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

Optimal. Leaf size=356 \[ \frac {a^{5/4} e^3 \sqrt {x} \left (\sqrt {a}+\sqrt {c} x\right ) \sqrt {\frac {a+c x^2}{\left (\sqrt {a}+\sqrt {c} x\right )^2}} \left (25 \sqrt {a} B-63 A \sqrt {c}\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{105 c^{9/4} \sqrt {e x} \sqrt {a+c x^2}}+\frac {6 a^{5/4} A e^3 \sqrt {x} \left (\sqrt {a}+\sqrt {c} x\right ) \sqrt {\frac {a+c x^2}{\left (\sqrt {a}+\sqrt {c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{5 c^{7/4} \sqrt {e x} \sqrt {a+c x^2}}-\frac {6 a A e^3 x \sqrt {a+c x^2}}{5 c^{3/2} \sqrt {e x} \left (\sqrt {a}+\sqrt {c} x\right )}+\frac {2 A e (e x)^{3/2} \sqrt {a+c x^2}}{5 c}-\frac {10 a B e^2 \sqrt {e x} \sqrt {a+c x^2}}{21 c^2}+\frac {2 B (e x)^{5/2} \sqrt {a+c x^2}}{7 c} \]

[Out]

2/5*A*e*(e*x)^(3/2)*(c*x^2+a)^(1/2)/c+2/7*B*(e*x)^(5/2)*(c*x^2+a)^(1/2)/c-6/5*a*A*e^3*x*(c*x^2+a)^(1/2)/c^(3/2
)/(a^(1/2)+x*c^(1/2))/(e*x)^(1/2)-10/21*a*B*e^2*(e*x)^(1/2)*(c*x^2+a)^(1/2)/c^2+6/5*a^(5/4)*A*e^3*(cos(2*arcta
n(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)+1/105*a^(5/4)*e^3*(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))*(25*B*a^(1/2)-63*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^(9/4)/(e*x)^(1/2)/(c*x^2+a)^(1/2)

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Rubi [A]  time = 0.41, antiderivative size = 356, normalized size of antiderivative = 1.00, number of steps used = 8, 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} \[ \frac {a^{5/4} e^3 \sqrt {x} \left (\sqrt {a}+\sqrt {c} x\right ) \sqrt {\frac {a+c x^2}{\left (\sqrt {a}+\sqrt {c} x\right )^2}} \left (25 \sqrt {a} B-63 A \sqrt {c}\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{105 c^{9/4} \sqrt {e x} \sqrt {a+c x^2}}+\frac {6 a^{5/4} A e^3 \sqrt {x} \left (\sqrt {a}+\sqrt {c} x\right ) \sqrt {\frac {a+c x^2}{\left (\sqrt {a}+\sqrt {c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{5 c^{7/4} \sqrt {e x} \sqrt {a+c x^2}}-\frac {6 a A e^3 x \sqrt {a+c x^2}}{5 c^{3/2} \sqrt {e x} \left (\sqrt {a}+\sqrt {c} x\right )}+\frac {2 A e (e x)^{3/2} \sqrt {a+c x^2}}{5 c}-\frac {10 a B e^2 \sqrt {e x} \sqrt {a+c x^2}}{21 c^2}+\frac {2 B (e x)^{5/2} \sqrt {a+c x^2}}{7 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-10*a*B*e^2*Sqrt[e*x]*Sqrt[a + c*x^2])/(21*c^2) + (2*A*e*(e*x)^(3/2)*Sqrt[a + c*x^2])/(5*c) + (2*B*(e*x)^(5/2
)*Sqrt[a + c*x^2])/(7*c) - (6*a*A*e^3*x*Sqrt[a + c*x^2])/(5*c^(3/2)*Sqrt[e*x]*(Sqrt[a] + Sqrt[c]*x)) + (6*a^(5
/4)*A*e^3*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^(5/4)*(25*Sqrt[a]*B - 63*A*Sqrt[c])*e^3*Sq
rt[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])/(105*c^(9/4)*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)^{5/2} (A+B x)}{\sqrt {a+c x^2}} \, dx &=\frac {2 B (e x)^{5/2} \sqrt {a+c x^2}}{7 c}+\frac {2 \int \frac {(e x)^{3/2} \left (-\frac {5}{2} a B e+\frac {7}{2} A c e x\right )}{\sqrt {a+c x^2}} \, dx}{7 c}\\ &=\frac {2 A e (e x)^{3/2} \sqrt {a+c x^2}}{5 c}+\frac {2 B (e x)^{5/2} \sqrt {a+c x^2}}{7 c}+\frac {4 \int \frac {\sqrt {e x} \left (-\frac {21}{4} a A c e^2-\frac {25}{4} a B c e^2 x\right )}{\sqrt {a+c x^2}} \, dx}{35 c^2}\\ &=-\frac {10 a B e^2 \sqrt {e x} \sqrt {a+c x^2}}{21 c^2}+\frac {2 A e (e x)^{3/2} \sqrt {a+c x^2}}{5 c}+\frac {2 B (e x)^{5/2} \sqrt {a+c x^2}}{7 c}+\frac {8 \int \frac {\frac {25}{8} a^2 B c e^3-\frac {63}{8} a A c^2 e^3 x}{\sqrt {e x} \sqrt {a+c x^2}} \, dx}{105 c^3}\\ &=-\frac {10 a B e^2 \sqrt {e x} \sqrt {a+c x^2}}{21 c^2}+\frac {2 A e (e x)^{3/2} \sqrt {a+c x^2}}{5 c}+\frac {2 B (e x)^{5/2} \sqrt {a+c x^2}}{7 c}+\frac {\left (8 \sqrt {x}\right ) \int \frac {\frac {25}{8} a^2 B c e^3-\frac {63}{8} a A c^2 e^3 x}{\sqrt {x} \sqrt {a+c x^2}} \, dx}{105 c^3 \sqrt {e x}}\\ &=-\frac {10 a B e^2 \sqrt {e x} \sqrt {a+c x^2}}{21 c^2}+\frac {2 A e (e x)^{3/2} \sqrt {a+c x^2}}{5 c}+\frac {2 B (e x)^{5/2} \sqrt {a+c x^2}}{7 c}+\frac {\left (16 \sqrt {x}\right ) \operatorname {Subst}\left (\int \frac {\frac {25}{8} a^2 B c e^3-\frac {63}{8} a A c^2 e^3 x^2}{\sqrt {a+c x^4}} \, dx,x,\sqrt {x}\right )}{105 c^3 \sqrt {e x}}\\ &=-\frac {10 a B e^2 \sqrt {e x} \sqrt {a+c x^2}}{21 c^2}+\frac {2 A e (e x)^{3/2} \sqrt {a+c x^2}}{5 c}+\frac {2 B (e x)^{5/2} \sqrt {a+c x^2}}{7 c}+\frac {\left (2 a^{3/2} \left (25 \sqrt {a} B-63 A \sqrt {c}\right ) e^3 \sqrt {x}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+c x^4}} \, dx,x,\sqrt {x}\right )}{105 c^2 \sqrt {e x}}+\frac {\left (6 a^{3/2} A e^3 \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 {10 a B e^2 \sqrt {e x} \sqrt {a+c x^2}}{21 c^2}+\frac {2 A e (e x)^{3/2} \sqrt {a+c x^2}}{5 c}+\frac {2 B (e x)^{5/2} \sqrt {a+c x^2}}{7 c}-\frac {6 a A e^3 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} A e^3 \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^{5/4} \left (25 \sqrt {a} B-63 A \sqrt {c}\right ) e^3 \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 )}{105 c^{9/4} \sqrt {e x} \sqrt {a+c x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.06, size = 133, normalized size = 0.37 \[ \frac {2 e^2 \sqrt {e x} \left (25 a^2 B \sqrt {\frac {c x^2}{a}+1} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};-\frac {c x^2}{a}\right )-\left (a+c x^2\right ) (25 a B-3 c x (7 A+5 B x))-21 a A c x \sqrt {\frac {c x^2}{a}+1} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-\frac {c x^2}{a}\right )\right )}{105 c^2 \sqrt {a+c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*e^2*Sqrt[e*x]*(-((a + c*x^2)*(25*a*B - 3*c*x*(7*A + 5*B*x))) + 25*a^2*B*Sqrt[1 + (c*x^2)/a]*Hypergeometric2
F1[1/4, 1/2, 5/4, -((c*x^2)/a)] - 21*a*A*c*x*Sqrt[1 + (c*x^2)/a]*Hypergeometric2F1[1/2, 3/4, 7/4, -((c*x^2)/a)
]))/(105*c^2*Sqrt[a + c*x^2])

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fricas [F]  time = 0.92, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (B e^{2} x^{3} + A e^{2} x^{2}\right )} \sqrt {e x}}{\sqrt {c x^{2} + a}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.14, size = 336, normalized size = 0.94 \[ \frac {\sqrt {e x}\, \left (30 B \,c^{3} x^{5}+42 A \,c^{3} x^{4}-20 B a \,c^{2} x^{3}+42 A a \,c^{2} x^{2}-126 \sqrt {2}\, \sqrt {-\frac {c x}{\sqrt {-a c}}}\, \sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {\frac {-c x +\sqrt {-a c}}{\sqrt {-a c}}}\, A \,a^{2} c \EllipticE \left (\sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}, \frac {\sqrt {2}}{2}\right )+63 \sqrt {2}\, \sqrt {-\frac {c x}{\sqrt {-a c}}}\, \sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {\frac {-c x +\sqrt {-a c}}{\sqrt {-a c}}}\, A \,a^{2} c \EllipticF \left (\sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}, \frac {\sqrt {2}}{2}\right )-50 B \,a^{2} c x +25 \sqrt {-a c}\, \sqrt {2}\, \sqrt {-\frac {c x}{\sqrt {-a c}}}\, \sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {\frac {-c x +\sqrt {-a c}}{\sqrt {-a c}}}\, B \,a^{2} \EllipticF \left (\sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}, \frac {\sqrt {2}}{2}\right )\right ) e^{2}}{105 \sqrt {c \,x^{2}+a}\, c^{3} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/105/x*e^2*(e*x)^(1/2)/(c*x^2+a)^(1/2)*(63*A*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))*((c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2)*((-c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^
(1/2)*a^2*c-126*A*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))*((c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2)*((-c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2)*a^2*c+25*B*(-a*c)^(1/2
)*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))*((c*x+(-a
*c)^(1/2))/(-a*c)^(1/2))^(1/2)*((-c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2)*a^2+30*B*c^3*x^5+42*A*c^3*x^4-20*B*a*c
^2*x^3+42*A*a*c^2*x^2-50*B*a^2*c*x)/c^3

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [C]  time = 20.01, size = 94, normalized size = 0.26 \[ \frac {A e^{\frac {5}{2}} x^{\frac {7}{2}} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {c x^{2} e^{i \pi }}{a}} \right )}}{2 \sqrt {a} \Gamma \left (\frac {11}{4}\right )} + \frac {B e^{\frac {5}{2}} x^{\frac {9}{2}} \Gamma \left (\frac {9}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {9}{4} \\ \frac {13}{4} \end {matrix}\middle | {\frac {c x^{2} e^{i \pi }}{a}} \right )}}{2 \sqrt {a} \Gamma \left (\frac {13}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

A*e**(5/2)*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)) +
B*e**(5/2)*x**(9/2)*gamma(9/4)*hyper((1/2, 9/4), (13/4,), c*x**2*exp_polar(I*pi)/a)/(2*sqrt(a)*gamma(13/4))

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