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

Optimal. Leaf size=119 \[ \frac{2}{5} x \left (c \sqrt{a+b x^2}\right )^{3/2}+\frac{6 a x \left (c \sqrt{a+b x^2}\right )^{3/2}}{5 \left (a+b x^2\right )}-\frac{6 \sqrt{a} \left (c \sqrt{a+b x^2}\right )^{3/2} E\left (\left .\frac{1}{2} \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )\right |2\right )}{5 \sqrt{b} \left (\frac{b x^2}{a}+1\right )^{3/4}} \]

[Out]

(2*x*(c*Sqrt[a + b*x^2])^(3/2))/5 + (6*a*x*(c*Sqrt[a + b*x^2])^(3/2))/(5*(a + b*x^2)) - (6*Sqrt[a]*(c*Sqrt[a +
b*x^2])^(3/2)*EllipticE[ArcTan[(Sqrt[b]*x)/Sqrt[a]]/2, 2])/(5*Sqrt[b]*(1 + (b*x^2)/a)^(3/4))

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Rubi [A]  time = 0.0522481, antiderivative size = 146, normalized size of antiderivative = 1.23, number of steps used = 5, number of rules used = 5, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.294, Rules used = {6720, 195, 229, 227, 196} \[ -\frac{6 a^{3/2} c \sqrt [4]{\frac{b x^2}{a}+1} \sqrt{c \sqrt{a+b x^2}} E\left (\left .\frac{1}{2} \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )\right |2\right )}{5 \sqrt{b} \sqrt{a+b x^2}}+\frac{6 a c x \sqrt{c \sqrt{a+b x^2}}}{5 \sqrt{a+b x^2}}+\frac{2}{5} c x \sqrt{a+b x^2} \sqrt{c \sqrt{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(c*Sqrt[a + b*x^2])^(3/2),x]

[Out]

(6*a*c*x*Sqrt[c*Sqrt[a + b*x^2]])/(5*Sqrt[a + b*x^2]) + (2*c*x*Sqrt[c*Sqrt[a + b*x^2]]*Sqrt[a + b*x^2])/5 - (6
*a^(3/2)*c*Sqrt[c*Sqrt[a + b*x^2]]*(1 + (b*x^2)/a)^(1/4)*EllipticE[ArcTan[(Sqrt[b]*x)/Sqrt[a]]/2, 2])/(5*Sqrt[
b]*Sqrt[a + b*x^2])

Rule 6720

Int[(u_.)*((a_.)*(v_)^(m_.))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a*v^m)^FracPart[p])/v^(m*FracPart[p]), Int
[u*v^(m*p), x], x] /; FreeQ[{a, m, p}, x] &&  !IntegerQ[p] &&  !FreeQ[v, x] &&  !(EqQ[a, 1] && EqQ[m, 1]) &&
!(EqQ[v, x] && EqQ[m, 1])

Rule 195

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 229

Int[((a_) + (b_.)*(x_)^2)^(-1/4), x_Symbol] :> Dist[(1 + (b*x^2)/a)^(1/4)/(a + b*x^2)^(1/4), Int[1/(1 + (b*x^2
)/a)^(1/4), x], x] /; FreeQ[{a, b}, x] && PosQ[a]

Rule 227

Int[((a_) + (b_.)*(x_)^2)^(-1/4), x_Symbol] :> Simp[(2*x)/(a + b*x^2)^(1/4), x] - Dist[a, Int[1/(a + b*x^2)^(5
/4), x], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b/a]

Rule 196

Int[((a_) + (b_.)*(x_)^2)^(-5/4), x_Symbol] :> Simp[(2*EllipticE[(1*ArcTan[Rt[b/a, 2]*x])/2, 2])/(a^(5/4)*Rt[b
/a, 2]), x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b/a]

Rubi steps

\begin{align*} \int \left (c \sqrt{a+b x^2}\right )^{3/2} \, dx &=\frac{\left (c \sqrt{c \sqrt{a+b x^2}}\right ) \int \left (a+b x^2\right )^{3/4} \, dx}{\sqrt [4]{a+b x^2}}\\ &=\frac{2}{5} c x \sqrt{c \sqrt{a+b x^2}} \sqrt{a+b x^2}+\frac{\left (3 a c \sqrt{c \sqrt{a+b x^2}}\right ) \int \frac{1}{\sqrt [4]{a+b x^2}} \, dx}{5 \sqrt [4]{a+b x^2}}\\ &=\frac{2}{5} c x \sqrt{c \sqrt{a+b x^2}} \sqrt{a+b x^2}+\frac{\left (3 a c \sqrt{c \sqrt{a+b x^2}} \sqrt [4]{1+\frac{b x^2}{a}}\right ) \int \frac{1}{\sqrt [4]{1+\frac{b x^2}{a}}} \, dx}{5 \sqrt{a+b x^2}}\\ &=\frac{6 a c x \sqrt{c \sqrt{a+b x^2}}}{5 \sqrt{a+b x^2}}+\frac{2}{5} c x \sqrt{c \sqrt{a+b x^2}} \sqrt{a+b x^2}-\frac{\left (3 a c \sqrt{c \sqrt{a+b x^2}} \sqrt [4]{1+\frac{b x^2}{a}}\right ) \int \frac{1}{\left (1+\frac{b x^2}{a}\right )^{5/4}} \, dx}{5 \sqrt{a+b x^2}}\\ &=\frac{6 a c x \sqrt{c \sqrt{a+b x^2}}}{5 \sqrt{a+b x^2}}+\frac{2}{5} c x \sqrt{c \sqrt{a+b x^2}} \sqrt{a+b x^2}-\frac{6 a^{3/2} c \sqrt{c \sqrt{a+b x^2}} \sqrt [4]{1+\frac{b x^2}{a}} E\left (\left .\frac{1}{2} \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )\right |2\right )}{5 \sqrt{b} \sqrt{a+b x^2}}\\ \end{align*}

Mathematica [C]  time = 0.0066299, size = 52, normalized size = 0.44 \[ \frac{x \left (c \sqrt{a+b x^2}\right )^{3/2} \, _2F_1\left (-\frac{3}{4},\frac{1}{2};\frac{3}{2};-\frac{b x^2}{a}\right )}{\left (\frac{b x^2}{a}+1\right )^{3/4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.006, size = 0, normalized size = 0. \begin{align*} \int \left ( c\sqrt{b{x}^{2}+a} \right ) ^{{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\sqrt{b x^{2} + a} \sqrt{\sqrt{b x^{2} + a} c} c, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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