3.3.56 \(\int \frac {A+B x^2}{x^{11/2} \sqrt {b x^2+c x^4}} \, dx\) [256]

Optimal. Leaf size=204 \[ -\frac {2 A \sqrt {b x^2+c x^4}}{11 b x^{13/2}}-\frac {2 (11 b B-9 A c) \sqrt {b x^2+c x^4}}{77 b^2 x^{9/2}}+\frac {10 c (11 b B-9 A c) \sqrt {b x^2+c x^4}}{231 b^3 x^{5/2}}+\frac {5 c^{7/4} (11 b B-9 A c) x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{231 b^{13/4} \sqrt {b x^2+c x^4}} \]

[Out]

-2/11*A*(c*x^4+b*x^2)^(1/2)/b/x^(13/2)-2/77*(-9*A*c+11*B*b)*(c*x^4+b*x^2)^(1/2)/b^2/x^(9/2)+10/231*c*(-9*A*c+1
1*B*b)*(c*x^4+b*x^2)^(1/2)/b^3/x^(5/2)+5/231*c^(7/4)*(-9*A*c+11*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)/b^(13/4)/(c*x^4+b*x^2)^(1/2)

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Rubi [A]
time = 0.22, antiderivative size = 204, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {2063, 2050, 2057, 335, 226} \begin {gather*} \frac {5 c^{7/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} (11 b B-9 A c) F\left (2 \text {ArcTan}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{231 b^{13/4} \sqrt {b x^2+c x^4}}+\frac {10 c \sqrt {b x^2+c x^4} (11 b B-9 A c)}{231 b^3 x^{5/2}}-\frac {2 \sqrt {b x^2+c x^4} (11 b B-9 A c)}{77 b^2 x^{9/2}}-\frac {2 A \sqrt {b x^2+c x^4}}{11 b x^{13/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*A*Sqrt[b*x^2 + c*x^4])/(11*b*x^(13/2)) - (2*(11*b*B - 9*A*c)*Sqrt[b*x^2 + c*x^4])/(77*b^2*x^(9/2)) + (10*c
*(11*b*B - 9*A*c)*Sqrt[b*x^2 + c*x^4])/(231*b^3*x^(5/2)) + (5*c^(7/4)*(11*b*B - 9*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])/(231*b^(13/4)*
Sqrt[b*x^2 + c*x^4])

Rule 226

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)]/(2*q*Sqrt[a + b*x^4]))*EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 335

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 2050

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 2057

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[c^IntPart[m]*(c*x)^FracPa
rt[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 2063

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 {A+B x^2}{x^{11/2} \sqrt {b x^2+c x^4}} \, dx &=-\frac {2 A \sqrt {b x^2+c x^4}}{11 b x^{13/2}}-\frac {\left (2 \left (-\frac {11 b B}{2}+\frac {9 A c}{2}\right )\right ) \int \frac {1}{x^{7/2} \sqrt {b x^2+c x^4}} \, dx}{11 b}\\ &=-\frac {2 A \sqrt {b x^2+c x^4}}{11 b x^{13/2}}-\frac {2 (11 b B-9 A c) \sqrt {b x^2+c x^4}}{77 b^2 x^{9/2}}-\frac {(5 c (11 b B-9 A c)) \int \frac {1}{x^{3/2} \sqrt {b x^2+c x^4}} \, dx}{77 b^2}\\ &=-\frac {2 A \sqrt {b x^2+c x^4}}{11 b x^{13/2}}-\frac {2 (11 b B-9 A c) \sqrt {b x^2+c x^4}}{77 b^2 x^{9/2}}+\frac {10 c (11 b B-9 A c) \sqrt {b x^2+c x^4}}{231 b^3 x^{5/2}}+\frac {\left (5 c^2 (11 b B-9 A c)\right ) \int \frac {\sqrt {x}}{\sqrt {b x^2+c x^4}} \, dx}{231 b^3}\\ &=-\frac {2 A \sqrt {b x^2+c x^4}}{11 b x^{13/2}}-\frac {2 (11 b B-9 A c) \sqrt {b x^2+c x^4}}{77 b^2 x^{9/2}}+\frac {10 c (11 b B-9 A c) \sqrt {b x^2+c x^4}}{231 b^3 x^{5/2}}+\frac {\left (5 c^2 (11 b B-9 A c) x \sqrt {b+c x^2}\right ) \int \frac {1}{\sqrt {x} \sqrt {b+c x^2}} \, dx}{231 b^3 \sqrt {b x^2+c x^4}}\\ &=-\frac {2 A \sqrt {b x^2+c x^4}}{11 b x^{13/2}}-\frac {2 (11 b B-9 A c) \sqrt {b x^2+c x^4}}{77 b^2 x^{9/2}}+\frac {10 c (11 b B-9 A c) \sqrt {b x^2+c x^4}}{231 b^3 x^{5/2}}+\frac {\left (10 c^2 (11 b B-9 A c) x \sqrt {b+c x^2}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {b+c x^4}} \, dx,x,\sqrt {x}\right )}{231 b^3 \sqrt {b x^2+c x^4}}\\ &=-\frac {2 A \sqrt {b x^2+c x^4}}{11 b x^{13/2}}-\frac {2 (11 b B-9 A c) \sqrt {b x^2+c x^4}}{77 b^2 x^{9/2}}+\frac {10 c (11 b B-9 A c) \sqrt {b x^2+c x^4}}{231 b^3 x^{5/2}}+\frac {5 c^{7/4} (11 b B-9 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 )}{231 b^{13/4} \sqrt {b x^2+c x^4}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 10.06, size = 84, normalized size = 0.41 \begin {gather*} -\frac {2 \left (7 A \left (b+c x^2\right )+(11 b B-9 A c) x^2 \sqrt {1+\frac {c x^2}{b}} \, _2F_1\left (-\frac {7}{4},\frac {1}{2};-\frac {3}{4};-\frac {c x^2}{b}\right )\right )}{77 b x^{9/2} \sqrt {x^2 \left (b+c x^2\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(7*A*(b + c*x^2) + (11*b*B - 9*A*c)*x^2*Sqrt[1 + (c*x^2)/b]*Hypergeometric2F1[-7/4, 1/2, -3/4, -((c*x^2)/b
)]))/(77*b*x^(9/2)*Sqrt[x^2*(b + c*x^2)])

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Maple [A]
time = 0.41, size = 274, normalized size = 1.34

method	result	size
risch	\(-\frac {2 \left (c \,x^{2}+b \right ) \left (45 A \,c^{2} x^{4}-55 x^{4} b B c -27 A b c \,x^{2}+33 b^{2} B \,x^{2}+21 b^{2} A \right )}{231 b^{3} x^{\frac {9}{2}} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}-\frac {5 c \left (9 A c -11 B b \right ) \sqrt {-b c}\, \sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {x c}{\sqrt {-b c}}}\, \EllipticF \left (\sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {x}\, \sqrt {x \left (c \,x^{2}+b \right )}}{231 b^{3} \sqrt {c \,x^{3}+b x}\, \sqrt {x^{2} \left (c \,x^{2}+b \right )}}\)	\(216\)
default	\(-\frac {45 A \sqrt {-b c}\, \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {x c}{\sqrt {-b c}}}\, \EllipticF \left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) c^{2} x^{5}-55 B \sqrt {-b c}\, \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {x c}{\sqrt {-b c}}}\, \EllipticF \left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) b c \,x^{5}+90 A \,c^{3} x^{6}-110 x^{6} B b \,c^{2}+36 A b \,c^{2} x^{4}-44 x^{4} B \,b^{2} c -12 A \,b^{2} c \,x^{2}+66 x^{2} B \,b^{3}+42 A \,b^{3}}{231 \sqrt {x^{4} c +b \,x^{2}}\, x^{\frac {9}{2}} b^{3}}\)	\(274\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/231/(c*x^4+b*x^2)^(1/2)/x^(9/2)*(45*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)*(-x*c/(-b*c)^(1/2))^(1/2)*EllipticF(((c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2),1
/2*2^(1/2))*c^2*x^5-55*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)*(-x*c/(-b*c)^(1/2))^(1/2)*EllipticF(((c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2),1/2*2^(1/2))*b*c*
x^5+90*A*c^3*x^6-110*x^6*B*b*c^2+36*A*b*c^2*x^4-44*x^4*B*b^2*c-12*A*b^2*c*x^2+66*x^2*B*b^3+42*A*b^3)/b^3

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.56, size = 96, normalized size = 0.47 \begin {gather*} \frac {2 \, {\left (5 \, {\left (11 \, B b c - 9 \, A c^{2}\right )} \sqrt {c} x^{7} {\rm weierstrassPInverse}\left (-\frac {4 \, b}{c}, 0, x\right ) + {\left (5 \, {\left (11 \, B b c - 9 \, A c^{2}\right )} x^{4} - 21 \, A b^{2} - 3 \, {\left (11 \, B b^{2} - 9 \, A b c\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}} \sqrt {x}\right )}}{231 \, b^{3} x^{7}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/231*(5*(11*B*b*c - 9*A*c^2)*sqrt(c)*x^7*weierstrassPInverse(-4*b/c, 0, x) + (5*(11*B*b*c - 9*A*c^2)*x^4 - 21
*A*b^2 - 3*(11*B*b^2 - 9*A*b*c)*x^2)*sqrt(c*x^4 + b*x^2)*sqrt(x))/(b^3*x^7)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {B\,x^2+A}{x^{11/2}\,\sqrt {c\,x^4+b\,x^2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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