3.477 \(\int \frac{(e x)^{7/2} (A+B x)}{(a+c x^2)^{5/2}} \, dx\)

Optimal. Leaf size=339 \[ \frac{e^4 \sqrt{x} \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} \left (21 \sqrt{a} B+5 A \sqrt{c}\right ) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right ),\frac{1}{2}\right )}{12 \sqrt [4]{a} c^{11/4} \sqrt{e x} \sqrt{a+c x^2}}-\frac{e^3 \sqrt{e x} (5 A+7 B x)}{6 c^2 \sqrt{a+c x^2}}-\frac{e (e x)^{5/2} (A+B x)}{3 c \left (a+c x^2\right )^{3/2}}+\frac{7 B e^4 x \sqrt{a+c x^2}}{2 c^{5/2} \sqrt{e x} \left (\sqrt{a}+\sqrt{c} x\right )}-\frac{7 \sqrt [4]{a} B e^4 \sqrt{x} \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{2 c^{11/4} \sqrt{e x} \sqrt{a+c x^2}} \]

[Out]

-(e*(e*x)^(5/2)*(A + B*x))/(3*c*(a + c*x^2)^(3/2)) - (e^3*Sqrt[e*x]*(5*A + 7*B*x))/(6*c^2*Sqrt[a + c*x^2]) + (
7*B*e^4*x*Sqrt[a + c*x^2])/(2*c^(5/2)*Sqrt[e*x]*(Sqrt[a] + Sqrt[c]*x)) - (7*a^(1/4)*B*e^4*Sqrt[x]*(Sqrt[a] + S
qrt[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]) + ((21*Sqrt[a]*B + 5*A*Sqrt[c])*e^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])/(12*a^(1/4)*c^(11/4)*Sqrt
[e*x]*Sqrt[a + c*x^2])

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Rubi [A]  time = 0.33838, antiderivative size = 339, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {819, 842, 840, 1198, 220, 1196} \[ -\frac{e^3 \sqrt{e x} (5 A+7 B x)}{6 c^2 \sqrt{a+c x^2}}+\frac{e^4 \sqrt{x} \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} \left (21 \sqrt{a} B+5 A \sqrt{c}\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{12 \sqrt [4]{a} c^{11/4} \sqrt{e x} \sqrt{a+c x^2}}-\frac{e (e x)^{5/2} (A+B x)}{3 c \left (a+c x^2\right )^{3/2}}+\frac{7 B e^4 x \sqrt{a+c x^2}}{2 c^{5/2} \sqrt{e x} \left (\sqrt{a}+\sqrt{c} x\right )}-\frac{7 \sqrt [4]{a} B e^4 \sqrt{x} \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{2 c^{11/4} \sqrt{e x} \sqrt{a+c x^2}} \]

Antiderivative was successfully verified.

[In]

Int[((e*x)^(7/2)*(A + B*x))/(a + c*x^2)^(5/2),x]

[Out]

-(e*(e*x)^(5/2)*(A + B*x))/(3*c*(a + c*x^2)^(3/2)) - (e^3*Sqrt[e*x]*(5*A + 7*B*x))/(6*c^2*Sqrt[a + c*x^2]) + (
7*B*e^4*x*Sqrt[a + c*x^2])/(2*c^(5/2)*Sqrt[e*x]*(Sqrt[a] + Sqrt[c]*x)) - (7*a^(1/4)*B*e^4*Sqrt[x]*(Sqrt[a] + S
qrt[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]) + ((21*Sqrt[a]*B + 5*A*Sqrt[c])*e^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])/(12*a^(1/4)*c^(11/4)*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 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)^{7/2} (A+B x)}{\left (a+c x^2\right )^{5/2}} \, dx &=-\frac{e (e x)^{5/2} (A+B x)}{3 c \left (a+c x^2\right )^{3/2}}+\frac{\int \frac{(e x)^{3/2} \left (\frac{5}{2} a A e^2+\frac{7}{2} a B e^2 x\right )}{\left (a+c x^2\right )^{3/2}} \, dx}{3 a c}\\ &=-\frac{e (e x)^{5/2} (A+B x)}{3 c \left (a+c x^2\right )^{3/2}}-\frac{e^3 \sqrt{e x} (5 A+7 B x)}{6 c^2 \sqrt{a+c x^2}}+\frac{\int \frac{\frac{5}{4} a^2 A e^4+\frac{21}{4} a^2 B e^4 x}{\sqrt{e x} \sqrt{a+c x^2}} \, dx}{3 a^2 c^2}\\ &=-\frac{e (e x)^{5/2} (A+B x)}{3 c \left (a+c x^2\right )^{3/2}}-\frac{e^3 \sqrt{e x} (5 A+7 B x)}{6 c^2 \sqrt{a+c x^2}}+\frac{\sqrt{x} \int \frac{\frac{5}{4} a^2 A e^4+\frac{21}{4} a^2 B e^4 x}{\sqrt{x} \sqrt{a+c x^2}} \, dx}{3 a^2 c^2 \sqrt{e x}}\\ &=-\frac{e (e x)^{5/2} (A+B x)}{3 c \left (a+c x^2\right )^{3/2}}-\frac{e^3 \sqrt{e x} (5 A+7 B x)}{6 c^2 \sqrt{a+c x^2}}+\frac{\left (2 \sqrt{x}\right ) \operatorname{Subst}\left (\int \frac{\frac{5}{4} a^2 A e^4+\frac{21}{4} a^2 B e^4 x^2}{\sqrt{a+c x^4}} \, dx,x,\sqrt{x}\right )}{3 a^2 c^2 \sqrt{e x}}\\ &=-\frac{e (e x)^{5/2} (A+B x)}{3 c \left (a+c x^2\right )^{3/2}}-\frac{e^3 \sqrt{e x} (5 A+7 B x)}{6 c^2 \sqrt{a+c x^2}}-\frac{\left (7 \sqrt{a} B e^4 \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{\left (2 \left (\frac{21}{4} a^2 B e^4+\frac{5}{4} a^{3/2} A \sqrt{c} e^4\right ) \sqrt{x}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+c x^4}} \, dx,x,\sqrt{x}\right )}{3 a^{3/2} c^{5/2} \sqrt{e x}}\\ &=-\frac{e (e x)^{5/2} (A+B x)}{3 c \left (a+c x^2\right )^{3/2}}-\frac{e^3 \sqrt{e x} (5 A+7 B x)}{6 c^2 \sqrt{a+c x^2}}+\frac{7 B e^4 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} B e^4 \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{\left (21 \sqrt{a} B+5 A \sqrt{c}\right ) e^4 \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 \sqrt [4]{a} c^{11/4} \sqrt{e x} \sqrt{a+c x^2}}\\ \end{align*}

Mathematica [C]  time = 0.119102, size = 139, normalized size = 0.41 \[ \frac{e^3 \sqrt{e x} \left (5 A \left (a+c x^2\right ) \sqrt{\frac{c x^2}{a}+1} \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{5}{4};-\frac{c x^2}{a}\right )-5 a A+7 B x \left (a+c x^2\right ) \sqrt{\frac{c x^2}{a}+1} \, _2F_1\left (\frac{1}{2},\frac{3}{4};\frac{7}{4};-\frac{c x^2}{a}\right )-7 a B x-7 A c x^2-9 B c x^3\right )}{6 c^2 \left (a+c x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[((e*x)^(7/2)*(A + B*x))/(a + c*x^2)^(5/2),x]

[Out]

(e^3*Sqrt[e*x]*(-5*a*A - 7*a*B*x - 7*A*c*x^2 - 9*B*c*x^3 + 5*A*(a + c*x^2)*Sqrt[1 + (c*x^2)/a]*Hypergeometric2
F1[1/4, 1/2, 5/4, -((c*x^2)/a)] + 7*B*x*(a + c*x^2)*Sqrt[1 + (c*x^2)/a]*Hypergeometric2F1[1/2, 3/4, 7/4, -((c*
x^2)/a)]))/(6*c^2*(a + c*x^2)^(3/2))

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Maple [A]  time = 0.034, size = 584, normalized size = 1.7 \begin{align*}{\frac{{e}^{3}}{12\,x{c}^{3}} \left ( 5\,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}{x}^{2}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}^{2}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}^{2}ac+5\,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 ) a-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 ){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 ){a}^{2}-18\,B{c}^{2}{x}^{4}-14\,A{c}^{2}{x}^{3}-14\,aBc{x}^{2}-10\,aAcx \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)^(7/2)*(B*x+A)/(c*x^2+a)^(5/2),x)

[Out]

1/12*(5*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^2*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)*Ell
ipticF(((c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2),1/2*2^(1/2))*x^2*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^2*a*c+5*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))*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))*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)*Ellipti
cE(((c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2),1/2*2^(1/2))*a^2-18*B*c^2*x^4-14*A*c^2*x^3-14*a*B*c*x^2-10*a*A*c*x)
*e^3/x*(e*x)^(1/2)/c^3/(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{7}{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)^(7/2)*(B*x+A)/(c*x^2+a)^(5/2),x, algorithm="maxima")

[Out]

integrate((B*x + A)*(e*x)^(7/2)/(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^{3} x^{4} + A e^{3} x^{3}\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)^(7/2)*(B*x+A)/(c*x^2+a)^(5/2),x, algorithm="fricas")

[Out]

integral((B*e^3*x^4 + A*e^3*x^3)*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)**(7/2)*(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{7}{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)^(7/2)*(B*x+A)/(c*x^2+a)^(5/2),x, algorithm="giac")

[Out]

integrate((B*x + A)*(e*x)^(7/2)/(c*x^2 + a)^(5/2), x)