3.228 \(\int \frac {(A+B x^2) \sqrt {b x^2+c x^4}}{x^{11/2}} \, dx\)

Optimal. Leaf size=167 \[ \frac {2 c^{3/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} (7 b B-A c) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{21 b^{5/4} \sqrt {b x^2+c x^4}}-\frac {2 \sqrt {b x^2+c x^4} (7 b B-A c)}{21 b x^{5/2}}-\frac {2 A \left (b x^2+c x^4\right )^{3/2}}{7 b x^{13/2}} \]

[Out]

-2/7*A*(c*x^4+b*x^2)^(3/2)/b/x^(13/2)-2/21*(-A*c+7*B*b)*(c*x^4+b*x^2)^(1/2)/b/x^(5/2)+2/21*c^(3/4)*(-A*c+7*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*ar
ctan(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^(5/4
)/(c*x^4+b*x^2)^(1/2)

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Rubi [A]  time = 0.25, antiderivative size = 167, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {2038, 2020, 2032, 329, 220} \[ \frac {2 c^{3/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} (7 b B-A c) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{21 b^{5/4} \sqrt {b x^2+c x^4}}-\frac {2 \sqrt {b x^2+c x^4} (7 b B-A c)}{21 b x^{5/2}}-\frac {2 A \left (b x^2+c x^4\right )^{3/2}}{7 b x^{13/2}} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x^2)*Sqrt[b*x^2 + c*x^4])/x^(11/2),x]

[Out]

(-2*(7*b*B - A*c)*Sqrt[b*x^2 + c*x^4])/(21*b*x^(5/2)) - (2*A*(b*x^2 + c*x^4)^(3/2))/(7*b*x^(13/2)) + (2*c^(3/4
)*(7*b*B - 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])/(21*b^(5/4)*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]

Rule 2038

Int[((e_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(jn_.))^(p_)*((c_) + (d_.)*(x_)^(n_.)), x_Symbol] :> Sim
p[(c*e^(j - 1)*(e*x)^(m - j + 1)*(a*x^j + b*x^(j + n))^(p + 1))/(a*(m + j*p + 1)), x] + Dist[(a*d*(m + j*p + 1
) - b*c*(m + n + p*(j + n) + 1))/(a*e^n*(m + j*p + 1)), Int[(e*x)^(m + n)*(a*x^j + b*x^(j + n))^p, x], x] /; F
reeQ[{a, b, c, d, e, j, p}, x] && EqQ[jn, j + n] &&  !IntegerQ[p] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && (LtQ[m
+ j*p, -1] || (IntegersQ[m - 1/2, p - 1/2] && LtQ[p, 0] && LtQ[m, -(n*p) - 1])) && (GtQ[e, 0] || IntegersQ[j,
n]) && NeQ[m + j*p + 1, 0] && NeQ[m - n + j*p + 1, 0]

Rubi steps

\begin {align*} \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x^{11/2}} \, dx &=-\frac {2 A \left (b x^2+c x^4\right )^{3/2}}{7 b x^{13/2}}-\frac {\left (2 \left (-\frac {7 b B}{2}+\frac {A c}{2}\right )\right ) \int \frac {\sqrt {b x^2+c x^4}}{x^{7/2}} \, dx}{7 b}\\ &=-\frac {2 (7 b B-A c) \sqrt {b x^2+c x^4}}{21 b x^{5/2}}-\frac {2 A \left (b x^2+c x^4\right )^{3/2}}{7 b x^{13/2}}+\frac {(2 c (7 b B-A c)) \int \frac {\sqrt {x}}{\sqrt {b x^2+c x^4}} \, dx}{21 b}\\ &=-\frac {2 (7 b B-A c) \sqrt {b x^2+c x^4}}{21 b x^{5/2}}-\frac {2 A \left (b x^2+c x^4\right )^{3/2}}{7 b x^{13/2}}+\frac {\left (2 c (7 b B-A c) x \sqrt {b+c x^2}\right ) \int \frac {1}{\sqrt {x} \sqrt {b+c x^2}} \, dx}{21 b \sqrt {b x^2+c x^4}}\\ &=-\frac {2 (7 b B-A c) \sqrt {b x^2+c x^4}}{21 b x^{5/2}}-\frac {2 A \left (b x^2+c x^4\right )^{3/2}}{7 b x^{13/2}}+\frac {\left (4 c (7 b B-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 )}{21 b \sqrt {b x^2+c x^4}}\\ &=-\frac {2 (7 b B-A c) \sqrt {b x^2+c x^4}}{21 b x^{5/2}}-\frac {2 A \left (b x^2+c x^4\right )^{3/2}}{7 b x^{13/2}}+\frac {2 c^{3/4} (7 b B-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 )}{21 b^{5/4} \sqrt {b x^2+c x^4}}\\ \end {align*}

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Mathematica [C]  time = 0.06, size = 98, normalized size = 0.59 \[ -\frac {2 \sqrt {x^2 \left (b+c x^2\right )} \left (x^2 (7 b B-A c) \, _2F_1\left (-\frac {3}{4},-\frac {1}{2};\frac {1}{4};-\frac {c x^2}{b}\right )+3 A \left (b+c x^2\right ) \sqrt {\frac {c x^2}{b}+1}\right )}{21 b x^{9/2} \sqrt {\frac {c x^2}{b}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x^2)*Sqrt[b*x^2 + c*x^4])/x^(11/2),x]

[Out]

(-2*Sqrt[x^2*(b + c*x^2)]*(3*A*(b + c*x^2)*Sqrt[1 + (c*x^2)/b] + (7*b*B - A*c)*x^2*Hypergeometric2F1[-3/4, -1/
2, 1/4, -((c*x^2)/b)]))/(21*b*x^(9/2)*Sqrt[1 + (c*x^2)/b])

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fricas [F]  time = 0.80, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {c x^{4} + b x^{2}} {\left (B x^{2} + A\right )}}{x^{\frac {11}{2}}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(c*x^4+b*x^2)^(1/2)/x^(11/2),x, algorithm="fricas")

[Out]

integral(sqrt(c*x^4 + b*x^2)*(B*x^2 + A)/x^(11/2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {c x^{4} + b x^{2}} {\left (B x^{2} + A\right )}}{x^{\frac {11}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(c*x^4+b*x^2)^(1/2)/x^(11/2),x, algorithm="giac")

[Out]

integrate(sqrt(c*x^4 + b*x^2)*(B*x^2 + A)/x^(11/2), x)

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maple [A]  time = 0.09, size = 255, normalized size = 1.53 \[ -\frac {2 \sqrt {c \,x^{4}+b \,x^{2}}\, \left (2 A \,c^{2} x^{4}+7 B b c \,x^{4}+\sqrt {-b c}\, \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, A c \,x^{3} \EllipticF \left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )-7 \sqrt {-b c}\, \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, B b \,x^{3} \EllipticF \left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )+5 A b c \,x^{2}+7 B \,b^{2} x^{2}+3 A \,b^{2}\right )}{21 \left (c \,x^{2}+b \right ) b \,x^{\frac {9}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^2+A)*(c*x^4+b*x^2)^(1/2)/x^(11/2),x)

[Out]

-2/21*(c*x^4+b*x^2)^(1/2)/x^(9/2)/(c*x^2+b)*(A*(-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^3*c-7*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^3*b+2*A*c^2*x^4+7*B*b*c*x^4+5*A*b*c*x^2+7*B*b^2*x^2+3*A*b^2)/b

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {c x^{4} + b x^{2}} {\left (B x^{2} + A\right )}}{x^{\frac {11}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(c*x^4+b*x^2)^(1/2)/x^(11/2),x, algorithm="maxima")

[Out]

integrate(sqrt(c*x^4 + b*x^2)*(B*x^2 + A)/x^(11/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\left (B\,x^2+A\right )\,\sqrt {c\,x^4+b\,x^2}}{x^{11/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x^2)*(b*x^2 + c*x^4)^(1/2))/x^(11/2),x)

[Out]

int(((A + B*x^2)*(b*x^2 + c*x^4)^(1/2))/x^(11/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {x^{2} \left (b + c x^{2}\right )} \left (A + B x^{2}\right )}{x^{\frac {11}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x**2+A)*(c*x**4+b*x**2)**(1/2)/x**(11/2),x)

[Out]

Integral(sqrt(x**2*(b + c*x**2))*(A + B*x**2)/x**(11/2), x)

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