3.359 \(\int \frac {\sqrt {b x^2+c x^4}}{x^{7/2}} \, dx\)

Optimal. Leaf size=118 \[ \frac {2 c^{3/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{3 \sqrt [4]{b} \sqrt {b x^2+c x^4}}-\frac {2 \sqrt {b x^2+c x^4}}{3 x^{5/2}} \]

[Out]

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

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Rubi [A]  time = 0.13, antiderivative size = 118, 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 = {2020, 2032, 329, 220} \[ \frac {2 c^{3/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{3 \sqrt [4]{b} \sqrt {b x^2+c x^4}}-\frac {2 \sqrt {b x^2+c x^4}}{3 x^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[b*x^2 + c*x^4]/x^(7/2),x]

[Out]

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

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 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 2020

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

Rule 2032

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[(c^IntPart[m]*(c*x)^FracP
art[m]*(a*x^j + b*x^n)^FracPart[p])/(x^(FracPart[m] + j*FracPart[p])*(a + b*x^(n - j))^FracPart[p]), Int[x^(m
+ j*p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[n, j] && PosQ[n
- j]

Rubi steps

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

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Mathematica [C]  time = 0.01, size = 57, normalized size = 0.48 \[ -\frac {2 \sqrt {x^2 \left (b+c x^2\right )} \, _2F_1\left (-\frac {3}{4},-\frac {1}{2};\frac {1}{4};-\frac {c x^2}{b}\right )}{3 x^{5/2} \sqrt {\frac {c x^2}{b}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[b*x^2 + c*x^4]/x^(7/2),x]

[Out]

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

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fricas [F]  time = 0.76, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {c x^{4} + b x^{2}}}{x^{\frac {7}{2}}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2)^(1/2)/x^(7/2),x, algorithm="fricas")

[Out]

integral(sqrt(c*x^4 + b*x^2)/x^(7/2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {c x^{4} + b x^{2}}}{x^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2)^(1/2)/x^(7/2),x, algorithm="giac")

[Out]

integrate(sqrt(c*x^4 + b*x^2)/x^(7/2), x)

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maple [A]  time = 0.03, size = 125, normalized size = 1.06 \[ \frac {2 \sqrt {c \,x^{4}+b \,x^{2}}\, \left (-c \,x^{2}+\sqrt {-b c}\, \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, x \EllipticF \left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )-b \right )}{3 \left (c \,x^{2}+b \right ) x^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^4+b*x^2)^(1/2)/x^(7/2),x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {c x^{4} + b x^{2}}}{x^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2)^(1/2)/x^(7/2),x, algorithm="maxima")

[Out]

integrate(sqrt(c*x^4 + b*x^2)/x^(7/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sqrt {c\,x^4+b\,x^2}}{x^{7/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2 + c*x^4)^(1/2)/x^(7/2),x)

[Out]

int((b*x^2 + c*x^4)^(1/2)/x^(7/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {x^{2} \left (b + c x^{2}\right )}}{x^{\frac {7}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**4+b*x**2)**(1/2)/x**(7/2),x)

[Out]

Integral(sqrt(x**2*(b + c*x**2))/x**(7/2), x)

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