3.442 \(\int (e x)^{5/2} (A+B x) (a+c x^2)^{3/2} \, dx\)

Optimal. Leaf size=438 \[ \frac {4 a^{13/4} e^3 \sqrt {x} \left (\sqrt {a}+\sqrt {c} x\right ) \sqrt {\frac {a+c x^2}{\left (\sqrt {a}+\sqrt {c} x\right )^2}} \left (65 \sqrt {a} B-231 A \sqrt {c}\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{15015 c^{9/4} \sqrt {e x} \sqrt {a+c x^2}}+\frac {8 a^{13/4} A e^3 \sqrt {x} \left (\sqrt {a}+\sqrt {c} x\right ) \sqrt {\frac {a+c x^2}{\left (\sqrt {a}+\sqrt {c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{65 c^{7/4} \sqrt {e x} \sqrt {a+c x^2}}-\frac {8 a^3 A e^3 x \sqrt {a+c x^2}}{65 c^{3/2} \sqrt {e x} \left (\sqrt {a}+\sqrt {c} x\right )}+\frac {4 a^2 e^2 \sqrt {e x} \sqrt {a+c x^2} (65 a B-231 A c x)}{15015 c^2}+\frac {2 a e^2 \sqrt {e x} \left (a+c x^2\right )^{3/2} (13 a B-77 A c x)}{3003 c^2}+\frac {2 A e (e x)^{3/2} \left (a+c x^2\right )^{5/2}}{13 c}-\frac {2 a B e^2 \sqrt {e x} \left (a+c x^2\right )^{5/2}}{33 c^2}+\frac {2 B (e x)^{5/2} \left (a+c x^2\right )^{5/2}}{15 c} \]

[Out]

2/13*A*e*(e*x)^(3/2)*(c*x^2+a)^(5/2)/c+2/15*B*(e*x)^(5/2)*(c*x^2+a)^(5/2)/c+2/3003*a*e^2*(-77*A*c*x+13*B*a)*(c
*x^2+a)^(3/2)*(e*x)^(1/2)/c^2-2/33*a*B*e^2*(c*x^2+a)^(5/2)*(e*x)^(1/2)/c^2-8/65*a^3*A*e^3*x*(c*x^2+a)^(1/2)/c^
(3/2)/(a^(1/2)+x*c^(1/2))/(e*x)^(1/2)+4/15015*a^2*e^2*(-231*A*c*x+65*B*a)*(e*x)^(1/2)*(c*x^2+a)^(1/2)/c^2+8/65
*a^(13/4)*A*e^3*(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)))*Ellipt
icE(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^(7/4)/(e*x)^(1/2)/(c*x^2+a)^(1/2)+4/15015*a^(13/4)*e^3*(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))
*(65*B*a^(1/2)-231*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^(9/4)/(e*x
)^(1/2)/(c*x^2+a)^(1/2)

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Rubi [A]  time = 0.58, antiderivative size = 438, 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 = {833, 815, 842, 840, 1198, 220, 1196} \[ \frac {4 a^2 e^2 \sqrt {e x} \sqrt {a+c x^2} (65 a B-231 A c x)}{15015 c^2}+\frac {4 a^{13/4} e^3 \sqrt {x} \left (\sqrt {a}+\sqrt {c} x\right ) \sqrt {\frac {a+c x^2}{\left (\sqrt {a}+\sqrt {c} x\right )^2}} \left (65 \sqrt {a} B-231 A \sqrt {c}\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{15015 c^{9/4} \sqrt {e x} \sqrt {a+c x^2}}-\frac {8 a^3 A e^3 x \sqrt {a+c x^2}}{65 c^{3/2} \sqrt {e x} \left (\sqrt {a}+\sqrt {c} x\right )}+\frac {8 a^{13/4} A e^3 \sqrt {x} \left (\sqrt {a}+\sqrt {c} x\right ) \sqrt {\frac {a+c x^2}{\left (\sqrt {a}+\sqrt {c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{65 c^{7/4} \sqrt {e x} \sqrt {a+c x^2}}+\frac {2 a e^2 \sqrt {e x} \left (a+c x^2\right )^{3/2} (13 a B-77 A c x)}{3003 c^2}+\frac {2 A e (e x)^{3/2} \left (a+c x^2\right )^{5/2}}{13 c}-\frac {2 a B e^2 \sqrt {e x} \left (a+c x^2\right )^{5/2}}{33 c^2}+\frac {2 B (e x)^{5/2} \left (a+c x^2\right )^{5/2}}{15 c} \]

Antiderivative was successfully verified.

[In]

Int[(e*x)^(5/2)*(A + B*x)*(a + c*x^2)^(3/2),x]

[Out]

(-8*a^3*A*e^3*x*Sqrt[a + c*x^2])/(65*c^(3/2)*Sqrt[e*x]*(Sqrt[a] + Sqrt[c]*x)) + (4*a^2*e^2*Sqrt[e*x]*(65*a*B -
231*A*c*x)*Sqrt[a + c*x^2])/(15015*c^2) + (2*a*e^2*Sqrt[e*x]*(13*a*B - 77*A*c*x)*(a + c*x^2)^(3/2))/(3003*c^2
) - (2*a*B*e^2*Sqrt[e*x]*(a + c*x^2)^(5/2))/(33*c^2) + (2*A*e*(e*x)^(3/2)*(a + c*x^2)^(5/2))/(13*c) + (2*B*(e*
x)^(5/2)*(a + c*x^2)^(5/2))/(15*c) + (8*a^(13/4)*A*e^3*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])/(65*c^(7/4)*Sqrt[e*x]*Sqrt[a + c*x^2]) +
(4*a^(13/4)*(65*Sqrt[a]*B - 231*A*Sqrt[c])*e^3*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])/(15015*c^(9/4)*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 815

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x)^(
m + 1)*(c*e*f*(m + 2*p + 2) - g*c*d*(2*p + 1) + g*c*e*(m + 2*p + 1)*x)*(a + c*x^2)^p)/(c*e^2*(m + 2*p + 1)*(m
+ 2*p + 2)), x] + Dist[(2*p)/(c*e^2*(m + 2*p + 1)*(m + 2*p + 2)), Int[(d + e*x)^m*(a + c*x^2)^(p - 1)*Simp[f*a
*c*e^2*(m + 2*p + 2) + a*c*d*e*g*m - (c^2*f*d*e*(m + 2*p + 2) - g*(c^2*d^2*(2*p + 1) + a*c*e^2*(m + 2*p + 1)))
*x, x], x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && NeQ[c*d^2 + a*e^2, 0] && GtQ[p, 0] && (IntegerQ[p] ||  !R
ationalQ[m] || (GeQ[m, -1] && LtQ[m, 0])) &&  !ILtQ[m + 2*p, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*
m, 2*p])

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 (e x)^{5/2} (A+B x) \left (a+c x^2\right )^{3/2} \, dx &=\frac {2 B (e x)^{5/2} \left (a+c x^2\right )^{5/2}}{15 c}+\frac {2 \int (e x)^{3/2} \left (-\frac {5}{2} a B e+\frac {15}{2} A c e x\right ) \left (a+c x^2\right )^{3/2} \, dx}{15 c}\\ &=\frac {2 A e (e x)^{3/2} \left (a+c x^2\right )^{5/2}}{13 c}+\frac {2 B (e x)^{5/2} \left (a+c x^2\right )^{5/2}}{15 c}+\frac {4 \int \sqrt {e x} \left (-\frac {45}{4} a A c e^2-\frac {65}{4} a B c e^2 x\right ) \left (a+c x^2\right )^{3/2} \, dx}{195 c^2}\\ &=-\frac {2 a B e^2 \sqrt {e x} \left (a+c x^2\right )^{5/2}}{33 c^2}+\frac {2 A e (e x)^{3/2} \left (a+c x^2\right )^{5/2}}{13 c}+\frac {2 B (e x)^{5/2} \left (a+c x^2\right )^{5/2}}{15 c}+\frac {8 \int \frac {\left (\frac {65}{8} a^2 B c e^3-\frac {495}{8} a A c^2 e^3 x\right ) \left (a+c x^2\right )^{3/2}}{\sqrt {e x}} \, dx}{2145 c^3}\\ &=\frac {2 a e^2 \sqrt {e x} (13 a B-77 A c x) \left (a+c x^2\right )^{3/2}}{3003 c^2}-\frac {2 a B e^2 \sqrt {e x} \left (a+c x^2\right )^{5/2}}{33 c^2}+\frac {2 A e (e x)^{3/2} \left (a+c x^2\right )^{5/2}}{13 c}+\frac {2 B (e x)^{5/2} \left (a+c x^2\right )^{5/2}}{15 c}+\frac {32 \int \frac {\left (\frac {585}{16} a^3 B c^2 e^5-\frac {3465}{16} a^2 A c^3 e^5 x\right ) \sqrt {a+c x^2}}{\sqrt {e x}} \, dx}{45045 c^4 e^2}\\ &=\frac {4 a^2 e^2 \sqrt {e x} (65 a B-231 A c x) \sqrt {a+c x^2}}{15015 c^2}+\frac {2 a e^2 \sqrt {e x} (13 a B-77 A c x) \left (a+c x^2\right )^{3/2}}{3003 c^2}-\frac {2 a B e^2 \sqrt {e x} \left (a+c x^2\right )^{5/2}}{33 c^2}+\frac {2 A e (e x)^{3/2} \left (a+c x^2\right )^{5/2}}{13 c}+\frac {2 B (e x)^{5/2} \left (a+c x^2\right )^{5/2}}{15 c}+\frac {128 \int \frac {\frac {2925}{32} a^4 B c^3 e^7-\frac {10395}{32} a^3 A c^4 e^7 x}{\sqrt {e x} \sqrt {a+c x^2}} \, dx}{675675 c^5 e^4}\\ &=\frac {4 a^2 e^2 \sqrt {e x} (65 a B-231 A c x) \sqrt {a+c x^2}}{15015 c^2}+\frac {2 a e^2 \sqrt {e x} (13 a B-77 A c x) \left (a+c x^2\right )^{3/2}}{3003 c^2}-\frac {2 a B e^2 \sqrt {e x} \left (a+c x^2\right )^{5/2}}{33 c^2}+\frac {2 A e (e x)^{3/2} \left (a+c x^2\right )^{5/2}}{13 c}+\frac {2 B (e x)^{5/2} \left (a+c x^2\right )^{5/2}}{15 c}+\frac {\left (128 \sqrt {x}\right ) \int \frac {\frac {2925}{32} a^4 B c^3 e^7-\frac {10395}{32} a^3 A c^4 e^7 x}{\sqrt {x} \sqrt {a+c x^2}} \, dx}{675675 c^5 e^4 \sqrt {e x}}\\ &=\frac {4 a^2 e^2 \sqrt {e x} (65 a B-231 A c x) \sqrt {a+c x^2}}{15015 c^2}+\frac {2 a e^2 \sqrt {e x} (13 a B-77 A c x) \left (a+c x^2\right )^{3/2}}{3003 c^2}-\frac {2 a B e^2 \sqrt {e x} \left (a+c x^2\right )^{5/2}}{33 c^2}+\frac {2 A e (e x)^{3/2} \left (a+c x^2\right )^{5/2}}{13 c}+\frac {2 B (e x)^{5/2} \left (a+c x^2\right )^{5/2}}{15 c}+\frac {\left (256 \sqrt {x}\right ) \operatorname {Subst}\left (\int \frac {\frac {2925}{32} a^4 B c^3 e^7-\frac {10395}{32} a^3 A c^4 e^7 x^2}{\sqrt {a+c x^4}} \, dx,x,\sqrt {x}\right )}{675675 c^5 e^4 \sqrt {e x}}\\ &=\frac {4 a^2 e^2 \sqrt {e x} (65 a B-231 A c x) \sqrt {a+c x^2}}{15015 c^2}+\frac {2 a e^2 \sqrt {e x} (13 a B-77 A c x) \left (a+c x^2\right )^{3/2}}{3003 c^2}-\frac {2 a B e^2 \sqrt {e x} \left (a+c x^2\right )^{5/2}}{33 c^2}+\frac {2 A e (e x)^{3/2} \left (a+c x^2\right )^{5/2}}{13 c}+\frac {2 B (e x)^{5/2} \left (a+c x^2\right )^{5/2}}{15 c}+\frac {\left (8 a^{7/2} \left (65 \sqrt {a} B-231 A \sqrt {c}\right ) e^3 \sqrt {x}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+c x^4}} \, dx,x,\sqrt {x}\right )}{15015 c^2 \sqrt {e x}}+\frac {\left (8 a^{7/2} A e^3 \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 )}{65 c^{3/2} \sqrt {e x}}\\ &=-\frac {8 a^3 A e^3 x \sqrt {a+c x^2}}{65 c^{3/2} \sqrt {e x} \left (\sqrt {a}+\sqrt {c} x\right )}+\frac {4 a^2 e^2 \sqrt {e x} (65 a B-231 A c x) \sqrt {a+c x^2}}{15015 c^2}+\frac {2 a e^2 \sqrt {e x} (13 a B-77 A c x) \left (a+c x^2\right )^{3/2}}{3003 c^2}-\frac {2 a B e^2 \sqrt {e x} \left (a+c x^2\right )^{5/2}}{33 c^2}+\frac {2 A e (e x)^{3/2} \left (a+c x^2\right )^{5/2}}{13 c}+\frac {2 B (e x)^{5/2} \left (a+c x^2\right )^{5/2}}{15 c}+\frac {8 a^{13/4} A e^3 \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 )}{65 c^{7/4} \sqrt {e x} \sqrt {a+c x^2}}+\frac {4 a^{13/4} \left (65 \sqrt {a} B-231 A \sqrt {c}\right ) e^3 \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 )}{15015 c^{9/4} \sqrt {e x} \sqrt {a+c x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.15, size = 137, normalized size = 0.31 \[ \frac {2 e^2 \sqrt {e x} \sqrt {a+c x^2} \left (65 a^3 B \, _2F_1\left (-\frac {3}{2},\frac {1}{4};\frac {5}{4};-\frac {c x^2}{a}\right )-165 a^2 A c x \, _2F_1\left (-\frac {3}{2},\frac {3}{4};\frac {7}{4};-\frac {c x^2}{a}\right )-\left (a+c x^2\right )^2 \sqrt {\frac {c x^2}{a}+1} (65 a B-11 c x (15 A+13 B x))\right )}{2145 c^2 \sqrt {\frac {c x^2}{a}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[(e*x)^(5/2)*(A + B*x)*(a + c*x^2)^(3/2),x]

[Out]

(2*e^2*Sqrt[e*x]*Sqrt[a + c*x^2]*(-((a + c*x^2)^2*Sqrt[1 + (c*x^2)/a]*(65*a*B - 11*c*x*(15*A + 13*B*x))) + 65*
a^3*B*Hypergeometric2F1[-3/2, 1/4, 5/4, -((c*x^2)/a)] - 165*a^2*A*c*x*Hypergeometric2F1[-3/2, 3/4, 7/4, -((c*x
^2)/a)]))/(2145*c^2*Sqrt[1 + (c*x^2)/a])

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fricas [F]  time = 1.05, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (B c e^{2} x^{5} + A c e^{2} x^{4} + B a e^{2} x^{3} + A a e^{2} x^{2}\right )} \sqrt {c x^{2} + a} \sqrt {e x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((B*c*e^2*x^5 + A*c*e^2*x^4 + B*a*e^2*x^3 + A*a*e^2*x^2)*sqrt(c*x^2 + a)*sqrt(e*x), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (c x^{2} + a\right )}^{\frac {3}{2}} {\left (B x + A\right )} \left (e x\right )^{\frac {5}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^(5/2)*(B*x+A)*(c*x^2+a)^(3/2),x, algorithm="giac")

[Out]

integrate((c*x^2 + a)^(3/2)*(B*x + A)*(e*x)^(5/2), x)

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maple [A]  time = 0.10, size = 384, normalized size = 0.88 \[ \frac {2 \sqrt {e x}\, \left (1001 B \,c^{5} x^{9}+1155 A \,c^{5} x^{8}+2548 B a \,c^{4} x^{7}+3080 A a \,c^{4} x^{6}+1703 B \,a^{2} c^{3} x^{5}+2233 A \,a^{2} c^{3} x^{4}-104 B \,a^{3} c^{2} x^{3}+308 A \,a^{3} c^{2} x^{2}-924 \sqrt {2}\, \sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {\frac {-c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {-\frac {c x}{\sqrt {-a c}}}\, A \,a^{4} c \EllipticE \left (\sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}, \frac {\sqrt {2}}{2}\right )+462 \sqrt {2}\, \sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {\frac {-c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {-\frac {c x}{\sqrt {-a c}}}\, A \,a^{4} c \EllipticF \left (\sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}, \frac {\sqrt {2}}{2}\right )-260 B \,a^{4} c x +130 \sqrt {2}\, \sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {\frac {-c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {-\frac {c x}{\sqrt {-a c}}}\, \sqrt {-a c}\, B \,a^{4} \EllipticF \left (\sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}, \frac {\sqrt {2}}{2}\right )\right ) e^{2}}{15015 \sqrt {c \,x^{2}+a}\, c^{3} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x)^(5/2)*(B*x+A)*(c*x^2+a)^(3/2),x)

[Out]

2/15015/x*e^2*(e*x)^(1/2)/(c*x^2+a)^(1/2)*(1001*B*x^9*c^5+1155*A*x^8*c^5+2548*B*x^7*a*c^4+3080*A*x^6*a*c^4+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)*EllipticF(((c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2),1/2*2^(1/2))*a^4*c-924*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))*a^4*c+130*B*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))*(-a*c)^(1/2)*a^4+1703*B*x^5*a^2*c^3+2233*A*x^4*a^2*c^3-104*B*x^3*a^3*c^2+308*A*x^2*a^3*c^2-26
0*B*x*a^4*c)/c^3

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (c x^{2} + a\right )}^{\frac {3}{2}} {\left (B x + A\right )} \left (e x\right )^{\frac {5}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^(5/2)*(B*x+A)*(c*x^2+a)^(3/2),x, algorithm="maxima")

[Out]

integrate((c*x^2 + a)^(3/2)*(B*x + A)*(e*x)^(5/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\left (e\,x\right )}^{5/2}\,{\left (c\,x^2+a\right )}^{3/2}\,\left (A+B\,x\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x)^(5/2)*(a + c*x^2)^(3/2)*(A + B*x),x)

[Out]

int((e*x)^(5/2)*(a + c*x^2)^(3/2)*(A + B*x), x)

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sympy [C]  time = 68.10, size = 199, normalized size = 0.45 \[ \frac {A a^{\frac {3}{2}} e^{\frac {5}{2}} x^{\frac {7}{2}} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {c x^{2} e^{i \pi }}{a}} \right )}}{2 \Gamma \left (\frac {11}{4}\right )} + \frac {A \sqrt {a} c e^{\frac {5}{2}} x^{\frac {11}{2}} \Gamma \left (\frac {11}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {11}{4} \\ \frac {15}{4} \end {matrix}\middle | {\frac {c x^{2} e^{i \pi }}{a}} \right )}}{2 \Gamma \left (\frac {15}{4}\right )} + \frac {B a^{\frac {3}{2}} e^{\frac {5}{2}} x^{\frac {9}{2}} \Gamma \left (\frac {9}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {9}{4} \\ \frac {13}{4} \end {matrix}\middle | {\frac {c x^{2} e^{i \pi }}{a}} \right )}}{2 \Gamma \left (\frac {13}{4}\right )} + \frac {B \sqrt {a} c e^{\frac {5}{2}} x^{\frac {13}{2}} \Gamma \left (\frac {13}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {13}{4} \\ \frac {17}{4} \end {matrix}\middle | {\frac {c x^{2} e^{i \pi }}{a}} \right )}}{2 \Gamma \left (\frac {17}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)**(5/2)*(B*x+A)*(c*x**2+a)**(3/2),x)

[Out]

A*a**(3/2)*e**(5/2)*x**(7/2)*gamma(7/4)*hyper((-1/2, 7/4), (11/4,), c*x**2*exp_polar(I*pi)/a)/(2*gamma(11/4))
+ A*sqrt(a)*c*e**(5/2)*x**(11/2)*gamma(11/4)*hyper((-1/2, 11/4), (15/4,), c*x**2*exp_polar(I*pi)/a)/(2*gamma(1
5/4)) + B*a**(3/2)*e**(5/2)*x**(9/2)*gamma(9/4)*hyper((-1/2, 9/4), (13/4,), c*x**2*exp_polar(I*pi)/a)/(2*gamma
(13/4)) + B*sqrt(a)*c*e**(5/2)*x**(13/2)*gamma(13/4)*hyper((-1/2, 13/4), (17/4,), c*x**2*exp_polar(I*pi)/a)/(2
*gamma(17/4))

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