3.475 \(\int \frac {(e x)^{11/2} (A+B x)}{(a+c x^2)^{5/2}} \, dx\)

Optimal. Leaf size=398 \[ -\frac {a^{3/4} e^6 \sqrt {x} \left (\sqrt {a}+\sqrt {c} x\right ) \sqrt {\frac {a+c x^2}{\left (\sqrt {a}+\sqrt {c} x\right )^2}} \left (77 \sqrt {a} B+25 A \sqrt {c}\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{20 c^{15/4} \sqrt {e x} \sqrt {a+c x^2}}+\frac {77 a^{5/4} B e^6 \sqrt {x} \left (\sqrt {a}+\sqrt {c} x\right ) \sqrt {\frac {a+c x^2}{\left (\sqrt {a}+\sqrt {c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{10 c^{15/4} \sqrt {e x} \sqrt {a+c x^2}}-\frac {e^3 (e x)^{5/2} (9 A+11 B x)}{6 c^2 \sqrt {a+c x^2}}-\frac {e (e x)^{9/2} (A+B x)}{3 c \left (a+c x^2\right )^{3/2}}+\frac {5 A e^5 \sqrt {e x} \sqrt {a+c x^2}}{2 c^3}-\frac {77 a B e^6 x \sqrt {a+c x^2}}{10 c^{7/2} \sqrt {e x} \left (\sqrt {a}+\sqrt {c} x\right )}+\frac {77 B e^4 (e x)^{3/2} \sqrt {a+c x^2}}{30 c^3} \]

[Out]

-1/3*e*(e*x)^(9/2)*(B*x+A)/c/(c*x^2+a)^(3/2)-1/6*e^3*(e*x)^(5/2)*(11*B*x+9*A)/c^2/(c*x^2+a)^(1/2)+77/30*B*e^4*
(e*x)^(3/2)*(c*x^2+a)^(1/2)/c^3-77/10*a*B*e^6*x*(c*x^2+a)^(1/2)/c^(7/2)/(a^(1/2)+x*c^(1/2))/(e*x)^(1/2)+5/2*A*
e^5*(e*x)^(1/2)*(c*x^2+a)^(1/2)/c^3+77/10*a^(5/4)*B*e^6*(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)/c^(15/4)/(e*x)^(1/2)/(c*x^2+a)^(1/2)-1/20*a^(3/4)*e^6*(c
os(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))*(77*B*a^(1/2)+25*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)/c^(15/4)/(e*x)^(1/2)/(c*x^2+a)^(1/2)

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Rubi [A]  time = 0.49, antiderivative size = 398, normalized size of antiderivative = 1.00, 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 = {819, 833, 842, 840, 1198, 220, 1196} \[ -\frac {a^{3/4} e^6 \sqrt {x} \left (\sqrt {a}+\sqrt {c} x\right ) \sqrt {\frac {a+c x^2}{\left (\sqrt {a}+\sqrt {c} x\right )^2}} \left (77 \sqrt {a} B+25 A \sqrt {c}\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{20 c^{15/4} \sqrt {e x} \sqrt {a+c x^2}}+\frac {77 a^{5/4} B e^6 \sqrt {x} \left (\sqrt {a}+\sqrt {c} x\right ) \sqrt {\frac {a+c x^2}{\left (\sqrt {a}+\sqrt {c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{10 c^{15/4} \sqrt {e x} \sqrt {a+c x^2}}-\frac {e^3 (e x)^{5/2} (9 A+11 B x)}{6 c^2 \sqrt {a+c x^2}}-\frac {e (e x)^{9/2} (A+B x)}{3 c \left (a+c x^2\right )^{3/2}}+\frac {5 A e^5 \sqrt {e x} \sqrt {a+c x^2}}{2 c^3}-\frac {77 a B e^6 x \sqrt {a+c x^2}}{10 c^{7/2} \sqrt {e x} \left (\sqrt {a}+\sqrt {c} x\right )}+\frac {77 B e^4 (e x)^{3/2} \sqrt {a+c x^2}}{30 c^3} \]

Antiderivative was successfully verified.

[In]

Int[((e*x)^(11/2)*(A + B*x))/(a + c*x^2)^(5/2),x]

[Out]

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

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Mathematica [C]  time = 0.13, size = 165, normalized size = 0.41 \[ \frac {e^5 \sqrt {e x} \left (75 a^2 A+77 a^2 B x-75 a A \left (a+c x^2\right ) \sqrt {\frac {c x^2}{a}+1} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};-\frac {c x^2}{a}\right )+105 a A c x^2-77 a B x \left (a+c x^2\right ) \sqrt {\frac {c x^2}{a}+1} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-\frac {c x^2}{a}\right )+99 a B c x^3+20 A c^2 x^4+12 B c^2 x^5\right )}{30 c^3 \left (a+c x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[((e*x)^(11/2)*(A + B*x))/(a + c*x^2)^(5/2),x]

[Out]

(e^5*Sqrt[e*x]*(75*a^2*A + 77*a^2*B*x + 105*a*A*c*x^2 + 99*a*B*c*x^3 + 20*A*c^2*x^4 + 12*B*c^2*x^5 - 75*a*A*(a
+ c*x^2)*Sqrt[1 + (c*x^2)/a]*Hypergeometric2F1[1/4, 1/2, 5/4, -((c*x^2)/a)] - 77*a*B*x*(a + c*x^2)*Sqrt[1 + (
c*x^2)/a]*Hypergeometric2F1[1/2, 3/4, 7/4, -((c*x^2)/a)]))/(30*c^3*(a + c*x^2)^(3/2))

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fricas [F]  time = 0.66, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (B e^{5} x^{6} + A e^{5} x^{5}\right )} \sqrt {c x^{2} + a} \sqrt {e x}}{c^{3} x^{6} + 3 \, a c^{2} x^{4} + 3 \, a^{2} c x^{2} + a^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^(11/2)*(B*x+A)/(c*x^2+a)^(5/2),x, algorithm="fricas")

[Out]

integral((B*e^5*x^6 + A*e^5*x^5)*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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B x + A\right )} \left (e x\right )^{\frac {11}{2}}}{{\left (c x^{2} + a\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^(11/2)*(B*x+A)/(c*x^2+a)^(5/2),x, algorithm="giac")

[Out]

integrate((B*x + A)*(e*x)^(11/2)/(c*x^2 + a)^(5/2), x)

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maple [A]  time = 0.13, size = 615, normalized size = 1.55 \[ -\frac {\left (-24 B \,c^{3} x^{6}-40 A \,c^{3} x^{5}-198 B a \,c^{2} x^{4}-210 A a \,c^{2} x^{3}+462 \sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {\frac {-c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {-\frac {c x}{\sqrt {-a c}}}\, \sqrt {2}\, B \,a^{2} c \,x^{2} \EllipticE \left (\sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}, \frac {\sqrt {2}}{2}\right )-231 \sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {\frac {-c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {-\frac {c x}{\sqrt {-a c}}}\, \sqrt {2}\, B \,a^{2} c \,x^{2} \EllipticF \left (\sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}, \frac {\sqrt {2}}{2}\right )+75 \sqrt {-a c}\, \sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {\frac {-c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {-\frac {c x}{\sqrt {-a c}}}\, \sqrt {2}\, A a c \,x^{2} \EllipticF \left (\sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}, \frac {\sqrt {2}}{2}\right )-154 B \,a^{2} c \,x^{2}-150 A \,a^{2} c x +462 \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {-\frac {c x}{\sqrt {-a c}}}\, \sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}\, B \,a^{3} \EllipticE \left (\sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}, \frac {\sqrt {2}}{2}\right )-231 \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {-\frac {c x}{\sqrt {-a c}}}\, \sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}\, B \,a^{3} \EllipticF \left (\sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}, \frac {\sqrt {2}}{2}\right )+75 \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {-\frac {c x}{\sqrt {-a c}}}\, \sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {-a c}\, A \,a^{2} \EllipticF \left (\sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}, \frac {\sqrt {2}}{2}\right )\right ) \sqrt {e x}\, e^{5}}{60 \left (c \,x^{2}+a \right )^{\frac {3}{2}} c^{4} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x)^(11/2)*(B*x+A)/(c*x^2+a)^(5/2),x)

[Out]

-1/60*(75*A*(-a*c)^(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))*2^(1/2)*x^2*a*c+462*B*(
(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)*El
lipticE(((c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2),1/2*2^(1/2))*2^(1/2)*x^2*a^2*c-231*B*((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))*2^(1/2)*x^2*a^2*c-24*B*c^3*x^6+75*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)*(-a*c)^(1/2)*a^2-40*A*c^3*x^5+462*B*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)*a^3-231*B*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)*a
^3-198*B*a*c^2*x^4-210*A*a*c^2*x^3-154*B*a^2*c*x^2-150*A*a^2*c*x)*e^5/x*(e*x)^(1/2)/c^4/(c*x^2+a)^(3/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B x + A\right )} \left (e x\right )^{\frac {11}{2}}}{{\left (c x^{2} + a\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^(11/2)*(B*x+A)/(c*x^2+a)^(5/2),x, algorithm="maxima")

[Out]

integrate((B*x + A)*(e*x)^(11/2)/(c*x^2 + a)^(5/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (e\,x\right )}^{11/2}\,\left (A+B\,x\right )}{{\left (c\,x^2+a\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((e*x)^(11/2)*(A + B*x))/(a + c*x^2)^(5/2),x)

[Out]

int(((e*x)^(11/2)*(A + B*x))/(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((e*x)**(11/2)*(B*x+A)/(c*x**2+a)**(5/2),x)

[Out]

Timed out

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