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

Optimal. Leaf size=296 \[ \frac {e^2 \sqrt {x} \left (\sqrt {a}+\sqrt {c} x\right ) \sqrt {\frac {a+c x^2}{\left (\sqrt {a}+\sqrt {c} x\right )^2}} \left (3 \sqrt {a} B+A \sqrt {c}\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{2 \sqrt [4]{a} c^{7/4} \sqrt {e x} \sqrt {a+c x^2}}-\frac {e \sqrt {e x} (A+B x)}{c \sqrt {a+c x^2}}+\frac {3 B e^2 x \sqrt {a+c x^2}}{c^{3/2} \sqrt {e x} \left (\sqrt {a}+\sqrt {c} x\right )}-\frac {3 \sqrt [4]{a} B e^2 \sqrt {x} \left (\sqrt {a}+\sqrt {c} x\right ) \sqrt {\frac {a+c x^2}{\left (\sqrt {a}+\sqrt {c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{c^{7/4} \sqrt {e x} \sqrt {a+c x^2}} \]

[Out]

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

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Rubi [A]  time = 0.27, antiderivative size = 296, 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 = {819, 842, 840, 1198, 220, 1196} \[ \frac {e^2 \sqrt {x} \left (\sqrt {a}+\sqrt {c} x\right ) \sqrt {\frac {a+c x^2}{\left (\sqrt {a}+\sqrt {c} x\right )^2}} \left (3 \sqrt {a} B+A \sqrt {c}\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{2 \sqrt [4]{a} c^{7/4} \sqrt {e x} \sqrt {a+c x^2}}-\frac {e \sqrt {e x} (A+B x)}{c \sqrt {a+c x^2}}+\frac {3 B e^2 x \sqrt {a+c x^2}}{c^{3/2} \sqrt {e x} \left (\sqrt {a}+\sqrt {c} x\right )}-\frac {3 \sqrt [4]{a} B e^2 \sqrt {x} \left (\sqrt {a}+\sqrt {c} x\right ) \sqrt {\frac {a+c x^2}{\left (\sqrt {a}+\sqrt {c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{c^{7/4} \sqrt {e x} \sqrt {a+c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((e*Sqrt[e*x]*(A + B*x))/(c*Sqrt[a + c*x^2])) + (3*B*e^2*x*Sqrt[a + c*x^2])/(c^(3/2)*Sqrt[e*x]*(Sqrt[a] + Sqr
t[c]*x)) - (3*a^(1/4)*B*e^2*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^(7/4)*Sqrt[e*x]*Sqrt[a + c*x^2]) + ((3*Sqrt[a]*B + A*Sqrt[c])*e^
2*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^(1/4)*c^(7/4)*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 819

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 + 1)*(a*(e*f + d*g) - (c*d*f - a*e*g)*x))/(2*a*c*(p + 1)), x] - Dist[1/(2*a*c*(p + 1)),
Int[(d + e*x)^(m - 2)*(a + c*x^2)^(p + 1)*Simp[a*e*(e*f*(m - 1) + d*g*m) - c*d^2*f*(2*p + 3) + e*(a*e*g*m - c*
d*f*(m + 2*p + 2))*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, 1] && (EqQ[d, 0] || (EqQ[m, 2] && EqQ[p, -3] && RationalQ[a, c, d, e, f, g]) ||  !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 {(e x)^{3/2} (A+B x)}{\left (a+c x^2\right )^{3/2}} \, dx &=-\frac {e \sqrt {e x} (A+B x)}{c \sqrt {a+c x^2}}+\frac {\int \frac {\frac {1}{2} a A e^2+\frac {3}{2} a B e^2 x}{\sqrt {e x} \sqrt {a+c x^2}} \, dx}{a c}\\ &=-\frac {e \sqrt {e x} (A+B x)}{c \sqrt {a+c x^2}}+\frac {\sqrt {x} \int \frac {\frac {1}{2} a A e^2+\frac {3}{2} a B e^2 x}{\sqrt {x} \sqrt {a+c x^2}} \, dx}{a c \sqrt {e x}}\\ &=-\frac {e \sqrt {e x} (A+B x)}{c \sqrt {a+c x^2}}+\frac {\left (2 \sqrt {x}\right ) \operatorname {Subst}\left (\int \frac {\frac {1}{2} a A e^2+\frac {3}{2} a B e^2 x^2}{\sqrt {a+c x^4}} \, dx,x,\sqrt {x}\right )}{a c \sqrt {e x}}\\ &=-\frac {e \sqrt {e x} (A+B x)}{c \sqrt {a+c x^2}}-\frac {\left (3 \sqrt {a} B e^2 \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 )}{c^{3/2} \sqrt {e x}}+\frac {\left (2 \left (\frac {3}{2} a B e^2+\frac {1}{2} \sqrt {a} A \sqrt {c} e^2\right ) \sqrt {x}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+c x^4}} \, dx,x,\sqrt {x}\right )}{\sqrt {a} c^{3/2} \sqrt {e x}}\\ &=-\frac {e \sqrt {e x} (A+B x)}{c \sqrt {a+c x^2}}+\frac {3 B e^2 x \sqrt {a+c x^2}}{c^{3/2} \sqrt {e x} \left (\sqrt {a}+\sqrt {c} x\right )}-\frac {3 \sqrt [4]{a} B e^2 \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^{7/4} \sqrt {e x} \sqrt {a+c x^2}}+\frac {\left (3 \sqrt {a} B+A \sqrt {c}\right ) e^2 \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 \sqrt [4]{a} c^{7/4} \sqrt {e x} \sqrt {a+c x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.05, size = 102, normalized size = 0.34 \[ \frac {e \sqrt {e x} \left (A \sqrt {\frac {c x^2}{a}+1} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};-\frac {c x^2}{a}\right )+B x \sqrt {\frac {c x^2}{a}+1} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-\frac {c x^2}{a}\right )-A-B x\right )}{c \sqrt {a+c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.72, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (B e x^{2} + A e x\right )} \sqrt {c x^{2} + a} \sqrt {e x}}{c^{2} x^{4} + 2 \, a c x^{2} + a^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((B*e*x^2 + A*e*x)*sqrt(c*x^2 + 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 \[ \int \frac {{\left (B x + A\right )} \left (e x\right )^{\frac {3}{2}}}{{\left (c x^{2} + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.10, size = 290, normalized size = 0.98 \[ \frac {\sqrt {e x}\, \left (-2 B c \,x^{2}-2 A c x +6 \sqrt {2}\, \sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {\frac {-c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {-\frac {c x}{\sqrt {-a c}}}\, B a \EllipticE \left (\sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}, \frac {\sqrt {2}}{2}\right )-3 \sqrt {2}\, \sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {\frac {-c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {-\frac {c x}{\sqrt {-a c}}}\, B a \EllipticF \left (\sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}, \frac {\sqrt {2}}{2}\right )+\sqrt {2}\, \sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {\frac {-c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {-\frac {c x}{\sqrt {-a c}}}\, \sqrt {-a c}\, A \EllipticF \left (\sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}, \frac {\sqrt {2}}{2}\right )\right ) e}{2 \sqrt {c \,x^{2}+a}\, c^{2} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [C]  time = 28.55, size = 94, normalized size = 0.32 \[ \frac {A e^{\frac {3}{2}} x^{\frac {5}{2}} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{4}, \frac {3}{2} \\ \frac {9}{4} \end {matrix}\middle | {\frac {c x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {3}{2}} \Gamma \left (\frac {9}{4}\right )} + \frac {B e^{\frac {3}{2}} x^{\frac {7}{2}} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {c x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {3}{2}} \Gamma \left (\frac {11}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

A*e**(3/2)*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)) + B
*e**(3/2)*x**(7/2)*gamma(7/4)*hyper((3/2, 7/4), (11/4,), c*x**2*exp_polar(I*pi)/a)/(2*a**(3/2)*gamma(11/4))

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