3.7.14 \(\int \frac {(c x)^{5/2}}{\sqrt {a+b x^2}} \, dx\) [614]

Optimal. Leaf size=273 \[ \frac {2 c (c x)^{3/2} \sqrt {a+b x^2}}{5 b}-\frac {6 a c^2 \sqrt {c x} \sqrt {a+b x^2}}{5 b^{3/2} \left (\sqrt {a}+\sqrt {b} x\right )}+\frac {6 a^{5/4} c^{5/2} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a} \sqrt {c}}\right )|\frac {1}{2}\right )}{5 b^{7/4} \sqrt {a+b x^2}}-\frac {3 a^{5/4} c^{5/2} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a} \sqrt {c}}\right )|\frac {1}{2}\right )}{5 b^{7/4} \sqrt {a+b x^2}} \]

[Out]

2/5*c*(c*x)^(3/2)*(b*x^2+a)^(1/2)/b-6/5*a*c^2*(c*x)^(1/2)*(b*x^2+a)^(1/2)/b^(3/2)/(a^(1/2)+x*b^(1/2))+6/5*a^(5
/4)*c^(5/2)*(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)))*EllipticE(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^(7/4)/(b*x^2+a)^(1/2)-3/5*a^(5/4)*c^(5/2)*(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^
(7/4)/(b*x^2+a)^(1/2)

________________________________________________________________________________________

Rubi [A]
time = 0.14, antiderivative size = 273, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {327, 335, 311, 226, 1210} \begin {gather*} -\frac {3 a^{5/4} c^{5/2} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} F\left (2 \text {ArcTan}\left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a} \sqrt {c}}\right )|\frac {1}{2}\right )}{5 b^{7/4} \sqrt {a+b x^2}}+\frac {6 a^{5/4} c^{5/2} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} E\left (2 \text {ArcTan}\left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a} \sqrt {c}}\right )|\frac {1}{2}\right )}{5 b^{7/4} \sqrt {a+b x^2}}-\frac {6 a c^2 \sqrt {c x} \sqrt {a+b x^2}}{5 b^{3/2} \left (\sqrt {a}+\sqrt {b} x\right )}+\frac {2 c (c x)^{3/2} \sqrt {a+b x^2}}{5 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c*x)^(5/2)/Sqrt[a + b*x^2],x]

[Out]

(2*c*(c*x)^(3/2)*Sqrt[a + b*x^2])/(5*b) - (6*a*c^2*Sqrt[c*x]*Sqrt[a + b*x^2])/(5*b^(3/2)*(Sqrt[a] + Sqrt[b]*x)
) + (6*a^(5/4)*c^(5/2)*(Sqrt[a] + Sqrt[b]*x)*Sqrt[(a + b*x^2)/(Sqrt[a] + Sqrt[b]*x)^2]*EllipticE[2*ArcTan[(b^(
1/4)*Sqrt[c*x])/(a^(1/4)*Sqrt[c])], 1/2])/(5*b^(7/4)*Sqrt[a + b*x^2]) - (3*a^(5/4)*c^(5/2)*(Sqrt[a] + Sqrt[b]*
x)*Sqrt[(a + b*x^2)/(Sqrt[a] + Sqrt[b]*x)^2]*EllipticF[2*ArcTan[(b^(1/4)*Sqrt[c*x])/(a^(1/4)*Sqrt[c])], 1/2])/
(5*b^(7/4)*Sqrt[a + b*x^2])

Rule 226

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)]/(2*q*Sqrt[a + b*x^4]))*EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 311

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

Rule 327

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 335

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]

Rule 1210

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, Simp[(-d)*x*(Sqrt[a +
c*x^4]/(a*(1 + q^2*x^2))), x] + Simp[d*(1 + q^2*x^2)*(Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]/(q*Sqrt[a + c*x^4
]))*EllipticE[2*ArcTan[q*x], 1/2], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

Rubi steps

\begin {align*} \int \frac {(c x)^{5/2}}{\sqrt {a+b x^2}} \, dx &=\frac {2 c (c x)^{3/2} \sqrt {a+b x^2}}{5 b}-\frac {\left (3 a c^2\right ) \int \frac {\sqrt {c x}}{\sqrt {a+b x^2}} \, dx}{5 b}\\ &=\frac {2 c (c x)^{3/2} \sqrt {a+b x^2}}{5 b}-\frac {(6 a c) \text {Subst}\left (\int \frac {x^2}{\sqrt {a+\frac {b x^4}{c^2}}} \, dx,x,\sqrt {c x}\right )}{5 b}\\ &=\frac {2 c (c x)^{3/2} \sqrt {a+b x^2}}{5 b}-\frac {\left (6 a^{3/2} c^2\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+\frac {b x^4}{c^2}}} \, dx,x,\sqrt {c x}\right )}{5 b^{3/2}}+\frac {\left (6 a^{3/2} c^2\right ) \text {Subst}\left (\int \frac {1-\frac {\sqrt {b} x^2}{\sqrt {a} c}}{\sqrt {a+\frac {b x^4}{c^2}}} \, dx,x,\sqrt {c x}\right )}{5 b^{3/2}}\\ &=\frac {2 c (c x)^{3/2} \sqrt {a+b x^2}}{5 b}-\frac {6 a c^2 \sqrt {c x} \sqrt {a+b x^2}}{5 b^{3/2} \left (\sqrt {a}+\sqrt {b} x\right )}+\frac {6 a^{5/4} c^{5/2} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a} \sqrt {c}}\right )|\frac {1}{2}\right )}{5 b^{7/4} \sqrt {a+b x^2}}-\frac {3 a^{5/4} c^{5/2} \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 )}{5 b^{7/4} \sqrt {a+b x^2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 10.03, size = 69, normalized size = 0.25 \begin {gather*} \frac {2 c (c x)^{3/2} \left (a+b x^2-a \sqrt {1+\frac {b x^2}{a}} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-\frac {b x^2}{a}\right )\right )}{5 b \sqrt {a+b x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(c*x)^(5/2)/Sqrt[a + b*x^2],x]

[Out]

(2*c*(c*x)^(3/2)*(a + b*x^2 - a*Sqrt[1 + (b*x^2)/a]*Hypergeometric2F1[1/2, 3/4, 7/4, -((b*x^2)/a)]))/(5*b*Sqrt
[a + b*x^2])

________________________________________________________________________________________

Maple [A]
time = 0.05, size = 210, normalized size = 0.77

method	result	size
default	\(-\frac {c^{2} \sqrt {c x}\, \left (6 \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, \EllipticE \left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right ) a^{2}-3 \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, \EllipticF \left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right ) a^{2}-2 b^{2} x^{4}-2 a b \,x^{2}\right )}{5 x \sqrt {b \,x^{2}+a}\, b^{2}}\)	\(210\)
risch	\(\frac {2 x^{2} \sqrt {b \,x^{2}+a}\, c^{3}}{5 b \sqrt {c x}}-\frac {3 a \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, \left (-\frac {2 \sqrt {-a b}\, \EllipticE \left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}+\frac {\sqrt {-a b}\, \EllipticF \left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}\right ) c^{3} \sqrt {c x \left (b \,x^{2}+a \right )}}{5 b^{2} \sqrt {b c \,x^{3}+a c x}\, \sqrt {c x}\, \sqrt {b \,x^{2}+a}}\)	\(217\)
elliptic	\(\frac {\sqrt {c x}\, \sqrt {c x \left (b \,x^{2}+a \right )}\, \left (\frac {2 c^{2} x \sqrt {b c \,x^{3}+a c x}}{5 b}-\frac {3 c^{3} a \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, \left (-\frac {2 \sqrt {-a b}\, \EllipticE \left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}+\frac {\sqrt {-a b}\, \EllipticF \left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}\right )}{5 b^{2} \sqrt {b c \,x^{3}+a c x}}\right )}{c x \sqrt {b \,x^{2}+a}}\)	\(221\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x)^(5/2)/(b*x^2+a)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/5*c^2/x*(c*x)^(1/2)/(b*x^2+a)^(1/2)/b^2*(6*((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)*(-x*b/(-a*b)^(1/2))^(1/2)*EllipticE(((b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2),1/2*2^(1/
2))*a^2-3*((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)*(-x*b/(-a*b
)^(1/2))^(1/2)*EllipticF(((b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2),1/2*2^(1/2))*a^2-2*b^2*x^4-2*a*b*x^2)

________________________________________________________________________________________

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*x)^(5/2)/(b*x^2+a)^(1/2),x, algorithm="maxima")

[Out]

integrate((c*x)^(5/2)/sqrt(b*x^2 + a), x)

________________________________________________________________________________________

Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.23, size = 54, normalized size = 0.20 \begin {gather*} \frac {2 \, {\left (\sqrt {b x^{2} + a} \sqrt {c x} b c^{2} x + 3 \, \sqrt {b c} a c^{2} {\rm weierstrassZeta}\left (-\frac {4 \, a}{b}, 0, {\rm weierstrassPInverse}\left (-\frac {4 \, a}{b}, 0, x\right )\right )\right )}}{5 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x)^(5/2)/(b*x^2+a)^(1/2),x, algorithm="fricas")

[Out]

2/5*(sqrt(b*x^2 + a)*sqrt(c*x)*b*c^2*x + 3*sqrt(b*c)*a*c^2*weierstrassZeta(-4*a/b, 0, weierstrassPInverse(-4*a
/b, 0, x)))/b^2

________________________________________________________________________________________

Sympy [C] Result contains complex when optimal does not.
time = 3.50, size = 44, normalized size = 0.16 \begin {gather*} \frac {c^{\frac {5}{2}} x^{\frac {7}{2}} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \sqrt {a} \Gamma \left (\frac {11}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x)**(5/2)/(b*x**2+a)**(1/2),x)

[Out]

c**(5/2)*x**(7/2)*gamma(7/4)*hyper((1/2, 7/4), (11/4,), b*x**2*exp_polar(I*pi)/a)/(2*sqrt(a)*gamma(11/4))

________________________________________________________________________________________

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*x)^(5/2)/(b*x^2+a)^(1/2),x, algorithm="giac")

[Out]

integrate((c*x)^(5/2)/sqrt(b*x^2 + a), x)

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (c\,x\right )}^{5/2}}{\sqrt {b\,x^2+a}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x)^(5/2)/(a + b*x^2)^(1/2),x)

[Out]

int((c*x)^(5/2)/(a + b*x^2)^(1/2), x)

________________________________________________________________________________________