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

Optimal. Leaf size=327 \[ \frac {\sqrt [4]{c} \sqrt {x} \left (\sqrt {a}+\sqrt {c} x\right ) \sqrt {\frac {a+c x^2}{\left (\sqrt {a}+\sqrt {c} x\right )^2}} \left (3 \sqrt {a} B-A \sqrt {c}\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{3 a^{5/4} e^2 \sqrt {e x} \sqrt {a+c x^2}}-\frac {2 B \sqrt [4]{c} \sqrt {x} \left (\sqrt {a}+\sqrt {c} x\right ) \sqrt {\frac {a+c x^2}{\left (\sqrt {a}+\sqrt {c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{a^{3/4} e^2 \sqrt {e x} \sqrt {a+c x^2}}-\frac {2 A \sqrt {a+c x^2}}{3 a e (e x)^{3/2}}-\frac {2 B \sqrt {a+c x^2}}{a e^2 \sqrt {e x}}+\frac {2 B \sqrt {c} x \sqrt {a+c x^2}}{a e^2 \sqrt {e x} \left (\sqrt {a}+\sqrt {c} x\right )} \]

[Out]

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

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Rubi [A]  time = 0.32, antiderivative size = 327, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {835, 842, 840, 1198, 220, 1196} \[ \frac {\sqrt [4]{c} \sqrt {x} \left (\sqrt {a}+\sqrt {c} x\right ) \sqrt {\frac {a+c x^2}{\left (\sqrt {a}+\sqrt {c} x\right )^2}} \left (3 \sqrt {a} B-A \sqrt {c}\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{3 a^{5/4} e^2 \sqrt {e x} \sqrt {a+c x^2}}-\frac {2 B \sqrt [4]{c} \sqrt {x} \left (\sqrt {a}+\sqrt {c} x\right ) \sqrt {\frac {a+c x^2}{\left (\sqrt {a}+\sqrt {c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{a^{3/4} e^2 \sqrt {e x} \sqrt {a+c x^2}}-\frac {2 A \sqrt {a+c x^2}}{3 a e (e x)^{3/2}}-\frac {2 B \sqrt {a+c x^2}}{a e^2 \sqrt {e x}}+\frac {2 B \sqrt {c} x \sqrt {a+c x^2}}{a e^2 \sqrt {e x} \left (\sqrt {a}+\sqrt {c} x\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*A*Sqrt[a + c*x^2])/(3*a*e*(e*x)^(3/2)) - (2*B*Sqrt[a + c*x^2])/(a*e^2*Sqrt[e*x]) + (2*B*Sqrt[c]*x*Sqrt[a +
c*x^2])/(a*e^2*Sqrt[e*x]*(Sqrt[a] + Sqrt[c]*x)) - (2*B*c^(1/4)*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)*e^2*Sqrt[e*x]*Sqrt[a +
c*x^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^(5/4)*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 835

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((e*f - d*g)
*(d + e*x)^(m + 1)*(a + c*x^2)^(p + 1))/((m + 1)*(c*d^2 + a*e^2)), x] + Dist[1/((m + 1)*(c*d^2 + a*e^2)), Int[
(d + e*x)^(m + 1)*(a + c*x^2)^p*Simp[(c*d*f + a*e*g)*(m + 1) - c*(e*f - d*g)*(m + 2*p + 3)*x, x], x], x] /; Fr
eeQ[{a, c, d, e, f, g, p}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[m, -1] && (IntegerQ[m] || IntegerQ[p] || Integer
sQ[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 {A+B x}{(e x)^{5/2} \sqrt {a+c x^2}} \, dx &=-\frac {2 A \sqrt {a+c x^2}}{3 a e (e x)^{3/2}}-\frac {2 \int \frac {-\frac {3}{2} a B e+\frac {1}{2} A c e x}{(e x)^{3/2} \sqrt {a+c x^2}} \, dx}{3 a e^2}\\ &=-\frac {2 A \sqrt {a+c x^2}}{3 a e (e x)^{3/2}}-\frac {2 B \sqrt {a+c x^2}}{a e^2 \sqrt {e x}}+\frac {4 \int \frac {-\frac {1}{4} a A c e^2+\frac {3}{4} a B c e^2 x}{\sqrt {e x} \sqrt {a+c x^2}} \, dx}{3 a^2 e^4}\\ &=-\frac {2 A \sqrt {a+c x^2}}{3 a e (e x)^{3/2}}-\frac {2 B \sqrt {a+c x^2}}{a e^2 \sqrt {e x}}+\frac {\left (4 \sqrt {x}\right ) \int \frac {-\frac {1}{4} a A c e^2+\frac {3}{4} a B c e^2 x}{\sqrt {x} \sqrt {a+c x^2}} \, dx}{3 a^2 e^4 \sqrt {e x}}\\ &=-\frac {2 A \sqrt {a+c x^2}}{3 a e (e x)^{3/2}}-\frac {2 B \sqrt {a+c x^2}}{a e^2 \sqrt {e x}}+\frac {\left (8 \sqrt {x}\right ) \operatorname {Subst}\left (\int \frac {-\frac {1}{4} a A c e^2+\frac {3}{4} a B c e^2 x^2}{\sqrt {a+c x^4}} \, dx,x,\sqrt {x}\right )}{3 a^2 e^4 \sqrt {e x}}\\ &=-\frac {2 A \sqrt {a+c x^2}}{3 a e (e x)^{3/2}}-\frac {2 B \sqrt {a+c x^2}}{a e^2 \sqrt {e x}}-\frac {\left (2 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 )}{\sqrt {a} e^2 \sqrt {e x}}+\frac {\left (2 \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 a e^2 \sqrt {e x}}\\ &=-\frac {2 A \sqrt {a+c x^2}}{3 a e (e x)^{3/2}}-\frac {2 B \sqrt {a+c x^2}}{a e^2 \sqrt {e x}}+\frac {2 B \sqrt {c} x \sqrt {a+c x^2}}{a e^2 \sqrt {e x} \left (\sqrt {a}+\sqrt {c} x\right )}-\frac {2 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 )}{a^{3/4} e^2 \sqrt {e x} \sqrt {a+c x^2}}+\frac {\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 a^{5/4} 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 = 0.25 \[ -\frac {2 x \sqrt {\frac {c x^2}{a}+1} \left (A \, _2F_1\left (-\frac {3}{4},\frac {1}{2};\frac {1}{4};-\frac {c x^2}{a}\right )+3 B x \, _2F_1\left (-\frac {1}{4},\frac {1}{2};\frac {3}{4};-\frac {c x^2}{a}\right )\right )}{3 (e x)^{5/2} \sqrt {a+c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.96, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {c x^{2} + a} {\left (B x + A\right )} \sqrt {e x}}{c e^{3} x^{5} + a e^{3} x^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(c*x^2 + a)*(B*x + A)*sqrt(e*x)/(c*e^3*x^5 + a*e^3*x^3), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.07, size = 306, normalized size = 0.94 \[ -\frac {6 B c \,x^{3}+2 A c \,x^{2}-6 \sqrt {2}\, \sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {\frac {-c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {-\frac {c x}{\sqrt {-a c}}}\, B a x \EllipticE \left (\sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}, \frac {\sqrt {2}}{2}\right )+3 \sqrt {2}\, \sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {\frac {-c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {-\frac {c x}{\sqrt {-a c}}}\, B a x \EllipticF \left (\sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}, \frac {\sqrt {2}}{2}\right )+\sqrt {2}\, \sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {\frac {-c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {-\frac {c x}{\sqrt {-a c}}}\, \sqrt {-a c}\, A x \EllipticF \left (\sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}, \frac {\sqrt {2}}{2}\right )+6 B a x +2 A a}{3 \sqrt {c \,x^{2}+a}\, \sqrt {e x}\, a \,e^{2} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [C]  time = 9.10, size = 100, normalized size = 0.31 \[ \frac {A \Gamma \left (- \frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, \frac {1}{2} \\ \frac {1}{4} \end {matrix}\middle | {\frac {c x^{2} e^{i \pi }}{a}} \right )}}{2 \sqrt {a} e^{\frac {5}{2}} x^{\frac {3}{2}} \Gamma \left (\frac {1}{4}\right )} + \frac {B \Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {1}{2} \\ \frac {3}{4} \end {matrix}\middle | {\frac {c x^{2} e^{i \pi }}{a}} \right )}}{2 \sqrt {a} e^{\frac {5}{2}} \sqrt {x} \Gamma \left (\frac {3}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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