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

Optimal. Leaf size=152 \[ \frac {4 a^{3/2} \left (c \sqrt {a+b x^2}\right )^{3/2} E\left (\left .\frac {1}{2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{15 b^{3/2} \left (\frac {b x^2}{a}+1\right )^{3/4}}-\frac {4 a^2 x \left (c \sqrt {a+b x^2}\right )^{3/2}}{15 b \left (a+b x^2\right )}+\frac {2 a x \left (c \sqrt {a+b x^2}\right )^{3/2}}{15 b}+\frac {2}{9} x^3 \left (c \sqrt {a+b x^2}\right )^{3/2} \]

[Out]

2/15*a*x*(c*(b*x^2+a)^(1/2))^(3/2)/b+2/9*x^3*(c*(b*x^2+a)^(1/2))^(3/2)-4/15*a^2*x*(c*(b*x^2+a)^(1/2))^(3/2)/b/
(b*x^2+a)+4/15*a^(3/2)*(cos(1/2*arctan(x*b^(1/2)/a^(1/2)))^2)^(1/2)/cos(1/2*arctan(x*b^(1/2)/a^(1/2)))*Ellipti
cE(sin(1/2*arctan(x*b^(1/2)/a^(1/2))),2^(1/2))*(c*(b*x^2+a)^(1/2))^(3/2)/b^(3/2)/(1+b*x^2/a)^(3/4)

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Rubi [A]  time = 0.17, antiderivative size = 191, normalized size of antiderivative = 1.26, number of steps used = 6, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {6720, 279, 321, 229, 227, 196} \[ \frac {4 a^{5/2} c \sqrt [4]{\frac {b x^2}{a}+1} \sqrt {c \sqrt {a+b x^2}} E\left (\left .\frac {1}{2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{15 b^{3/2} \sqrt {a+b x^2}}-\frac {4 a^2 c x \sqrt {c \sqrt {a+b x^2}}}{15 b \sqrt {a+b x^2}}+\frac {2}{9} c x^3 \sqrt {a+b x^2} \sqrt {c \sqrt {a+b x^2}}+\frac {2 a c x \sqrt {a+b x^2} \sqrt {c \sqrt {a+b x^2}}}{15 b} \]

Antiderivative was successfully verified.

[In]

Int[x^2*(c*Sqrt[a + b*x^2])^(3/2),x]

[Out]

(-4*a^2*c*x*Sqrt[c*Sqrt[a + b*x^2]])/(15*b*Sqrt[a + b*x^2]) + (2*a*c*x*Sqrt[c*Sqrt[a + b*x^2]]*Sqrt[a + b*x^2]
)/(15*b) + (2*c*x^3*Sqrt[c*Sqrt[a + b*x^2]]*Sqrt[a + b*x^2])/9 + (4*a^(5/2)*c*Sqrt[c*Sqrt[a + b*x^2]]*(1 + (b*
x^2)/a)^(1/4)*EllipticE[ArcTan[(Sqrt[b]*x)/Sqrt[a]]/2, 2])/(15*b^(3/2)*Sqrt[a + b*x^2])

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]

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 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 279

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^p)/(c*(m +
n*p + 1)), x] + Dist[(a*n*p)/(m + n*p + 1), Int[(c*x)^m*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c, m}, x]
&& IGtQ[n, 0] && GtQ[p, 0] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

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])

Rubi steps

\begin {align*} \int x^2 \left (c \sqrt {a+b x^2}\right )^{3/2} \, dx &=\frac {\left (c \sqrt {c \sqrt {a+b x^2}}\right ) \int x^2 \left (a+b x^2\right )^{3/4} \, dx}{\sqrt [4]{a+b x^2}}\\ &=\frac {2}{9} c x^3 \sqrt {c \sqrt {a+b x^2}} \sqrt {a+b x^2}+\frac {\left (a c \sqrt {c \sqrt {a+b x^2}}\right ) \int \frac {x^2}{\sqrt [4]{a+b x^2}} \, dx}{3 \sqrt [4]{a+b x^2}}\\ &=\frac {2 a c x \sqrt {c \sqrt {a+b x^2}} \sqrt {a+b x^2}}{15 b}+\frac {2}{9} c x^3 \sqrt {c \sqrt {a+b x^2}} \sqrt {a+b x^2}-\frac {\left (2 a^2 c \sqrt {c \sqrt {a+b x^2}}\right ) \int \frac {1}{\sqrt [4]{a+b x^2}} \, dx}{15 b \sqrt [4]{a+b x^2}}\\ &=\frac {2 a c x \sqrt {c \sqrt {a+b x^2}} \sqrt {a+b x^2}}{15 b}+\frac {2}{9} c x^3 \sqrt {c \sqrt {a+b x^2}} \sqrt {a+b x^2}-\frac {\left (2 a^2 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}{15 b \sqrt {a+b x^2}}\\ &=-\frac {4 a^2 c x \sqrt {c \sqrt {a+b x^2}}}{15 b \sqrt {a+b x^2}}+\frac {2 a c x \sqrt {c \sqrt {a+b x^2}} \sqrt {a+b x^2}}{15 b}+\frac {2}{9} c x^3 \sqrt {c \sqrt {a+b x^2}} \sqrt {a+b x^2}+\frac {\left (2 a^2 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}{15 b \sqrt {a+b x^2}}\\ &=-\frac {4 a^2 c x \sqrt {c \sqrt {a+b x^2}}}{15 b \sqrt {a+b x^2}}+\frac {2 a c x \sqrt {c \sqrt {a+b x^2}} \sqrt {a+b x^2}}{15 b}+\frac {2}{9} c x^3 \sqrt {c \sqrt {a+b x^2}} \sqrt {a+b x^2}+\frac {4 a^{5/2} 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 )}{15 b^{3/2} \sqrt {a+b x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.06, size = 68, normalized size = 0.45 \[ \frac {2 x \left (c \sqrt {a+b x^2}\right )^{3/2} \left (-\frac {a \, _2F_1\left (-\frac {3}{4},\frac {1}{2};\frac {3}{2};-\frac {b x^2}{a}\right )}{\left (\frac {b x^2}{a}+1\right )^{3/4}}+a+b x^2\right )}{9 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x*(c*Sqrt[a + b*x^2])^(3/2)*(a + b*x^2 - (a*Hypergeometric2F1[-3/4, 1/2, 3/2, -((b*x^2)/a)])/(1 + (b*x^2)/a
)^(3/4)))/(9*b)

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fricas [F]  time = 0.97, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {b x^{2} + a} \sqrt {\sqrt {b x^{2} + a} c} c x^{2}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(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^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.01, size = 0, normalized size = 0.00 \[ \int \left (\sqrt {b \,x^{2}+a}\, c \right )^{\frac {3}{2}} x^{2}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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