3.793 \(\int \frac {\sqrt {a+b x^2} (A+B x^2)}{x^{13/2}} \, dx\)

Optimal. Leaf size=187 \[ \frac {2 b^{7/4} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} (5 A b-11 a B) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{231 a^{9/4} \sqrt {a+b x^2}}+\frac {4 b \sqrt {a+b x^2} (5 A b-11 a B)}{231 a^2 x^{3/2}}+\frac {2 \sqrt {a+b x^2} (5 A b-11 a B)}{77 a x^{7/2}}-\frac {2 A \left (a+b x^2\right )^{3/2}}{11 a x^{11/2}} \]

[Out]

-2/11*A*(b*x^2+a)^(3/2)/a/x^(11/2)+2/77*(5*A*b-11*B*a)*(b*x^2+a)^(1/2)/a/x^(7/2)+4/231*b*(5*A*b-11*B*a)*(b*x^2
+a)^(1/2)/a^2/x^(3/2)+2/231*b^(7/4)*(5*A*b-11*B*a)*(cos(2*arctan(b^(1/4)*x^(1/2)/a^(1/4)))^2)^(1/2)/cos(2*arct
an(b^(1/4)*x^(1/2)/a^(1/4)))*EllipticF(sin(2*arctan(b^(1/4)*x^(1/2)/a^(1/4))),1/2*2^(1/2))*(a^(1/2)+x*b^(1/2))
*((b*x^2+a)/(a^(1/2)+x*b^(1/2))^2)^(1/2)/a^(9/4)/(b*x^2+a)^(1/2)

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Rubi [A]  time = 0.11, antiderivative size = 187, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {453, 277, 325, 329, 220} \[ \frac {2 b^{7/4} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} (5 A b-11 a B) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{231 a^{9/4} \sqrt {a+b x^2}}+\frac {4 b \sqrt {a+b x^2} (5 A b-11 a B)}{231 a^2 x^{3/2}}+\frac {2 \sqrt {a+b x^2} (5 A b-11 a B)}{77 a x^{7/2}}-\frac {2 A \left (a+b x^2\right )^{3/2}}{11 a x^{11/2}} \]

Antiderivative was successfully verified.

[In]

Int[(Sqrt[a + b*x^2]*(A + B*x^2))/x^(13/2),x]

[Out]

(2*(5*A*b - 11*a*B)*Sqrt[a + b*x^2])/(77*a*x^(7/2)) + (4*b*(5*A*b - 11*a*B)*Sqrt[a + b*x^2])/(231*a^2*x^(3/2))
- (2*A*(a + b*x^2)^(3/2))/(11*a*x^(11/2)) + (2*b^(7/4)*(5*A*b - 11*a*B)*(Sqrt[a] + Sqrt[b]*x)*Sqrt[(a + b*x^2
)/(Sqrt[a] + Sqrt[b]*x)^2]*EllipticF[2*ArcTan[(b^(1/4)*Sqrt[x])/a^(1/4)], 1/2])/(231*a^(9/4)*Sqrt[a + b*x^2])

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 277

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^p)/(c*(m +
1)), x] - Dist[(b*n*p)/(c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] &&
IGtQ[n, 0] && GtQ[p, 0] && LtQ[m, -1] &&  !ILtQ[(m + n*p + n + 1)/n, 0] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

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 453

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(c*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(a*e*(m + 1)), x] + Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -1]

Rubi steps

\begin {align*} \int \frac {\sqrt {a+b x^2} \left (A+B x^2\right )}{x^{13/2}} \, dx &=-\frac {2 A \left (a+b x^2\right )^{3/2}}{11 a x^{11/2}}-\frac {\left (2 \left (\frac {5 A b}{2}-\frac {11 a B}{2}\right )\right ) \int \frac {\sqrt {a+b x^2}}{x^{9/2}} \, dx}{11 a}\\ &=\frac {2 (5 A b-11 a B) \sqrt {a+b x^2}}{77 a x^{7/2}}-\frac {2 A \left (a+b x^2\right )^{3/2}}{11 a x^{11/2}}-\frac {(2 b (5 A b-11 a B)) \int \frac {1}{x^{5/2} \sqrt {a+b x^2}} \, dx}{77 a}\\ &=\frac {2 (5 A b-11 a B) \sqrt {a+b x^2}}{77 a x^{7/2}}+\frac {4 b (5 A b-11 a B) \sqrt {a+b x^2}}{231 a^2 x^{3/2}}-\frac {2 A \left (a+b x^2\right )^{3/2}}{11 a x^{11/2}}+\frac {\left (2 b^2 (5 A b-11 a B)\right ) \int \frac {1}{\sqrt {x} \sqrt {a+b x^2}} \, dx}{231 a^2}\\ &=\frac {2 (5 A b-11 a B) \sqrt {a+b x^2}}{77 a x^{7/2}}+\frac {4 b (5 A b-11 a B) \sqrt {a+b x^2}}{231 a^2 x^{3/2}}-\frac {2 A \left (a+b x^2\right )^{3/2}}{11 a x^{11/2}}+\frac {\left (4 b^2 (5 A b-11 a B)\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b x^4}} \, dx,x,\sqrt {x}\right )}{231 a^2}\\ &=\frac {2 (5 A b-11 a B) \sqrt {a+b x^2}}{77 a x^{7/2}}+\frac {4 b (5 A b-11 a B) \sqrt {a+b x^2}}{231 a^2 x^{3/2}}-\frac {2 A \left (a+b x^2\right )^{3/2}}{11 a x^{11/2}}+\frac {2 b^{7/4} (5 A b-11 a B) \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{231 a^{9/4} \sqrt {a+b x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.11, size = 80, normalized size = 0.43 \[ \frac {2 \sqrt {a+b x^2} \left (\frac {x^2 (5 A b-11 a B) \, _2F_1\left (-\frac {7}{4},-\frac {1}{2};-\frac {3}{4};-\frac {b x^2}{a}\right )}{\sqrt {\frac {b x^2}{a}+1}}-7 A \left (a+b x^2\right )\right )}{77 a x^{11/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sqrt[a + b*x^2]*(A + B*x^2))/x^(13/2),x]

[Out]

(2*Sqrt[a + b*x^2]*(-7*A*(a + b*x^2) + ((5*A*b - 11*a*B)*x^2*Hypergeometric2F1[-7/4, -1/2, -3/4, -((b*x^2)/a)]
)/Sqrt[1 + (b*x^2)/a]))/(77*a*x^(11/2))

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fricas [F]  time = 0.89, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (B x^{2} + A\right )} \sqrt {b x^{2} + a}}{x^{\frac {13}{2}}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(b*x^2+a)^(1/2)/x^(13/2),x, algorithm="fricas")

[Out]

integral((B*x^2 + A)*sqrt(b*x^2 + a)/x^(13/2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B x^{2} + A\right )} \sqrt {b x^{2} + a}}{x^{\frac {13}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(b*x^2+a)^(1/2)/x^(13/2),x, algorithm="giac")

[Out]

integrate((B*x^2 + A)*sqrt(b*x^2 + a)/x^(13/2), x)

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maple [A]  time = 0.04, size = 270, normalized size = 1.44 \[ \frac {\frac {20 A \,b^{3} x^{6}}{231}-\frac {4 B a \,b^{2} x^{6}}{21}+\frac {10 \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \sqrt {-a b}\, A \,b^{2} x^{5} \EllipticF \left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{231}-\frac {2 \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \sqrt {-a b}\, B a b \,x^{5} \EllipticF \left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{21}+\frac {8 A a \,b^{2} x^{4}}{231}-\frac {10 B \,a^{2} b \,x^{4}}{21}-\frac {18 A \,a^{2} b \,x^{2}}{77}-\frac {2 B \,a^{3} x^{2}}{7}-\frac {2 A \,a^{3}}{11}}{\sqrt {b \,x^{2}+a}\, a^{2} x^{\frac {11}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^2+A)*(b*x^2+a)^(1/2)/x^(13/2),x)

[Out]

2/231/(b*x^2+a)^(1/2)*(5*A*((b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2)*2^(1/2)*((-b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^
(1/2)*(-1/(-a*b)^(1/2)*b*x)^(1/2)*EllipticF(((b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2),1/2*2^(1/2))*(-a*b)^(1/2)*
x^5*b^2-11*B*((b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2)*2^(1/2)*((-b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2)*(-1/(-a*
b)^(1/2)*b*x)^(1/2)*EllipticF(((b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2),1/2*2^(1/2))*(-a*b)^(1/2)*x^5*a*b+10*A*b
^3*x^6-22*B*a*b^2*x^6+4*A*a*b^2*x^4-55*B*a^2*b*x^4-27*A*a^2*b*x^2-33*B*a^3*x^2-21*A*a^3)/x^(11/2)/a^2

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B x^{2} + A\right )} \sqrt {b x^{2} + a}}{x^{\frac {13}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(b*x^2+a)^(1/2)/x^(13/2),x, algorithm="maxima")

[Out]

integrate((B*x^2 + A)*sqrt(b*x^2 + a)/x^(13/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\left (B\,x^2+A\right )\,\sqrt {b\,x^2+a}}{x^{13/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x^2)*(a + b*x^2)^(1/2))/x^(13/2),x)

[Out]

int(((A + B*x^2)*(a + b*x^2)^(1/2))/x^(13/2), x)

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sympy [C]  time = 135.65, size = 100, normalized size = 0.53 \[ \frac {A \sqrt {a} \Gamma \left (- \frac {11}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {11}{4}, - \frac {1}{2} \\ - \frac {7}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 x^{\frac {11}{2}} \Gamma \left (- \frac {7}{4}\right )} + \frac {B \sqrt {a} \Gamma \left (- \frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {7}{4}, - \frac {1}{2} \\ - \frac {3}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 x^{\frac {7}{2}} \Gamma \left (- \frac {3}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x**2+A)*(b*x**2+a)**(1/2)/x**(13/2),x)

[Out]

A*sqrt(a)*gamma(-11/4)*hyper((-11/4, -1/2), (-7/4,), b*x**2*exp_polar(I*pi)/a)/(2*x**(11/2)*gamma(-7/4)) + B*s
qrt(a)*gamma(-7/4)*hyper((-7/4, -1/2), (-3/4,), b*x**2*exp_polar(I*pi)/a)/(2*x**(7/2)*gamma(-3/4))

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