3.3.57 \(\int \frac {(c \sqrt {a+b x^2})^{3/2}}{x^2} \, dx\) [257]

Optimal. Leaf size=115 \[ -\frac {\left (c \sqrt {a+b x^2}\right )^{3/2}}{x}+\frac {3 b x \left (c \sqrt {a+b x^2}\right )^{3/2}}{a+b x^2}-\frac {3 \sqrt {b} \left (c \sqrt {a+b x^2}\right )^{3/2} E\left (\left .\frac {1}{2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{\sqrt {a} \left (1+\frac {b x^2}{a}\right )^{3/4}} \]

[Out]

-(c*(b*x^2+a)^(1/2))^(3/2)/x+3*b*x*(c*(b*x^2+a)^(1/2))^(3/2)/(b*x^2+a)-3*(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)))*EllipticE(sin(1/2*arctan(x*b^(1/2)/a^(1/2))),2^(1/2))*b^(1/2)*(c*(b
*x^2+a)^(1/2))^(3/2)/(1+b*x^2/a)^(3/4)/a^(1/2)

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Rubi [A]
time = 0.03, antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {1973, 283, 233, 202} \begin {gather*} -\frac {3 \sqrt {b} \left (c \sqrt {a+b x^2}\right )^{3/2} E\left (\left .\frac {1}{2} \text {ArcTan}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{\sqrt {a} \left (\frac {b x^2}{a}+1\right )^{3/4}}-\frac {\left (c \sqrt {a+b x^2}\right )^{3/2}}{x}+\frac {3 b x \left (c \sqrt {a+b x^2}\right )^{3/2}}{a+b x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c*Sqrt[a + b*x^2])^(3/2)/x^2,x]

[Out]

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

Rule 202

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

Rule 233

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 283

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

Rule 1973

Int[(u_.)*((c_.)*((a_) + (b_.)*(x_)^(n_.))^(q_))^(p_), x_Symbol] :> Dist[Simp[(c*(a + b*x^n)^q)^p/(1 + b*(x^n/
a))^(p*q)], Int[u*(1 + b*(x^n/a))^(p*q), x], x] /; FreeQ[{a, b, c, n, p, q}, x] &&  !GeQ[a, 0]

Rubi steps

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

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 7.72, size = 55, normalized size = 0.48 \begin {gather*} -\frac {\left (c \sqrt {a+b x^2}\right )^{3/2} \, _2F_1\left (-\frac {3}{4},-\frac {1}{2};\frac {1}{2};-\frac {b x^2}{a}\right )}{x \left (1+\frac {b x^2}{a}\right )^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*(b*x^2+a)^(1/2))^(3/2)/x^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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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (c \sqrt {a + b x^{2}}\right )^{\frac {3}{2}}}{x^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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 \begin {gather*} \int \frac {{\left (c\,\sqrt {b\,x^2+a}\right )}^{3/2}}{x^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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