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

Optimal. Leaf size=338 \[ \frac {2 \sqrt [4]{b} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} (5 a B+A b) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt [4]{a} \sqrt {e}}\right )|\frac {1}{2}\right )}{5 a^{3/4} e^{7/2} \sqrt {a+b x^2}}-\frac {4 \sqrt [4]{b} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} (5 a B+A b) E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt [4]{a} \sqrt {e}}\right )|\frac {1}{2}\right )}{5 a^{3/4} e^{7/2} \sqrt {a+b x^2}}+\frac {4 \sqrt {b} \sqrt {e x} \sqrt {a+b x^2} (5 a B+A b)}{5 a e^4 \left (\sqrt {a}+\sqrt {b} x\right )}-\frac {2 \sqrt {a+b x^2} (5 a B+A b)}{5 a e^3 \sqrt {e x}}-\frac {2 A \left (a+b x^2\right )^{3/2}}{5 a e (e x)^{5/2}} \]

[Out]

-2/5*A*(b*x^2+a)^(3/2)/a/e/(e*x)^(5/2)-2/5*(A*b+5*B*a)*(b*x^2+a)^(1/2)/a/e^3/(e*x)^(1/2)+4/5*(A*b+5*B*a)*b^(1/
2)*(e*x)^(1/2)*(b*x^2+a)^(1/2)/a/e^4/(a^(1/2)+x*b^(1/2))-4/5*b^(1/4)*(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*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^(
3/4)/e^(7/2)/(b*x^2+a)^(1/2)+2/5*b^(1/4)*(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^(3/4)/e^(7/2)/(b*x^2+a)^(1/2)

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Rubi [A]  time = 0.26, antiderivative size = 338, 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, 277, 329, 305, 220, 1196} \[ \frac {2 \sqrt [4]{b} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} (5 a B+A b) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt [4]{a} \sqrt {e}}\right )|\frac {1}{2}\right )}{5 a^{3/4} e^{7/2} \sqrt {a+b x^2}}-\frac {4 \sqrt [4]{b} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} (5 a B+A b) E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt [4]{a} \sqrt {e}}\right )|\frac {1}{2}\right )}{5 a^{3/4} e^{7/2} \sqrt {a+b x^2}}-\frac {2 \sqrt {a+b x^2} (5 a B+A b)}{5 a e^3 \sqrt {e x}}+\frac {4 \sqrt {b} \sqrt {e x} \sqrt {a+b x^2} (5 a B+A b)}{5 a e^4 \left (\sqrt {a}+\sqrt {b} x\right )}-\frac {2 A \left (a+b x^2\right )^{3/2}}{5 a e (e x)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [C]  time = 0.07, size = 95, normalized size = 0.28 \[ -\frac {2 x \sqrt {a+b x^2} \left (x^2 (5 a B+A b) \, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};-\frac {b x^2}{a}\right )+A \left (a+b x^2\right ) \sqrt {\frac {b x^2}{a}+1}\right )}{5 a (e x)^{7/2} \sqrt {\frac {b x^2}{a}+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.68, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (B x^{2} + A\right )} \sqrt {b x^{2} + a} \sqrt {e x}}{e^{4} x^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(b*x^2+a)^(1/2)/(e*x)^(7/2),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 = 0.05, size = 417, normalized size = 1.23 \[ \frac {-\frac {4 A \,b^{2} x^{4}}{5}-2 B a b \,x^{4}+\frac {4 \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, A a b \,x^{2} \EllipticE \left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{5}-\frac {2 \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, A a b \,x^{2} \EllipticF \left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{5}+4 \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, B \,a^{2} x^{2} \EllipticE \left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )-2 \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, B \,a^{2} x^{2} \EllipticF \left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )-\frac {6 A a b \,x^{2}}{5}-2 B \,a^{2} x^{2}-\frac {2 A \,a^{2}}{5}}{\sqrt {b \,x^{2}+a}\, \sqrt {e x}\, a \,e^{3} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5/x^2*(2*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-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))*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^2-2*A*b^2*x^4-5*B*x^4*a*b-3*A*a*b*x^2-5*B*a^2*x^2-A*a^2)/(b*x^2+a)^(1/2)/e^3/(e*x)^(1/2)/a

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(b*x^2+a)^(1/2)/(e*x)^(7/2),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 \[ \int \frac {\left (B\,x^2+A\right )\,\sqrt {b\,x^2+a}}{{\left (e\,x\right )}^{7/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [C]  time = 22.34, size = 107, normalized size = 0.32 \[ \frac {A \sqrt {a} \Gamma \left (- \frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{4}, - \frac {1}{2} \\ - \frac {1}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 e^{\frac {7}{2}} x^{\frac {5}{2}} \Gamma \left (- \frac {1}{4}\right )} + \frac {B \sqrt {a} \Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, - \frac {1}{4} \\ \frac {3}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 e^{\frac {7}{2}} \sqrt {x} \Gamma \left (\frac {3}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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