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

Optimal. Leaf size=287 \[ -\frac {\sqrt [4]{a} e \sqrt {x} \left (\sqrt {a}+\sqrt {c} x\right ) \sqrt {\frac {a+c x^2}{\left (\sqrt {a}+\sqrt {c} x\right )^2}} \left (\sqrt {a} B-3 A \sqrt {c}\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{3 c^{5/4} \sqrt {e x} \sqrt {a+c x^2}}-\frac {2 \sqrt [4]{a} A e \sqrt {x} \left (\sqrt {a}+\sqrt {c} x\right ) \sqrt {\frac {a+c x^2}{\left (\sqrt {a}+\sqrt {c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{c^{3/4} \sqrt {e x} \sqrt {a+c x^2}}+\frac {2 A e x \sqrt {a+c x^2}}{\sqrt {c} \sqrt {e x} \left (\sqrt {a}+\sqrt {c} x\right )}+\frac {2 B \sqrt {e x} \sqrt {a+c x^2}}{3 c} \]

[Out]

2*A*e*x*(c*x^2+a)^(1/2)/c^(1/2)/(a^(1/2)+x*c^(1/2))/(e*x)^(1/2)+2/3*B*(e*x)^(1/2)*(c*x^2+a)^(1/2)/c-2*a^(1/4)*
A*e*(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)))*EllipticE(sin(2*ar
ctan(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
)/c^(3/4)/(e*x)^(1/2)/(c*x^2+a)^(1/2)-1/3*a^(1/4)*e*(cos(2*arctan(c^(1/4)*x^(1/2)/a^(1/4)))^2)^(1/2)/cos(2*arc
tan(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))*(B*a^(1/2)-3*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)/c^(5/4)/(e*x)^(1/2)/(c*x^2+a)^(1/2)

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Rubi [A]  time = 0.25, antiderivative size = 287, 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 = {833, 842, 840, 1198, 220, 1196} \[ -\frac {\sqrt [4]{a} e \sqrt {x} \left (\sqrt {a}+\sqrt {c} x\right ) \sqrt {\frac {a+c x^2}{\left (\sqrt {a}+\sqrt {c} x\right )^2}} \left (\sqrt {a} B-3 A \sqrt {c}\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{3 c^{5/4} \sqrt {e x} \sqrt {a+c x^2}}-\frac {2 \sqrt [4]{a} A e \sqrt {x} \left (\sqrt {a}+\sqrt {c} x\right ) \sqrt {\frac {a+c x^2}{\left (\sqrt {a}+\sqrt {c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{c^{3/4} \sqrt {e x} \sqrt {a+c x^2}}+\frac {2 A e x \sqrt {a+c x^2}}{\sqrt {c} \sqrt {e x} \left (\sqrt {a}+\sqrt {c} x\right )}+\frac {2 B \sqrt {e x} \sqrt {a+c x^2}}{3 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*B*Sqrt[e*x]*Sqrt[a + c*x^2])/(3*c) + (2*A*e*x*Sqrt[a + c*x^2])/(Sqrt[c]*Sqrt[e*x]*(Sqrt[a] + Sqrt[c]*x)) -
(2*a^(1/4)*A*e*Sqrt[x]*(Sqrt[a] + Sqrt[c]*x)*Sqrt[(a + c*x^2)/(Sqrt[a] + Sqrt[c]*x)^2]*EllipticE[2*ArcTan[(c^(
1/4)*Sqrt[x])/a^(1/4)], 1/2])/(c^(3/4)*Sqrt[e*x]*Sqrt[a + c*x^2]) - (a^(1/4)*(Sqrt[a]*B - 3*A*Sqrt[c])*e*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*c^(5/4)*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 833

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(g*(d + e*x)
^m*(a + c*x^2)^(p + 1))/(c*(m + 2*p + 2)), x] + Dist[1/(c*(m + 2*p + 2)), Int[(d + e*x)^(m - 1)*(a + c*x^2)^p*
Simp[c*d*f*(m + 2*p + 2) - a*e*g*m + c*(e*f*(m + 2*p + 2) + d*g*m)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g, p
}, x] && NeQ[c*d^2 + a*e^2, 0] && GtQ[m, 0] && NeQ[m + 2*p + 2, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ
[2*m, 2*p]) &&  !(IGtQ[m, 0] && EqQ[f, 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 {\sqrt {e x} (A+B x)}{\sqrt {a+c x^2}} \, dx &=\frac {2 B \sqrt {e x} \sqrt {a+c x^2}}{3 c}+\frac {2 \int \frac {-\frac {1}{2} a B e+\frac {3}{2} A c e x}{\sqrt {e x} \sqrt {a+c x^2}} \, dx}{3 c}\\ &=\frac {2 B \sqrt {e x} \sqrt {a+c x^2}}{3 c}+\frac {\left (2 \sqrt {x}\right ) \int \frac {-\frac {1}{2} a B e+\frac {3}{2} A c e x}{\sqrt {x} \sqrt {a+c x^2}} \, dx}{3 c \sqrt {e x}}\\ &=\frac {2 B \sqrt {e x} \sqrt {a+c x^2}}{3 c}+\frac {\left (4 \sqrt {x}\right ) \operatorname {Subst}\left (\int \frac {-\frac {1}{2} a B e+\frac {3}{2} A c e x^2}{\sqrt {a+c x^4}} \, dx,x,\sqrt {x}\right )}{3 c \sqrt {e x}}\\ &=\frac {2 B \sqrt {e x} \sqrt {a+c x^2}}{3 c}-\frac {\left (2 \sqrt {a} \left (\sqrt {a} B-3 A \sqrt {c}\right ) e \sqrt {x}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+c x^4}} \, dx,x,\sqrt {x}\right )}{3 c \sqrt {e x}}-\frac {\left (2 \sqrt {a} A e \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 )}{\sqrt {c} \sqrt {e x}}\\ &=\frac {2 B \sqrt {e x} \sqrt {a+c x^2}}{3 c}+\frac {2 A e x \sqrt {a+c x^2}}{\sqrt {c} \sqrt {e x} \left (\sqrt {a}+\sqrt {c} x\right )}-\frac {2 \sqrt [4]{a} A e \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 )}{c^{3/4} \sqrt {e x} \sqrt {a+c x^2}}-\frac {\sqrt [4]{a} \left (\sqrt {a} B-3 A \sqrt {c}\right ) e \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 c^{5/4} \sqrt {e x} \sqrt {a+c x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.04, size = 109, normalized size = 0.38 \[ \frac {2 \sqrt {e x} \left (A c x \sqrt {\frac {c x^2}{a}+1} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-\frac {c x^2}{a}\right )-a B \sqrt {\frac {c x^2}{a}+1} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};-\frac {c x^2}{a}\right )+B \left (a+c x^2\right )\right )}{3 c \sqrt {a+c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.98, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (B x + A\right )} \sqrt {e x}}{\sqrt {c x^{2} + a}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.06, size = 295, normalized size = 1.03 \[ -\frac {\sqrt {e x}\, \left (-2 B \,c^{2} x^{3}-6 \sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {\frac {-c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {-\frac {c x}{\sqrt {-a c}}}\, \sqrt {2}\, A a c \EllipticE \left (\sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}, \frac {\sqrt {2}}{2}\right )+3 \sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {\frac {-c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {-\frac {c x}{\sqrt {-a c}}}\, \sqrt {2}\, A a c \EllipticF \left (\sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}, \frac {\sqrt {2}}{2}\right )-2 B a c x +\sqrt {-a c}\, \sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {\frac {-c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {-\frac {c x}{\sqrt {-a c}}}\, \sqrt {2}\, B a \EllipticF \left (\sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}, \frac {\sqrt {2}}{2}\right )\right )}{3 \sqrt {c \,x^{2}+a}\, c^{2} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(e*x)^(1/2)/(c*x^2+a)^(1/2)*(3*A*((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))*2^(1/2)*a*
c-6*A*((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))*2^(1/2)*a*c+B*(-a*c)^(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))*2^(1/2)*a-2*B*c^2*x^3-2*B*a*c*x)/x/c^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [C]  time = 3.75, size = 92, normalized size = 0.32 \[ \frac {A \left (e x\right )^{\frac {3}{2}} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {c x^{2} e^{i \pi }}{a}} \right )}}{2 \sqrt {a} e \Gamma \left (\frac {7}{4}\right )} + \frac {B \left (e x\right )^{\frac {5}{2}} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {\frac {c x^{2} e^{i \pi }}{a}} \right )}}{2 \sqrt {a} e^{2} \Gamma \left (\frac {9}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

A*(e*x)**(3/2)*gamma(3/4)*hyper((1/2, 3/4), (7/4,), c*x**2*exp_polar(I*pi)/a)/(2*sqrt(a)*e*gamma(7/4)) + B*(e*
x)**(5/2)*gamma(5/4)*hyper((1/2, 5/4), (9/4,), c*x**2*exp_polar(I*pi)/a)/(2*sqrt(a)*e**2*gamma(9/4))

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