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

Optimal. Leaf size=393 \[ -\frac {c^{3/4} \sqrt {x} \left (\sqrt {a}+\sqrt {c} x\right ) \sqrt {\frac {a+c x^2}{\left (\sqrt {a}+\sqrt {c} x\right )^2}} \left (25 \sqrt {a} B+63 A \sqrt {c}\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{30 a^{11/4} e^3 \sqrt {e x} \sqrt {a+c x^2}}+\frac {21 A c^{5/4} \sqrt {x} \left (\sqrt {a}+\sqrt {c} x\right ) \sqrt {\frac {a+c x^2}{\left (\sqrt {a}+\sqrt {c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{5 a^{11/4} e^3 \sqrt {e x} \sqrt {a+c x^2}}-\frac {21 A c^{3/2} x \sqrt {a+c x^2}}{5 a^3 e^3 \sqrt {e x} \left (\sqrt {a}+\sqrt {c} x\right )}+\frac {21 A c \sqrt {a+c x^2}}{5 a^3 e^3 \sqrt {e x}}-\frac {7 A \sqrt {a+c x^2}}{5 a^2 e (e x)^{5/2}}-\frac {5 B \sqrt {a+c x^2}}{3 a^2 e^2 (e x)^{3/2}}+\frac {A+B x}{a e (e x)^{5/2} \sqrt {a+c x^2}} \]

[Out]

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

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Rubi [A]  time = 0.50, antiderivative size = 393, 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 = {823, 835, 842, 840, 1198, 220, 1196} \[ -\frac {c^{3/4} \sqrt {x} \left (\sqrt {a}+\sqrt {c} x\right ) \sqrt {\frac {a+c x^2}{\left (\sqrt {a}+\sqrt {c} x\right )^2}} \left (25 \sqrt {a} B+63 A \sqrt {c}\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{30 a^{11/4} e^3 \sqrt {e x} \sqrt {a+c x^2}}-\frac {21 A c^{3/2} x \sqrt {a+c x^2}}{5 a^3 e^3 \sqrt {e x} \left (\sqrt {a}+\sqrt {c} x\right )}+\frac {21 A c^{5/4} \sqrt {x} \left (\sqrt {a}+\sqrt {c} x\right ) \sqrt {\frac {a+c x^2}{\left (\sqrt {a}+\sqrt {c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{5 a^{11/4} e^3 \sqrt {e x} \sqrt {a+c x^2}}+\frac {21 A c \sqrt {a+c x^2}}{5 a^3 e^3 \sqrt {e x}}-\frac {7 A \sqrt {a+c x^2}}{5 a^2 e (e x)^{5/2}}-\frac {5 B \sqrt {a+c x^2}}{3 a^2 e^2 (e x)^{3/2}}+\frac {A+B x}{a e (e x)^{5/2} \sqrt {a+c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(A + B*x)/(a*e*(e*x)^(5/2)*Sqrt[a + c*x^2]) - (7*A*Sqrt[a + c*x^2])/(5*a^2*e*(e*x)^(5/2)) - (5*B*Sqrt[a + c*x^
2])/(3*a^2*e^2*(e*x)^(3/2)) + (21*A*c*Sqrt[a + c*x^2])/(5*a^3*e^3*Sqrt[e*x]) - (21*A*c^(3/2)*x*Sqrt[a + c*x^2]
)/(5*a^3*e^3*Sqrt[e*x]*(Sqrt[a] + Sqrt[c]*x)) + (21*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])/(5*a^(11/4)*e^3*Sqrt[e*x]*Sqrt[a
+ c*x^2]) - ((25*Sqrt[a]*B + 63*A*Sqrt[c])*c^(3/4)*Sqrt[x]*(Sqrt[a] + Sqrt[c]*x)*Sqrt[(a + c*x^2)/(Sqrt[a] + S
qrt[c]*x)^2]*EllipticF[2*ArcTan[(c^(1/4)*Sqrt[x])/a^(1/4)], 1/2])/(30*a^(11/4)*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 )^{3/2}} \, dx &=\frac {A+B x}{a e (e x)^{5/2} \sqrt {a+c x^2}}-\frac {\int \frac {-\frac {7}{2} a A c e^2-\frac {5}{2} a B c e^2 x}{(e x)^{7/2} \sqrt {a+c x^2}} \, dx}{a^2 c e^2}\\ &=\frac {A+B x}{a e (e x)^{5/2} \sqrt {a+c x^2}}-\frac {7 A \sqrt {a+c x^2}}{5 a^2 e (e x)^{5/2}}+\frac {2 \int \frac {\frac {25}{4} a^2 B c e^3-\frac {21}{4} a A c^2 e^3 x}{(e x)^{5/2} \sqrt {a+c x^2}} \, dx}{5 a^3 c e^4}\\ &=\frac {A+B x}{a e (e x)^{5/2} \sqrt {a+c x^2}}-\frac {7 A \sqrt {a+c x^2}}{5 a^2 e (e x)^{5/2}}-\frac {5 B \sqrt {a+c x^2}}{3 a^2 e^2 (e x)^{3/2}}-\frac {4 \int \frac {\frac {63}{8} a^2 A c^2 e^4+\frac {25}{8} a^2 B c^2 e^4 x}{(e x)^{3/2} \sqrt {a+c x^2}} \, dx}{15 a^4 c e^6}\\ &=\frac {A+B x}{a e (e x)^{5/2} \sqrt {a+c x^2}}-\frac {7 A \sqrt {a+c x^2}}{5 a^2 e (e x)^{5/2}}-\frac {5 B \sqrt {a+c x^2}}{3 a^2 e^2 (e x)^{3/2}}+\frac {21 A c \sqrt {a+c x^2}}{5 a^3 e^3 \sqrt {e x}}+\frac {8 \int \frac {-\frac {25}{16} a^3 B c^2 e^5-\frac {63}{16} a^2 A c^3 e^5 x}{\sqrt {e x} \sqrt {a+c x^2}} \, dx}{15 a^5 c e^8}\\ &=\frac {A+B x}{a e (e x)^{5/2} \sqrt {a+c x^2}}-\frac {7 A \sqrt {a+c x^2}}{5 a^2 e (e x)^{5/2}}-\frac {5 B \sqrt {a+c x^2}}{3 a^2 e^2 (e x)^{3/2}}+\frac {21 A c \sqrt {a+c x^2}}{5 a^3 e^3 \sqrt {e x}}+\frac {\left (8 \sqrt {x}\right ) \int \frac {-\frac {25}{16} a^3 B c^2 e^5-\frac {63}{16} a^2 A c^3 e^5 x}{\sqrt {x} \sqrt {a+c x^2}} \, dx}{15 a^5 c e^8 \sqrt {e x}}\\ &=\frac {A+B x}{a e (e x)^{5/2} \sqrt {a+c x^2}}-\frac {7 A \sqrt {a+c x^2}}{5 a^2 e (e x)^{5/2}}-\frac {5 B \sqrt {a+c x^2}}{3 a^2 e^2 (e x)^{3/2}}+\frac {21 A c \sqrt {a+c x^2}}{5 a^3 e^3 \sqrt {e x}}+\frac {\left (16 \sqrt {x}\right ) \operatorname {Subst}\left (\int \frac {-\frac {25}{16} a^3 B c^2 e^5-\frac {63}{16} a^2 A c^3 e^5 x^2}{\sqrt {a+c x^4}} \, dx,x,\sqrt {x}\right )}{15 a^5 c e^8 \sqrt {e x}}\\ &=\frac {A+B x}{a e (e x)^{5/2} \sqrt {a+c x^2}}-\frac {7 A \sqrt {a+c x^2}}{5 a^2 e (e x)^{5/2}}-\frac {5 B \sqrt {a+c x^2}}{3 a^2 e^2 (e x)^{3/2}}+\frac {21 A c \sqrt {a+c x^2}}{5 a^3 e^3 \sqrt {e x}}-\frac {\left (\left (25 \sqrt {a} B+63 A \sqrt {c}\right ) c \sqrt {x}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+c x^4}} \, dx,x,\sqrt {x}\right )}{15 a^{5/2} e^3 \sqrt {e x}}+\frac {\left (21 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 )}{5 a^{5/2} e^3 \sqrt {e x}}\\ &=\frac {A+B x}{a e (e x)^{5/2} \sqrt {a+c x^2}}-\frac {7 A \sqrt {a+c x^2}}{5 a^2 e (e x)^{5/2}}-\frac {5 B \sqrt {a+c x^2}}{3 a^2 e^2 (e x)^{3/2}}+\frac {21 A c \sqrt {a+c x^2}}{5 a^3 e^3 \sqrt {e x}}-\frac {21 A c^{3/2} x \sqrt {a+c x^2}}{5 a^3 e^3 \sqrt {e x} \left (\sqrt {a}+\sqrt {c} x\right )}+\frac {21 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 )}{5 a^{11/4} e^3 \sqrt {e x} \sqrt {a+c x^2}}-\frac {\left (25 \sqrt {a} B+63 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 )}{30 a^{11/4} e^3 \sqrt {e x} \sqrt {a+c x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.06, size = 107, normalized size = 0.27 \[ \frac {x \left (-21 A \sqrt {\frac {c x^2}{a}+1} \, _2F_1\left (-\frac {5}{4},\frac {1}{2};-\frac {1}{4};-\frac {c x^2}{a}\right )-25 B x \sqrt {\frac {c x^2}{a}+1} \, _2F_1\left (-\frac {3}{4},\frac {1}{2};\frac {1}{4};-\frac {c x^2}{a}\right )+15 (A+B x)\right )}{15 a (e x)^{7/2} \sqrt {a+c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.79, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {c x^{2} + a} {\left (B x + A\right )} \sqrt {e x}}{c^{2} e^{4} x^{8} + 2 \, a c e^{4} x^{6} + a^{2} e^{4} x^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(c*x^2 + a)*(B*x + A)*sqrt(e*x)/(c^2*e^4*x^8 + 2*a*c*e^4*x^6 + a^2*e^4*x^4), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.08, size = 331, normalized size = 0.84 \[ -\frac {-126 A \,c^{2} x^{4}+126 \sqrt {2}\, \sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {\frac {-c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {-\frac {c x}{\sqrt {-a c}}}\, A a c \,x^{2} \EllipticE \left (\sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}, \frac {\sqrt {2}}{2}\right )-63 \sqrt {2}\, \sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {\frac {-c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {-\frac {c x}{\sqrt {-a c}}}\, A a c \,x^{2} \EllipticF \left (\sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}, \frac {\sqrt {2}}{2}\right )+50 B a c \,x^{3}-84 A a c \,x^{2}+25 \sqrt {2}\, \sqrt {-a c}\, \sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {\frac {-c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {-\frac {c x}{\sqrt {-a c}}}\, B a \,x^{2} \EllipticF \left (\sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}, \frac {\sqrt {2}}{2}\right )+20 B \,a^{2} x +12 A \,a^{2}}{30 \sqrt {c \,x^{2}+a}\, \sqrt {e x}\, a^{3} e^{3} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/30/x^2*(126*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^2*a*c-63*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)*Ell
ipticF(((c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2),1/2*2^(1/2))*x^2*a*c+25*B*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)*(-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^2*a-126*A*c^2*x^4+50*B*a*c*x^3-84*A*a*c*x^2+20*B*a^2*x+12*A*a^
2)/(c*x^2+a)^(1/2)/e^3/(e*x)^(1/2)/a^3

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [C]  time = 150.96, size = 104, normalized size = 0.26 \[ \frac {A \Gamma \left (- \frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{4}, \frac {3}{2} \\ - \frac {1}{4} \end {matrix}\middle | {\frac {c x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {3}{2}} e^{\frac {7}{2}} x^{\frac {5}{2}} \Gamma \left (- \frac {1}{4}\right )} + \frac {B \Gamma \left (- \frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, \frac {3}{2} \\ \frac {1}{4} \end {matrix}\middle | {\frac {c x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {3}{2}} e^{\frac {7}{2}} x^{\frac {3}{2}} \Gamma \left (\frac {1}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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