3.363 \(\int \frac {\sqrt {b x^2+c x^4}}{x^{15/2}} \, dx\)

Optimal. Leaf size=176 \[ \frac {10 c^{11/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{231 b^{9/4} \sqrt {b x^2+c x^4}}+\frac {20 c^2 \sqrt {b x^2+c x^4}}{231 b^2 x^{5/2}}-\frac {4 c \sqrt {b x^2+c x^4}}{77 b x^{9/2}}-\frac {2 \sqrt {b x^2+c x^4}}{11 x^{13/2}} \]

[Out]

-2/11*(c*x^4+b*x^2)^(1/2)/x^(13/2)-4/77*c*(c*x^4+b*x^2)^(1/2)/b/x^(9/2)+20/231*c^2*(c*x^4+b*x^2)^(1/2)/b^2/x^(
5/2)+10/231*c^(11/4)*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^(9/4)/(c*x^4+b*x^2)^(1/2)

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Rubi [A]  time = 0.23, antiderivative size = 176, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {2020, 2025, 2032, 329, 220} \[ \frac {20 c^2 \sqrt {b x^2+c x^4}}{231 b^2 x^{5/2}}+\frac {10 c^{11/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{231 b^{9/4} \sqrt {b x^2+c x^4}}-\frac {4 c \sqrt {b x^2+c x^4}}{77 b x^{9/2}}-\frac {2 \sqrt {b x^2+c x^4}}{11 x^{13/2}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[b*x^2 + c*x^4]/x^(15/2),x]

[Out]

(-2*Sqrt[b*x^2 + c*x^4])/(11*x^(13/2)) - (4*c*Sqrt[b*x^2 + c*x^4])/(77*b*x^(9/2)) + (20*c^2*Sqrt[b*x^2 + c*x^4
])/(231*b^2*x^(5/2)) + (10*c^(11/4)*x*(Sqrt[b] + Sqrt[c]*x)*Sqrt[(b + c*x^2)/(Sqrt[b] + Sqrt[c]*x)^2]*Elliptic
F[2*ArcTan[(c^(1/4)*Sqrt[x])/b^(1/4)], 1/2])/(231*b^(9/4)*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 2025

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(c^(j - 1)*(c*x)^(m - j +
1)*(a*x^j + b*x^n)^(p + 1))/(a*(m + j*p + 1)), x] - Dist[(b*(m + n*p + n - j + 1))/(a*c^(n - j)*(m + j*p + 1)
), Int[(c*x)^(m + n - j)*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, p}, x] &&  !IntegerQ[p] && LtQ[0, j,
n] && (IntegersQ[j, n] || GtQ[c, 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^{15/2}} \, dx &=-\frac {2 \sqrt {b x^2+c x^4}}{11 x^{13/2}}+\frac {1}{11} (2 c) \int \frac {1}{x^{7/2} \sqrt {b x^2+c x^4}} \, dx\\ &=-\frac {2 \sqrt {b x^2+c x^4}}{11 x^{13/2}}-\frac {4 c \sqrt {b x^2+c x^4}}{77 b x^{9/2}}-\frac {\left (10 c^2\right ) \int \frac {1}{x^{3/2} \sqrt {b x^2+c x^4}} \, dx}{77 b}\\ &=-\frac {2 \sqrt {b x^2+c x^4}}{11 x^{13/2}}-\frac {4 c \sqrt {b x^2+c x^4}}{77 b x^{9/2}}+\frac {20 c^2 \sqrt {b x^2+c x^4}}{231 b^2 x^{5/2}}+\frac {\left (10 c^3\right ) \int \frac {\sqrt {x}}{\sqrt {b x^2+c x^4}} \, dx}{231 b^2}\\ &=-\frac {2 \sqrt {b x^2+c x^4}}{11 x^{13/2}}-\frac {4 c \sqrt {b x^2+c x^4}}{77 b x^{9/2}}+\frac {20 c^2 \sqrt {b x^2+c x^4}}{231 b^2 x^{5/2}}+\frac {\left (10 c^3 x \sqrt {b+c x^2}\right ) \int \frac {1}{\sqrt {x} \sqrt {b+c x^2}} \, dx}{231 b^2 \sqrt {b x^2+c x^4}}\\ &=-\frac {2 \sqrt {b x^2+c x^4}}{11 x^{13/2}}-\frac {4 c \sqrt {b x^2+c x^4}}{77 b x^{9/2}}+\frac {20 c^2 \sqrt {b x^2+c x^4}}{231 b^2 x^{5/2}}+\frac {\left (20 c^3 x \sqrt {b+c x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {b+c x^4}} \, dx,x,\sqrt {x}\right )}{231 b^2 \sqrt {b x^2+c x^4}}\\ &=-\frac {2 \sqrt {b x^2+c x^4}}{11 x^{13/2}}-\frac {4 c \sqrt {b x^2+c x^4}}{77 b x^{9/2}}+\frac {20 c^2 \sqrt {b x^2+c x^4}}{231 b^2 x^{5/2}}+\frac {10 c^{11/4} 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 )}{231 b^{9/4} \sqrt {b x^2+c x^4}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 57, normalized size = 0.32 \[ -\frac {2 \sqrt {x^2 \left (b+c x^2\right )} \, _2F_1\left (-\frac {11}{4},-\frac {1}{2};-\frac {7}{4};-\frac {c x^2}{b}\right )}{11 x^{13/2} \sqrt {\frac {c x^2}{b}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[b*x^2 + c*x^4]/x^(15/2),x]

[Out]

(-2*Sqrt[x^2*(b + c*x^2)]*Hypergeometric2F1[-11/4, -1/2, -7/4, -((c*x^2)/b)])/(11*x^(13/2)*Sqrt[1 + (c*x^2)/b]
)

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fricas [F]  time = 0.90, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {c x^{4} + b x^{2}}}{x^{\frac {15}{2}}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2)^(1/2)/x^(15/2),x, algorithm="fricas")

[Out]

integral(sqrt(c*x^4 + b*x^2)/x^(15/2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {c x^{4} + b x^{2}}}{x^{\frac {15}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2)^(1/2)/x^(15/2),x, algorithm="giac")

[Out]

integrate(sqrt(c*x^4 + b*x^2)/x^(15/2), x)

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maple [A]  time = 0.04, size = 156, normalized size = 0.89 \[ \frac {2 \sqrt {c \,x^{4}+b \,x^{2}}\, \left (10 c^{3} x^{6}+5 \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \sqrt {-b c}\, c^{2} x^{5} \EllipticF \left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )+4 b \,c^{2} x^{4}-27 b^{2} c \,x^{2}-21 b^{3}\right )}{231 \left (c \,x^{2}+b \right ) b^{2} x^{\frac {13}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^4+b*x^2)^(1/2)/x^(15/2),x)

[Out]

2/231*(c*x^4+b*x^2)^(1/2)/x^(13/2)/(c*x^2+b)*(5*((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))*(-b*c)^(1/2)*x^5*c^2+10*c^3*x^6+4*b*c^2*x^4-27*b^2*c*x^2-21*b^3)/b^2

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {c x^{4} + b x^{2}}}{x^{\frac {15}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2)^(1/2)/x^(15/2),x, algorithm="maxima")

[Out]

integrate(sqrt(c*x^4 + b*x^2)/x^(15/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sqrt {c\,x^4+b\,x^2}}{x^{15/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2 + c*x^4)^(1/2)/x^(15/2),x)

[Out]

int((b*x^2 + c*x^4)^(1/2)/x^(15/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**4+b*x**2)**(1/2)/x**(15/2),x)

[Out]

Timed out

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