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

Optimal. Leaf size=251 \[ -\frac {15 b^{7/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} (13 b B-11 A c) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{154 c^{17/4} \sqrt {b x^2+c x^4}}+\frac {15 b \sqrt {b x^2+c x^4} (13 b B-11 A c)}{77 c^4 \sqrt {x}}-\frac {9 x^{3/2} \sqrt {b x^2+c x^4} (13 b B-11 A c)}{77 c^3}+\frac {x^{7/2} \sqrt {b x^2+c x^4} (13 b B-11 A c)}{11 b c^2}-\frac {x^{15/2} (b B-A c)}{b c \sqrt {b x^2+c x^4}} \]

[Out]

-(-A*c+B*b)*x^(15/2)/b/c/(c*x^4+b*x^2)^(1/2)-9/77*(-11*A*c+13*B*b)*x^(3/2)*(c*x^4+b*x^2)^(1/2)/c^3+1/11*(-11*A
*c+13*B*b)*x^(7/2)*(c*x^4+b*x^2)^(1/2)/b/c^2+15/77*b*(-11*A*c+13*B*b)*(c*x^4+b*x^2)^(1/2)/c^4/x^(1/2)-15/154*b
^(7/4)*(-11*A*c+13*B*b)*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^(17/4)/(c*x^4+b*x^2)^(1/2)

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Rubi [A]  time = 0.39, antiderivative size = 251, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {2037, 2024, 2032, 329, 220} \[ -\frac {15 b^{7/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} (13 b B-11 A c) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{154 c^{17/4} \sqrt {b x^2+c x^4}}+\frac {x^{7/2} \sqrt {b x^2+c x^4} (13 b B-11 A c)}{11 b c^2}-\frac {9 x^{3/2} \sqrt {b x^2+c x^4} (13 b B-11 A c)}{77 c^3}+\frac {15 b \sqrt {b x^2+c x^4} (13 b B-11 A c)}{77 c^4 \sqrt {x}}-\frac {x^{15/2} (b B-A c)}{b c \sqrt {b x^2+c x^4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((b*B - A*c)*x^(15/2))/(b*c*Sqrt[b*x^2 + c*x^4])) + (15*b*(13*b*B - 11*A*c)*Sqrt[b*x^2 + c*x^4])/(77*c^4*Sqr
t[x]) - (9*(13*b*B - 11*A*c)*x^(3/2)*Sqrt[b*x^2 + c*x^4])/(77*c^3) + ((13*b*B - 11*A*c)*x^(7/2)*Sqrt[b*x^2 + c
*x^4])/(11*b*c^2) - (15*b^(7/4)*(13*b*B - 11*A*c)*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])/(154*c^(17/4)*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 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]

Rule 2037

Int[((e_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(jn_.))^(p_)*((c_) + (d_.)*(x_)^(n_.)), x_Symbol] :> -Si
mp[(e^(j - 1)*(b*c - a*d)*(e*x)^(m - j + 1)*(a*x^j + b*x^(j + n))^(p + 1))/(a*b*n*(p + 1)), x] - Dist[(e^j*(a*
d*(m + j*p + 1) - b*c*(m + n + p*(j + n) + 1)))/(a*b*n*(p + 1)), Int[(e*x)^(m - j)*(a*x^j + b*x^(j + n))^(p +
1), x], x] /; FreeQ[{a, b, c, d, e, j, m, n}, x] && EqQ[jn, j + n] &&  !IntegerQ[p] && NeQ[b*c - a*d, 0] && Lt
Q[p, -1] && GtQ[j, 0] && LeQ[j, m] && (GtQ[e, 0] || IntegerQ[j])

Rubi steps

\begin {align*} \int \frac {x^{17/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^{3/2}} \, dx &=-\frac {(b B-A c) x^{15/2}}{b c \sqrt {b x^2+c x^4}}+\frac {\left (\frac {13 b B}{2}-\frac {11 A c}{2}\right ) \int \frac {x^{13/2}}{\sqrt {b x^2+c x^4}} \, dx}{b c}\\ &=-\frac {(b B-A c) x^{15/2}}{b c \sqrt {b x^2+c x^4}}+\frac {(13 b B-11 A c) x^{7/2} \sqrt {b x^2+c x^4}}{11 b c^2}-\frac {(9 (13 b B-11 A c)) \int \frac {x^{9/2}}{\sqrt {b x^2+c x^4}} \, dx}{22 c^2}\\ &=-\frac {(b B-A c) x^{15/2}}{b c \sqrt {b x^2+c x^4}}-\frac {9 (13 b B-11 A c) x^{3/2} \sqrt {b x^2+c x^4}}{77 c^3}+\frac {(13 b B-11 A c) x^{7/2} \sqrt {b x^2+c x^4}}{11 b c^2}+\frac {(45 b (13 b B-11 A c)) \int \frac {x^{5/2}}{\sqrt {b x^2+c x^4}} \, dx}{154 c^3}\\ &=-\frac {(b B-A c) x^{15/2}}{b c \sqrt {b x^2+c x^4}}+\frac {15 b (13 b B-11 A c) \sqrt {b x^2+c x^4}}{77 c^4 \sqrt {x}}-\frac {9 (13 b B-11 A c) x^{3/2} \sqrt {b x^2+c x^4}}{77 c^3}+\frac {(13 b B-11 A c) x^{7/2} \sqrt {b x^2+c x^4}}{11 b c^2}-\frac {\left (15 b^2 (13 b B-11 A c)\right ) \int \frac {\sqrt {x}}{\sqrt {b x^2+c x^4}} \, dx}{154 c^4}\\ &=-\frac {(b B-A c) x^{15/2}}{b c \sqrt {b x^2+c x^4}}+\frac {15 b (13 b B-11 A c) \sqrt {b x^2+c x^4}}{77 c^4 \sqrt {x}}-\frac {9 (13 b B-11 A c) x^{3/2} \sqrt {b x^2+c x^4}}{77 c^3}+\frac {(13 b B-11 A c) x^{7/2} \sqrt {b x^2+c x^4}}{11 b c^2}-\frac {\left (15 b^2 (13 b B-11 A c) x \sqrt {b+c x^2}\right ) \int \frac {1}{\sqrt {x} \sqrt {b+c x^2}} \, dx}{154 c^4 \sqrt {b x^2+c x^4}}\\ &=-\frac {(b B-A c) x^{15/2}}{b c \sqrt {b x^2+c x^4}}+\frac {15 b (13 b B-11 A c) \sqrt {b x^2+c x^4}}{77 c^4 \sqrt {x}}-\frac {9 (13 b B-11 A c) x^{3/2} \sqrt {b x^2+c x^4}}{77 c^3}+\frac {(13 b B-11 A c) x^{7/2} \sqrt {b x^2+c x^4}}{11 b c^2}-\frac {\left (15 b^2 (13 b B-11 A c) x \sqrt {b+c x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {b+c x^4}} \, dx,x,\sqrt {x}\right )}{77 c^4 \sqrt {b x^2+c x^4}}\\ &=-\frac {(b B-A c) x^{15/2}}{b c \sqrt {b x^2+c x^4}}+\frac {15 b (13 b B-11 A c) \sqrt {b x^2+c x^4}}{77 c^4 \sqrt {x}}-\frac {9 (13 b B-11 A c) x^{3/2} \sqrt {b x^2+c x^4}}{77 c^3}+\frac {(13 b B-11 A c) x^{7/2} \sqrt {b x^2+c x^4}}{11 b c^2}-\frac {15 b^{7/4} (13 b B-11 A c) 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 )}{154 c^{17/4} \sqrt {b x^2+c x^4}}\\ \end {align*}

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Mathematica [C]  time = 0.24, size = 134, normalized size = 0.53 \[ \frac {x^{3/2} \left (15 b^2 \sqrt {\frac {c x^2}{b}+1} (11 A c-13 b B) \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};-\frac {c x^2}{b}\right )+b^2 \left (78 B c x^2-165 A c\right )-2 b c^2 x^2 \left (33 A+13 B x^2\right )+2 c^3 x^4 \left (11 A+7 B x^2\right )+195 b^3 B\right )}{77 c^4 \sqrt {x^2 \left (b+c x^2\right )}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^(3/2)*(195*b^3*B + 2*c^3*x^4*(11*A + 7*B*x^2) - 2*b*c^2*x^2*(33*A + 13*B*x^2) + b^2*(-165*A*c + 78*B*c*x^2)
+ 15*b^2*(-13*b*B + 11*A*c)*Sqrt[1 + (c*x^2)/b]*Hypergeometric2F1[1/4, 1/2, 5/4, -((c*x^2)/b)]))/(77*c^4*Sqrt
[x^2*(b + c*x^2)])

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fricas [F]  time = 0.89, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (B x^{6} + A x^{4}\right )} \sqrt {c x^{4} + b x^{2}} \sqrt {x}}{c^{2} x^{4} + 2 \, b c x^{2} + b^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((B*x^6 + A*x^4)*sqrt(c*x^4 + b*x^2)*sqrt(x)/(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 \[ \int \frac {{\left (B x^{2} + A\right )} x^{\frac {17}{2}}}{{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.15, size = 281, normalized size = 1.12 \[ \frac {\left (c \,x^{2}+b \right ) \left (28 B \,c^{4} x^{7}+44 A \,c^{4} x^{5}-52 B b \,c^{3} x^{5}-132 A b \,c^{3} x^{3}+156 B \,b^{2} c^{2} x^{3}-330 A \,b^{2} c^{2} x +390 B \,b^{3} c x +165 \sqrt {-b c}\, \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, A \,b^{2} c \EllipticF \left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )-195 \sqrt {-b c}\, \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, B \,b^{3} \EllipticF \left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )\right ) x^{\frac {5}{2}}}{154 \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} c^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/154/(c*x^4+b*x^2)^(3/2)*x^(5/2)*(c*x^2+b)*(28*B*c^4*x^7+165*A*(-b*c)^(1/2)*EllipticF(((c*x+(-b*c)^(1/2))/(-b
*c)^(1/2))^(1/2),1/2*2^(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)*b^2*c+44*A*c^4*x^5-195*B*(-b*c)^(1/2)*EllipticF(((c*x+(-b*c)^(1/2))/(-b*c
)^(1/2))^(1/2),1/2*2^(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)*b^3-52*B*b*c^3*x^5-132*A*b*c^3*x^3+156*B*b^2*c^2*x^3-330*A*b^2*c^2*x+390*B*
x*b^3*c)/c^5

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((B*x^2 + A)*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 = -0.00 \[ \int \frac {x^{17/2}\,\left (B\,x^2+A\right )}{{\left (c\,x^4+b\,x^2\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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