3.433 \(\int (e x)^{7/2} (A+B x) \sqrt {a+c x^2} \, dx\)

Optimal. Leaf size=427 \[ \frac {2 a^{11/4} e^4 \sqrt {x} \left (\sqrt {a}+\sqrt {c} x\right ) \sqrt {\frac {a+c x^2}{\left (\sqrt {a}+\sqrt {c} x\right )^2}} \left (539 \sqrt {a} B+325 A \sqrt {c}\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{15015 c^{11/4} \sqrt {e x} \sqrt {a+c x^2}}-\frac {28 a^{13/4} B e^4 \sqrt {x} \left (\sqrt {a}+\sqrt {c} x\right ) \sqrt {\frac {a+c x^2}{\left (\sqrt {a}+\sqrt {c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{195 c^{11/4} \sqrt {e x} \sqrt {a+c x^2}}+\frac {28 a^3 B e^4 x \sqrt {a+c x^2}}{195 c^{5/2} \sqrt {e x} \left (\sqrt {a}+\sqrt {c} x\right )}+\frac {2 a^2 e^3 \sqrt {e x} \sqrt {a+c x^2} (325 A+539 B x)}{15015 c^2}-\frac {10 a A e^3 \sqrt {e x} \left (a+c x^2\right )^{3/2}}{77 c^2}+\frac {2 A e (e x)^{5/2} \left (a+c x^2\right )^{3/2}}{11 c}-\frac {14 a B e^2 (e x)^{3/2} \left (a+c x^2\right )^{3/2}}{117 c^2}+\frac {2 B (e x)^{7/2} \left (a+c x^2\right )^{3/2}}{13 c} \]

[Out]

-14/117*a*B*e^2*(e*x)^(3/2)*(c*x^2+a)^(3/2)/c^2+2/11*A*e*(e*x)^(5/2)*(c*x^2+a)^(3/2)/c+2/13*B*(e*x)^(7/2)*(c*x
^2+a)^(3/2)/c-10/77*a*A*e^3*(c*x^2+a)^(3/2)*(e*x)^(1/2)/c^2+28/195*a^3*B*e^4*x*(c*x^2+a)^(1/2)/c^(5/2)/(a^(1/2
)+x*c^(1/2))/(e*x)^(1/2)+2/15015*a^2*e^3*(539*B*x+325*A)*(e*x)^(1/2)*(c*x^2+a)^(1/2)/c^2-28/195*a^(13/4)*B*e^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*arcta
n(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
^(11/4)/(e*x)^(1/2)/(c*x^2+a)^(1/2)+2/15015*a^(11/4)*e^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))*(539*B*a^(1/2
)+325*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^(11/4)/(e*x)^(1/2)/(c*x
^2+a)^(1/2)

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Rubi [A]  time = 0.59, antiderivative size = 427, 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} \[ \frac {2 a^2 e^3 \sqrt {e x} \sqrt {a+c x^2} (325 A+539 B x)}{15015 c^2}+\frac {2 a^{11/4} e^4 \sqrt {x} \left (\sqrt {a}+\sqrt {c} x\right ) \sqrt {\frac {a+c x^2}{\left (\sqrt {a}+\sqrt {c} x\right )^2}} \left (539 \sqrt {a} B+325 A \sqrt {c}\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{15015 c^{11/4} \sqrt {e x} \sqrt {a+c x^2}}+\frac {28 a^3 B e^4 x \sqrt {a+c x^2}}{195 c^{5/2} \sqrt {e x} \left (\sqrt {a}+\sqrt {c} x\right )}-\frac {28 a^{13/4} B e^4 \sqrt {x} \left (\sqrt {a}+\sqrt {c} x\right ) \sqrt {\frac {a+c x^2}{\left (\sqrt {a}+\sqrt {c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{195 c^{11/4} \sqrt {e x} \sqrt {a+c x^2}}-\frac {10 a A e^3 \sqrt {e x} \left (a+c x^2\right )^{3/2}}{77 c^2}+\frac {2 A e (e x)^{5/2} \left (a+c x^2\right )^{3/2}}{11 c}-\frac {14 a B e^2 (e x)^{3/2} \left (a+c x^2\right )^{3/2}}{117 c^2}+\frac {2 B (e x)^{7/2} \left (a+c x^2\right )^{3/2}}{13 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^2*e^3*Sqrt[e*x]*(325*A + 539*B*x)*Sqrt[a + c*x^2])/(15015*c^2) + (28*a^3*B*e^4*x*Sqrt[a + c*x^2])/(195*c^
(5/2)*Sqrt[e*x]*(Sqrt[a] + Sqrt[c]*x)) - (10*a*A*e^3*Sqrt[e*x]*(a + c*x^2)^(3/2))/(77*c^2) - (14*a*B*e^2*(e*x)
^(3/2)*(a + c*x^2)^(3/2))/(117*c^2) + (2*A*e*(e*x)^(5/2)*(a + c*x^2)^(3/2))/(11*c) + (2*B*(e*x)^(7/2)*(a + c*x
^2)^(3/2))/(13*c) - (28*a^(13/4)*B*e^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])/(195*c^(11/4)*Sqrt[e*x]*Sqrt[a + c*x^2]) + (2*a^(11/4)*(
539*Sqrt[a]*B + 325*A*Sqrt[c])*e^4*Sqrt[x]*(Sqrt[a] + Sqrt[c]*x)*Sqrt[(a + c*x^2)/(Sqrt[a] + Sqrt[c]*x)^2]*Ell
ipticF[2*ArcTan[(c^(1/4)*Sqrt[x])/a^(1/4)], 1/2])/(15015*c^(11/4)*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)^{7/2} (A+B x) \sqrt {a+c x^2} \, dx &=\frac {2 B (e x)^{7/2} \left (a+c x^2\right )^{3/2}}{13 c}+\frac {2 \int (e x)^{5/2} \left (-\frac {7}{2} a B e+\frac {13}{2} A c e x\right ) \sqrt {a+c x^2} \, dx}{13 c}\\ &=\frac {2 A e (e x)^{5/2} \left (a+c x^2\right )^{3/2}}{11 c}+\frac {2 B (e x)^{7/2} \left (a+c x^2\right )^{3/2}}{13 c}+\frac {4 \int (e x)^{3/2} \left (-\frac {65}{4} a A c e^2-\frac {77}{4} a B c e^2 x\right ) \sqrt {a+c x^2} \, dx}{143 c^2}\\ &=-\frac {14 a B e^2 (e x)^{3/2} \left (a+c x^2\right )^{3/2}}{117 c^2}+\frac {2 A e (e x)^{5/2} \left (a+c x^2\right )^{3/2}}{11 c}+\frac {2 B (e x)^{7/2} \left (a+c x^2\right )^{3/2}}{13 c}+\frac {8 \int \sqrt {e x} \left (\frac {231}{8} a^2 B c e^3-\frac {585}{8} a A c^2 e^3 x\right ) \sqrt {a+c x^2} \, dx}{1287 c^3}\\ &=-\frac {10 a A e^3 \sqrt {e x} \left (a+c x^2\right )^{3/2}}{77 c^2}-\frac {14 a B e^2 (e x)^{3/2} \left (a+c x^2\right )^{3/2}}{117 c^2}+\frac {2 A e (e x)^{5/2} \left (a+c x^2\right )^{3/2}}{11 c}+\frac {2 B (e x)^{7/2} \left (a+c x^2\right )^{3/2}}{13 c}+\frac {16 \int \frac {\left (\frac {585}{16} a^2 A c^2 e^4+\frac {1617}{16} a^2 B c^2 e^4 x\right ) \sqrt {a+c x^2}}{\sqrt {e x}} \, dx}{9009 c^4}\\ &=\frac {2 a^2 e^3 \sqrt {e x} (325 A+539 B x) \sqrt {a+c x^2}}{15015 c^2}-\frac {10 a A e^3 \sqrt {e x} \left (a+c x^2\right )^{3/2}}{77 c^2}-\frac {14 a B e^2 (e x)^{3/2} \left (a+c x^2\right )^{3/2}}{117 c^2}+\frac {2 A e (e x)^{5/2} \left (a+c x^2\right )^{3/2}}{11 c}+\frac {2 B (e x)^{7/2} \left (a+c x^2\right )^{3/2}}{13 c}+\frac {64 \int \frac {\frac {2925}{32} a^3 A c^3 e^6+\frac {4851}{32} a^3 B c^3 e^6 x}{\sqrt {e x} \sqrt {a+c x^2}} \, dx}{135135 c^5 e^2}\\ &=\frac {2 a^2 e^3 \sqrt {e x} (325 A+539 B x) \sqrt {a+c x^2}}{15015 c^2}-\frac {10 a A e^3 \sqrt {e x} \left (a+c x^2\right )^{3/2}}{77 c^2}-\frac {14 a B e^2 (e x)^{3/2} \left (a+c x^2\right )^{3/2}}{117 c^2}+\frac {2 A e (e x)^{5/2} \left (a+c x^2\right )^{3/2}}{11 c}+\frac {2 B (e x)^{7/2} \left (a+c x^2\right )^{3/2}}{13 c}+\frac {\left (64 \sqrt {x}\right ) \int \frac {\frac {2925}{32} a^3 A c^3 e^6+\frac {4851}{32} a^3 B c^3 e^6 x}{\sqrt {x} \sqrt {a+c x^2}} \, dx}{135135 c^5 e^2 \sqrt {e x}}\\ &=\frac {2 a^2 e^3 \sqrt {e x} (325 A+539 B x) \sqrt {a+c x^2}}{15015 c^2}-\frac {10 a A e^3 \sqrt {e x} \left (a+c x^2\right )^{3/2}}{77 c^2}-\frac {14 a B e^2 (e x)^{3/2} \left (a+c x^2\right )^{3/2}}{117 c^2}+\frac {2 A e (e x)^{5/2} \left (a+c x^2\right )^{3/2}}{11 c}+\frac {2 B (e x)^{7/2} \left (a+c x^2\right )^{3/2}}{13 c}+\frac {\left (128 \sqrt {x}\right ) \operatorname {Subst}\left (\int \frac {\frac {2925}{32} a^3 A c^3 e^6+\frac {4851}{32} a^3 B c^3 e^6 x^2}{\sqrt {a+c x^4}} \, dx,x,\sqrt {x}\right )}{135135 c^5 e^2 \sqrt {e x}}\\ &=\frac {2 a^2 e^3 \sqrt {e x} (325 A+539 B x) \sqrt {a+c x^2}}{15015 c^2}-\frac {10 a A e^3 \sqrt {e x} \left (a+c x^2\right )^{3/2}}{77 c^2}-\frac {14 a B e^2 (e x)^{3/2} \left (a+c x^2\right )^{3/2}}{117 c^2}+\frac {2 A e (e x)^{5/2} \left (a+c x^2\right )^{3/2}}{11 c}+\frac {2 B (e x)^{7/2} \left (a+c x^2\right )^{3/2}}{13 c}-\frac {\left (28 a^{7/2} 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 )}{195 c^{5/2} \sqrt {e x}}+\frac {\left (4 a^3 \left (539 \sqrt {a} B+325 A \sqrt {c}\right ) e^4 \sqrt {x}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+c x^4}} \, dx,x,\sqrt {x}\right )}{15015 c^{5/2} \sqrt {e x}}\\ &=\frac {2 a^2 e^3 \sqrt {e x} (325 A+539 B x) \sqrt {a+c x^2}}{15015 c^2}+\frac {28 a^3 B e^4 x \sqrt {a+c x^2}}{195 c^{5/2} \sqrt {e x} \left (\sqrt {a}+\sqrt {c} x\right )}-\frac {10 a A e^3 \sqrt {e x} \left (a+c x^2\right )^{3/2}}{77 c^2}-\frac {14 a B e^2 (e x)^{3/2} \left (a+c x^2\right )^{3/2}}{117 c^2}+\frac {2 A e (e x)^{5/2} \left (a+c x^2\right )^{3/2}}{11 c}+\frac {2 B (e x)^{7/2} \left (a+c x^2\right )^{3/2}}{13 c}-\frac {28 a^{13/4} 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 )}{195 c^{11/4} \sqrt {e x} \sqrt {a+c x^2}}+\frac {2 a^{11/4} \left (539 \sqrt {a} B+325 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 )}{15015 c^{11/4} \sqrt {e x} \sqrt {a+c x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.13, size = 142, normalized size = 0.33 \[ \frac {2 e^3 \sqrt {e x} \sqrt {a+c x^2} \left (585 a^2 A \, _2F_1\left (-\frac {1}{2},\frac {1}{4};\frac {5}{4};-\frac {c x^2}{a}\right )+539 a^2 B x \, _2F_1\left (-\frac {1}{2},\frac {3}{4};\frac {7}{4};-\frac {c x^2}{a}\right )-\left (a+c x^2\right ) \sqrt {\frac {c x^2}{a}+1} \left (a (585 A+539 B x)-63 c x^2 (13 A+11 B x)\right )\right )}{9009 c^2 \sqrt {\frac {c x^2}{a}+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*e^3*Sqrt[e*x]*Sqrt[a + c*x^2]*(-((a + c*x^2)*Sqrt[1 + (c*x^2)/a]*(-63*c*x^2*(13*A + 11*B*x) + a*(585*A + 53
9*B*x))) + 585*a^2*A*Hypergeometric2F1[-1/2, 1/4, 5/4, -((c*x^2)/a)] + 539*a^2*B*x*Hypergeometric2F1[-1/2, 3/4
, 7/4, -((c*x^2)/a)]))/(9009*c^2*Sqrt[1 + (c*x^2)/a])

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fricas [F]  time = 1.01, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (B e^{3} x^{4} + A e^{3} x^{3}\right )} \sqrt {c x^{2} + a} \sqrt {e x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^(7/2)*(B*x+A)*(c*x^2+a)^(1/2),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^(7/2)*(B*x+A)*(c*x^2+a)^(1/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.24, size = 368, normalized size = 0.86 \[ \frac {2 \sqrt {e x}\, \left (3465 B \,c^{4} x^{8}+4095 A \,c^{4} x^{7}+4235 B a \,c^{3} x^{6}+5265 A a \,c^{3} x^{5}-308 B \,a^{2} c^{2} x^{4}-780 A \,a^{2} c^{2} x^{3}-1078 B \,a^{3} c \,x^{2}-1950 A \,a^{3} c x +3234 \sqrt {2}\, \sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {\frac {-c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {-\frac {c x}{\sqrt {-a c}}}\, B \,a^{4} \EllipticE \left (\sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}, \frac {\sqrt {2}}{2}\right )-1617 \sqrt {2}\, \sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {\frac {-c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {-\frac {c x}{\sqrt {-a c}}}\, B \,a^{4} \EllipticF \left (\sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}, \frac {\sqrt {2}}{2}\right )+975 \sqrt {2}\, \sqrt {-a c}\, \sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {\frac {-c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {-\frac {c x}{\sqrt {-a c}}}\, A \,a^{3} \EllipticF \left (\sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}, \frac {\sqrt {2}}{2}\right )\right ) e^{3}}{45045 \sqrt {c \,x^{2}+a}\, c^{3} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x)^(7/2)*(B*x+A)*(c*x^2+a)^(1/2),x)

[Out]

2/45045*e^3/x*(e*x)^(1/2)/(c*x^2+a)^(1/2)/c^3*(3465*B*c^4*x^8+4095*A*c^4*x^7+4235*B*a*c^3*x^6+975*A*2^(1/2)*(-
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)*(-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^3+5265*A*a*c^3*x^5+3234*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)*(-x*c/(-a*c)^(1/2))^(1/2)*Ellipt
icE(((c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2),1/2*2^(1/2))*a^4-1617*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)*(-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^4-308*B*a^2*c^2*x^4-780*A*a^2*c^2*x^3-1078*B*a^3*c*x^2-1950*A*a^3*c*x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^(7/2)*(B*x+A)*(c*x^2+a)^(1/2),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x)^(7/2)*(a + c*x^2)^(1/2)*(A + B*x),x)

[Out]

int((e*x)^(7/2)*(a + c*x^2)^(1/2)*(A + B*x), x)

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sympy [C]  time = 77.12, size = 97, normalized size = 0.23 \[ \frac {A \sqrt {a} e^{\frac {7}{2}} x^{\frac {9}{2}} \Gamma \left (\frac {9}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {9}{4} \\ \frac {13}{4} \end {matrix}\middle | {\frac {c x^{2} e^{i \pi }}{a}} \right )}}{2 \Gamma \left (\frac {13}{4}\right )} + \frac {B \sqrt {a} e^{\frac {7}{2}} x^{\frac {11}{2}} \Gamma \left (\frac {11}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {11}{4} \\ \frac {15}{4} \end {matrix}\middle | {\frac {c x^{2} e^{i \pi }}{a}} \right )}}{2 \Gamma \left (\frac {15}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)**(7/2)*(B*x+A)*(c*x**2+a)**(1/2),x)

[Out]

A*sqrt(a)*e**(7/2)*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)*e**(7/2)*x**(11/2)*gamma(11/4)*hyper((-1/2, 11/4), (15/4,), c*x**2*exp_polar(I*pi)/a)/(2*gamma(15/4
))

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