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

Optimal. Leaf size=402 \[ \frac{\sqrt [4]{c} \sqrt{x} \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} \left (7 \sqrt{a} B-5 A \sqrt{c}\right ) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right ),\frac{1}{2}\right )}{4 a^{13/4} e^2 \sqrt{e x} \sqrt{a+c x^2}}+\frac{9 A+7 B x}{6 a^2 e (e x)^{3/2} \sqrt{a+c x^2}}-\frac{5 A \sqrt{a+c x^2}}{2 a^3 e (e x)^{3/2}}-\frac{7 B \sqrt{a+c x^2}}{2 a^3 e^2 \sqrt{e x}}+\frac{7 B \sqrt{c} x \sqrt{a+c x^2}}{2 a^3 e^2 \sqrt{e x} \left (\sqrt{a}+\sqrt{c} x\right )}-\frac{7 B \sqrt [4]{c} \sqrt{x} \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{2 a^{11/4} e^2 \sqrt{e x} \sqrt{a+c x^2}}+\frac{A+B x}{3 a e (e x)^{3/2} \left (a+c x^2\right )^{3/2}} \]

[Out]

(A + B*x)/(3*a*e*(e*x)^(3/2)*(a + c*x^2)^(3/2)) + (9*A + 7*B*x)/(6*a^2*e*(e*x)^(3/2)*Sqrt[a + c*x^2]) - (5*A*S
qrt[a + c*x^2])/(2*a^3*e*(e*x)^(3/2)) - (7*B*Sqrt[a + c*x^2])/(2*a^3*e^2*Sqrt[e*x]) + (7*B*Sqrt[c]*x*Sqrt[a +
c*x^2])/(2*a^3*e^2*Sqrt[e*x]*(Sqrt[a] + Sqrt[c]*x)) - (7*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])/(2*a^(11/4)*e^2*Sqrt[e*x]*Sq
rt[a + c*x^2]) + ((7*Sqrt[a]*B - 5*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])/(4*a^(13/4)*e^2*Sqrt[e*x]*Sqrt[a + c*x^2]
)

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Rubi [A]  time = 0.498318, antiderivative size = 402, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.292, Rules used = {823, 835, 842, 840, 1198, 220, 1196} \[ \frac{\sqrt [4]{c} \sqrt{x} \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} \left (7 \sqrt{a} B-5 A \sqrt{c}\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{4 a^{13/4} e^2 \sqrt{e x} \sqrt{a+c x^2}}+\frac{9 A+7 B x}{6 a^2 e (e x)^{3/2} \sqrt{a+c x^2}}-\frac{5 A \sqrt{a+c x^2}}{2 a^3 e (e x)^{3/2}}-\frac{7 B \sqrt{a+c x^2}}{2 a^3 e^2 \sqrt{e x}}+\frac{7 B \sqrt{c} x \sqrt{a+c x^2}}{2 a^3 e^2 \sqrt{e x} \left (\sqrt{a}+\sqrt{c} x\right )}-\frac{7 B \sqrt [4]{c} \sqrt{x} \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{2 a^{11/4} e^2 \sqrt{e x} \sqrt{a+c x^2}}+\frac{A+B x}{3 a e (e x)^{3/2} \left (a+c x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(A + B*x)/(3*a*e*(e*x)^(3/2)*(a + c*x^2)^(3/2)) + (9*A + 7*B*x)/(6*a^2*e*(e*x)^(3/2)*Sqrt[a + c*x^2]) - (5*A*S
qrt[a + c*x^2])/(2*a^3*e*(e*x)^(3/2)) - (7*B*Sqrt[a + c*x^2])/(2*a^3*e^2*Sqrt[e*x]) + (7*B*Sqrt[c]*x*Sqrt[a +
c*x^2])/(2*a^3*e^2*Sqrt[e*x]*(Sqrt[a] + Sqrt[c]*x)) - (7*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])/(2*a^(11/4)*e^2*Sqrt[e*x]*Sq
rt[a + c*x^2]) + ((7*Sqrt[a]*B - 5*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])/(4*a^(13/4)*e^2*Sqrt[e*x]*Sqrt[a + c*x^2]
)

Rule 823

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(
m + 1)*(f*a*c*e - a*g*c*d + c*(c*d*f + a*e*g)*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), x] + Di
st[1/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*Simp[f*(c^2*d^2*(2*p + 3) + a*c*e^2*
(m + 2*p + 3)) - a*c*d*e*g*m + c*e*(c*d*f + a*e*g)*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x]
&& NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

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 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 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 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]

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 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]

Rubi steps

\begin{align*} \int \frac{A+B x}{(e x)^{5/2} \left (a+c x^2\right )^{5/2}} \, dx &=\frac{A+B x}{3 a e (e x)^{3/2} \left (a+c x^2\right )^{3/2}}-\frac{\int \frac{-\frac{9}{2} a A c e^2-\frac{7}{2} a B c e^2 x}{(e x)^{5/2} \left (a+c x^2\right )^{3/2}} \, dx}{3 a^2 c e^2}\\ &=\frac{A+B x}{3 a e (e x)^{3/2} \left (a+c x^2\right )^{3/2}}+\frac{9 A+7 B x}{6 a^2 e (e x)^{3/2} \sqrt{a+c x^2}}+\frac{\int \frac{\frac{45}{4} a^2 A c^2 e^4+\frac{21}{4} a^2 B c^2 e^4 x}{(e x)^{5/2} \sqrt{a+c x^2}} \, dx}{3 a^4 c^2 e^4}\\ &=\frac{A+B x}{3 a e (e x)^{3/2} \left (a+c x^2\right )^{3/2}}+\frac{9 A+7 B x}{6 a^2 e (e x)^{3/2} \sqrt{a+c x^2}}-\frac{5 A \sqrt{a+c x^2}}{2 a^3 e (e x)^{3/2}}-\frac{2 \int \frac{-\frac{63}{8} a^3 B c^2 e^5+\frac{45}{8} a^2 A c^3 e^5 x}{(e x)^{3/2} \sqrt{a+c x^2}} \, dx}{9 a^5 c^2 e^6}\\ &=\frac{A+B x}{3 a e (e x)^{3/2} \left (a+c x^2\right )^{3/2}}+\frac{9 A+7 B x}{6 a^2 e (e x)^{3/2} \sqrt{a+c x^2}}-\frac{5 A \sqrt{a+c x^2}}{2 a^3 e (e x)^{3/2}}-\frac{7 B \sqrt{a+c x^2}}{2 a^3 e^2 \sqrt{e x}}+\frac{4 \int \frac{-\frac{45}{16} a^3 A c^3 e^6+\frac{63}{16} a^3 B c^3 e^6 x}{\sqrt{e x} \sqrt{a+c x^2}} \, dx}{9 a^6 c^2 e^8}\\ &=\frac{A+B x}{3 a e (e x)^{3/2} \left (a+c x^2\right )^{3/2}}+\frac{9 A+7 B x}{6 a^2 e (e x)^{3/2} \sqrt{a+c x^2}}-\frac{5 A \sqrt{a+c x^2}}{2 a^3 e (e x)^{3/2}}-\frac{7 B \sqrt{a+c x^2}}{2 a^3 e^2 \sqrt{e x}}+\frac{\left (4 \sqrt{x}\right ) \int \frac{-\frac{45}{16} a^3 A c^3 e^6+\frac{63}{16} a^3 B c^3 e^6 x}{\sqrt{x} \sqrt{a+c x^2}} \, dx}{9 a^6 c^2 e^8 \sqrt{e x}}\\ &=\frac{A+B x}{3 a e (e x)^{3/2} \left (a+c x^2\right )^{3/2}}+\frac{9 A+7 B x}{6 a^2 e (e x)^{3/2} \sqrt{a+c x^2}}-\frac{5 A \sqrt{a+c x^2}}{2 a^3 e (e x)^{3/2}}-\frac{7 B \sqrt{a+c x^2}}{2 a^3 e^2 \sqrt{e x}}+\frac{\left (8 \sqrt{x}\right ) \operatorname{Subst}\left (\int \frac{-\frac{45}{16} a^3 A c^3 e^6+\frac{63}{16} a^3 B c^3 e^6 x^2}{\sqrt{a+c x^4}} \, dx,x,\sqrt{x}\right )}{9 a^6 c^2 e^8 \sqrt{e x}}\\ &=\frac{A+B x}{3 a e (e x)^{3/2} \left (a+c x^2\right )^{3/2}}+\frac{9 A+7 B x}{6 a^2 e (e x)^{3/2} \sqrt{a+c x^2}}-\frac{5 A \sqrt{a+c x^2}}{2 a^3 e (e x)^{3/2}}-\frac{7 B \sqrt{a+c x^2}}{2 a^3 e^2 \sqrt{e x}}-\frac{\left (7 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 )}{2 a^{5/2} e^2 \sqrt{e x}}+\frac{\left (\left (7 \sqrt{a} B-5 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 )}{2 a^3 e^2 \sqrt{e x}}\\ &=\frac{A+B x}{3 a e (e x)^{3/2} \left (a+c x^2\right )^{3/2}}+\frac{9 A+7 B x}{6 a^2 e (e x)^{3/2} \sqrt{a+c x^2}}-\frac{5 A \sqrt{a+c x^2}}{2 a^3 e (e x)^{3/2}}-\frac{7 B \sqrt{a+c x^2}}{2 a^3 e^2 \sqrt{e x}}+\frac{7 B \sqrt{c} x \sqrt{a+c x^2}}{2 a^3 e^2 \sqrt{e x} \left (\sqrt{a}+\sqrt{c} x\right )}-\frac{7 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 )}{2 a^{11/4} e^2 \sqrt{e x} \sqrt{a+c x^2}}+\frac{\left (7 \sqrt{a} B-5 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 )}{4 a^{13/4} e^2 \sqrt{e x} \sqrt{a+c x^2}}\\ \end{align*}

Mathematica [C]  time = 0.0971501, size = 137, normalized size = 0.34 \[ \frac{x \left (-15 A \left (a+c x^2\right ) \sqrt{\frac{c x^2}{a}+1} \, _2F_1\left (-\frac{3}{4},\frac{1}{2};\frac{1}{4};-\frac{c x^2}{a}\right )+11 a A-21 B x \left (a+c x^2\right ) \sqrt{\frac{c x^2}{a}+1} \, _2F_1\left (-\frac{1}{4},\frac{1}{2};\frac{3}{4};-\frac{c x^2}{a}\right )+9 a B x+9 A c x^2+7 B c x^3\right )}{6 a^2 (e x)^{5/2} \left (a+c x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(11*a*A + 9*a*B*x + 9*A*c*x^2 + 7*B*c*x^3 - 15*A*(a + c*x^2)*Sqrt[1 + (c*x^2)/a]*Hypergeometric2F1[-3/4, 1/
2, 1/4, -((c*x^2)/a)] - 21*B*x*(a + c*x^2)*Sqrt[1 + (c*x^2)/a]*Hypergeometric2F1[-1/4, 1/2, 3/4, -((c*x^2)/a)]
))/(6*a^2*(e*x)^(5/2)*(a + c*x^2)^(3/2))

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Maple [A]  time = 0.044, size = 602, normalized size = 1.5 \begin{align*} -{\frac{1}{12\,x{e}^{2}{a}^{3}} \left ( 15\,A\sqrt{-ac}\sqrt{{\frac{cx+\sqrt{-ac}}{\sqrt{-ac}}}}\sqrt{2}\sqrt{{\frac{-cx+\sqrt{-ac}}{\sqrt{-ac}}}}\sqrt{-{\frac{cx}{\sqrt{-ac}}}}{\it EllipticF} \left ( \sqrt{{\frac{cx+\sqrt{-ac}}{\sqrt{-ac}}}},1/2\,\sqrt{2} \right ){x}^{3}c+21\,B\sqrt{{\frac{cx+\sqrt{-ac}}{\sqrt{-ac}}}}\sqrt{2}\sqrt{{\frac{-cx+\sqrt{-ac}}{\sqrt{-ac}}}}\sqrt{-{\frac{cx}{\sqrt{-ac}}}}{\it EllipticF} \left ( \sqrt{{\frac{cx+\sqrt{-ac}}{\sqrt{-ac}}}},1/2\,\sqrt{2} \right ){x}^{3}ac-42\,B\sqrt{{\frac{cx+\sqrt{-ac}}{\sqrt{-ac}}}}\sqrt{2}\sqrt{{\frac{-cx+\sqrt{-ac}}{\sqrt{-ac}}}}\sqrt{-{\frac{cx}{\sqrt{-ac}}}}{\it EllipticE} \left ( \sqrt{{\frac{cx+\sqrt{-ac}}{\sqrt{-ac}}}},1/2\,\sqrt{2} \right ){x}^{3}ac+15\,A\sqrt{{\frac{cx+\sqrt{-ac}}{\sqrt{-ac}}}}\sqrt{2}\sqrt{{\frac{-cx+\sqrt{-ac}}{\sqrt{-ac}}}}\sqrt{-{\frac{cx}{\sqrt{-ac}}}}{\it EllipticF} \left ( \sqrt{{\frac{cx+\sqrt{-ac}}{\sqrt{-ac}}}},1/2\,\sqrt{2} \right ) \sqrt{-ac}xa+21\,B\sqrt{{\frac{cx+\sqrt{-ac}}{\sqrt{-ac}}}}\sqrt{2}\sqrt{{\frac{-cx+\sqrt{-ac}}{\sqrt{-ac}}}}\sqrt{-{\frac{cx}{\sqrt{-ac}}}}{\it EllipticF} \left ( \sqrt{{\frac{cx+\sqrt{-ac}}{\sqrt{-ac}}}},1/2\,\sqrt{2} \right ) x{a}^{2}-42\,B\sqrt{{\frac{cx+\sqrt{-ac}}{\sqrt{-ac}}}}\sqrt{2}\sqrt{{\frac{-cx+\sqrt{-ac}}{\sqrt{-ac}}}}\sqrt{-{\frac{cx}{\sqrt{-ac}}}}{\it EllipticE} \left ( \sqrt{{\frac{cx+\sqrt{-ac}}{\sqrt{-ac}}}},1/2\,\sqrt{2} \right ) x{a}^{2}+42\,B{c}^{2}{x}^{5}+30\,A{c}^{2}{x}^{4}+70\,aBc{x}^{3}+42\,aAc{x}^{2}+24\,{a}^{2}Bx+8\,A{a}^{2} \right ){\frac{1}{\sqrt{ex}}} \left ( c{x}^{2}+a \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(15*A*(-a*c)^(1/2)*((c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2)*2^(1/2)*((-c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1
/2)*(-x*c/(-a*c)^(1/2))^(1/2)*EllipticF(((c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2),1/2*2^(1/2))*x^3*c+21*B*((c*x+
(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2)*2^(1/2)*((-c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2)*(-x*c/(-a*c)^(1/2))^(1/2)*E
llipticF(((c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2),1/2*2^(1/2))*x^3*a*c-42*B*((c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(
1/2)*2^(1/2)*((-c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2)*(-x*c/(-a*c)^(1/2))^(1/2)*EllipticE(((c*x+(-a*c)^(1/2))/
(-a*c)^(1/2))^(1/2),1/2*2^(1/2))*x^3*a*c+15*A*((c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2)*2^(1/2)*((-c*x+(-a*c)^(1
/2))/(-a*c)^(1/2))^(1/2)*(-x*c/(-a*c)^(1/2))^(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*a+21*B*((c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2)*2^(1/2)*((-c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(
1/2)*(-x*c/(-a*c)^(1/2))^(1/2)*EllipticF(((c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2),1/2*2^(1/2))*x*a^2-42*B*((c*x
+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2)*2^(1/2)*((-c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2)*(-x*c/(-a*c)^(1/2))^(1/2)*
EllipticE(((c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2),1/2*2^(1/2))*x*a^2+42*B*c^2*x^5+30*A*c^2*x^4+70*a*B*c*x^3+42
*a*A*c*x^2+24*a^2*B*x+8*A*a^2)/x/e^2/(e*x)^(1/2)/a^3/(c*x^2+a)^(3/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{B x + A}{{\left (c x^{2} + a\right )}^{\frac{5}{2}} \left (e x\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{c x^{2} + a}{\left (B x + A\right )} \sqrt{e x}}{c^{3} e^{3} x^{9} + 3 \, a c^{2} e^{3} x^{7} + 3 \, a^{2} c e^{3} x^{5} + a^{3} e^{3} x^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(c*x^2 + a)*(B*x + A)*sqrt(e*x)/(c^3*e^3*x^9 + 3*a*c^2*e^3*x^7 + 3*a^2*c*e^3*x^5 + a^3*e^3*x^3),
x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{B x + A}{{\left (c x^{2} + a\right )}^{\frac{5}{2}} \left (e x\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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