3.466 \(\int \frac {(e x)^{7/2} (A+B x)}{(a+c x^2)^{3/2}} \, dx\)

Optimal. Leaf size=360 \[ -\frac {a^{3/4} e^4 \sqrt {x} \left (\sqrt {a}+\sqrt {c} x\right ) \sqrt {\frac {a+c x^2}{\left (\sqrt {a}+\sqrt {c} x\right )^2}} \left (63 \sqrt {a} B+25 A \sqrt {c}\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{30 c^{11/4} \sqrt {e x} \sqrt {a+c x^2}}+\frac {21 a^{5/4} B e^4 \sqrt {x} \left (\sqrt {a}+\sqrt {c} x\right ) \sqrt {\frac {a+c x^2}{\left (\sqrt {a}+\sqrt {c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{5 c^{11/4} \sqrt {e x} \sqrt {a+c x^2}}-\frac {e (e x)^{5/2} (A+B x)}{c \sqrt {a+c x^2}}+\frac {5 A e^3 \sqrt {e x} \sqrt {a+c x^2}}{3 c^2}-\frac {21 a B e^4 x \sqrt {a+c x^2}}{5 c^{5/2} \sqrt {e x} \left (\sqrt {a}+\sqrt {c} x\right )}+\frac {7 B e^2 (e x)^{3/2} \sqrt {a+c x^2}}{5 c^2} \]

[Out]

-e*(e*x)^(5/2)*(B*x+A)/c/(c*x^2+a)^(1/2)+7/5*B*e^2*(e*x)^(3/2)*(c*x^2+a)^(1/2)/c^2-21/5*a*B*e^4*x*(c*x^2+a)^(1
/2)/c^(5/2)/(a^(1/2)+x*c^(1/2))/(e*x)^(1/2)+5/3*A*e^3*(e*x)^(1/2)*(c*x^2+a)^(1/2)/c^2+21/5*a^(5/4)*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*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^(11/4
)/(e*x)^(1/2)/(c*x^2+a)^(1/2)-1/30*a^(3/4)*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))*(63*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^(11/4)/(e*x)^(1/2)/(c*x^2+a)^(1/2)

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Rubi [A]  time = 0.40, antiderivative size = 360, normalized size of antiderivative = 1.00, number of steps used = 8, 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} \[ -\frac {a^{3/4} e^4 \sqrt {x} \left (\sqrt {a}+\sqrt {c} x\right ) \sqrt {\frac {a+c x^2}{\left (\sqrt {a}+\sqrt {c} x\right )^2}} \left (63 \sqrt {a} B+25 A \sqrt {c}\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{30 c^{11/4} \sqrt {e x} \sqrt {a+c x^2}}+\frac {21 a^{5/4} B e^4 \sqrt {x} \left (\sqrt {a}+\sqrt {c} x\right ) \sqrt {\frac {a+c x^2}{\left (\sqrt {a}+\sqrt {c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{5 c^{11/4} \sqrt {e x} \sqrt {a+c x^2}}-\frac {e (e x)^{5/2} (A+B x)}{c \sqrt {a+c x^2}}+\frac {5 A e^3 \sqrt {e x} \sqrt {a+c x^2}}{3 c^2}-\frac {21 a B e^4 x \sqrt {a+c x^2}}{5 c^{5/2} \sqrt {e x} \left (\sqrt {a}+\sqrt {c} x\right )}+\frac {7 B e^2 (e x)^{3/2} \sqrt {a+c x^2}}{5 c^2} \]

Antiderivative was successfully verified.

[In]

Int[((e*x)^(7/2)*(A + B*x))/(a + c*x^2)^(3/2),x]

[Out]

-((e*(e*x)^(5/2)*(A + B*x))/(c*Sqrt[a + c*x^2])) + (5*A*e^3*Sqrt[e*x]*Sqrt[a + c*x^2])/(3*c^2) + (7*B*e^2*(e*x
)^(3/2)*Sqrt[a + c*x^2])/(5*c^2) - (21*a*B*e^4*x*Sqrt[a + c*x^2])/(5*c^(5/2)*Sqrt[e*x]*(Sqrt[a] + Sqrt[c]*x))
+ (21*a^(5/4)*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])/(5*c^(11/4)*Sqrt[e*x]*Sqrt[a + c*x^2]) - (a^(3/4)*(63*Sqrt[a]*B + 25*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)*S
qrt[x])/a^(1/4)], 1/2])/(30*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 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)^{7/2} (A+B x)}{\left (a+c x^2\right )^{3/2}} \, dx &=-\frac {e (e x)^{5/2} (A+B x)}{c \sqrt {a+c x^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 )}{\sqrt {a+c x^2}} \, dx}{a c}\\ &=-\frac {e (e x)^{5/2} (A+B x)}{c \sqrt {a+c x^2}}+\frac {7 B e^2 (e x)^{3/2} \sqrt {a+c x^2}}{5 c^2}+\frac {2 \int \frac {\sqrt {e x} \left (-\frac {21}{4} a^2 B e^3+\frac {25}{4} a A c e^3 x\right )}{\sqrt {a+c x^2}} \, dx}{5 a c^2}\\ &=-\frac {e (e x)^{5/2} (A+B x)}{c \sqrt {a+c x^2}}+\frac {5 A e^3 \sqrt {e x} \sqrt {a+c x^2}}{3 c^2}+\frac {7 B e^2 (e x)^{3/2} \sqrt {a+c x^2}}{5 c^2}+\frac {4 \int \frac {-\frac {25}{8} a^2 A c e^4-\frac {63}{8} a^2 B c e^4 x}{\sqrt {e x} \sqrt {a+c x^2}} \, dx}{15 a c^3}\\ &=-\frac {e (e x)^{5/2} (A+B x)}{c \sqrt {a+c x^2}}+\frac {5 A e^3 \sqrt {e x} \sqrt {a+c x^2}}{3 c^2}+\frac {7 B e^2 (e x)^{3/2} \sqrt {a+c x^2}}{5 c^2}+\frac {\left (4 \sqrt {x}\right ) \int \frac {-\frac {25}{8} a^2 A c e^4-\frac {63}{8} a^2 B c e^4 x}{\sqrt {x} \sqrt {a+c x^2}} \, dx}{15 a c^3 \sqrt {e x}}\\ &=-\frac {e (e x)^{5/2} (A+B x)}{c \sqrt {a+c x^2}}+\frac {5 A e^3 \sqrt {e x} \sqrt {a+c x^2}}{3 c^2}+\frac {7 B e^2 (e x)^{3/2} \sqrt {a+c x^2}}{5 c^2}+\frac {\left (8 \sqrt {x}\right ) \operatorname {Subst}\left (\int \frac {-\frac {25}{8} a^2 A c e^4-\frac {63}{8} a^2 B c e^4 x^2}{\sqrt {a+c x^4}} \, dx,x,\sqrt {x}\right )}{15 a c^3 \sqrt {e x}}\\ &=-\frac {e (e x)^{5/2} (A+B x)}{c \sqrt {a+c x^2}}+\frac {5 A e^3 \sqrt {e x} \sqrt {a+c x^2}}{3 c^2}+\frac {7 B e^2 (e x)^{3/2} \sqrt {a+c x^2}}{5 c^2}+\frac {\left (21 a^{3/2} 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 )}{5 c^{5/2} \sqrt {e x}}-\frac {\left (a \left (63 \sqrt {a} B+25 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 )}{15 c^{5/2} \sqrt {e x}}\\ &=-\frac {e (e x)^{5/2} (A+B x)}{c \sqrt {a+c x^2}}+\frac {5 A e^3 \sqrt {e x} \sqrt {a+c x^2}}{3 c^2}+\frac {7 B e^2 (e x)^{3/2} \sqrt {a+c x^2}}{5 c^2}-\frac {21 a B e^4 x \sqrt {a+c x^2}}{5 c^{5/2} \sqrt {e x} \left (\sqrt {a}+\sqrt {c} x\right )}+\frac {21 a^{5/4} 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 )}{5 c^{11/4} \sqrt {e x} \sqrt {a+c x^2}}-\frac {a^{3/4} \left (63 \sqrt {a} B+25 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 )}{30 c^{11/4} \sqrt {e x} \sqrt {a+c x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.07, size = 127, normalized size = 0.35 \[ \frac {e^3 \sqrt {e x} \left (-25 a A \sqrt {\frac {c x^2}{a}+1} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};-\frac {c x^2}{a}\right )+25 a A-21 a B x \sqrt {\frac {c x^2}{a}+1} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-\frac {c x^2}{a}\right )+21 a B x+10 A c x^2+6 B c x^3\right )}{15 c^2 \sqrt {a+c x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[((e*x)^(7/2)*(A + B*x))/(a + c*x^2)^(3/2),x]

[Out]

(e^3*Sqrt[e*x]*(25*a*A + 21*a*B*x + 10*A*c*x^2 + 6*B*c*x^3 - 25*a*A*Sqrt[1 + (c*x^2)/a]*Hypergeometric2F1[1/4,
1/2, 5/4, -((c*x^2)/a)] - 21*a*B*x*Sqrt[1 + (c*x^2)/a]*Hypergeometric2F1[1/2, 3/4, 7/4, -((c*x^2)/a)]))/(15*c
^2*Sqrt[a + c*x^2])

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fricas [F]  time = 0.98, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (B e^{3} x^{4} + A e^{3} x^{3}\right )} \sqrt {c x^{2} + a} \sqrt {e x}}{c^{2} x^{4} + 2 \, a c x^{2} + a^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B x + A\right )} \left (e x\right )^{\frac {7}{2}}}{{\left (c x^{2} + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^(7/2)*(B*x+A)/(c*x^2+a)^(3/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.13, size = 318, normalized size = 0.88 \[ -\frac {\sqrt {e x}\, \left (-12 B \,c^{2} x^{4}-20 A \,c^{2} x^{3}-42 B a c \,x^{2}-50 A a c x +126 \sqrt {2}\, \sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {\frac {-c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {-\frac {c x}{\sqrt {-a c}}}\, B \,a^{2} \EllipticE \left (\sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}, \frac {\sqrt {2}}{2}\right )-63 \sqrt {2}\, \sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {\frac {-c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {-\frac {c x}{\sqrt {-a c}}}\, B \,a^{2} \EllipticF \left (\sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}, \frac {\sqrt {2}}{2}\right )+25 \sqrt {2}\, \sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {\frac {-c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {-\frac {c x}{\sqrt {-a c}}}\, \sqrt {-a c}\, A a \EllipticF \left (\sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}, \frac {\sqrt {2}}{2}\right )\right ) e^{3}}{30 \sqrt {c \,x^{2}+a}\, c^{3} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x)^(7/2)*(B*x+A)/(c*x^2+a)^(3/2),x)

[Out]

-1/30*e^3/x*(e*x)^(1/2)*(25*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*c)^(1/
2)*a+126*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)*EllipticE(((c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2),1/2*2^(1/2))*a^2-63*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^2-12*B*c^2*x^4-20*A*c^2*x^3-42*B*a*c*x^2-50*A*a*c*x)/(c*x^
2+a)^(1/2)/c^3

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B x + A\right )} \left (e x\right )^{\frac {7}{2}}}{{\left (c x^{2} + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^(7/2)*(B*x+A)/(c*x^2+a)^(3/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (e\,x\right )}^{7/2}\,\left (A+B\,x\right )}{{\left (c\,x^2+a\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((e*x)^(7/2)*(A + B*x))/(a + c*x^2)^(3/2),x)

[Out]

int(((e*x)^(7/2)*(A + B*x))/(a + c*x^2)^(3/2), 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)**(7/2)*(B*x+A)/(c*x**2+a)**(3/2),x)

[Out]

Timed out

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