3.469 \(\int \frac {\sqrt {e x} (A+B x)}{(a+c x^2)^{3/2}} \, dx\)

Optimal. Leaf size=298 \[ \frac {e \sqrt {x} \left (\sqrt {a}+\sqrt {c} x\right ) \sqrt {\frac {a+c x^2}{\left (\sqrt {a}+\sqrt {c} x\right )^2}} \left (\sqrt {a} B-A \sqrt {c}\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{2 a^{3/4} c^{5/4} \sqrt {e x} \sqrt {a+c x^2}}+\frac {A e \sqrt {x} \left (\sqrt {a}+\sqrt {c} x\right ) \sqrt {\frac {a+c x^2}{\left (\sqrt {a}+\sqrt {c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{a^{3/4} c^{3/4} \sqrt {e x} \sqrt {a+c x^2}}-\frac {\sqrt {e x} (a B-A c x)}{a c \sqrt {a+c x^2}}-\frac {A e x \sqrt {a+c x^2}}{a \sqrt {c} \sqrt {e x} \left (\sqrt {a}+\sqrt {c} x\right )} \]

[Out]

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

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Rubi [A]  time = 0.24, 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 = {821, 842, 840, 1198, 220, 1196} \[ \frac {e \sqrt {x} \left (\sqrt {a}+\sqrt {c} x\right ) \sqrt {\frac {a+c x^2}{\left (\sqrt {a}+\sqrt {c} x\right )^2}} \left (\sqrt {a} B-A \sqrt {c}\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{2 a^{3/4} c^{5/4} \sqrt {e x} \sqrt {a+c x^2}}+\frac {A e \sqrt {x} \left (\sqrt {a}+\sqrt {c} x\right ) \sqrt {\frac {a+c x^2}{\left (\sqrt {a}+\sqrt {c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{a^{3/4} c^{3/4} \sqrt {e x} \sqrt {a+c x^2}}-\frac {\sqrt {e x} (a B-A c x)}{a c \sqrt {a+c x^2}}-\frac {A e x \sqrt {a+c x^2}}{a \sqrt {c} \sqrt {e x} \left (\sqrt {a}+\sqrt {c} x\right )} \]

Antiderivative was successfully verified.

[In]

Int[(Sqrt[e*x]*(A + B*x))/(a + c*x^2)^(3/2),x]

[Out]

-((Sqrt[e*x]*(a*B - A*c*x))/(a*c*Sqrt[a + c*x^2])) - (A*e*x*Sqrt[a + c*x^2])/(a*Sqrt[c]*Sqrt[e*x]*(Sqrt[a] + S
qrt[c]*x)) + (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])/(a^(3/4)*c^(3/4)*Sqrt[e*x]*Sqrt[a + c*x^2]) + ((Sqrt[a]*B - 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])/(2*a^(3/4)*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 821

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

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

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Mathematica [C]  time = 0.06, size = 113, normalized size = 0.38 \[ \frac {\sqrt {e x} \left (-A c x \sqrt {\frac {c x^2}{a}+1} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-\frac {c x^2}{a}\right )+3 a B \sqrt {\frac {c x^2}{a}+1} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};-\frac {c x^2}{a}\right )-3 a B+3 A c x\right )}{3 a c \sqrt {a+c x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sqrt[e*x]*(A + B*x))/(a + c*x^2)^(3/2),x]

[Out]

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

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fricas [F]  time = 1.18, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {c x^{2} + a} {\left (B x + A\right )} \sqrt {e x}}{c^{2} x^{4} + 2 \, a c x^{2} + a^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^(1/2)*(B*x+A)/(c*x^2+a)^(3/2),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^(1/2)*(B*x+A)/(c*x^2+a)^(3/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.09, size = 297, normalized size = 1.00 \[ \frac {\sqrt {e x}\, \left (2 A \,c^{2} x^{2}-2 \sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {\frac {-c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {-\frac {c x}{\sqrt {-a c}}}\, \sqrt {2}\, A a c \EllipticE \left (\sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}, \frac {\sqrt {2}}{2}\right )+\sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {\frac {-c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {-\frac {c x}{\sqrt {-a c}}}\, \sqrt {2}\, A a c \EllipticF \left (\sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}, \frac {\sqrt {2}}{2}\right )-2 B a c x +\sqrt {-a c}\, \sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {\frac {-c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {-\frac {c x}{\sqrt {-a c}}}\, \sqrt {2}\, B a \EllipticF \left (\sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}, \frac {\sqrt {2}}{2}\right )\right )}{2 \sqrt {c \,x^{2}+a}\, a \,c^{2} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x)^(1/2)*(B*x+A)/(c*x^2+a)^(3/2),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^(1/2)*(B*x+A)/(c*x^2+a)^(3/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\sqrt {e\,x}\,\left (A+B\,x\right )}{{\left (c\,x^2+a\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((e*x)^(1/2)*(A + B*x))/(a + c*x^2)^(3/2),x)

[Out]

int(((e*x)^(1/2)*(A + B*x))/(a + c*x^2)^(3/2), x)

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sympy [C]  time = 13.19, size = 94, normalized size = 0.32 \[ \frac {A \sqrt {e} x^{\frac {3}{2}} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {3}{2} \\ \frac {7}{4} \end {matrix}\middle | {\frac {c x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {3}{2}} \Gamma \left (\frac {7}{4}\right )} + \frac {B \sqrt {e} x^{\frac {5}{2}} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{4}, \frac {3}{2} \\ \frac {9}{4} \end {matrix}\middle | {\frac {c x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {3}{2}} \Gamma \left (\frac {9}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)**(1/2)*(B*x+A)/(c*x**2+a)**(3/2),x)

[Out]

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

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