrt{\frac{a+b (c x)^{3/2}}{a}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.053, size = 0, normalized size = 0. \begin{align*} \int \left ( dx \right ) ^{m}\sqrt{a+b \left ( cx \right ) ^{{\frac{3}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{\left (c x\right )^{\frac{3}{2}} b + a} \left (d x\right )^{m}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: UnboundLocalError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: UnboundLocalError

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (d x\right )^{m} \sqrt{a + b \left (c x\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{\left (c x\right )^{\frac{3}{2}} b + a} \left (d x\right )^{m}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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