3.476 \(\int \frac{(e x)^{9/2} (A+B x)}{(a+c x^2)^{5/2}} \, dx\)

Optimal. Leaf size=368 \[ -\frac{\sqrt [4]{a} e^5 \sqrt{x} \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} \left (5 \sqrt{a} B-7 A \sqrt{c}\right ) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right ),\frac{1}{2}\right )}{4 c^{13/4} \sqrt{e x} \sqrt{a+c x^2}}-\frac{e^3 (e x)^{3/2} (7 A+9 B x)}{6 c^2 \sqrt{a+c x^2}}-\frac{e (e x)^{7/2} (A+B x)}{3 c \left (a+c x^2\right )^{3/2}}+\frac{7 A e^5 x \sqrt{a+c x^2}}{2 c^{5/2} \sqrt{e x} \left (\sqrt{a}+\sqrt{c} x\right )}-\frac{7 \sqrt [4]{a} A e^5 \sqrt{x} \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{2 c^{11/4} \sqrt{e x} \sqrt{a+c x^2}}+\frac{5 B e^4 \sqrt{e x} \sqrt{a+c x^2}}{2 c^3} \]

[Out]

-(e*(e*x)^(7/2)*(A + B*x))/(3*c*(a + c*x^2)^(3/2)) - (e^3*(e*x)^(3/2)*(7*A + 9*B*x))/(6*c^2*Sqrt[a + c*x^2]) +
(5*B*e^4*Sqrt[e*x]*Sqrt[a + c*x^2])/(2*c^3) + (7*A*e^5*x*Sqrt[a + c*x^2])/(2*c^(5/2)*Sqrt[e*x]*(Sqrt[a] + Sqr
t[c]*x)) - (7*a^(1/4)*A*e^5*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*c^(11/4)*Sqrt[e*x]*Sqrt[a + c*x^2]) - (a^(1/4)*(5*Sqrt[a]*B - 7*
A*Sqrt[c])*e^5*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*c^(13/4)*Sqrt[e*x]*Sqrt[a + c*x^2])

________________________________________________________________________________________

Rubi [A]  time = 0.422963, antiderivative size = 368, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.292, Rules used = {819, 833, 842, 840, 1198, 220, 1196} \[ -\frac{e^3 (e x)^{3/2} (7 A+9 B x)}{6 c^2 \sqrt{a+c x^2}}-\frac{\sqrt [4]{a} e^5 \sqrt{x} \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} \left (5 \sqrt{a} B-7 A \sqrt{c}\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{4 c^{13/4} \sqrt{e x} \sqrt{a+c x^2}}-\frac{e (e x)^{7/2} (A+B x)}{3 c \left (a+c x^2\right )^{3/2}}+\frac{7 A e^5 x \sqrt{a+c x^2}}{2 c^{5/2} \sqrt{e x} \left (\sqrt{a}+\sqrt{c} x\right )}-\frac{7 \sqrt [4]{a} A e^5 \sqrt{x} \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{2 c^{11/4} \sqrt{e x} \sqrt{a+c x^2}}+\frac{5 B e^4 \sqrt{e x} \sqrt{a+c x^2}}{2 c^3} \]

Antiderivative was successfully verified.

[In]

Int[((e*x)^(9/2)*(A + B*x))/(a + c*x^2)^(5/2),x]

[Out]

-(e*(e*x)^(7/2)*(A + B*x))/(3*c*(a + c*x^2)^(3/2)) - (e^3*(e*x)^(3/2)*(7*A + 9*B*x))/(6*c^2*Sqrt[a + c*x^2]) +
(5*B*e^4*Sqrt[e*x]*Sqrt[a + c*x^2])/(2*c^3) + (7*A*e^5*x*Sqrt[a + c*x^2])/(2*c^(5/2)*Sqrt[e*x]*(Sqrt[a] + Sqr
t[c]*x)) - (7*a^(1/4)*A*e^5*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*c^(11/4)*Sqrt[e*x]*Sqrt[a + c*x^2]) - (a^(1/4)*(5*Sqrt[a]*B - 7*
A*Sqrt[c])*e^5*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*c^(13/4)*Sqrt[e*x]*Sqrt[a + c*x^2])

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 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 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{(e x)^{9/2} (A+B x)}{\left (a+c x^2\right )^{5/2}} \, dx &=-\frac{e (e x)^{7/2} (A+B x)}{3 c \left (a+c x^2\right )^{3/2}}+\frac{\int \frac{(e x)^{5/2} \left (\frac{7}{2} a A e^2+\frac{9}{2} a B e^2 x\right )}{\left (a+c x^2\right )^{3/2}} \, dx}{3 a c}\\ &=-\frac{e (e x)^{7/2} (A+B x)}{3 c \left (a+c x^2\right )^{3/2}}-\frac{e^3 (e x)^{3/2} (7 A+9 B x)}{6 c^2 \sqrt{a+c x^2}}+\frac{\int \frac{\sqrt{e x} \left (\frac{21}{4} a^2 A e^4+\frac{45}{4} a^2 B e^4 x\right )}{\sqrt{a+c x^2}} \, dx}{3 a^2 c^2}\\ &=-\frac{e (e x)^{7/2} (A+B x)}{3 c \left (a+c x^2\right )^{3/2}}-\frac{e^3 (e x)^{3/2} (7 A+9 B x)}{6 c^2 \sqrt{a+c x^2}}+\frac{5 B e^4 \sqrt{e x} \sqrt{a+c x^2}}{2 c^3}+\frac{2 \int \frac{-\frac{45}{8} a^3 B e^5+\frac{63}{8} a^2 A c e^5 x}{\sqrt{e x} \sqrt{a+c x^2}} \, dx}{9 a^2 c^3}\\ &=-\frac{e (e x)^{7/2} (A+B x)}{3 c \left (a+c x^2\right )^{3/2}}-\frac{e^3 (e x)^{3/2} (7 A+9 B x)}{6 c^2 \sqrt{a+c x^2}}+\frac{5 B e^4 \sqrt{e x} \sqrt{a+c x^2}}{2 c^3}+\frac{\left (2 \sqrt{x}\right ) \int \frac{-\frac{45}{8} a^3 B e^5+\frac{63}{8} a^2 A c e^5 x}{\sqrt{x} \sqrt{a+c x^2}} \, dx}{9 a^2 c^3 \sqrt{e x}}\\ &=-\frac{e (e x)^{7/2} (A+B x)}{3 c \left (a+c x^2\right )^{3/2}}-\frac{e^3 (e x)^{3/2} (7 A+9 B x)}{6 c^2 \sqrt{a+c x^2}}+\frac{5 B e^4 \sqrt{e x} \sqrt{a+c x^2}}{2 c^3}+\frac{\left (4 \sqrt{x}\right ) \operatorname{Subst}\left (\int \frac{-\frac{45}{8} a^3 B e^5+\frac{63}{8} a^2 A c e^5 x^2}{\sqrt{a+c x^4}} \, dx,x,\sqrt{x}\right )}{9 a^2 c^3 \sqrt{e x}}\\ &=-\frac{e (e x)^{7/2} (A+B x)}{3 c \left (a+c x^2\right )^{3/2}}-\frac{e^3 (e x)^{3/2} (7 A+9 B x)}{6 c^2 \sqrt{a+c x^2}}+\frac{5 B e^4 \sqrt{e x} \sqrt{a+c x^2}}{2 c^3}-\frac{\left (\sqrt{a} \left (5 \sqrt{a} B-7 A \sqrt{c}\right ) e^5 \sqrt{x}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+c x^4}} \, dx,x,\sqrt{x}\right )}{2 c^3 \sqrt{e x}}-\frac{\left (7 \sqrt{a} A e^5 \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 c^{5/2} \sqrt{e x}}\\ &=-\frac{e (e x)^{7/2} (A+B x)}{3 c \left (a+c x^2\right )^{3/2}}-\frac{e^3 (e x)^{3/2} (7 A+9 B x)}{6 c^2 \sqrt{a+c x^2}}+\frac{5 B e^4 \sqrt{e x} \sqrt{a+c x^2}}{2 c^3}+\frac{7 A e^5 x \sqrt{a+c x^2}}{2 c^{5/2} \sqrt{e x} \left (\sqrt{a}+\sqrt{c} x\right )}-\frac{7 \sqrt [4]{a} A e^5 \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 c^{11/4} \sqrt{e x} \sqrt{a+c x^2}}-\frac{\sqrt [4]{a} \left (5 \sqrt{a} B-7 A \sqrt{c}\right ) e^5 \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 c^{13/4} \sqrt{e x} \sqrt{a+c x^2}}\\ \end{align*}

Mathematica [C]  time = 0.129381, size = 156, normalized size = 0.42 \[ \frac{e^4 \sqrt{e x} \left (15 a^2 B+7 A c x \left (a+c x^2\right ) \sqrt{\frac{c x^2}{a}+1} \, _2F_1\left (\frac{1}{2},\frac{3}{4};\frac{7}{4};-\frac{c x^2}{a}\right )-7 a A c x-15 a B \left (a+c x^2\right ) \sqrt{\frac{c x^2}{a}+1} \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{5}{4};-\frac{c x^2}{a}\right )+21 a B c x^2-9 A c^2 x^3+4 B c^2 x^4\right )}{6 c^3 \left (a+c x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[((e*x)^(9/2)*(A + B*x))/(a + c*x^2)^(5/2),x]

[Out]

(e^4*Sqrt[e*x]*(15*a^2*B - 7*a*A*c*x + 21*a*B*c*x^2 - 9*A*c^2*x^3 + 4*B*c^2*x^4 - 15*a*B*(a + c*x^2)*Sqrt[1 +
(c*x^2)/a]*Hypergeometric2F1[1/4, 1/2, 5/4, -((c*x^2)/a)] + 7*A*c*x*(a + c*x^2)*Sqrt[1 + (c*x^2)/a]*Hypergeome
tric2F1[1/2, 3/4, 7/4, -((c*x^2)/a)]))/(6*c^3*(a + c*x^2)^(3/2))

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Maple [A]  time = 0.022, size = 607, normalized size = 1.7 \begin{align*} -{\frac{{e}^{4}}{12\,x{c}^{4}} \left ( 21\,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}-42\,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}+15\,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 ) \sqrt{-ac}{x}^{2}ac+21\,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-42\,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+15\,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 ) \sqrt{-ac}{a}^{2}-8\,B{c}^{3}{x}^{5}+18\,A{c}^{3}{x}^{4}-42\,aB{c}^{2}{x}^{3}+14\,aA{c}^{2}{x}^{2}-30\,{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)^(9/2)*(B*x+A)/(c*x^2+a)^(5/2),x)

[Out]

-1/12*(21*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^2-42*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+15*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+21*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-42*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+15*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)*Ellip
ticF(((c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2),1/2*2^(1/2))*(-a*c)^(1/2)*a^2-8*B*c^3*x^5+18*A*c^3*x^4-42*a*B*c^2
*x^3+14*a*A*c^2*x^2-30*a^2*B*c*x)*e^4/x*(e*x)^(1/2)/c^4/(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 )} \left (e x\right )^{\frac{9}{2}}}{{\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)^(9/2)*(B*x+A)/(c*x^2+a)^(5/2),x, algorithm="maxima")

[Out]

integrate((B*x + A)*(e*x)^(9/2)/(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{{\left (B e^{4} x^{5} + A e^{4} x^{4}\right )} \sqrt{c x^{2} + a} \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)^(9/2)*(B*x+A)/(c*x^2+a)^(5/2),x, algorithm="fricas")

[Out]

integral((B*e^4*x^5 + A*e^4*x^4)*sqrt(c*x^2 + 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 [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((e*x)**(9/2)*(B*x+A)/(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{{\left (B x + A\right )} \left (e x\right )^{\frac{9}{2}}}{{\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)^(9/2)*(B*x+A)/(c*x^2+a)^(5/2),x, algorithm="giac")

[Out]

integrate((B*x + A)*(e*x)^(9/2)/(c*x^2 + a)^(5/2), x)