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

Optimal. Leaf size=432 \[ -\frac {c^{3/4} \sqrt {x} \left (\sqrt {a}+\sqrt {c} x\right ) \sqrt {\frac {a+c x^2}{\left (\sqrt {a}+\sqrt {c} x\right )^2}} \left (25 \sqrt {a} B+77 A \sqrt {c}\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{20 a^{15/4} e^3 \sqrt {e x} \sqrt {a+c x^2}}+\frac {77 A c^{5/4} \sqrt {x} \left (\sqrt {a}+\sqrt {c} x\right ) \sqrt {\frac {a+c x^2}{\left (\sqrt {a}+\sqrt {c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{10 a^{15/4} e^3 \sqrt {e x} \sqrt {a+c x^2}}-\frac {77 A c^{3/2} x \sqrt {a+c x^2}}{10 a^4 e^3 \sqrt {e x} \left (\sqrt {a}+\sqrt {c} x\right )}+\frac {77 A c \sqrt {a+c x^2}}{10 a^4 e^3 \sqrt {e x}}-\frac {77 A \sqrt {a+c x^2}}{30 a^3 e (e x)^{5/2}}-\frac {5 B \sqrt {a+c x^2}}{2 a^3 e^2 (e x)^{3/2}}+\frac {11 A+9 B x}{6 a^2 e (e x)^{5/2} \sqrt {a+c x^2}}+\frac {A+B x}{3 a e (e x)^{5/2} \left (a+c x^2\right )^{3/2}} \]

[Out]

1/3*(B*x+A)/a/e/(e*x)^(5/2)/(c*x^2+a)^(3/2)+1/6*(9*B*x+11*A)/a^2/e/(e*x)^(5/2)/(c*x^2+a)^(1/2)-77/30*A*(c*x^2+
a)^(1/2)/a^3/e/(e*x)^(5/2)-5/2*B*(c*x^2+a)^(1/2)/a^3/e^2/(e*x)^(3/2)+77/10*A*c*(c*x^2+a)^(1/2)/a^4/e^3/(e*x)^(
1/2)-77/10*A*c^(3/2)*x*(c*x^2+a)^(1/2)/a^4/e^3/(a^(1/2)+x*c^(1/2))/(e*x)^(1/2)+77/10*A*c^(5/4)*(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)))*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)/a^(15/4)/e^3/(e*x
)^(1/2)/(c*x^2+a)^(1/2)-1/20*c^(3/4)*(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)+77*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)/a^(15/4)/e^3/(e*x)^(1/2)/(c*x^2+a)^(1/2)

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Rubi [A]  time = 0.58, antiderivative size = 432, normalized size of antiderivative = 1.00, number of steps used = 10, 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} \[ -\frac {c^{3/4} \sqrt {x} \left (\sqrt {a}+\sqrt {c} x\right ) \sqrt {\frac {a+c x^2}{\left (\sqrt {a}+\sqrt {c} x\right )^2}} \left (25 \sqrt {a} B+77 A \sqrt {c}\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{20 a^{15/4} e^3 \sqrt {e x} \sqrt {a+c x^2}}+\frac {11 A+9 B x}{6 a^2 e (e x)^{5/2} \sqrt {a+c x^2}}-\frac {77 A c^{3/2} x \sqrt {a+c x^2}}{10 a^4 e^3 \sqrt {e x} \left (\sqrt {a}+\sqrt {c} x\right )}+\frac {77 A c^{5/4} \sqrt {x} \left (\sqrt {a}+\sqrt {c} x\right ) \sqrt {\frac {a+c x^2}{\left (\sqrt {a}+\sqrt {c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{10 a^{15/4} e^3 \sqrt {e x} \sqrt {a+c x^2}}+\frac {77 A c \sqrt {a+c x^2}}{10 a^4 e^3 \sqrt {e x}}-\frac {77 A \sqrt {a+c x^2}}{30 a^3 e (e x)^{5/2}}-\frac {5 B \sqrt {a+c x^2}}{2 a^3 e^2 (e x)^{3/2}}+\frac {A+B x}{3 a e (e x)^{5/2} \left (a+c x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(A + B*x)/(3*a*e*(e*x)^(5/2)*(a + c*x^2)^(3/2)) + (11*A + 9*B*x)/(6*a^2*e*(e*x)^(5/2)*Sqrt[a + c*x^2]) - (77*A
*Sqrt[a + c*x^2])/(30*a^3*e*(e*x)^(5/2)) - (5*B*Sqrt[a + c*x^2])/(2*a^3*e^2*(e*x)^(3/2)) + (77*A*c*Sqrt[a + c*
x^2])/(10*a^4*e^3*Sqrt[e*x]) - (77*A*c^(3/2)*x*Sqrt[a + c*x^2])/(10*a^4*e^3*Sqrt[e*x]*(Sqrt[a] + Sqrt[c]*x)) +
(77*A*c^(5/4)*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])/(10*a^(15/4)*e^3*Sqrt[e*x]*Sqrt[a + c*x^2]) - ((25*Sqrt[a]*B + 77*A*Sqrt[c])*c^(
3/4)*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])/(20*a^(15/4)*e^3*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 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 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 {A+B x}{(e x)^{7/2} \left (a+c x^2\right )^{5/2}} \, dx &=\frac {A+B x}{3 a e (e x)^{5/2} \left (a+c x^2\right )^{3/2}}-\frac {\int \frac {-\frac {11}{2} a A c e^2-\frac {9}{2} a B c e^2 x}{(e x)^{7/2} \left (a+c x^2\right )^{3/2}} \, dx}{3 a^2 c e^2}\\ &=\frac {A+B x}{3 a e (e x)^{5/2} \left (a+c x^2\right )^{3/2}}+\frac {11 A+9 B x}{6 a^2 e (e x)^{5/2} \sqrt {a+c x^2}}+\frac {\int \frac {\frac {77}{4} a^2 A c^2 e^4+\frac {45}{4} a^2 B c^2 e^4 x}{(e x)^{7/2} \sqrt {a+c x^2}} \, dx}{3 a^4 c^2 e^4}\\ &=\frac {A+B x}{3 a e (e x)^{5/2} \left (a+c x^2\right )^{3/2}}+\frac {11 A+9 B x}{6 a^2 e (e x)^{5/2} \sqrt {a+c x^2}}-\frac {77 A \sqrt {a+c x^2}}{30 a^3 e (e x)^{5/2}}-\frac {2 \int \frac {-\frac {225}{8} a^3 B c^2 e^5+\frac {231}{8} a^2 A c^3 e^5 x}{(e x)^{5/2} \sqrt {a+c x^2}} \, dx}{15 a^5 c^2 e^6}\\ &=\frac {A+B x}{3 a e (e x)^{5/2} \left (a+c x^2\right )^{3/2}}+\frac {11 A+9 B x}{6 a^2 e (e x)^{5/2} \sqrt {a+c x^2}}-\frac {77 A \sqrt {a+c x^2}}{30 a^3 e (e x)^{5/2}}-\frac {5 B \sqrt {a+c x^2}}{2 a^3 e^2 (e x)^{3/2}}+\frac {4 \int \frac {-\frac {693}{16} a^3 A c^3 e^6-\frac {225}{16} a^3 B c^3 e^6 x}{(e x)^{3/2} \sqrt {a+c x^2}} \, dx}{45 a^6 c^2 e^8}\\ &=\frac {A+B x}{3 a e (e x)^{5/2} \left (a+c x^2\right )^{3/2}}+\frac {11 A+9 B x}{6 a^2 e (e x)^{5/2} \sqrt {a+c x^2}}-\frac {77 A \sqrt {a+c x^2}}{30 a^3 e (e x)^{5/2}}-\frac {5 B \sqrt {a+c x^2}}{2 a^3 e^2 (e x)^{3/2}}+\frac {77 A c \sqrt {a+c x^2}}{10 a^4 e^3 \sqrt {e x}}-\frac {8 \int \frac {\frac {225}{32} a^4 B c^3 e^7+\frac {693}{32} a^3 A c^4 e^7 x}{\sqrt {e x} \sqrt {a+c x^2}} \, dx}{45 a^7 c^2 e^{10}}\\ &=\frac {A+B x}{3 a e (e x)^{5/2} \left (a+c x^2\right )^{3/2}}+\frac {11 A+9 B x}{6 a^2 e (e x)^{5/2} \sqrt {a+c x^2}}-\frac {77 A \sqrt {a+c x^2}}{30 a^3 e (e x)^{5/2}}-\frac {5 B \sqrt {a+c x^2}}{2 a^3 e^2 (e x)^{3/2}}+\frac {77 A c \sqrt {a+c x^2}}{10 a^4 e^3 \sqrt {e x}}-\frac {\left (8 \sqrt {x}\right ) \int \frac {\frac {225}{32} a^4 B c^3 e^7+\frac {693}{32} a^3 A c^4 e^7 x}{\sqrt {x} \sqrt {a+c x^2}} \, dx}{45 a^7 c^2 e^{10} \sqrt {e x}}\\ &=\frac {A+B x}{3 a e (e x)^{5/2} \left (a+c x^2\right )^{3/2}}+\frac {11 A+9 B x}{6 a^2 e (e x)^{5/2} \sqrt {a+c x^2}}-\frac {77 A \sqrt {a+c x^2}}{30 a^3 e (e x)^{5/2}}-\frac {5 B \sqrt {a+c x^2}}{2 a^3 e^2 (e x)^{3/2}}+\frac {77 A c \sqrt {a+c x^2}}{10 a^4 e^3 \sqrt {e x}}-\frac {\left (16 \sqrt {x}\right ) \operatorname {Subst}\left (\int \frac {\frac {225}{32} a^4 B c^3 e^7+\frac {693}{32} a^3 A c^4 e^7 x^2}{\sqrt {a+c x^4}} \, dx,x,\sqrt {x}\right )}{45 a^7 c^2 e^{10} \sqrt {e x}}\\ &=\frac {A+B x}{3 a e (e x)^{5/2} \left (a+c x^2\right )^{3/2}}+\frac {11 A+9 B x}{6 a^2 e (e x)^{5/2} \sqrt {a+c x^2}}-\frac {77 A \sqrt {a+c x^2}}{30 a^3 e (e x)^{5/2}}-\frac {5 B \sqrt {a+c x^2}}{2 a^3 e^2 (e x)^{3/2}}+\frac {77 A c \sqrt {a+c x^2}}{10 a^4 e^3 \sqrt {e x}}-\frac {\left (\left (25 \sqrt {a} B+77 A \sqrt {c}\right ) c \sqrt {x}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+c x^4}} \, dx,x,\sqrt {x}\right )}{10 a^{7/2} e^3 \sqrt {e x}}+\frac {\left (77 A c^{3/2} \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 )}{10 a^{7/2} e^3 \sqrt {e x}}\\ &=\frac {A+B x}{3 a e (e x)^{5/2} \left (a+c x^2\right )^{3/2}}+\frac {11 A+9 B x}{6 a^2 e (e x)^{5/2} \sqrt {a+c x^2}}-\frac {77 A \sqrt {a+c x^2}}{30 a^3 e (e x)^{5/2}}-\frac {5 B \sqrt {a+c x^2}}{2 a^3 e^2 (e x)^{3/2}}+\frac {77 A c \sqrt {a+c x^2}}{10 a^4 e^3 \sqrt {e x}}-\frac {77 A c^{3/2} x \sqrt {a+c x^2}}{10 a^4 e^3 \sqrt {e x} \left (\sqrt {a}+\sqrt {c} x\right )}+\frac {77 A c^{5/4} \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 )}{10 a^{15/4} e^3 \sqrt {e x} \sqrt {a+c x^2}}-\frac {\left (25 \sqrt {a} B+77 A \sqrt {c}\right ) c^{3/4} \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 )}{20 a^{15/4} e^3 \sqrt {e x} \sqrt {a+c x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.09, size = 137, normalized size = 0.32 \[ \frac {x \left (-77 A \left (a+c x^2\right ) \sqrt {\frac {c x^2}{a}+1} \, _2F_1\left (-\frac {5}{4},\frac {1}{2};-\frac {1}{4};-\frac {c x^2}{a}\right )+65 a A-75 B x \left (a+c x^2\right ) \sqrt {\frac {c x^2}{a}+1} \, _2F_1\left (-\frac {3}{4},\frac {1}{2};\frac {1}{4};-\frac {c x^2}{a}\right )+55 a B x+55 A c x^2+45 B c x^3\right )}{30 a^2 (e x)^{7/2} \left (a+c x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(65*a*A + 55*a*B*x + 55*A*c*x^2 + 45*B*c*x^3 - 77*A*(a + c*x^2)*Sqrt[1 + (c*x^2)/a]*Hypergeometric2F1[-5/4,
1/2, -1/4, -((c*x^2)/a)] - 75*B*x*(a + c*x^2)*Sqrt[1 + (c*x^2)/a]*Hypergeometric2F1[-3/4, 1/2, 1/4, -((c*x^2)
/a)]))/(30*a^2*(e*x)^(7/2)*(a + c*x^2)^(3/2))

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fricas [F]  time = 1.06, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {c x^{2} + a} {\left (B x + A\right )} \sqrt {e x}}{c^{3} e^{4} x^{10} + 3 \, a c^{2} e^{4} x^{8} + 3 \, a^{2} c e^{4} x^{6} + a^{3} e^{4} x^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(e*x)^(7/2)/(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^4*x^10 + 3*a*c^2*e^4*x^8 + 3*a^2*c*e^4*x^6 + a^3*e^4*x^4),
x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {B x + A}{{\left (c x^{2} + a\right )}^{\frac {5}{2}} \left (e x\right )^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.26, size = 632, normalized size = 1.46 \[ -\frac {-462 A \,c^{3} x^{6}+462 \sqrt {2}\, \sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {\frac {-c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {-\frac {c x}{\sqrt {-a c}}}\, A a \,c^{2} x^{4} \EllipticE \left (\sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}, \frac {\sqrt {2}}{2}\right )-231 \sqrt {2}\, \sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {\frac {-c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {-\frac {c x}{\sqrt {-a c}}}\, A a \,c^{2} x^{4} \EllipticF \left (\sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}, \frac {\sqrt {2}}{2}\right )+150 B a \,c^{2} x^{5}-770 A a \,c^{2} x^{4}+75 \sqrt {-a c}\, \sqrt {2}\, \sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {\frac {-c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {-\frac {c x}{\sqrt {-a c}}}\, B a c \,x^{4} \EllipticF \left (\sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}, \frac {\sqrt {2}}{2}\right )+462 \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {-\frac {c x}{\sqrt {-a c}}}\, \sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}\, A \,a^{2} c \,x^{2} \EllipticE \left (\sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}, \frac {\sqrt {2}}{2}\right )-231 \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {-\frac {c x}{\sqrt {-a c}}}\, \sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}\, A \,a^{2} c \,x^{2} \EllipticF \left (\sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}, \frac {\sqrt {2}}{2}\right )+210 B \,a^{2} c \,x^{3}-264 A \,a^{2} c \,x^{2}+75 \sqrt {-a c}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {-\frac {c x}{\sqrt {-a c}}}\, \sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}\, B \,a^{2} x^{2} \EllipticF \left (\sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}, \frac {\sqrt {2}}{2}\right )+40 B \,a^{3} x +24 A \,a^{3}}{60 \sqrt {e x}\, \left (c \,x^{2}+a \right )^{\frac {3}{2}} a^{4} e^{3} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*(462*A*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)*(-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))*x^4*a*c^2-231*A*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)*(-1/(-a*c)^(1/2)*c*x)^(1/2)*Elli
pticF(((c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2),1/2*2^(1/2))*x^4*a*c^2+75*B*(-a*c)^(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)*(-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))*x^4*a*c+462*A*2^(1/2)*((-c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(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)*x^2*a^2*c-231*A*2^(1/2)*((-c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(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)*x^2
*a^2*c+75*B*(-a*c)^(1/2)*2^(1/2)*((-c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2)*(-1/(-a*c)^(1/2)*c*x)^(1/2)*Elliptic
F(((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)*x^2*a^2-462*A*c
^3*x^6+150*B*a*c^2*x^5-770*A*a*c^2*x^4+210*B*a^2*c*x^3-264*A*a^2*c*x^2+40*B*a^3*x+24*A*a^3)/x^2/a^4/e^3/(e*x)^
(1/2)/(c*x^2+a)^(3/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {B x + A}{{\left (c x^{2} + a\right )}^{\frac {5}{2}} \left (e x\right )^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {A+B\,x}{{\left (e\,x\right )}^{7/2}\,{\left (c\,x^2+a\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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