3.439 \(\int \frac {(A+B x) \sqrt {a+c x^2}}{(e x)^{5/2}} \, dx\)

Optimal. Leaf size=298 \[ \frac {2 \sqrt [4]{c} \sqrt {x} \left (\sqrt {a}+\sqrt {c} x\right ) \sqrt {\frac {a+c x^2}{\left (\sqrt {a}+\sqrt {c} x\right )^2}} \left (3 \sqrt {a} B+A \sqrt {c}\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{3 \sqrt [4]{a} e^2 \sqrt {e x} \sqrt {a+c x^2}}-\frac {2 \sqrt {a+c x^2} (A+3 B x)}{3 e (e x)^{3/2}}+\frac {4 B \sqrt {c} x \sqrt {a+c x^2}}{e^2 \sqrt {e x} \left (\sqrt {a}+\sqrt {c} x\right )}-\frac {4 \sqrt [4]{a} B \sqrt [4]{c} \sqrt {x} \left (\sqrt {a}+\sqrt {c} x\right ) \sqrt {\frac {a+c x^2}{\left (\sqrt {a}+\sqrt {c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{e^2 \sqrt {e x} \sqrt {a+c x^2}} \]

[Out]

-2/3*(3*B*x+A)*(c*x^2+a)^(1/2)/e/(e*x)^(3/2)+4*B*x*c^(1/2)*(c*x^2+a)^(1/2)/e^2/(a^(1/2)+x*c^(1/2))/(e*x)^(1/2)
-4*a^(1/4)*B*c^(1/4)*(cos(2*arctan(c^(1/4)*x^(1/2)/a^(1/4)))^2)^(1/2)/cos(2*arctan(c^(1/4)*x^(1/2)/a^(1/4)))*E
llipticE(sin(2*arctan(c^(1/4)*x^(1/2)/a^(1/4))),1/2*2^(1/2))*(a^(1/2)+x*c^(1/2))*x^(1/2)*((c*x^2+a)/(a^(1/2)+x
*c^(1/2))^2)^(1/2)/e^2/(e*x)^(1/2)/(c*x^2+a)^(1/2)+2/3*c^(1/4)*(cos(2*arctan(c^(1/4)*x^(1/2)/a^(1/4)))^2)^(1/2
)/cos(2*arctan(c^(1/4)*x^(1/2)/a^(1/4)))*EllipticF(sin(2*arctan(c^(1/4)*x^(1/2)/a^(1/4))),1/2*2^(1/2))*(3*B*a^
(1/2)+A*c^(1/2))*(a^(1/2)+x*c^(1/2))*x^(1/2)*((c*x^2+a)/(a^(1/2)+x*c^(1/2))^2)^(1/2)/a^(1/4)/e^2/(e*x)^(1/2)/(
c*x^2+a)^(1/2)

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Rubi [A]  time = 0.29, antiderivative size = 298, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {811, 842, 840, 1198, 220, 1196} \[ \frac {2 \sqrt [4]{c} \sqrt {x} \left (\sqrt {a}+\sqrt {c} x\right ) \sqrt {\frac {a+c x^2}{\left (\sqrt {a}+\sqrt {c} x\right )^2}} \left (3 \sqrt {a} B+A \sqrt {c}\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{3 \sqrt [4]{a} e^2 \sqrt {e x} \sqrt {a+c x^2}}-\frac {2 \sqrt {a+c x^2} (A+3 B x)}{3 e (e x)^{3/2}}+\frac {4 B \sqrt {c} x \sqrt {a+c x^2}}{e^2 \sqrt {e x} \left (\sqrt {a}+\sqrt {c} x\right )}-\frac {4 \sqrt [4]{a} B \sqrt [4]{c} \sqrt {x} \left (\sqrt {a}+\sqrt {c} x\right ) \sqrt {\frac {a+c x^2}{\left (\sqrt {a}+\sqrt {c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{e^2 \sqrt {e x} \sqrt {a+c x^2}} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*Sqrt[a + c*x^2])/(e*x)^(5/2),x]

[Out]

(-2*(A + 3*B*x)*Sqrt[a + c*x^2])/(3*e*(e*x)^(3/2)) + (4*B*Sqrt[c]*x*Sqrt[a + c*x^2])/(e^2*Sqrt[e*x]*(Sqrt[a] +
Sqrt[c]*x)) - (4*a^(1/4)*B*c^(1/4)*Sqrt[x]*(Sqrt[a] + Sqrt[c]*x)*Sqrt[(a + c*x^2)/(Sqrt[a] + Sqrt[c]*x)^2]*El
lipticE[2*ArcTan[(c^(1/4)*Sqrt[x])/a^(1/4)], 1/2])/(e^2*Sqrt[e*x]*Sqrt[a + c*x^2]) + (2*(3*Sqrt[a]*B + A*Sqrt[
c])*c^(1/4)*Sqrt[x]*(Sqrt[a] + Sqrt[c]*x)*Sqrt[(a + c*x^2)/(Sqrt[a] + Sqrt[c]*x)^2]*EllipticF[2*ArcTan[(c^(1/4
)*Sqrt[x])/a^(1/4)], 1/2])/(3*a^(1/4)*e^2*Sqrt[e*x]*Sqrt[a + c*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 811

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Simp[((d + e*x)^
(m + 1)*(a + c*x^2)^p*((d*g - e*f*(m + 2))*(c*d^2 + a*e^2) - 2*c*d^2*p*(e*f - d*g) - e*(g*(m + 1)*(c*d^2 + a*e
^2) + 2*c*d*p*(e*f - d*g))*x))/(e^2*(m + 1)*(m + 2)*(c*d^2 + a*e^2)), x] - Dist[p/(e^2*(m + 1)*(m + 2)*(c*d^2
+ a*e^2)), Int[(d + e*x)^(m + 2)*(a + c*x^2)^(p - 1)*Simp[2*a*c*e*(e*f - d*g)*(m + 2) - c*(2*c*d*(d*g*(2*p + 1
) - e*f*(m + 2*p + 2)) - 2*a*e^2*g*(m + 1))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2
, 0] && GtQ[p, 0] && LtQ[m, -2] && LtQ[m + 2*p, 0] &&  !ILtQ[m + 2*p + 3, 0]

Rule 840

Int[((f_) + (g_.)*(x_))/(Sqrt[x_]*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[2, Subst[Int[(f + g*x^2)/Sqrt[
a + c*x^4], x], x, Sqrt[x]], x] /; FreeQ[{a, c, f, g}, x]

Rule 842

Int[((f_) + (g_.)*(x_))/(Sqrt[(e_)*(x_)]*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[Sqrt[x]/Sqrt[e*x], Int[
(f + g*x)/(Sqrt[x]*Sqrt[a + c*x^2]), x], x] /; FreeQ[{a, c, e, f, g}, x]

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]

Rule 1198

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

Rubi steps

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

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Mathematica [C]  time = 0.03, size = 82, normalized size = 0.28 \[ -\frac {2 x \sqrt {a+c x^2} \left (A \, _2F_1\left (-\frac {3}{4},-\frac {1}{2};\frac {1}{4};-\frac {c x^2}{a}\right )+3 B x \, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};-\frac {c x^2}{a}\right )\right )}{3 (e x)^{5/2} \sqrt {\frac {c x^2}{a}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*Sqrt[a + c*x^2])/(e*x)^(5/2),x]

[Out]

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

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fricas [F]  time = 0.99, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {c x^{2} + a} {\left (B x + A\right )} \sqrt {e x}}{e^{3} x^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+a)^(1/2)/(e*x)^(5/2),x, algorithm="fricas")

[Out]

integral(sqrt(c*x^2 + a)*(B*x + A)*sqrt(e*x)/(e^3*x^3), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {c x^{2} + a} {\left (B x + A\right )}}{\left (e x\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+a)^(1/2)/(e*x)^(5/2),x, algorithm="giac")

[Out]

integrate(sqrt(c*x^2 + a)*(B*x + A)/(e*x)^(5/2), x)

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maple [A]  time = 0.10, size = 303, normalized size = 1.02 \[ \frac {-2 B c \,x^{3}-\frac {2 A c \,x^{2}}{3}+4 \sqrt {2}\, \sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {\frac {-c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {-\frac {c x}{\sqrt {-a c}}}\, B a x \EllipticE \left (\sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}, \frac {\sqrt {2}}{2}\right )-2 \sqrt {2}\, \sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {\frac {-c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {-\frac {c x}{\sqrt {-a c}}}\, B a x \EllipticF \left (\sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}, \frac {\sqrt {2}}{2}\right )+\frac {2 \sqrt {2}\, \sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {\frac {-c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {-\frac {c x}{\sqrt {-a c}}}\, \sqrt {-a c}\, A x \EllipticF \left (\sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}, \frac {\sqrt {2}}{2}\right )}{3}-2 B a x -\frac {2 A a}{3}}{\sqrt {c \,x^{2}+a}\, \sqrt {e x}\, e^{2} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(c*x^2+a)^(1/2)/(e*x)^(5/2),x)

[Out]

2/3/x*(A*2^(1/2)*((c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2)*((-c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2)*(-1/(-a*c)^(
1/2)*c*x)^(1/2)*EllipticF(((c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2),1/2*2^(1/2))*(-a*c)^(1/2)*x-3*B*2^(1/2)*((c*
x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2)*((-c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2)*(-1/(-a*c)^(1/2)*c*x)^(1/2)*Ellip
ticF(((c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2),1/2*2^(1/2))*x*a+6*B*2^(1/2)*((c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1
/2)*((-c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2)*(-1/(-a*c)^(1/2)*c*x)^(1/2)*EllipticE(((c*x+(-a*c)^(1/2))/(-a*c)^
(1/2))^(1/2),1/2*2^(1/2))*x*a-3*B*c*x^3-A*c*x^2-3*B*a*x-a*A)/(c*x^2+a)^(1/2)/e^2/(e*x)^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {c x^{2} + a} {\left (B x + A\right )}}{\left (e x\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+a)^(1/2)/(e*x)^(5/2),x, algorithm="maxima")

[Out]

integrate(sqrt(c*x^2 + a)*(B*x + A)/(e*x)^(5/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\sqrt {c\,x^2+a}\,\left (A+B\,x\right )}{{\left (e\,x\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + c*x^2)^(1/2)*(A + B*x))/(e*x)^(5/2),x)

[Out]

int(((a + c*x^2)^(1/2)*(A + B*x))/(e*x)^(5/2), x)

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sympy [C]  time = 7.50, size = 104, normalized size = 0.35 \[ \frac {A \sqrt {a} \Gamma \left (- \frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, - \frac {1}{2} \\ \frac {1}{4} \end {matrix}\middle | {\frac {c x^{2} e^{i \pi }}{a}} \right )}}{2 e^{\frac {5}{2}} x^{\frac {3}{2}} \Gamma \left (\frac {1}{4}\right )} + \frac {B \sqrt {a} \Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, - \frac {1}{4} \\ \frac {3}{4} \end {matrix}\middle | {\frac {c x^{2} e^{i \pi }}{a}} \right )}}{2 e^{\frac {5}{2}} \sqrt {x} \Gamma \left (\frac {3}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x**2+a)**(1/2)/(e*x)**(5/2),x)

[Out]

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

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