3.632 \(\int \frac {\sqrt {c x}}{(a+b x^2)^{5/2}} \, dx\)

Optimal. Leaf size=302 \[ -\frac {\sqrt {c} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} \operatorname {EllipticF}\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a} \sqrt {c}}\right ),\frac {1}{2}\right )}{4 a^{7/4} b^{3/4} \sqrt {a+b x^2}}+\frac {\sqrt {c} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a} \sqrt {c}}\right )|\frac {1}{2}\right )}{2 a^{7/4} b^{3/4} \sqrt {a+b x^2}}+\frac {(c x)^{3/2}}{2 a^2 c \sqrt {a+b x^2}}-\frac {\sqrt {c x} \sqrt {a+b x^2}}{2 a^2 \sqrt {b} \left (\sqrt {a}+\sqrt {b} x\right )}+\frac {(c x)^{3/2}}{3 a c \left (a+b x^2\right )^{3/2}} \]

[Out]

1/3*(c*x)^(3/2)/a/c/(b*x^2+a)^(3/2)+1/2*(c*x)^(3/2)/a^2/c/(b*x^2+a)^(1/2)-1/2*(c*x)^(1/2)*(b*x^2+a)^(1/2)/a^2/
b^(1/2)/(a^(1/2)+x*b^(1/2))+1/2*(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))*c^(1/2)*((b*x^2+a)/(a^(1/2)+x*b^(1/2))^2)^(1/2)/a^(7/4)/b^(3/4)/(b*x^2+a)^(1/2)-1/4*(cos(2*a
rctan(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)))*Ellipt
icF(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))*c^(1/2)*((b*x^2+a)/(a^
(1/2)+x*b^(1/2))^2)^(1/2)/a^(7/4)/b^(3/4)/(b*x^2+a)^(1/2)

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Rubi [A]  time = 0.22, antiderivative size = 302, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {290, 329, 305, 220, 1196} \[ -\frac {\sqrt {c} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a} \sqrt {c}}\right )|\frac {1}{2}\right )}{4 a^{7/4} b^{3/4} \sqrt {a+b x^2}}+\frac {\sqrt {c} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a} \sqrt {c}}\right )|\frac {1}{2}\right )}{2 a^{7/4} b^{3/4} \sqrt {a+b x^2}}+\frac {(c x)^{3/2}}{2 a^2 c \sqrt {a+b x^2}}-\frac {\sqrt {c x} \sqrt {a+b x^2}}{2 a^2 \sqrt {b} \left (\sqrt {a}+\sqrt {b} x\right )}+\frac {(c x)^{3/2}}{3 a c \left (a+b x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[c*x]/(a + b*x^2)^(5/2),x]

[Out]

(c*x)^(3/2)/(3*a*c*(a + b*x^2)^(3/2)) + (c*x)^(3/2)/(2*a^2*c*Sqrt[a + b*x^2]) - (Sqrt[c*x]*Sqrt[a + b*x^2])/(2
*a^2*Sqrt[b]*(Sqrt[a] + Sqrt[b]*x)) + (Sqrt[c]*(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])/(2*a^(7/4)*b^(3/4)*Sqrt[a + b*x^2]) - (Sqrt[
c]*(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])/(4*a^(7/4)*b^(3/4)*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 290

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

Rule 305

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 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]

Rule 1196

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)]*EllipticE[2*ArcTan[
q*x], 1/2])/(q*Sqrt[a + c*x^4]), x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

Rubi steps

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

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Mathematica [C]  time = 0.01, size = 59, normalized size = 0.20 \[ \frac {2 x \sqrt {c x} \sqrt {\frac {b x^2}{a}+1} \, _2F_1\left (\frac {3}{4},\frac {5}{2};\frac {7}{4};-\frac {b x^2}{a}\right )}{3 a^2 \sqrt {a+b x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.79, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {b x^{2} + a} \sqrt {c x}}{b^{3} x^{6} + 3 \, a b^{2} x^{4} + 3 \, a^{2} b x^{2} + a^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(b*x^2 + a)*sqrt(c*x)/(b^3*x^6 + 3*a*b^2*x^4 + 3*a^2*b*x^2 + a^3), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {c x}}{{\left (b x^{2} + a\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.02, size = 382, normalized size = 1.26 \[ -\frac {\left (-6 b^{2} x^{4}+6 \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, a b \,x^{2} \EllipticE \left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )-3 \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, a b \,x^{2} \EllipticF \left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )-10 a b \,x^{2}+6 \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, a^{2} \EllipticE \left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )-3 \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, a^{2} \EllipticF \left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )\right ) \sqrt {c x}}{12 \left (b \,x^{2}+a \right )^{\frac {3}{2}} a^{2} b x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(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)*(-1/(-a*b)^(
1/2)*b*x)^(1/2)*EllipticE(((b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2),1/2*2^(1/2))*x^2*a*b-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)*(-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))*x^2*a*b+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)*(-1/(-a*b)^(1/2)*b*x)^(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)*(-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^2-6*b^2*x^4-1
0*a*b*x^2)*(c*x)^(1/2)/b/a^2/x/(b*x^2+a)^(3/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {c x}}{{\left (b x^{2} + a\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x)^(1/2)/(a + b*x^2)^(5/2),x)

[Out]

int((c*x)^(1/2)/(a + b*x^2)^(5/2), x)

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sympy [C]  time = 2.82, size = 44, normalized size = 0.15 \[ \frac {\sqrt {c} x^{\frac {3}{2}} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {5}{2} \\ \frac {7}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {5}{2}} \Gamma \left (\frac {7}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(c)*x**(3/2)*gamma(3/4)*hyper((3/4, 5/2), (7/4,), b*x**2*exp_polar(I*pi)/a)/(2*a**(5/2)*gamma(7/4))

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