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

Optimal. Leaf size=326 \[ -\frac {a^{3/4} e^2 \sqrt {x} \left (\sqrt {a}+\sqrt {c} x\right ) \sqrt {\frac {a+c x^2}{\left (\sqrt {a}+\sqrt {c} x\right )^2}} \left (9 \sqrt {a} B+5 A \sqrt {c}\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{15 c^{7/4} \sqrt {e x} \sqrt {a+c x^2}}+\frac {6 a^{5/4} B e^2 \sqrt {x} \left (\sqrt {a}+\sqrt {c} x\right ) \sqrt {\frac {a+c x^2}{\left (\sqrt {a}+\sqrt {c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{5 c^{7/4} \sqrt {e x} \sqrt {a+c x^2}}+\frac {2 A e \sqrt {e x} \sqrt {a+c x^2}}{3 c}-\frac {6 a B e^2 x \sqrt {a+c x^2}}{5 c^{3/2} \sqrt {e x} \left (\sqrt {a}+\sqrt {c} x\right )}+\frac {2 B (e x)^{3/2} \sqrt {a+c x^2}}{5 c} \]

[Out]

2/5*B*(e*x)^(3/2)*(c*x^2+a)^(1/2)/c-6/5*a*B*e^2*x*(c*x^2+a)^(1/2)/c^(3/2)/(a^(1/2)+x*c^(1/2))/(e*x)^(1/2)+2/3*
A*e*(e*x)^(1/2)*(c*x^2+a)^(1/2)/c+6/5*a^(5/4)*B*e^2*(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)))*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)-1/15*a^(3/4)*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))*(9*B*a^(1/2)+5*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^(7/4)/(e*x)^(1/2)/(c*x^2+a)^(1/2)

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Rubi [A]  time = 0.34, antiderivative size = 326, 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 = {833, 842, 840, 1198, 220, 1196} \[ -\frac {a^{3/4} e^2 \sqrt {x} \left (\sqrt {a}+\sqrt {c} x\right ) \sqrt {\frac {a+c x^2}{\left (\sqrt {a}+\sqrt {c} x\right )^2}} \left (9 \sqrt {a} B+5 A \sqrt {c}\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{15 c^{7/4} \sqrt {e x} \sqrt {a+c x^2}}+\frac {6 a^{5/4} B e^2 \sqrt {x} \left (\sqrt {a}+\sqrt {c} x\right ) \sqrt {\frac {a+c x^2}{\left (\sqrt {a}+\sqrt {c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{5 c^{7/4} \sqrt {e x} \sqrt {a+c x^2}}+\frac {2 A e \sqrt {e x} \sqrt {a+c x^2}}{3 c}-\frac {6 a B e^2 x \sqrt {a+c x^2}}{5 c^{3/2} \sqrt {e x} \left (\sqrt {a}+\sqrt {c} x\right )}+\frac {2 B (e x)^{3/2} \sqrt {a+c x^2}}{5 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*A*e*Sqrt[e*x]*Sqrt[a + c*x^2])/(3*c) + (2*B*(e*x)^(3/2)*Sqrt[a + c*x^2])/(5*c) - (6*a*B*e^2*x*Sqrt[a + c*x^
2])/(5*c^(3/2)*Sqrt[e*x]*(Sqrt[a] + Sqrt[c]*x)) + (6*a^(5/4)*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])/(5*c^(7/4)*Sqrt[e*x]*Sqrt[a
+ c*x^2]) - (a^(3/4)*(9*Sqrt[a]*B + 5*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])/(15*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 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 {(e x)^{3/2} (A+B x)}{\sqrt {a+c x^2}} \, dx &=\frac {2 B (e x)^{3/2} \sqrt {a+c x^2}}{5 c}+\frac {2 \int \frac {\sqrt {e x} \left (-\frac {3}{2} a B e+\frac {5}{2} A c e x\right )}{\sqrt {a+c x^2}} \, dx}{5 c}\\ &=\frac {2 A e \sqrt {e x} \sqrt {a+c x^2}}{3 c}+\frac {2 B (e x)^{3/2} \sqrt {a+c x^2}}{5 c}+\frac {4 \int \frac {-\frac {5}{4} a A c e^2-\frac {9}{4} a B c e^2 x}{\sqrt {e x} \sqrt {a+c x^2}} \, dx}{15 c^2}\\ &=\frac {2 A e \sqrt {e x} \sqrt {a+c x^2}}{3 c}+\frac {2 B (e x)^{3/2} \sqrt {a+c x^2}}{5 c}+\frac {\left (4 \sqrt {x}\right ) \int \frac {-\frac {5}{4} a A c e^2-\frac {9}{4} a B c e^2 x}{\sqrt {x} \sqrt {a+c x^2}} \, dx}{15 c^2 \sqrt {e x}}\\ &=\frac {2 A e \sqrt {e x} \sqrt {a+c x^2}}{3 c}+\frac {2 B (e x)^{3/2} \sqrt {a+c x^2}}{5 c}+\frac {\left (8 \sqrt {x}\right ) \operatorname {Subst}\left (\int \frac {-\frac {5}{4} a A c e^2-\frac {9}{4} a B c e^2 x^2}{\sqrt {a+c x^4}} \, dx,x,\sqrt {x}\right )}{15 c^2 \sqrt {e x}}\\ &=\frac {2 A e \sqrt {e x} \sqrt {a+c x^2}}{3 c}+\frac {2 B (e x)^{3/2} \sqrt {a+c x^2}}{5 c}+\frac {\left (6 a^{3/2} 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 )}{5 c^{3/2} \sqrt {e x}}-\frac {\left (2 a \left (9 \sqrt {a} B+5 A \sqrt {c}\right ) e^2 \sqrt {x}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+c x^4}} \, dx,x,\sqrt {x}\right )}{15 c^{3/2} \sqrt {e x}}\\ &=\frac {2 A e \sqrt {e x} \sqrt {a+c x^2}}{3 c}+\frac {2 B (e x)^{3/2} \sqrt {a+c x^2}}{5 c}-\frac {6 a B e^2 x \sqrt {a+c x^2}}{5 c^{3/2} \sqrt {e x} \left (\sqrt {a}+\sqrt {c} x\right )}+\frac {6 a^{5/4} 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 )}{5 c^{7/4} \sqrt {e x} \sqrt {a+c x^2}}-\frac {a^{3/4} \left (9 \sqrt {a} B+5 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 )}{15 c^{7/4} \sqrt {e x} \sqrt {a+c x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.05, size = 118, normalized size = 0.36 \[ \frac {2 e \sqrt {e x} \left (\left (a+c x^2\right ) (5 A+3 B x)-5 a A \sqrt {\frac {c x^2}{a}+1} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};-\frac {c x^2}{a}\right )-3 a B x \sqrt {\frac {c x^2}{a}+1} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-\frac {c x^2}{a}\right )\right )}{15 c \sqrt {a+c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*e*Sqrt[e*x]*((5*A + 3*B*x)*(a + c*x^2) - 5*a*A*Sqrt[1 + (c*x^2)/a]*Hypergeometric2F1[1/4, 1/2, 5/4, -((c*x^
2)/a)] - 3*a*B*x*Sqrt[1 + (c*x^2)/a]*Hypergeometric2F1[1/2, 3/4, 7/4, -((c*x^2)/a)]))/(15*c*Sqrt[a + c*x^2])

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fricas [F]  time = 1.00, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (B e x^{2} + A e x\right )} \sqrt {e x}}{\sqrt {c x^{2} + a}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.09, size = 316, normalized size = 0.97 \[ -\frac {\sqrt {e x}\, \left (-6 B \,c^{2} x^{4}-10 A \,c^{2} x^{3}-6 B a c \,x^{2}-10 A a c x +18 \sqrt {2}\, \sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {\frac {-c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {-\frac {c x}{\sqrt {-a c}}}\, B \,a^{2} \EllipticE \left (\sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}, \frac {\sqrt {2}}{2}\right )-9 \sqrt {2}\, \sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {\frac {-c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {-\frac {c x}{\sqrt {-a c}}}\, B \,a^{2} \EllipticF \left (\sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}, \frac {\sqrt {2}}{2}\right )+5 \sqrt {2}\, \sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {\frac {-c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {-\frac {c x}{\sqrt {-a c}}}\, \sqrt {-a c}\, A a \EllipticF \left (\sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}, \frac {\sqrt {2}}{2}\right )\right ) e}{15 \sqrt {c \,x^{2}+a}\, c^{2} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*e/x*(e*x)^(1/2)/(c*x^2+a)^(1/2)/c^2*(5*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)*a+18*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^2-9*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^2-6*B*c^2*x^4-10*A*c^2*x^3-6*B*a*c*x^2-10*A
*a*c*x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [C]  time = 6.71, size = 94, normalized size = 0.29 \[ \frac {A e^{\frac {3}{2}} x^{\frac {5}{2}} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {\frac {c x^{2} e^{i \pi }}{a}} \right )}}{2 \sqrt {a} \Gamma \left (\frac {9}{4}\right )} + \frac {B e^{\frac {3}{2}} x^{\frac {7}{2}} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {c x^{2} e^{i \pi }}{a}} \right )}}{2 \sqrt {a} \Gamma \left (\frac {11}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

A*e**(3/2)*x**(5/2)*gamma(5/4)*hyper((1/2, 5/4), (9/4,), c*x**2*exp_polar(I*pi)/a)/(2*sqrt(a)*gamma(9/4)) + B*
e**(3/2)*x**(7/2)*gamma(7/4)*hyper((1/2, 7/4), (11/4,), c*x**2*exp_polar(I*pi)/a)/(2*sqrt(a)*gamma(11/4))

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