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

Optimal. Leaf size=127 \[ \frac {2 c \sqrt {c x} \sqrt {a+b x^2}}{3 b}-\frac {a^{3/4} c^{3/2} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} \operatorname {EllipticF}\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a} \sqrt {c}}\right ),\frac {1}{2}\right )}{3 b^{5/4} \sqrt {a+b x^2}} \]

[Out]

2/3*c*(c*x)^(1/2)*(b*x^2+a)^(1/2)/b-1/3*a^(3/4)*c^(3/2)*(cos(2*arctan(b^(1/4)*(c*x)^(1/2)/a^(1/4)/c^(1/2)))^2)
^(1/2)/cos(2*arctan(b^(1/4)*(c*x)^(1/2)/a^(1/4)/c^(1/2)))*EllipticF(sin(2*arctan(b^(1/4)*(c*x)^(1/2)/a^(1/4)/c
^(1/2))),1/2*2^(1/2))*(a^(1/2)+x*b^(1/2))*((b*x^2+a)/(a^(1/2)+x*b^(1/2))^2)^(1/2)/b^(5/4)/(b*x^2+a)^(1/2)

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Rubi [A]  time = 0.07, antiderivative size = 127, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {321, 329, 220} \[ \frac {2 c \sqrt {c x} \sqrt {a+b x^2}}{3 b}-\frac {a^{3/4} c^{3/2} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a} \sqrt {c}}\right )|\frac {1}{2}\right )}{3 b^{5/4} \sqrt {a+b x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(c*x)^(3/2)/Sqrt[a + b*x^2],x]

[Out]

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

Rule 220

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[((1 + q^2*x^2)*Sqrt[(a + b*x^4)/(a*(
1 + q^2*x^2)^2)]*EllipticF[2*ArcTan[q*x], 1/2])/(2*q*Sqrt[a + b*x^4]), x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

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 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rubi steps

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

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Mathematica [C]  time = 0.02, size = 69, normalized size = 0.54 \[ \frac {2 c \sqrt {c x} \left (-a \sqrt {\frac {b x^2}{a}+1} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};-\frac {b x^2}{a}\right )+a+b x^2\right )}{3 b \sqrt {a+b x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*c*Sqrt[c*x]*(a + b*x^2 - a*Sqrt[1 + (b*x^2)/a]*Hypergeometric2F1[1/4, 1/2, 5/4, -((b*x^2)/a)]))/(3*b*Sqrt[a
+ b*x^2])

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fricas [F]  time = 0.93, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {c x} c x}{\sqrt {b x^{2} + a}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 125, normalized size = 0.98 \[ -\frac {\sqrt {c x}\, \left (-2 b^{2} x^{3}-2 a b x +\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \sqrt {-a b}\, a \EllipticF \left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )\right ) c}{3 \sqrt {b \,x^{2}+a}\, b^{2} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x)^(3/2)/(a + b*x^2)^(1/2),x)

[Out]

int((c*x)^(3/2)/(a + b*x^2)^(1/2), x)

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sympy [C]  time = 1.93, size = 44, normalized size = 0.35 \[ \frac {c^{\frac {3}{2}} x^{\frac {5}{2}} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \sqrt {a} \Gamma \left (\frac {9}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c**(3/2)*x**(5/2)*gamma(5/4)*hyper((1/2, 5/4), (9/4,), b*x**2*exp_polar(I*pi)/a)/(2*sqrt(a)*gamma(9/4))

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