3.391 \(\int \frac {x^{17/2}}{(b x^2+c x^4)^{3/2}} \, dx\)

Optimal. Leaf size=174 \[ \frac {15 b^{7/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{14 c^{13/4} \sqrt {b x^2+c x^4}}-\frac {15 b \sqrt {b x^2+c x^4}}{7 c^3 \sqrt {x}}+\frac {9 x^{3/2} \sqrt {b x^2+c x^4}}{7 c^2}-\frac {x^{11/2}}{c \sqrt {b x^2+c x^4}} \]

[Out]

-x^(11/2)/c/(c*x^4+b*x^2)^(1/2)+9/7*x^(3/2)*(c*x^4+b*x^2)^(1/2)/c^2-15/7*b*(c*x^4+b*x^2)^(1/2)/c^3/x^(1/2)+15/
14*b^(7/4)*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)/c^(13/4)/(c*x^4+b*x^2)^(1/2)

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Rubi [A]  time = 0.24, antiderivative size = 174, 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 = {2022, 2024, 2032, 329, 220} \[ \frac {15 b^{7/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{14 c^{13/4} \sqrt {b x^2+c x^4}}+\frac {9 x^{3/2} \sqrt {b x^2+c x^4}}{7 c^2}-\frac {15 b \sqrt {b x^2+c x^4}}{7 c^3 \sqrt {x}}-\frac {x^{11/2}}{c \sqrt {b x^2+c x^4}} \]

Antiderivative was successfully verified.

[In]

Int[x^(17/2)/(b*x^2 + c*x^4)^(3/2),x]

[Out]

-(x^(11/2)/(c*Sqrt[b*x^2 + c*x^4])) - (15*b*Sqrt[b*x^2 + c*x^4])/(7*c^3*Sqrt[x]) + (9*x^(3/2)*Sqrt[b*x^2 + c*x
^4])/(7*c^2) + (15*b^(7/4)*x*(Sqrt[b] + Sqrt[c]*x)*Sqrt[(b + c*x^2)/(Sqrt[b] + Sqrt[c]*x)^2]*EllipticF[2*ArcTa
n[(c^(1/4)*Sqrt[x])/b^(1/4)], 1/2])/(14*c^(13/4)*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 2022

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n +
1)*(a*x^j + b*x^n)^(p + 1))/(b*(n - j)*(p + 1)), x] - Dist[(c^n*(m + j*p - n + j + 1))/(b*(n - j)*(p + 1)), I
nt[(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] && (I
ntegersQ[j, n] || GtQ[c, 0]) && LtQ[p, -1] && GtQ[m + j*p + 1, n - j]

Rule 2024

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n +
1)*(a*x^j + b*x^n)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^(n - j)*(m + j*p - n + j + 1))/(b*(m + n*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]) && GtQ[m + j*p + 1 - n + j, 0] && NeQ[m + n*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 {x^{17/2}}{\left (b x^2+c x^4\right )^{3/2}} \, dx &=-\frac {x^{11/2}}{c \sqrt {b x^2+c x^4}}+\frac {9 \int \frac {x^{9/2}}{\sqrt {b x^2+c x^4}} \, dx}{2 c}\\ &=-\frac {x^{11/2}}{c \sqrt {b x^2+c x^4}}+\frac {9 x^{3/2} \sqrt {b x^2+c x^4}}{7 c^2}-\frac {(45 b) \int \frac {x^{5/2}}{\sqrt {b x^2+c x^4}} \, dx}{14 c^2}\\ &=-\frac {x^{11/2}}{c \sqrt {b x^2+c x^4}}-\frac {15 b \sqrt {b x^2+c x^4}}{7 c^3 \sqrt {x}}+\frac {9 x^{3/2} \sqrt {b x^2+c x^4}}{7 c^2}+\frac {\left (15 b^2\right ) \int \frac {\sqrt {x}}{\sqrt {b x^2+c x^4}} \, dx}{14 c^3}\\ &=-\frac {x^{11/2}}{c \sqrt {b x^2+c x^4}}-\frac {15 b \sqrt {b x^2+c x^4}}{7 c^3 \sqrt {x}}+\frac {9 x^{3/2} \sqrt {b x^2+c x^4}}{7 c^2}+\frac {\left (15 b^2 x \sqrt {b+c x^2}\right ) \int \frac {1}{\sqrt {x} \sqrt {b+c x^2}} \, dx}{14 c^3 \sqrt {b x^2+c x^4}}\\ &=-\frac {x^{11/2}}{c \sqrt {b x^2+c x^4}}-\frac {15 b \sqrt {b x^2+c x^4}}{7 c^3 \sqrt {x}}+\frac {9 x^{3/2} \sqrt {b x^2+c x^4}}{7 c^2}+\frac {\left (15 b^2 x \sqrt {b+c x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {b+c x^4}} \, dx,x,\sqrt {x}\right )}{7 c^3 \sqrt {b x^2+c x^4}}\\ &=-\frac {x^{11/2}}{c \sqrt {b x^2+c x^4}}-\frac {15 b \sqrt {b x^2+c x^4}}{7 c^3 \sqrt {x}}+\frac {9 x^{3/2} \sqrt {b x^2+c x^4}}{7 c^2}+\frac {15 b^{7/4} 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 )}{14 c^{13/4} \sqrt {b x^2+c x^4}}\\ \end {align*}

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Mathematica [C]  time = 0.03, size = 86, normalized size = 0.49 \[ \frac {x^{3/2} \left (15 b^2 \sqrt {\frac {c x^2}{b}+1} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};-\frac {c x^2}{b}\right )-15 b^2-6 b c x^2+2 c^2 x^4\right )}{7 c^3 \sqrt {x^2 \left (b+c x^2\right )}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(17/2)/(b*x^2 + c*x^4)^(3/2),x]

[Out]

(x^(3/2)*(-15*b^2 - 6*b*c*x^2 + 2*c^2*x^4 + 15*b^2*Sqrt[1 + (c*x^2)/b]*Hypergeometric2F1[1/4, 1/2, 5/4, -((c*x
^2)/b)]))/(7*c^3*Sqrt[x^2*(b + c*x^2)])

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fricas [F]  time = 0.77, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {c x^{4} + b x^{2}} x^{\frac {9}{2}}}{c^{2} x^{4} + 2 \, b c x^{2} + b^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(17/2)/(c*x^4+b*x^2)^(3/2),x, algorithm="fricas")

[Out]

integral(sqrt(c*x^4 + b*x^2)*x^(9/2)/(c^2*x^4 + 2*b*c*x^2 + b^2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{\frac {17}{2}}}{{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(17/2)/(c*x^4+b*x^2)^(3/2),x, algorithm="giac")

[Out]

integrate(x^(17/2)/(c*x^4 + b*x^2)^(3/2), x)

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maple [A]  time = 0.04, size = 144, normalized size = 0.83 \[ \frac {\left (c \,x^{2}+b \right ) \left (4 c^{3} x^{5}-12 b \,c^{2} x^{3}-30 b^{2} c x +15 \sqrt {-b c}\, \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, b^{2} \EllipticF \left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )\right ) x^{\frac {5}{2}}}{14 \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} c^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(17/2)/(c*x^4+b*x^2)^(3/2),x)

[Out]

1/14/(c*x^4+b*x^2)^(3/2)*x^(5/2)*(c*x^2+b)*(15*b^2*(-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))+4*c^3*x^5-12*b*c^2*x^3-30*b^2*c*x)/c^4

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{\frac {17}{2}}}{{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(17/2)/(c*x^4+b*x^2)^(3/2),x, algorithm="maxima")

[Out]

integrate(x^(17/2)/(c*x^4 + b*x^2)^(3/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^{17/2}}{{\left (c\,x^4+b\,x^2\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(17/2)/(b*x^2 + c*x^4)^(3/2),x)

[Out]

int(x^(17/2)/(b*x^2 + c*x^4)^(3/2), 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(x**(17/2)/(c*x**4+b*x**2)**(3/2),x)

[Out]

Timed out

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