3.806 \(\int \frac {A+B x^2}{(e x)^{7/2} \sqrt {a+b x^2}} \, dx\)

Optimal. Leaf size=342 \[ -\frac {\sqrt [4]{b} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} (3 A b-5 a B) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt [4]{a} \sqrt {e}}\right )|\frac {1}{2}\right )}{5 a^{7/4} e^{7/2} \sqrt {a+b x^2}}+\frac {2 \sqrt [4]{b} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} (3 A b-5 a B) E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt [4]{a} \sqrt {e}}\right )|\frac {1}{2}\right )}{5 a^{7/4} e^{7/2} \sqrt {a+b x^2}}-\frac {2 \sqrt {b} \sqrt {e x} \sqrt {a+b x^2} (3 A b-5 a B)}{5 a^2 e^4 \left (\sqrt {a}+\sqrt {b} x\right )}+\frac {2 \sqrt {a+b x^2} (3 A b-5 a B)}{5 a^2 e^3 \sqrt {e x}}-\frac {2 A \sqrt {a+b x^2}}{5 a e (e x)^{5/2}} \]

[Out]

-2/5*A*(b*x^2+a)^(1/2)/a/e/(e*x)^(5/2)+2/5*(3*A*b-5*B*a)*(b*x^2+a)^(1/2)/a^2/e^3/(e*x)^(1/2)-2/5*(3*A*b-5*B*a)
*b^(1/2)*(e*x)^(1/2)*(b*x^2+a)^(1/2)/a^2/e^4/(a^(1/2)+x*b^(1/2))+2/5*b^(1/4)*(3*A*b-5*B*a)*(cos(2*arctan(b^(1/
4)*(e*x)^(1/2)/a^(1/4)/e^(1/2)))^2)^(1/2)/cos(2*arctan(b^(1/4)*(e*x)^(1/2)/a^(1/4)/e^(1/2)))*EllipticE(sin(2*a
rctan(b^(1/4)*(e*x)^(1/2)/a^(1/4)/e^(1/2))),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^(7/4)/e^(7/2)/(b*x^2+a)^(1/2)-1/5*b^(1/4)*(3*A*b-5*B*a)*(cos(2*arctan(b^(1/4)*(e*x)^(1/2)/a^(1/4)/e^(
1/2)))^2)^(1/2)/cos(2*arctan(b^(1/4)*(e*x)^(1/2)/a^(1/4)/e^(1/2)))*EllipticF(sin(2*arctan(b^(1/4)*(e*x)^(1/2)/
a^(1/4)/e^(1/2))),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^(7/4)/e^(7/2)/(b*
x^2+a)^(1/2)

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Rubi [A]  time = 0.25, antiderivative size = 342, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {453, 325, 329, 305, 220, 1196} \[ \frac {2 \sqrt {a+b x^2} (3 A b-5 a B)}{5 a^2 e^3 \sqrt {e x}}-\frac {2 \sqrt {b} \sqrt {e x} \sqrt {a+b x^2} (3 A b-5 a B)}{5 a^2 e^4 \left (\sqrt {a}+\sqrt {b} x\right )}-\frac {\sqrt [4]{b} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} (3 A b-5 a B) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt [4]{a} \sqrt {e}}\right )|\frac {1}{2}\right )}{5 a^{7/4} e^{7/2} \sqrt {a+b x^2}}+\frac {2 \sqrt [4]{b} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} (3 A b-5 a B) E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt [4]{a} \sqrt {e}}\right )|\frac {1}{2}\right )}{5 a^{7/4} e^{7/2} \sqrt {a+b x^2}}-\frac {2 A \sqrt {a+b x^2}}{5 a e (e x)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x^2)/((e*x)^(7/2)*Sqrt[a + b*x^2]),x]

[Out]

(-2*A*Sqrt[a + b*x^2])/(5*a*e*(e*x)^(5/2)) + (2*(3*A*b - 5*a*B)*Sqrt[a + b*x^2])/(5*a^2*e^3*Sqrt[e*x]) - (2*Sq
rt[b]*(3*A*b - 5*a*B)*Sqrt[e*x]*Sqrt[a + b*x^2])/(5*a^2*e^4*(Sqrt[a] + Sqrt[b]*x)) + (2*b^(1/4)*(3*A*b - 5*a*B
)*(Sqrt[a] + Sqrt[b]*x)*Sqrt[(a + b*x^2)/(Sqrt[a] + Sqrt[b]*x)^2]*EllipticE[2*ArcTan[(b^(1/4)*Sqrt[e*x])/(a^(1
/4)*Sqrt[e])], 1/2])/(5*a^(7/4)*e^(7/2)*Sqrt[a + b*x^2]) - (b^(1/4)*(3*A*b - 5*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[e*x])/(a^(1/4)*Sqrt[e])], 1/2])/(5*a^(7
/4)*e^(7/2)*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 305

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 2]}, Dist[1/q, Int[1/Sqrt[a + b*x^4], x],
x] - Dist[1/q, Int[(1 - q*x^2)/Sqrt[a + b*x^4], x], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

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]

Rule 1196

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, -Simp[(d*x*Sqrt[a + c
*x^4])/(a*(1 + q^2*x^2)), x] + Simp[(d*(1 + q^2*x^2)*Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]*EllipticE[2*ArcTan[
q*x], 1/2])/(q*Sqrt[a + c*x^4]), x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

Rubi steps

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

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Mathematica [C]  time = 0.05, size = 82, normalized size = 0.24 \[ -\frac {2 x \left (x^2 \sqrt {\frac {b x^2}{a}+1} (5 a B-3 A b) \, _2F_1\left (-\frac {1}{4},\frac {1}{2};\frac {3}{4};-\frac {b x^2}{a}\right )+A \left (a+b x^2\right )\right )}{5 a (e x)^{7/2} \sqrt {a+b x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x^2)/((e*x)^(7/2)*Sqrt[a + b*x^2]),x]

[Out]

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

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fricas [F]  time = 0.83, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (B x^{2} + A\right )} \sqrt {b x^{2} + a} \sqrt {e x}}{b e^{4} x^{6} + a e^{4} x^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/(e*x)^(7/2)/(b*x^2+a)^(1/2),x, algorithm="fricas")

[Out]

integral((B*x^2 + A)*sqrt(b*x^2 + a)*sqrt(e*x)/(b*e^4*x^6 + a*e^4*x^4), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {B x^{2} + A}{\sqrt {b x^{2} + a} \left (e x\right )^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/(e*x)^(7/2)/(b*x^2+a)^(1/2),x, algorithm="giac")

[Out]

integrate((B*x^2 + A)/(sqrt(b*x^2 + a)*(e*x)^(7/2)), x)

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maple [A]  time = 0.02, size = 417, normalized size = 1.22 \[ -\frac {-6 A \,b^{2} x^{4}+10 B a b \,x^{4}+6 \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, A a b \,x^{2} \EllipticE \left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )-3 \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, A a b \,x^{2} \EllipticF \left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )-10 \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, B \,a^{2} x^{2} \EllipticE \left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )+5 \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, B \,a^{2} x^{2} \EllipticF \left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )-4 A a b \,x^{2}+10 B \,a^{2} x^{2}+2 A \,a^{2}}{5 \sqrt {b \,x^{2}+a}\, \sqrt {e x}\, a^{2} e^{3} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^2+A)/(e*x)^(7/2)/(b*x^2+a)^(1/2),x)

[Out]

-1/5/x^2*(6*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)*EllipticE(((b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2),1/2*2^(1/2))*x^2*a*b-3*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)*Ellipti
cF(((b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2),1/2*2^(1/2))*x^2*a*b-10*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)*EllipticE(((b*x+(-a*b)^(1/2))/(-a*
b)^(1/2))^(1/2),1/2*2^(1/2))*x^2*a^2+5*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))
*x^2*a^2-6*A*b^2*x^4+10*B*a*b*x^4-4*A*a*b*x^2+10*B*a^2*x^2+2*A*a^2)/(b*x^2+a)^(1/2)/e^3/(e*x)^(1/2)/a^2

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {B x^{2} + A}{\sqrt {b x^{2} + a} \left (e x\right )^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/(e*x)^(7/2)/(b*x^2+a)^(1/2),x, algorithm="maxima")

[Out]

integrate((B*x^2 + A)/(sqrt(b*x^2 + a)*(e*x)^(7/2)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {B\,x^2+A}{{\left (e\,x\right )}^{7/2}\,\sqrt {b\,x^2+a}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x^2)/((e*x)^(7/2)*(a + b*x^2)^(1/2)),x)

[Out]

int((A + B*x^2)/((e*x)^(7/2)*(a + b*x^2)^(1/2)), x)

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sympy [C]  time = 30.89, size = 104, normalized size = 0.30 \[ \frac {A \Gamma \left (- \frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{4}, \frac {1}{2} \\ - \frac {1}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \sqrt {a} e^{\frac {7}{2}} x^{\frac {5}{2}} \Gamma \left (- \frac {1}{4}\right )} + \frac {B \Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {1}{2} \\ \frac {3}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \sqrt {a} e^{\frac {7}{2}} \sqrt {x} \Gamma \left (\frac {3}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x**2+A)/(e*x)**(7/2)/(b*x**2+a)**(1/2),x)

[Out]

A*gamma(-5/4)*hyper((-5/4, 1/2), (-1/4,), b*x**2*exp_polar(I*pi)/a)/(2*sqrt(a)*e**(7/2)*x**(5/2)*gamma(-1/4))
+ B*gamma(-1/4)*hyper((-1/4, 1/2), (3/4,), b*x**2*exp_polar(I*pi)/a)/(2*sqrt(a)*e**(7/2)*sqrt(x)*gamma(3/4))

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