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

Optimal. Leaf size=342 \[ \frac{e \sqrt{x} \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} \left (\sqrt{a} B-3 A \sqrt{c}\right ) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right ),\frac{1}{2}\right )}{12 a^{7/4} c^{5/4} \sqrt{e x} \sqrt{a+c x^2}}+\frac{\sqrt{e x} (a B+3 A c x)}{6 a^2 c \sqrt{a+c x^2}}+\frac{A e \sqrt{x} \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{2 a^{7/4} c^{3/4} \sqrt{e x} \sqrt{a+c x^2}}-\frac{A e x \sqrt{a+c x^2}}{2 a^2 \sqrt{c} \sqrt{e x} \left (\sqrt{a}+\sqrt{c} x\right )}-\frac{\sqrt{e x} (a B-A c x)}{3 a c \left (a+c x^2\right )^{3/2}} \]

[Out]

-(Sqrt[e*x]*(a*B - A*c*x))/(3*a*c*(a + c*x^2)^(3/2)) + (Sqrt[e*x]*(a*B + 3*A*c*x))/(6*a^2*c*Sqrt[a + c*x^2]) -
(A*e*x*Sqrt[a + c*x^2])/(2*a^2*Sqrt[c]*Sqrt[e*x]*(Sqrt[a] + Sqrt[c]*x)) + (A*e*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^(7/4)*c^(3
/4)*Sqrt[e*x]*Sqrt[a + c*x^2]) + ((Sqrt[a]*B - 3*A*Sqrt[c])*e*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])/(12*a^(7/4)*c^(5/4)*Sqrt[e*x]*Sqr
t[a + c*x^2])

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Rubi [A]  time = 0.34599, antiderivative size = 342, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.292, Rules used = {821, 823, 842, 840, 1198, 220, 1196} \[ \frac{e \sqrt{x} \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} \left (\sqrt{a} B-3 A \sqrt{c}\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{12 a^{7/4} c^{5/4} \sqrt{e x} \sqrt{a+c x^2}}+\frac{\sqrt{e x} (a B+3 A c x)}{6 a^2 c \sqrt{a+c x^2}}+\frac{A e \sqrt{x} \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{2 a^{7/4} c^{3/4} \sqrt{e x} \sqrt{a+c x^2}}-\frac{A e x \sqrt{a+c x^2}}{2 a^2 \sqrt{c} \sqrt{e x} \left (\sqrt{a}+\sqrt{c} x\right )}-\frac{\sqrt{e x} (a B-A c x)}{3 a c \left (a+c x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[e*x]*(a*B - A*c*x))/(3*a*c*(a + c*x^2)^(3/2)) + (Sqrt[e*x]*(a*B + 3*A*c*x))/(6*a^2*c*Sqrt[a + c*x^2]) -
(A*e*x*Sqrt[a + c*x^2])/(2*a^2*Sqrt[c]*Sqrt[e*x]*(Sqrt[a] + Sqrt[c]*x)) + (A*e*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^(7/4)*c^(3
/4)*Sqrt[e*x]*Sqrt[a + c*x^2]) + ((Sqrt[a]*B - 3*A*Sqrt[c])*e*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])/(12*a^(7/4)*c^(5/4)*Sqrt[e*x]*Sqr
t[a + c*x^2])

Rule 821

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

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 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{\sqrt{e x} (A+B x)}{\left (a+c x^2\right )^{5/2}} \, dx &=-\frac{\sqrt{e x} (a B-A c x)}{3 a c \left (a+c x^2\right )^{3/2}}+\frac{\int \frac{\frac{a B e}{2}+\frac{3}{2} A c e x}{\sqrt{e x} \left (a+c x^2\right )^{3/2}} \, dx}{3 a c}\\ &=-\frac{\sqrt{e x} (a B-A c x)}{3 a c \left (a+c x^2\right )^{3/2}}+\frac{\sqrt{e x} (a B+3 A c x)}{6 a^2 c \sqrt{a+c x^2}}-\frac{\int \frac{-\frac{1}{4} a^2 B c e^3+\frac{3}{4} a A c^2 e^3 x}{\sqrt{e x} \sqrt{a+c x^2}} \, dx}{3 a^3 c^2 e^2}\\ &=-\frac{\sqrt{e x} (a B-A c x)}{3 a c \left (a+c x^2\right )^{3/2}}+\frac{\sqrt{e x} (a B+3 A c x)}{6 a^2 c \sqrt{a+c x^2}}-\frac{\sqrt{x} \int \frac{-\frac{1}{4} a^2 B c e^3+\frac{3}{4} a A c^2 e^3 x}{\sqrt{x} \sqrt{a+c x^2}} \, dx}{3 a^3 c^2 e^2 \sqrt{e x}}\\ &=-\frac{\sqrt{e x} (a B-A c x)}{3 a c \left (a+c x^2\right )^{3/2}}+\frac{\sqrt{e x} (a B+3 A c x)}{6 a^2 c \sqrt{a+c x^2}}-\frac{\left (2 \sqrt{x}\right ) \operatorname{Subst}\left (\int \frac{-\frac{1}{4} a^2 B c e^3+\frac{3}{4} a A c^2 e^3 x^2}{\sqrt{a+c x^4}} \, dx,x,\sqrt{x}\right )}{3 a^3 c^2 e^2 \sqrt{e x}}\\ &=-\frac{\sqrt{e x} (a B-A c x)}{3 a c \left (a+c x^2\right )^{3/2}}+\frac{\sqrt{e x} (a B+3 A c x)}{6 a^2 c \sqrt{a+c x^2}}+\frac{\left (\left (\sqrt{a} B-3 A \sqrt{c}\right ) e \sqrt{x}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+c x^4}} \, dx,x,\sqrt{x}\right )}{6 a^{3/2} c \sqrt{e x}}+\frac{\left (A e \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^{3/2} \sqrt{c} \sqrt{e x}}\\ &=-\frac{\sqrt{e x} (a B-A c x)}{3 a c \left (a+c x^2\right )^{3/2}}+\frac{\sqrt{e x} (a B+3 A c x)}{6 a^2 c \sqrt{a+c x^2}}-\frac{A e x \sqrt{a+c x^2}}{2 a^2 \sqrt{c} \sqrt{e x} \left (\sqrt{a}+\sqrt{c} x\right )}+\frac{A e \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^{7/4} c^{3/4} \sqrt{e x} \sqrt{a+c x^2}}+\frac{\left (\sqrt{a} B-3 A \sqrt{c}\right ) e \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 )}{12 a^{7/4} c^{5/4} \sqrt{e x} \sqrt{a+c x^2}}\\ \end{align*}

Mathematica [C]  time = 0.134459, size = 145, normalized size = 0.42 \[ \frac{\sqrt{e x} \left (-a^2 B-A c x \left (a+c x^2\right ) \sqrt{\frac{c x^2}{a}+1} \, _2F_1\left (\frac{1}{2},\frac{3}{4};\frac{7}{4};-\frac{c x^2}{a}\right )+5 a A c x+a B \left (a+c x^2\right ) \sqrt{\frac{c x^2}{a}+1} \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{5}{4};-\frac{c x^2}{a}\right )+a B c x^2+3 A c^2 x^3\right )}{6 a^2 c \left (a+c x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[e*x]*(-(a^2*B) + 5*a*A*c*x + a*B*c*x^2 + 3*A*c^2*x^3 + a*B*(a + c*x^2)*Sqrt[1 + (c*x^2)/a]*Hypergeometri
c2F1[1/4, 1/2, 5/4, -((c*x^2)/a)] - A*c*x*(a + c*x^2)*Sqrt[1 + (c*x^2)/a]*Hypergeometric2F1[1/2, 3/4, 7/4, -((
c*x^2)/a)]))/(6*a^2*c*(a + c*x^2)^(3/2))

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Maple [A]  time = 0.017, size = 596, normalized size = 1.7 \begin{align*}{\frac{1}{12\,{a}^{2}x{c}^{2}} \left ( 3\,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 ){x}^{2}a{c}^{2}-6\,A\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}^{2}a{c}^{2}+B\sqrt{{ \left ( cx+\sqrt{-ac} \right ){\frac{1}{\sqrt{-ac}}}}}\sqrt{2}\sqrt{{ \left ( -cx+\sqrt{-ac} \right ){\frac{1}{\sqrt{-ac}}}}}\sqrt{-{cx{\frac{1}{\sqrt{-ac}}}}}{\it EllipticF} \left ( \sqrt{{ \left ( cx+\sqrt{-ac} \right ){\frac{1}{\sqrt{-ac}}}}},{\frac{\sqrt{2}}{2}} \right ) \sqrt{-ac}{x}^{2}ac+3\,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 ){a}^{2}c-6\,A\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 ){a}^{2}c+B\sqrt{{ \left ( cx+\sqrt{-ac} \right ){\frac{1}{\sqrt{-ac}}}}}\sqrt{2}\sqrt{{ \left ( -cx+\sqrt{-ac} \right ){\frac{1}{\sqrt{-ac}}}}}\sqrt{-{cx{\frac{1}{\sqrt{-ac}}}}}{\it EllipticF} \left ( \sqrt{{ \left ( cx+\sqrt{-ac} \right ){\frac{1}{\sqrt{-ac}}}}},{\frac{\sqrt{2}}{2}} \right ) \sqrt{-ac}{a}^{2}+6\,A{c}^{3}{x}^{4}+2\,aB{c}^{2}{x}^{3}+10\,aA{c}^{2}{x}^{2}-2\,{a}^{2}Bcx \right ) \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((e*x)^(1/2)*(B*x+A)/(c*x^2+a)^(5/2),x)

[Out]

1/12*(3*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))*x^2*a*c^2-6*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)*EllipticE(((c
*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2),1/2*2^(1/2))*x^2*a*c^2+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))*(-a*c)^(1/2)*x^2*a*c+3*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^2*c-6*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)*EllipticE(((c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2),1/2*2^(1/2))*a^2*c+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))*(-a*c)^(1/2)*a^2+6*A*c^3*x^4+2*a*B*c^2*x^3+10*a*A*c^2*x^2-2*a^2
*B*c*x)*(e*x)^(1/2)/a^2/x/c^2/(c*x^2+a)^(3/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((B*x + A)*sqrt(e*x)/(c*x^2 + a)^(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} x^{6} + 3 \, a c^{2} x^{4} + 3 \, a^{2} c x^{2} + a^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(c*x^2 + a)*(B*x + A)*sqrt(e*x)/(c^3*x^6 + 3*a*c^2*x^4 + 3*a^2*c*x^2 + a^3), x)

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Sympy [C]  time = 177.461, size = 94, normalized size = 0.27 \begin{align*} \frac{A \sqrt{e} x^{\frac{3}{2}} \Gamma \left (\frac{3}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{3}{4}, \frac{5}{2} \\ \frac{7}{4} \end{matrix}\middle |{\frac{c x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac{5}{2}} \Gamma \left (\frac{7}{4}\right )} + \frac{B \sqrt{e} x^{\frac{5}{2}} \Gamma \left (\frac{5}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{5}{4}, \frac{5}{2} \\ \frac{9}{4} \end{matrix}\middle |{\frac{c x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac{5}{2}} \Gamma \left (\frac{9}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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